Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(Darren Dugan) #1

1.1 SIMPLE FUNCTIONS AND EQUATIONS


We now note thatf 1 (x)=0ifxis set equal to 2. Thusx−2isafactoroff 1 (x), which
therefore can be written as


f 1 (x)=(x−2)f 2 (x)=(x−2)(c 2 x^2 +c 1 x+c 0 )

with


c 2 =3,
c 1 − 2 c 2 =− 4 ,
c 0 − 2 c 1 =− 6 ,
− 2 c 0 =4.

These equations determinef 2 (x)as3x^2 +2x−2. Sincef 2 (x) = 0 is a quadratic equation,
its solutions can be written explicitly as


x=

− 1 ±



1+6


3


.


Thus the four roots off(x)=0are− 1 , 2 ,^13 (−1+



7) and^13 (− 1 −


7).


1.1.3 Properties of roots

From the fact that a polynomial equation can be written in any of the alternative


forms


f(x)=anxn+an− 1 xn−^1 +···+a 1 x+a 0 =0,

f(x)=an(x−α 1 )m^1 (x−α 2 )m^2 ···(x−αr)mr=0,

f(x)=an(x−α 1 )(x−α 2 )···(x−αn)=0,

it follows that it must be possible to express the coefficientsaiin terms of the


rootsαk. To take the most obvious example, comparison of the constant terms


(formally the coefficient ofx^0 ) in the first and third expressions shows that


an(−α 1 )(−α 2 )···(−αn)=a 0 ,

or, using the product notation,


∏n

k=1

αk=(−1)n

a 0
an

. (1.12)


Only slightly less obvious is a result obtained by comparing the coefficients of


xn−^1 in the same two expressions of the polynomial:


∑n

k=1

αk=−

an− 1
an

. (1.13)


Comparing the coefficients of other powers ofxyields further results, though

they are of less general use than the two just given. One such, which the reader


may wish to derive, is


∑n

j=1

∑n

k>j

αjαk=

an− 2
an

. (1.14)

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