PARTIAL FRACTIONS: for clear understanding..

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

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

$\sum_j \frac{f_j(x)}{g_j(x)}$

where g_j (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 fractions, that produces a single rational fraction with a numerator and denominator usually of high 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. Thus, 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. To decrease the degree of the numerator directly, the Euclidean algorithm can be used, but in fact if ƒ already has lower degree than g this isn't helpful.

The main motivation to decompose a rational function into a sum of simpler fractions is that it makes it simpler to perform linear operations on it. Therefore the problem of computing derivatives, antiderivatives, integrals, power series expansions, Fourier series, residues, and 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.

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[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) = 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

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

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:

$\frac {G(x)}{F(x)^n}$

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. The result is the following theorem:

Let ƒ and g be nonzero polynomials over a field K. 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_ij 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 ƒ < deg g, then b = 0.

Therefore when the field K is the complex numbers, we can assume that each p_i has degree 1 (by the fundamental theorem of algebra) the numerators will be constant. When K is the real numbers, some of the p_i might be quadratic, so in the partial fraction decomposition 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]Procedure

Given two polynomials

P (x)

and $Q(x) = (x-\alpha_1)(x-\alpha_2) \cdots (x-\alpha_n)$ , where the α_i 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 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.)

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

If deg P $\ge$ deg Q, then it is necessary to perform the division

$P(x) \div Q(x) = D(x) + \frac{R(x)}{Q(x)},$

via polynomial long division or otherwise, and then seek partial fractions for the remainder fraction (which by definition has deg R < deg Q).

If Q(x) contains 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 deg N < deg F, rather than as a constant. For example, take the following decomposition over R:

$\frac{x^2 + 1}{(x+2)(x-1)\color{Blue}(x^2+x+1)} = \frac{a}{x+2} + \frac{b}{x-1} + \frac{\color{OliveGreen}cx + d}{\color{Blue}x^2 + x + 1}.$

Suppose Q(x) = (x − α)^rS(x) and S(α) ≠ 0. Then Q(x) has a zero α of multiplicity r, and in the partial fraction decomposition, r of the partial fractions will involve the powers of (x − α). For illustration, take S(x) = 1 to get the following decomposition:

$\frac{P(x)}{Q(x)} = \frac{P(x)}{(x-\alpha)^r} = \frac{c_1}{x-\alpha} + \frac{c_2}{(x-\alpha)^2} + \cdots + \frac{c_r}{(x-\alpha)^r}.$

[edit]Illustration

In an example application of this procedure, (3x + 5)/(1 − 2x)² can be decomposed in the form

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

Clearing denominators shows that 3x + 5 = A + B(1 − 2x). Expanding and equating the coefficients of powers of x gives

5 = A + B and 3x = −2Bx

Solving for A and B yields A = 13/2 and B = −3/2. Hence,

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

[edit]Residue method

[edit]Over the reals

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. For applications of partial fraction decomposition over the reals, see

[edit]General result

Let ƒ(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 ƒ(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.

[edit]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 into this equation gives A = -1/4, and substituting x = 1 gives B = 1/4, so that

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

[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

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

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. 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 ƒ(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]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 uyu

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

Define also

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

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 $\textstyle\lambda_i$ :

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

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 1 ≤ i ≤ 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]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:

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

PARTIAL FRACTIONS

Monday, 4 July 2011

for clear understanding..

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