84 Higher Engineering Mathematics
δ 2 ≈
−11.217 + 24.090 − 6.2084 − 7
21.681 − 31.042 + 4
≈
−0.3354
−5.361
≈ 0.06256
Thus x 3 ≈ 1.5521 + 0.06256 ≈ 1.6147
(f) Values of x 4 and x 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
giving δ 3 ≈ 0.003175 and x 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
giving δ 4 ≈ 0.0000417, and x 5 ≈ 1.62, correct to
3 significant figures.
Since x 4 and x 5 are the same when expressed to
the required degree of accuracy, then the required
root is 1.62, correct to 3 significant figures.
Now try the following exercise
Exercise 36 Further problems on solving
equations by an algebraic method of
successive approximations
Use an algebraic method of successive approximation to solve the following equations to the
accuracy stated.
1. 3x 2 + 5x − 17 = 0, correct to 3 significant
figures.
[−3.36, 1.69]
2. x 3 − 2x + 14 =0, correct to 3 decimal places.
[−2.686]
3. x 4 − 3x 3 + 7x − 5.5 = 0, correct to 3 significant figures.
[−1.53, 1.68]
4. x 4 + 12x 3 − 13 = 0, correct to 4 significant
figures.
[−12.01, 1.000]
9.4 The Newton-Raphson method
The Newton-Raphson formula, often just referred to as
Newton’s method, may be stated as follows:
If r 1 is the approximate value of a real root of the
equation f (x) = 0, then a closer approximation to the
root r 2 is given by:
r 2 = r 1 −
f(r 1 )
f
(r 1 )
The advantages of Newton’s method over the algebraic method of successive approximations is that it
can be used for any type of mathematical equation
(i.e. ones containing trigonometric, exponential, logarithmic, hyperbolic and algebraic functions), and it is
usually easier to apply than the algebraic method.
Problem 6. Use Newton’s method to determine
the positive root of the quadratic equation
5x 2 + 11x − 17 =0, correct to 3 significant figures.
Check the value of the root by using the quadratic
formula.
The functional notation method is used to determine the
first approximation to the root.
f (x) = 5x
2
+ 11x − 17
f (0) = 5(0)
2
+ 11(0) − 17 = −17
f (1) = 5(1)
2
+ 11(1) − 17 = −1
f (2) = 5(2)
2
+ 11(2) − 17 = 25
This shows that the value of the root is close to x = 1.
Let the first approximation to the root, r 1 , be 1.
Newton’s formula states that a closer approximation,
r 2 = r 1 −
f (r 1 )
f (r 1 )
f (x) = 5x 2 + 11x − 17,
thus, f (r 1 ) = 5(r 1 ) 2 + 11(r 1 ) − 17
= 5(1) 2 + 11(1) − 17 = −1
f (x) is the differential coefficient of f (x),
i.e. f (x) = 10x + 11.
Thus f (r 1 ) = 10(r 1 ) + 11
= 10(1) + 11 = 21
δ 2 ≈
−11.217 + 24.090 − 6.2084 − 7
21.681 − 31.042 + 4
≈
−0.3354
−5.361
≈ 0.06256
Thus x 3 ≈ 1.5521 + 0.06256 ≈ 1.6147
(f) Values of x 4 and x 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
giving δ 3 ≈ 0.003175 and x 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
giving δ 4 ≈ 0.0000417, and x 5 ≈ 1.62, correct to
3 significant figures.
Since x 4 and x 5 are the same when expressed to
the required degree of accuracy, then the required
root is 1.62, correct to 3 significant figures.
Now try the following exercise
Exercise 36 Further problems on solving
equations by an algebraic method of
successive approximations
Use an algebraic method of successive approximation to solve the following equations to the
accuracy stated.
1. 3x 2 + 5x − 17 = 0, correct to 3 significant
figures.
[−3.36, 1.69]
2. x 3 − 2x + 14 =0, correct to 3 decimal places.
[−2.686]
3. x 4 − 3x 3 + 7x − 5.5 = 0, correct to 3 significant figures.
[−1.53, 1.68]
4. x 4 + 12x 3 − 13 = 0, correct to 4 significant
figures.
[−12.01, 1.000]
9.4 The Newton-Raphson method
The Newton-Raphson formula, often just referred to as
Newton’s method, may be stated as follows:
If r 1 is the approximate value of a real root of the
equation f (x) = 0, then a closer approximation to the
root r 2 is given by:
r 2 = r 1 −
f(r 1 )
f
(r 1 )
The advantages of Newton’s method over the algebraic method of successive approximations is that it
can be used for any type of mathematical equation
(i.e. ones containing trigonometric, exponential, logarithmic, hyperbolic and algebraic functions), and it is
usually easier to apply than the algebraic method.
Problem 6. Use Newton’s method to determine
the positive root of the quadratic equation
5x 2 + 11x − 17 =0, correct to 3 significant figures.
Check the value of the root by using the quadratic
formula.
The functional notation method is used to determine the
first approximation to the root.
f (x) = 5x
2
+ 11x − 17
f (0) = 5(0)
2
+ 11(0) − 17 = −17
f (1) = 5(1)
2
+ 11(1) − 17 = −1
f (2) = 5(2)
2
+ 11(2) − 17 = 25
This shows that the value of the root is close to x = 1.
Let the first approximation to the root, r 1 , be 1.
Newton’s formula states that a closer approximation,
r 2 = r 1 −
f (r 1 )
f (r 1 )
f (x) = 5x 2 + 11x − 17,
thus, f (r 1 ) = 5(r 1 ) 2 + 11(r 1 ) − 17
= 5(1) 2 + 11(1) − 17 = −1
f (x) is the differential coefficient of f (x),
i.e. f (x) = 10x + 11.
Thus f (r 1 ) = 10(r 1 ) + 11
= 10(1) + 11 = 21