Higher Engineering Mathematics

82 NUMBER AND ALGEBRA

[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 thef(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 figures accuracy in this type of

calculation. If a small error is made in calculations,

the only 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. let

x 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 of δ 2

gives:

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

+ 6. 2084 + 4 δ 2 + 7 ≈ 0

δ 2 ≈

− 11. 217 + 24. 090 − 6. 2084 − 7

21. 681 − 31. 042 + 4

≈

− 0. 3354

− 5. 361

≈ 0. 06256

Thusx 3 ≈ 1. 5521 + 0. 06256 ≈ 1. 6147

(f) Values ofx 4 andx 5 are found in a similar way.

f(x 3 +δ 3 )=3(1. 6147 +δ 3 )^3 −10(1. 6147

+δ 3 )^2 +4(1. 6147 +δ 3 )+ 7 = 0

givingδ 3 ≈ 0 .003175 andx 4 ≈ 1 .618, i.e. 1.62

correct to 3 significant figures

f(x 4 +δ 4 )=3(1. 618 +δ 4 )^3 −10(1. 618

+δ 4 )^2 +4(1. 618 +δ 4 )+ 7 = 0

givingδ 4 ≈ 0 .0000417, andx 5 ≈ 1 .62, correct

to 3 significant figures.