Solving equations by iterative methods 83
Neglecting terms containing products of δ 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, correct 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 − 6x − 7 = 0 is −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 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
3x 3 − 10x 2 + 4x + 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) = 3x
3
− 10x
2
+ 4x + 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) Let f (x 1 + δ 1 ) = 0, then since x 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 of δ 1 and
using the binomial series gives:
3[1.8
3
+ 3(1.8)
2
δ 1 ] − 10[1.8
2
+ (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
Thus x 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) Let f (x 2 + δ 2 ) = 0, then since x 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
Neglecting terms containing products of δ 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, correct 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 − 6x − 7 = 0 is −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 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
3x 3 − 10x 2 + 4x + 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) = 3x
3
− 10x
2
+ 4x + 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) Let f (x 1 + δ 1 ) = 0, then since x 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 of δ 1 and
using the binomial series gives:
3[1.8
3
+ 3(1.8)
2
δ 1 ] − 10[1.8
2
+ (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
Thus x 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) Let f (x 2 + δ 2 ) = 0, then since x 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