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 a full 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.

In symbols, one can use partial fraction expansion to change a rational function in the form

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

where ƒ and g are polynomials, into a function of the form

$\sum \frac{a_n}{h_n(x)}$

where h_n are polynomials that are factors of g(x), and are in general of lower degree. The full decomposition pushes the reduction as far as it will go: in other words, the factorization of g is used as much as possible. To decrease the degree of the numerator directly, the Euclidean algorithm can be used, but in fact if f already has lower degree than g this isn't helpful.

Thus the partial fraction decomposition may be seen as the inverse procedure of the more elementary operation of addition of fractions, that produces a single rational fraction with a numerator and denominator usually of high degree.

The main motivation to decompose a rational function into a sum of simpler fractions is that it makes simpler to perform linear operations on it. Therefore the problem of computing derivatives, antiderivatives, integrals, power expansions, Fourier series, residues, linear functional transformations, of rational functions can be reduced, via partial fraction decomposition, to making the computation on each single element used in the decomposition. See e.g. partial fractions in integration for an account of the use of the partial fractions in finding antiderivatives. 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, or a finite field, then some higher-degree polynomials are irreducible.

[edit] Some examples

The basic practical idea used for working with partial fractions is to work backwards. To separate a rational function as a sum, assume that there exist possible values of (unknown) parameters within a certain form. Then solve for these unknown by assuming equivalence to the initial function.

[edit] Simple example

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

Write as

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

where : $A\, ,B\,$ and $a\,$ are constants

Multiply both sides by x + a

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

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

means A and B must simultaneously solve:

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

because the coefficients of both x and 1 have to agree, giving rise to the two equations.

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:

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

By solving, A = −70, B = 80, C = −63. Therefore, the solution becomes

${10x^2-63x+29 \over x^3-11x^2+40x-48}={-70 \over x-3}+{80 \over x-4}+{-63 \over (x-4)^2}.$

[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. On the other hand, the existence of a decomposition of a certain kind is an assumption in practical cases, and the principles should explain which assumptions are justified.

Assume a rational function R(x) in one indeterminate x has a 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] The role of the Taylor polynomial

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

$P(x), Q(x), A_1(x),\dots, A_r(x)$

be real or complex polynomials; assume that

$\textstyle Q=\prod_{j=1}^{r}(x-\lambda_j)^{\nu_j},$

that

$\textstyle\deg(P)<\deg(Q)=\sum_{j=1}^{r}\nu_j,$

and that

$\textstyle\deg A_j<\nu_j$ for $j=1,\dots,r.$

Define also

$\textstyle Q_i=\prod_{j\neq i}(x-\lambda_j)^{\nu_j}=\frac{Q}{(x-\lambda_i)^{\nu_i}}$ for $i=1,\dots,r.$

Then we have

$\frac{P}{Q}=\sum_{j=1}^{r}\frac{A_j}{(x-\lambda_j)^{\nu_j}}$

if, and only if, for each $\textstyle i$ the polynomial $\textstyle A_i(x)$ is the Taylor polynomial of $\textstyle\frac{P}{Q_i}$ of order $\textstyle\nu_i-1$ at the point $\textstyle\lambda_j$ :

$A_i(x):=\sum_{k=0}^{\nu_i-1} \frac{1}{k!}\left(\frac{P}{Q_i}\right)^{(k)}(\lambda_i)\ (x-\lambda_i)^k \ .$

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 $\textstyle 1\leq i\leq r$ , a polynomial expansion

$\frac{P}{Q_i}=A_i + O((x-\lambda_i)^{\nu_i})\qquad$ , as $x\to\lambda_i$ ;

so $\textstyle A_i$ is the Taylor polynomial of $\textstyle\frac{P}{Q_i}$ , because of the unicity of the polynomial expansion of order $\textstyle\nu_i-1$ , and by assumption $\textstyle\deg A_i<\nu_i$ .

Conversely, if the $\textstyle A_i$ are the Taylor polynomials, the above expansions at each $\textstyle\lambda_i$ hold, therefore we also have

$P-Q_i A_i = O((x-\lambda_i)^{\nu_i})\qquad$ , as $x\to\lambda_i$ ,

which implies that the polynomial $\textstyle P-Q_iA_i$ is divisible by $\textstyle (x-\lambda_i)^{\nu_i}.$

For $\textstyle j\neq i$ also $\textstyle Q_jA_j$ is divisible by $\textstyle (x-\lambda_i)^{\nu_i}$ , so we have in turn that $\textstyle P- \sum_{j=1}^{r}Q_jA_j$ is divisible by $\textstyle Q$ . Since $\textstyle\deg\left( P- \sum_{j=1}^{r}Q_jA_j \right) < \deg(Q)$ we then have $\textstyle P- \sum_{j=1}^{r}Q_jA_j=0$ , and we find the partial fraction decomposition dividing by $\textstyle Q$ .

[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:

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

or

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

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,

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

or

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

[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] Partial Fraction Decomposition

Partial fraction decomposition is a theorem in algebra which states that a rational function over a field can be decomposed into a polynomial plus a sum of proper fractions, each of which has the form $p (x) / q (x) n$ , where the degree of $p (x)$ is smaller than the degree of $q (x)$ and $q (x)$ is irreducible.

[edit] Statement of theorem

Let f and g be nonzero polynomials. Write g as a product of powers of distinct irreducible polynomials:

$g=\prod_{i=1}^k p_i^{n_i}.$

There are (unique) polynomials b and $a i j$ with $\deg\ a_{ij} < \deg\ p_i$ such that

$\frac{f}{g}=b+\sum_{i=1}^k\sum_{j=1}^{n_i}\frac{a_{ij}}{p_i^j}.$

If $\deg\ f < \deg\ g$ , then $b = 0$ .

[edit] Outline of proof

[edit] Lemma 1

Let f, g and h be nonzero polynomials with f and g coprime. There are polynomials a and b such that

$\frac{h}{fg}=\frac{a}{f}+\frac{b}{g}.$

[edit] Lemma 2

Let f and g be nonzero polynomials and let n be a positive integer. There exist polynomials b and $a i$ with $\deg\ a_i<\deg\ g$ such that $\frac{f}{g^n}=b+\sum_{j=1}^n\frac{a_j}{g^j}.$

[edit] Over the reals

In mathematics, 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.

This article describes the general method for obtaining the partial fraction decomposition of any real rational function. The proof of the existence of the partial fraction decomposition over an arbitrary field is not given here. For a sketch of the general proof, see partial fraction.

For applications of partial fraction decomposition over the reals, see

First we state the general result, and then we offer several examples to illustrate the result and show how the method is used in practice.

[edit] General result

Let f(x) be any rational function over the real numbers. In other words, suppose there exist real polynomials p(x) and q(x)≠ 0, such that

$f(x) = \frac{p(x)}{q(x)}$

By removing the leading coefficient of q(x), we may assume without loss of generality that q(x) is monic. By the fundamental theorem of algebra, we can write

$q(x) = (x-a_1)^{j_1}\cdots(x-a_m)^{j_m}(x^2+b_1x+c_1)^{k_1}\cdots(x^2+b_nx+c_n)^{k_n}$

where a₁,..., a_m, b₁,..., b_n, c₁,..., c_n are real numbers with b_i² - 4c_i < 0, and j₁,..., j_m, k₁,..., 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_i² + b_ix + c_i) are the irreducible quadratic factors of q(x) which correspond to pairs of complex conjugate roots of q(x).

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

$f(x) = \frac{p(x)}{q(x)} = P(x) + \sum_{i=1}^m\sum_{r=1}^{j_i} \frac{A_{ir}}{(x-a_i)^r} + \sum_{i=1}^n\sum_{r=1}^{k_i} \frac{B_{ir}x+C_{ir}}{(x^2+b_ix+c_i)^r}$

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. However, this is often not the best way to go when computing by pencil and paper, and there are other ways ("tricks") to obtain the constants.

[edit] The algorithm

The partial fraction decomposition of a real rational function can be obtained by the following procedure:

Remove the leading coefficient of the denominator q(x) if necessary.
Reduce the rational function as much as possible; i.e. cancel any obvious common factors of the numerator and denominator.
Make sure the degree of the numerator p(x) is strictly smaller than the degree of the denominator q(x). If it is not, then use the division algorithm for polynomials and long-divide the denominator into the numerator. You will get a (nonzero) polynomial quotient and a remainder whose degree is now strictly smaller than the degree of the denominator. Take this remainder as your new numerator and continue with the following steps. (Don't forget the polynomial quotient when you're done!)
Factor the denominator q(x) into linear and irreducible quadratic factors. This is not a straightforward task; however, you have some tools to use. If you find any real root a of q(x), then you know that (x - a) is a factor of q(x). Also, if you differentiate q(x) and use the Euclidean algorithm to find the greatest common divisor of q ' (x) and q(x), then any real root a of the greatest common divisor gives you a factor (x - a). You can look for ways to complete the square in different powers of x. Most importantly, you can use the rational root test to determine all possible rational roots of q(x).
Express the partial fraction decomposition of f(x) as the sum of double sums above.
Find the constants A_ir, B_ir, and C_ir by whatever methods possible. (These methods are described below.) Once you have found all the constants, you now have the partial fraction decomposition. (Don't forget the polynomial quotient!)

[edit] Some examples

[edit] Example 1

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

Here, the denominator splits into two distinct linear factors:

q (x) = x 2 + 2 x - 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 and x = 1 into this equation gives A = -1/4 and B = 1/4, so that

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

[edit] Example 2

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

After 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

4 x 2 - 8 x + 16 = A (x 2 - 4 x + 8) + (B x + 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,

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

The following example illustrates almost all the "tricks" one would need to use short of consulting a computer algebra system.

[edit] Example 3

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

After long-division and factoring, we have

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

The partial fraction decomposition takes the form

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

Multiplying through by (x - 1)³(x² + 1)² we have the polynomial identity

x 5 - 2 x 4 + 5 x 3 - 5 x 2 + 6 x - 1

= A (x - 1) 2 (x 2 + 1) 2 + B (x - 1)(x 2 + 1) 2 + C (x 2 + 1) 2 + (D x + E)(x - 1) 3 (x 2 + 1) + (F x + G)(x - 1) 3

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. We now have the identity

x 5 - 2 x 4 + 5 x 3 - 5 x 2 + 6 x - 1

= A (x - 1) 2 (x 2 + 1) 2 + B (x - 1)(x 2 + 1) 2 + (x 2 + 1) 2 + (D x + E)(x - 1) 3 (x 2 + 1) + (x - 1) 3

Taking constant terms gives E = A - B + 1, taking leading coefficients gives A = -D, and taking x-coefficients gives B = 3 - D - 3E. Putting all of this together, E = A - B + 1 = -D - (3 - D - 3E) + 1 = 3E - 2, so E = 1 and A = B = -D. Now,

x 5 - 2 x 4 + 5 x 3 - 5 x 2 + 6 x - 1

= A (x - 1) 2 (x 2 + 1) 2 + A (x - 1)(x 2 + 1) 2 + (x 2 + 1) 2 + ( - A x + 1)(x - 1) 3 (x 2 + 1) + (x - 1) 3

Taking x = -1 gives -20 = -8A - 20, so A = B = D = 0. The partial fraction decomposition of f(x) is thus

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

[edit] General methods for finding coefficients

We present in the sequel a list of methods for finding the coefficients occurring in the r.h.s. sums of the above equation for f(x)=p(x)/q(x)=..., to which we shall refer to by (PFD).

For some of the methods it would be useful if p and/or q were already written in factored form.

[edit] Simple poles of multiplicity 1

If a_i is a simple pole of f, i.e. j_i = 1, then one multiplies (PFD) by (x − a_i) and takes the limit x→a_i. On the r.h.s. only A_i1 "survives", which is therefore given by

A_i1 = lim_x→ai (x &minuns; a_i) f(x) = p(a_i)/q_i(a_i)

where q_i is q divided (simplified) by the factor (x − a_i). It holds $q i (a i) = q'(a i)$ .

[edit] Leading term of a pole

Using the same idea, we determine the coefficients $A_{ij_i}$ (multiplying in the highest negative power), by just replacing (x − a_i) by its j_ith power.

[edit] Leading term of a quadratic factor

Still using the same idea, but going to the complex root of x² + bx + c, one gets in the same way coefficients $B_{ik_i},C_{ik_i}$ (from real and imaginary part of the previous equation for A which is now an equation for Bx + C).

[edit] Other coefficients A_ik

All coefficients relative to one pole can be obtained "in one step" by making a change of variables t = x - a_i (i.e. replacing x by t + a_i) and then (long) dividing the "new" p by the "new" q, with the "pure" power t^j_i removed.

[edit] Method of derivatives

Instead of making the change of variables and using long division, one can obtain a recursive formula involving derivatives of p and q_i at x = a_i.

[edit] General methods

Besides the above methods to obtain a particular coefficient, there are several general methods to get one or more equations involving one or several of the coefficients, which should then allow to determine the remaining unknown coefficients:

method of particular values: this just consists in putting x equal to particular fixed values (0, ±1, ±2...), and thus obtaining numerical equations for the coefficients.
method of limits: multiply both sides of (PFD), after subtracting known terms (especially the polynomial), by the highest possible power of x such that lim(x→∞) still is finite, and take this limit
method of identification: the "final rescue", put all of the r.h.s. on the common denominator q(x) and identify coefficients of alike powers in the numerator on the r.h.s. and p(x) on the l.h.s.

[edit] See also

[edit] Notes

^ 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] External links

[0] 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.

[1] Page 188, Vedic Mathematics

[2] Page 186, Vedic Mathematics

[3] Pages 188-189, Vedic Mathematics

[4] Page 189, Vedic Mathematics

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[2]

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[5]

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] The role of the Taylor polynomial

[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] Partial Fraction Decomposition

[edit] Statement of theorem

[edit] Outline of proof

[edit] Lemma 1

[edit] Lemma 2

[edit] Over the reals

[edit] General result

[edit] The algorithm

[edit] Some examples

[edit] Example 1

[edit] Example 2

[edit] Example 3

[edit] General methods for finding coefficients

[edit] Simple poles of multiplicity 1

[edit] Leading term of a pole

[edit] Leading term of a quadratic factor

[edit] Other coefficients Aik

[edit] Method of derivatives

[edit] General methods

[edit] See also

[edit] Notes

[edit] External links

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[edit] Other coefficients A_ik