Mathematical Methods for Physics and Engineering : A Comprehensive Guide

PRELIMINARY ALGEBRA

In this form, allx-dependence has disappeared from the numerators but at the

expense ofp−1 additional terms; the total number of constants to be determined

remains unchanged, as it must.

When describing possible methods of determining the constants in a partial

fraction expansion, we noted that method (iii), p. 20, which avoids the need to

solve simultaneous equations, is restricted to terms involving non-repeated roots.

In fact, it can be applied in repeated-root situations, when the expansion is put

in the form (1.48), but only to find the constant in the term involving the largest

inverse power ofx−α,i.e.Dpin (1.48).

We conclude this section with a more protracted worked example that contains

all three of the complications discussed.

Resolve the following expressionF(x)into partial fractions:

F(x)=

x^5 − 2 x^4 −x^3 +5x^2 − 46 x+ 100

(x^2 +6)(x−2)^2

.

We note that the degree of the denominator (4) is not greater than that of the numerator
(5), and so we must start by dividing the latter by the former. It follows, from the difference
in degrees and the coefficients of the highest powers in each, that the result will be a linear
expressions 1 x+s 0 with the coefficients 1 equal to 1. Thus the numerator ofF(x)mustbe
expressible as

(x+s 0 )(x^4 − 4 x^3 +10x^2 − 24 x+ 24) + (r 3 x^3 +r 2 x^2 +r 1 x+r 0 ),

where the second factor in parentheses is the denominator ofF(x) written as a polynomial.
Equating the coefficients ofx^4 gives−2=−4+s 0 and fixess 0 as 2. Equating the coefficients
of powers less than 4 gives equations involving the coefficientsrias follows:

−1=−8+10+r 3 ,

5=−24+20+r 2 ,

−46 = 24−48 +r 1 ,

100 = 48 +r 0.

Thus the remainder polynomialr(x) can be constructed andF(x) written as

F(x)=x+2+

− 3 x^3 +9x^2 − 22 x+52

(x^2 +6)(x−2)^2

≡x+2+f(x).

The polynomial ratiof(x) can now be expressed in partial fraction form, noting that its
denominator contains both a term of the formx^2 +a^2 and a repeated root. Thus

f(x)=

Bx+C

x^2 +6

+

D 1

x− 2

+

D 2

(x−2)^2

.

We could now put the RHS of this equation over the common denominator (x^2 +6)(x−2)^2
and findB, C, D 1 andD 2 by equating coefficients of powers ofx. It is quicker, however,
to use methods (iii) and (ii). Method (iii) givesD 2 as (−24 + 36−44 + 52)/(4 + 6) = 2.
We choose to evaluate the other coefficients by method (ii), and settingx=0,x=1and