MATHEMATICS FOR ENGINEERING STUDENTS: NUMERICAL METHOD

Given the following polynomial equation 3 − 9 + 14 = 0 .
Find:
i. The initial approximation value, 0by using False Position

Method.
ii. The approximate real root correct to 3 decimal places by

using Newton Raphson method.

Based on Graph 1, the root is located between -4 to -3.

False Position Method

= 3 − 9 + 14 1 1 1
x -4 -3 0 = 2 − 1 2 2
y -14 14
1 −4 −14
0 = 14 − −14 −3 14

0 = −3.5

Let = 0, find ′( ) Substitute f(x) and f’(x) into the formula
= 3 − 9 + 14 = 0
Let = 3 − 9 + 14 3 − 9 + 14
′ = 3 2 − 6 +1 = − 3 2 − 6

Take the initial value 0 from51 False Position Method.
Therefore, 0 = −3.5

1 = −3.5 − −3.5 3−9 −3.5 +14 = −3.545
3 −3.5 2−6 −3.5

2 = −3.545 − −3.545 3−9 −3.545 +14 = −3.568
3 −3.545 2−6 −3.545

3 = −3.568 − −3.568 3−9 −3.568 +14 = −3.580
3 −3.568 2−6 −3.568

4 = −3.580 − −3.580 3−9 −3.580 +14 = −3.580
3 −3.580 2−6 −3.580

∴ Since the root of x is repeated, then the root − 3.580

45

y Graph 1

When 3 − 9 + 14 = 0
Split the equation into 2 equation

3 = 9 − 14

For =
When = 0, = 0 (0,0)

For = −
When = 0, = −14 (0, −14)

When = 0, = 14/9 14 @1.556 , 0

9

-4 -3 (1.556,0) x
(0,-14)
52

Root located
between -4 to -3

46

0 = 2.5 3 − 9 + 14

+1 = − 3 2 − 6

53 −3.545(3 . )
−3.568(3 . )
−3.580(3 . )
−3.580(3 . )

47

Solve the equation

3 + 3 2 − 2 = 0, by using Newton Raphson method. Give the

answer correct to three decimal places with an initial guess of

0 = 1. The root is 0.732

Determine the root of the function
= + + − , correct to three decimal places using

Newton Raphson method with the initial guess of = .

The root is 1.106

Show that = − − , has a root in the interval [1,2]. The
solve the equation by using Newton Raphson Method with the

initial guess of = . and give the answer correct to 4
decimal places.

The root is 1.3247

Show that the equation of − − = has a root in the
interval [1,2]. Then solve the equation by using Newton-
Raphson Method and give the answer correct to 3 decimal

places.

The root is 1.135

Given the non-linear equation (polynomial equation) is
− + − = . Calculate:

i. The initial approximation value, 0 by using False
Position Method.

ii. The approximate real root correct to 3 decimal
places by using Newton Raphson method.

i. 0=0.4286
ii. The root is 0.4081

48

The root finding problem ( )=0 can always be
transformed into another form, = ( ), known
as the fixed point problem.

STEP 0 To show a root exist, substitute the given 2
points into the equation. The answer must be
1 positive and 1 negative value to prove that
there is a root between the given range.

STEP 1 Rearrange the equation of = 0 into the
STEP 2 form of = ( ) where x is the subject.

Find ′( ) and check whether the iterative
sequence is55 converges to the root ′ 0 < 1

STEP 3  Choose a suitable initial value 0 and
compute +1 = .

 Substitute 0 into the iterative function to get
a closer approximation to a solution.

 Perform iterations until find the repeated root.

 Therefore, x is the root of the polynomial.

49

Show that the equation f x = 4 − 5 − 2 has a root between 1.0
and 2.0. Then, solve using Fixed Iteration method with the initial
value of 0 = 1.5. Give the answer correct to 3 decimal places.

= 4 − 5 − 2

When = 1.0, When = 2.0,

= (1.0)4−5 1.0 − 2 = −6 = (2.0)4−5 2.0 − 2 = 4

∴ Since f(1)<0 and f(2)>0, there is a root in [1.0,2.0].

Rearrange the equation of Find ′( )and check whether the iterative
f(x)=0 into the form x = g(x) sequence is converges to the root ′ 0 < 1

Let = 0, 1

4 − 5 − 2 = 0 = (5 + 2)4

4 = 5 + 2 ′ =1 5 + 2 − 3 × 5 = − 5 (5 + 2)−43
4
= 4 5 + 2 44

1 Let 0 = 1.5,

∴ = (5 + 2)4 ′ 1.5 = − 5 (5 1.5 + 2)−34

4

′ = 0.232 < 1 ∴ the iteration will converge

The iterative sequence, +1 1

= (5 + 2)4

Given that 0 = 1.5,

1

1 = (5 1.5 + 2)4= 1.756

1

2 = (5 1.756 + 2)4= 1.812

1

3 = (5 1.812 + 2)4= 1.824

1

4 = (5 1.824 + 2)4= 1.826

1

5 = (5 1.826 + 2)4= 1.827

1

6 = (5 1.827 + 2)4= 1.827

∴ Since the root of x is repeated, then the root 1.827

50

0 = 1.5 1

+1 = (5 + 2)4

1.756 (3 . )

57

1.812(3 . )
1.824(3 . )
1.826(3 . )
1.827(3 . )
1.827(3 . )

51

Determine the approximate of the = 2 − 5 − 7 using fixed
Point Iteration Method. Given 0 = 6 and then give your answer
in three decimal places.

Rearrange the equation of f(x)=0 into the form x = g(x)
Let = 0,
2 − 5 − 7 = 0
2 = 5 + 7

= 5 + 7

1

∴ = (5 + 7)2

Find ′( )and check whether the iterative sequence is converges
to the root ′ 0 < 1

1

= (5 + 7)2

′ = 1 (5 + 7)− 1 5 = 5 (5 + 7)− 1
2
2×
22
Let 0 = ,
1
′ 6 =5 56 +7 − 2

2
′ 6 = 0.411 < 1 ∴ the iteration will converge

The iterative sequence, +1 1
Given that 0 = 6,
= (5 + 7)2

1

1 = (5 6 + 7)2= 6.083
1

2 = (5 6.083 + 7)2= 6.117
1

3 = (5 6.117 + 7)2= 6.131
1

4 = (5 6.131 + 7)2= 6.136
1

5 = (5 6.136 + 7)2= 6.138
1

6 = (5 6.138 + 7)2= 6.139
1

7 = (5 6.139 + 7)2= 6.140
1

8 = (5 6.140 + 7)2= 6.140

∴ Since the root of x is repeated, then the root 6.140 52

0 = 6 1

+1 = (5 + 7)2

6.083 (3 . )
59 6.117 (3 . )

6.131 (3 . )
6.136 (3 . )
6.138 (3 . )
6.139 (3 . )
6.140 (3 . )
6.140 (3 . )

53

Find the root of the equation 2 3 − 3 = 1 with 0 = 1.5 by using
Fixed Iteration method. Round off the answer correct to 2 decimal
places.

Rearrange the equation of f(x)=0 into the form x = g(x)

Let = 0,

2 3 − 3 − 1 = 0

3 = 3 +1
2

= 3 3 +1 1
3
2

∴ = 3 +1

2

Find ′( )and check whether the iterative sequence is converges

to the root ′ 0 < 1

1

= 3 +1 3

2

′ =1 3 +1 − 2 × 3
3

32 2

′ =1 3 +1 − 2
3

22

Let 0 = . ,

′ 1.5 =1 3 1.5 +1 − 2
3

22

′ 1.5 = 0.26 < 1 ∴ the iteration will converge

1

The iterative sequence, +1 = 3 +1 3
2

Given that 0 = 1.5,

1

1 = 3 1.5 +1 3 = 1.40
2

1

2 = 3 1.40 +1 3 = 1.38
2

1

3 = 3 1.38 +1 3 = 1.37
2

1

4 = 3 1.37 +1 3 = 1.37
2

∴ Since the root of x is repeated, then the root 1.37 54

0 = 1.5 1

3 + 1 3
2
+1 =

1.40 (2 . )
1.38 (2 . )
1.37 (2 . )
1.37 (2 . )

55

Find the root of the equation 3 + 3 = 1 with 0 = 0.5 by using
Fixed Iteration method. Round off the answer correct to 2 decimal
places.

Rearrange the equation of f(x)=0 into the form x = g(x)

Let = 0,

3 + 3 − 1 = 0 3 + 3 − 1 = 0

3 = 1 − 3 = 1 − 3
=
1 ∴ 1 3 3
−
= 1 − 3 3
3
1

∴ = 1 − 3 3

Find ′( )and check whether the iterative sequence is converges

to the root ′ 0 < 1 1 − 3
1 = 3
1 3
= 1 − 3 3 = 3− 3

′ = 1 1 − 3 −32 × −3 ′ = − 2

3

′ = − 1 − 3 −23

Let 0 = . , Let 0 = . ,
′ 1.5 = − 1 − 3(0.5) −32 ′ 0.5 = − 0.5 2
′ 0.5 = −0.25
2
′ 0 = 0.25 < 1
′ 1.5 = −1/ 1 − 3(0.5) 3
′(1.5) = −1.587

′ 0 = 1.587 > 1 ∴ the iteration will converge
∴ the iteration will not converge

The iterative sequence, +1 = 1− 3

Given that 0 = 0.5, 3

1 = 1−(0.5)3 = 0.29
3

2 = 1−(0.29)3 = 0.33
3

3 = 1−(0.33)3 = 0.32
3

4 = 1−(0.32)3 = 0.32
3
∴ Since the root of x is repeated, then the root 0.32
56

0 =0.5 1 − 3
+1 = 3

0.29 (2 . )
63 0.33 (2 . )

0.32 (2 . )
0.32 (2 . )

57

Given that the polynomial equation 3 − − 1 = 0 with the
iterative function +1 = 3 1 + . Find
i. The initial approximation value, 0 by using False Position

Method.
ii. The approximate real root correct to 3 decimal places by

using Fixed Iteration method.
.

Based on Graph 2, the root is located between 1 to 2.

= 3 − − 1 False Position Method

x1 2 1 1 1
y -1 5 0 = 2 − 1 2 2

1 1 −1
0 = 5 − −1 2 5

0 = 1.167

1

The iterative sequence, +1 = 1 + 3

Given that 0 = 1.167,
1

1 = 1 + 1.167 3 = 1.294

1

2 = 1 + 1.294 3 = 1.319

1

3 = 1 + 1.319 3 = 1.324

1

4 = 1 + 1.324 3 = 1.325

1

5 = 1 + 1.325 3 = 1.325

∴ Since the root of x is repeated, then the root 1.325

58

y Graph 2

When 3 − − 1 = 0
Split the equation into 2 equation

3 = + 1

For =
When = 0, = 0 (0,0)

For = +
When = 0, = 1 (0,1)

When = 0, = −1 −1,0

(0,1) 1 Root located
(-1,0) between 1 to 2

-1 (0,0) 1 x

2

65

59

0 = 1.167 1

+1 = 1 + 3

1.294 (3 . )
1.319 (3 . )
1.324 (3 . )
1.325 (3 . )
1.325 (3 . )

60

Solve the equation

= + , by using Fixed Iteration method. Give the

answer correct to four decimal places with an initial guess of

= . . The root is 3.1698

Determine the root of the function
= − − , correct to 4 decimal places using Fixed

Iteration method with the initial guess of = .

The root is 1.3247

Show that = 4 − 4 − , has a root in the interval [1,2].
Solve the equation by using Fixed Iteration method with the

initial guess of = . and give the answer correct to 3
decimal places.

The root is 1.633

Show that the equation of − − = has a root in the
interval [1,2]. Then solve the equation by using Fixed Iteration

method and give the answer correct to 3 decimal places.

The root is 1.135

Given that the polynomial equation is − − = and
has a root in the interval [1,2]. Find:

i. The initial approximation value, 0 by using False
Position Method.

ii. The fixed point iterative function for − − = ,

+1
iii. The approximate real root correct to 3 decimal places by

using Fixed Iteration method. i. 0=1.417 1

ii. +1 = 2 +5 3
2

iii. The root is 1.600

61

REFERENCES

Bird, J. (2017). Higher Engineering Mathematics (7th Edition). UK.
Routledge
Casio fx-570MS User Manual.
https://www.manualslib.com/products/Casio-Fx-570ms-
182843.html
Croft, A. and Davison, R. (2019). Mathematics for Engineers (5th
Edition). Pearson Education.

68

MATHEMATICS FOR
ENGINEERING STUDENTS

Mathematics For Engineering Student : Numerical Method,
First Edition includes chapters covered in the Electrical
Engineering Mathematics 3 (DBM 30043) courses for
polytechnic students. This e-book has been designed to
help students to seek a good reference source for their
better understanding and stimulate the students’ interest
in numerical method. This e-book also shown the step by
step how to use the Calculator Casio FX-570MS in
Numerical Methods.

This e-book is using simple words so that the delivery of
the concepts, methods and notes can be represented in
the convenience way so that students can manage to
understand easily. Example questions are provided for
every subtopic to show the solution of question step by
step. Exercises questions to ensure students understand
the mathematical concept and gain skills to solve
mathematics problem. E69xample question and exercises
focuses on examination question.

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