# Partial fraction

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In algebra, the partial fraction decomposition or partial fraction expansion is a procedure used to reduce the degree of either the numerator or the denominator of a rational function (also known as a rational algebraic fraction).

In symbols, one can use partial fraction expansion to change a rational function in the form

${\frac {f(x)}{g(x)}}$ where ƒ and g are polynomials, into a function of the form

$\sum _{j}{\frac {f_{j}(x)}{g_{j}(x)}}$ where gj (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 algebraic fractions, that produces a single rational function with a numerator and denominator usually of high degree. The full decomposition pushes the reduction as far as it will 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 function as a sum of fractions, where:

• the denominator of each term is a power of an irreducible (not factorable) polynomial and
• the numerator is a polynomial of smaller degree than that irreducible polynomial. To decrease the degree of the numerator directly, the Euclidean algorithm can be used, but in fact if ƒ already has lower degree than g this isn't helpful.

The main motivation to decompose a rational function into a sum of simpler fractions is that it makes it simpler to perform linear operations on it. Therefore the problem of computing derivatives, antiderivatives, integrals, power series expansions, Fourier series, residues, and linear functional transformations of rational functions can be reduced, via partial fraction decomposition, to making the computation on each single element used in the decomposition. See e.g. partial fractions in integration for an account of the use of the partial fractions in finding antiderivatives. Just which polynomials are irreducible depends on which field of scalars one adopts. Thus if one allows only real numbers, then irreducible polynomials are of degree either 1 or 2. If complex numbers are allowed, only 1st-degree polynomials can be irreducible. If one allows only rational numbers, or a finite field, then some higher-degree polynomials are irreducible.

## Basic principles

The basic principles involved are quite simple; it is the algorithmic aspects that require attention in particular cases. On the other hand, the existence of a decomposition of a certain kind is an assumption in practical cases, and the principles should explain which assumptions are justified.

Assume a rational function R(x) = ƒ(x)/g(x) in one indeterminate x has a denominator that factors as

$g(x)=P(x)\cdot Q(x)\,$ over a field K (we can take this to be real numbers, or complex numbers). If P and Q have no common factor, then R may be written as

${\frac {A}{P}}+{\frac {B}{Q}}$ for some polynomials A(x) and B(x) over K. The existence of such a decomposition is a consequence of the fact that the polynomial ring over K is a principal ideal domain, so that

$CP+DQ=1\,$ for some polynomials C(x) and D(x) (see Bézout's identity).

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:

${\frac {G(x)}{F(x)^{n}}}$ 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:

Let ƒ 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 ƒ < deg g, then b = 0.

Therefore when the field K is the complex numbers, we can assume that each pi has degree 1 (by the fundamental theorem of algebra) the numerators will be constant. When K is the real numbers, some of the pi might be quadratic, so in the partial fraction decomposition a quotient of a linear polynomial by a power of a quadratic will occur.

In the preceding theorem, one may replace "distinct irreducible polynomials" by "pairwise coprime polynomials that are coprime with their derivative". For example, the pi may be the factors of the square-free 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

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:

$g=\prod _{i=1}^{k}p_{i}^{n_{i}}.$ There are (unique) polynomials b and c ij with deg c ij < deg p i such that

${\frac {f}{g}}=b+\sum _{i=1}^{k}\sum _{j=2}^{n_{i}}\left({\frac {c_{ij}}{p_{i}^{j-1}}}\right)'+\sum _{i=1}^{k}{\frac {c_{i1}}{p_{i}}}.$ This reduces the computation of the antiderivative of a rational function to the integration of the last sum, with is called the logarithmic part, because its antiderivative is a linear combination of logarithms.

## Procedure

Given two polynomials $P(x)$ and $Q(x)=(x-\alpha _{1})(x-\alpha _{2})\cdots (x-\alpha _{n})$ , where the αi are distinct constants and deg P < n, partial fractions are generally 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 ci 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

${\frac {P(x)}{Q(x)}}=\sum _{i=1}^{n}{\frac {P(\alpha _{i})}{Q'(\alpha _{i})}}{\frac {1}{(x-\alpha _{i})}}$ This approach does not account for several other cases, but can be modified accordingly:

${\frac {P(x)}{Q(x)}}=E(x)+{\frac {R(x)}{Q(x)}},$ and then seek partial fractions for the remainder fraction (which by definition satisfies deg R < deg Q).
• If Q(x) contains 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) = (xα)rS(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 (xα). For illustration, take S(x) = 1 to get the following decomposition:
${\frac {P(x)}{Q(x)}}={\frac {P(x)}{(x-\alpha )^{r}}}={\frac {c_{1}}{x-\alpha }}+{\frac {c_{2}}{(x-\alpha )^{2}}}+\cdots +{\frac {c_{r}}{(x-\alpha )^{r}}}.$ ### Illustration

In an example application of this procedure, (3x + 5)/(1 − 2x)2 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

5 = A + B and 3x = −2Bx

Solving for A and B yields A = 13/2 and B = −3/2. Hence,

${\frac {3x+5}{(1-2x)^{2}}}={\frac {13/2}{(1-2x)^{2}}}+{\frac {-3/2}{(1-2x)}}.$ ### Residue method

{{#invoke:see also|seealso}} Over the complex numbers, suppose ƒ(x) is a rational proper fraction, and can be decomposed into

$f(x)=\sum _{i}\left({\frac {a_{i1}}{x-x_{i}}}+{\frac {a_{i2}}{(x-x_{i})^{2}}}+\cdots +{\frac {a_{ik_{i}}}{(x-x_{i})^{k_{i}}}}\right).$ Let

$g_{ij}(x)=(x-x_{i})^{j-1}f(x),$ then according to the uniqueness of Laurent series, aij is the coefficient of the term (x − xi)−1 in the Laurent expansion of gij(x) about the point xi, 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 xi is a simple root,

$a_{i1}={\frac {P(x_{i})}{Q'(x_{i})}},$ when

$f(x)={\frac {P(x)}{Q(x)}}.$ Note that P(x) and Q(x) may or may not be polynomials.

## 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

### General result

Let ƒ(x) be any rational function over the real numbers. In other words, suppose there exist real polynomials p(x) and q(x)≠ 0, such that

$f(x)={\frac {p(x)}{q(x)}}$ By removing 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

$q(x)=(x-a_{1})^{j_{1}}\cdots (x-a_{m})^{j_{m}}(x^{2}+b_{1}x+c_{1})^{k_{1}}\cdots (x^{2}+b_{n}x+c_{n})^{k_{n}}$ where a1,..., am, b1,..., bn, c1,..., cn are real numbers with bi2 - 4ci < 0, and j1,..., jm, k1,..., kn are positive integers. The terms (x - ai) are the linear factors of q(x) which correspond to real roots of q(x), and the terms (xi2 + bix + ci) 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:

$f(x)={\frac {p(x)}{q(x)}}=P(x)+\sum _{i=1}^{m}\sum _{r=1}^{j_{i}}{\frac {A_{ir}}{(x-a_{i})^{r}}}+\sum _{i=1}^{n}\sum _{r=1}^{k_{i}}{\frac {B_{ir}x+C_{ir}}{(x^{2}+b_{i}x+c_{i})^{r}}}$ Here, P(x) is a (possibly zero) 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(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 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.

## Examples

### Example 1

$f(x)={\frac {1}{x^{2}+2x-3}}$ 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 x2 + 2x - 3, we have the polynomial identity

$1=A(x-1)+B(x+3)$ Substituting x = -3 into this equation gives A = -1/4, and substituting x = 1 gives B = 1/4, so that

$f(x)={\frac {1}{x^{2}+2x-3}}={\frac {1}{4}}\left({\frac {-1}{x+3}}+{\frac {1}{x-1}}\right)$ ### Example 2

$f(x)={\frac {x^{3}+16}{x^{3}-4x^{2}+8x}}$ After long-division, we have

$f(x)=1+{\frac {4x^{2}-8x+16}{x^{3}-4x^{2}+8x}}=1+{\frac {4x^{2}-8x+16}{x(x^{2}-4x+8)}}$ Since (−4)2 − 4(8) = −16 < 0, x2 − 4x + 8 is irreducible, and so

${\frac {4x^{2}-8x+16}{x(x^{2}-4x+8)}}={\frac {A}{x}}+{\frac {Bx+C}{x^{2}-4x+8}}$ Multiplying through by x3 − 4x2 + 8x, we have the polynomial identity

$4x^{2}-8x+16=A(x^{2}-4x+8)+(Bx+C)x$ 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,

$f(x)=1+2\left({\frac {1}{x}}+{\frac {x}{x^{2}-4x+8}}\right)$ The following example illustrates almost all the "tricks" one would need to use short of consulting a computer algebra system.

### Example 3

$f(x)={\frac {x^{9}-2x^{6}+2x^{5}-7x^{4}+13x^{3}-11x^{2}+12x-4}{x^{7}-3x^{6}+5x^{5}-7x^{4}+7x^{3}-5x^{2}+3x-1}}$ After long-division and factoring, we have

$f(x)=x^{2}+3x+4+{\frac {2x^{6}-4x^{5}+5x^{4}-3x^{3}+x^{2}+3x}{(x-1)^{3}(x^{2}+1)^{2}}}$ The partial fraction decomposition takes the form

${\frac {2x^{6}-4x^{5}+5x^{4}-3x^{3}+x^{2}+3x}{(x-1)^{3}(x^{2}+1)^{2}}}={\frac {A}{x-1}}+{\frac {B}{(x-1)^{2}}}+{\frac {C}{(x-1)^{3}}}+{\frac {Dx+E}{x^{2}+1}}+{\frac {Fx+G}{(x^{2}+1)^{2}}}$ Multiplying through by (x − 1)3(x2 + 1)2 we have the polynomial identity

{\begin{aligned}&{}\quad 2x^{6}-4x^{5}+5x^{4}-3x^{3}+x^{2}+3x\\&=A(x-1)^{2}(x^{2}+1)^{2}+B(x-1)(x^{2}+1)^{2}+C(x^{2}+1)^{2}+(Dx+E)(x-1)^{3}(x^{2}+1)+(Fx+G)(x-1)^{3}\end{aligned}} 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

{\begin{aligned}&{}2x^{6}-4x^{5}+5x^{4}-3x^{3}+x^{2}+3x\\&=A(x-1)^{2}(x^{2}+1)^{2}+B(x-1)(x^{2}+1)^{2}+(x^{2}+1)^{2}+(Dx+(A-B))(x-1)^{3}(x^{2}+1)+(x-1)^{3}\\&=A((x-1)^{2}(x^{2}+1)^{2}+(x-1)^{3}(x^{2}+1))+B((x-1)(x^{2}+1)-(x-1)^{3}(x^{2}+1))+(x^{2}+1)^{2}+Dx(x-1)^{3}(x^{2}+1)+(x-1)^{3}\end{aligned}} Expanding and sorting by exponents of x we get

{\begin{aligned}&{}2x^{6}-4x^{5}+5x^{4}-3x^{3}+x^{2}+3x\\&=(A+D)x^{6}+(-A-3D)x^{5}+(2B+4D+1)x^{4}+(-2B-4D+1)x^{3}+(-A+2B+3D-1)x^{2}+(A-2B-D+3)x\end{aligned}} We can now compare the coefficients and see that

{\begin{aligned}A+D&=&2\\-A-3D&=&-4\\2B+4D+1&=&5\\-2B-4D+1&=&-3\\-A+2B+3D-1&=&1\\A-2B-D+3&=&3,\end{aligned}} with A = 2 - D 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

$f(x)=x^{2}+3x+4+{\frac {1}{(x-1)}}+{\frac {1}{(x-1)^{3}}}+{\frac {x+1}{x^{2}+1}}+{\frac {1}{(x^{2}+1)^{2}}}.$ 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)mp(x) vanishes if m > 1 and it is just p(a) if m=1.) Thus, for instance the first derivative at x=1 gives

$2\cdot 6-4\cdot 5+5\cdot 4-3\cdot 3+2+3=A\cdot (0+0)+B\cdot (2+0)+8+D\cdot 0$ that is 8 = 2B + 8 so B=0.

### Example 4 (residue method)

$f(z)={\frac {z^{2}-5}{(z^{2}-1)(z^{2}+1)}}={\frac {z^{2}-5}{(z+1)(z-1)(z+i)(z-i)}}$ 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

${\frac {P(z_{i})}{Q'(z_{i})}}={\frac {z_{i}^{2}-5}{4z_{i}^{3}}}$ ,

are

$1,-1,{\tfrac {3i}{2}},-{\tfrac {3i}{2}}$ ,

respectively, and

$f(z)={\frac {1}{z+1}}-{\frac {1}{z-1}}+{\frac {3i}{2}}{\frac {1}{z+i}}-{\frac {3i}{2}}{\frac {1}{z-i}}$ .

## The role of the Taylor polynomial

The partial fraction decomposition of a rational function can be related to Taylor's theorem as follows. Let

$P(x),Q(x),A_{1}(x),\dots ,A_{r}(x)$ be real or complex polynomials; assume that

$\textstyle Q=\prod _{j=1}^{r}(x-\lambda _{j})^{\nu _{j}},$ that

$\textstyle \deg(P)<\deg(Q)=\sum _{j=1}^{r}\nu _{j},$ and that

$\textstyle \deg A_{j}<\nu _{j}{\text{ for }}j=1,\dots ,r.$ Define also

$\textstyle Q_{i}=\prod _{j\neq i}(x-\lambda _{j})^{\nu _{j}}={\frac {Q}{(x-\lambda _{i})^{\nu _{i}}}}{\text{ for }}i=1,\dots ,r.$ Then we have

${\frac {P}{Q}}=\sum _{j=1}^{r}{\frac {A_{j}}{(x-\lambda _{j})^{\nu _{j}}}}$ $A_{i}(x):=\sum _{k=0}^{\nu _{i}-1}{\frac {1}{k!}}\left({\frac {P}{Q_{i}}}\right)^{(k)}(\lambda _{i})\ (x-\lambda _{i})^{k}.$ 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

${\frac {P}{Q_{i}}}=A_{i}+O((x-\lambda _{i})^{\nu _{i}})\qquad$ , as $x\to \lambda _{i};$ Conversely, if the $\textstyle A_{i}$ are the Taylor polynomials, the above expansions at each $\textstyle \lambda _{i}$ hold, therefore we also have

$P-Q_{i}A_{i}=O((x-\lambda _{i})^{\nu _{i}})\qquad$ , as $x\to \lambda _{i},$ ## 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:

${\frac {1}{18}}={\frac {1}{2}}-{\frac {1}{3}}-{\frac {1}{3^{2}}}.$ 