Partial Fraction

Partial fraction Distinct first-degree factors in the denominator Suppose it is desired to decompose the rational function

into partial fractions. The denominator factors as

and so we seek scalars A and B such that

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

Collecting like terms gives

Equating coefficients of like terms then yields:

The solution is A = 11/13, B = 2/13. Thus we have the partial fraction decomposition

An irreducible quadratic factor in the denominator In order to decompose

into partial fractions, first observe that

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

When we clear fractions, we get

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

i.e., 84 = 12A, so A = 7, and we have

Next, substitution of 0 for x yields

and so C = 4. We now have

Substitution of 1 for x yields

and so B = 3. Our partial fraction decomposition is therefore:

A repeated first-degree factor in the denominator Consider the rational function

The denominator factors thus:

The multiplicity of the first-degree factor (x − 4) is more than 1. In such cases, the partial fraction decomposition takes the following form:

Repeated factors in the denominator generally For rational functions of the form

(where the " " may be any polynomial of sufficiently small degree) the partial fraction decomposition looks like this:

The general pattern may be quickly guessed. For rational functions of the form

with the irreducible quadratic factor x2 + 1 in the denominator (where again, the " " may be any polynomial of sufficiently small degree), the partial fraction decomposition looks like this:

and a similar pattern holds for any other irreducible quadratic factor. [Still to be added: high-degree polynomials in the numerator.]

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 B x + = 2 ( x + 1) ( x − 1) ( x − 1) where A and B are constants. In more explicit form A( x − 1) + B( x + 1) = x 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

i.e.

B − A = 0, 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)

1

=

1 2 + 2 ( x + 1) ( x − 1)

which holds true for all x not = ±1.

Partial fraction decomposition over the reals Example 1

Here, the denominator splits into two distinct linear factors: q(x) = x2 + 2x − 3 = (x + 3)(x − 1) so we have the partial fraction decomposition

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

Example 2

After long-division, we have

1 x − 4 x + 8 x x + 16 3

3

>>

( x3 + 16) − ( x3 − 4 x 2 + 8 x)

>> 4 x 2 − 8 x + 16 = 1+

4 x 2 − 8 x + 16 x3 − 4 x + 8x

Since (-4)2-4(8) = -16 < 0, x2 - 4x + 8 is irreducible, and so

Multiplying through by

x( x 2 − 4 x + 8) we have the polynomial identity

4x2 − 8x + 16 = A(x2 − 4x + 8) + (Bx + C)x 4x2 − 8x + 16 = Ax2 − 4Ax + 8A + Bx 2 − Cx 4 x 2 − 8 x + 16 = ( A + B ) x 2 + (C − 4 A) x + 8 A

We see that 16 = 8A , so A = 2 4 = A + B = 2 + B, so B = 2 -8 = C-4A = C-4*2 = C-8 , so C = 0 Finally:

f (x ) = 1 +

2 2x + 0 + x x2 − 4x + 8

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

Example 3

After long-division and factoring, we have

The partial fraction decomposition takes the form

Multiplying through by (x - 1)3(x2 + 1)2 we have the polynomial identity x5 − 2x4 + 5x3 − 5x2 + 6x − 1 = A(x − 1)2(x2 + 1)2 + B(x − 1)(x2 + 1)2 + C(x2 + 1)2 + (Dx + E)(x − 1)3(x2 + 1) + (Fx + 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 x5 − 2x4 + 5x3 − 5x2 + 6x − 1 = A(x − 1)2(x2 + 1)2 + B(x − 1)(x2 + 1)2 + (x2 + 1)2 + (Dx + E)(x − 1)3(x2 + 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, x5 − 2x4 + 5x3 − 5x2 + 6x − 1 = A(x − 1)2(x2 + 1)2 + A(x − 1)(x2 + 1)2 + (x2 + 1)2 + ( − Ax + 1)(x − 1)3(x2 + 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

General methods for finding coefficients We present in the sequel a list of methods for finding the coefficients occuring 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.

Simple poles of multiplicity 1 If ai is a simple pole of f, i.e. ji=1, then one multiplies (PFD) by (x-ai) and takes the limit x→ai. On the r.h.s. only Ai1 "survives", which is therefore given by Ai1 = limx→ai (x-ai) f(x) = p(ai)/qi(ai) where qi is q divided (simplified) by the factor (x - ai).

Leading term of a pole Using the same idea, we determine the coefficients

(multiplying in the

highest negative power), by just replacing (x-ai) by its ji-th power.

Leading term of a quadratic factor Still using the same idea, but going to the complex root of x²+b x+c, one gets in the same way coefficients

(from real and imaginary part of the

previous equation for A which is now an equation for B x+C).

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

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 qi at x = ai.

General methods Besides the above methods to obtain a particular coefficient, there are several general methods to get one ore 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.