Partial fraction
From Academic Kids

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In algebra, the partial fraction decomposition of a rational function expresses the function as a sum of fractions, in each term of which, the denominator is an irreducible (i.e., prime, 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 inverse Laplace transforms.
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 1stdegree polynomials can be irreducible. If one allows only rational numbers, then some higherdegree polynomials are irreducible.
Contents 
Some examples
Distinct firstdegree factors in the denominator
Suppose it is desired to decompose the rational function <math>{x+3 \over x^23x40}\,<math>
into partial fractions. The denominator factors as
 <math>(x8)(x+5)\,<math>
and so we seek scalars A and B such that
 <math>{x+3 \over x^23x40}={x+3 \over (x8)(x+5)}={A \over x8}+{B \over x+5}.<math>
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
 <math>x+3=A(x+5)+B(x8).\,<math>
Collecting like terms gives
 <math>x+3=(A+B)x+(5A8B).\,<math>
Equating coefficients of like terms then yields:
 <math>
\begin{matrix} A & + & B & = & 1 \\ 5A &  & 8B & = & 3 \end{matrix} <math>
The solution is A = 11/13, B = 2/13. Thus we have the partial fraction decomposition
 <math>{x+3 \over x^23x40}={11/13 \over x8}+{2/13 \over x+5}.<math>
An irreducible quadratic factor in the denominator
In order to decompose <math>{10x^2+12x+20 \over x^38}<math>
into partial fractions, first observe that
 <math>x^38=(x2)(x^2+2x+4).\,<math>
The fact that x^{2} + 2x + 4 cannot be factored using real numbers can be seen by observing that the discriminant 2^{2</sub> − 4(1)(4) is negative. Thus we seek scalars A, B, C such that }
 <math>{10x^2+12x+20 \over x^38}={10x^2+12x+20 \over (x2)(x^2+2x+4)}={A \over x2}+{Bx+C \over x^2+2x+4}.<math>
When we clear fractions, we get
 <math>10x^2+12x+20=A(x^2+2x+4)+(Bx+C)(x2).\,<math>
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
 <math>10\cdot 2^2+12\cdot 2+20=A(2^2+2\cdot 2+4),\,<math>
i.e., 84 = 12A, so A = 7, and we have
 <math>10x^2+12x+20=7(x^2+2x+4)+(Bx+C)(x2).\,<math>
Next, substitution of 0 for x yields
 <math>20=7(4)+C(2),\,<math>
and so C = 4. We now have
 <math>10x^2+12x+20=7(x^2+2x+4)+(Bx+4)(x2).\,<math>
Substitution of 1 for x yields
 <math>10+12+20=7(1+2+4)+(B+4)(12),\,<math>
and so B = 3. Our partial fraction decomposition is therefore:
 <math>{10x^2+12x+20 \over x^38}={7 \over x2}+{3x+4 \over x^2+2x+4}.<math>
A repeated firstdegree factor in the denominator
Consider the rational function
 <math>{10x^263x+29 \over x^311x^2+40x48}.<math>
The denominator factors thus:
 <math>x^311x^2+40x48=(x3)(x4)^2.\,<math>
The multiplicity of the firstdegree factor (x − 4) is more than 1. In such cases, the partial fraction decomposition takes the following form:
 <math>{10x^263x+29 \over x^311x^2+40x48}={10x^263x+29 \over (x3)(x4)^2}={A \over x3}+{B \over x4}+{C \over (x4)^2}.<math>
Repeated factors in the denominator generally
For rational functions of the form
 <math>{\bullet \over (x+2)(x+3)^5}<math>
(where the "<math>\bullet<math>" may be any polynomial of sufficiently small degree) the partial fraction decomposition looks like this:
 <math>{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}.<math>
The general pattern may be quickly guessed.
For rational functions of the form
 <math>{\bullet \over (x+2)(x^2+1)^5}<math>
with the irreducible quadratic factor x^{2} + 1 in the denominator (where again, the "<math>\bullet<math>" may be any polynomial of sufficiently small degree), the partial fraction decomposition looks like this:
 <math>{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},<math>
and a similar pattern holds for any other irreducible quadratic factor.
[Still to be added: highdegree polynomials in the numerator.]
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 factorises 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).
Examples
 As an introductory example we take the rational function
 x/(x^{2} − 1).
 This by the difference of two squares identity can also be written as
 x/{(x + 1)(x − 1)},
 which can be transformed further. Consider an identity
 A/(x + 1) + B/(x − 1) = x/(x^{2} − 1),
 where A and B are constants. In more explicit form
 Ax − A + Bx + B = 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, so 2A = 1 and A = ½ = B.
 Finally we find
 x/(x^{2} − 1) = ½·1/(x + 1) + ½·1/(x − 1) = 1/2(x + 1) + 1/2(x − 1)
 which holds true for all x not = ±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
 <math> Q(x)=\prod_{i=1}^n (xx_i) <math>
 where all <math> x_i <math> are pairwise different. In other words Q has simple roots (over K). If P(x) is any polynomial of degree <math> \le n1 <math> then according to the Lagrange interpolation formula (see Lagrange form) P(x) can be uniquely written as a sum (the Lagrange form representation)
 <math> P(x)=\sum_{j=1}^n P(x_j)L_j(x;x_j) <math>
 where <math>\, L_j(x;x_j) <math> is the Lagrange polynomial
 <math> L_j(x;x_j)=\prod_{k\le n,\, k\ne j} {{(xx_k)}\over {(x_jx_k)}} \ .<math>
 Dividing the Lagrange representation on the right side termwise by the polynomial Q(x) in its factored form one obtains
 <math> {P(x)\over Q(x)} =\sum_{j=1}^n {P(x_j)\over {\prod_{k \le n, \, k\ne j} (x_jx_k)}} \,\cdot {1 \over {xx_j}} \ .<math>
 This is the partial fraction decomposition
 <math> {P(x)\over Q(x)} =\sum_{j=1}^n c_j \cdot {1 \over {xx_j}} <math>
 of the rational function <math>\, R(x)=P(x)/Q(x) <math> with coefficients
 <math> c_j= {P(x_j)\over {\prod_{k \le n, \, k\ne j} (x_jx_k)}} \ .<math>
 The first example can be obtained as the special case <math> Q(x)=(x1)(x+1), \; P(x)=x <math>.de:Partialbruchzerlegung