Partial fraction decomposition

 ( Algebra ) 

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 ) 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 for various computations with rational functions, including the explicit computation of ,Horowitz, Ellis. " Algorithms for partial fraction decomposition and rational function integration." Proceedings of the second ACM symposium on Symbolic and algebraic manipulation. ACM, 1971. Taylor series, Z-transform, 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 $\backslash frac\{f(x)\}\{g(x)\},$ where and are polynomials, is the expression of the rational fraction as

$$\backslash frac\{f(x)\}\{g(x)\}=p(x)\; +\; \backslash sum\_j\; \backslash frac\{f\_j(x)\}\{g\_j(x)\}$$

where

is a polynomial, and, for each ,
     

the denominator is a Exponentiation of an irreducible polynomial (i.e. not factorizable into polynomials of positive degrees), and the numerator 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 or .

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Basic principles Let $$R(x)\; =\; \backslash frac\; FG$$ be a rational fraction, where and are univariate polynomials in the indeterminate over a field. The existence of the partial fraction can be proved by applying inductively the following reduction steps.

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Polynomial part There exist two polynomials and such that $$\backslash frac\; FG=E+\backslash frac\{F\_1\}G,$$ and where $\backslash deg\; P$ denotes the degree of the polynomial .

This results immediately from the Euclidean division of by , which asserts the existence of and such that $F\; =\; EG\; +\; F\_1$ and

This allows supposing in the next steps that

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Factors of the denominator If and $$G\; =\; G\_1\; G\_2,$$ where and are coprime polynomials, then there exist polynomials $F\_1$ and $F\_2$ such that $$\backslash frac\; FG=\backslash frac\{F\_1\}\{G\_1\}+\backslash frac\{F\_2\}\{G\_2\},$$ and

This can be proved as follows. Bézout's identity asserts the existence of polynomials and such that $$CG\_1\; +\; DG\_2\; =\; 1$$ (by hypothesis, is a greatest common divisor of and ).

Let $DF=G\_1Q+F\_1$ with be the Euclidean division of by $G\_1.$ Setting $F\_2=CF+QG\_2,$ one gets It remains to show that By reducing the last sum of fractions to a common denominator, one gets $F=F\_2G\_1+F\_1G\_2,$ and thus

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Powers in the denominator Using the preceding decomposition inductively one gets fractions of the form $\backslash frac\; F\; \{G^k\},$ with where is an irreducible polynomial. If , 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 $G\text{'}$ is the derivative of , Bézout's identity provides polynomials and such that $CG\; +\; DG\text{'}\; =\; 1$ and thus $F=FCG+FDG\text{'}.$ Euclidean division of $FDG\text{'}$ by $G$ gives polynomials $H\_k$ and $Q$ such that $FDG\text{'}\; =\; QG\; +\; H\_k$ and Setting $F\_\{k-1\}=FC+Q,$ one gets $$\backslash frac\; F\; \{G^k\}\; =\; \backslash frac\{H\_k\}\{G^k\}+\backslash frac\{F\_\{k-1\}\}\{G^\{k-1\}\},$$ with

Iterating this process with $\backslash frac\{F\_\{k-1\}\}\{G^\{k-1\}\}$ in place of $\backslash frac\; F\{G^k\}$ leads eventually to the following theorem.

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Statement {p_i^j}.

The uniqueness can be proved as follows. Let . All together, and the have coefficients. The shape of the decomposition defines a linear map from coefficient vectors to polynomials of degree less than . The existence proof means that this map is surjective. As the two 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 is the field of , the fundamental theorem of algebra implies that all have degree one, and all numerators $a\_\{ij\}$ are constants. When is the field of , some of the 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 may be the factors of the square-free factorization of . When is the field of , 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.

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Application to symbolic integration For the purpose of symbolic integration, the preceding result may be refined into

{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, 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 Charles 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.

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Procedure Given two polynomials $P(x)$ and $Q(x)\; =\; (x-\backslash alpha\_1)(x-\backslash alpha\_2)\; \backslash cdots\; (x-\backslash alpha\_n)$, where the α _n are distinct constants and , explicit expressions for partial fractions can be obtained by supposing that $$\backslash frac\{P(x)\}\{Q(x)\}\; =\; \backslash frac\{c\_1\}\{x-\backslash alpha\_1\}\; +\; \backslash frac\{c\_2\}\{x-\backslash alpha\_2\}\; +\; \backslash cdots\; +\; \backslash frac\{c\_n\}\{x-\backslash alpha\_n\}$$ 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 $$\backslash frac\{P(x)\}\{Q(x)\}\; =\; \backslash sum\_\{i=1\}^n\; \backslash frac\{P(\backslash alpha\_i)\}\{Q\text{'}(\backslash alpha\_i)\}\backslash frac\{1\}\{(x-\backslash alpha\_i)\}$$ where $Q\text{'}$ is the derivative of the polynomial $Q$. The coefficients of $\backslash tfrac\{1\}\{x-\backslash alpha\_j\}$ are called the residues of f/g.

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

• If $\backslash deg\; P\; \backslash geq\; \backslash deg\; Q,$ then it is necessary to perform the Euclidean division of P by Q, using polynomial long division, giving with . Dividing by Q( x) this gives $$\backslash frac\{P(x)\}\{Q(x)\}\; =\; E(x)\; +\; \backslash frac\{R(x)\}\{Q(x)\},$$ and then seek partial fractions for the remainder fraction (which by definition satisfies ).
• 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 , rather than as a constant. For example, take the following decomposition over R: $$\backslash frac\{x^2\; +\; 1\}\{(x+2)(x-1)\backslash color\{Blue\}(x^2+x+1)\}\; =\; \backslash frac\{a\}\{x+2\}\; +\; \backslash frac\{b\}\{x-1\}\; +\; \backslash frac\{\backslash color\{OliveGreen\}cx\; +\; d\}\{\backslash color\{Blue\}x^2\; +\; x\; +\; 1\}.$$
• Suppose and , that is is a root of of multiplicity . In the partial fraction decomposition, the first powers of will occur as denominators of the partial fractions (possibly with a zero numerator). For example, if the partial fraction decomposition has the form $$\backslash frac\{P(x)\}\{Q(x)\}\; =\; \backslash frac\{P(x)\}\{(x-\backslash alpha)^r\}\; =\; \backslash frac\{c\_1\}\{x-\backslash alpha\}\; +\; \backslash frac\{c\_2\}\{(x-\backslash alpha)^2\}\; +\; \backslash cdots\; +\; \backslash frac\{c\_r\}\{(x-\backslash alpha)^r\}.$$

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Illustration In an example application of this procedure, can be decomposed in the form

$$\backslash frac\{3x\; +\; 5\}\{(1-2x)^2\}\; =\; \backslash frac\{A\}\{(1-2x)^2\}\; +\; \backslash frac\{B\}\{(1-2x)\}.$$

Clearing denominators shows that . Expanding and equating the coefficients of powers of gives

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

$$\backslash frac\{3x\; +\; 5\}\{(1-2x)^2\}\; =\; \backslash frac\{13/2\}\{(1-2x)^2\}\; +\; \backslash frac\{-3/2\}\{(1-2x)\}.$$

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

$$f(x)\; =\; \backslash sum\_i\; \backslash left(\; \backslash frac\{a\_\{i1\}\}\{x\; -\; x\_i\}\; +\; \backslash frac\{a\_\{i2\}\}\{(\; x\; -\; x\_i)^2\}\; +\; \backslash cdots\; +\; \backslash frac\{a\_\{i\; k\_i\}\}\{(x\; -\; x\_i)^\{k\_i\}\}\; \backslash right).$$

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 in the Laurent expansion of g _ij( x) about the point x _i, i.e., its residue $$a\_\{ij\}\; =\; \backslash operatorname\{Res\}(g\_\{ij\},x\_i).$$

This is given directly by the formula $$a\_\{ij\}\; =\; \backslash frac\; 1\; \{(k\_i-j)!\}\backslash lim\_\{x\backslash to\; x\_i\}\backslash frac\{d^\{k\_i-j\}\}\{dx^\{k\_i-j\}\}\; \backslash left((x-x\_i)^\{k\_i\}\; f(x)\backslash right),$$ or in the special case when x _i is a simple root, $$a\_\{i1\}=\backslash frac\{P(x\_i)\}\{Q\text{'}(x\_i)\},$$ when $$f(x)=\backslash frac\{P(x)\}\{Q(x)\}.$$

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Over the reals Partial fractions are used in real number integral calculus to find real-valued 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

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

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 polynomial. By the fundamental theorem of algebra, we can write

$$q(x)\; =\; (x-a\_1)^\{j\_1\}\backslash cdots(x-a\_m)^\{j\_m\}(x^2+b\_1x+c\_1)^\{k\_1\}\backslash cdots(x^2\; +\; b\_n\; x\; +\; c\_n)^\{k\_n\}$$

where $a\_1,\; \backslash dots,\; a\_m$, $b\_1,\; \backslash dots,\; b\_n$, $c\_1,\; \backslash dots,\; c\_n$ are real numbers with , and $j\_1,\; \backslash dots,\; j\_m$, $k\_1,\; \backslash dots,\; k\_n$ 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 number conjugate roots of $q(x)$.

Then the partial fraction decomposition of $f(x)$ is the following:

$$f(x)\; =\; \backslash frac\{p(x)\}\{q(x)\}\; =\; P(x)\; +\; \backslash sum\_\{i=1\}^m\backslash sum\_\{r=1\}^\{j\_i\}\; \backslash frac\{A\_\{ir\}\}\{(x-a\_i)^r\}\; +\; \backslash sum\_\{i=1\}^n\backslash sum\_\{r=1\}^\{k\_i\}\; \backslash frac\{B\_\{ir\}x+C\_\{ir\}\}\{(x^2+b\_ix+c\_i)^r\}$$

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.

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).

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Examples

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Example 1 $$f(x)=\backslash frac\{1\}\{x^2+2x-3\}$$

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

$$f(x)=\backslash frac\{1\}\{x^2+2x-3\}\; =\backslash frac\{A\}\{x+3\}+\backslash frac\{B\}\{x-1\}$$

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

$$1=A(x-1)+B(x+3)$$

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

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

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Example 2 $$f(x)=\backslash frac\{x^3+16\}\{x^3-4x^2+8x\}$$

After long division, we have

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

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

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

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

$$4x^2-8x+16\; =\; A\; \backslash left(x^2-4x+8\backslash right)\; +\; \backslash left(Bx+C\backslash right)x$$

Taking x = 0, we see that 16 = 8 A, 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 = −4 A + C = −8 + C, so C = 0. Altogether,

$$f(x)=1+2\backslash left(\backslash frac\{1\}\{x\}+\backslash frac\{x\}\{x^2-4x+8\}\backslash right)$$

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:

$$\backslash frac\{x\}\{x^2-4x+8\}=\backslash frac\{D\}\{x-(2+2i)\}+\backslash frac\{E\}\{x-(2-2i)\}$$

Multiplying through by the denominator gives:

$$x=D(x-(2-2i))+E(x-(2+2i))$$

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

$$D=\backslash frac\{1+i\}\{2i\}=\backslash frac\{1-i\}\{2\},\; \backslash qquad\; E=\backslash frac\{1-i\}\{-2i\}=\backslash frac\{1+i\}\{2\}.$$

Thus we have a complete decomposition:

$$f(x)=\backslash frac\{x^3+16\}\{x^3-4x^2+8x\}=1+\backslash frac\{2\}\{x\}+\backslash frac\{1-i\}\{x-(2+2i)\}+\backslash frac\{1+i\}\{x-(2-2i)\}$$

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

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Example 3 This example illustrates almost all the "tricks" we might need to use, short of consulting a computer algebra system.

$$f(x)=\backslash frac\{x^9-2x^6+2x^5-7x^4+13x^3-11x^2+12x-4\}\{x^7-3x^6+5x^5-7x^4+7x^3-5x^2+3x-1\}$$

After long division and factoring the denominator, we have

$$f(x)=x^2+3x+4+\backslash frac\{2x^6-4x^5+5x^4-3x^3+x^2+3x\}\{(x-1)^3(x^2+1)^2\}$$

The partial fraction decomposition takes the form

$$\backslash frac\{2x^6-4x^5+5x^4-3x^3+x^2+3x\}\{(x-1)^3(x^2+1)^2\}\; =\; \backslash frac\{A\}\{x-1\}+\backslash frac\{B\}\{(x-1)^2\}+\backslash frac\{C\}\{(x-1)^3\}+\backslash frac\{Dx+E\}\{x^2+1\}+\backslash frac\{Fx+G\}\{(x^2+1)^2\}.$$

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:

$$\backslash begin\{cases\}\; C\; =\; 1\; \backslash \backslash \; F\; =0,\; G\; =1\; \backslash \backslash \; E\; =\; A-B\backslash end\{cases\}$$

Using these values we can write:

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

$$\backslash begin\{cases\}\; A+D=2\; \backslash \backslash \; -A-3D\; =\; -4\; \backslash end\{cases\}\; \backslash quad\; \backslash Rightarrow\; \backslash quad\; A=\; D\; =\; 1.$$

Therefore:

$$2x^6-4x^5+5x^4-3x^3+x^2+3x\; =\; 2x^6\; -4x^5\; +\; (2B\; +\; 5)\; x^4\; +\; (-2B\; -\; 3)\; x^3\; +\; (2B\; +1)\; x^2\; +\; (-\; 2B\; +\; 3)\; x$$

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

$$f(x)=x^2+3x+4+\backslash frac\{1\}\{(x-1)\}\; +\; \backslash frac\{1\}\{(x\; -\; 1)^3\}\; +\; \backslash frac\{x\; +\; 1\}\{x^2+1\}+\backslash frac\{1\}\{(x^2+1)^2\}.$$

Alternatively, instead of expanding, one can obtain other linear dependences on the coefficients computing some derivatives at $x\; =\; 1,\; \backslash imath$ 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

$$2\backslash cdot6-4\backslash cdot5+5\backslash cdot4-3\backslash cdot3+2+3\; =\; A\backslash cdot(0+0)\; +\; B\backslash cdot(\; 4+\; 0)\; +\; 8\; +\; D\backslash cdot0$$

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

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Example 4 (residue method) $$f(z)=\backslash frac\{z^\{2\}-5\}\{(z^2-1)(z^2+1)\}=\backslash frac\{z^\{2\}-5\}\{(z+1)(z-1)(z+i)(z-i)\}$$

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 $$\backslash frac\{P(z\_i)\}\{Q\text{'}(z\_i)\}\; =\; \backslash frac\{z\_i^2\; -\; 5\}\{4z\_i^3\},$$ are $$1,\; -1,\; \backslash tfrac\{3i\}\{2\},\; -\backslash tfrac\{3i\}\{2\},$$ respectively, and

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

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Example 5 (limit method) Limits can be used to find a partial fraction decomposition. Consider the following example:

$$\backslash frac\{1\}\{x^3\; -\; 1\}$$

First, factor the denominator which determines the decomposition:

$$\backslash frac\{1\}\{x^3\; -\; 1\}\; =\; \backslash frac\{1\}\{(x\; -\; 1)(x^2\; +\; x\; +\; 1)\}\; =\; \backslash frac\{A\}\{x\; -\; 1\}\; +\; \backslash frac\{Bx\; +\; C\}\{x^2\; +\; x\; +\; 1\}.$$

Multiplying everything by $x-1$, and taking the limit when $x\; \backslash to\; 1$, we get

$$\backslash lim\_\{x\; \backslash to\; 1\}\; \backslash left((x-1)\backslash left\; (\; \backslash frac\{A\}\{x-1\}\; +\; \backslash frac\{Bx\; +\; C\}\{x^2\; +\; x\; +\; 1\}\; \backslash right\; )\backslash right)\; =\; \backslash lim\_\{x\; \backslash to\; 1\}\; A\; +\; \backslash lim\_\{x\; \backslash to\; 1\}\backslash frac\{(x-1)(Bx\; +\; C)\}\{x^2\; +\; x\; +\; 1\}\; =A.$$

On the other hand,

$$\backslash lim\_\{x\; \backslash to\; 1\}\; \backslash frac\{(x-1)\}\{(x\; -\; 1)(x^2\; +\; x\; +\; 1)\}\; =\; \backslash lim\_\{x\; \backslash to\; 1\}\backslash frac\{1\}\{x^2\; +\; x\; +\; 1\}\; =\; \backslash frac13,$$

and thus:

$$A\; =\; \backslash frac\{1\}\{3\}.$$

Multiplying by and taking the limit when $x\; \backslash to\; \backslash infty$, we have

$$\backslash lim\_\{x\; \backslash to\; \backslash infty\}\; x\backslash left(\; \backslash frac\{A\}\{x-1\}\; +\; \backslash frac\{Bx\; +\; C\}\{x^2\; +\; x\; +\; 1\}\; \backslash right\; )=\; \backslash lim\_\{x\; \backslash to\; \backslash infty\}\; \backslash frac\{Ax\}\{x-1\}\; +\; \backslash lim\_\{x\; \backslash to\; \backslash infty\}\; \backslash frac\{Bx^2+Cx\}\{x^2+x+1\}=\; A+B,$$

and

$$\backslash lim\_\{x\; \backslash to\; \backslash infty\}\; \backslash frac\{x\}\{(x\; -\; 1)(x^2\; +\; x\; +\; 1)\}\; =0.$$

This implies and so $B\; =\; -\backslash frac\{1\}\{3\}$.

For , we get $-1\; =\; -A\; +\; C,$ and thus $C\; =\; -\backslash tfrac\{2\}\{3\}$.

Putting everything together, we get the decomposition

$$\backslash frac\{1\}\{x^3\; -1\}\; =\; \backslash frac\{1\}\{3\}\; \backslash left(\; \backslash frac\{1\}\{x\; -\; 1\}\; +\; \backslash frac\{-x\; -2\}\{x^2\; +\; x\; +\; 1\}\; \backslash right\; ).$$

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Example 6 (integral) Suppose we have the indefinite integral:

$$\backslash int\; \backslash frac\{x^4+x^3+x^2+1\}\{x^2+x-2\}\; \backslash ,dx$$

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

$$\backslash int\; \backslash left(x^2\; +\; 3\; +\; \backslash frac\{-3x+7\}\{(x+2)(x-1)\}\backslash right)\; dx$$

Upon this, we may now perform partial fraction decomposition.

$$\backslash int\; \backslash left(x^2+3+\; \backslash frac\{-3x+7\}\{(x+2)(x-1)\}\backslash right)\; dx\; =\; \backslash int\; \backslash left(x^2+3+\; \backslash frac\{A\}\{(x+2)\}+\backslash frac\{B\}\{(x-1)\}\backslash right)\; dx$$ so: $$A(x-1)+B(x+2)=-3x+7$$. 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:

$$A=\backslash frac\{-13\}\{3\}\; \backslash \; ,\; B=\backslash frac\{4\}\{3\}$$

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

$$\backslash int\; \backslash left(x^2+3+\; \backslash frac\{-13/3\}\{(x+2)\}+\backslash frac\{4/3\}\{(x-1)\}\backslash right)\; \backslash ,dx\; =\; \backslash frac\{x^3\}\{3\}\; \backslash \; +\; 3x-\backslash frac\{13\}\{3\}\; \backslash ln(|x+2|)+\backslash frac\{4\}\{3\}\; \backslash ln(|x-1|)+C$$

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

$$P(x),\; Q(x),\; A\_1(x),\backslash ldots,\; A\_r(x)$$

be real or complex polynomials assume that

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

satisfies

Also define

$$Q\_i=\backslash prod\_\{j\backslash neq\; i\}(x-\backslash lambda\_j)^\{\backslash nu\_j\}=\backslash frac\{Q\}\{(x-\backslash lambda\_i)^\{\backslash nu\_i\}\},\; \backslash qquad\; 1\; \backslash leqslant\; i\; \backslash leqslant\; r.$$

Then we have

$$\backslash frac\{P\}\{Q\}=\backslash sum\_\{j=1\}^\{r\}\backslash frac\{A\_j\}\{(x-\backslash lambda\_j)^\{\backslash nu\_j\}\}$$

if, and only if, each polynomial $A\_i(x)$ is the Taylor polynomial of $\backslash tfrac\{P\}\{Q\_i\}$ of order $\backslash nu\_i-1$ at the point $\backslash lambda\_i$:

$$A\_i(x):=\backslash sum\_\{k=0\}^\{\backslash nu\_i-1\}\; \backslash frac\{1\}\{k!\}\backslash left(\backslash frac\{P\}\{Q\_i\}\backslash right)^\{(k)\}(\backslash lambda\_i)\backslash \; (x-\backslash lambda\_i)^k.$$

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.

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Sketch of the proof The above partial fraction decomposition implies, for each 1 ≤  i ≤  r, a polynomial expansion

$$\backslash frac\{P\}\{Q\_i\}=A\_i\; +\; O((x-\backslash lambda\_i)^\{\backslash nu\_i\}),\; \backslash qquad\; \backslash text\{for\; \}\; x\backslash to\backslash lambda\_i,$$

so $A\_i$ is the Taylor polynomial of $\backslash tfrac\{P\}\{Q\_i\}$, because of the unicity of the polynomial expansion of order $\backslash nu\_i-1$, and by assumption .

Conversely, if the $A\_i$ are the Taylor polynomials, the above expansions at each $\backslash lambda\_i$ hold, therefore we also have

$$P-Q\_i\; A\_i\; =\; O((x-\backslash lambda\_i)^\{\backslash nu\_i\}),\; \backslash qquad\; \backslash text\{for\; \}\; x\backslash to\backslash lambda\_i,$$

which implies that the polynomial $P-Q\_iA\_i$ is divisible by $(x-\backslash lambda\_i)^\{\backslash nu\_i\}.$

For $j\backslash neq\; i,\; Q\_jA\_j$ is also divisible by $(x-\backslash lambda\_i)^\{\backslash nu\_i\}$, so

$$P-\; \backslash sum\_\{j=1\}^\{r\}Q\_jA\_j$$

is divisible by $Q$. Since

we then have

$$P-\; \backslash sum\_\{j=1\}^\{r\}Q\_jA\_j=0,$$

and we find the partial fraction decomposition dividing by $Q$.

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Fractions of integers The idea of partial fractions can be generalized to other , say the ring of where prime numbers take the role of irreducible denominators. For example:

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

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Notes

• (1990). 9780673386380, Addison-Wesley Educational Publishers, Inc.. . ISBN 9780673386380
• (2025). 9783540260448 ISBN 9783540260448

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External links

• Make partial fraction decompositions with Scilab.

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