Partial fraction decomposition
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In algebra, the partial fraction decomposition or partial fraction expansion of a rational function (that is a fraction such that the numerator and the denominator are both polynomials) is the operation that consists in expressing the fraction as a sum of a polynomial (possibly zero) and one or several fractions with a simpler denominator.
The importance of the partial fraction decomposition lies in the fact that it provides an algorithm for computing the antiderivative of a rational function.
In symbols, one can use partial fraction expansion to change a rational fraction in the form
where ƒ and g are polynomials, into an expression of the form
where g_{j} (x) are polynomials that are factors of g(x), and are in general of lower degree. Thus, the partial fraction decomposition may be seen as the inverse procedure of the more elementary operation of addition of rational fractions, which produces a single rational fraction with a numerator and denominator usually of high degree. The full decomposition pushes the reduction as far as it can go: in other words, the factorization of g is used as much as possible. Thus, the outcome of a full partial fraction expansion expresses that fraction as a sum of a polynomial and one of several fractions, such that:
 the denominator of each fraction is a power of an irreducible (not factorable) polynomial and
 the numerator is a polynomial of smaller degree than this irreducible polynomial.
As factorization of polynomials may be difficult, a coarser decomposition is often preferred, which consists of replacing factorization by squarefree factorization. This amounts to replace "irreducible" by "squarefree" in the preceding description of the outcome.
Contents
Basic principles
Assume a rational function in one indeterminate x has a denominator that factors as
over a field K (we can take this to be real numbers, or complex numbers). Assume further that P and Q have no common factor. By Bézout's identity for polynomials, there exist polynomials C(x) and D(x) such that
Thus and hence R may be written as
where all numerators are polynomials.
Using this idea inductively we can write R(x) as a sum with denominators powers of irreducible polynomials. To take this further, if required, write:
as a sum with denominators powers of F and numerators of degree less than F, plus a possible extra polynomial. This can be done by the Euclidean algorithm, polynomial case. The result is the following theorem:
Theorem — Let ƒ 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 ƒ < deg g, then b = 0.
Therefore, when the field K is the complex numbers, we can assume that each p_{i} has degree 1 (by the fundamental theorem of algebra) the numerators will be constant. When K is the real numbers, some of the p_{i} might be quadratic, so, in the partial fraction decomposition, a quotient of a linear polynomial by a power of a quadratic might 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 squarefree factorization of g. When K is the field of the rational numbers, as it is typically the case in computer algebra, this allows to replace factorization by greatest common divisor to compute the partial fraction decomposition.
Application to symbolic integration
For the purpose of symbolic integration, the preceding result may be refined into
Theorem — Let ƒ 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. In fact, we have
There are various methods to compute above decomposition. The one that is the simplest to describe is probably the socalled Hermite's method. As the degree of c_{ij} is bounded by the degree of p_{i}, and the degree of b is the difference of the degrees of f and g (if this difference is non negative; otherwise, b=0), 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 α_{i} are distinct constants and deg P < n, partial fractions are generally 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.)
A more direct computation, which is strongly related with Lagrange interpolation consists in writing
where is the derivative of the polynomial .
This approach does not account for several other cases, but can be modified accordingly:
 If deg P deg Q, 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. Then Q(x) has a zero α of multiplicity r, and in the partial fraction decomposition, r of the partial fractions will involve the powers of (x − α). For illustration, take S(x) = 1 to get the following decomposition:
Illustration
In an example application of this procedure, (3x + 5)/(1 − 2x)^{2} can be decomposed in the form
Clearing denominators shows that 3x + 5 = A + B(1 − 2x). Expanding and equating the coefficients of powers of x gives
 5 = A + B and 3x = −2Bx
Solving for A and B yields A = 13/2 and B = −3/2. Hence,
Residue method
Over the complex numbers, suppose ƒ(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
Note that P(x) and Q(x) may or may not be polynomials.
Over the reals
Partial fractions are used in realvariable integral calculus to find realvalued 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
Let ƒ(x) be any rational function over the real numbers. In other words, suppose there exist real polynomials functions p(x) and q(x)≠ 0, such that
By dividing both the numerator and the denominator by the leading coefficient of q(x), we may assume without loss of generality that q(x) is monic. By the fundamental theorem of algebra, we can write
where a_{1},..., a_{m}, b_{1},..., b_{n}, c_{1},..., c_{n} are real numbers with b_{i}^{2} − 4c_{i} < 0, and j_{1},..., j_{m}, k_{1},..., k_{n} are positive integers. The terms (x − a_{i}) are the linear factors of q(x) which correspond to real roots of q(x), and the terms (x_{i}^{2} + b_{i}x + c_{i}) are the irreducible quadratic factors of q(x) which correspond to pairs of complex conjugate roots of q(x).
Then the partial fraction decomposition of ƒ(x) 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 lefthand side is simply p(x) and whose righthand 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
Example 1
Here, the denominator splits into two distinct linear factors:
so we have the partial fraction decomposition
Multiplying through by x^{2} + 2x − 3, we have 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 longdivision, we have
Since (−4)^{2} − 4×8 = −16 < 0, the factor x^{2} − 4x + 8 is irreducible, and the partial fraction decomposition over the reals has the shape
Multiplying through by x^{3} − 4x^{2} + 8x, we have the polynomial identity
Taking x = 0, we see that 16 = 8A, 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 = −4A + C = −8 + C, so C = 0. Altogether,
Example 3
This example illustrates almost all the "tricks" we might need to use, short of consulting a computer algebra system.
After longdivision and factoring the denominator, we have
The partial fraction decomposition takes the form
Multiplying through by (x − 1)^{3}(x^{2} + 1)^{2} we have the polynomial identity
Taking x = 1 gives 4 = 4C, so C = 1. Similarly, taking x = i gives 2 + 2i = (Fi + G)(2 + 2i), so Fi + G = 1, so F = 0 and G = 1 by equating real and imaginary parts. With C = G = 1 and F = 0, taking x = 0 we get A − B + 1 − E − 1 = 0, thus E = A − B.
We now have the identity
Expanding and sorting by exponents of x we get
We can now compare the coefficients and see that
with A = 2 − D and −A −3 D =−4 we get A = D = 1 and so B = 0, furthermore is C = 1, E = A − B = 1, F = 0 and G = 1.
The partial fraction decomposition of ƒ(x) is thus
Alternatively, instead of expanding, one can obtain other linear dependences on the coefficients computing some derivatives at x=1 and at x=i 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 it is just p(a) if m=1.) Thus, for instance the first derivative at x=1 gives
that is 8 = 4B + 8 so B=0.
Example 4 (residue method)
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)
Limits can be used to find a partial fraction decomposition.^{[1]}
First, factor the denominator:
The decomposition takes the form of
As , the A term dominates, so the righthand side approaches . Thus, we have
As , the righthand side is
Thus, .
At , . Therefore, .
The decomposition is thus .
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
that
and that
Define also
Then we have
if, and only if, for each the 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: The above partial fraction decomposition implies, for each 1 ≤ i ≤ r, a polynomial expansion
 , as
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
 , as
which implies that the polynomial is divisible by
For also is divisible by , so we have in turn that 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:
Notes
 ↑ Bluman, George W. (1984). Problem Book for First Year Calculus. New York: SpringerVerlag. pp. 250–251.<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles>
References
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 Miller, Charles D.; Lial, Margaret L.; Schneider, David I. (1990). Fundamentals of College Algebra (3rd ed.). AddisonWesley Educational Publishers, Inc. pp. 364–370. ISBN 0673386384.<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles>
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 Kudryavtsev, L. D. (2001), "Undetermined coefficients, method of", in Hazewinkel, Michiel, Encyclopedia of Mathematics, Springer, ISBN 9781556080104<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles>
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External links
 Weisstein, Eric W., "Partial Fraction Decomposition", MathWorld.
 Blake, Sam. "StepbyStep Partial Fractions".<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles>
 [1] Make partial fraction decompositions with Scilab.