partial fraction

In more traditional treatments of algebra, great emphasis has been placed on the computation of the partial fraction decomposition of a rational function. This is because of applications, in particular partial fractions in integration and their use in calculating inverse Laplace transforms. 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 Bzout'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/(x² − 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/(x² − 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/(x² − 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.