Partial fraction decomposition

Template:Short description Template:More footnotes 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.<ref>Template:Cite book</ref>

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,<ref>Horowitz, Ellis. "Algorithms for partial fraction decomposition and rational function integration." Proceedings of the second ACM symposium on Symbolic and algebraic manipulation. ACM, 1971.</ref> Taylor series expansions, inverse Z-transforms, and inverse Laplace transforms. The concept was discovered independently in 1702 by both Johann Bernoulli and Gottfried Leibniz.<ref>Template:Cite book</ref>

In symbols, the partial fraction decomposition of a rational fraction of the form <math display="inline"> \frac{f(x)}{g(x)}, </math> where Template:Math and Template:Math are polynomials, is the expression of the rational fraction as

<math display="block">\frac{f(x)}{g(x)}=p(x) + \sum_j \frac{f_j(x)}{g_j(x)} </math>

where Template:Math is a polynomial, and, for each Template:Mvar, the denominator Template:Math is a power of an irreducible polynomial (i.e. not factorizable into polynomials of positive degrees), and the numerator Template:Math 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 <math display="block">R(x) = \frac FG</math> be a rational fraction, where Template:Mvar and Template:Mvar are univariate polynomials in the indeterminate Template:Math over a field. The existence of the partial fraction can be proved by applying inductively the following reduction steps.

Polynomial part

There exist two polynomials Template:Mvar and Template:Math such that <math display="block">\frac FG=E+\frac{F_1}G,</math> and <math display="block">\deg F_1 <\deg G,</math> where <math>\deg P</math> denotes the degree of the polynomial Template:Mvar.

This results immediately from the Euclidean division of Template:Mvar by Template:Mvar, which asserts the existence of Template:Mvar and Template:Math such that <math>F = EG + F_1</math> and <math>\deg F_1 < \deg G.</math>

This allows supposing in the next steps that <math>\deg F <\deg G.</math>

Factors of the denominator

If <math>\deg F < \deg G,</math> and <math display="block">G = G_1 G_2,</math> where Template:Math and Template:Math are coprime polynomials, then there exist polynomials <math>F_1</math> and <math>F_2</math> such that <math display="block">\frac FG=\frac{F_1}{G_1}+\frac{F_2}{G_2},</math> and <math display="block">\deg F_1 < \deg G_1\quad\text{and}\quad\deg F_2 < \deg G_2.</math>

This can be proved as follows. Bézout's identity asserts the existence of polynomials Template:Math and Template:Math such that <math display="block">CG_1 + DG_2 = 1</math> (by hypothesis, Template:Math is a greatest common divisor of Template:Math and Template:Math).

Let <math>DF=G_1Q+F_1</math> with <math>\deg F_1 < \deg G_1</math> be the Euclidean division of Template:Mvar by <math>G_1.</math> Setting <math>F_2=CF+QG_2,</math> one gets <math display="block">\begin{align} \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{align}</math> It remains to show that <math>\deg F_2 < \deg G_2.</math> By reducing the last sum of fractions to a common denominator, one gets <math>F=F_2G_1+F_1G_2,</math> and thus <math display="block">\begin{align} \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{align}</math>

Powers in the denominator

Using the preceding decomposition inductively one gets fractions of the form <math>\frac F {G^k},</math> with <math>\deg F < \deg G^k= k\deg G,</math> where Template:Mvar is an irreducible polynomial. If Template:Math, one can decompose further, by using that an irreducible polynomial is a square-free polynomial, that is, <math>1</math> is a greatest common divisor of the polynomial and its derivative. If <math>G'</math> is the derivative of Template:Mvar, Bézout's identity provides polynomials Template:Mvar and Template:Mvar such that <math>CG + DG' = 1</math> and thus <math>F=FCG+FDG'.</math> Euclidean division of <math>FDG'</math> by <math>G</math> gives polynomials <math>H_k</math> and <math>Q</math> such that <math>FDG' = QG + H_k</math> and <math>\deg H_k < \deg G.</math> Setting <math>F_{k-1}=FC+Q,</math> one gets <math display="block">\frac F {G^k} = \frac{H_k}{G^k}+\frac{F_{k-1}}{G^{k-1}},</math> with <math>\deg H_k <\deg G.</math>

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

Statement

Template:Math theorem{p_i^j}.</math>

If Template:Math, then Template:Math.}}

The uniqueness can be proved as follows. Let Template:Math. All together, Template:Math and the Template:Math have Template:Mvar coefficients. The shape of the decomposition defines a linear map from coefficient vectors to polynomials Template:Mvar of degree less than Template:Mvar. 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 Template:Math is the field of complex numbers, the fundamental theorem of algebra implies that all Template:Math have degree one, and all numerators <math>a_{ij}</math> are constants. When Template:Math is the field of real numbers, some of the Template:Math 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 Template:Math may be the factors of the square-free factorization of Template:Math. When Template:Math 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

Template:Math theorem{p_i^{j-1}}\right)' + \sum_{i=1}^k \frac{c_{i1}}{p_i}.</math> where <math> X'</math> denotes the derivative of <math>X.</math>}}

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

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

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

• If <math>\deg P \geq \deg Q, </math> then it is necessary to perform the Euclidean division of P by Q, using polynomial long division, giving Template:Math with Template:Math. Dividing by Q(x) this gives <math display="block">\frac{P(x)}{Q(x)} = E(x) + \frac{R(x)}{Q(x)},</math> and then seek partial fractions for the remainder fraction (which by definition satisfies Template:Math).
• 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 Template:Math, rather than as a constant. For example, take the following decomposition over R: <math display="block">\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}.</math>
• Suppose Template:Math and Template:Math, that is Template:Math is a root of Template:Math of multiplicity Template:Mvar. In the partial fraction decomposition, the Template:Mvar first powers of Template:Math will occur as denominators of the partial fractions (possibly with a zero numerator). For example, if Template:Math the partial fraction decomposition has the form <math display="block">\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}.</math>

Illustration

In an example application of this procedure, Template:Math can be decomposed in the form

<math display="block">\frac{3x + 5}{(1-2x)^2} = \frac{A}{(1-2x)^2} + \frac{B}{(1-2x)}.</math>

Clearing denominators shows that Template:Math. Expanding and equating the coefficients of powers of Template:Math gives Template:Block indent

Solving this system of linear equations for Template:Math and Template:Math yields Template:Math. Hence,

<math display="block">\frac{3x + 5}{(1-2x)^2} = \frac{13/2}{(1-2x)^2} + \frac{-3/2}{(1-2x)}.</math>

Residue method

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

<math display="block">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). </math>

Let <math display="block"> g_{ij}(x) = (x - x_i)^{j-1}f(x),</math> then according to the uniqueness of Laurent series, a_ij is the coefficient of the term Template:Math in the Laurent expansion of g_ij(x) about the point x_i, i.e., its residue <math display="block">a_{ij} = \operatorname{Res}(g_{ij},x_i).</math>

This is given directly by the formula <math display="block">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),</math> or in the special case when x_i is a simple root, <math display="block">a_{i1}=\frac{P(x_i)}{Q'(x_i)},</math> when <math display="block">f(x)=\frac{P(x)}{Q(x)}.</math>

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

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

General result

Let <math>f(x)</math> be any rational function over the real numbers. In other words, suppose there exist real polynomials functions <math>p(x)</math> and <math>q(x) \neq 0</math>, such that <math display="block">f(x) = \frac{p(x)}{q(x)}</math>

By dividing both the numerator and the denominator by the leading coefficient of <math>q(x)</math>, we may assume without loss of generality that <math>q(x)</math> is monic. By the fundamental theorem of algebra, we can write

<math display="block">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}</math>

where <math>a_1, \dots, a_m</math>, <math>b_1, \dots, b_n</math>, <math>c_1, \dots, c_n</math> are real numbers with <math>b_{i}^{2} -4c_{i} < 0</math>, and <math>j_1, \dots, j_m </math>, <math>k_1, \dots, k_n </math> are positive integers. The terms <math> (x -a_i) </math> are the linear factors of <math>q(x)</math> which correspond to real roots of <math>q(x)</math>, and the terms <math> ( x_i^2 + b_ix + c_i ) </math> are the irreducible quadratic factors of <math>q(x)</math> which correspond to pairs of complex conjugate roots of <math>q(x)</math>.

Then the partial fraction decomposition of <math>f(x)</math> is the following:

<math display="block">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}</math>

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

Example 1

<math display="block">f(x)=\frac{1}{x^2+2x-3}</math>

Here, the denominator splits into two distinct linear factors:

<math display="block">q(x)=x^2+2x-3=(x+3)(x-1)</math>

so we have the partial fraction decomposition

<math display="block">f(x)=\frac{1}{x^2+2x-3} =\frac{A}{x+3}+\frac{B}{x-1}</math>

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

<math display="block">1=A(x-1)+B(x+3)</math>

Substituting x = −3 into this equation gives A = −1/4, and substituting x = 1 gives B = 1/4, so that

<math display="block">f(x) =\frac{1}{x^2+2x-3} =\frac{1}{4}\left(\frac{-1}{x+3}+\frac{1}{x-1}\right)</math>

Example 2

<math display="block">f(x)=\frac{x^3+16}{x^3-4x^2+8x}</math>

After long division, we have

<math display="block">f(x)=1+\frac{4x^2-8x+16}{x^3-4x^2+8x}=1+\frac{4x^2-8x+16}{x(x^2-4x+8)}</math>

The factor x² − 4x + 8 is irreducible over the reals, as its discriminant Template:Math is negative. Thus the partial fraction decomposition over the reals has the shape

<math display="block">\frac{4x^2-8x+16}{x(x^2-4x+8)}=\frac{A}{x}+\frac{Bx+C}{x^2-4x+8}</math>

Multiplying through by x³ − 4x² + 8x, we have the polynomial identity

<math display="block">4x^2-8x+16 = A \left(x^2-4x+8\right) + \left(Bx+C\right)x</math>

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,

<math display="block">f(x)=1+2\left(\frac{1}{x}+\frac{x}{x^2-4x+8}\right)</math>

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:

<math display="block">\frac{x}{x^2-4x+8}=\frac{D}{x-(2+2i)}+\frac{E}{x-(2-2i)}</math>

Multiplying through by the denominator gives:

<math display="block">x=D(x-(2-2i))+E(x-(2+2i)) </math>

Equating the coefficients of Template:Math and the constant (with respect to Template:Math) coefficients of both sides of this equation, one gets a system of two linear equations in Template:Math and Template:Math, whose solution is

<math display="block">D=\frac{1+i}{2i}=\frac{1-i}{2}, \qquad E=\frac{1-i}{-2i}=\frac{1+i}{2}.</math>

Thus we have a complete decomposition:

<math display="block">f(x)=\frac{x^3+16}{x^3-4x^2+8x}=1+\frac{2}{x}+\frac{1-i}{x-(2+2i)}+\frac{1+i}{x-(2-2i)}</math>

One may also compute directly Template:Math and Template:Math with the residue method (see also example 4 below).

Example 3

This example illustrates almost all the "tricks" we might need to use, short of consulting a computer algebra system.

<math display="block">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}</math>

After long division and factoring the denominator, we have

<math display="block">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}</math>

The partial fraction decomposition takes the form

<math display="block">\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}.</math>

Multiplying through by the denominator on the left-hand side we have the polynomial identity

<math display="block">\begin{align} &2x^6 - 4x^5 + 5x^4 - 3x^3 + x^2 + 3x \\[4pt] ={}&A\left(x-1\right)^2 \left(x^2+1\right)^2+B\left(x-1\right)\left(x^2+1\right)^2 +C\left(x^2+1\right)^2 + \left(Dx+E\right)\left(x-1\right)^3\left(x^2+1\right)+\left(Fx+G\right)\left(x-1\right)^3 \end{align}</math>

Now we use different values of x to compute the coefficients:

<math display="block">\begin{cases} 4 = 4C & x =1 \\ 2 + 2i = (Fi + G) (2+ 2i) & x = i \\ 0 = A- B +C - E - G & x = 0 \end{cases}</math>

Solving this we have:

<math display="block">\begin{cases} C = 1 \\ F =0, G =1 \\ E = A-B\end{cases}</math>

Using these values we can write:

<math display="block">\begin{align} &2x^6-4x^5+5x^4-3x^3+x^2+3x \\[4pt] ={}& A\left(x-1\right)^2 \left(x^2+1\right)^2 + B\left(x-1\right)\left(x^2+1\right)^2 + \left(x^2 + 1\right)^2 + \left(Dx + \left(A-B\right)\right)\left(x-1\right)^3 \left(x^2+1\right) + \left(x-1\right)^3 \\[4pt] ={}& \left(A + D\right) x^6 + \left(-A - 3D\right) x^5 + \left(2B + 4D + 1\right) x^4 + \left(-2B - 4D + 1\right) x^3 + \left(-A + 2B + 3D - 1\right) x^2 + \left(A - 2B - D + 3\right) x \end{align}</math>

We compare the coefficients of x⁶ and x⁵ on both side and we have:

<math display="block">\begin{cases} A+D=2 \\ -A-3D = -4 \end{cases} \quad \Rightarrow \quad A= D = 1.</math>

Therefore:

<math display="block">2x^6-4x^5+5x^4-3x^3+x^2+3x = 2x^6 -4x^5 + (2B + 5) x^4 + (-2B - 3) x^3 + (2B +1) x^2 + (- 2B + 3) x</math>

which gives us B = 0. Thus the partial fraction decomposition is given by:

<math display="block">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}.</math>

Alternatively, instead of expanding, one can obtain other linear dependences on the coefficients computing some derivatives at <math>x = 1, \imath</math> 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 is just p(a) for m = 1.) For instance the first derivative at x = 1 gives

<math display="block"> 2\cdot6-4\cdot5+5\cdot4-3\cdot3+2+3 = A\cdot(0+0) + B\cdot( 4+ 0) + 8 + D\cdot0 </math>

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

Example 4 (residue method)

<math display="block"> f(z)=\frac{z^{2}-5}{(z^2-1)(z^2+1)}=\frac{z^{2}-5}{(z+1)(z-1)(z+i)(z-i)}</math>

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 <math display="block">\frac{P(z_i)}{Q'(z_i)} = \frac{z_i^2 - 5}{4z_i^3},</math> are <math display="block"> 1, -1, \tfrac{3i}{2}, -\tfrac{3i}{2},</math> respectively, and

<math display="block"> f(z)=\frac{1}{z+1}-\frac{1}{z-1}+\frac{3i}{2}\frac{1}{z+i}-\frac{3i}{2}\frac{1}{z-i}. </math>

Example 5 (limit method)

Limits can be used to find a partial fraction decomposition.<ref>Template:Cite book</ref> Consider the following example:

<math display="block"> \frac{1}{x^3 - 1}</math>

First, factor the denominator which determines the decomposition:

<math display="block"> \frac{1}{x^3 - 1} = \frac{1}{(x - 1)(x^2 + x + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1}.</math>

Multiplying everything by <math>x-1</math>, and taking the limit when <math>x \to 1</math>, we get

<math display="block">\lim_{x \to 1} \left((x-1)\left ( \frac{A}{x-1} + \frac{Bx + C}{x^2 + x + 1} \right )\right) = \lim_{x \to 1} A + \lim_{x \to 1}\frac{(x-1)(Bx + C)}{x^2 + x + 1} =A.</math>

On the other hand,

<math display="block">\lim_{x \to 1} \frac{(x-1)}{(x - 1)(x^2 + x + 1)} = \lim_{x \to 1}\frac{1}{x^2 + x + 1} = \frac13,</math>

and thus:

<math display="block">A = \frac{1}{3}.</math>

Multiplying by Template:Math and taking the limit when <math>x \to \infty</math>, we have

<math display="block">\lim_{x \to \infty} x\left( \frac{A}{x-1} + \frac{Bx + C}{x^2 + x + 1} \right )= \lim_{x \to \infty} \frac{Ax}{x-1} + \lim_{x \to \infty} \frac{Bx^2+Cx}{x^2+x+1}= A+B,</math>

and

<math display="block">\lim_{x \to \infty} \frac{x}{(x - 1)(x^2 + x + 1)} =0.</math>

This implies Template:Math and so <math>B = -\frac{1}{3}</math>.

For Template:Math, we get <math>-1 = -A + C,</math> and thus <math>C = -\tfrac{2}{3}</math>.

Putting everything together, we get the decomposition

<math display="block">\frac{1}{x^3 -1} = \frac{1}{3} \left( \frac{1}{x - 1} + \frac{-x -2}{x^2 + x + 1} \right ).</math>

Example 6 (integral)

Suppose we have the indefinite integral:

<math display="block">\int \frac{x^4+x^3+x^2+1}{x^2+x-2} \,dx</math>

Before performing decomposition, it is obvious we must perform polynomial long division and factor the denominator. Doing this would result in:

<math display="block">\int \left(x^2 + 3 + \frac{-3x+7}{(x+2)(x-1)}\right) dx</math>

Upon this, we may now perform partial fraction decomposition.

<math display="block">\int \left(x^2+3+ \frac{-3x+7}{(x+2)(x-1)}\right) dx = \int \left(x^2+3+ \frac{A}{(x+2)}+\frac{B}{(x-1)}\right) dx</math> so: <math display="block">A(x-1)+B(x+2)=-3x+7</math>. 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:

<math display="block">A=\frac{-13}{3} \ , B=\frac{4}{3} </math>

Plugging all of this back into our integral allows us to find the answer:

<math display="block">\int \left(x^2+3+ \frac{-13/3}{(x+2)}+\frac{4/3}{(x-1)}\right) \,dx = \frac{x^3}{3} \ + 3x-\frac{13}{3} \ln(|x+2|)+\frac{4}{3} \ln(|x-1|)+C </math>

The role of the Taylor polynomial

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

<math display="block">P(x), Q(x), A_1(x),\ldots, A_r(x)</math>

be real or complex polynomials assume that

<math display="block">Q=\prod_{j=1}^{r}(x-\lambda_j)^{\nu_j},</math>

satisfies <math display="block">\deg A_1<\nu_1, \ldots, \deg A_r<\nu_r, \quad \text{and} \quad \deg(P)<\deg(Q)=\sum_{j=1}^{r}\nu_j.</math>

Also define

<math display="block">Q_i=\prod_{j\neq i}(x-\lambda_j)^{\nu_j}=\frac{Q}{(x-\lambda_i)^{\nu_i}}, \qquad 1 \leqslant i \leqslant r.</math>

Then we have

<math display="block">\frac{P}{Q}=\sum_{j=1}^{r}\frac{A_j}{(x-\lambda_j)^{\nu_j}}</math>

if, and only if, each polynomial <math>A_i(x)</math> is the Taylor polynomial of <math>\tfrac{P}{Q_i}</math> of order <math>\nu_i-1</math> at the point <math>\lambda_i</math>:

<math display="block">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. </math>

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

<math display="block">\frac{P}{Q_i}=A_i + O((x-\lambda_i)^{\nu_i}), \qquad \text{for } x\to\lambda_i,</math>

so <math>A_i</math> is the Taylor polynomial of <math>\tfrac{P}{Q_i}</math>, because of the unicity of the polynomial expansion of order <math>\nu_i-1</math>, and by assumption <math>\deg A_i<\nu_i</math>.

Conversely, if the <math>A_i</math> are the Taylor polynomials, the above expansions at each <math>\lambda_i</math> hold, therefore we also have

<math display="block">P-Q_i A_i = O((x-\lambda_i)^{\nu_i}), \qquad \text{for } x\to\lambda_i,</math>

which implies that the polynomial <math> P-Q_iA_i</math> is divisible by <math> (x-\lambda_i)^{\nu_i}.</math>

For <math> j\neq i, Q_jA_j</math> is also divisible by <math>(x-\lambda_i)^{\nu_i}</math>, so

<math display="block"> P- \sum_{j=1}^{r}Q_jA_j</math>

is divisible by <math>Q</math>. Since

<math display="block"> \deg\left( P- \sum_{j=1}^{r}Q_jA_j \right) < \deg(Q)</math>

we then have

<math display="block"> P- \sum_{j=1}^{r}Q_jA_j=0,</math>

and we find the partial fraction decomposition dividing by <math> Q</math>.

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:

<math display="block">\frac{1}{18} = \frac{1}{2} - \frac{1}{3} - \frac{1}{3^2}. </math>

References

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External links

• {{#invoke:Template wrapper|{{#if:|list|wrap}}|_template=cite web

|_exclude=urlname, _debug, id |url = https://mathworld.wolfram.com/{{#if:PartialFractionDecomposition%7CPartialFractionDecomposition.html}} |title = Partial Fraction Decomposition |author = Weisstein, Eric W. |website = MathWorld |access-date = |ref = Template:SfnRef }}

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• Make partial fraction decompositions with Scilab.