# Partial fraction decomposition

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In algebra, the **partial fraction decomposition** or **partial fraction expansion** of a rational fraction (that is, a fraction such that the numerator and the denominator are both polynomials) is an operation that consists of expressing the fraction as a sum of a polynomial (possibly zero) and one or several fractions with a simpler denominator.^{[1]}

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,^{[2]} Taylor series expansions, inverse Z-transforms, and inverse Laplace transforms. The concept was discovered independently in 1702 by both Johann Bernoulli and Gottfried Leibniz.^{[3]}

In symbols, the *partial fraction decomposition* of a rational fraction of the form where *f* and *g* are polynomials, is the expression of the rational fraction as

where
*p*(*x*) is a polynomial, and, for each j,
the denominator *g*_{j} (*x*) is a power of an irreducible polynomial (i.e. not factorizable into polynomials of positive degrees), and
the numerator *f*_{j} (*x*) 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

[edit]Let
be a rational fraction, where F and G are univariate polynomials in the indeterminate *x* over a field. The existence of the partial fraction can be proved by applying inductively the following reduction steps.

### Polynomial part

[edit]There exist two polynomials E and *F*_{1} such that
and
where denotes the degree of the polynomial P.

This results immediately from the Euclidean division of F by G, which asserts the existence of E and *F*_{1} such that and

This allows supposing in the next steps that

### Factors of the denominator

[edit]If and
where *G*_{1} and *G*_{2} are coprime polynomials, then there exist polynomials and such that
and

This can be proved as follows. Bézout's identity asserts the existence of polynomials *C* and *D* such that
(by hypothesis, 1 is a greatest common divisor of *G*_{1} and *G*_{2}).

Let with be the Euclidean division of DF by Setting one gets It remains to show that By reducing the last sum of fractions to a common denominator, one gets and thus

### Powers in the denominator

[edit]Using the preceding decomposition inductively one gets fractions of the form with where G is an irreducible polynomial. If *k* > 1, 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 G, Bézout's identity provides polynomials C and D such that and thus 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

[edit]**Theorem** — Let *f* and *g* be nonzero polynomials over a field *K*. Write *g* as a product of powers of distinct irreducible polynomials :

There are (unique) polynomials *b* and *a*_{ij} with deg *a*_{ij} < deg *p*_{i} such that

If deg *f* < deg *g*, then *b* = 0.

The uniqueness can be proved as follows. Let *d* = max(1 + deg *f*, deg *g*). All together, *b* and the *a*_{ij} have d coefficients. The shape of the decomposition defines a linear map from coefficient vectors to polynomials f of degree less than d. The existence proof means that this map is surjective. As the two vector spaces have the same dimension, the map is also injective, which means uniqueness of the decomposition. By the way, this proof induces an algorithm for computing the decomposition through linear algebra.

If *K* is the field of complex numbers, the fundamental theorem of algebra implies that all *p*_{i} have degree one, and all numerators are constants. When *K* is the field of real numbers, some of the *p*_{i} 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 *p*_{i} may be the factors of the square-free factorization of *g*. When *K* 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

[edit]For the purpose of symbolic integration, the preceding result may be refined into

**Theorem** — Let *f* and *g* be nonzero polynomials over a field *K*. Write *g* as a product of powers of pairwise coprime polynomials which have no multiple root in an algebraically closed field:

There are (unique) polynomials *b* and *c*_{ij} with deg *c*_{ij} < deg *p*_{i} such that
where denotes the derivative of

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.

There are various methods to compute decomposition in the Theorem. One simple way is called Hermite's method. First, *b* is immediately computed by Euclidean division of *f* by *g*, reducing to the case where deg(*f*) < deg(*g*). Next, one knows deg(*c*_{ij}) < deg(*p*_{i}), so one may write each *c _{ij}* as a polynomial with unknown coefficients. Reducing the sum of fractions in the Theorem to a common denominator, and equating the coefficients of each power of

*x*in the two numerators, one gets a system of linear equations which can be solved to obtain the desired (unique) values for the unknown coefficients.

## Procedure

[edit]Given two polynomials and , where the *α*_{n} are distinct constants and deg *P* < *n*, explicit expressions for partial fractions can be obtained by supposing that
and solving for the *c*_{i} constants, by substitution, by equating the coefficients of terms involving the powers of *x*, or otherwise. (This is a variant of the method of undetermined coefficients. After both sides of the equation are multiplied by Q(x), one side of the equation is a specific polynomial, and the other side is a polynomial with undetermined coefficients. The equality is possible only when the coefficients of like powers of *x* are equal. This yields n equations in n unknowns, the c_{k}.)

A more direct computation, which is strongly related to Lagrange interpolation, consists of writing
where is the derivative of the polynomial . The coefficients of are called the residues of *f/g*.

This approach does not account for several other cases, but can be modified accordingly:

- If then it is necessary to perform the Euclidean division of
*P*by*Q*, using polynomial long division, giving*P*(*x*) =*E*(*x*)*Q*(*x*) +*R*(*x*) with deg*R*<*n*. Dividing by*Q*(*x*) this gives and then seek partial fractions for the remainder fraction (which by definition satisfies deg*R*< deg*Q*). - If
*Q*(*x*) contains factors which are irreducible over the given field, then the numerator*N*(*x*) of each partial fraction with such a factor*F*(*x*) in the denominator must be sought as a polynomial with deg*N*< deg*F*, rather than as a constant. For example, take the following decomposition over**R**: - Suppose
*Q*(*x*) = (*x*−*α*)^{r}*S*(*x*) and*S*(*α*) ≠ 0, that is*α*is a root of*Q*(*x*) of multiplicity r. In the partial fraction decomposition, the r first powers of (*x*−*α*) will occur as denominators of the partial fractions (possibly with a zero numerator). For example, if*S*(*x*) = 1 the partial fraction decomposition has the form

### Illustration

[edit]In an example application of this procedure, (3*x* + 5)/(1 − 2*x*)^{2} can be decomposed in the form

Clearing denominators shows that 3*x* + 5 = *A* + *B*(1 − 2*x*). Expanding and equating the coefficients of powers of *x* gives

*A*+

*B*and 3

*x*= −2

*Bx*

Solving this system of linear equations for *A* and *B* yields *A* = 13/2 and *B* = −3/2. Hence,

### Residue method

[edit]Over the complex numbers, suppose *f*(*x*) is a rational proper fraction, and can be decomposed into

Let
then according to the uniqueness of Laurent series, *a*_{ij} is the coefficient of the term (*x* − *x*_{i})^{−1} in the Laurent expansion of *g*_{ij}(*x*) about the point *x*_{i}, i.e., its residue

This is given directly by the formula
or in the special case when *x*_{i} is a simple root,
when

## Over the reals

[edit]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

### General result

[edit]Let be any rational function over the real numbers. In other words, suppose there exist real polynomials functions and , such that

By dividing both the numerator and the denominator by the leading coefficient of , we may assume without loss of generality that is monic. By the fundamental theorem of algebra, we can write

where , , are real numbers with , and , are positive integers. The terms are the *linear factors* of which correspond to real roots of , and the terms are the *irreducible quadratic factors* of which correspond to pairs of complex conjugate roots of .

Then the partial fraction decomposition of is the following:

Here, *P*(*x*) is a (possibly zero) polynomial, and the *A*_{ir}, *B*_{ir}, and *C*_{ir} 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*(*x*). We then obtain an equation of polynomials whose left-hand side is simply *p*(*x*) and whose right-hand side has coefficients which are linear expressions of the constants *A*_{ir}, *B*_{ir}, and *C*_{ir}. 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 (see Example 5).

## Examples

[edit]### Example 1

[edit]

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

[edit]

After long division, we have

The factor *x*^{2} − 4*x* + 8 is irreducible over the reals, as its discriminant (−4)^{2} − 4×8 = −16 is negative. Thus the partial fraction decomposition over the reals has the shape

Multiplying through by *x*^{3} − 4*x*^{2} + 8*x*, we have the polynomial identity

Taking *x* = 0, we see that 16 = 8*A*, so *A* = 2. Comparing the *x*^{2} coefficients, we see that 4 = *A* + *B* = 2 + *B*, so *B* = 2. Comparing linear coefficients, we see that −8 = −4*A* + *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* (complex) roots (some of which can be repeated). The second fraction can be decomposed to:

Multiplying through by the denominator gives:

Equating the coefficients of *x* and the constant (with respect to *x*) coefficients of both sides of this equation, one gets a system of two linear equations in *D* and *E*, whose solution is

Thus we have a complete decomposition:

One may also compute directly *A*, *D* and *E* with the residue method (see also example 4 below).

### Example 3

[edit]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 *x*^{6} and *x*^{5} 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. (To this end, recall that the derivative at *x* = *a* of (*x* − *a*)^{m}*p*(*x*) vanishes if *m* > 1 and is just *p*(*a*) for *m* = 1.) For instance the first derivative at *x* = 1 gives

that is 8 = 4*B* + 8 so *B* = 0.

### Example 4 (residue method)

[edit]

Thus, *f*(*z*) 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)

[edit]Limits can be used to find a partial fraction decomposition.^{[4]} 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 *x* and taking the limit when , we have

and

This implies *A* + *B* = 0 and so .

For *x* = 0, we get and thus .

Putting everything together, we get the decomposition

### Example 6 (integral)

[edit]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

[edit]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 (in the real or complex case) then provides a proof of the existence and uniqueness of the partial fraction decomposition, and a characterization of the coefficients.

### Sketch of the proof

[edit]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

[edit]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:

## Notes

[edit]**^**Larson, Ron (2016).*Algebra & Trigonometry*. Cengage Learning. ISBN 9781337271172.**^**Horowitz, Ellis. "Algorithms for partial fraction decomposition and rational function integration." Proceedings of the second ACM symposium on Symbolic and algebraic manipulation. ACM, 1971.**^**Grosholz, Emily (2000).*The Growth of Mathematical Knowledge*. Kluwer Academic Publilshers. p. 179. ISBN 978-90-481-5391-6.**^**Bluman, George W. (1984).*Problem Book for First Year Calculus*. New York: Springer-Verlag. pp. 250–251.

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