Partial fraction

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In algebra, the partial fraction decomposition or (partial fraction expansion) is used to reduce the degree of either the numerator or the denominator of a rational function. The outcome of 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 the denominator.

See partial fractions in integration for an account of their use in finding antiderivatives. They are also used in calculating the inverse of transforms; such as the Laplace transform, or the Z-transform.

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, then some higher-degree polynomials are irreducible.

1 Uses
2 Some examples
3 Basic principles
4 Examples
5 See also
6 External links

[edit] Uses

One can use partial fraction expansion for one of two main uses:

To change a function in the form $\frac{f(x)}{x+a}$ into a function of the form

$\frac{A}{x+a} + \sum \frac{g_n(x)}{1}$

To change a function in the form $\frac{f(x)}{g(x)}$ into a function of the form

$\sum \frac{A_n}{g(x)/h_n(x)}$

[edit] Some examples

[edit] Simple example

If one wants to decompose x/(x+a), then one can follow these steps:

$\frac{x}{x + a} = A + \frac{B}{x + a}$

where A, B, and a are constants

$x + 0 = A(x + a) + B \$

$x = Ax \$ and $0 = Aa + B \$

therefore

$A = 1 \$ and $B = -a \$

[edit] Distinct first-degree factors in the denominator

Suppose it is desired to decompose the rational function

${x+3 \over x^2-3x-40}\,$

into partial fractions. The denominator factors as

$(x-8)(x+5)\,$

and so we seek scalars A and B such that

${x+3 \over x^2-3x-40}={x+3 \over (x-8)(x+5)}={A \over x-8}+{B \over x+5}.$

One way of finding A and B begins by "clearing fractions", i.e., multiplying both sides by the common denominator (x − 8)(x + 5). This yields

$x+3=A(x+5)+B(x-8).\,$

Collecting like terms gives

$x+3=(A+B)x+(5A-8B).\,$

Equating coefficients of like terms then yields:

$\begin{matrix} A & + & B & = & 1 \\ 5A & - & 8B & = & 3 \end{matrix}$

The solution is A = 11/13, B = 2/13. Thus we have the partial fraction decomposition

${x+3 \over x^2-3x-40}={11/13 \over x-8}+{2/13 \over x+5}.$

[edit] An irreducible quadratic factor in the denominator

In order to decompose

${10x^2+12x+20 \over x^3-8}$

into partial fractions, first observe that

$x^3-8=(x-2)(x^2+2x+4).\,$

The fact that x² + 2x + 4 cannot be factored using real numbers can be seen by observing that the discriminant 2² − 4(1)(4) is negative. Thus we seek scalars A, B, C such that

${10x^2+12x+20 \over x^3-8}={10x^2+12x+20 \over (x-2)(x^2+2x+4)}={A \over x-2}+{Bx+C \over x^2+2x+4}.$

When we clear fractions, we get

$10x^2+12x+20=A(x^2+2x+4)+(Bx+C)(x-2).\,$

We could proceed as in the previous example, getting three linear equations in three variables A, B, and C. However, since solving such systems becomes onerous as the number of variables grows, we try a different method. Substitution of 2 for x in the identity above makes the entire second term vanish, and we get

$10\cdot 2^2+12\cdot 2+20=A(2^2+2\cdot 2+4),\,$

i.e., 84 = 12A, so A = 7, and we have

$10x^2+12x+20=7(x^2+2x+4)+(Bx+C)(x-2).\,$

Next, substitution of 0 for x yields

$20=7(4)+C(-2),\,$

and so C = 4. We now have

$10x^2+12x+20=7(x^2+2x+4)+(Bx+4)(x-2).\,$

Substitution of 1 for x yields

$10+12+20=7(1+2+4)+(B+4)(1-2),\,$

and so B = 3. Our partial fraction decomposition is therefore:

${10x^2+12x+20 \over x^3-8}={7 \over x-2}+{3x+4 \over x^2+2x+4}.$

[edit] A repeated first-degree factor in the denominator

Consider the rational function

${10x^2-63x+29 \over x^3-11x^2+40x-48}.$

The denominator factors thus:

$x^3-11x^2+40x-48=(x-3)(x-4)^2.\,$

The multiplicity of the first-degree factor (x − 4) is more than 1. In such cases, the partial fraction decomposition takes the following form:

${10x^2-63x+29 \over x^3-11x^2+40x-48}={10x^2-63x+29 \over (x-3)(x-4)^2}={A \over x-3}+{B \over x-4}+{C \over (x-4)^2}.$

[edit] Repeated factors in the denominator generally

For rational functions of the form

${\bullet \over (x+2)(x+3)^5}$

(where the " $\bullet$ " may be any polynomial of sufficiently small degree) the partial fraction decomposition looks like this:

${A \over x+2}+{B \over x+3}+{C \over (x+3)^2}+{D \over (x+3)^3}+{E \over (x+3)^4}+{F \over (x+3)^5}.$

The general pattern may be quickly guessed.

For rational functions of the form

${\bullet \over (x+2)(x^2+1)^5}$

with the irreducible quadratic factor x² + 1 in the denominator (where again, the " $\bullet$ " may be any polynomial of sufficiently small degree), the partial fraction decomposition looks like this:

${A \over x+2}+{Bx+C \over x^2+1}+{Dx+E \over (x^2+1)^2}+{Fx+G \over (x^2+1)^3}+{Hx+I \over (x^2+1)^4}+{Jx+K \over (x^2+1)^5},$

and a similar pattern holds for any other irreducible quadratic factor.

[edit] High-degree polynomials in the numerator

When you need to apply the partial fraction decomposition to a polynomial division like

${a_nx^n+...+a_1x+a_0 \over b_mx^m+...+b_1x+b_0}$

where $n \geq m$ , you just need to make the Polynomial long division procedure first, and apply the partial fraction decomposition to the remainder.

[edit] Use in deriving the logistic general equation

In many beginning calculus courses, partial fractions are introduced as a way to derive the general equation for a logistic function.

Logistic functions model a population which grows until it reaches a limit. The rate of change for the function is proportional (constant k) to both the population reached (P) and the fraction of the total carrying capacity (M) remaining. Thus:

$\frac{dP}{dt}=kP\left({M-P \over M}\right)$

$\int {1 \over P(M-P)}\, dP = \int {k \over M}\, dt$

${1 \over P(M-P)} = {A \over P} + {B \over (M-P)}$

1 = A (M - P) + B P

1 = A M - A P + B P

$A = B, A= {1 \over M}, B = {1 \over M}$

$\int {{1 \over MP} + {1 \over M(M-P)}}\,dP = \int {k \over M}\, dt$

$\int {{1 \over P} + {1 \over (M-P)}}\,dP = \int {k}\, dt$

$\ln{P \over (P-M)} = kt + C$

${P \over P-M} = e^{kt + C}$

${M-P \over P} = e^{-kt - C}$

${M \over P} - 1 = e^{-kt - C}$

$P = {M \over {e^{-kt - C} + 1}}$

$P = {M \over {Ae^{-kt} + 1}}$

[edit] Basic principles

The basic principles involved are quite simple; it is the algorithmic aspects that require attention in particular cases.

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

P(x)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

A/P + 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

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.

Therefore when K is the complex numbers and we can assume F has degree 1 (by the fundamental theorem of algebra) the numerators will be constant. When K is the real numbers we can have the case of

degree F = 2,

and a quotient of a linear polynomial by a power of a quadratic will occur. This therefore is a case that requires discussion, in the systematic theory of integration (for example in computer algebra).

[edit] Examples

As an introductory example we take the rational function

$\frac{x}{x^2-1}.$

This by the difference of two squares identity can also be written as

$\frac{x}{(x+1)(x-1)},$

which can be transformed further. Consider an identity

$\frac{A}{x+1}+\frac{B}{x-1}=\frac{x}{x^2-1},$

where A and B are constants. In more explicit form,

$\!\, A(x-1)+B(x+1)=x.$

We know that the constants on one side of an expression must equal those on the other side. On the left hand side, the constants are −A and B, and on the right, the constant is simply 0. So, comparing constants on both sides of the expression, we can see that

$B-A=0,\,$

i.e.

A = B .

Now, in the same way, we know that the number of x terms on the left must equal the number of x's on the right. Therefore, looking at x terms on both sides,

$Ax+Bx=x,\,$

therefore

$A+B=1,\,$

and so, given that

A = B

, we can say that

$A + A$	$= 1$
$2 A$	$= 1$
$A$	$= B = 0.5$

Finally we find:

$\frac{x}{x^2-1}=\frac{1}{2}\cdot\frac{1}{x+1}+\frac{1}{2}\cdot\frac{1}{x-1}$

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

which holds true for all x ≠ ±1.

The preceding example can be generalized to the following situation:

Assume that Q(x) is a monic polynomial of some degree n which over the underlying field K decomposes into linear factors

$Q(x)=\prod_{i=1}^n (x-x_i),$

where all

x i

are pairwise different. In other words Q has simple roots (over K). If P(x) is any polynomial of degree $\le n-1$ then according to the Lagrange interpolation formula (see Lagrange form) P(x) can be uniquely written as a sum (the Lagrange form representation)

$P(x)=\sum_{j=1}^n P(x_j)L_j(x;x_j),$