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 that irreducible polynomial.

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.

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

$\frac{A}{x+a} + g(x)$

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 = A(x + a) + B \$

$x = A\cdot x + A\cdot a + B \$

means $A\,$ and $B\,$ must simultaneously solve:

$\begin{cases} x = A\cdot x \\ 0 = A\cdot a + B \end{cases}$

therefore

$\begin{cases} A = 1 \\ B = -a \end{cases}$

Finally, the decomposed form is:

$\frac{x}{x + a} = 1 + \frac{-a}{x + a}$

[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

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

Alternatively, take the original equation

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

multiply by (x-8) to get

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

evaluate at x=8 to solve for A as

${11 \over 13}={A}.$

multiply the original equation by (x+5) to get

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

evaluate at x=-5 to solve for B as

${-2 \over -13}={2 \over 13}={B}.$

[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

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

(where the $p (x)$ 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

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

with the irreducible quadratic factor x² + 1 in the denominator (where again, the $p (x)$ 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

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

where $\scriptstyle n \geq m$ , you just need to apply 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:

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

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

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

$1=A(M-P) + BP \,$

$1=AM - AP + BP \,$

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

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

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

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

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

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

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

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

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

[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 a denominator that factors as

R(x) = 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] Algorithms

[edit] Lagrange interpolation

Partial fraction decomposition can be derived using Lagrange interpolation.

[edit] Example

As an introductory example we take the rational function

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

By the difference of two squares identity, this 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, we have the relation of the numerators,

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

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

$\begin{align} A + A & {} = 1 \\ 2A & {} = 1 \\ A & {} = B = 0.5. \end{align}$

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.

[edit] Derivation

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)

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

where $\, L_j(x;x_j)$ is the Lagrange polynomial

$L_j(x;x_j)=\prod_{k\le n,\, k\ne j} {{(x-x_k)}\over {(x_j-x_k)}}. \$

Dividing the Lagrange representation on the right side termwise by the polynomial Q(x) in its factored form one obtains

${P(x)\over Q(x)} =\sum_{j=1}^n {P(x_j)\over {\prod_{k \le n, \, k\ne j} (x_j-x_k)}} \,\cdot {1 \over {x-x_j}}. \$

This is the partial fraction decomposition

${P(x)\over Q(x)} =\sum_{j=1}^n \frac{P(x_j)}{Q'(x_j)} \cdot {1 \over {x-x_j}}$

of the rational function $\, R(x)=P(x)/Q(x)$ with the derivative of $Q$ given by

$Q'(x_j) = \sum_{l \le n} {\prod_{k \le n, \, k\ne l} (x_j-x_k)} = \prod_{k \le n, \, k\ne j} (x_j-x_k)$ .

The first example can be obtained as the special case $Q(x)=(x-1)\cdot(x+1), \; P(x)=x$ .

Note the close relationship to divided differences.

[edit] Parāvartya Sūtra

Separation of a fractional algebraic expression into partial fractions is the reverse of the process of combining fractions by converting each fraction to the lowest common denominator (LCM) and adding the numerators. This separation is accomplished by a mental, one-line Vedic formula called the Parāvartya Sūtra^[1]. Case one has fractional expressions where factors in the denominator are unique. Case two has fractional expressions where some factors may repeat as powers of a binomial.

In integral calculus we would want to write a fractional algebraic expression as the sum of its partial fractions in order to take the integral of each simple fraction separately. Once the original denominator, D₀, has been factored we set up a fraction for each factor in the denominator. We may use a subscripted D to represent the denominator of the respective partial fractions which are the factors in D₀. Letters A, B, C, D, E, and so on will represent the numerators of the respective partial fractions.

We calculate each respective numerator by (1) calculating the Parāvartya value of the denominator (which is the value of the variable making that binomial factor equal to zero) and (2) then substituting this value into the original expression but ignoring that factor in the denominator. Each Parāvartya value for the variable is the value which would give an undefined value to the expression since we do not divide by zero.

General formula:

$\frac{\ell x^2 + mx + n}{(x-a)(x-b)(x-c)} = \frac{A}{(x-a)} + \frac{B}{(x-b)} + \frac{C}{(x-c)}$

Here, a, b, c, $\scriptstyle\ell$ , m, and n are given integer values.

Where x = a and

$A =\frac{\ell a^2 + ma + n}{(a-b)(a-c)};$

and where x = b and

$B = \frac{\ell b^2 + mb + n}{(b-c)(b-a)};$

and where x = c and

$C = \frac{\ell c^2 + mc + n}{(c-a)(c-b)}.$ ^[2]

[edit] Case one

Factorize the expression in the denominator. Set up a partial fraction for each factor in the denominator. Apply the Parāvartya Sūtra to solve for the new numerator of each partial fraction.

[edit] Example

$\frac{3x^2 + 12x + 11}{(x+1)(x+2)(x+3)} = \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{x+3}$

Set up a partial fraction for each factor in the denominator. With this framework we apply the Sūtra to solve for A, B, and C by mental math.

1. D₁ is x + 1; set it equal to zero. This gives the Parāvartya value for A when x = −1.

2. Next, substitute this value of x into the fractional expression, but without D₁.

3. Put this value down as the value of A.

Proceed similarly for B and C.

D₂ is x + 2; For Parāvartya B use x = −2.

D₃ is x + 3; For Parāvartya C use x = −3.

Thus, to solve for A, use x = −1 in the expression but without D₁:

$\frac{3x^2 + 12x + 11}{(x+2)(x+3)} = \frac{3 -12 +11}{(1)(2)} = \frac{2}{2} = 1 = A.$

Thus, to solve for B, use x = −2 in the expression but without D₂:

$\frac{3x^2 + 12x + 11}{(x+1)(x+3)} = \frac{12 -24 +11}{(-1)(1)} = \frac{-1}{(-1)} = +1 = B.$

Thus, to solve for C, use x = −3 in the expression but without D₃:

$\frac{3x^2 + 12x + 11}{(x+1)(x+2)} = \frac{27 -36 +11}{(-2)(-1)} = \frac{2}{(+2)} = +1 = C.$

Thus,

$\frac{3x^2 + 12x + 11}{(x+1)(x+2)(x+3)} = \frac{1}{x+1} + \frac{1}{x+2} + \frac{1}{x+3}$ ^[3]

[edit] Case two

When factors of the denominator include powers of one expression we (1) Set up a partial fraction for each unique factor and each lower power of D; (2) We set up an equation showing the relation of the numerators if all were converted to the LCD. From the equation of numerators we solve for each numerator, A, B, C, D, and so on. This equation of the numerators is an absolute identity, true for all values of x. So, we may select any value of x and solve for the numerator.^[4]

[edit] Example

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

Here, we set up a partial fraction for each descending power of the denominator. Then we solve for the numerators, A and B. As the Parāvartya value for A and B will be the same, x = ½, we need an additional relation in order to solve for both. To write the relation of numerators the second fraction needs another factor of (1-2x) to convert it to the LCD, giving us 3x + 5 = A + B(1 − 2x).

To solve for A: Set the denominator of the first fraction to zero, 1 − 2x = 0. Solving for x gives the Parāvartya value for A, when x = ½. When we substitute this value, x = ½, into the relation of numerators we have 3(1/2) + 5 = A + B(0). Solving for A gives us A = 3/2 + 5 = 13/2. Hence, numerator A equals six and one-half.^[5]

To solve for B: Since the equation of the numerators, here, 3x + 5 = A + B(1 − 2x), is true for all values of x, pick a value for x and use it to solve for B. As we have solved for the value of A above, A = 13/2, we may use that value to solve for B.

We may pick x = 0, use A = 13/2, and then solve for B.

3x + 5 = A + B(1 − 2x)

0 + 5 = 13/2 + B(1 + 0)

10/2 = 13/2 + B

−3/2 = B

We may pick x = 1. Then solve for B:

3x + 5 = A + B(1 − 2x)

3 + 5 = 13/2 + B(1 − 2)

8 = 13/2 + B(−1)

16/2 = 13/2 − B

B = −3/2

We may pick x = −1. Solve for B:

3x + 5 = A + B(1 − 2x)

−3 + 5 = 13/2 + B(1 + 2)

4/2 = 13/2 + 3B

−9/2 = 3B

−3/2 = B

Hence,

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

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

[edit] Technique three

A third technique is an analysis of the expanded relation of the numerators. Just match up the x-terms of each degree. Then one can see the coefficients of the matching terms and solve for the missing numerator.

[edit] Example

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

Converting each fraction to the LCD we have the relation in the numerators: 3x + 5 = A + B(1 − 2x). As A and B are constants, we can expand and match up the constant terms and the x-terms. The values of A and B will then be apparent.

3x + 5 = A + B(1 − 2x)

3x + 5 = A + B − 2Bx

Hence, the constant terms are set equal and the x-terms are set equal:

5 = A + B and 3x = −2Bx

Therefore, by setting the coefficients equal, we may solve for B:

3 = −2B

−3/2 = B

And then solve for A:

5 = A − 3/2

5 + 3/2 = A

13/2 = A

Hence,

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

[edit] Fractions of integers

The idea of partial fractions can be generalized to other rings, say the ring of integers where prime numbers take the role of irreducible denominators. E.g., it is:

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

[edit] References

^ Vedic Mathematics: Sixteen Simple Mathematical Formulae from the Vedas, by Swami Sankaracarya (1884-1960), Motilal Banarsidass Indological Publishers and Booksellers, Varnasi, India, 1965; reprinted in Delhi, India, 1975, 1978. 367 pages.
^ Page 188, Vedic Mathematics
^ Page 186, Vedic Mathematics
^ Pages 188-189, Vedic Mathematics
^ Page 189, Vedic Mathematics

[edit] See also

[edit] External links

Categories: Algebra | Partial fractions | Elementary algebra

Hidden category: Articles to be merged since June 2007

Partial fraction

From Wikipedia, the free encyclopedia

Contents

[edit] Some examples

[edit] Simple example

[edit] Distinct first-degree factors in the denominator

[edit] An irreducible quadratic factor in the denominator

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

[edit] Repeated factors in the denominator generally

[edit] High-degree polynomials in the numerator

[edit] Use in deriving the logistic general equation

[edit] Basic principles

[edit] Algorithms

[edit] Lagrange interpolation

[edit] Example

[edit] Derivation

[edit] Parāvartya Sūtra

[edit] Case one

[edit] Example

[edit] Case two

[edit] Example

[edit] Technique three

[edit] Example

[edit] Fractions of integers

[edit] References

[edit] See also

[edit] External links

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