Higher Engineering Mathematics

84 NUMBER AND ALGEBRA

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

nificant 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 −3 sinx+2ln(x+1)− 3. 5

f(r 1 )=f(2)= 22 −3 sin 2+2ln3− 3 .5,

where sin2 means the sine of 2 radians

= 4 − 2. 7279 + 2. 1972 − 3. 5

=− 0. 0307

f′(x)= 2 x−3 cosx+

2

x+ 1

f′(r 1 )=f′(2)=2(2)−3 cos 2+

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 −3 sin 2. 005 +2ln3. 005 − 3 .5]

[

2(2.005)−3 cos 2. 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 andr 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+5 cos

x

3

=9,

correct to 3 significant figures.

The functional notational method is used to deter-

mine the approximate value of the root.

f(x)=(x+4)^3 −e^1.^92 x+5 cos

x

3

− 9

f(0)=(0+4)^3 −e^0 +5 cos 0− 9 = 59

f(1)= 53 −e^1.^92 +5 cos

1

3

− 9 ≈ 114

f(2)= 63 −e^3.^84 +5 cos

2

3

− 9 ≈ 164

f(3)= 73 −e^5.^76 +5 cos 1− 9 ≈ 19

f(4)= 83 −e^7.^68 +5 cos

4

3

− 9 ≈− 1660

From these results, let a first approximation to the

root ber 1 =3.

Newton’s formula states that a better approximation

to the root,

r 2 =r 1 −

f(r 1 )

f′(r 1 )

f(r 1 )=f(3)= 73 −e^5.^76 +5 cos 1− 9

= 19. 35

f′(x)=3(x+4)^2 − 1 .92e^1.^92 x−

5

3

sin

x

3

f′(r 1 )=f′(3)=3(7)^2 − 1 .92e^5.^76 −

5

3

sin 1

=− 463. 7