Higher Engineering Mathematics, Sixth Edition

Solving equations by iterative methods 85

By Newton’s formula, a better approximation to the
root is:

r 2 = 1 −

− 1

21

= 1 −(− 0. 048 )= 1. 05 ,

correct to 3 significant figures.

A still better approximation to the root,r 3 , is given by:

r 3 =r 2 −

f(r 2 )

f′(r 2 )

= 1. 05 −

[5( 1. 05 )^2 + 11 ( 1. 05 )−17]

[10( 1. 05 )+11]

= 1. 05 −

0. 0625

21. 5

= 1. 05 − 0. 003 = 1. 047 ,

i.e. 1.05, correct to 3 significant figures.

Since the values of r 2 and r 3 are the same when
expressed to the required degree of accuracy, the
required root is1.05, correct to 3 significant figures.
Checking, using the quadratic equation formula,

x=

− 11 ±

√

[121− 4 ( 5 )(− 17 )]

( 2 )( 5 )

=

− 11 ± 21. 47

10

The positive root is 1.047, i.e.1.05, correct to 3 signi-
ficant figures (This root was determined in Problem 1
using the bisection method; Newton’s method is clearly
quicker).

Problem 7. Taking the first approximation as 2,

determine the root of the equation

x^2 −3sinx+2ln(x+ 1 )= 3 .5, correct to 3

significant figures, by using Newton’s method.

Newton’s formula states thatr 2 =r 1 −

f(r 1 )

f′(r 1 )

,where

r 1 is a first approximation to the root andr 2 is a better
approximation to the root.

Since f(x)=x^2 −3sinx+2ln(x+ 1 )− 3. 5

f(r 1 )=f( 2 )= 22 −3sin2+2ln3− 3. 5 ,

where sin2 means the sine of 2 radians

= 4 − 2. 7279 + 2. 1972 − 3. 5

=− 0. 0307

f′(x)= 2 x−3cosx+

2

x+ 1

f′(r 1 )=f′( 2 )= 2 ( 2 )−3cos2+

2

3

= 4 + 1. 2484 + 0. 6667

= 5. 9151

Hence, r 2 =r 1 −

f(r 1 )

f′(r 1 )

= 2 −

− 0. 0307

5. 9151

= 2 .005 or 2. 01 ,correct to

3 significant figures.

A still better approximation to the root,r 3 , is given by:

r 3 =r 2 −

f(r 2 )

f′(r 2 )

= 2. 005 −

[( 2. 005 )^2 −3sin2. 005 +2ln3. 005 − 3 .5]

[

2 ( 2. 005 )−3cos2. 005 +

2

2. 005 + 1

]

= 2. 005 −

(− 0. 00104 )

5. 9376

= 2. 005 + 0. 000175

i.e.r 3 = 2 .01, correct to 3 significant figures.

Since the values ofr 2 and r 3 are the same when

expressed to the required degree of accuracy, then the

required root is2.01, correct to 3 significant figures.

Problem 8. Use Newton’s method to find the

positive root of:

(x+ 4 )^3 −e^1.^92 x+5cos

x

3

= 9 ,

correct to 3 significant figures.

The functional notational method is used to determine

the approximate value of the root.

f(x)=(x+ 4 )^3 −e^1.^92 x+5cos

x

3

− 9

f( 0 )=( 0 + 4 )^3 −e^0 +5cos0− 9 = 59