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 is 1.05, 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 sin x + 2 ln(x + 1) = 3.5, correct to 3
significant figures, by using Newton’s method.
Newton’s formula states that r 2 = r 1 −
f (r 1 )
f (r 1 )
, where
r 1 is a first approximation to the root and r 2 is a better
approximation to the root.
Since f (x) = x
2
− 3 sin x + 2 ln (x + 1) − 3.5
f (r 1 ) = f (2) = 2
2
− 3 sin 2 + 2 ln3 − 3.5,
where sin2 means the sine of 2 radians
= 4 − 2.7279 + 2.1972 − 3.5
= −0.0307
f
(x) = 2x − 3 cos x +
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 sin2.005 + 2 ln3.005 − 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 of r 2 and r 3 are the same when
expressed to the required degree of accuracy, then the
required root is 2.01, correct to 3 significant figures.
Problem 8. Use Newton’s method to find the
positive root of:
(x + 4)
3
− e
1.92x
+ 5 cos
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
1.92x
+ 5 cos
x
3
− 9
f (0) = (0 + 4)
3
− e
0
+ 5 cos 0 − 9 = 59
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 is 1.05, 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 sin x + 2 ln(x + 1) = 3.5, correct to 3
significant figures, by using Newton’s method.
Newton’s formula states that r 2 = r 1 −
f (r 1 )
f (r 1 )
, where
r 1 is a first approximation to the root and r 2 is a better
approximation to the root.
Since f (x) = x
2
− 3 sin x + 2 ln (x + 1) − 3.5
f (r 1 ) = f (2) = 2
2
− 3 sin 2 + 2 ln3 − 3.5,
where sin2 means the sine of 2 radians
= 4 − 2.7279 + 2.1972 − 3.5
= −0.0307
f
(x) = 2x − 3 cos x +
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 sin2.005 + 2 ln3.005 − 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 of r 2 and r 3 are the same when
expressed to the required degree of accuracy, then the
required root is 2.01, correct to 3 significant figures.
Problem 8. Use Newton’s method to find the
positive root of:
(x + 4)
3
− e
1.92x
+ 5 cos
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
1.92x
+ 5 cos
x
3
− 9
f (0) = (0 + 4)
3
− e
0
+ 5 cos 0 − 9 = 59