Partial fraction decomposition


In algebra, the partial fraction decomposition or partial fraction expansion of a rational fraction is an operation that consists of expressing the fraction as a sum of a polynomial and one or several fractions with a simpler denominator.
The importance of the partial fraction decomposition lies in the fact that it provides algorithms for various computations with rational functions, including the explicit computation of antiderivatives, Taylor series expansions, inverse Z-transforms, inverse Laplace transforms. The concept was discovered independently in 1702 by both Johann Bernoulli and Gottfried Leibniz.
In symbols, the partial fraction decomposition of a rational fraction of the form
where and are polynomials, is its expression as
where
is a polynomial, and, for each,
the denominator is a power of an irreducible polynomial, and
the numerator is a polynomial of a smaller degree than the degree of this irreducible polynomial.
When explicit computation is involved, a coarser decomposition is often preferred, which consists of replacing "irreducible polynomial" by "square-free polynomial" in the description of the outcome. This allows replacing polynomial factorization by the much easier to compute square-free factorization. This is sufficient for most applications, and avoids introducing irrational coefficients when the coefficients of the input polynomials are integers or rational numbers.

Basic principles

Let
be a rational fraction, where and are univariate polynomials in the indeterminate. The existence of the partial fraction can be proved by applying inductively the following reduction steps.

Polynomial part

There exists two polynomials and such that
and
where denotes the degree of the polynomial.
This results immediately from the Euclidean division of by, which asserts the existence of and such that and
This allows supposing in the next steps that

Factors of the denominator

If and
where and are coprime polynomials, then there exist polynommials and such that
and
This can be proved as follows. Bézout's identity asserts the existence of polynomials and such that
. Let with be the Euclidean division of by Setting one gets
It remains to show that By reducing to the same denominator the last sum of fractions, one gets
and thus

Powers in the denominator

Using the preceding decomposition inductively one gets fractions of the form with where is an irreducible polynomial. If, one can decompose further, by using that an irreducible polynomial is a square-free polynomial, that is, is a greatest common divisor of the polynomial and its derivative. If is the derivative of, Bézout's identity provides polynomials and such that and thus The Euclidean division of `by gives polynomials and such that and Setting one gets
with
Iterating this process with in place of leads eventually to the following theorem.

Statement

If is field of complex numbers, the fundamental theorem of algebra implies that all have degree one, and all numerators are constants. When is the field of real numbers, some of the may be quadratic, so, in the partial fraction decomposition, quotients of linear polynomials by powers of quadratic polynomials may also occur.
In the preceding theorem, one may replace "distinct irreducible polynomials" by "pairwise coprime polynomials that are coprime with their derivative". For example, the may be the factors of the square-free factorization of. When is the field of rational numbers, as it is typically the case in computer algebra, this allows to replace factorization by greatest common divisor computation for computing a partial fraction decomposition.

[|Application to symbolic integration]

For the purpose of symbolic integration, the preceding result may be refined into
This reduces the computation of the antiderivative of a rational function to the integration of the last sum, which is called the logarithmic part, because its antiderivative is a linear combination of logarithms. In fact, we have
There are various methods to compute above decomposition. The one that is the simplest to describe is probably the so-called Hermite's method. As the degree of cij is bounded by the degree of pi, and the degree of b is the difference of the degrees of f and g, one may write these unknowns polynomials as polynomials with unknown coefficients. Reducing the two members of above formula to the same denominator and writing that the coefficients of each power of x are the same in the two numerators, one gets a system of linear equations which can be solved to obtain the desired values for the unknowns coefficients.

Procedure

Given two polynomials and, where the αi are distinct constants and deg P < n, partial fractions are generally obtained by supposing that
and solving for the ci constants, by substitution, by equating the coefficients of terms involving the powers of x, or otherwise.
A more direct computation, which is strongly related with Lagrange interpolation consists of writing
where is the derivative of the polynomial.
This approach does not account for several other cases, but can be modified accordingly:
In an example application of this procedure, can be decomposed in the form
Clearing denominators shows that. Expanding and equating the coefficients of powers of gives
Solving this system of linear equations for and yields. Hence,

Residue method

Over the complex numbers, suppose f is a rational proper fraction, and can be decomposed into
Let
then according to the uniqueness of Laurent series, aij is the coefficient of the term −1 in the Laurent expansion of gij about the point xi, i.e., its residue
This is given directly by the formula
or in the special case when xi is a simple root,
when

Over the reals

Partial fractions are used in real-variable integral calculus to find real-valued antiderivatives of rational functions. Partial fraction decomposition of real rational functions is also used to find their Inverse Laplace transforms. For applications of partial fraction decomposition over the reals, see
Let f be any rational function over the real numbers. In other words, suppose there exist real polynomials functions p and q≠ 0, such that
By dividing both the numerator and the denominator by the leading coefficient of q, we may assume without loss of generality that q is monic. By the fundamental theorem of algebra, we can write
where a1,..., am, b1,..., bn, c1,..., cn are real numbers with bi2 − 4ci < 0, and j1,..., jm, k1,..., kn are positive integers. The terms are the linear factors of q which correspond to real roots of q, and the terms are the irreducible quadratic factors of q which correspond to pairs of complex conjugate roots of q.
Then the partial fraction decomposition of f is the following:
Here, P is a polynomial, and the Air, Bir, and Cir are real constants. There are a number of ways the constants can be found.
The most straightforward method is to multiply through by the common denominator q. We then obtain an equation of polynomials whose left-hand side is simply p and whose right-hand side has coefficients which are linear expressions of the constants Air, Bir, and Cir. Since two polynomials are equal if and only if their corresponding coefficients are equal, we can equate the coefficients of like terms. In this way, a system of linear equations is obtained which always has a unique solution. This solution can be found using any of the standard methods of linear algebra. It can also be found with limits.

Examples

Example 1

Here, the denominator splits into two distinct linear factors:
so we have the partial fraction decomposition
Multiplying through by the denominator on the left-hand side gives us the polynomial identity
Substituting x = −3 into this equation gives A = −1/4, and substituting x = 1 gives B = 1/4, so that

Example 2

After long-division, we have
The factor x2 − 4x + 8 is irreducible over the reals, as its discriminant is negative. Thus the partial fraction decomposition over the reals has the shape
Multiplying through by x3 − 4x2 + 8x, we have the polynomial identity
Taking x = 0, we see that 16 = 8A, so A = 2. Comparing the x2 coefficients, we see that 4 = A + B = 2 + B, so B = 2. Comparing linear coefficients, we see that −8 = −4A + C = −8 + C, so C = 0. Altogether,
The fraction can be completely decomposed using complex numbers. According to the fundamental theorem of algebra every complex polynomial of degree n has n roots. The second fraction can be decomposed to:
Multiplying through by the denominator gives:
Equating the coefficients of and the constant coefficients of both sides of this equation, one gets a system of two linear equations in and, whose solution is
Thus we have a complete decomposition:
One may also compute directly and with the residue method.

Example 3

This example illustrates almost all the "tricks" we might need to use, short of consulting a computer algebra system.
After long-division and factoring the denominator, we have
The partial fraction decomposition takes the form
Multiplying through by the denominator on the left-hand side we have the polynomial identity
Now we use different values of x to compute the coefficients:
Solving this we have:
Using these values we can write:
We compare the coefficients of x6 and x5 on both side and we have:
Therefore:
which gives us B = 0. Thus the partial fraction decomposition is given by:
Alternatively, instead of expanding, one can obtain other linear dependences on the coefficients computing some derivatives at in the above polynomial identity. mp vanishes if m > 1 and is just p For instance the first derivative at x = 1 gives
that is 8 = 4B + 8 so B = 0.

Example 4 (residue method)

Thus, f can be decomposed into rational functions whose denominators are z+1, z−1, z+i, z−i. Since each term is of power one, −1, 1, −i and i are simple poles.
Hence, the residues associated with each pole, given by
are
respectively, and

[|Example 5] (limit method)

can be used to find a partial fraction decomposition. Consider the following example:
First, factor the denominator which determines the decomposition:
Multiplying everything by, and taking the limit when, we get
On the other hand,
and thus:
Multiplying by and taking the limit when, we have
and
This implies and so.
For, we get and thus .
Putting everything together, we get the decomposition

Example 6 (integral)

Suppose we have the indefinite integral:
Before performing decomposition, it is obvious we must perform polynomial long division and factor the denominator. Doing this would result in:
Upon this, we may now perform partial fraction decomposition.
so:
Upon substituting our values, in this case, where x=1 to solve for B and x=-2 to solve for A, we will result in:
Plugging all of this back into our integral allows us to find the answer:

The role of the Taylor polynomial

The partial fraction decomposition of a rational function can be related to Taylor's theorem as follows. Let
be real or complex polynomials
assume that
satisfies
Also define
Then we have
if, and only if, each polynomial is the Taylor polynomial of of order at the point :
Taylor's theorem then provides a proof of the existence and uniqueness of the partial fraction decomposition, and a characterization of the coefficients.

Sketch of the proof

The above partial fraction decomposition implies, for each 1 ≤ i ≤ r, a polynomial expansion
so is the Taylor polynomial of, because of the unicity of the polynomial expansion of order, and by assumption.
Conversely, if the are the Taylor polynomials, the above expansions at each hold, therefore we also have
which implies that the polynomial is divisible by
For is also divisible by, so
is divisible by. Since
we then have
and we find the partial fraction decomposition dividing by.

Fractions of integers

The idea of partial fractions can be generalized to other integral domains, say the ring of integers where prime numbers take the role of irreducible denominators. For example: