## FANDOM

160 Pages

The partial fraction decomposition or also commonly known as partial fraction expansion 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

$\frac{f(x)}{g(x)}$

where ƒ and g are polynomials, into an expression 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 rational fractions, which produces a single rational fraction with a numerator and denominator usually of high degree. Thus, the outcome of a full partial fraction expansion expresses that fraction 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.

## Basic principles Edit

The basic principles involved are quite simple,

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

$\frac{A}{P} + \frac{B}{Q}$

## Procedure Edit

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, consists in 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 $Q$.

## Examples Edit

### Example 1 Edit

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

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 Edit

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

After performing 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,

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

### Example 3 Edit

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

After performing long-division and factoring the denominator, 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{align} & {} \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{align}

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 AB + 1 − E − 1 = 0, thus E = AB.

We now have the identity

\begin{align} & {} 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{align}

Expanding and sorting by exponents of x we get

\begin{align} & {} 2 x^6 -4 x^5 +5 x^4 -3 x^3 + x^2 +3 x \\ & = (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{align}

We can now compare the coefficients and see that

\begin{align} 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{align}

with A = 2 − D and −A −3 D =−4 we get A = D = 1 and so B = 0, furthermore is C = 1, E = AB = 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

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

that is 8 = 2B + 8 so B=0.