Partial fraction decomposition

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.

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, and 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 \(\frac{f(x)}{g(x)},\) where f and g are polynomials, is the expression of the rational fraction as

\[\frac{f(x)}{g(x)}=p(x) + \sum_j \frac{f_j(x)}{g_j(x)}\]

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.

Let \[R(x) = \frac FG\] be a rational fraction, where F and G are univariate polynomials in the indeterminate x over a field. The existence of the partial fraction decomposition can be proved by applying inductively the following reduction steps.

There exist two polynomials E and F₁ such that \[\frac FG=E+\frac{F_1}G,\] and \[\deg F_1 <\deg G,\] where \(\deg P\) 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 F₁ such that \(F = EG + F_1\) and \(\deg F_1 < \deg G.\)

This allows supposing in the next steps that \(\deg F <\deg G.\)

If \(\deg F < \deg G,\) and \[G = G_1 G_2,\] where G₁ and G₂ are coprime polynomials, then there exist polynomials \(F_1\) and \(F_2\) such that \[\frac FG=\frac{F_1}{G_1}+\frac{F_2}{G_2},\] and \[\deg F_1 < \deg G_1\quad\text{and}\quad\deg F_2 < \deg G_2.\]

This can be proved as follows. Bézout's identity asserts the existence of polynomials C and D such that \[CG_1 + DG_2 = 1\] (by hypothesis, 1 is a greatest common divisor of G₁ and G₂).

Let \(DF=G_1Q+F_1\) with \(\deg F_1 < \deg G_1\) be the Euclidean division of DF by \(G_1.\) Setting \(F_2=CF+QG_2,\) one gets \[\begin{aligned} \frac FG&=\frac{F(CG_1 + DG_2)}{G_1G_2} =\frac{D F}{G_1}+\frac{CF}{G_2}\\ &=\frac{F_1+G_1Q}{G_1}+\frac{F_2-G_2Q}{G_2}\\ &=\frac{F_1}{G_1} + Q + \frac{F_2}{G_2} - Q\\ &=\frac{F_1}{G_1}+\frac{F_2}{G_2}. \end{aligned}\] It remains to show that \(\deg F_2 < \deg G_2.\) By reducing the last sum of fractions to a common denominator, one gets \(F=F_2G_1+F_1G_2,\) and thus \[\begin{aligned} \deg F_2 &=\deg(F-F_1G_2)-\deg G_1 \le \max(\deg F,\deg (F_1G_2))-\deg G_1\\ &< \max(\deg G,\deg(G_1G_2))-\deg G_1= \deg G_2 \end{aligned}\]

Using the preceding decomposition inductively one gets fractions of the form \(\frac F {G^k},\) with \(\deg F < \deg G^k= k\deg G,\) 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, \(1\) is a greatest common divisor of the polynomial and its derivative. If \(G'\) is the derivative of G, Bézout's identity provides polynomials C and D such that \(CG + DG' = 1\) and thus \(F=FCG+FDG'.\) Euclidean division of \(FDG'\) by \(G\) gives polynomials \(H_k\) and \(Q\) such that \(FDG' = QG + H_k\) and \(\deg H_k < \deg G.\) Setting \(F_{k-1}=FC+Q,\) one gets \[\frac F {G^k} = \frac{H_k}{G^k}+\frac{F_{k-1}}{G^{k-1}},\] with \(\deg H_k <\deg G.\)

Iterating this process with \(\frac{F_{k-1}}{G^{k-1}}\) in place of \(\frac F{G^k}\) leads eventually to the following theorem.

Theorem, Let f and g be nonzero polynomials over a field K. Write g as a product of powers of distinct irreducible polynomials : \[g=\prod_{i=1}^k p_i^{n_i}.\]

There are (unique) polynomials b and a_ij with deg a_ij < deg p_i such that \[\frac{f}{g}=b+\sum_{i=1}^k\sum_{j=1}^{n_i}\frac{a_{ij}}{p_i^j}.\]

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 \(a_{ij}\) 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.

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:

\[g=\prod_{i=1}^k p_i^{n_i}.\]

There are (unique) polynomials b and c_ij with deg c_ij < deg p_i such that \[\frac{f}{g} = b+\sum_{i=1}^k\sum_{j=2}^{n_i}\left(\frac{c_{ij}}{p_i^{j-1}}\right)' + \sum_{i=1}^k \frac{c_{i1}}{p_i}.\] where \(X'\) denotes the derivative of \(X.\)

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.

Given two polynomials \(P(x)\) and \(Q(x) = (x-\alpha_1)(x-\alpha_2) \cdots (x-\alpha_n)\), where the α_n are distinct constants and deg P < n, explicit expressions for partial fractions can be obtained by supposing that \[\frac{P(x)}{Q(x)} = \frac{c_1}{x-\alpha_1} + \frac{c_2}{x-\alpha_2} + \cdots + \frac{c_n}{x-\alpha_n}\] 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 \[\frac{P(x)}{Q(x)} = \sum_{i=1}^n \frac{P(\alpha_i)}{Q'(\alpha_i)}\frac{1}{(x-\alpha_i)}\] where \(Q'\) is the derivative of the polynomial \(Q\). The coefficients of \(\tfrac{1}{x-\alpha_j}\) are called the residues of f/g.

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

• If \(\deg P \geq \deg Q,\) 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 \[\frac{P(x)}{Q(x)} = E(x) + \frac{R(x)}{Q(x)},\] and then seek partial fractions for the remainder fraction (which by definition satisfies deg R < deg Q).
• If Q(x) contains nonlinear 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: \[\frac{x^2 + 1}{(x+2)(x-1)\color{Blue}(x^2+x+1)} = \frac{a}{x+2} + \frac{b}{x-1} + \frac{\color{OliveGreen}cx + d}{\color{Blue}x^2 + x + 1}.\]
• Suppose Q(x) = (x − α) 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 \[\frac{P(x)}{Q(x)} = \frac{P(x)}{(x-\alpha)^r} = \frac{c_1}{x-\alpha} + \frac{c_2}{(x-\alpha)^2} + \cdots + \frac{c_r}{(x-\alpha)^r}.\]

In an example application of this procedure, (3x + 5)/(1 − 2x) can be decomposed in the form

\[\frac{3x + 5}{(1-2x)^2} = \frac{A}{(1-2x)^2} + \frac{B}{(1-2x)}.\]

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

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

\[\frac{3x + 5}{(1-2x)^2} = \frac{13/2}{(1-2x)^2} + \frac{-3/2}{(1-2x)}.\]

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

\[f(x) = \sum_i \left( \frac{a_{i1}}{x - x_i} + \frac{a_{i2}}{( x - x_i)^2} + \cdots + \frac{a_{i k_i}}{(x - x_i)^{k_i}} \right).\]

Let \[g_{ij}(x) = (x - x_i)^{j-1}f(x),\] then according to the uniqueness of Laurent series, a_ij is the coefficient of the term (x − x_i) in the Laurent expansion of g_ij(x) about the point x_i, i.e., its residue \[a_{ij} = \operatorname{Res}(g_{ij},x_i).\]

This is given directly by the formula \[a_{ij} = \frac 1 {(k_i-j)!}\lim_{x\to x_i}\frac{d^{k_i-j}}{dx^{k_i-j}} \left((x-x_i)^{k_i} f(x)\right),\] or in the special case when x_i is a simple root, \[a_{i1}=\frac{P(x_i)}{Q'(x_i)},\] when \[f(x)=\frac{P(x)}{Q(x)}.\]

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

• Application to symbolic integration, above
• Partial fractions in Laplace transforms

Let \(f(x)\) be any rational function over the real numbers. In other words, suppose there exist real polynomials functions \(p(x)\) and \(q(x) \neq 0\), such that \[f(x) = \frac{p(x)}{q(x)}\]

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

\[q(x) = (x-a_1)^{j_1}\cdots(x-a_m)^{j_m}(x^2+b_1x+c_1)^{k_1}\cdots(x^2 + b_n x + c_n)^{k_n}\]

where \(a_1, \dots, a_m\), \(b_1, \dots, b_n\), \(c_1, \dots, c_n\) are real numbers with \(b_{i}^{2} -4c_{i} < 0\), and \(j_1, \dots, j_m\), \(k_1, \dots, k_n\) 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 \(( x_i^2 + b_ix + c_i )\) 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 \(f(x)\) is the following:

\[f(x) = \frac{p(x)}{q(x)} = P(x) + \sum_{i=1}^m\sum_{r=1}^{j_i} \frac{A_{ir}}{(x-a_i)^r} + \sum_{i=1}^n\sum_{r=1}^{k_i} \frac{B_{ir}x+C_{ir}}{(x^2+b_ix+c_i)^r}\]

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.

Condensed: the full section is in Wikipedia.

\[f(x)=\frac{1}{x^2+2x-3}\]

Here, the denominator splits into two distinct linear factors:

\[q(x)=x^2+2x-3=(x+3)(x-1)\]

so we have the partial fraction decomposition

\[f(x)=\frac{1}{x^2+2x-3} =\frac{A}{x+3}+\frac{B}{x-1}\]

Multiplying through by the denominator on the left-hand side gives us the polynomial identity

\[1=A(x-1)+B(x+3)\]

Condensed: the full section is in Wikipedia.

\[f(x)=\frac{x^3+16}{x^3-4x^2+8x}\]

After long division, we have

\[f(x)=1+\frac{4x^2-8x+16}{x^3-4x^2+8x}=1+\frac{4x^2-8x+16}{x(x^2-4x+8)}\]

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

\[\frac{4x^2-8x+16}{x(x^2-4x+8)}=\frac{A}{x}+\frac{Bx+C}{x^2-4x+8}\]

Multiplying through by x − 4x + 8x, we have the polynomial identity

\[4x^2-8x+16 = A \left(x^2-4x+8\right) + \left(Bx+C\right)x\]

Condensed: the full section is in Wikipedia.