Mathematical Methods for Physics and Engineering : A Comprehensive Guide

PRELIMINARY ALGEBRA

numerators on both sides of the equation. This leads to

4 x+2=A 1 (x+2)+A 2 (x+1),

4=A 1 +A 2 2=2A 1 +A 2.

Solving the simultaneous equations forA 1 andA 2 givesA 1 =−2and

A 2 =6.

(ii) A second method is to substitute two (or more generallyn) different

values ofxinto each side of (1.44) and so obtain two (orn) simultaneous

equations for the two (orn)constantsAi. To justify this practical way of

proceeding it is necessary, strictly speaking, to appeal to method (i) above,

which establishes that there are unique values forA 1 andA 2 valid for

all values ofx. It is normally very convenient to take zero as one of the

values ofx, but of course any set will do. Suppose in the present case that

we use the valuesx= 0 andx= 1 and substitute in (1.44). The resulting

equations are

2

2

=

A 1

1

+

A 2

2

,

6

6

=

A 1

2

+

A 2

3

,

which on solution giveA 1 =−2andA 2 = 6, as before. The reader can

easily verify that any other pair of values forx(except for a pair that

includesα 1 orα 2 ) gives the same values forA 1 andA 2.

(iii) The very reason why method (ii) fails ifxis chosen as one of the roots

αiofh(x) = 0 can be made the basis for determining the values of theAi

corresponding to non-multiple roots without having to solve simultaneous

equations. The method is conceptually more difficult than the other meth-

ods presented here, and needs results from the theory of complex variables

(chapter 24) to justify it. However, we give a practical ‘cookbook’ recipe

for determining the coefficients.

(a) To determine the coefficientAk, imagine the denominatorh(x)

written as the product (x−α 1 )(x−α 2 )···(x−αn), with anym-fold

repeated root giving rise tomfactors in parentheses.

(b) Now setxequal toαkand evaluate the expression obtained after

omitting the factor that readsαk−αk.

(c) Divide the value so obtained intog(αk); the result is the required

coefficientAk.

For our specific example we find that in step (a) thath(x)=(x+1)(x+2)

and that in evaluatingA 1 step (b) yields−1 + 2, i.e. 1. Sinceg(−1) =

4(−1) + 2 =−2, step (c) givesA 1 as (−2)/(1), i.e in agreement with our

other evaluations. In a similar wayA 2 is evaluated as (−6)/(−1) = 6.