Higher Engineering Mathematics, Sixth Edition

Solving equations by iterative methods 83

Neglecting terms containing products ofδ 3 gives:

2. 3759 − 6. 1656 δ 3 + 4. 6242 − 6 δ 3 − 7 ≈ 0

i.e.δ 3 ≈

− 2. 3759 − 4. 6242 + 7

− 6. 1656 − 6

≈

− 0. 0001

− 12. 156

≈+ 0. 00000822

Thus,x 4 , the fourth approximation to the root is

(− 0. 7707 + 0. 00000822 ),i.e.x 4 =− 0 .7707, cor-

rect to 4 significant figures, and−0.771, correct to

3 significant figures.

Since the values of the roots are the same on

two consecutive approximations, when stated to

the required degree of accuracy, then the negative

root of 4x^2 − 6 x− 7 =0is−0.771, correct to 3

significant figures.

[Checking, using the quadratic formula:

x=

−(− 6 )±

√

[(− 6 )^2 −( 4 )( 4 )(− 7 )]

( 2 )( 4 )

=

6 ± 12. 166

8

=−0.771 and 2.27,

correct to 3 significant figures]

[Note on accuracy and errors. Depending on the
accuracy of evaluating the f(x+δ)terms, one or two
iterations (i.e. successive approximations) might be
saved. However, it is not usual to work to more than
about 4 significant figuresaccuracy in this type of cal-
culation.If asmall error is madein calculations,theonly
likely effect is to increase the number of iterations.]

Problem 5. Determine the value of the

smallest positive root of the equation

3 x^3 − 10 x^2 + 4 x+ 7 =0, correct to 3 significant

figures, using an algebraic method of successive

approximations.

The functional notation method is used to find the value
of the first approximation.

f(x)= 3 x^3 − 10 x^2 + 4 x+ 7

f( 0 )= 3 ( 0 )^3 − 10 ( 0 )^2 + 4 ( 0 )+ 7 = 7

f( 1 )= 3 ( 1 )^3 − 10 ( 1 )^2 + 4 ( 1 )+ 7 = 4

f( 2 )= 3 ( 2 )^3 − 10 ( 2 )^2 + 4 ( 2 )+ 7 =− 1

Following the above procedure:

First approximation

(a) Let the first approximation be such that it divides

the interval 1 to 2 in the ratio of 4 to−1, i.e. letx 1

be 1.8.

Second approximation

(b) Let the true value of the root,x 2 ,be(x 1 +δ 1 ).

(c) Letf(x 1 +δ 1 )=0, then sincex 1 = 1 .8,

3 ( 1. 8 +δ 1 )^3 − 10 ( 1. 8 +δ 1 )^2

+ 4 ( 1. 8 +δ 1 )+ 7 = 0

Neglecting terms containing products ofδ 1 and

using the binomial series gives:

3[1. 83 + 3 ( 1. 8 )^2 δ 1 ]−10[1. 82 +( 2 )( 1. 8 )δ 1 ]

+ 4 ( 1. 8 +δ 1 )+ 7 ≈ 0

3 ( 5. 832 + 9. 720 δ 1 )− 32. 4 − 36 δ 1

+ 7. 2 + 4 δ 1 + 7 ≈ 0

17. 496 + 29. 16 δ 1 − 32. 4 − 36 δ 1

+ 7. 2 + 4 δ 1 + 7 ≈ 0

δ 1 ≈

− 17. 496 + 32. 4 − 7. 2 − 7

29. 16 − 36 + 4

≈−

0. 704

2. 84

≈− 0. 2479

Thusx 2 ≈ 1. 8 − 0. 2479 = 1. 5521

Third approximation

(d) Let the true value of the root,x 3 ,be(x 2 +δ 2 ).

(e) Letf(x 2 +δ 2 )=0, then sincex 2 = 1 .5521,

3 ( 1. 5521 +δ 2 )^3 − 10 ( 1. 5521 +δ 2 )^2

+ 4 ( 1. 5521 +δ 2 )+ 7 = 0

Neglecting terms containing products ofδ 2 gives:

11. 217 + 21. 681 δ 2 − 24. 090 − 31. 042 δ 2

+ 6. 2084 + 4 δ 2 + 7 ≈ 0