PREFACE
ADVANCED ENGINEERING
MATHEMATICS
A Simplified Textbook in Advanced Mathematics for Engineering
Harold Jan R. Terano, ECE, MET
Faculty Member
College of Engineering and Graduate School
Camarines Sur Polytechnic Colleges
Nabua, Camarines Sur
Any copy of this book not bearing the original signature of the author/publisher on this
page shall be considered as coming from an illegal source.
Any entity (individual or corporation) who violates this copyright will be punished to the
fullest extent of the law.
Author/Publisher
vii
PREFACE
Copyright © 2017 by:
HAROLD JAN R. TERANO
Camarines Sur Polytechnic Colleges
Nabua, Camarines Sur
haroldterano@cspc.edu.ph
ALL RIGHTS RESERVED. No part of this book may be reproduced, photocopied or distributed,
or stored in a database or retrieval system in any form or by any means, electronic, mechanical,
recording or scanning without the prior written permission of the author.
viii
Author: Harold Jan R. Terano, ECE, ME
PREFACE
Preface
Mathematics is the most important foundations of engineering. From the basics
of Algebra up to the studies of Calculus, all of these are much important in engineering
studies.
Advanced Engineering Mathematics focuses on the study and analysis of the
advanced topics in mathematics that are very useful in the study of engineering fields.
This book covers the discussions on a simplified approach the topics covered in
advanced engineering mathematics. This consists of ten chapters that discuss the basic
of advanced engineering mathematics.
Chapter 1 covers the discussions on complex numbers and complex variables.
These include discussions on properties and forms of complex numbers.
Chapter 2 covers Laplace transforms and inverse Laplace transforms. These
also include applications of Laplace transforms in solving linear differential equations.
Chapter 3 covers the discussions on infinite series. These include sequences
and series and tests for convergent and divergent series.
Chapter 4 covers the discussions on power series. These include topics on
Taylor and Maclaurin series. Also, applications on power series solutions to linear
differential equations were also covered.
Chapter 5 covers the discussions on Fourier series. These include Fourier series
of periodic and non-periodic functions.
Chapter 6 covers the discussions on matrices and determinants.
Chapter 7 covers the topics on vector analysis which include algebraic
operations of vectors, dot product and cross product.
Chapter 8 covers the topics on numerical methods. These include numerical
solutions to equations, systems of linear equations, interpolation and numerical
integration.
Chapter 9 covers the topics on the introduction to partial differential equations.
This book was entitled Advanced Engineering Mathematics: A Simplified
Textbook in Advanced Mathematics for Engineering.
ix
PREFACE
ACKNOWLEDGMENTS
I would like to express my sincere gratitude and thanks to all my co-instructors
and professors from Camarines Sur Polytechnic Colleges, Nabua, Camarines Sur, to
my students, friends and relatives, to my family and most of all to the Almighty God for
their continual supports and encouragement for the completion of this work.
Without them, this work will never been in reality.
To all, my sincere thanks and May God bless you!!!
HAROLD JAN R. TERANO, ECE, MET
Faculty Member
College of Engineering and Graduate School
Camarines Sur Polytechnic Colleges, Nabua, Camarines Sur
Philippines
2016
x
Author: Harold Jan R. Terano, ECE, ME
TABLE OF CONTENTS
Table of Contents
Page
Preface
i
Chapter 1: COMPLEX NUMBERS AND COMPLEX VARIABLES
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
Complex Numbers
Exercise 1.1
Operations on Complex Numbers
Exercise 1.2
Complex Equations
Exercise 1.3
Polar Form of a Complex Number
Exercise 1.4
Multiplication and Division in Polar Form
Exercise 1.5
De Moivre’s Theorem
Exercise 1.6
Roots of a Complex Number
Exercise 1.7
Exponential Form of a Complex Number
Exercise 1.8
Chapter 2: LAPLACE TRANSFORMS
2.1
2.2
2.3
The Laplace Transforms
Exercise 2.1
Inverse Laplace Transforms
Exercise 2.2
Solutions to Linear Differential Equations using
Laplace Transforms
Exercise 2.3
Chapter 3: INFINITE SERIES
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
1
2
2
3
7
7
9
9
12
12
13
13
15
15
17
17
18
19
20
24
24
26
27
29
31
Sequence and Series
Limit of a Sequence
Exercise 3.1
Infinite Series
Convergent Series and Divergent Series
Exercise 3.2
nth Term Test for Divergence
Exercise 3.3
Comparison Test
Exercise 3.4
Integral Test
Exercise 3.5
Root Test
Exercise 3.6
Ratio Test
vii
32
32
33
33
34
36
36
37
37
39
39
41
41
42
42
TABLE OF CONTENTS
3.10
Exercise 3.7
Alternating Series
Exercise 3.8
44
45
46
Chapter 4: POWER SERIES
4.1
4.2
4.3
4.4
4.5
47
Power Series
Radius and Interval of Convergence
Exercise 4.1
Geometric Power Series
Exercise 4.2
Taylor Series and Maclaurin Series
Exercise 4.3
Power Series Solutions to Ordinary Differential Equations
Exercise 4.4
Chapter 5: FOURIER SERIES
5.1
5.2
5.3
59
Fourier Series for Periodic Functions of Period 2𝜋
Exercise 5.1
Fourier Series for a Non-Periodic Function over Range 2𝜋
Fourier Series over Any Range
Exercise 5.2
Chapter 6: MATRICES AND DETERMINANTS
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
Matrices
Determinants
Exercise 6.1
Operations on Matrices
Exercise 6.2
Algebraic Operations of Matrices
Exercise 6.3
Expansion of Minors
Exercise 6.4
Pivotal Element Method
Exercise 6.5
Solutions of Linear Systems using Inverse of a Matrix
Exercise 6.6
Cramer’s Rule
Cramer’s Rule for a System of Three Linear Equations
Exercise 6.7
Chapter 7: VECTOR ANALYSIS
7.1
7.2
7.3
7.4
7.5
7.6
60
64
65
67
70
71
72
73
76
76
80
81
84
85
88
89
91
91
95
95
97
99
101
Vector and Scalar
Properties of Vectors
Laws of Vector Algebra
Components of a Vector
Exercise 7.1
Dot Product
Exercise 7.2
Cross Product
Exercise 7.3
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Author: Harold Jan R. Terano, ECE, ME
48
48
50
50
52
53
55
55
58
102
102
103
104
105
106
108
109
113
TABLE OF CONTENTS
Chapter 8: NUMERICAL METHODS
8.1
8.2
8.3
8.4
8.5
Numerical Methods
Solutions of Equations
Exercise 8.1
Exercise 8.2
Numerical Solution of Linear Systems of Equations
Exercise 8.3
Exercise 8.4
Interpolation
Exercise 8.5
Numerical Integration
Exercise 8.6
Chapter 9: INTRODUCTION TO
PARTIAL DIFFERENTIAL EQUATIONS
9.1
9.2
Partial Differential Equations
Exercise 9.1
Solution by Direct Partial Integration
Exercise 9.2
References
115
116
116
119
121
121
124
128
129
132
133
138
141
142
143
143
146
147
ix
CHAPTER 1
COMPLEX NUMBERS AND
COMPLEX VARIABLES
Chapter 1
COMPLEX NUMBERS AND
COMPLEX VARIABLES
Chapter Outline:
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
Complex Numbers
Operations on Complex Numbers
Complex Equations
Polar Form of a Complex Number
Multiplication and Division of Polar Forms
De Moivre’s Theorem
Roots of a Complex Number
Exponential Form of a Complex Number
Learning Objectives:
At the end of the lesson, the students will be able to:
1. Define complex number.
2. Simplify complex numbers.
3. Solve algebraic operations on complex numbers.
4. Solve complex equations.
5. Convert complex numbers into different forms.
6. Explain De Moivre’s Theorem.
7. Apply De Moivre’s Theorem in solving complex numbers.
Overview:
This chapter deals with complex numbers. This include operations on complex
numbers, complex equations and different forms of complex numbers.
1
Author: Harold Jan R. Terano, ECE, ME
COMPLEX NUMBERS AND
COMPLEX VARIABLES
CHAPTER 1
1.1 Complex Numbers
A complex number is given by,
𝑧 = 𝑎 + 𝑏𝑖
where 𝑎 is the real part and 𝑏𝑖 is the imaginary part of the complex number. 𝑖 ( 𝑗,
commonly used symbol in applications) is the imaginary unit defined as
𝑖 2 = −1 or 𝑖 = √−1
Example 1:
Simplify 𝑖 3 + 1.
Solution:
𝑖 3 + 1 = (𝑖 2 )(𝑖 ) + 1 = (−1)𝑖 + 1 = 1 − 𝑖
Example 2:
Simplify 2𝑖 6 − 3𝑖 4 − 8𝑖 2 + 4𝑖.
Solution:
2𝑖 6 − 3𝑖 4 + 8𝑖 2 + 4 = 2(𝑖 2 )3 − 3(𝑖 2 )2 − 8(𝑖 2 ) + 4𝑖
= 2(−1)3 − 3(−1)2 − 8(−1) + 4𝑖
= 2(−1) − 3(1) + 8 + 4𝑖
= 3 + 4𝑖
Example 3:
Simplify 3𝑖 125 − 2𝑖 52 + 𝑖 23 − 4𝑖 15 − 1.
Solution:
3𝑖 125 − 2𝑖 52 + 𝑖 23 − 4𝑖 15 − 1 = 3(𝑖 124)(𝑖 ) − 2𝑖 52 + (𝑖 22 )(𝑖 ) − 4(𝑖 14 )(𝑖 ) − 1
= 3(𝑖 2 )62 (𝑖 ) − 2(𝑖 2 )26 + (𝑖 2 )11 (𝑖 ) − 4(𝑖 2 )7 (𝑖 ) − 1
= 3(−1)62 (𝑖) − 2(−1)26 + (−1)11 (𝑖) − 4(−1)7 (𝑖) − 1
= 3(1)(𝑖 ) − 2(1) + (−1)(𝑖 ) − 4(−1)(𝑖 ) − 1
= 3𝑖 − 2 − 𝑖 + 4𝑖 − 1
= −3 + 6𝑖
Exercise 1.1
Simplify the following complex numbers.
1. 2𝑖 4 − 2𝑖 3 + 12 − 𝑖 + 2
2. 2 − 3𝑖 3 − 2𝑖 5
3. 2𝑖 6 − 2𝑖 4 + 6𝑖 3 − 2𝑖 + 1
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Author: Harold Jan R. Teran
CHAPTER 1
COMPLEX NUMBERS AND
COMPLEX VARIABLES
4. −𝑖 2 + 𝑖 5 − 2𝑖 7
5. 5𝑖 5 − 6𝑖 4 + 𝑖 3 − 3𝑖 + 4
1
2
1
6. 2 𝑖 6 − 3 𝑖 5 − 3𝑖 3 − 3 𝑖 − 1
7. 2𝑖 43 − 3𝑖 23 + 5𝑖 13 − 6𝑖 9 − 4𝑖 + 3
8. −5𝑖 245 + 6𝑖 125 − 3𝑖 69 + 4𝑖 20 − 3𝑖 8 + 𝑖
9. 3𝑖 553 − 𝑖 342 + 𝑖 114 − 5𝑖 84 + 6𝑖 32
10. 𝑖 2015 − 2𝑖 2012 + 3𝑖 2001 − 5𝑖 1995
1.2 Operations on Complex Numbers
Let 𝑧1 = 𝑎 + 𝑏𝑖 and 𝑧2 = 𝑐 + 𝑑𝑖 be complex numbers.
Addition and Subtraction
𝑧1 + 𝑧2 = (𝑎 + 𝑏𝑖 ) + (𝑐 + 𝑑𝑖 ) = (𝑎 + 𝑐 ) + (𝑏 + 𝑑 )𝑖
𝑧1 − 𝑧2 = (𝑎 + 𝑏𝑖 ) − (𝑐 + 𝑑𝑖 ) = (𝑎 − 𝑐 ) + (𝑏 − 𝑑 )𝑖
Example 1:
If 𝑧1 = 3 + 2𝑖 and 𝑧2 = 4 + 𝑖, find 𝑧1 + 𝑧2.
Solution:
𝑧1 + 𝑧2 = (3 + 2𝑖 ) + (4 + 𝑖)
= (3 + 4) + (2 + 1)𝑖
= 7 + 3𝑖
Example 2:
If 𝑧1 = 5 − 10𝑖 and 𝑧2 = 3 + 7𝑖, find 𝑧1 − 𝑧2 .
Solution:
𝑧1 − 𝑧2 = (5 − 10𝑖 ) − (3 + 7𝑖)
= (5 − 3) + (−10 − 7)𝑖
= 2 − 17𝑖
Example 3:
If 𝑧1 = −2 − 5𝑖 and 𝑧2 = −1 + 𝑖, find 𝑧1 + 𝑧2 .
Solution:
𝑧1 + 𝑧2 = (−2 − 5𝑖 ) + (−1 + 𝑖)
= (−2 − 1) + (−5 + 1)𝑖
= −3 − 4𝑖
3
Author: Harold Jan R. Terano, ECE, ME
COMPLEX NUMBERS AND
COMPLEX VARIABLES
CHAPTER 1
Example 4:
If 𝑧1 = 5 + 4𝑖, 𝑧2 = −3 + 𝑖 and 𝑧3 = 1 + 3𝑖, find 𝑧1 + 𝑧2 + 𝑧3.
Solution:
𝑧1 + 𝑧2 + 𝑧3 = (5 + 4𝑖 ) + (−3 + 𝑖 ) + (1 + 3𝑖 )
= (5 − 3 + 1) + (4 + 1 + 3)𝑖
= 3 + 8𝑖
Example 5:
If 𝑧1 = 2 + 𝑖, 𝑧2 = −4 − 2𝑖 and 𝑧3 = 5 + 5𝑖, find 𝑧1 − 𝑧2 − 𝑧3.
Solution:
𝑧1 − 𝑧2 − 𝑧3 = (2 + 𝑖 ) − (−4 − 2𝑖 ) − (5 + 5𝑖)
= [2 − (−4) − 5] + [1 − (−2) − 5]𝑖
= 1 − 2𝑖
Multiplication of Complex Numbers
𝑧1𝑧2 = (𝑎 + 𝑏𝑖 )(𝑐 + 𝑑𝑖 ) = 𝑎𝑐 + 𝑎𝑑𝑖 + 𝑏𝑐𝑖 + 𝑏𝑑𝑖 2 = (𝑎𝑐 − 𝑏𝑑 ) + (𝑎𝑑 + 𝑏𝑐 )𝑖
Example 1:
Let 𝑧1 = 1 + 2𝑖 and 𝑧2 = 3 + 2𝑖, find 𝑧1 𝑧2.
Solution:
𝑧1 𝑧2 = (1 + 2𝑖 )(3 + 2𝑖 )
= 3 + 2𝑖 + 6𝑖 + 4𝑖 2
= 3 + 8𝑖 − 4
= −1 + 8𝑖
Example 2:
Let 𝑧1 = −4 + 2𝑖 and 𝑧2 = 3 − 2𝑖, find 𝑧1𝑧2 .
Solution:
𝑧1 𝑧2 = (−4 + 2𝑖 )(3 − 2𝑖)
= −12 + 8𝑖 + 6𝑖 − 4𝑖 2
= −8 + 14𝑖
Example 3:
Let 𝑧1 = 3 + 𝑖, 𝑧2 = −2 + 4𝑖 and 𝑧3 = −5 − 2𝑖, find 𝑧1 𝑧2 𝑧3.
Solution:
𝑧1 𝑧2𝑧3 = (3 + 𝑖 )(−2 + 4𝑖 )(−5 − 2𝑖 )
= (−6 + 12𝑖 − 2𝑖 + 4𝑖 2 )(−5 − 2𝑖 )
= (−10 + 10𝑖 )(−5 − 2𝑖 )
4
Author: Harold Jan R. Teran
CHAPTER 1
COMPLEX NUMBERS AND
COMPLEX VARIABLES
= 50 + 20𝑖 − 50𝑖 − 20𝑖 2
= 70 − 30𝑖
Example 4:
Let 𝑧1 = 4 + 2𝑖 and 𝑧2 = −3 + 5𝑖, find 2𝑧1 − 5𝑧2 .
Solution:
2𝑧1 − 5𝑧2 = 2(4 + 2𝑖 ) − 5(−3 + 5𝑖)
= 8 + 4𝑖 + 15 − 25𝑖
= 23 − 21𝑖
Example 5:
Let 𝑧1 = 5 − 3𝑖 and 𝑧2 = 4 + 3𝑖, find 3𝑧1 𝑧2 − 𝑧1 + 2𝑧2 .
Solution:
3𝑧1 𝑧2 − 𝑧1 + 2𝑧2 = 3(5 − 3𝑖 )(4 + 3𝑖 ) − (5 − 3𝑖 ) + 2(4 + 3𝑖)
= 3(29 + 3𝑖 ) − 5 + 3𝑖 + 8 + 6𝑖
= 90 + 18𝑖
Division of Complex Numbers
𝑧1
𝑧2
𝑧
̅̅̅
𝑧
2
2
2
= 𝑧1 × ̅̅̅
𝑧
where 𝑧̅2 is the conjugate of the complex number 𝑧2 . The conjugate of a complex
nuimber 𝑎 + 𝑏𝑖 is 𝑎 − 𝑏𝑖.
Example 1:
Let 𝑧1 = 5 − 2𝑖 and 𝑧2 = 4 + 3𝑖, find
𝑧1
𝑧2
.
Solution:
𝑧1
5−2𝑖
4−3𝑖
= 4+3𝑖 × 4−3𝑖 =
𝑧
20−15𝑖−8+6𝑖 2
2
20−23𝑖+6(−1)
=
16−9(−1)
14−23𝑖
=
𝑧1
𝑧2
16−9𝑖 2
25
14
23
= 25 − 25 𝑖
Example 2:
2𝑧
Let 𝑧1 = 3 + 𝑖 and 𝑧2 = −2 + 2𝑖, find 𝑧 +𝑧1 .
1
2
Solution:
2𝑧1
2(3+𝑖)
𝑧1 +𝑧2
= (3+𝑖)+(−2+2𝑖)
6+2𝑖
1−3𝑖
= 1+3𝑖 × 1−3𝑖
5
Author: Harold Jan R. Terano, ECE, ME
COMPLEX NUMBERS AND
COMPLEX VARIABLES
12−16𝑖
=
2𝑧1
CHAPTER 1
10
6
8
= 5 − 5𝑖
𝑧1 +𝑧2
Example 3:
𝑧
Let 𝑧1 = −10 + 3𝑖 and 𝑧2 = −2𝑖, find (𝑧 1)5.
2
Solution:
𝑧1
−10+3𝑖
−10+3𝑖
= (−2𝑖)5 =
(𝑧2 )5
−32𝑖 5
=
−10+3𝑖
−10+3𝑖
=
−32𝑖(𝑖 2 )4
−32𝑖(−1)4
=
−10+3𝑖
32𝑖
×
−32𝑖
32𝑖
=
−96−320𝑖
1024
𝑧1
3
5
=− − 𝑖
(𝑧2 )5
32
16
Example 4:
𝑧
Let 𝑧1 = 3 − 2𝑖 and 𝑧2 = −1 + 3𝑖, find (𝑧 2)2.
1
Solution:
𝑧2
−1+3𝑖
= (3−2𝑖)2
(𝑧1 )2
−1+3𝑖
5+12𝑖
= 5−12𝑖 × 5+12𝑖
=
−41+3𝑖
169
𝑧2
41
3
=
−
+
𝑖
2
(𝑧1 )
169
169
Example 5:
Let 𝑧1 = −4 + 6𝑖 and 𝑧2 = 1 − 4𝑖, find
Solution:
𝑧1 +𝑧2
𝑧2 𝑧2
(−4+6𝑖)+(1−4𝑖)
= (−4+6𝑖)(1−4𝑖)
−3+2𝑖
20−22𝑖
= 20+22𝑖 × 20−22𝑖
=
𝑧1 +𝑧2
𝑧2 𝑧2
−16+106𝑖
884
4
53
= − 221 + 442 𝑖
6
Author: Harold Jan R. Teran
𝑧1 +𝑧2
𝑧2 𝑧2
.
CHAPTER 1
COMPLEX NUMBERS AND
COMPLEX VARIABLES
Exercise 1.2
Given 𝑧1 = 2 + 𝑖, 𝑧2 = −3 + 4𝑖, 𝑧3 = 6 − 8𝑖 and 𝑧4 = −3 − 4𝑖, find:
1. 𝑧1 + 𝑧2 − 𝑧3
2. 3𝑧3 − 2𝑧2 + 4𝑧1
𝑧
3. 𝑧2
4
4. (𝑧2 + 𝑧3 )(2𝑧4 − 𝑧1 )
5.
(𝑧2 +2𝑧1 )
𝑧4
6. 2(𝑧2 − 3𝑧3 )(𝑧1 + 2𝑧4 )
𝑧
2𝑧
3
4
7. 𝑧1 − 9𝑧3
8. 𝑧4 + (2𝑧3 − 5𝑧1 )
9. (𝑧1 − 4𝑧2 ) + 3(𝑧3 + 2𝑧4 )
10.
(𝑧3−2𝑧1 )
2(𝑧4 +1)
1.3 Complex Equations
If two complex numbers are equal, then, their real parts are equal and their
imaginary parts are equal. Thus, if 𝑎 + 𝑏𝑖 = 𝑐 + 𝑑𝑖, then 𝑎 = 𝑐 and 𝑏 = 𝑑.
Example 1:
Solve (2 + 3𝑖)(−2 + 𝑖) = 4𝑥 + 2𝑦𝑖.
Solution:
(2 + 3𝑖)(−2 + 𝑖) = 4𝑥 + 2𝑦𝑖
(−4 + 2𝑖 − 6𝑖 + 3𝑖 2 ) = 4𝑥 + 2𝑦𝑖
−7 − 4𝑖 = 4𝑥 + 2𝑦𝑖
Equating the real parts,
−7 = 4𝑥
𝑥=−
7
4
Equating the imaginary parts,
−4 = 2𝑦
𝑦 = −2
Example 2:
Solve √𝑎 + 3𝑏𝑖 = (3 + 2𝑖).
7
Author: Harold Jan R. Terano, ECE, ME
COMPLEX NUMBERS AND
COMPLEX VARIABLES
CHAPTER 1
Solution:
√𝑎 + 3𝑏𝑖 = (3 + 2𝑖)
𝑎 + 3𝑏𝑖 = (3 + 2𝑖)2
𝑎 + 3𝑏𝑖 = 5 + 12𝑖
Equating the real parts,
𝑎=5
Equating the imaginary parts,
3𝑏 = 12
𝑏=4
Example 3:
Solve the complex equation 2𝑥 − 3𝑦𝑖 = (−4 + 𝑖 )(3 − 6𝑖 ).
Solution:
2𝑥 − 3𝑦𝑖 = (−4 + 𝑖 )(3 − 6𝑖 )
2𝑥 − 3𝑦𝑖 = −6 + 27𝑖
Equating the real parts and the imaginary parts, we have,
2𝑥 = −6;
𝑥 = −3
−3𝑦 = 27;
𝑦 = −9
The values are 𝑥 = −3 and 𝑦 = −9
Example 4:
Solve 2(𝑥 + 3𝑦𝑖) − 3(2𝑥 − 4𝑦𝑖) = 3 − 2𝑖.
Solution:
2(𝑥 + 3𝑦𝑖) − 3(2𝑥𝑖 − 4𝑦) = 3 − 2𝑖
2𝑥 + 6𝑦𝑖 − 6𝑥𝑖 + 12𝑦 = 3 − 2𝑖
Equating the real parts,
2𝑥 + 12𝑦 = 3
Equating the imaginary parts,
6𝑦 − 6𝑥 = −2
3𝑥 − 3𝑦 = 1
Solving the two equations,
𝑥=
1
2
and 𝑦 =
1
6
Example 5:
Solve
2+𝑖
1−𝑖
= 𝑖(3𝑎 − 5𝑏𝑖).
Solution:
2+𝑖
= 𝑖(3𝑎 − 5𝑏𝑖)
1−𝑖
2+𝑖
1+𝑖
×
= 3𝑎𝑖 − 5𝑏𝑖 2
1−𝑖
1+𝑖
8
Author: Harold Jan R. Teran
CHAPTER 1
COMPLEX NUMBERS AND
COMPLEX VARIABLES
1
2
3
+ 𝑖 = 3𝑎𝑖 + 5𝑏
2
Equating the real parts,
1
= 5𝑏
2
𝑏=
1
10
Equating the imaginary parts,
3
= 3𝑎
2
𝑎=
1
2
Exercise 1.3
Solve the following complex equations.
1. (2 + 𝑖 )(−1 − 𝑖 ) = 2𝑥 + 3𝑦𝑖
2. (3 + 2𝑖)2 = 4𝑎 + 3𝑏𝑖
3. (1 − 2𝑖)(1 + 2𝑖) = √2𝑥 + 4𝑦𝑖
4.
2+3𝑖
= 𝑥 + 2𝑦𝑖
1+𝑖
5.
4+3𝑖
= (2𝑤 − 4𝑧𝑖)(1 − 𝑖)
2+𝑖
6. (1 − 2𝑖)(2 + 𝑖) = (−2𝑎 + 3𝑏𝑖)(2 − 3𝑖)
7. (– 𝑥 + 𝑦𝑖) + 3(𝑥𝑖 − 2𝑦) = 5 − 𝑖
8. −2(𝑎 + 4𝑏𝑖) − (2𝑎𝑖 + 𝑏) = (3 + 2𝑖)2
9. (5 − 2𝑖)(4 + 𝑖) = −3(𝑥 − 𝑦𝑖) + 4(4𝑥 − 3𝑦𝑖)
10.
4−3𝑖
1+2𝑖
= 5(𝑎 − 2𝑏𝑖) − 2(2𝑎𝑖 − 𝑏)
1.4 Polar Form of a Complex Number
Consider the figure,
𝑟
𝑏
𝜃
𝑎
9
Author: Harold Jan R. Terano, ECE, ME
COMPLEX NUMBERS AND
COMPLEX VARIABLES
CHAPTER 1
From the figure,
𝑟 = √𝑎2 + 𝑏2
where 𝑟 is called the modulus or the magnitude of 𝑧. 𝜃 is called the argument or
amplitude of 𝑧 (where 𝑧 = 𝑎 + 𝑏𝑖) and is written as
𝑏
𝐴𝑟𝑔 𝑧 = 𝜃 = Arctan 𝑎.
From trigonometry,
𝑎 = 𝑟 cos 𝜃 and 𝑏 = 𝑟 sin 𝜃
Since,
𝑧 = 𝑎 + 𝑏𝑖
Then,
𝑧 = 𝑟 cos 𝜃 + (𝑟 sin 𝜃 )𝑖
𝑧 = 𝑟(cos 𝜃 + 𝑖 sin 𝜃 )
This is the polar form (also known as trigonometric form) of the complex number
𝑧 = 𝑎 + 𝑏𝑖.
Common abbreviation is,
𝑧 = 𝑟∠𝜃
Example 1:
Express the following complex numbers in polar form.
a. 3 + 2𝑖
c. −3 − 2𝑖
b. −3 + 2𝑖
d. 3 − 2𝑖
Solution:
Consider the figure below,
(−3 + 2𝑖)
(3 + 2𝑖)
𝜽
𝜽
𝜽
𝜽
(−3 − 2𝑖)
(3 − 2𝑖)
10
Author: Harold Jan R. Teran
CHAPTER 1
COMPLEX NUMBERS AND
COMPLEX VARIABLES
a.) 3 + 2𝑖 lies on the first quadrant.
2
The modulus, 𝑟 = √(3)2 + (2)2 = √13 and argument 𝜃 = Arctan 3 = 33.690 .
Hence, 3 + 2𝑖 = √13 ∠33.690 .
b.) −3 + 2𝑖 lies on the second quadrant.
2
The modulus, 𝑟 = √(−3)2 + (2)2 = √13 and argument 𝜃 = Arctan 3 = 33.690 .
The argument must be measured from the positive real axis, thus, 𝜃 = 1800 − 33.690 =
146.310 . Hence, −3 + 2𝑖 = √13 ∠146.310 .
c.) −3 − 2𝑖 lies on the third quadrant.
2
The modulus, 𝑟 = √(−3)2 + (−2)2 = √13 and argument 𝜃 = Arctan 3 = 33.690 .
When the argument is measured from the positive real axis, thus, 𝜃 = 1800 + 33.690 =
213.690 . This is also the same as 𝜃 = −146.310 (clockwise direction). By convention,
the principal value is normally used (least value) such that −𝜋 < 𝜃 < 𝜋. Hence, −3 −
2𝑖 = √13 ∠ − 146.310 .
d.) 3 − 2𝑖 lies on the fourth quadrant.
2
The modulus, 𝑟 = √(3)2 + (−2)2 = √13 and argument 𝜃 = Arctan 3 = 33.690 .
Measuring from the positive real axis, we have 𝜃 = −33.690 , hence, 3 − 2𝑖 = √13 ∠ −
33.690 .
Example 2:
Convert 5∠53.130 into rectangular form.
Solution:
Using the relations, 𝑎 = 𝑟 cos 𝜃 and 𝑏 = 𝑟 sin 𝜃, then,
𝑎 = 5 cos 53.130 = 3
𝑏 = 5 sin 53.130 = 4
Hence, 5∠53.130 = 3 + 4𝑖
Example 3:
Convert 10∠ − 370 into rectangular form.
Solution:
Using the relations,
𝑎 = 10 cos(−370 ) = 8
𝑏 = 10 sin(−370 ) = −6
Hence, 10∠ − 370 = 8 − 6𝑖
11
Author: Harold Jan R. Terano, ECE, ME
COMPLEX NUMBERS AND
COMPLEX VARIABLES
CHAPTER 1
Exercise 1.4
Solve the following.
1. Convert the following complex numbers in polar forms.
a.) 4 + 2𝑖
d.) −5 − 3𝑖
b.) 10 − 2𝑖
e.) −6 + 3𝑖
c.) 12 + 5𝑖
2. Convert the following complex numbers in rectangular forms.
a.) 12∠ − 300
d.) 10∠490
b.) 5∠ − 1500
e.) 16∠ − 1120
c.) 20∠1200
1.5 Multiplication and Division in Polar Form
If 𝑧1 = 𝑟1 ∠𝜃1 and 𝑧2 = 𝑟2 ∠𝜃2, then,
𝑧1 𝑧2 = (𝑟1 ∠𝜃1 )(𝑟2 ∠𝜃2 ) = 𝑟1 𝑟2 ∠(𝜃1 + 𝜃2 ) and
𝑧1
𝑧2
𝑟 ∠𝜃1
= 1
𝑟2 ∠𝜃2
𝑟
= 1 ∠(𝜃1 − 𝜃2 )
𝑟2
Example 1:
Given 𝑧1 = 2∠200 and 𝑧2 = 4∠300 , find:
a.) 𝑧1 𝑧2
𝑧
b.) 𝑧1
2
Solution:
a.) 𝑧1 𝑧2 = (2∠200 )(4∠300 ) = (2)(4)∠(200 + 300 ) = 8∠500
2∠200
𝑧
2
1
b.) 𝑧1 = 4∠300 = 4 ∠(200 − 300 ) = 2 ∠ − 100
2
Example 2:
Given 𝑧1 = 12∠350 , 𝑧2 = 3∠−1200 and 𝑧3 = 15∠1100 , find:
a.) 𝑧1 𝑧2𝑧3
b.)
𝑧1 𝑧2
𝑧3
Solution:
a.) 𝑧1 𝑧2𝑧3 = (12∠350 )(3∠−1200 )(15∠1100 ) = (12)(3)(15)∠(350 − 1200 +
1100 )
= 540∠250
b.)
𝑧1 𝑧2
𝑧3
=
(12∠350)(3∠−1200)
15∠1100
36∠−850
12
Author: Harold Jan R. Teran
12
= 15∠1100 = 5 ∠ − 1950 or
12
5
∠1650
CHAPTER 1
COMPLEX NUMBERS AND
COMPLEX VARIABLES
Example 3:
Simplify 3∠250 + 5∠ − 1250 − 4∠350. Express in polar form.
Solution:
Addition and subtraction in polar form is not possible. Convert first the
polar forms into its rectangular forms. Thus,
3∠250 = 3(cos 250 + 𝑖 sin 250 ) = 2.72 + 1.27𝑖
5∠ − 1250 = 5[cos(−1250 ) + 𝑖 sin(−1250 )] = −2.87 − 4.10𝑖
4∠350 = 4(cos 350 + 𝑖 sin 350 ) = 3.28 + 2.29𝑖
Then,
3∠250 + 5∠ − 1250 − 4∠350 = (2.72 + 1.27𝑖 ) + (−2.87 − 4.10𝑖 ) −
(3.28 + 2.29𝑖 ) = −3.43 − 5.12𝑖
In polar form,
−3.43 − 5.12𝑖 = 6.16∠ − 123.770
Exercise 1.5
1. Given 𝑧1 = 3∠250 , 𝑧2 = 10∠ − 1250 and 𝑧3 = 5∠1560 , simplify the following and
express the answer in polar forms:
a.) 𝑧1 𝑧2 𝑧3
𝑧
𝑧 +𝑧
b.) 𝑧1
e.) 2𝑧 1
2
c.)
𝑧2
𝑧1
d.) 2(𝑧1 𝑧2 + 𝑧3 )
3
+
𝑧3
𝑧2
f.)
𝑧2
𝑧3
+ 𝑧1𝑧3
1.6 De Moivre’s Theorem
From the multiplication of polar forms,
(𝑟∠𝜃)(𝑟∠𝜃 ) = 𝑟 2 ∠2𝜃
Also, (𝑟∠𝜃)(𝑟∠𝜃 )(𝑟∠𝜃 ) = 𝑟 3 ∠3𝜃 and so on.
Generally,
[𝑟∠𝜃 ]𝑛 = 𝑟 𝑛 ∠𝑛𝜃
is the De Moivre’s Theorem.
Example 1:
Simplify in polar form [3∠350 ]4.
13
Author: Harold Jan R. Terano, ECE, ME
COMPLEX NUMBERS AND
COMPLEX VARIABLES
CHAPTER 1
Solution:
[3∠350 ]4 = 34 ∠4(35) = 81∠1400
Example 2:
Simplify in polar form (4 − 𝑖 )5 .
Solution:
In polar form,
1
(4 − 𝑖) = √(4)2 + (−1)2 ∠Arctan − 4
= √17 ∠ − 14.040
By De Moivre’s Theorem,
(4 − 𝑖 )5 = [√17 ∠ − 14.040 ]
5
5
= (√17 ) ∠5(−14.040 )
= 1191.58∠ − 70.200
Example 3:
Determine the value of (−2 + 3𝑖 )6 in rectangular and polar forms.
Solution:
In polar form,
3
(−2 + 3𝑖 ) = √(−2)2 + (3)2 ∠Arctan
−2
= √13 ∠123.690 since
(−2 + 3𝑖) lies on the second quadrant.
By De Moivre’s Theorem,
6
(−2 + 3𝑖 )6 = [√13 ∠123.690 ]
6
= (√13 ) ∠6(123.690 )
= 2197∠742.140
Since 742.140 = 742.140 − 3600 = 382.140 and
382.140 = 382.140 − 3600 = 22.140 , therefore,
(−2 + 3𝑖 )6 = 2197∠22.140
In rectangular form,
(−2 + 3𝑖 )6 = 2197(cos 22.140 + 𝑖 sin 22.140 )
= 2035 + 828𝑖
14
Author: Harold Jan R. Teran
CHAPTER 1
COMPLEX NUMBERS AND
COMPLEX VARIABLES
Exercise 1.6
I. Simplify the following in polar forms.
1. [1∠10]4
1
2. [2 ∠250 ]
5
3. [10∠ − 760 ]3
4. [2∠ − 120]10
5. [8.5∠155]8
II. Simplify the following and express the answer in rectangular forms.
1. (2 + 4𝑖 )8
2. (−1 + 5𝑖 )10
3. (−3 − 𝑖 )6
1
4. (3 + 2 𝑖)
20
5. (1 + 𝑖 )10,000
1.7 Roots of a Complex Number
Given a complex number,
𝑧 = 𝑟∠𝜃
The root of 𝑧 is given by,
1
1
𝑧 𝑛 = [𝑟∠𝜃 ]𝑛
By De Moivre’s Theorem,
1
1
𝑧 𝑛 = 𝑟 𝑛∠
𝜃+2𝑘𝜋
𝑛
where 𝑘 = 0,1,2, … , 𝑛 − 1 and 𝜋 = 1800 .
Example 1:
1
Find the roots of [√2 ∠450 ]2 .
Solution:
The first root is,
1
1
[√2 ∠450 ]2 = (√2 )2 ∠
450+0
2
0
= 1.19∠22.5
The second root is,
= 1.19∠
450+2(1)(1800 )
2
0
= 1.19∠202.5
15
Author: Harold Jan R. Terano, ECE, ME
COMPLEX NUMBERS AND
COMPLEX VARIABLES
CHAPTER 1
Example 2:
1
Find the roots of (−2 + 3𝑖 )3 .
Solution:
1
1
(−2 + 3𝑖 )3 = [√13 ∠123.690 ]3
The first root is,
= 1.53∠
123.690 +0
3
= 1.53∠41.230
The second root is,
= 1.53∠
123.690 +2(1)(1800)
3
0
= 1.53∠161.23
The third root is,
= 1.53∠
123.690 +2(2)(1800)
3
= 1.53∠281.230
Example 3:
1
Find the roots of (−3 − 5𝑖 )4 .
Solution:
1
1
4
0 ]4
(−3 − 5𝑖 ) = [√34 ∠ − 120.96
The first root is,
= 1.55∠
−120.960 +0
4
= 1.55∠ − 30.240
The second root is,
= 1.55∠
−120.960 +2(1)(1800 )
4
= 1.55∠59.760
The third root is,
= 1.55∠
−120.960 +2(2)(1800 )
4
0
= 1.55∠149.76
The fourth root is,
= 1.55∠
−120.960 +2(3)(1800 )
4
0
= 1.55∠239.76
16
Author: Harold Jan R. Teran
CHAPTER 1
COMPLEX NUMBERS AND
COMPLEX VARIABLES
Exercise 1.7
Find the roots of the following complex numbers.
1
1. [2∠250 ]2
1
2. [√5 ∠−63.430 ]4
1
3. (2 − 5𝑖 )3
1
4. (−5 + 𝑖 )2
1
5. (−2 − 4𝑖 )3
1.8 Exponential Form of a Complex Number
The polar form of a complex number 𝑧 = 𝑎 + 𝑏𝑖 is,
𝑧 = 𝑟∠𝜃 = 𝑟 (cos 𝜃 + 𝑖 sin 𝜃 )
From the Euler’s formula,
𝑒 𝑖𝜃 = cos 𝜃 + 𝑖 sin 𝜃
Therefore, the exponential form is,
𝑧 = 𝑟 𝑒 𝑖𝜃
Example 1:
Convert 𝑧 = 1 + 2𝑖 into exponential form.
Solution:
0
𝑧 = 1 + 2𝑖 = √5 ∠63.430 = √5 𝑒 63.43 𝑖 or = √5 𝑒 1.11𝑖
Example 2:
Convert 𝑧 = −2 + 3𝑖 into exponential form.
Solution:
0
𝑧 = −2 + 3𝑖 = √13 ∠123.690 = √13 𝑒 123.69 𝑖 or = √13 𝑒 2.16𝑖
Example 3:
Solve ln(3 + 5𝑖).
Solution:
ln(3 + 5𝑖 ) = ln(√34 𝑒 1.03𝑖 )
By properties of logarithm,
ln(3 + 5𝑖 ) = ln √34 + ln 𝑒 1.03𝑖
= 1.76 + 1.03𝑖
Example 4:
Solve ln(−5 + 2𝑖).
17
Author: Harold Jan R. Terano, ECE, ME
COMPLEX NUMBERS AND
COMPLEX VARIABLES
CHAPTER 1
Solution:
ln(−5 + 2𝑖) = ln(√29 𝑒 2.76𝑖 )
= ln √29 + ln 𝑒 2.76𝑖
= 1.68 + 2.76𝑖
Exercise 1.8
I. Convert the following into exponential form
1. 5 + 2𝑖
2. −2 − 𝑖
3. 8 + 3𝑖
4. −5 + 4𝑖
5. 10 + 5𝑖
II. Given 𝑧1 = 5 − 2𝑖, 𝑧2 = 6 + 2𝑖 and 𝑧3 = −3 − 8𝑖, simplify:
1. ln(𝑧1 )
2. ln(𝑧1 𝑧2 )
3. ln(𝑧1 + 𝑧2 + 𝑧3 )
𝑧
4. ln (𝑧1 )
2
5. ln(2𝑧1 + 𝑧2 − 3𝑧3 )
18
Author: Harold Jan R. Teran
CHAPTER 2
LAPLACE AND
INVERSE LAPLACE TRANSFORMS
Chapter 2
LAPLACE TRANSFORMS
Chapter Outline:
2.1
2.2
2.3
The Laplace Transforms
Inverse Laplace Transforms
Solutions to Linear Differential Equations Using Laplace Transforms
Learning Objectives:
At the end of the lesson, the students will be able to:
1. Define Laplace Transforms.
2. Find the Laplace transform of a given function.
3. Find the inverse Laplace transform of a given function.
4. Apply Laplace transforms to solve solutions to linear differential equations.
Overview:
Laplace transforms is a powerful method for solving mathematical solutions
which involves solutions to differential equations. This chapter will introduce an
introduction to Laplace transforms, method of finding the Laplace transforms of a certain
function, inverse Laplace transforms and its solution to problems in differential
equations.
19
Author: Harold Jan R. Terano, ECE, MET
LAPLACE AND
INVERSE LAPLACE TRANSFORMS
2.1
CHAPTER 2
The Laplace Transforms
Given the function 𝑓(𝑥), the Laplace transform is given by the Laplace integral,
∞
𝓛{𝒇(𝒙)} = 𝑭(𝒔) = ∫𝟎 𝒇(𝒙) 𝒆−𝒔𝒙 𝒅𝒙
Example 1:
Find the Laplace transform of the function 𝑓(𝑥 ) = 1.
Solution:
By the Laplace integral,
∞
𝓛{𝑓(𝑥)} = 𝐹(𝑠) = ∫0 𝑓(𝑥) 𝑒 −𝑠𝑥 𝑑𝑥
∞
𝓛{1} = 𝐹(𝑠) = ∫0 𝑒 −𝑠𝑥 𝑑𝑥
1
∞
𝑠
0
𝓛{1} = 𝐹 (𝑠) = |− 𝑒 −𝑠𝑥 |
Since 𝑒 −∞ = 0, therefore,
1
𝓛{1} = 𝐹 (𝑠) = (0) − [− 𝑠 𝑒 −𝑠(0) ]
𝓛{1} = 𝐹 (𝑠) =
1
𝑠
Example 2:
Find the Laplace transform of the function 𝑓(𝑥 ) = 𝑥.
Solution:
By the Laplace integral,
∞
𝓛{𝑓(𝑥)} = 𝐹(𝑠) = ∫0 𝑓(𝑥) 𝑒 −𝑠𝑥 𝑑𝑥
∞
𝓛{𝑥 } = 𝐹(𝑠) = ∫0 𝑥 𝑒 −𝑠𝑥 𝑑𝑥
For ∫ 𝑥 𝑒 −𝑠𝑥 𝑑𝑥, use integration by parts,
let 𝑢 = 𝑥 and 𝑑𝑣 = 𝑒 −𝑠𝑥 𝑑𝑥, then
1
𝑑𝑢 = 𝑑𝑥 and 𝑣 = − 𝑠 𝑒 −𝑠𝑥
1
1
∫ 𝑥 𝑒 −𝑠𝑥 𝑑𝑥 = − 𝑠 𝑥𝑒 −𝑠𝑥 − (− 𝑠 ) ∫ 𝑒 −𝑠𝑥 𝑑𝑥
1
1
= − 𝑠 𝑥𝑒 −𝑠𝑥 − 𝑠 2 𝑒 −𝑠𝑥
Thus,
∞
1
∞
1
∫0 𝑥 𝑒 −𝑠𝑥 𝑑𝑥 = |− 𝑠 𝑥𝑒 −𝑠𝑥 − 𝑠 2 𝑒 −𝑠𝑥 |
0
Since 𝑒
−∞
= 0, therefore,
∞
1
1
∫0 𝑥 𝑒 −𝑠𝑥 𝑑𝑥 = (0) − [− 𝑠 (0)𝑒 −𝑠(0) − 𝑠 2 𝑒 −𝑠(0) ]
1
= 𝑠2
Therefore,
1
𝓛{𝑥 } = 𝐹(𝑠) = 2
𝑠
20
Author: Harold Jan R. Teran
CHAPTER 2
LAPLACE AND
INVERSE LAPLACE TRANSFORMS
Example 3:
Find the Laplace transform of the function 𝑓(𝑥 ) = 𝑒 2𝑥 .
Solution:
By the Laplace integral,
∞
𝓛{𝑓(𝑥)} = 𝐹(𝑠) = ∫0 𝑓(𝑥) 𝑒 −𝑠𝑥 𝑑𝑥
∞
𝓛{𝑒 2𝑥 } = 𝐹(𝑠) = ∫0 𝑒 2𝑥 𝑒 −𝑠𝑥 𝑑𝑥
∞
𝓛{𝑒 2𝑥 } = 𝐹(𝑠) = ∫0 𝑒 −(𝑠−2)𝑥 𝑑𝑥
∞
1
𝓛{𝑒 2𝑥 } = 𝐹(𝑠) = |− 𝑠−2 𝑒 −(𝑠−2)𝑥 |
0
Since 𝑒 −∞ = 0, therefore,
𝓛{𝑒 2𝑥 } = 𝐹 (𝑠) = (0) − [−
1
𝑠−2
𝑒 −(𝑠−2)(0)]
1
𝓛{𝑒 2𝑥 } = 𝐹 (𝑠) = 𝑠−2
Example 4:
Find the Laplace transform of the function 𝑓(𝑥 ) = sin 3𝑥
Solution:
By the Laplace integral,
∞
𝓛{𝑓(𝑥)} = 𝐹 (𝑠) = ∫0 𝑓 (𝑥 ) 𝑒 −𝑠𝑥 𝑑𝑥
∞
𝓛{sin 3𝑥 } = 𝐹(𝑠) = ∫0 (sin 3𝑥) 𝑒 −𝑠𝑥 𝑑𝑥
For ∫ 𝑒 −𝑠𝑥 sin 3𝑥 𝑑𝑥, use integration by parts,
let 𝑢 = 𝑒 −𝑠𝑥 and 𝑑𝑣 = sin 3𝑥 𝑑𝑥, then
1
𝑑𝑢 = −𝑠𝑒 −𝑠𝑥 𝑑𝑥 and 𝑣 = − 3 cos 3𝑥
1
𝑠
∫ 𝑒 −𝑠𝑥 sin 3𝑥 𝑑𝑥 = − 3 𝑒 −𝑠𝑥 cos 3𝑥 − 3 ∫ 𝑒 −𝑠𝑥 cos 3𝑥 𝑑𝑥
For ∫ 𝑒 −𝑠𝑥 cos 3𝑥 𝑑𝑥, use integration by parts,
let 𝑢 = 𝑒 −𝑠𝑥 and 𝑑𝑣 = cos 3𝑥 𝑑𝑥, then
1
𝑑𝑢 = −𝑠𝑒 −𝑠𝑥 𝑑𝑥 and 𝑣 = sin 3𝑥
3
1
𝑠
∫ 𝑒 −𝑠𝑥 cos 3𝑥 𝑑𝑥 = 3 𝑒 −𝑠𝑥 sin 3𝑥 + 3 ∫ 𝑒 −𝑠𝑥 sin 3𝑥 𝑑𝑥
Thus,
1
𝑠 1
1
𝑠
𝑠
∫ 𝑒 −𝑠𝑥 sin 3𝑥 𝑑𝑥 = − 3 𝑒 −𝑠𝑥 cos 3𝑥 − 3 (3 𝑒 −𝑠𝑥 sin 3𝑥 + 3 ∫ 𝑒 −𝑠𝑥 sin 3𝑥 𝑑𝑥)
𝑠2
∫ 𝑒 −𝑠𝑥 sin 3𝑥 𝑑𝑥 = − 3 𝑒 −𝑠𝑥 cos 3𝑥 − 9 𝑒 −𝑠𝑥 sin 3𝑥 − 9 ∫ 𝑒 −𝑠𝑥 sin 3𝑥 𝑑𝑥
𝑠2
1
𝑠
9+𝑠2
1
𝑠
(1 + 9 ) ∫ 𝑒 −𝑠𝑥 sin 3𝑥 𝑑𝑥 = − 3 𝑒 −𝑠𝑥 cos 3𝑥 − 9 𝑒 −𝑠𝑥 sin 3𝑥
( 9 ) ∫ 𝑒 −𝑠𝑥 sin 3𝑥 𝑑𝑥 = − 3 𝑒 −𝑠𝑥 cos 3𝑥 − 9 𝑒 −𝑠𝑥 sin 3𝑥
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CHAPTER 2
3
𝑠
∫ 𝑒 −𝑠𝑥 sin 3𝑥 𝑑𝑥 = − (9+𝑠 2) 𝑒 −𝑠𝑥 cos 3𝑥 − (9+𝑠2) 𝑒 −𝑠𝑥 sin 3𝑥
Thus,
∞
3
𝑠
∫0 𝑒 −𝑠𝑥 sin 3𝑥 𝑑𝑥 = |− (9+𝑠2 ) 𝑒 −𝑠𝑥 cos 3𝑥 − (9+𝑠2 ) 𝑒 −𝑠𝑥 sin 3𝑥|
Since 𝑒 −∞ = 0, therefore,
3
∞
0
𝑠
𝓛{sin 3𝑥 } = (0) − [− (9+𝑠2 ) 𝑒 −𝑠(0) cos 3(0) − (9+𝑠2) 𝑒 −𝑠(0) sin 3(0)]
3
3
𝓛{sin 3𝑥 } = 𝐹 (𝑠) = 9+𝑠2 = 𝑠 2+9
Table of the Laplace Transforms of Common Functions
𝒇(𝒙)
𝓛{𝒇(𝒙)} = 𝑭(𝒔)
1
1
𝑠
𝑥
1
𝑠2
𝑥𝑛
𝑛!
𝑠𝑛+1
𝑒 𝑎𝑥
1
𝑠−𝑎
𝑎
sin 𝑎𝑥
𝑠2 + 𝑎2
𝑠
cos 𝑎𝑥
𝑠2 + 𝑎2
𝑎
sinh 𝑎𝑥
𝑠2 − 𝑎2
𝑠
cosh 𝑎𝑥
𝑠2 − 𝑎2
𝑥 sin 𝑎𝑥
2𝑎𝑠
(𝑠2 + 𝑎2 )2
𝑥 cos 𝑎𝑥
𝑠2 − 𝑎2
(𝑠2 + 𝑎2 )2
𝑥 𝑛 𝑒 𝑎𝑥
𝑛!
(𝑠 − 𝑎)𝑛+1
𝑒 𝑏𝑥 sin 𝑎𝑥
𝑎
(𝑠 − 𝑏)2 + 𝑎2
𝑒 𝑏𝑥 cos 𝑎𝑥
𝑠−𝑏
(𝑠 − 𝑏)2 + 𝑎2
Example 1:
Use the table to find the Laplace transform of the function 𝑓(𝑥 ) = 4𝑥 2 .
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CHAPTER 2
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Solution:
From the table,
𝑛!
𝓛{𝑥 𝑛 } = 𝑠 𝑛+1
Thus,
𝓛{4𝑥 2 } = 4 L [𝑥 2 ]
2!
8
𝓛{4𝑥 2 } = 4 [𝑠2+1 ] = 𝑠 3
Example 2:
Use the table to find the Laplace transform of the function 𝑓(𝑥 ) = 𝑒 −4𝑥 .
Solution:
From the table,
1
𝓛{𝑒 𝑎𝑥 } = 𝑠−𝑎
Thus,
1
1
𝓛{𝑒 −4𝑥 } = [𝑠−(−4)] = 𝑠+4
Example 3:
Use the table to find the Laplace transform of the function 𝑓(𝑥 ) = sin 8𝑥.
Solution:
From the table,
𝑎
𝓛{sin 𝑎𝑥 } = 𝑠 2 +𝑎2
Thus,
8
8
𝓛{sin 8𝑥 } = 𝑠 2+(8)2 = 𝑠 2+64
Example 4:
Use the table to find the Laplace transform of the function 𝑓(𝑥 ) = 𝑥 4 𝑒 −3𝑥 .
Solution:
From the table,
𝑛!
𝓛{𝑥 𝑛 𝑒 𝑎𝑥 } = (𝑠−𝑎)𝑛+1
Thus,
4!
24
𝓛{𝑥 4 𝑒 −3𝑥 } = [𝑠−(−3)]4+1 = (𝑠+3)5
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Author: Harold Jan R. Terano, ECE, MET
LAPLACE AND
INVERSE LAPLACE TRANSFORMS
CHAPTER 2
Exercise 2.1
I. Find the Laplace transforms of the following functions by using the Laplace integral.
1. 𝑓(𝑥 ) = cos 5𝑥
2. 𝑓(𝑥 ) = 𝑒 3𝑥 sin 3𝑥
3. 𝑓(𝑥 ) = 2𝑥 2 𝑒 3𝑥
4. 𝑓(𝑥 ) = cosh 3𝑥
5. 𝑓(𝑥 ) = 𝑒 −2𝑥 cos 3𝑥
II. Find the Laplace transforms of the following functions by using the table of the
Laplace transforms.
1. 𝑓 (𝑥 ) = 𝑥 2 𝑒 −2𝑥
2. 𝑓 (𝑥 ) = (𝑥 + 2)2
3. 𝑓 (𝑥 ) = 𝑥 (3cos 5𝑥 − 5 sin 3𝑥)
4. 𝑓 (𝑥 ) = sin 𝑥 cos 𝑥
5. 𝑓 (𝑥 ) = 𝑒 −3𝑥 (4 sin 4𝑥 − cos 2𝑥)
2.2
Inverse Laplace Transforms
The inverse Laplace transforms of 𝐹(𝑠) is designated by 𝓛–1[𝐹(𝑠)], in which
𝓛{𝑓(𝑥)} = 𝐹(𝑠).
The simplest way to obtain the inverse Laplace transform is to recognize the
formulas in the table as discussed in the previous section.
Example 1:
1
Find 𝓛–1{𝑠+6}.
Solution:
From the table,
1
𝓛{𝑒 𝑎𝑥 } = 𝑠−𝑎
Thus,
1
𝓛–1{𝑠+6} = 𝑒 −6𝑥
Example 2:
1
Find 𝓛–1{𝑠 2+4𝑠+9}.
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CHAPTER 2
LAPLACE AND
INVERSE LAPLACE TRANSFORMS
Solution:
No function of this form appears in the table, thus, by manipulating the
denominator by completing the square, we have,
1
1
𝓛–1{𝑠 2+4𝑠+9} = 𝓛–1{(𝑠+2)2 +5}
1
= 𝓛–1{
2
(𝑠+2)2 +(√5 )
}
From the table,
𝑎
𝓛{𝑒 𝑏𝑥 sin 𝑎𝑥 } = (𝑠−𝑏)2 +𝑎2
And 𝑎 = √5 and 𝑏 = −2, the numerator does not contain the factor √5 , thus,
√5
multiply both the numerator and denominator by
1
𝓛–1{
2
(𝑠+2)2 +(√5 )
} = 𝓛–1{
1
= 𝓛–1 {
1
2
(𝑠+2)2 +(√5 )
1
√5
𝓛–1{
√5
√5
√5
2
(𝑠+2)2 +(√5 )
=
(
√5
(
)}
)}
√5
√5
2
(𝑠+2)2 +(√5 )
, then,
}
Therefore,
1
𝓛–1{
2
(𝑠+2)2 +(√5 )
}=
1
𝑒 −2𝑥 sin √5 𝑥
√5
Example 3:
𝑠+3
Find 𝓛–1{ 2
𝑠 +2𝑠+4
}.
Solution:
By completing the square of the denominator,
𝑠+3
𝑠+3
𝓛–1{𝑠 2+2𝑠+4} = 𝓛–1{(𝑠+1)2 +3}
Rewrite the numerator as (𝑠 + 3) = (𝑠 + 1) + 2, then,
(𝑠+1)+2
𝑠+3
𝓛–1{𝑠 2+2𝑠+4} = 𝓛–1{(𝑠+1)2 +3}
Decompose the fraction to obtain,
𝑠+3
𝑠+1
2
𝓛–1{𝑠 2+2𝑠+4} = 𝓛–1{(𝑠+1)2 +3} + 𝓛–1{(𝑠+1)2 +3}
From the table,
𝑎
𝑠−𝑏
𝓛{𝑒 𝑏𝑥 sin 𝑎𝑥 } = (𝑠−𝑏)2 +𝑎2 and 𝓛{𝑒 𝑏𝑥 cos 𝑎𝑥 } = (𝑠−𝑏)2 +𝑎2
Then,
𝓛–1{
𝑠+1
2
(𝑠+1)2 +(√3)
} = 𝑒 −𝑥 cos √3 𝑥 and
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Author: Harold Jan R. Terano, ECE, MET
LAPLACE AND
INVERSE LAPLACE TRANSFORMS
CHAPTER 2
2
1
𝓛–1{(𝑠+1)2+3} = 2 𝓛–1 {
√3
× 3}
2
(𝑠+1)2 +(√3 )
2
=
√3
2
=
√3
√3
𝓛–1 {
2
(𝑠+1)2 +(√3 )
√
}
𝑒 −𝑥 sin √3 𝑥
Therefore,
𝑠+3
2
𝓛–1{𝑠 2+2𝑠+4} = 𝑒 −𝑥 cos √3 𝑥 + 3 𝑒 −𝑥 sin √3 𝑥
√
Example 4:
(𝑠−1)
Find 𝓛–1{𝑠 2 −4 }.
Solution:
(𝑠−1)
(𝑠−1)
𝓛–1{𝑠 2 −4 } = 𝓛–1{(𝑠−2)(𝑠+2) }
By partial fraction decomposition,
(𝑠−1)
(𝑠−2)(𝑠+2)
𝐴
𝐵
= 𝑠−2 + 𝑠+2
𝑠 − 1 = 𝐴(𝑠 + 2) + 𝐵(𝑠 − 2)
1
When 𝑠 = 2, 𝐴 = 4,
3
When 𝑠 = −2, 𝐵 = 4.
Then,
(𝑠−1)
1
1
1
3
3
𝐵
𝓛–1{𝑠 2 −4 } = 4 𝓛–1{𝑠−2} + 4 𝓛–1{𝑠+2}
= 4 𝑒 2𝑥 + 4 𝑒 −2𝑥
Exercise 2.2
Find the inverse Laplace transforms of the following functions.
1
1. 𝐹 (𝑠) = 𝑠 2 +9
4
2. 𝐹 (𝑠) = 𝑠 2 −2𝑠+5
2
3. 𝐹 (𝑠) = 𝑠 2 −6𝑠+34
𝑠+4
4. 𝐹 (𝑠) = 𝑠 2 −5𝑠+6
5. 𝐹 (𝑠) =
3
𝑠(𝑠2 −8𝑠+16)
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CHAPTER 2
LAPLACE AND
INVERSE LAPLACE TRANSFORMS
2.3 Solutions to Linear Differential Equations using Laplace
Transforms
The Laplace transforms of an nth-derivative is,
𝒅𝒏 𝒚
𝓛{𝒅𝒙𝒏 } = 𝒔𝒏 𝒀(𝒔) − 𝒔𝒏−𝟏 𝒚(𝟎) − 𝒔𝒏−𝟐 𝒚′(𝟎) − ⋯ − 𝒔𝒚(𝒏−𝟐)(𝟎) − 𝒚(𝒏−𝟏) (𝟎)
where 𝑌(𝑠) is denoted as 𝓛[𝑦(𝑥)] and 𝑦(0), 𝑦 ′ (0), … are the given initial conditions.
If the initial conditions are,
𝑦(0) = 𝑐0 , 𝑦 ′ (0) = 𝑐1 , … , 𝑦 (𝑛−1) (0) = 𝑐𝑛−1
Then,
𝒅𝒏 𝒚
𝓛{𝒅𝒙𝒏 } = 𝒔𝒏 𝒀(𝒔) − 𝒄𝟎 𝒔𝒏−𝟏 − 𝒄𝟏 𝒔𝒏−𝟐 − ⋯ − 𝒄𝒏−𝟐 𝒔 − 𝒄𝒏−𝟏
For special cases where 𝑛 = 1, 𝑛 = 2 and 𝑛 = 3,
𝓛{𝑦 ′ (𝑥 )} = 𝑠𝑌(𝑠) − 𝑐0
𝓛{𝑦 ′′ (𝑥 )} = 𝑠 2 𝑌(𝑠) − 𝑐0 𝑠 − 𝑐1
𝓛{𝑦 ′′′ (𝑥 )} = 𝑠 3 𝑌(𝑠) − 𝑐0 𝑠 2 − 𝑐1 𝑠 − 𝑐2
Example 1:
Solve 𝑦 ′ − 3𝑦 = 0.
Solution:
Taking the Laplace transforms of the equation, we have,
𝓛{𝑦 ′} − 3 L {𝑦} = 𝓛{0}
[𝑠𝑌(𝑠) − 𝑐0] − 3𝑌(𝑠) = 0
𝑠𝑌(𝑠) − 3𝑌(𝑠) = 𝑐0
𝑌(𝑠)[𝑠 − 3] = 𝑐0
𝑐
0
𝑌(𝑠) = 𝑠−3
Taking the inverse Laplace transforms,
1
𝓛–1 {𝑌(𝑠)} = 𝑐0 𝓛 –1 {𝑠−3}
Then,
𝑦(𝑥 ) = 𝑐0𝑒 3𝑥 or 𝑦(𝑥 ) = 𝑐𝑒 3𝑥
Example 2:
Solve 𝑦 ′ + 3𝑦 + 𝑒 3𝑥 = 0, that satisfies the initial condition 𝑦(0) = 1.
Solution:
Taking the Laplace transforms of the equation, we have,
𝓛{𝑦 ′ } + 3 L {𝑦} + 𝓛{𝑒 3𝑥 } = 𝓛{0}
[𝑠𝑌(𝑠) − 𝑐0] + 3𝑌 (𝑠) +
1
𝑠−3
=0
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LAPLACE AND
INVERSE LAPLACE TRANSFORMS
CHAPTER 2
Since 𝑐0 = 1, then,
[𝑠𝑌(𝑠) − 1] + 3𝑌 (𝑠) +
(𝑠 + 3)𝑌 (𝑠) = 1 −
(𝑠 + 3)𝑌 (𝑠) =
1
𝑠−3
=0
1
𝑠−3
𝑠−4
𝑠−3
𝑠−4
𝑌 (𝑠) = (𝑠+3)(𝑠−3)
Taking the inverse Laplace transforms,
𝑠−4
𝓛 –1 {𝑌(𝑠)} = 𝓛–1 {(𝑠+3)(𝑠−3)}
By partial fraction decomposition,
𝑠−4
𝐴
𝐵
=
+
(𝑠+3)(𝑠−3)
𝑠+3
𝑠−3
𝑠 − 4 = 𝐴(𝑠 − 3) + 𝐵(𝑠 + 3)
7
when 𝑠 = −3, 𝐴 =
6
1
when 𝑠 = 3, 𝐵 = − 6
Taking the inverse Laplace transforms,
7
1
1
1
𝓛–1 {𝑌(𝑠)} = 6 𝓛–1 {𝑠+3} − 6 𝓛–1 {𝑠−3}
Then,
7
1
𝑦(𝑥 ) = 6 𝑒 −3𝑥 − 6 𝑒 3𝑥
Example 3:
Solve 𝑦 ′ + 𝑦 = cos 𝑥, that satisfies the initial condition 𝑦(0) = 1.
Solution:
Taking the Laplace transforms of the equation, we have,
𝓛{𝑦 ′ } + 𝓛 {𝑦} = 𝓛{cos 𝑥 }
𝑠
[𝑠𝑌(𝑠) − 𝑐0] + 𝑌 (𝑠) = 2
𝑠 +1
Since 𝑐0 = 1, then,
𝑠
[𝑠𝑌(𝑠) − 1] + 𝑌 (𝑠) = 2
𝑠 +1
𝑠
(𝑠 + 1)𝑌 (𝑠) = 1 + 2
𝑠 +1
𝑠 2 +𝑠+1
𝑌 (𝑠) = (𝑠+1)(𝑠2 +1)
Taking the inverse Laplace transforms,
𝑠 2+𝑠+1
𝓛–1 {𝑌(𝑠)} = 𝓛–1 {(𝑠+1)(𝑠2 +1)}
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CHAPTER 2
LAPLACE AND
INVERSE LAPLACE TRANSFORMS
By partial fraction decomposition,
𝑠 2 +𝑠+1
𝐴
𝐵𝑠+𝐶
= 𝑠+1 + 𝑠 2+1
(𝑠+1)(𝑠2 +1)
𝑠 2 + 𝑠 + 1 = 𝐴(𝑠 2 + 1) + 𝐵𝑠(𝑠 + 1) + 𝐶 (𝑠 + 1)
𝑠 2 + 𝑠 + 1 = 𝐴(𝑠 2 + 1) + 𝐵(𝑠 2 + 𝑠) + 𝐶 (𝑠 + 1)
when 𝑠 = −1, 𝐴 =
1
2
coefficients of 𝑠 2 :
1 = 𝐴(1) + 𝐵(1) + 𝐶 (0)
1
1=2+𝐵
1
𝐵=2
coefficients of 𝑠:
1 = 𝐴(0) + 𝐵(1) + 𝐶 (1)
1
1=2+𝐶
𝐶=
1
2
Thus,
1
1
1
𝑠
1
1
𝓛–1 {𝑌(𝑠)} = 2 𝓛–1 {𝑠+1} + 2 𝓛–1 {𝑠 2 +1} + 2 𝓛–1 {𝑠 2 +1}
1
1
1
𝑦(𝑥 ) = 2 𝑒 −𝑥 + 2 cos 𝑥 + 2 sin 𝑥
Exercise 2.3
Solve the following linear differential equations using Laplace transforms.
1. 𝑦 ′ + 3𝑦 = 𝑥
2. 𝑦 ′′ − 4𝑦 ′ + 4𝑦 = 0
3. 𝑦 ′ − 5𝑦 − 𝑒 2𝑥 = 0
4. 𝑦 ′ + 2𝑦 = cos 𝑥 , 𝑦(0) = 1
5. 𝑦 ′′′ − 𝑦 ′′ − 4𝑦 ′ + 4𝑦 = 3𝑒 3𝑥 , 𝑦(0) = 𝑦 ′ (0) = 𝑦 ′′ (0) = 0
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LAPLACE AND
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CHAPTER 2
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Author: Harold Jan R. Teran
CHAPTER 3
INFINITE SERIES
Chapter 3
INFINITE SERIES
Chapter Outline:
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
3.10
Sequences and Series
Limit of the Sequence
Infinite Series
Convergent Series and Divergent Series
nth Term Test for Divergence
Comparison Test
Integral Test
Root Test
Ratio Test
Alternating Series
Learning Objectives:
At the end of the lesson, the students will be able to:
1. Differentiate between sequences and series.
2. Evaluate limit of a sequence.
3. Test series for convergence and divergence.
Overview:
This chapter will introduce the differences between sequence and series. This
includes limit of a sequence and infinite series. Also, this chapter will cover various
methods to test the convergence or divergence of series.
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INFINITE SERIES
CHAPTER 3
3.1 Sequence and Series
A sequence is an ordered list. It contains member, also known as elements or
terms.
Given the sequence,
𝒂𝟏 , 𝒂𝟐 , 𝒂𝟑 , … , 𝒂𝒏
where 𝑎1 , 𝑎2 , 𝑎3 , … , 𝑎𝑛 are the terms of the sequence.
The sum of the sequence is given by,
∑𝒏𝒏=𝟏 𝒂𝒏 = 𝒂𝟏 + 𝒂𝟐 + 𝒂𝟑 + ⋯ + 𝒂𝒏
this is known as a series.
3.2 Limit of a Sequence
The limit of a sequence {𝑎𝑛 } is given by
𝐥𝐢𝐦 𝒂𝒏 = 𝑳
𝒏→∞
If the limit 𝐿 of the sequence exists, then the sequence converges to 𝐿. If the
limit of the sequence does not exist, then the sequence diverges.
Example 1:
Find the limit of the sequence 𝑎𝑛 =
𝑛+1
𝑛
Solution:
𝑛+1
1
1+
1
lim ( 𝑛 × 𝑛1 ) = lim ( 1 𝑛)
𝑛→∞
𝑛→∞
𝑛
1
Since, lim (𝑛) = 0, therefore,
𝑛→∞
𝑛+1
lim ( 𝑛 ) = 1
𝑛→∞
𝑛+1
∞
Or, since, lim ( 𝑛 ) = ∞, the limit is indeterminate form, therefore, by L’hopitals
𝑛→∞
rule,
𝑛+1
1
lim ( 𝑛 ) = lim (1) = 1
𝑛→∞
𝑛→∞
Example 2:
𝑛
Find the limit of the sequence 𝑎𝑛 = 𝑛2−2.
Solution:
𝑛
1
𝑛
1
𝑛
lim (𝑛2 −2 × ) = lim (
𝑛→∞
𝑛→∞
1
2
𝑛
𝑛−
)=0
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CHAPTER 3
INFINITE SERIES
Or, since the limit is indeterminate, therefore, apply the L’hopital’s rule.
𝑛
1
1
lim (𝑛2 −2) = lim (2𝑛) = ∞ = 0
𝑛→∞
𝑛→∞
Example 3:
𝑛
Determine whether the sequence 𝑎𝑛 = 2𝑛+1 converges or diverges.
Solution:
1
𝑛
lim (2𝑛+1 × 𝑛1 ) = lim (
𝑛→∞
𝑛
𝑛
1
1
𝑛
𝑛→∞
2+
1
1
)
1
lim (2𝑛+1) = 2
𝑛→∞
Or by L’hopital’s rule,
lim (
𝑛
𝑛→∞ 2𝑛+1
) = lim (2) = 2
𝑛→∞
1
The sequence converges to 2.
Exercise 3.1
Find the limit of the given sequences and determine whether it converges or diverges.
1
1. 𝑎𝑛 = 3𝑛
1
2. 𝑎𝑛 = 𝑛4
1
3. 𝑎𝑛 = 𝑛𝑛
𝑛+1 𝑛
4. 𝑎𝑛 = (
𝑛
)
2
5. 𝑎𝑛 = 𝑒𝑛
𝑛3
6. 𝑎𝑛 = 𝑛2 +1
2−𝑛2
7. 𝑎𝑛 = 1+𝑛2
ln 𝑛
8. 𝑎𝑛 = 𝑛2
9. 𝑎𝑛 = Arctan 𝑛
𝑛
10. 𝑎𝑛 = 22
3.3 Infinite Series
Given a sequence {𝑎𝑛 }, the infinite series is given by the infinite summation,
∑∞
𝒏=𝟏 𝒂𝒏 = 𝒂𝟏 + 𝒂𝟐 + 𝒂𝟑 + ⋯ + 𝒂𝒏 + ⋯
this is infinite series or simply a series.
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INFINITE SERIES
CHAPTER 3
3.4 Convergent Series and Divergent Series
For the infinite series ∑∞
𝑛=1 𝑎𝑛 , the 𝑛th partial sum is
𝑺𝒏 = 𝒂𝟏 + 𝒂𝟐 + 𝒂𝟑 + ⋯ + 𝒂𝒏
If the partial sum of the sequence {𝑆𝑛 } converges to 𝑆, then the series ∑∞
𝑛=1 𝑎𝑛
converges.
If {𝑆𝑛 } diverges, then the series diverges.
I. Geometric Series
Geometric series has the form,
𝑛
2
∑∞
𝑛=0 𝑎𝑟 = 𝑎 + 𝑎𝑟 + 𝑎𝑟 + ⋯
where 𝑎 and 𝑟 are nonzero integer. The first term is 𝑎 and each succeding term is
obtained by multiplying the preceeding term by 𝑟.
Let us find for 𝑆𝑛 .
𝑆𝑛 = 𝑎 + 𝑎𝑟 + 𝑎𝑟 2 +. . +𝑎𝑟 𝑛−1
Multiply both sides by 𝑟,
𝑟𝑆𝑛 = 𝑎𝑟 + 𝑎𝑟 2 + 𝑎𝑟 3 + ⋯ + 𝑎𝑟 𝑛
Subtract 𝑟𝑆𝑛 from 𝑆𝑛 ,
𝑆𝑛 − 𝑟𝑆𝑛 = 𝑎 + 𝑎𝑟 + 𝑎𝑟 2 +. . +𝑎𝑟 𝑛−1 − ( 𝑎𝑟 + 𝑎𝑟 2 + 𝑎𝑟 3 + ⋯ + 𝑎𝑟 𝑛−1 + 𝑎𝑟 𝑛 )
𝑆𝑛 − 𝑟𝑆𝑛 = 𝑎 − 𝑎𝑟 𝑛
𝑆𝑛 (1 − 𝑟) = 𝑎(1 − 𝑟 𝑛 )
𝑆𝑛 =
𝑎(1−𝑟𝑛 )
1−𝑟
This is the sum of a geometric progression. For
infinity.
For infinite geometric progression,
𝑆 = lim 𝑆𝑛
𝑛→∞
𝑆 = lim [
𝑎(1−𝑟𝑛 )
1−𝑟
𝑛→∞
]
For 0 < 𝑟 < 1,
lim 𝑟 𝑛 = 0
𝑛→∞
Therefore,
𝑆 = lim [
𝑛→∞
𝑎(1−𝑟𝑛 )
1−𝑟
]=
𝑎
1−𝑟
𝑎
𝑆 = 1−𝑟
34
Author: Harold Jan R. Teran
𝑟 < 1, the sum diverges to
CHAPTER 3
INFINITE SERIES
This is the sum of infinite geometric progression. The sum converges to 𝑆.
Thus,
𝑛
∑∞
𝑛=0 𝑎𝑟 =
𝑎
for 0 < 𝑟 < 1 and diverges for |𝑟| ≥ 1.
1−𝑟
Example 1:
1 𝑛
Show that ∑∞
𝑛=0 (2)
is a geometric series and find its sum.
Solution:
1 𝑛
1
1
∑∞
𝑛=0 (2) = 1 + 2 + 4 + ⋯
1
This is a geometric series with the first term 𝑎 = 1 and 𝑟 = 2. Therefore, the
sum is,
1 𝑛
1
2
1−
∑∞
𝑛=0 ( ) =
1
2
=2
Example 2:
2 𝑛
Show that ∑∞
𝑛=0 2 (3)
geometric series and find its sum.
Solution:
2 𝑛
4
8
∑∞
𝑛=0 2 (3) = 2 + 3 + 9 + ⋯
2
The series is geometric with the first term 𝑎 = 2 and 𝑟 = 3. The sum is,
2 𝑛
∑∞
𝑛=0 2 (3) =
2
2
3
1−
=6
II. Harmonic Series
Harmonic series is given by,
𝟏
𝟏
𝟏
∑∞
𝒏=𝟏 𝒏 = 𝟏 + 𝟐 + 𝟑 + ⋯
Harmonic series diverges to infinity.
III. The p-Series
The p-series is given by,
𝟏
∑∞
𝒏=𝟏 𝒑
𝒏
The series converges if 𝑝 > 1 and diverges if 𝑝 ≤ 1.
Examples:
1
∑∞
𝑛=1 2 is a p-series with 𝑝 = 2, therefore, the series converges.
𝑛
∑∞
𝑛=1
1
√𝑛
1
is a p-series with 𝑝 = 2, therefore, the series diverges.
35
Author: Harold Jan R. Terano, ECE, MET
INFINITE SERIES
∑∞
𝑛=1
1
𝑛
CHAPTER 3
is a p-series with 𝑝 = 1, therefore, the series diverges. Also, it is a
harmonic series.
Exercise 3.2
Determine whether the given series converges or diverges. If converge, find its sum.
𝑛
1. ∑∞
𝑛=1 3
1
2. ∑∞
𝑛=1 4𝑛
1
3. ∑∞
𝑛=1 𝑛4
4. ∑∞
𝑛=1 3
1
√𝑛
5
5. ∑∞
𝑛=1 𝑛
1
6. ∑∞
𝑛=1 1
𝑛6
2 𝑛
7. ∑∞
𝑛=1 2 (− 3)
5
8. ∑∞
𝑛=1 4𝑛
1
9. ∑∞
𝑛=1 2𝑛
−𝑛
10. ∑∞
𝑛=1 5
3.5 𝒏th Term Test for Divergence
If 𝐥𝐢𝐦 𝒂𝒏 ≠ 𝟎, then the series ∑∞
𝒏=𝟏 𝒂𝒏 diverges.
𝒏→∞
Example 1:
𝑛
Test the series ∑∞
𝑛=1 4 .
Solution:
lim 4𝑛 = ∞, the series diverges.
𝑛→∞
Example 2:
𝑛
Test the series ∑∞
𝑛=1 3𝑛+1.
Solution:
lim
𝑛
𝑛→∞ 3𝑛+1
𝑛
= lim 3𝑛+1 ×
𝑛→∞
1
𝑛
1
𝑛
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CHAPTER 3
INFINITE SERIES
lim
1
1
1
𝑛→∞ 3+𝑛
= 3, the series diverges
Example 3:
1
Test the series ∑∞
𝑛=1 𝑛2 .
Solution:
lim
1
𝑛→∞ 𝑛2
=0
Since the limit is 0, the nth term test for divergence does not apply and
you
cannot draw any conclusion about convergence or divergence.
Looking at the series, it is a p-series in which 𝑝 = 2, therefore it is a convergent
series.
Exercise 3.3
Use the nth term test for divergence to test the given series.
1. ∑∞
𝑛=1
2. ∑∞
𝑛=1
𝑛
𝑛−1
(𝑛+1)2
𝑛
1
3. ∑∞
𝑛=1 𝑛2 +𝑛−2
−𝑛
4. ∑∞
𝑛=1 𝑒
𝑛
5. ∑∞
𝑛=1 𝑛+1
√
3.6 Comparison Test
Consider the series,
∑∞
𝑛=1 𝑢𝑛 = 𝑢1 + 𝑢2 + 𝑢3 + ⋯ + 𝑢𝑛 + ⋯
be a series to be tested.
If a series
∑∞
𝒏=𝟏 𝒂𝒏 = 𝒂𝟏 + 𝒂𝟐 + 𝒂𝟑 + ⋯ + 𝒂𝟒 + ⋯
is a known convergent series, then,
𝒖𝒏 ≤ 𝒂𝒏
∞
the series ∑𝒏=𝟏 𝒖𝒏 is convergent.
If a series
∑∞
𝒏=𝟏 𝒂𝒏 = 𝒂𝟏 + 𝒂𝟐 + 𝒂𝟑 + ⋯ + 𝒂𝟒 + ⋯
is a known divergent series, then,
𝒖𝒏 ≥ 𝒂𝒏
∞
the series ∑𝒏=𝟏 𝒖𝒏 is divergent.
37
Author: Harold Jan R. Terano, ECE, MET
INFINITE SERIES
CHAPTER 3
Example 1:
1
Test the series ∑∞
𝑛=1 𝑛2 +𝑛−1.
Solution:
1
Let 𝑢𝑛 = 𝑛2 +𝑛−1, the series to be tested.
1
Let 𝑎𝑛 =
𝑛2
, a known convergent series (p-series with 𝑝 = 2).
Then,
𝑢𝑛 ≤ 𝑎𝑛
1
1
≤ 𝑛2
𝑛2 +𝑛−1
Therefore, the series is convergent.
Example 2:
1
Test the series ∑∞
𝑛=1 1+2𝑛 .
Solution:
1
Let 𝑢𝑛 = 1+2𝑛, the series to be tested.
1
Let 𝑎𝑛 = 2𝑛, a known convergent series (a geometric series).
Then,
𝑢𝑛 ≤ 𝑎𝑛
1
1
1+2𝑛
≤ 2𝑛
Therefore, the series is convergent.
Example 3:
1
Test the series ∑∞
𝑛=1 3+ 𝑛 .
√
Solution:
1
Let 𝑢𝑛 =
3+√𝑛
, the series to be tested.
1
Let 𝑎𝑛 =
1
, a known divergent series (a p-series with 𝑝 = 2).
𝑛
√
Then,
𝑢𝑛 ≥ 𝑎𝑛
1
3+√𝑛
≤
1
√𝑛
which does not satisfy the comparison test for divergence. Use another comparison, let
1
𝑎𝑛 = , a known divergent series, thus,
𝑛
𝑢𝑛 ≥ 𝑎𝑛
1
3+√𝑛
1
≥ 𝑛 for 𝑛 ≥ 6
The series is divergent.
38
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CHAPTER 3
INFINITE SERIES
Exercise 3.4
Use comparison test to determine whether the given series converge or diverge.
1
1. ∑∞
𝑛=1 𝑛3 +1
1
2. ∑∞
𝑛=1 1+4𝑛
1
3. ∑∞
𝑛=1 2𝑛+1
√
1
4. ∑∞
𝑛=1 2𝑛2+3𝑛−1
2+𝑛
5. ∑∞
𝑛=1 𝑛2
3.7 Integral Test
A series ∑∞
𝟏 𝒂𝒏 is given with 𝒂𝒏 ≥ 𝟎. If 𝒇 is positive, continuous and
decreasing for 𝒙 ≥ 𝟎 and 𝒂𝒏 = 𝒇(𝒙), then,
∞
∑∞
𝟏 𝒂𝒏 and ∫𝟏 𝒇(𝒙) 𝒅𝒙
both converge or both diverge.
Example 1:
𝑛
Test the series ∑∞
𝑛=1 𝑛2 +2.
Solution:
The function is positive and continuous for 𝑥 ≥ 1. To determine if
𝑓 is
decreasing, find the derivative.
𝑓 ′ (𝑥 ) =
(𝑥 2 +2)(1)−𝑥(2𝑥)
(𝑥 2 +2)2
−𝑥 2 +2
= (𝑥2 +2)2
So, 𝑓′(𝑥) < 0 for 𝑥 > 1, therefore, it satisfies the condition for integral test.
Then,
𝑡
𝑡 𝑥
lim ∫ 𝑓 (𝑥 ) 𝑑𝑥 = lim ∫1 𝑥2 +2 𝑑𝑥
𝑡→∞ 1
𝑡→∞
1
= lim [2 ln(𝑥 2 + 2)]
𝑡→∞
𝑡
1
1
= 2 lim [ln(𝑡 2 + 2) − ln(3)]
𝑡→∞
𝑡
lim ∫1 𝑓 (𝑥 ) 𝑑𝑥 = ∞
𝑡→∞
The series diverges.
39
Author: Harold Jan R. Terano, ECE, MET
INFINITE SERIES
CHAPTER 3
Example 2:
1
Test the series ∑∞
𝑛=1 𝑛4.
Solution:
The function is positive and continuous for 𝑥 ≥ 1. To determine if
𝑓 is
decreasing, find the derivative.
4
𝑓 ′ (𝑥 ) = − 𝑛 5
So, 𝑓′(𝑥) < 0 for 𝑥 > 1, therefore, it satisfies the condition for integral test.
Then,
𝑡
𝑡 1
lim ∫1 𝑓 (𝑥 ) 𝑑𝑥 = lim ∫1 𝑥4 𝑑𝑥
𝑡→∞
𝑡→∞
= lim [−
𝑡→∞
= lim [−
𝑡→∞
1
3𝑥 3
]
1
3𝑥 3
𝑡
𝑡
1
1
+ ]
3
1
lim ∫1 𝑓 (𝑥 ) 𝑑𝑥 = 3
𝑡→∞
The series converges.
Example 3:
2
Test the series ∑∞
𝑛=1 1+𝑛2.
Solution:
The function is positive and continuous for 𝑥 ≥ 1. To determine if
𝑓 is
decreasing, find the derivative.
4𝑛
𝑓 ′ (𝑥 ) = − (1+𝑛2 )2
So, 𝑓′(𝑥) < 0 for 𝑥 > 1, therefore, it satisfies the condition for integral test.
Then,
𝑡
𝑡 2
lim ∫ 𝑓 (𝑥 ) 𝑑𝑥 = lim ∫1 1+𝑥2 𝑑𝑥
𝑡→∞ 1
𝑡→∞
= lim[2 Arctan 𝑥 ]1𝑡
𝑡→∞
= 2 lim (Arctan 𝑡 − Arctan 1)
𝑡→∞
𝜋
𝜋
= 2 (2 − 4 )
𝑡
𝜋
lim ∫1 𝑓 (𝑥 ) 𝑑𝑥 = 2
𝑡→∞
The series converges.
40
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CHAPTER 3
INFINITE SERIES
Exercise 3.5
Use integral test to determine whether the given series converge or diverge.
1
1. ∑∞
𝑛=1 (𝑛+1)3
1
2. ∑∞
𝑛=2 𝑛 ln 𝑛
𝑛
3. ∑∞
𝑛=1 √𝑛2
+1
1
4. ∑∞
𝑛=1 1+ 𝑛
√
−𝑛
5. ∑∞
𝑛=1 𝑛𝑒
3.8 Root Test
Given the series ∑∞
𝒏=𝟏 𝒂𝒏 , then,
𝐥𝐢𝐦 𝒏√𝒂𝒏 = 𝑳
𝒏→∞
If 𝑳 < 𝟏, the series converges.
If 𝑳 > 𝟏 (or ∞), the series diverges.
If 𝑳 = 𝟏, the test fails.
Example 1:
Test the series ∑∞
𝑛=1 (
𝑛2+1
3𝑛−2
𝑛
) .
Solution:
𝑛
𝑛2 +1 𝑛
lim 𝑛√𝑎𝑛 = lim √(3𝑛−2)
𝑛→∞
𝑛→∞
𝑛2 +1
= lim 3𝑛−2
𝑛→∞
𝑛2 +1
1
𝑛
1
𝑛
= lim (3𝑛−2 × )
𝑛→∞
1
𝑛
2
𝑛→∞ 3−𝑛
= lim
𝑛+
∞
lim 𝑛√𝑎𝑛 = 3 = ∞ > 1, the series diverges.
𝑛→∞
Example 2:
4𝑛
Test the series ∑∞
𝑛=1 𝑛 .
41
Author: Harold Jan R. Terano, ECE, MET
INFINITE SERIES
CHAPTER 3
Solution:
𝑛
4𝑛
lim 𝑛√𝑎𝑛 = lim √ 𝑛
𝑛→∞
𝑛→∞
= lim
4
4
= 𝑛0
1
𝑛→∞ 𝑛𝑛
lim 𝑛√𝑎𝑛 = 4 > 1, the series diverges.
𝑛→∞
Example 3:
2𝑛−𝑛3
𝑛
Test the series ∑∞
𝑛=1 ( 1+4𝑛3 ) .
Solution:
𝑛
2𝑛−𝑛3 𝑛
lim 𝑛√𝑎𝑛 = lim √(1+4𝑛3 )
𝑛→∞
𝑛→∞
2𝑛−𝑛3
1
3
= lim (1+4𝑛3 × 𝑛1 )
𝑛→∞
𝑛3
2
2 −1
= lim (𝑛1
𝑛→∞
𝑛3
+4
)
1
lim 𝑛√𝑎𝑛 = − < 1, the series converges.
4
𝑛→∞
Exercise 3.6
Use root test to determine whether the given series converge or diverge.
𝑛
1. ∑∞
𝑛=1 4𝑛
2. ∑∞
𝑛=1
𝑛
√𝑛2 +1
𝑛𝑛
3. ∑∞
𝑛=1 23𝑛+1
4𝑛−𝑛3
𝑛
4. ∑∞
𝑛=1 ( 5𝑛3 +3 )
5. ∑∞
𝑛=1
(−3)𝑛
𝑛
3.9 Ratio Test
Given the series ∑∞
𝒏=𝟏 𝒂𝒏 , then,
𝒂
𝐥𝐢𝐦 𝒏+𝟏 = 𝑳
𝒏→∞ 𝒂𝒏
If 𝑳 < 1, the series converges.
If 𝑳 > 1 (or ∞), the series diverges.
If 𝑳 = 𝟏, the test fails.
42
Author: Harold Jan R. Teran
CHAPTER 3
INFINITE SERIES
Example 1:
2𝑛
Test the series ∑∞
𝑛=1 𝑛 .
Solution:
𝑎𝑛+1
lim ( 𝑎 ) = lim (
𝑛→∞
𝑛→∞
𝑛
2𝑛+1
𝑛+1
2𝑛
𝑛
)
2𝑛+1
𝑛
2𝑛 ∙21
𝑛
= lim ( 𝑛+1 × 2𝑛 )
𝑛→∞
= lim ( 𝑛+1 × 2𝑛 )
𝑛→∞
2𝑛
= lim (𝑛+1)
𝑛→∞
1
𝑛
1
𝑛
2𝑛
= lim (𝑛+1 × )
𝑛→∞
= lim (
2
1
𝑛
1+
𝑛→∞
)
𝑎
lim ( 𝑎𝑛+1 ) = 2 > 1, the series diverges.
𝑛→∞
𝑛
Example 2:
𝑛𝑛
Test the series ∑∞
𝑛=1 𝑛! .
Solution:
𝑎𝑛+1
lim (
𝑛→∞
𝑎𝑛
) = lim [
𝑛→∞
(𝑛+1)𝑛+1
(𝑛+1)!
𝑛𝑛
]
𝑛!
(𝑛+1)𝑛+1
𝑛!
= lim [ (𝑛+1)! × 𝑛𝑛 ]
𝑛→∞
= lim [
(𝑛+1)𝑛+1
𝑛𝑛
𝑛→∞
𝑛!
Since, (𝑛+1)! =
𝑛!
(𝑛+1)(𝑛!)
= lim [
=
, then,
𝑛+1
(𝑛+1)𝑛 (𝑛+1)
𝑛𝑛
𝑛→∞
= lim [
1
𝑛!
× (𝑛+1)!]
(𝑛+1)𝑛
𝑛→∞
𝑛𝑛
1
× 𝑛+1]
]
𝑛+1 𝑛
= lim ( 𝑛 )
𝑛→∞
= 1∞ (indeterminate form)
𝑛+1 𝑛
𝑛+1 𝑛
𝑛+1
For lim ( 𝑛 ) , let 𝑦 = ( 𝑛 ) ; ln 𝑦 = 𝑛 ln ( 𝑛 ), thus,
𝑛→∞
𝑛+1
lim ln 𝑦 = lim 𝑛 ln ( 𝑛 )
𝑛→∞
𝑛→∞
43
Author: Harold Jan R. Terano, ECE, MET
INFINITE SERIES
CHAPTER 3
𝑛+1
)
𝑛
1
𝑛
ln(
= lim
𝑛→∞
∞
= ∞(indeterminate form)
By L’hopital’s Rule,
lim
𝑛+1
)
𝑛
1
𝑛
ln(
𝑛→∞
𝑛+1
)]
𝑛
1
𝑑( )
𝑛
𝑑[ln(
= lim
𝑛→∞
= lim
𝑛→∞
1
𝑛(𝑛+1)
1
− 2
𝑛
−
𝑛
= lim (𝑛+1)
𝑛→∞
=1
Since ln 𝑦 = 1, then, 𝑦 = 𝑒 1 = 𝑒, therefore,
𝑛+1 𝑛
lim ( 𝑛 ) = 𝑒
𝑛→∞
Thus,
𝑎
lim ( 𝑎𝑛+1 ) = 𝑒 > 1, the series diverges.
𝑛→∞
𝑛
Example 3:
1
Test the series ∑∞
𝑛=1 𝑛5.
Solution:
𝑎𝑛+1
lim ( 𝑎 ) = lim [
𝑛→∞
𝑛
𝑛→∞
1
(𝑛+1)5
1
𝑛5
]
𝑛5
= lim [(𝑛+1)5 ]
𝑛→∞
By L’hopital’s rule,
𝑛5
lim [(𝑛+1)5 ] = 1
𝑛→∞
Therefore, the test fails.
Exercise 3.7
Use ratio test to determine whether the given series converge or diverge.
𝑛3
1. ∑∞
𝑛=1 𝑛!
3𝑛
2. ∑∞
𝑛=1 𝑛!
−7
3. ∑∞
𝑛=1 𝑛
𝑛6
4. ∑∞
𝑛=1 𝑒 𝑛
2𝑛+1
5. ∑∞
𝑛=1 4𝑛
44
Author: Harold Jan R. Teran
CHAPTER 3
INFINITE SERIES
3.10 Alternating Series
An alternating series is a series whose terms alternate in sign. This series is
given by,
𝑛−1
∑∞
𝑎𝑛 = 𝑎1 − 𝑎2 + 𝑎3 − 𝑎4 + ⋯
𝑛=1(−1)
𝑛
or ∑∞
𝑛=1(−1) 𝑎𝑛 if the first term happens to be negative.
Alternating Series Test
Given {𝒂𝒏 } a nonincreasing sequence of positive term. If,
𝐥𝐢𝐦 𝒂𝒏 = 𝟎 and
𝒏→∞
𝒂𝒏+𝟏 ≤ 𝒂𝒏 for all 𝒏
𝒏−𝟏
are satisfied, then the alternating series ∑∞
𝒂𝒏 converges.
𝒏=𝟏(−𝟏)
∞
If ∑∞
𝒏=𝟏|𝒂𝒏 | converges, then ∑𝒏=𝟏 𝒂𝒏 is absolutely convergent.
If ∑∞
converges but
𝒏=𝟏 𝒂𝒏
conditionally convergent.
∑∞
𝒏=𝟏|𝒂𝒏 |
diverges, then
∑∞
𝒏=𝟏 𝒂𝒏
is
Example 1:
Test the series ∑∞
𝑛=1
(−1)𝑛−1
𝑛
.
Solution:
1
lim 𝑎𝑛 = lim 𝑛 = 0
𝑛→∞
𝑛→∞
The first condition is satisfied.
1
1
𝑎𝑛+1 = 𝑛+1 ≤ 𝑛 = 𝑎𝑛 , for all 𝑛.
Therefore, the series converges.
Since,
1
∑∞
𝑛=1 | | is a harmonic series which diverges,
𝑛
Then, the series is conditionally convergent.
Example 2:
Test the series ∑∞
𝑛=1
(−1)𝑛−1
𝑛2
.
Solution:
1
lim 𝑎𝑛 = lim 𝑛2 = 0
𝑛→∞
𝑛→∞
The first condition is satisfied. For the second condition,
1
1
𝑎𝑛+1 = (𝑛+1)2 ≤ 𝑛2 = 𝑎𝑛 , for all 𝑛.
Therefore, the series converges.
45
Author: Harold Jan R. Terano, ECE, MET
INFINITE SERIES
CHAPTER 3
Since,
1
∑∞
𝑛=1 | 2 | is a p-series in which is convergent.
𝑛
Then, the series is absolutely convergent.
Example 3:
Test the series ∑∞
𝑛=1
(−1)𝑛−1𝑛
𝑛+1
.
Solution:
For the first condition,
lim
𝑛
=1
𝑛→∞ 𝑛+1
For the second condition,
𝑛+1
𝑛+2
𝑛
≥ 𝑛+1
The series does not satisfy any of the two conditions, therefore, the series
diverges.
Example 4:
Test the series ∑∞
𝑛=1
(−1)𝑛
𝑛!
.
Solution:
1
lim 𝑎𝑛 = 𝑛! = 0
𝑛→∞
The first condition is satisfied.
1
1
≤ 𝑛!
(𝑛+1)!
The second condition is satisfied. Therefore, the series converges.
Since,
1
∑∞
𝑛=1 | | is convergent (by ratio test).
𝑛!
Then, the series is absolutely convergent.
Exercise 3.8
Determine whether the given series converge or diverge.
(−1)𝑛
1. ∑∞
𝑛=1
(2𝑛+1)
2. ∑∞
𝑛=1
(−1)𝑛−1
𝑛4
𝑛
3. ∑∞
𝑛=1 (−3)𝑛−1
(−1)𝑛
4. ∑∞
𝑛=1 𝑛2 +𝑛−1
5. ∑∞
𝑛=1
(−1)𝑛𝑛!
𝑛
46
Author: Harold Jan R. Teran
CHAPTER 4
POWER SERIES
Chapter 4
POWER SERIES
Chapter Outline:
4.1
4.2
4.3
4.4
4.5
Power Series
Radius and Interval of Convergence
Geometric Power Series
Taylor Series and Maclaurin Series
Power Series Solutions to Ordinary Differential Equations
Learning Objectives:
At the end of the lesson, the students will be able to:
1. Define power series.
2. Determine the radius and interval of convergence.
3. Obtain geometric power series of a given function.
4. Differentiate between Taylor and Maclaurin series.
5. Obtain the Taylor and Maclaurin series of a given function.
6. Solve ordinary differential equations using power series.
Overview:
This chapter will introduce the concept of power series. This includes the
determination of the radius and interval of convergence of the series. Functions can
obtain its geometric power series which will be cover in this chapter. Taylor series and
Maclaurin series are common types of series in obtaining series expansion of functions.
Also, solutions to ordinary differential equations using power series will be discussed.
47
Author: Harold Jan R. Terano, ECE, MET
POWER SERIES
CHAPTER 4
4.1 Power Series
A finite sum of the form
𝑎0 + 𝑎1 (𝑥 − 𝑥0 ) + 𝑎2 (𝑥 − 𝑥0 ) 2 + ⋯ + 𝑎𝑛 (𝑥 − 𝑥0 ) 𝑛
is a polynomial in (𝑥 − 𝑎). An infinite sum is given by,
𝑛
2
𝑛
∑∞
𝑛=0 𝑎𝑛 (𝑥 − 𝑥0 ) = 𝑎0 + 𝑎1 (𝑥 − 𝑥0 ) + 𝑎2 (𝑥 − 𝑥0 ) + ⋯ + 𝑎𝑛 (𝑥 − 𝑥0 ) + ⋯
which is called as a power series in (𝑥 − 𝑥0 ).
If 𝑥0 = 0, the power series is,
𝑛
2
∑∞
𝑛=0 𝑎𝑛 𝑥 = 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 + ⋯
4.2 Radius and Interval of Convergence
The radius of convergence 𝑟 is a nonnegative real number or ∞ such that the
series converges if
|𝑥 − 𝑥0 | < 𝑟
and diverges if
|𝑥 − 𝑥0 | > 𝑟
In the interval
(−𝑟, 𝑟), there is some positive real number
𝑟
such that it
converges absolutely for all 𝑥 and diverges for 𝑥 outside of the interval [−𝑟, 𝑟].
The number 𝑟 is called the radius of convergence of the series. Set 𝑟 = 0 if the
series converges only for 𝑥 = 0 and set 𝑟 = ∞ if it converges for all 𝑥. All the points
for which the series converges is called the interval of convergence.
Example 1:
𝑛
Find the radius and interval of convergence of the power series ∑∞
𝑛=1 𝑥 .
Solution:
By ratio test:
𝑥 𝑛+1
lim | 𝑥𝑛 | = lim |𝑥 |
𝑛→∞
𝑛→∞
The series converges for |𝑥 | < 1 and diverges for |𝑥 | > 1. Therefore, −1 < 𝑥 <
1, but at
𝑥 = −1, the series is convergent (alternating series), thus the interval of
convergence is [−1,1). The radius of convergence is 𝑟 = 1.
Example 2:
𝑥𝑛
Find the radius and interval of convergence of ∑∞
𝑛=1 2𝑛 .
48
Author: Harold Jan R. Teran
CHAPTER 4
POWER SERIES
Solution:
By ratio test:
lim |
𝑛→∞
𝑥𝑛+1
2𝑛+1
𝑥𝑛
2𝑛
𝑥 𝑛+1
2𝑛
𝑥
𝑥
𝑛→∞ 2
| = lim | 𝑛+1 × 𝑛 | = lim | |
𝑛→∞ 2
𝑥
𝑥
𝑥
The series converges at |2| < 1 and diverges at |2| > 1. Therefore, −1 < 2 < 1,
−2 < 𝑥 < 2. At 𝑥 = −2 and 𝑥 = 2, the series both diverge, thus, the interval of
convergence is (−2,2). The radius of convergence is 𝑟 = 2.
Example 3:
Find the radius and interval of convergence of ∑∞
𝑛=1
(5𝑥)𝑛
𝑛!
.
Solution:
By ratio test:
lim |
𝑛→∞
(5𝑥)𝑛+1
(𝑛+1)!
(5𝑥)𝑛
𝑛!
𝑛!
(5𝑥)𝑛+1
𝑛!
5𝑥𝑛!
| = lim | (𝑛+1)! × (5𝑥)𝑛 | = lim |(𝑛+1)!|
𝑛→∞
𝑛→∞
1
Since (𝑛+1)! = (𝑛+1),
lim |
5𝑥𝑛!
| = lim |
𝑛→∞ (𝑛+1)!
5𝑥
𝑛→∞ 𝑛+1
|=0
for any value of 𝑥, thus, 𝑟 = ∞ and the series converges for all real 𝑥. Therefore, the
interval of convergence is (−∞, ∞). The radius of convergence is 𝑟 = ∞.
Example 4:
Find the radius and interval of convergence of ∑∞
𝑛=1
(𝑥−2)𝑛
𝑛
.
Solution:
By ratio test:
lim |
𝑛→∞
(𝑥−2)𝑛+1
𝑛+1
(𝑥−2)𝑛
𝑛
| = lim |
𝑛→∞
(𝑥−2)𝑛+1
𝑛+1
(𝑥−2)𝑛
𝑛
× (𝑥−2)𝑛| = lim |
𝑛→∞
𝑛+1
| = |𝑥 − 2|
This is a power series in (𝑥 − 𝑥0 ) with 𝑥0 = 2. The series is centered at 2. The
interval of convergence is (1,3). But at 𝑥 = 1, the series is convergent and divergent at
𝑥 = 3 since it is a harmonic series. Therefore, the interval of convergence is [1,3). The
radius of convergence is 𝑟 = 1.
49
Author: Harold Jan R. Terano, ECE, MET
POWER SERIES
CHAPTER 4
Exercise 4.1
Find the radius and interval of convergence for the following series.
𝑥𝑛
1. ∑∞
𝑛=1 𝑛+1
𝑛 𝑛
2. ∑∞
𝑛=1(−1) 𝑥
3. ∑∞
𝑛=1
𝑥𝑛
𝑛
𝑛
4. ∑∞
𝑛=1 2(𝑥 − 3)
𝑥𝑛
5. ∑∞
𝑛=1 𝑛2
6. ∑∞
𝑛=1
(−1)𝑛𝑥 𝑛+1
(𝑛+1)!
𝑥𝑛
7. ∑∞
𝑛=1 𝑛!
8. ∑∞
𝑛=1
(−1)𝑛(𝑥−2)𝑛
9. ∑∞
𝑛=1
(𝑛+1)
2𝑛
𝑥𝑛
𝑛
10. ∑∞
𝑛=1 𝑛 (𝑥 + 5)
4.3 Geometric Power Series
Consider the function,
1
𝑓 (𝑥 ) = 1−𝑥
From the sum of geometric series,
𝑛
∑∞
𝑛=0 𝑎𝑟 =
𝑎
1−𝑟
, 0 < |𝑟 | < 1
When 𝑎 = 1 and 𝑟 = 𝑥, the power series representation for
1
1−𝑥
centered at 0
is,
1
1−𝑥
𝑛
= ∑∞
𝑛=0 𝑎𝑟
𝑛
= ∑∞
𝑛=0 𝑥
= 1 + 𝑥 + 𝑥 2 + 𝑥 3 + ⋯, |𝑥 | < 1
1
The series represents 𝑓(𝑥 ) = 1−𝑥 on the interval (−1,1) and 𝑓(𝑥) is not
defined for all 𝑥 ≠ 1.
Example 1:
4
Find a power series for 𝑓(𝑥 ) = 𝑥+4 centered at 0.
50
Author: Harold Jan R. Teran
CHAPTER 4
POWER SERIES
Solution:
𝑎
Writing 𝑓 (𝑥 ) in the form 1−𝑟,
4
=
𝑥+4
1
𝑥
4
1−(− )
𝑥
then, 𝑎 = 1 and 𝑟 = − 4.
The power series for 𝑓(𝑥) is,
4
𝑛
= ∑∞
𝑛=0 𝑎𝑟
𝑥+4
𝑥 𝑛
= ∑∞
𝑛=0(1) (− 4 )
𝑥2
𝑥
𝑥3
= 1 − 4 + 16 − 64 + ⋯
The power series converges when
𝑥
|− | < 1, the interval of convergence is
4
(−4,4).
Example 2:
1
Find a power series for 𝑓(𝑥 ) = 𝑥 centered at 1.
Solution:
Writing 𝑓(𝑥) in the form
1
𝑥
𝑎
1−𝑟
,
1
= 1−(−𝑥+1)
then, 𝑎 = 1 and 𝑟 = 1 − 𝑥 = −(𝑥 − 1). The power series 𝑓(𝑥) is,
1
𝑥
𝑛
= ∑∞
𝑛=0 𝑎𝑟
𝑛
= ∑∞
𝑛=0(1)[−(𝑥 − 1)]
𝑛
𝑛
= ∑∞
𝑛=0(−1) (𝑥 − 1)
= 1 − (𝑥 − 1) + (𝑥 − 1)2 − (𝑥 − 1)3 + ⋯
The power series converges when
|𝑥 − 1| < 1, therefore, the interval of
convergence is (0,2).
Example 3:
𝑛
2
Find a function that is represented by the series ∑∞
𝑛=0 𝑥 = 1 + 𝑥 + 𝑥 + ⋯.
Solution:
By ratio test:
𝑥 𝑛+1
lim | 𝑥𝑛 | = |𝑥 |
𝑛→∞
The series converges at |𝑥 | < 1. The interval of convergence is (−1,1). The
1
radius of convergence 𝑟 = 1. Since 1−𝑥, the sum is,
𝑛
∑∞
𝑛=0 𝑥 =
1
1−𝑥
𝑥 in (−1,1)
51
Author: Harold Jan R. Terano, ECE, MET
POWER SERIES
CHAPTER 4
Example 4:
𝑥𝑛
𝑥
𝑥2
Find a function that is represented by the series ∑∞
𝑛=0 2𝑛 = 1 + 2 + 4 + ⋯.
Solution:
By ratio test:
lim |
𝑛→∞
𝑥𝑛+1
2𝑛+1
𝑥𝑛
2𝑛
𝑥
|=| |
2
𝑥
The series converges at |2| < 1. The interval of convergence is (−2,2). The sum
is,
1
2
𝑥
1−( )
2
= 2−𝑥
Therefore,
𝑥𝑛
2
2
2−𝑥
∑∞
𝑛=0 𝑛 =
𝑥 in (−2,2)
Example 5:
Find a function that is represented by the series
𝑛
2
∑∞
𝑛=0(2 − 𝑥 ) = 1 + (2 − 𝑥 ) + (2 − 𝑥 ) + ⋯.
Solution:
By ratio test:
(2−𝑥)𝑛+1
lim | (2−𝑥)𝑛 | = |2 − 𝑥 |
𝑥→∞
The series converges at |2 − 𝑥 | < 1. The interval of convergence is (1,3). The
sum is,
1
1−(2−𝑥)
=
1
𝑥−1
Therefore,
𝑛
∑∞
𝑛=0(2 − 𝑥 ) =
1
𝑥−1
𝑥 in (1,3).
Exercise 4.2
I. Find a power series for the given functions.
2
1. 𝑓 (𝑥 ) = 2−𝑥2 centered at 0.
−𝑥
2. 𝑓 (𝑥 ) = 1+𝑥 centered at 0.
1
3. 𝑓 (𝑥 ) = 𝑥−3 centered at 4.
1
4. 𝑓 (𝑥 ) = 𝑥−1 centered at 2.
2
5. 𝑓 (𝑥 ) = 𝑥 centered at 2.
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Author: Harold Jan R. Teran
CHAPTER 4
POWER SERIES
II. Find a function that is represented by the given series.
𝑛+2
1. ∑∞
𝑛=0 𝑥
𝑥 𝑛+2
2. ∑∞
𝑛=0 4𝑛
𝑛 𝑛+1
3. ∑∞
𝑛=0(−1) 𝑥
𝑛
4. ∑∞
𝑛=0(3 − 𝑥 )
𝑥𝑛
5. ∑∞
𝑛=0 2𝑛+1
4.4 Taylor Series and Maclaurin Series
The Taylor series of a function 𝑓(𝑥) is given by
1
(𝑛) ( )
𝑛
𝑓 (𝑥 ) = ∑ ∞
𝑥0
𝑛=0 𝑎𝑛 (𝑥 − 𝑥0 ) where 𝑎𝑛 = 𝑛! 𝑓
Therefore,
1
(𝑛) ( ) (
𝑓 (𝑥 ) = ∑ ∞
𝑥0 𝑥 − 𝑥0 )𝑛
𝑛=0 𝑛! 𝑓
1
1
1
𝑓 (𝑥 ) = 𝑓 (𝑥0 ) + 1! 𝑓 ′ (𝑥0 )(𝑥 − 𝑥0 ) + 2! 𝑓 ′′ (𝑥0 )(𝑥 − 𝑥0 )2 + 3! 𝑓 ′′′ (𝑥0 )(𝑥 − 𝑥0 )3 + ⋯
𝒇 (𝒙 ) = 𝒇 ( 𝒙 𝟎 ) +
(𝒙 − 𝒙 𝟎 ) ′
(𝒙 − 𝒙𝟎 )𝟐 ′′
𝟏
𝒇 (𝒙 𝟎 ) +
𝒇 (𝒙𝟎 ) + 𝒇′′′(𝒙𝟎 )(𝒙 − 𝒙𝟎 )𝟑 + ⋯
𝟏!
𝟐!
𝟑!
This is called Taylor’s Formula.
If 𝑥0 = 0, then the Taylor series becomes,
𝒇 (𝒙 ) = 𝒇 (𝟎) +
𝒙 ′
𝒙𝟐 ′′
𝒙𝟑 ′′′
(
)
(
)
𝒇 𝟎 + 𝒇 𝟎 + 𝒇 (𝟎 ) + ⋯
𝟏!
𝟐!
𝟑!
This is the Maclaurin Series.
Example 1:
Find the Taylor series and Maclaurin series of the function 𝑓 (𝑥 ) = 𝑒 𝑥 .
Solution:
𝑓 (𝑥 ) = 𝑒 𝑥
𝑓 ′ (𝑥 ) = 𝑓 ′′ (𝑥 ) = 𝑓 ′′′ (𝑥 ) = ⋯ = 𝑓 (𝑛)(𝑥 ) = 𝑒 𝑥
The Taylor series is,
𝑒𝑎
𝑒𝑎
𝑓 (𝑥 ) = 𝑒 𝑎 + 𝑒 𝑎 (𝑥 − 𝑎) + 2! (𝑥 − 𝑎)2 + 3! (𝑥 − 𝑎)3 + ⋯
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Author: Harold Jan R. Terano, ECE, MET
POWER SERIES
CHAPTER 4
When 𝑥0 = 0, the Maclaurin series is,
𝑥2
𝑥3
𝑓 (𝑥 ) = 1 + 𝑥 + 2! + 3! + ⋯
In summation notation,
𝑥𝑛
𝑓 (𝑥 ) = ∑ ∞
𝑛=0 𝑛!
Example 2:
Find the Maclaurin series of the function 𝑓 (𝑥 ) = sin 𝑥.
Solution:
𝑓 (𝑥 ) = sin 𝑥
𝑓 (0) = 0
𝑓 ′ (𝑥 ) = cos 𝑥
𝑓 ′ (0) = 1
𝑓 ′′ (𝑥 ) = − sin 𝑥
𝑓 ′′ (0) = 0
𝑓 ′′′ (𝑥 ) = − cos 𝑥
𝑓 ′′′ (0) = −1
𝑓 ′′′′ (𝑥 ) = sin 𝑥
𝑓 ′′′′ (0) = 0
The Maclaurin series is,
𝑓 (𝑥 ) = 0 + 1(𝑥 ) + 0 −
𝑥3
𝑥5
1(𝑥)3
3!
+0+
1(𝑥)5
5!
+0−
1(𝑥)7
7!
+⋯
𝑥7
𝑓 (𝑥 ) = 𝑥 − 3! + 5! − 7! + ⋯
In summation notation,
(−1)𝑛
2𝑛+1
𝑓 (𝑥 ) = ∑ ∞
𝑛=0 (2𝑛+1)! 𝑥
Example 3:
1
Find the Taylor series of the function 𝑓 (𝑥 ) = 𝑥 about the point 𝑥0 = 1.
Solution:
1
𝑓 (𝑥 ) = 𝑥
𝑓 (1) = 1
1
𝑓 ′ (𝑥 ) = − 𝑥 2
2
𝑓 ′′ (𝑥 ) = 𝑥3
𝑓 (1) = −1
𝑓 (1) = 2
6
𝑓 ′′′ (𝑥 ) = − 𝑥4
24
𝑓 ′′′′ (𝑥 ) = 𝑥5
𝑓 (1) = −6
𝑓 (1) = 24
The Taylor series at 𝑥0 = 1 is,
2
6
24
𝑓 (𝑥 ) = 1 − 1(𝑥 − 1) + 2! (𝑥 − 1)2 − 3! (𝑥 − 1)3 + 4! (𝑥 − 1)4 − ⋯
𝑓 (𝑥 ) = 1 − (𝑥 − 1) + (𝑥 − 1)2 − (𝑥 − 1)3 + (𝑥 − 1)4 − ⋯
In summation notation,
𝑛
𝑛
𝑓 (𝑥 ) = ∑ ∞
𝑛=0(−1) (𝑥 − 1)
54
Author: Harold Jan R. Teran
CHAPTER 4
POWER SERIES
Exercise 4.3
I. Find the Taylor series of the given functions at the point 𝑥0 .
1. 𝑓 (𝑥 ) = ln 𝑥 at 𝑥0 = 1
2. 𝑓 (𝑥 ) = 𝑒 𝑥 at 𝑥0 = −1
3. 𝑓 (𝑥 ) = 𝑒 −𝑥 at 𝑥0 = −2
𝜋
4. 𝑓 (𝑥 ) = sin 𝑥 at 𝑥0 = 2
5. 𝑓 (𝑥 ) = 𝑥𝑒 𝑥 at 𝑥0 = 1
II. Find the Maclaurin series of the given functions.
1. 𝑓 (𝑥 ) =
1
1−𝑥
2. 𝑓 (𝑥 ) = cos 𝑥
3. 𝑓 (𝑥 ) = sin 2𝑥
4. 𝑓 (𝑥 ) = Arctan 𝑥
5. 𝑓 (𝑥 ) = 𝑥 sin 𝑥
4.5 Power Series Solutions to Ordinary Differential Equations
Given the series
𝑛
𝑦(𝑥 ) = 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 + 𝑎3 𝑥 3 + 𝑎4 𝑥 4 + ⋯ = ∑∞
𝑛=0 𝑎𝑛 𝑥
The derivatives are,
𝒏−𝟏
𝒚′ (𝒙) = 𝒂𝟏 + 𝟐𝒂𝟐 𝒙 + 𝟑𝒂𝟑 𝒙𝟐 + 𝟒𝒂𝟒 𝒙𝟑 + ⋯ = ∑∞
𝒏=𝟏 𝒏𝒂𝒏 𝒙
𝒏−𝟐
𝒚′′ (𝒙) = 𝟐𝒂𝟐 + 𝟔𝒂𝟑 𝒙 + 𝟏𝟐𝒂𝟒 𝒙𝟐 + ⋯ = ∑∞
𝒏=𝟐 𝒏(𝒏 − 𝟏)𝒂𝒏 𝒙
Example 1:
𝑑𝑦
Solve the differential equation 𝑑𝑥 = 𝑦.
Solution:
𝑑𝑦
𝑑𝑥
=𝑦
𝑦′ = 𝑦
𝑎1 + 2𝑎2 𝑥 + 3𝑎3 𝑥 2 + 4𝑎4 𝑥 3 + ⋯ = 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 + 𝑎3 𝑥 3 + 𝑎4 𝑥 4 + ⋯
Equating the coefficients of 𝑥 0 ,
𝑎1 = 𝑎0
Equating the coefficients of 𝑥 1 ,
2𝑎2 = 𝑎1
1
1
𝑎2 = 2 𝑎1 = 2 𝑎0
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Author: Harold Jan R. Terano, ECE, MET
POWER SERIES
CHAPTER 4
Equating the coefficients of 𝑥 2 ,
3𝑎3 = 𝑎2
1
1 1
1
𝑎3 = 3 𝑎2 = 3 (2 𝑎0 ) = 6 𝑎0
Equating the coefficients of 𝑥 3 ,
4𝑎4 = 𝑎3
1
1 1
1
𝑎4 = 4 𝑎3 = 4 (6 𝑎0 ) = 24 𝑎0
And so on…
Since,
𝑦(𝑥 ) = 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 + 𝑎3 𝑥 3 + 𝑎4 𝑥 4 + ⋯
1
1
1
𝑦(𝑥 ) = 𝑎0 + 𝑎0 𝑥 + 2 𝑎0 𝑥 2 + 6 𝑎0 𝑥 3 + 24 𝑎0 𝑥 4 + ⋯
1
1
1
2
6
24
1
1
1
𝑦(𝑥 ) = 𝑎0 (1 + 𝑥 + 𝑥 2 + 𝑥 3 +
𝑥4 + ⋯ )
𝑦(𝑥 ) = 𝑎0 (1 + 𝑥 + 2! 𝑥 2 + 3! 𝑥 3 + 4! 𝑥 4 + ⋯ )
And, by Maclaurin Series,
1
1
1
𝑒 𝑥 = 1 + 𝑥 + 2! 𝑥 2 + 3! 𝑥 3 + 4! 𝑥 4 + ⋯
Therefore,
𝑦(𝑥 ) = 𝑎0 𝑒 𝑥
Example 2:
Solve the differential equation 𝑦 ′′ + 𝑦 = 0.
Solution:
𝑦 ′′ + 𝑦 = 0
𝑛−2
𝑛
∑∞
+ ∑∞
𝑛=2 𝑛 (𝑛 − 1)𝑎𝑛 𝑥
𝑛=0 𝑎𝑛 𝑥 = 0
2𝑎2 + 6𝑎3 𝑥 + 12𝑎4 𝑥 2 + 20𝑎5 𝑥 3 … + 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 + 𝑎3 𝑥 3 + 𝑎4 𝑥 4 + ⋯ = 0
Equating the coefficients of 𝑥 0 ,
2𝑎2 + 𝑎0 = 0
1
𝑎2 = − 2 𝑎0
Equating the coefficients of 𝑥 1 ,
6𝑎3 + 𝑎1 = 0
1
𝑎3 = − 𝑎1
6
Equating the coefficients of 𝑥 2 ,
12𝑎4 + 𝑎2 = 0
1
1
1
1
𝑎4 = − 12 𝑎2 = − 12 (− 2 𝑎0 ) = 24 𝑎0
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Author: Harold Jan R. Teran
CHAPTER 4
POWER SERIES
Equating the coefficients of 𝑥 3 ,
20𝑎5 + 𝑎3 = 0
1
1
1
1
𝑎5 = − 20 𝑎3 = − 20 (− 6 𝑎1 ) = 120 𝑎1
𝑎0 and 𝑎1 remains arbitrary. Therefore,
𝑦(𝑥 ) = 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 + 𝑎3 𝑥 3 + 𝑎4 𝑥 4 + ⋯
1
1
1
1
1
1
1
1
𝑦(𝑥 ) = 𝑎0 + 𝑎1 𝑥 − 2 𝑎0 𝑥 2 − 6 𝑎1 𝑥 3 + 24 𝑎0 𝑥 4 + 120 𝑎1 𝑥 5 + ⋯
𝑦(𝑥 ) = 𝑎0 + 𝑎1 𝑥 − 2! 𝑎0 𝑥 2 − 3! 𝑎1 𝑥 3 + 4! 𝑎0 𝑥 4 + 5! 𝑎1 𝑥 5 + ⋯
𝑥2
𝑥4
𝑥3
𝑥5
𝑦(𝑥 ) = 𝑎0 (1 − 2! + 4! − ⋯ ) + 𝑎1 (𝑥 − 3! + 5! − ⋯ )
By Maclaurin series,
sin 𝑥 = 𝑥 −
cos 𝑥 = 1 −
𝑥3
3!
𝑥2
2!
+
+
𝑥5
5!
𝑥4
4!
−
−
𝑥7
7!
𝑥6
6!
+⋯
+⋯
Thus,
𝑦(𝑥 ) = 𝑎0 cos 𝑥 + 𝑎1 sin 𝑥
Example 3:
Solve the differential equation 𝑦 ′ + 𝑥𝑦 = 0.
Solution:
𝑦 ′ + 𝑥𝑦 = 0
𝑛−1
𝑛
∑∞
+ 𝑥 ∑∞
𝑛=1 𝑛𝑎𝑛 𝑥
𝑛=0 𝑎𝑛 𝑥 = 0
𝑎1 + 2𝑎2 𝑥 + 3𝑎3 𝑥 2 + 4𝑎4 𝑥 3 +. . +𝑥(𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 + 𝑎3 𝑥 3 + ⋯ ) = 0
𝑎1 + 2𝑎2 𝑥 + 3𝑎3 𝑥 2 + 4𝑎4 𝑥 3 +. . +𝑎0 𝑥 + 𝑎1 𝑥 2 + 𝑎2 𝑥 3 + 𝑎3 𝑥 4 + ⋯ = 0
Equating the coefficients of 𝑥 0 ,
𝑎1 = 0
Equating the coefficients of 𝑥 1 ,
2𝑎2 + 𝑎0 = 0
1
𝑎2 = − 2 𝑎0
Equating the coefficients of 𝑥 2 ,
3𝑎3 + 𝑎1 = 0
1
𝑎3 = − 𝑎1 = 0
3
Equating the coefficients of 𝑥 3 ,
4𝑎4 + 𝑎2 = 0
1
1
1
1
𝑎4 = − 4 𝑎2 = − 4 (− 2 𝑎0 ) = 8 𝑎0
57
Author: Harold Jan R. Terano, ECE, MET
POWER SERIES
CHAPTER 4
Equating the coefficients of 𝑥 4 ,
5𝑎5 + 𝑎3 = 0
1
1
𝑎5 = − 5 𝑎3 = − 5 (0) = 0
Equating the coefficients of 𝑥 5 ,
6𝑎6 + 𝑎4 = 0
1
1 1
1
𝑎6 = − 6 𝑎4 = − 6 (8 𝑎0 ) = − 48 𝑎0
Therefore,
𝑦(𝑥 ) = 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 + 𝑎3 𝑥 3 + ⋯
1
1
1
𝑦(𝑥 ) = 𝑎0 + 0 − 2 𝑎0 𝑥 2 + 0 + 8 𝑎0 𝑥 4 + 0 − 48 𝑎0 𝑥 6 …
𝑥2
𝑥4
𝑥6
𝑦(𝑥 ) = 𝑎0 (1 − 2 + 8 − 48 + ⋯ )
1 2
𝑥2
𝑥4
𝑥6
Since 𝑒 −2𝑥 = 1 − 2 + 8 − 48 + ⋯, thus,
1 2
𝑦(𝑥 ) = 𝑎0 𝑒 −2𝑥
Exercise 4.4
Solve the following ordinary differential equations by applying power series solution.
1. 𝑦 ′ − 𝑦 = 0
2. 𝑦 ′ + 2𝑦 = 0
3. 𝑦 ′ + 2(1 + 𝑥 )𝑦 = 0
4. 𝑦 ′ = 𝑥 + 𝑦
5. 𝑦 ′′ − 4𝑦 = 0
6. 𝑦 ′′ + 𝑥𝑦 ′ + 𝑦 = 0
7. (1 + 𝑥 )𝑦 ′ = 𝑦
8. 𝑦 ′′ − 𝑦 = 0
9. 𝑦 ′ = 𝑦 + 𝑦 2
10. 𝑦 ′′ + 2𝑥𝑦 ′ + 2𝑦 = 0
58
Author: Harold Jan R. Teran
CHAPTER 5
FOURIER SERIES
Chapter 5
FOURIER SERIES
Chapter Outline:
5.1
5.2
5.3
Fourier Series for Periodic Functions of Period 2π
Fourier Series for a Non-Periodic Function Over Range 2π
Fourier Series Over any Range
Learning Objectives:
At the end of the lesson, the students will be able to:
1. Define Fourier series.
2. Obtain Fourier series for periodic functions.
3. Obtain Fourier series for non-periodic functions.
4. Obtain Fourier series over any range.
Overview:
Fourier series are infinite series that represent general periodic functions in terms
of sines and cosines. This chapter will discuss techniques in obtaining Fourier series of
periodic functions and non-periodic functions.
59
Author: Harold Jan R. Terano, ECE, MET
FOURIER SERIES
CHAPTER 5
5.1 Fourier Series for Periodic Functions of Period 𝟐𝝅
A function 𝑓(𝑥) is periodic if 𝑓 (𝑥 + 𝑇) = 𝑓(𝑥) for all values of 𝑥, where 𝑇 is
some positive number that describes the interval between two successive repetitions
and is called the period of the functions 𝑓(𝑥).
Consider the figure,
𝒇(𝒙)
𝟏
−𝟐𝝅
−𝝅
𝝅
𝟎
𝒙
𝟐𝝅
−𝟏
The function is periodic of period 2𝜋 and is defined by:
𝑓 (𝑥 ) = {
−1, when − 𝜋 < 𝑥 < 0
+1, when
0<𝑥<𝜋
A Fourier series representation of a function 𝑓(𝑥) over the interval
−𝜋 ≤ 𝑥 ≤ 𝜋 is given by the expression,
𝒇(𝒙) = 𝒂𝟎 + ∑∞
𝒏=𝟏(𝒂𝒏 𝐜𝐨𝐬 𝒏𝒙 + 𝒃𝒏 𝐬𝐢𝐧 𝒏𝒙)
𝟏
𝝅
𝟏
𝝅
𝟏
𝝅
𝒂𝟎 = 𝟐𝝅 ∫−𝝅 𝒇(𝒙) 𝒅𝒙
where:
𝒂𝒏 = ∫−𝝅 𝒇(𝒙) 𝐜𝐨𝐬 𝒏𝒙 𝒅𝒙, (𝒏 = 𝟏, 𝟐, 𝟑, … )
𝝅
𝒃𝒏 = 𝝅 ∫−𝝅 𝒇(𝒙) 𝐬𝐢𝐧 𝒏𝒙 𝒅𝒙, (𝒏 = 𝟏, 𝟐, 𝟑, … )
(𝑎0 , 𝑎𝑛 and 𝑏𝑛 ) are the called the Fourier coefficients.
Example 1:
Obtain a Fourier series for the periodic function 𝑓(𝑥) defined as:
𝑓 (𝑥 ) = {
−1, when − 𝜋 < 𝑥 < 0
.
+1, when
0<𝑥<𝜋
Solution:
For 𝑎0 ,
0
1
1
𝜋
1
1
𝑎0 = 2𝜋 ∫−𝜋(−1)𝑑𝑥 + 2𝜋 ∫0 (+1)𝑑𝑥 = − 2𝜋 |𝑥 |0−𝜋 + 2𝜋 |𝑥 |𝜋0
1
1
𝑎0 = − 2𝜋 [𝜋] + 2𝜋 [𝜋] = 0
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Author: Harold Jan R. Teran
CHAPTER 5
FOURIER SERIES
For 𝑎𝑛 ,
0
1
𝜋
1
𝑎𝑛 = 𝜋 ∫−𝜋(−1) cos 𝑛𝑥 𝑑𝑥 + 𝜋 ∫0 (+1) cos 𝑛𝑥 𝑑𝑥
0
1 1
𝜋
1 1
𝑎𝑛 = − 𝜋 |𝑛 sin 𝑛𝑥|
+ 𝜋 |𝑛 sin 𝑛𝑥|
−𝜋
0
1 1
1 1
𝑎𝑛 = − 𝜋 {𝑛 [sin 0 − sin(−𝑛𝜋)]} + 𝜋 {𝑛 [sin(𝑛𝜋) − sin 0]}
1
1
1 1
𝑎𝑛 = − 𝜋 (𝑛 sin 𝑛𝜋) + 𝜋 (𝑛 sin 𝑛𝜋) = 0
For 𝑏𝑛 ,
0
1
𝜋
1
𝑏𝑛 = 𝜋 ∫−𝜋(−1) sin 𝑛𝑥 𝑑𝑥 + 𝜋 ∫0 (+1) sin 𝑛𝑥 𝑑𝑥
1
0
1
1
𝑏𝑛 = − 𝜋 |− 𝑛 cos 𝑛𝑥|
−𝜋
𝑏𝑛 =
1
𝑛𝜋
𝜋
1
+ 𝜋 |− 𝑛 cos 𝑛𝑥|
0
{[cos 0 − cos(−𝑛𝜋)]} −
1
1
1
2
2
2
1
𝑛𝜋
{[cos(𝑛𝜋) − cos 0]}
1
𝑏𝑛 = 𝑛𝜋 − 𝑛𝜋 cos 𝑛𝜋 − 𝑛𝜋 cos 𝑛𝜋 + 𝑛𝜋
𝑏𝑛 = 𝑛𝜋 − 𝑛𝜋 cos 𝑛𝜋 = 𝑛𝜋 (1 − cos 𝑛𝜋)
When 𝑛 is odd,
4
4
4
𝑏1 = 𝜋 , 𝑏3 = 3𝜋 , 𝑏5 = 5𝜋, and so on.
When 𝑛 is even,
𝑏𝑛 = 0
The Fourier series is,
𝑓 (𝑥 ) = 𝑎0 + ∑∞
𝑛=1(𝑎𝑛 cos 𝑛𝑥 + 𝑏𝑛 sin 𝑛𝑥 )
𝑓 (𝑥 ) = 0 + ∑ ∞
𝑛=1(0 + 𝑏𝑛 sin 𝑛𝑥 )
4
4
4
4
1
1
𝜋
3
5
𝑓 (𝑥 ) = 𝜋 sin 𝑥 + 3𝜋 sin 3𝑥 + 5𝜋 sin 5𝑥 + ⋯
𝑓 (𝑥 ) = (sin 𝑥 + sin 3𝑥 + sin 5𝑥 + ⋯ )
In summation notation,
4
1
𝑓 (𝑥 ) = 𝜋 ∑ ∞
𝑛=1 2𝑛−1 sin(2𝑛 − 1)𝑥
Example 2:
Obtain a Fourier series for the periodic function 𝑓(𝑥) defined as:
0, when − 𝜋 < 𝑥 < −
𝑓 (𝑥 ) =
𝜋
𝜋
𝜋
2
1, when − 2 < 𝑥 < 2
{0, when
𝜋
2
.
<𝑥<𝜋
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Author: Harold Jan R. Terano, ECE, MET
FOURIER SERIES
CHAPTER 5
Solution:
For 𝑎0 ,
𝜋
1
1
𝜋
1
𝜋
𝜋
1
1
𝑎0 = 2𝜋 ∫−2𝜋 𝑑𝑥 = 2𝜋 |𝑥 | 2 𝜋 = 2𝜋 [ 2 + 2 ] = 2𝜋 (𝜋) = 2
−
2
2
For 𝑎𝑛 ,
1
𝜋
2
𝜋
−
2
𝜋
2
1 1
𝑎𝑛 = 𝜋 ∫ cos 𝑛𝑥 𝑑𝑥 = 𝜋 |𝑛 sin 𝑛𝑥| 𝜋
−
1
𝑛𝜋
2
𝑛𝜋
2
𝑛𝜋
𝑎𝑛 = 𝑛𝜋 [sin ( 2 ) − sin (− 2 )]
𝑎𝑛 = 𝑛𝜋 sin ( 2 )
when 𝑛 is odd,
2
2
2
2
𝑎1 = 𝜋 , 𝑎3 = − 3𝜋 , 𝑎5 = 5𝜋 , 𝑎7 = − 7𝜋, and so on.
when 𝑛 is even,
𝑎𝑛 = 0
For 𝑏𝑛 ,
1
𝜋
2
𝜋
−
2
1
𝜋
2
1
𝑏𝑛 = 𝜋 ∫ sin 𝑛𝑥 𝑑𝑥 = 𝜋 |− 𝑛 cos 𝑛𝑥| 𝜋
−
1
𝑛𝜋
2
𝑛𝜋
𝑏𝑛 = − 𝑛𝜋 [cos ( 2 ) − cos (− 2 )]
𝑏𝑛 = 0
The Fourier series is,
𝑓 (𝑥 ) = 𝑎0 + ∑∞
𝑛=1(𝑎𝑛 cos 𝑛𝑥 + 𝑏𝑛 sin 𝑛𝑥 )
1
𝑓 (𝑥 ) = + ∑ ∞
𝑛=1(𝑎𝑛 cos 𝑛𝑥 + 0)
2
1
2
2
2
2
𝑓 (𝑥 ) = 2 + (𝜋 cos 𝑥 − 3𝜋 cos 3𝑥 + 5𝜋 cos 5𝑥 − 7𝜋 cos 7𝑥 + ⋯ )
1
2
1
1
1
𝑓 (𝑥 ) = 2 + 𝜋 (cos 𝑥 − 3 cos 3𝑥 + 5 cos 5𝑥 − 7 cos 7𝑥 + ⋯ )
In summation notation,
1
2
1
𝑛−1 [
𝑓 (𝑥 ) = 2 + 𝜋 ∑ ∞
cos(2𝑛 − 1)]
𝑛=1(−1)
(2𝑛−1)
Example 3:
Find the Fourier series of the given function which is assumed to have the period
2𝜋.
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𝑓(𝑥)
𝜋
−𝜋
−2𝜋
𝜋
0
𝑥
2𝜋
Solution:
The function can be defined as 𝑓 (𝑥 ) = {
0, when − 𝜋 < 𝑥 < 0
.
𝑥, when 0 < 𝑥 < 𝜋
For 𝑎0 ,
𝜋
1
𝑥2
1
𝜋
1
𝜋
𝑎0 = 2𝜋 ∫0 𝑥𝑑𝑥 = 2𝜋 | 2 | = 4𝜋 [(𝜋)2 − 0] = 4
0
For 𝑎𝑛 ,
𝜋
1
𝑎𝑛 = 𝜋 ∫0 𝑥 cos 𝑛𝑥 𝑑𝑥
By integration by parts,
𝑢=𝑥
𝑑𝑣 = cos 𝑛𝑥 𝑑𝑥
𝑑𝑢 = 𝑑𝑥
𝑣 = 𝑛 sin 𝑛𝑥
1
1 1
𝜋
1
𝑎𝑛 = 𝜋 |𝑛 𝑥 sin 𝑛𝑥 − 𝑛 ∫ sin 𝑛𝑥 𝑑𝑥|
0
1 1
1
𝑎𝑛 = 𝜋 |𝑛 𝑥 sin 𝑛𝑥 + 𝑛2 cos 𝑛𝑥|
1 𝜋
1
1 𝜋
1
𝜋
0
1
𝑎𝑛 = 𝜋 {𝑛 sin 𝑛𝜋 + 𝑛2 cos 𝑛𝜋 − [0 + 𝑛2 cos(0)]}
1
𝑎𝑛 = 𝜋 (𝑛 sin 𝑛𝜋 + 𝑛2 cos 𝑛𝜋 − 𝑛2 )
when 𝑛 is odd,
2
2
2
𝑎1 = − 𝜋 , 𝑎3 = − 9𝜋 , 𝑎5 = − 25𝜋 , … and so on.
when 𝑛 is even,
𝑎𝑛 = 0
For 𝑏𝑛 ,
1
𝜋
𝑏𝑛 = 𝜋 ∫0 𝑥 sin 𝑛𝑥 𝑑𝑥
By integration by parts,
𝑢=𝑥
𝑑𝑣 = sin 𝑛𝑥 𝑑𝑥
𝑑𝑢 = 𝑑𝑥
𝑣 = − 𝑛 cos 𝑛𝑥
1
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CHAPTER 5
1
1
𝜋
1
𝑏𝑛 = 𝜋 |− 𝑛 𝑥 cos 𝑛𝑥 + 𝑛 ∫ cos 𝑛𝑥 𝑑𝑥|
0
1
1
1
𝑏𝑛 = 𝜋 |− 𝑛 𝑥 cos 𝑛𝑥 + 𝑛2 sin 𝑛𝑥|
1
𝜋
1
𝜋
1
𝜋
𝑛
𝑛2
𝜋
0
1
1
𝑏𝑛 = 𝜋 {− 𝑛 cos 𝑛𝜋 + 𝑛2 sin 𝑛𝜋 − [0 + 𝑛2 sin(0)]}
𝑏𝑛 = (− cos 𝑛𝜋 +
sin 𝑛𝜋)
when 𝑛 is odd,
1
1
𝑏1 = 1, 𝑏3 = 3 , 𝑏5 = 5 , … and so on.
when 𝑛 is even,
1
1
1
𝑏2 = − 2 , 𝑏4 = − 4 , 𝑏6 = − 6 , … and so on.
The Fourier series is,
𝑓 (𝑥 ) = 𝑎0 + ∑∞
𝑛=1(𝑎𝑛 cos 𝑛𝑥 + 𝑏𝑛 sin 𝑛𝑥 )
𝜋
2
2
2
1
𝑓 (𝑥 ) = 4 + (− 𝜋 cos 𝑥 − 9𝜋 cos 3𝑥 − 25𝜋 cos 5𝑥 − ⋯ ) + (sin 𝑥 − 2 sin 2𝑥 +
1
3
1
1
1
4
5
6
sin 3𝑥 − sin 4𝑥 + sin 5𝑥 − sin 6𝑥 + ⋯ )
𝜋
2
1
1
1
𝑓 (𝑥 ) = 4 − 𝜋 (cos 𝑥 + 9 cos 3𝑥 + 25 cos 5𝑥 + ⋯ ) + (sin 𝑥 − 2 sin 2𝑥 +
1
3
1
1
1
sin 3𝑥 − 4 sin 4𝑥 + 5 sin 5𝑥 − 6 sin 6𝑥 + ⋯ )
In summation notation,
𝜋
2
1
∞
𝑛−1
𝑓 (𝑥 ) = 4 − 𝜋 ∑ ∞
𝑛=1 (2𝑛−1)2 cos(2𝑛 − 1)𝑥 + ∑𝑛=1(−1)
sin 𝑛𝑥
𝑛
Exercise 5.1
Solve the following problems.
−1, when − 𝜋 < 𝑥 < 0
1. Obtain the Fourier series for the periodic function 𝑓(𝑥 ) = {
.
0, when 0 < 𝑥 < 𝜋
−𝑥, when − 𝜋 < 𝑥 < 0
2. Obtain the Fourier series for the periodic function 𝑓(𝑥 ) = {
.
𝑥, when 0 < 𝑥 < 𝜋
3. Obtain the Fourier series for the periodic function
𝑓 (𝑥 ) = {
0, when − 𝜋 < 𝑥 < 0
.
sin 𝑥 , when 𝜃 < 𝑥 < 𝜋
4. Find the Fourier series of the given function from the figure below which is assumed
to have the period 2𝜋:
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𝑓(𝑥)
1
−2𝜋
−𝜋
𝜋
0
𝑥
2𝜋
5. Find the Fourier series of the given function from the figure below which is assumed
to have the period 2𝜋:
𝑓(𝑥)
𝜋
−2𝜋
−𝜋
𝜋
0
𝑥
2𝜋
5.2 Fourier Series for a Non-Periodic Function Over Range 𝟐𝝅
A function 𝑓(𝑥) that is not periodic can’t be expanded in a Fourier series for all
values of 𝑥. But, it is possible to determine a Fourier series to represent the function
over any range of width 2𝜋.
Let us consider the function 𝑓 (𝑥 ) = 𝑥 which is non-periodic. Given that function,
a new function may be constructed by taking the values of 𝑓(𝑥) in the given range and
then repeating them outside of the given range at intervals of 2𝜋.
The function 𝑓(𝑥 ) = 𝑥 is constructed so that it is periodic with period 2𝜋 as
shown in the figure
𝒇(𝒙)
𝒇(𝒙) = 𝒙
𝝅
−𝟐𝝅
−𝝅
𝟎
𝝅
𝟐𝝅
𝒙
−𝝅
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Author: Harold Jan R. Terano, ECE, MET
FOURIER SERIES
CHAPTER 5
The Fourier series of a non-periodic function over the range 2𝜋 is the same as
the formula for the Fourier series for periodic function.
Example 1:
Determine the Fourier series to represent 𝑓 (𝑥 ) = 𝑥 in the range −𝜋 to 𝜋.
Solution:
For 𝑎0 ,
𝜋
1
𝜋
1
𝑎0 = 2𝜋 ∫−𝜋 𝑥𝑑𝑥 = |2 𝑥 2 |
−𝜋
1
= 2𝜋 [(𝜋)2 − (−𝜋)2 ] = 0
For 𝑎𝑛 ,
𝜋
1
𝑎𝑛 = 𝜋 ∫−𝜋 𝑥 cos 𝑛𝑥 𝑑𝑥
By integration by parts,
𝑢=𝑥
𝑑𝑣 = cos 𝑛𝑥 𝑑𝑥
𝑑𝑢 = 𝑑𝑥
𝑣 = sin 𝑛𝑥
1
𝑛
1 1
𝜋
1
𝑎𝑛 = 𝜋 |𝑛 𝑥 sin 𝑛𝑥 − 𝑛 ∫ sin 𝑛𝑥 𝑑𝑥|
−𝜋
1 1
1
𝑎𝑛 = 𝜋 |𝑛 𝑥 sin 𝑛𝑥 + 𝑛2 cos 𝑛𝑥|
1 𝜋
1
1 𝜋
1
𝜋
−𝜋
𝜋
1
𝑎𝑛 = 𝜋 {𝑛 sin 𝑛𝜋 + 𝑛2 cos 𝑛𝜋 − [− 𝑛 sin(−𝑛𝜋) + 𝑛2 cos(−𝑛𝜋)]}
𝜋
1
𝑎𝑛 = 𝜋 [𝑛 sin 𝑛𝜋 + 𝑛2 cos 𝑛𝜋 − (𝑛 sin 𝑛𝜋 + 𝑛2 cos 𝑛𝜋)]
𝑎𝑛 = 0
For 𝑏𝑛 ,
1
𝜋
𝑏𝑛 = 𝜋 ∫−𝜋 𝑥 sin 𝑛𝑥 𝑑𝑥
By integration by parts,
𝑢=𝑥
𝑑𝑣 = sin 𝑛𝑥 𝑑𝑥
𝑑𝑢 = 𝑑𝑥
𝑣 = − cos 𝑛𝑥
1
1
𝑛
1
𝜋
1
𝑏𝑛 = 𝜋 |− 𝑛 𝑥 cos 𝑛𝑥 + 𝑛 ∫ cos 𝑛𝑥 𝑑𝑥|
−𝜋
1
1
1
𝑏𝑛 = 𝜋 |− 𝑛 𝑥 cos 𝑛𝑥 + 𝑛2 sin 𝑛𝑥|
𝜋
−𝜋
1
𝜋
1
𝜋
1
𝜋
1
𝜋
1
2𝜋
1
𝑏𝑛 = 𝜋 {− 𝑛 cos 𝑛𝜋 + 𝑛2 sin 𝑛𝜋 − [𝑛 cos(−𝑛𝜋) + 𝑛2 sin(−𝑛𝜋)]}
1
𝑏𝑛 = 𝜋 (− 𝑛 cos 𝑛𝜋 + 𝑛2 sin 𝑛𝜋 − 𝑛 cos 𝑛𝜋 + 𝑛2 sin 𝑛𝜋)
2
𝑏𝑛 = 𝜋 (− 𝑛 cos 𝑛𝜋 + 𝑛2 sin 𝑛𝜋)
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2
𝜋
1
𝑏𝑛 = 𝜋 (− 𝑛 cos 𝑛𝜋 + 𝑛2 sin 𝑛𝜋)
when 𝑛 is odd,
2
2
𝑏1 = 2, 𝑏3 = 3 , 𝑏5 = 5 , … and so on.
when 𝑛 is even,
1
1
2
3
𝑏2 = −1, 𝑏4 = − , 𝑏6 = − , … and so on.
The Fourier series is,
𝑓 (𝑥 ) = 𝑎0 + ∑∞
𝑛=1(𝑎𝑛 cos 𝑛𝑥 + 𝑏𝑛 sin 𝑛𝑥 )
2
1
2
1
𝑓 (𝑥 ) = 2 sin 𝑥 − sin 2𝑥 + 3 sin 3𝑥 − 2 sin 4𝑥 + 5 sin 5𝑥 − 3 sin 6𝑥 + ⋯
1
1
1
1
1
𝑓 (𝑥 ) = 2 (sin 𝑥 − 2 sin 2𝑥 + 3 sin 3𝑥 − 4 sin 4𝑥 + 5 sin 5𝑥 − 6 sin 6𝑥 + ⋯ )
In summation notation,
𝑓 (𝑥 ) = 2 ∑ ∞
𝑛=1
(−1)𝑛−1
𝑛
sin 𝑛𝑥
5.3 Fourier Series Over Any Range
The functions previously discussed had period of 2𝜋. For other applications, it is
generally of any other periods. To change from period 𝑝 = 2𝜋 to a period 𝑝 = 2𝐿,
consider the Fourier series with 𝑔(𝑢) instead of 𝑓(𝑥), then,
𝒇(𝒖) = 𝒂𝟎 + ∑∞
𝒏=𝟏(𝒂𝒏 𝐜𝐨𝐬 𝒏𝒖 + 𝒃𝒏 𝐬𝐢𝐧 𝒏𝒖)
𝟏
𝝅
𝟏
𝝅
𝟏
𝝅
𝒂𝟎 = 𝟐𝝅 ∫−𝝅 𝒇(𝒖) 𝒅𝒖
where:
𝒂𝒏 = 𝝅 ∫−𝝅 𝒇(𝒖) 𝐜𝐨𝐬 𝒏𝒖 𝒅𝒖, (𝒏 = 𝟏, 𝟐, 𝟑, … )
𝒃𝒏 = 𝝅 ∫−𝝅 𝒇(𝒖) 𝐬𝐢𝐧 𝒏𝒖 𝒅𝒖, (𝒏 = 𝟏, 𝟐, 𝟑, … )
Set 𝑢 = 𝑘𝑥 with 𝑘 such that the previous period 𝑢 = 2𝜋 gives for the new
𝜋
variable 𝑥 the new period 𝑥 = 2𝐿. Thus, 2𝜋 = 𝑘 (2𝐿), therefore, 𝑘 = 𝐿 . Since 𝑢 = 𝑘𝑥,
𝜋
𝜋
then 𝑢 = 𝐿 𝑥 and 𝑑𝑢 = 𝐿 𝑑𝑥. Substituting from the above formula, we have,
𝒏𝝅
𝒏𝝅
𝒇(𝒙) = 𝒂𝟎 + + ∑∞
𝒏=𝟏 [𝒂𝒏 𝐜𝐨𝐬 ( 𝑳 ) 𝒙 + 𝒃𝒏 𝐬𝐢𝐧 ( 𝑳 ) 𝒙]
where:
𝟏
𝑳
𝟏
𝑳
𝒏𝝅
𝟏
𝑳
𝒏𝝅
𝒂𝟎 = 𝟐𝑳 ∫−𝑳 𝒇(𝒙) 𝒅𝒙
𝒂𝒏 = 𝑳 ∫−𝑳 𝒇(𝒙) 𝐜𝐨𝐬 𝑳 𝒙 𝒅𝒙, (𝒏 = 𝟏, 𝟐, 𝟑, … )
𝒃𝒏 = 𝑳 ∫−𝑳 𝒇(𝒙) 𝐬𝐢𝐧 𝑳 𝒙 𝒅𝒙, (𝒏 = 𝟏, 𝟐, 𝟑, … )
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Example 1:
0, when − 2 < 𝑥 < 0
Find the Fourier series of the function 𝑓 (𝑥 ) = {
.
+1, when
0<𝑥<2
Solution:
𝑝 = 2𝐿
4 = 2𝐿; 𝐿 = 2
For 𝑎0 ,
2
1
1
1
1
𝑎0 = 2(2) ∫0 𝑑𝑥 = 4 |𝑥 |20 = 4 (2) = 2
For 𝑎𝑛 ,
1
2
𝑛𝜋
1
2
2
𝑛𝜋
𝑎𝑛 = 2 ∫0 cos ( 2 ) 𝑥 𝑑𝑥 = 2 (𝑛𝜋) |sin ( 2 ) 𝑥|
0
1
𝑎𝑛 = 𝑛𝜋 sin(𝑛𝜋)
when 𝑛 is odd,
𝑎1 = 0, 𝑎3 = 0, … and so on.
when 𝑛 is even,
𝑎2 = 0, 𝑎4 = 0, … and so on.
For 𝑏𝑛 ,
1
2
𝑛𝜋
1
2
2
𝑛𝜋
1
𝑏𝑛 = 2 ∫0 sin ( 2 ) 𝑥 𝑑𝑥 = 2 (𝑛𝜋) |−cos ( 2 ) 𝑥| = − 𝑛𝜋 [cos(𝑛𝜋) − 1]
0
1
𝑏𝑛 = 𝑛𝜋 [1 − cos(𝑛𝜋)]
when 𝑛 is odd,
2
2
2
𝑏1 = 𝜋 , 𝑏3 = 3𝜋 , 𝑏5 = 5𝜋 , … and so on.
when 𝑛 is even,
𝑏2 = 0, 𝑏4 = 0, … and so on.
The Fourier series is,
𝑛𝜋
𝑛𝜋
𝑓 (𝑥 ) = 𝑎0 + ∑∞
𝑛=1 [𝑎𝑛 cos ( 𝐿 ) 𝑥 + 𝑏𝑛 sin ( 𝐿 ) 𝑥]
1
2
𝜋
2
3𝜋
2
5𝜋
𝑓 (𝑥 ) = 2 + 𝜋 sin ( 2 ) 𝑥 + 3𝜋 sin ( 2 ) 𝑥 + 5𝜋 sin ( 2 ) 𝑥 + ⋯
In summation notation,
1
2
1
𝑓 (𝑥 ) = 2 + 𝜋 ∑ ∞
𝑛=1 (2𝑛−1) sin [
(2𝑛−1)𝜋
2
]𝑥
Example 2:
Find the Fourier series of the function 𝑓 (𝑥 ) = 𝑥 in the range 𝑥 = 0 to 𝑥 = 3.
Solution:
The function 𝑓(𝑥 ) = 𝑥 in the interval 0 to 3 as shown in the figure below. The
function is not periodic, but it can be constructed outside of this range so that it is
periodic of period 3.
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𝑓(𝑥)
3
−3
−6
3
0
𝑥
6
𝑝 = 2𝐿
3
3 = 2𝐿; 𝐿 = 2
For 𝑎0 ,
𝑎0 =
3
1
3
2( )
2
1 1
3
3
∫0 𝑥𝑑𝑥 = 3 |2 𝑥 2 | = 2
0
For 𝑎𝑛 ,
3
1
𝑛𝜋
3
2
2
𝑎𝑛 = 3 ∫0 𝑥 cos ( 3 ) 𝑥 𝑑𝑥 = 3 ∫0 𝑥 cos (3 𝑛𝜋) 𝑥𝑑𝑥
( )
2
2
By integration by parts,
2
𝑢=𝑥
𝑑𝑣 = cos (3 𝑛𝜋) 𝑥 𝑑𝑥
𝑑𝑢 = 𝑑𝑥
𝑣 = 2𝑛𝜋 sin (3 𝑛𝜋) 𝑥
2
𝑎𝑛 = |
3
3
3 2𝑛𝜋
𝑎𝑛 = |
2
3
2
3
2
9
3 2𝑛𝜋
2
2
𝑥 sin ( 𝑛𝜋) 𝑥 −
3
3
0
2
9
3
4𝑛2 𝜋2
𝑥 sin ( 𝑛𝜋) 𝑥 +
3
2
∫ sin (3 𝑛𝜋) 𝑥 𝑑𝑥|
2𝑛𝜋
2
3
3
0
cos ( 𝑛𝜋) 𝑥|
9
9
𝑎𝑛 = 3 {2𝑛𝜋 (3) sin(2𝑛𝜋) + 4𝑛2 𝜋2 cos(2𝑛𝜋) − (0 + 4𝑛2 𝜋2)}
9
9
𝑎𝑛 = 3 [2𝑛𝜋 sin(2𝑛𝜋) + 4𝑛2𝜋2 cos(2𝑛𝜋) − 4𝑛2𝜋2 ]
when 𝑛 is odd,
𝑎1 = 0, 𝑎3 = 0, … and so on.
when 𝑛 is even,
𝑎2 = 0, 𝑎4 = 0, … and so on.
For 𝑏𝑛 ,
1
3
𝑛𝜋
2
3
2
𝑏𝑛 = 3 ∫0 𝑥 sin ( 3 ) 𝑥 𝑑𝑥 = 3 ∫0 𝑥 sin (3 𝑛𝜋) 𝑥𝑑𝑥
( )
2
2
69
Author: Harold Jan R. Terano, ECE, MET
FOURIER SERIES
CHAPTER 5
By integration by parts,
2
𝑢=𝑥
𝑑𝑣 = sin (3 𝑛𝜋) 𝑥 𝑑𝑥
𝑑𝑢 = 𝑑𝑥
𝑣 = − 2𝑛𝜋 cos (3 𝑛𝜋) 𝑥
2
3
3
2
2
3
3
2
𝑏𝑛 = 3 |− 2𝑛𝜋 𝑥 cos (3 𝑛𝜋) 𝑥 + 2𝑛𝜋 ∫ cos (3 𝑛𝜋) 𝑥 𝑑𝑥|
0
2
3
2
9
3
2
𝑏𝑛 = 3 |− 2𝑛𝜋 𝑥 cos (3 𝑛𝜋) 𝑥 + 4𝑛2 𝜋2 sin (3 𝑛𝜋) 𝑥|
0
2
9
9
2
9
9
𝑏𝑛 = 3 {− 2𝑛𝜋 cos(2𝑛𝜋) + 4𝑛2 𝜋2 sin(2𝑛𝜋) − (−0 + 0)}
𝑏𝑛 = 3 [− 2𝑛𝜋 cos(2𝑛𝜋) + 4𝑛2𝜋2 sin(2𝑛𝜋)]
𝑏𝑛 =
3
𝑛𝜋
[− cos(2𝑛𝜋) +
1
2𝑛𝜋
sin(𝑛𝜋)]
when 𝑛 is odd,
3
3
𝜋
3𝜋
𝑏1 = − , 𝑏3 = −
, 𝑏5 = −
3
5𝜋
and so on.
when 𝑛 is even,
3
3
3
𝑏2 = − 2𝜋 , 𝑏4 = − 4𝜋 , 𝑏6 = − 6𝜋 , … and so on.
The Fourier series is,
𝑛𝜋
𝑛𝜋
𝑓 (𝑥 ) = 𝑎0 + ∑∞
𝑛=1 [𝑎𝑛 cos ( 𝐿 ) 𝑥 + 𝑏𝑛 sin ( 𝐿 ) 𝑥]
3
3
𝜋
3
2𝜋
3
3𝜋
𝑓 (𝑥 ) = 2 + [− 𝜋 sin ( 3 ) 𝑥 − 2𝜋 sin ( 3 ) 𝑥 − 3𝜋 sin ( 3 ) 𝑥 − ⋯ ]
2
2
2
3
3
2𝜋
1
4𝜋
1
6𝜋
2
𝜋
3
2
3
3
3
1
2𝑛𝜋
𝑓 (𝑥 ) = − [sin ( ) 𝑥 + sin ( ) 𝑥 + sin ( ) 𝑥 + ⋯]
In summation notation,
3
3
𝑓 (𝑥 ) = 2 − 𝜋 ∑ ∞
𝑛=1 𝑛 sin ( 3 )
Exercise 5.2
Solve the following problems.
1. Obtain the Fourier series of the function 𝑓 (𝑥 ) = 𝑥 2 in the range 0 to 2𝜋.
0, when 0 < 𝑥 < 𝜋
2. Obtain the Fourier series of the function defined by 𝑓(𝑥 ) = {
.
𝑥, when 𝜋 < 𝑥 < 2𝜋
3. Fourier series of the function 𝑓 (𝑥 ) = 𝑥 in the range 𝑥 = 0 to 𝑥 = 6.
−1, when − 2 < 𝑥 < 0
4. Obtain the Fourier series of the function defined by 𝑓(𝑥 ) = {
.
1, when 0 < 𝑥 < 2
0, when − 2 < 𝑥 < −1
5. Obtain the Fourier series of the function defined by 𝑓(𝑥 ) = { 2, when − 1 < 𝑥 < 1 .
0, when 1 < 𝑥 < 2
70
Author: Harold Jan R. Teran
CHAPTER 6
MATRICES AND DETERMINANTS
Chapter 6
MATRICES AND
DETERMINANTS
Chapter Outline:
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
Matrices
Determinants
Operations on Matrices
Algebraic Operations of Matrix
Expansion of Minors
Pivotal Element Method
Solutions to Linear Systems Using Inverse of a Matrix
Cramer’s Rule
Learning Objectives:
At the end of the lesson, the students will be able to:
1. Define matrix.
2. Obtain the determinant of a matrix.
3. Algebraically operate matrices.
4. Solve higher order determinant using expansion of minors and pivotal element
method.
5. Solve systems of linear equations using inverse of a matrix and Cramer’s rule.
Overview:
Matrix is a mathematical technique and a powerful tool in solving systems of
linear equations. This chapter begins with an introduction to matrix, its classifications
and operations. Determinant of a matrix will be covered. Reduction of higher-order
determinants will be discussed by using the expansions of minors and pivotal element
method. These topics are very important in using the Cramer’s rule to solve systems of
linear equations.
71
Author: Harold Jan R. Terano, ECE, MET
MATRICES AND DETERMINANTS
CHAPTER 6
6.1 Matrices
A matrix is a rectangular collection of variables or scalars contained within a set
of square [ ] or round ( ) brackets. A matrix consists of 𝑚 rows and 𝑛 columns. The
numbers are the entries or elements of the matrix.
Examples of matrices are:
1 −3
]
→
4
2
2 5
2. [−3 1]
→
0 6
−2
3. [ 3 ]
→
−2
2 −3
1
]→
4. [
5
2 −1
1. [
2 × 2 matrix
3 × 2 matrix
3 × 1 matrix
2 × 3 matrix
When a matrix has the same number of rows as columns, it is a square matrix.
𝑎11 𝑎12 𝑎13 … 𝑎1𝑛
𝑎21 𝑎22 𝑎23 … 𝑎2𝑛
.
.
.
.
𝐴= .
.
.
.
.
.
.
.
[𝑎𝑚1 𝑎𝑚2 𝑎𝑚3 … 𝑎𝑚𝑛 ]
The matrix
𝐴
is an
𝑚×𝑛
matrix. The entries in matrix
𝐴
are double
subscripted with the first number indicating the row of the entry and the second number
indicating the column of the entry. The general entry for the matrix is denoted by 𝑎𝑖𝑗 .
The matrix 𝐴 can be denoted by [𝑎𝑖𝑗 ].
Classification of Matrices
1. Square matrix
A matrix whose number of rows 𝑚 is equal to the number of columns 𝑛.
2. Diagonal matrix
A diagonal matrix is a square matrix with all zero values except for the 𝑎𝑖𝑗 value
for all 𝑖 = 𝑗.
1 0 0
[0 3 0 ]
0 0 6
3. Identity matrix
An identity matrix is a diagonal matrix with all non-zero entries equal to 1.
72
Author: Harold Jan R. Teran
CHAPTER 6
MATRICES AND DETERMINANTS
1 0 0
[0 1 0 ]
0 0 1
4. Scalar matrix
A scalar matrix is a diagonal matrix with all non-zero entries equal to some other
constant.
5 0 0
[0 5 0 ]
0 0 5
5. Triangular matrix
A triangular matrix has zeros in all positions above or below the diagonal.
6.2 Determinants
Determinant is a scalar calculated from a square matrix. The determinant of a
matrix is indicated by enclosing the matrix by vertical lines.
The following are some of the important properties of determinant.
1. If a matrix has a row or column of zeros, its determinant is zero.
2 0 1
|4 0 5 | = 0
1 0 2
0 0 0
|2 6 4 | = 0
5 1 3
2. If a matrix has two identical rows or columns, its determinant is zero.
1 3 7
|3 6 5 | = 0
1 3 7
1 3 3
|4 5 5 | = 0
6 4 4
3. If a matrix is triangular, its determinant is equal to the product of the diagonal.
1 0 0
|0 4 0| = (1)(4)(5) = 20
0 0 5
4. The value of the determinant is not changed if corresponding rows and columns are
interchanged.
1 3 6
1 4 5
|4 2 1| = |3 2 3|
5 3 7
6 1 7
5. If a row or a column of a determinant is multiplied by a nonzero constant 𝑘, the value
of the determinant is multiplied by 𝑘.
1 3
2 |4 5
6 1
1(2)
2
3| = | 4
4
6
3(2)
5
1
2(2)
2
3 | = |4
6
4
6 4
5 3|
1 4
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Author: Harold Jan R. Terano, ECE, MET
MATRICES AND DETERMINANTS
1 3
2 |4 5
6 1
CHAPTER 6
1(2)
2
3| = |4(2)
6(2)
4
3 2
2
5 3| = | 8
1 4
12
3 2
5 3|
1 4
6. If two rows or columns of a determinant are interchanged, the sign will be changed.
2
|5
1
2
|5
1
7
3
2
7
3
2
5
5
6 | = − |2
5
1
5
7
|
|
6 =− 3
5
2
3
7
2
2
5
1
6
5|
5
5
6|
5
Determinant of a 2x2 matrix
The symbol,
𝑎
| 1
𝑎2
𝑏1
|
𝑏2
consisting of the four numbers 𝑎1 , 𝑏1 , 𝑎2 and 𝑏2 arranged in two rows and two columns
is called a determinant of second order or determinant of order two. The four
numbers are called elements of the determinant.
By definition,
𝑎
| 1
𝑎2
𝑏1
| = 𝑎1 𝑏2 − 𝑎2 𝑏1
𝑏2
Example 1:
3 −2
|.
Evaluate |
1
3
Solution:
3 −2
|
| = (3)(3) − (1)(−2) = 9 + 2 = 11
1
3
Example 2:
−2 1
|.
Evaluate |
−4 5
Solution:
−2 1
|
| = (−2)(5) − (−4)(1) = −10 + 4 = −6
−4 5
Example 3:
6 −2
|.
Evaluate |
−5
7
Solution:
6 −2
|
| = (6)(7) − (−5)(−2) = 42 − 10 = 32
−5
7
74
Author: Harold Jan R. Teran
CHAPTER 6
MATRICES AND DETERMINANTS
Determinant of a 3x3 matrix
The symbol
𝑎1
𝑎
| 2
𝑎3
𝑏1
𝑏2
𝑏3
𝑐1
𝑐2 |
𝑐3
𝑎1
|𝑎2
𝑎3
𝑏1
𝑏2
𝑏3
𝑐1 𝑎1
𝑐2 | 𝑎2
𝑐3 𝑎3
By definition,
− − −
𝑏1
𝑏2 |
𝑏3
+ + +
= 𝑎1 𝑏2 𝑐3 + 𝑏1 𝑐2 𝑎3 + 𝑐1 𝑎2 𝑏3 − 𝑎3 𝑏2 𝑐1 − 𝑏3 𝑐2𝑎1 − 𝑐3 𝑎2 𝑏1
Example 1:
2 −1
3
Evaluate | 5 −2 −2|.
−3
1
4
Solution:
2 −1
3
2 −1
3 2 −1
| 5 −2 −2| = | 5 −2 −2| 5 −2|
−3
1
4
−3
1
4 −3
1
= (2)(−2)(4) + (−1)(−2)(−3) + (3)(5)(1) −
(−3)(−2)(3) − (1)(−2)(2) − (4)(5)(−1)
= −1
Example 2:
4
3 10
|
Evaluate −3
2
5 |.
0 −4 −8
Solution:
4
3
|−3
2
0 −4
10
4
3 10 4
3
5 | = |−3
2
5 | −3
2|
−8
0 −4 −8 0 −4
= (4)(2)(−8) + (3)(5)(0) + (10)(−3)(−4) −
(0)(2)(10) − (−4)(5)(4) − (−8)(−3)(3)
= 64
Example 3:
3
|
Evaluate
5
−2
−4
8
6
7|
5 −1
75
Author: Harold Jan R. Terano, ECE, MET
MATRICES AND DETERMINANTS
CHAPTER 6
Solution:
3
| 5
−2
−4
8
3 −4
8 3 −4
6
7| = | 5
6
7| 5
6|
5 −1
−2
5 −1 −2
5
= (3)(6)(−1) + (−4)(7)(−2) + (8)(5)(5) −
(−2)(6)(8) − (5)(7)(3) − (−1)(5)(−4)
= 209
Exercise 6.1
Evaluate the following determinants.
−2
3
|
1 −5
5 −5
|
2. |
−10
4
3 12
|
3. |
−2 −6
13 −4
|
4. |
2
9
2
4
2
5. |−4 −1
5|
2
3 −2
1
3 −6
6. |5 −5
4|
3
0
2
2 −4
8
7. |−3 15 −3|
9
2
6
−3
0
5
8. | 4 −2
9|
−7
3
−1
1. |
6.3 Operations on Matrices
1. Transpose of a Matrix
The transpose is an (𝑛 × 𝑚) matrix formed from the original (𝑚 × 𝑛) matrix by
taking the 𝑖 𝑡ℎ row and making it the 𝑖 𝑡ℎ column. The diagonal is unchanged in this
operation. The transpose of a matrix 𝐴 is symbolized by 𝐴𝑇 .
Example 1:
Determine the transpose of 𝐴 = [
2 1
].
−4 3
Solution:
𝐴𝑇 = [
2
1
−4
]
3
76
Author: Harold Jan R. Teran
CHAPTER 6
MATRICES AND DETERMINANTS
Example 2:
3 −1
0
Determine the transpose of 𝐴 = [−4
1 −2].
5
6 −2
Solution:
3 −4
5
𝐴𝑇 = [−1
1
6]
0 −2 −2
2. The Cofactor of an Entry in a Matrix
The cofactor of an entry in a matrix is the determinant of the matrix formed by
omitting the entry’s row and column in the original matrix. The sign of the cofactor is
determined from the following positional matrices:
+
[−
+
−
+
−
+
−]
+
or the sign of the cofactor can be determined by the relation (−1)𝑖+𝑗 , where 𝑖 is the
row number and 𝑗 is the column number.
Example 1:
−1
2 4
Find the cofactor of −3 in the matrix [−3 −4 5].
2
6 3
Solution:
−3 is at row 2, column 1. The resulting matrix is,
(−1)2+1 [2
6
4
]
3
The cofactor is:
2
−1 |
6
4
| = −[(2)(3) − (6)(4)] = 18
3
Example 2:
4 −2
7
Find the cofactor of 5 in the matrix [−1
5
1 ].
2
6 −5
Solution:
5 is at row 2, column 2. The resulting matrix is,
(−1)2+2 [4
2
The cofactor is:
4
1|
2
7
]
−5
7
| = (4)(−5) − (2)(7) = −34
−5
77
Author: Harold Jan R. Terano, ECE, MET
MATRICES AND DETERMINANTS
CHAPTER 6
3. Classical Adjoint
The classical adjoint is a matrix formed from the transposed cofactor matrix
with the conventional sign arrangement. The resulting matrix is represented as 𝐴𝑎𝑑𝑗 .
Example 1:
3 −1
2
Determine the classical adjoint of the matrix 𝐴 = [ 5
1 −4].
−3
4
6
Solution:
1 −4
| = (1)(6) − (4)(−4) = 22
The cofactor of 3 is + |
4
6
5 −4
| = −[(5)(6) − (−3)(−4)] = −18
The cofactor of −1 is − |
−3
6
5 1
| = (5)(4) − (−3)(1) = 23
The cofactor of 2 is + |
−3 4
−1 2
| = −[(−1)(6) − (4)(2)] = 14
The cofactor of 5 is − |
4 6
3 2
| = (3)(6) − (−3)(2) = 24
The cofactor of 1 is + |
−3 6
3 −1
| = −[(3)(4) − (−3)(−1) = −9
The cofactor of −4 is − |
−3
4
−1
2
| = (−1)(−4) − (1)(2) = 2
The cofactor of −3 is + |
1 −4
3
2
| = −[3(−4) − (5)(2)] = 22
The cofactor of 4 is − |
5 −4
3 −1
| = (3)(1) − (5)(−1) = 8
The cofactor of 6 is + |
5
1
The matrix of the cofactors is,
22 −18
[14
24
2
22
23
−9]
8
By taking the transpose of the above matrix, the classical adjoint is,
22 14
2
𝐴𝑎𝑑𝑗 = [−18 24 22]
23 −9 8
4. The Inverse Matrix
The inverse of a matrix 𝐴 is symbolized as 𝐴−1 and is given by,
𝑨𝒂𝒅𝒋
𝑨−𝟏 = |𝑨|
where 𝐴𝑎𝑑𝑗 is the classical adjoint of the matrix 𝐴 and |𝐴| is the determinant of the
matrix 𝐴.
78
Author: Harold Jan R. Teran
CHAPTER 6
MATRICES AND DETERMINANTS
Example 1:
3 −1
2
Find the inverse of the matrix 𝐴 = [ 5
1 −4].
−3
4
6
Solution:
The classical adjoint of the matrix 𝐴 (from the previous example) is
22 14
2
𝐴𝑎𝑑𝑗 = [−18 24 22]
23 −9 8
The determinant of the matrix 𝐴 is
3 −1
2 3 −1
|𝐴 | = | 5
1 −4| 5
1|
−3
4
6 −3
4
|𝐴| = (3)(1)(6) + (−1)(−4)(−3) + (2)(5)(4) − (−3)(1)(2) −
(4)(−4)(3) − (6)(5)(−1)
|𝐴| = 130
The inverse of the matrix 𝐴 is,
𝐴−1 =
22
[−18
23
14
24
−9
130
2
22]
8
11
7
1
65
9
65
12
65
11
23
65
9
65
4
= − 65
[
130
− 130
65]
Example 2:
0 2 1
Find the inverse of the matrix 𝐴 = [4 1 2].
6 2 4
Solution:
Starting from the first row:
1
The cofactor of 0 is + |
2
4
The cofactor of 2 is − |
6
4
The cofactor of 1 is + |
6
2
The cofactor of 4 is − |
2
0
The cofactor of 1 is + |
6
2
| = (1)(4) − (2)(2) = 0
4
2
| = −[(4)(4) − (6)(2)] = −4
4
1
| = (4)(2) − (6)(1) = 2
2
1
| = −[(2)(4) − (2)(1)] = −6
4
1
| = (0)(4) − (6)(1) = −6
4
The cofactor of 2 is − |0 2| = −[(0)(2) − (6)(2)] = 12
6 2
The cofactor of 6 is + |2 1| = (2)(2) − (1)(1) = 3
1 2
The cofactor of 2 is − |0 1| = −[(0)(2) − (4)(1)] = 4
4 2
79
Author: Harold Jan R. Terano, ECE, MET
MATRICES AND DETERMINANTS
CHAPTER 6
The cofactor of 4 is + |0 2| = (0)(1) − (4)(2) = −8
4 1
The matrix of the cofactors is,
0 −4
2
[−6 −6 12 ]
3
4 −8
The classical adjoint is,
0 −6
3
𝐴𝑎𝑑𝑗 = [−4 −6
4]
2 12 −8
The determinant of 𝐴 is,
0 2 1 0 2
|𝐴| = |4 1 2| 4 1|
6 2 4 6 2
|𝐴| = (0)(1)(4) + (2)(2)(6) + (1)(4)(2) − (6)(1)(1) −
(2)(2)(0) − (4)(4)(2)
|𝐴| = −6
The inverse of the matrix is,
0 −6
3
[−4 −6
4]
2 12 −8
−1
𝐴 =
−6
0
2
𝐴−1 =
3
1
[− 3
1
1
−2
1
−3
−2
2
4
3]
Exercise 6.2
Solve the following problems.
−1
2
], find,
1. Given 𝐴 = [
4 −5
a.) 𝐴𝑇
b.) 𝐴𝑎𝑑𝑗
c.) 𝐴−1
4
2. Given 𝐴 = [3
4
2 −1
−5
6 ], find,
−7
2
a.) cofactor of −5
d.) 𝐴𝑇
b.) cofactor of −1
e.) 𝐴𝑎𝑑𝑗
c.) cofactor of 2 on the 3rd column
f.) 𝐴−1
80
Author: Harold Jan R. Teran
CHAPTER 6
6.4
MATRICES AND DETERMINANTS
Algebraic Operations of Matrix
1. Equality of Matrices
Two matrices are equal if they have the same number of rows and columns and
their corresponding entries are also equal.
2. Addition and Subtraction of Matrices
Addition or subtraction of two matrices can be accomplished by adding or
subtracting the corresponding entries of two matrices which have the same shape.
Example 1:
If 𝐴 = [
−2 4
4 0
−2
1
] and 𝐵 = [
−3
3
−4 2
], find 𝐴 + 𝐵.
5 7
Solution:
−2
4
−1
𝐴+𝐵 =[
7
𝐴+𝐵 =[
4
0
0
5
−2 + 1 4 − 4 −2 + 2
−2
1 −4 2
]+[
]=[
]
4 + 3 0 + 5 −3 + 7
−3
3
5 7
0
]
4
Example 2:
−1
4 2
6 3 −2
If 𝐴 = [ 0 −3 5] and 𝐵 = [ 4 5
4 ], find 𝐴 + 𝐵.
2
8 2
−6 8
1
Solution:
−1 + 6
−1
4 2
6 3 −2
𝐴 + 𝐵 = [ 0 −3 5] + [ 4 5
4] = [ 0 + 4
2
8 2
−6 8
1
2−6
5 7 0
𝐴 + 𝐵 = [ 4 2 9]
−4 16 3
4+3 2−2
−3 + 5 5 + 4]
8+8 2+1
Example 3:
−2
5 10
−1 0
2
If 𝐴 = [ 4 −3 2 ] and 𝐵 = [−4 2 −4], find 𝐴 − 𝐵.
5 6 −3
−5 −4 1
Solution:
−2
5 10
−1 0
2
𝐴 − 𝐵 = [ 4 −3 2 ] − [−4 2 −4]
5 6 −3
−5 −4 1
−2 − (−1)
= [ 4 − (−4)
−5 − 5
−1
5
𝐴−𝐵 =[ 8
−5
−10 −10
5−0
−3 − 2
−4 − 6
10 − 2
2 − (−4) ]
1 − (−3)
8
6]
4
81
Author: Harold Jan R. Terano, ECE, MET
MATRICES AND DETERMINANTS
CHAPTER 6
3. Multiplication of Matrices
Multiplying a matrix by a scalar (real number) results in every entry in the matrix
being multiplied by the scalar.
Example 1
−2 3 −4
Multiply [ 6 5
0 ] by −2.
−8 3 −3
Solution:
(−2)(−2) (−2)(3)
−2 3 −4
(−2)(5)
−2 [ 6 5
0 ] = [ (−2)(6)
−8 3 −3
(−2)(−8) (−2)(3)
4
= [−12
16
(−2)(−4)
(−2)(0) ]
(−2)(−3)
−6 8
−10 0]
−. 6 6
Example 2:
4 −6
Simplify 3 [−2 −2
5
4
5
5 −4
−3 ] − 2 [−2 −1
1
1
7
−3
5].
−4
Solution:
4 −6
3 [−2 −2
5
4
5
5 −4 −3
12 −18
−3 ] − 2 [−2 −1
5] = [−6 −6
1
1
7 −4
15
12
12 − 10 −18 + 8 15 + 6
= [ −6 + 4 −6 + 2 −9 − 10]
15 − 2 12 − 14
3+8
2 −10
21
= [−2 −4 −19]
13 −2
11
15
−10
8
2
−9] + [ 4
−2 −14
3
6
−10]
8
Multiplication of two matrices can be done only if the number of columns of the
left-hand matrix is equal to the number of rows of the right-hand matrix. Multiplication is
accomplished by multiplying the elements in each right-hand matrix column, adding the
products and then placing the sum at the intersection point of the involved row and
column.
Example 1:
If 𝐴 = [
3
5
−2
−2
5
] and 𝐵 = [
], find 𝐴 × 𝐵.
1
6 −4
Solution:
3 −2 −2
5
][
]
5
1
6 −4
(3)(−2) + (−2)(6)
𝐴×𝐵 =[
(5)(−2) + (1)(6)
𝐴×𝐵 =[
(3)(5) + (−2)(−4)
]
(5)(5) + (1)(−4)
82
Author: Harold Jan R. Teran
CHAPTER 6
MATRICES AND DETERMINANTS
𝐴×𝐵 =[
−18 23
]
−4 21
Example 2:
3
If 𝐴 = [
4
2
1 2
] and 𝐵 = [5], find 𝐴 × 𝐵.
−2 3
1
Solution:
2
3
1 2
] [5]
4 −2 3
1
(3)(2) + (1)(5) + (2)(1)
]
𝐴×𝐵 =[
(4)(2) + (−2)(5) + (3)(1)
𝐴×𝐵 =[
𝐴×𝐵 =[
13
]
1
Example 3:
−1 3
If 𝐴 = [ 5 0
6 4
2
2
−2] and 𝐵 = [−5
7
1
6
1
4 −2], find 𝐴 × 𝐵.
−3
2
Solution:
−1 3
2
2
6
1
𝐴 × 𝐵 = [ 5 0 −2] [−5
4 −2]
6 4
7
1 −3
2
𝐴×𝐵 =
(−1)(2) + (3)(−5) + (2)(1)
[ (5)(2) + (0)(−5) + (−2)(1)
(6)(2) + (4)(−5) + (7)(1)
(−1)(6) + (3)(4) + (2)(−3)
(5)(6) + (0)(4) + (−2)(−3)
(6)(6) + (4)(4) + (7)(−3)
−15 0
𝐴 × 𝐵 = [ 8 36
−1 31
(−1)(1) + (3)(−2) + (2)(2)
(5)(1) + (0)(−2) + (−2)(2)]
(6)(1) + (4)(−2) + (7)(2)
−3
1]
12
4. Division of Matrices
Division of matrices can be accomplished only by multiplying the inverse of the
denominator matrix.
Example 1:
1
If 𝐴 = |1
4
2
4
2 −3 1
3 −2| and 𝐵 = | 0
4 2|, find 𝐴/𝐵.
0
3
−3 −2 4
Solution:
𝐴/𝐵 = 𝐴 × 𝐵−1
𝐵𝑎𝑑𝑗
Find for 𝐵−1 . Since 𝐵−1 = |𝐵|
The classical adjoint of matrix 𝐵 is
83
Author: Harold Jan R. Terano, ECE, MET
MATRICES AND DETERMINANTS
CHAPTER 6
20 10
𝐵𝑎𝑑𝑗 = |−6 11
12 13
−10
−4 |
8
The determinant of matrix 𝐵 is
|𝐵| = 70
Therefore,
2
1
7
3
7
11
6
70
13
35
70
𝐵−1 = ||− 35
1
−7
2
− 35||
4
35
Then,
1
𝐴/𝐵 = 𝐴 × 𝐵−1 = |1
4
2
1
7
3
2
4
3 −2| × ||− 35
0
3
6
7
11
35
70
4
6
1
5
11
5
17
5
19 |
58
70
79
35
70
𝐴/𝐵 = ||− 35
70
13
1
−7
2
− 35||
4
35
− 35|
8
− 35
Exercise 6.3
Solve the following matrices.
−2
0 4
−3
2 −5
[
]
[
1. Given 𝐴 = −3 −1 4 and 𝐵 = 8
0
3], find,
6
2 5
4 −6
7
a.) 𝐴 + 𝐵
d.) −2𝐴 + 3𝐵.
b.) 𝐴 − 𝐵
e.) 𝐴 × 𝐵
c.) 𝐴 − 4𝐵
f.) 𝐴/𝐵
−3 2 4
3
2. Evaluate [ 5 2 0] × [−4].
−1 1 3
2
4 −10 12
1 −4
3. Evaluate [−2
0
8 ] × [ 3 −2].
5 −6
7
−5
1
−2
3 6
2 3 −1
4. Divide [ 1
0 4] by [ 0 4 −5].
5 −4 3
−2 6
6
84
Author: Harold Jan R. Teran
CHAPTER 6
6.5
MATRICES AND DETERMINANTS
Expansion of Minors
The minor of an element in a square matrix can be obtained by deleting the row
and column in which that certain element appears, then the minor is determined by
taking the determinant of the resulting matrix.
2 −3 1
Consider the matrix 𝐴 = [ 5
1 4].
−2
3 6
Consider the elements of the first row: 2, −3 and 1. The minor of 2 is
2 −3 1
[ 5
1 4].
−2
3 6
1 4
|
| = (1)(6) − (3)(4) = −6
3 6
The minor of −3 is
2 −3 1
[ 5
1 4]
−2
3 6
5 4
|
| = (5)(6) − (−2)(4) = 38
−2 6
The minor of 1 is
2 −3 1
[ 5
1 4]
−2
3 6
5 1
|
| = (5)(3) − (−2)(1) = 17
−2 3
The cofactor of an entry in a matrix is the determinant of the matrix formed by
omitting the entry’s row and column in the original matrix. The sign of the cofactor is
determined from the following positional matrices:
+
[−
+
−
+
−
+
−]
+
or the sign of the cofactor can be determined by the relation (−1)𝑖+𝑗 , where 𝑖 is the
row number and 𝑗 is the column number.
Suppose the matrix in the above illustration. Consider the elements of the first
row: 2, −3 and 1. The cofactor of 2 is
(−1)1+1 |1
3
4
1
| = +|
6
3
4
| = (1)(6) − (3)(4) = −6
6
The cofactor of −3 is
(−1)1+2 | 5 4| = − | 5 4| = −[(5)(6) − (−2)(4)] = −38
−2 6
−2 6
85
Author: Harold Jan R. Terano, ECE, MET
MATRICES AND DETERMINANTS
CHAPTER 6
The cofactor of 1 is
(−1)1+3 | 5 1| = + | 5 1| = (5)(3) − (−2)(1) = 17
−2 3
−2 3
Consider the matrix
𝑎1
[
𝐴 = 𝑎2
𝑎3
𝑏1
𝑏2
𝑏3
𝑐1
𝑐2]
𝑐3
The product of 𝑎1 and its cofactor is
𝑏
𝑎1 ∙ (−1)1+1 | 2
𝑏3
𝑐2
𝑏
| = 𝑎1 | 2
𝑐3
𝑏3
The product of 𝑏1 and its cofactor is
𝑎2 𝑐2
𝑎2
𝑏1 ∙ (−1)1+2 |𝑎 𝑐 | = −𝑏1 |𝑎
3
3
3
𝑐2
|
𝑐3
𝑐2
𝑐3 |
The product of 𝑐1 and its cofactor is
𝑎
𝑐1 ∙ (−1)1+3 | 2
𝑎3
𝑏2
𝑎
| = 𝑐1 | 2
𝑏3
𝑎3
𝑏2
|
𝑏3
Thus, the determinant of 𝐴 is
𝑎1
|𝑎2
𝑎3
𝑏1
𝑏2
𝑏3
𝑐1
𝑏
𝑐2 | = 𝑎1 | 2
𝑏3
𝑐3
𝑎2
𝑐2
| −𝑏1 |𝑎
𝑐3
3
𝑐2
𝑎2
𝑐3| + 𝑐1 |𝑎3
𝑏2
|
𝑏3
Therefore, the third-order determinant is reduced to a second-order determinant.
Also, higher-order determinant is reducible to a lower-order determinant by using
the expansion of minors.
Example 1:
−3
8 2
Evaluate | 3 −2 0| by using the expansion of minors.
2
5 6
Solution:
Considering the elements of the first row,
−3
8 2
−2 0
3 0
3 −2
| 3 −2 0| = −3(−1)1+1 |
| + 8(−1)1+2 |
| + 2(−1)1+3 |
|
5 6
2 6
2
5
2
5 6
−2 0
3 0
3 −2
| − 8|
|+2|
|
= −3 |
5 6
2 6
2
5
= −3[(−2)(6) − (5)(0)] − 8[(3)(6) − (2)(0)] +
2[(3)(5) − (2)(−2)]
= −3(−12) − 8(18) + 2(19)
= −70
86
Author: Harold Jan R. Teran
CHAPTER 6
MATRICES AND DETERMINANTS
Example 2:
3 −4 −6
Evaluate |−4
1
8 | by using the expansion of minors considering the
7
5
2
elements of the last row.
Solution:
Considering the elements of the last row,
3 −4 −6
−4 −6
3 −6
3
|−4
| + 5(−1)3+2 |
| + 2(−1)3+3 |
1
8 | = 7(−1)3+1 |
1
8
−4
8
−4
7
5
2
−4 −6
3 −6
3 −4
| − 5|
| + 2|
|
= 7|
1
8
−4
8
−4
1
= 7[(−4)(8) − (1)(−6)] − 5[(3)(8) − (−4)(−6)] +
−4
|
1
2[(3)(1) − (−4)(−4)]
= 7(−26) − 5(0) + 2(−13)
= −208
Example 3:
−3
2 1 −1
Evaluate | 2 −1 5 0 | by using the expansion of minors.
4
5 3 6
−1 −3 4 −3
Solution:
Considering the elements of the first row,
−3
2 1 −1
−1 5
0
2 5
0
2
−1
5
0
1+1
1+2
|
| = −3(−1)
| 5 3
| 4 3
6 | + 2(−1)
6|
4
5 3
6
−3
4
−3
−1
4
−3
−1 −3 4 −3
2 −1
0
2 −1 5
1+3
1+4
| 4
| 4
+1(−1)
5
6 | − 1(−1)
5 3|
−1 −3 −3
−1 −3 4
−1 5
0
2 5
0
2 −1
0
2 −1 5
= −3 | 5 3
5
6| + | 4
6| − 2| 4 3
6| + | 4
5 3|
−1 −3 −3
−3 4 −3
−1 4 −3
−1 −3 4
Then, again, consider the elements of the first row,
3
6
5
6
5 3
3
6
4
6
| − 5|
| + 0|
|} − 2 {2 |
| − 5|
|+
= −3 {−1 |
4 −3
4 −3
−1 −3
−3 −3
−3 4
4 3
5
6
4
6
4
5
5 3
4 3
4
5
|} + 2 |
|+|
|+0|
| + 2|
|+|
|+5|
|
0|
−1 4
−1 −3
−1 4
−3 −3
−1 −3
−3 4
−1 −3
= −3{33 − 15 + 0} − 2{−66 + 30 + 0} + 6 − 6 + 0 + 58 + 19 − 35
= −54 + 72 + 6 − 6 + 58 + 19 − 35
= 60
87
Author: Harold Jan R. Terano, ECE, MET
MATRICES AND DETERMINANTS
CHAPTER 6
Example 4:
2 −2 −1
3
5
6
0
−1
|.
Evaluate |
4 −2
5
7
1
3 −3 −4
Solution:
Apply expansion of minors to reduce the fourth-order determinant into a thirdorder determinant and then solve.
Considering the elements of the first row,
2
|5
4
1
−2
6
−2
3
−1
3
6 0 −1
5 0 −1
0 −1| = 2(−1)1+1 |−2 5
2+1
(
)
|4 5
7 | − 2 −1
7|
5
7
3
−3
−4
1
−3
−4
−3 −4
5
6 −1
5
6
0
3+1
4+1
|4 −2
|4 −2
−1(−1)
7 | + 3(−1)
5|
1
3 −4
1
3 −3
6
0 −1
5
0 −1
5
6 −1
5
6
0
= 2 |−2
5
7 | + 2 |4
5
7 | − |4 −2
7 | − 3 |4 −2
5|
3 −3 −4
1 −3 −4
1
3 −4
1
3 −3
6
0 −1 6
0
5
0 −1 5
0
= 2 |−2
5
7 | −2
5 | + 2 |4
5
7| 4
5|
3 −3 −4 3 −3
1 −3 −4 1 −3
5
6 −1 5
6
5
6
0 5
6
|
|
|
|
− |4 −2
3
7 4 −2
4 −2
5 4 −2|
1
3 −4 1
3
1
3 −3 1
3
= 2[(6)(5)(−4) + (0)(7)(3) + (−1)(−2)(−3) − (3)(5)(−1) − (−3)(7)(6) +
(−4)(−2)(0)] + 2[(5)(5)(−4) + (0)(7)(1) + (−1)(4)(−3) − (1)(5)(−1) − (−3)(7)(5) −
(−4)(4)(0)] − [(5)(−2)(−4) + (6)(7)(1) + (−1)(4)(3) − (1)(−2)(−1) − (3)(7)(5) −
(−4)(4)(6)] − 3[(5)(−2)(−3) + (6)(5)(1) + (0)(4)(3) − (1)(−2)(0) − (3)(5)(5) −
(−3)(4)(6)]
= 2(15) + 2(22) − 59 − 3(57)
= −156
Exercise 6.4
Evaluate the following determinants.
5
1. | 1
−3
−3
2. | 1
3
9
3. |−1
8
−4
−2
4
2
−4
6
−3
−7
5
6
1|
8
0
−5|
7
1
−8|
3
88
Author: Harold Jan R. Teran
CHAPTER 6
MATRICES AND DETERMINANTS
−1
3
5 −3
2|
7 −1
4. | 6
−2 −5 −5 −4
2
−3
2 −4
3 −6 −2
1
−2
1
−4
−4
|
5. |
0 −3
3
5
5
1 −8 −7
6.6
Pivotal Element Method
Consider the determinant,
𝑎1
|𝑎2
𝑎3
𝑏1
𝑏2
𝑏3
𝑐1
𝑐2 |
𝑐3
By pivotal element method, choose pivot point (element that is unity), say 𝑎1
(where 𝑎1 = 1), the reduced determinant is a second order defined by,
𝑎1
|𝑎2
𝑎3
𝑏1
𝑏2
𝑏3
𝑐1
𝑏 − 𝑎2 𝑏1
𝑐2 | = (−1)1+1 | 2
𝑏3 − 𝑎3 𝑏1
𝑐3
𝑐2 − 𝑎2 𝑏1
|
𝑐3 − 𝑎2 𝑏1
Also for a 4th order determinant, given,
𝑎1
𝑎
| 2
𝑎3
𝑎4
𝑏1
𝑏2
𝑏3
𝑏4
𝑐1
𝑐2
𝑐3
𝑐4
𝑑1
𝑑2
|
𝑑3
𝑑4
Say 𝑎1 = 1, the reduced determinant is a third order defined by,
𝑎1
𝑎
| 2
𝑎3
𝑎4
𝑏1
𝑏2
𝑏3
𝑏4
𝑐1
𝑐2
𝑐3
𝑐4
𝑑1
𝑏2 − 𝑎2 𝑏1
𝑑2
1+1
| = (1)
|𝑏3 − 𝑎3 𝑏1
𝑑3
𝑏4 − 𝑎4 𝑏1
𝑑4
𝑐2 − 𝑎2 𝑐1
𝑐3 − 𝑎3 𝑐1
𝑐4 − 𝑎4 𝑐1
𝑑2 − 𝑎2 𝑑1
𝑑3 − 𝑎3 𝑑1 |
𝑑4 − 𝑎4 𝑑1
Example 1:
3 4 −1
Evaluate |−2 1
3 | by pivotal element method.
5 2
6
Solution:
Consider the element on the second row, second column since it is unity, thus,
3 4 −1
3 − (4)(−2) −1 − (4)(3)
|−2 1
|
3 | = (−1)2+2 |
5 − (2)(−2)
6 − (2)(3)
5 2
6
11 −13
| = (11)(0) − (9)(−13) = 117
= +1 |
9
0
89
Author: Harold Jan R. Terano, ECE, MET
MATRICES AND DETERMINANTS
CHAPTER 6
Example 2:
−2
3 −3 −4
5
4
0
2 | by pivotal element method.
Evaluate |
8
1 −8 −1
−4 −5
7 −9
Solution:
Consider the element on the third row, second column since it is unity, thus,
−2
3 −3 −4
−2 − (3)(8)
−3 − (3)(−8)
4
0
2 | = (−1)3+2 | 5 − (4)(8)
0 − (4)(−8)
| 5
8
1 −8 −1
−4 − (−5)(8)
7 − (−5)(−8)
−4 −5
7 −9
−26
21
−1
= −1 |−27
32
6|
36 −33 −14
−26
21
−1 −26
21
|
= −1 |−27
32
6 −27
32 |
36 −33 −14 36 −33
−4 − (3)(−1)
2 − (4)(−1) |
−9 − (−5)(−1)
= −1[(11,648 + 4,536 − 891) − (−1,152 + 5,148 + 7,938)]
= −3,359
Example 3:
2 −5
0 −10
2 −4 −2 | by pivotal element method.
Evaluate | −3
3
2 −6 −5
3 −2
8 −8
Solution:
There is no unity element. Consider the element on the first row, first column and
that is 2.
1
2
1
2
Multiply the determinant by ( ). This becomes,
2 −5
2
| −3
2 −6
3 −2
0 −10
2 −5
0 −10
1
−4 −2 | = ( 2 ) | −3
2 −4 −2 |
1
3
3
−5
2
−6
−5
2
8 −8
3 −2
8 −8
1 −5/2
0 −5
−3
2
−4
−2|
= 2|
3
2
−6
−5
3 −2
8 −8
We can now choose the element on the first row, first column since it is now
unity. Thus,
1 −5/2
0 −5
2 − (−3)(−5/2)
−2
−3
2
−4
−6
− (2)(−5/2)
2|
| = 2|
3
2
−6
−5
−2 − (3)(−5/2)
3 −2
8 −8
90
Author: Harold Jan R. Teran
−4 − (−3)(0)
−5 − (2)(0)
8 − (3)(0)
−2 − (−3)(−5)
3 − (2)(−5) |
−8 − (3)(−5)
CHAPTER 6
MATRICES AND DETERMINANTS
−11/2
= 2 | −1
11/2
−11/2
= 2 | −1
11/2
385
= 2 [(
2
−4 −17
−5
13 |
8
7
−4 −17 −11/2
−5
13 | −1
8
7 11/2
935
− 286 + 136) − (
2
−4
−5|
8
− 572 + 28)]
= 238
Exercise 6.5
Evaluate the following determinants using pivotal element method.
−1
1. |−3
2
4
2. |−2
−2
1
3. |−4
6
−1
10
4. |−2
−4
8
2
5. | −3
0
6
6.7
1
2
−7
3|
6 −2
−2 −4
−3 −2|
8
4
−2
5
5
−1 −3
4|
−3 −6 −3
2 −4
9
−2
5
6
5 −1
2|
−2
3 −3
4
5 −5
5 −3
3
−1 −9 −5 |
−2
7
7
5 −3 −2
Solutions of Linear Systems using Inverse of a Matrix
In the following example we show how the inverse of a matrix is used to solve a
system of linear equations.
Example 1:
5𝑥 + 3𝑦 = 6
Solve the following system of equations {
.
−2𝑥 − 3𝑦 = 12
Solution:
If
𝐴=[
5
3
]
−2 −3
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Author: Harold Jan R. Terano, ECE, MET
MATRICES AND DETERMINANTS
CHAPTER 6
Then, in matrix form, the system is,
𝑥
6
𝐴 [𝑦] = [ ]
12
Multiply both sides of the equation by 𝐴−1, thus,
𝑥
6
[𝑦] = 𝐴−1 [ ]
12
Then solve for 𝐴−1, since,
𝐴𝑎𝑑𝑗
𝐴−1 = |𝐴|
Solve for the classical adjoint.
The cofactor of 5 is +|−3| = −3
The cofactor of 3 is −|−2| = 2
The cofactor of −2 is −|3| = −3
The cofactor of −3 is |5| = 5
The matrix of the cofactors is,
[
−3 2
]
−3 5
By taking the transpose of the above matrix, the classical adjoint is,
𝐴𝑎𝑑𝑗 = [
−3 −3
]
2
5
The determinant of 𝐴 is,
3
|𝐴 | = | 5
| = (5)(−3) − (−2)(3) = −9
−2 −3
Therefore,
𝐴𝑎𝑑𝑗
𝐴−1 = |𝐴|
𝐴
−1
−3 −3
[
]
= 2 5
−9
1
1
3
𝐴−1 = [ 32
5]
−9 −9
Then,
𝑥
6
[𝑦] = 𝐴−1 [ ]
12
1
1
𝑥
6
3
[𝑦] = [ 32
5] [12]
−9 −9
1
1
( ) (6) + (3) (12)
𝑥
[𝑦] = [ 32
]
5
(− 9) (6) + (− 9) (12)
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CHAPTER 6
MATRICES AND DETERMINANTS
𝑥
6
[𝑦 ] = [ ]
−8
The values are 𝑥 = 6 and 𝑦 = −8.
Example 2:
−2𝑥 − 3𝑦 + 5𝑧 = −5
Solve the following system of equations { 2𝑥 − 6𝑦 − 2𝑧 = −4 .
−𝑥 + 6𝑦 − 2𝑧 = 8
Solution:
If
−2 −3
5
𝐴 = [ 2 −6 −2]
−1
6 −2
Then, in matrix form, the system is,
𝑥
−5
𝐴 [𝑦] = [−4]
𝑧
8
Multiply both sides of the equation by 𝐴−1, thus,
𝑥
−5
[𝑦] = 𝐴−1 [−4]
𝑧
8
Then solve for 𝐴−1, since,
𝐴𝑎𝑑𝑗
𝐴−1 = |𝐴|
Solve for the classical adjoint, starting from the first row:
−6 −2
| = (−6)(−2) − (6)(−2) = 24
The cofactor of −2 is + |
6 −2
2 −2
| = −[(2)(−2) − (−1)(−2)] = 6
The cofactor of −3 is − |
−1 −2
2 −6
| = (2)(6) − (−1)(−6) = 6
The cofactor of 5 is + |
−1
6
−3
5
| = −[(−3)(−2) − (6)(5)] = 24
The cofactor of 2 is − |
6 −2
−2
5 | ( )( ) ( )( )
The cofactor of −6 is + |
= −2 −2 − −1 5 = 9
−1 −2
−2 −3
| = −[(−2)(6) − (−1)(−3)] = 15
The cofactor of −2 is − |
−1
6
−3
5
| = (−3)(−2) − (−6)(5) = 36
The cofactor of −1 is + |
−6 −2
−2
5|
The cofactor of 6 is − |
= −[(−2)(−2) − (2)(5)] = 6
2 −2
−2 −3
| = (−2)(−6) + (2)(−3) = 18
The cofactor of −2 is + |
2 −6
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MATRICES AND DETERMINANTS
CHAPTER 6
The matrix of the cofactors is,
24 6
[24 9
36 6
6
15]
18
By taking the transpose of the above matrix, the classical adjoint is,
24 24
𝐴𝑎𝑑𝑗 = [ 6
9
6 15
36
6]
18
The determinant of 𝐴 is,
−2 −3
5 −2 −3
|𝐴| = | 2 −6 −2| 2 −6|
6
−1
6 −2 −1
|𝐴| = (−2)(−6)(−2) + (−3)(−2)(−1) + (5)(2)(6) −
(−1)(−6)(5) − (6)(−2)(−2) − (−2)(2)(−3)
|𝐴| = −36
Therefore,
𝐴𝑎𝑑𝑗
𝐴−1 = |𝐴|
24 24 36
[6
9
6]
6 15 18
−1
𝐴 =
−36
2
2
−3
−3
1
1
𝐴−1 = − 6
1
[− 6
Then,
−1
1
−4
−6
5
1
− 12 − 2]
𝑥
−5
−1
[𝑦] = 𝐴 [−4]
𝑧
8
2
2
−3
−3
−1
𝑥
−5
1
1
1
[𝑦] = − 6 − 4 − 6 [−4]
𝑧
8
1
5
1
[− 6 − 12 − 2]
2
2
(− 3) (−5) + (− 3) (−4) + (−1)(8)
𝑥
1
1
1
[𝑦] = (− ) (−5) + (− ) (−4) + (− ) (8)
6
4
6
𝑧
1
5
1
[(− 6) (−5) + (− 12) (−4) + (− 2) (8)]
−2
𝑥
1
[𝑦] = [ 2 ]
3
𝑧
−2
1
3
The values are 𝑥 = −2, 𝑦 = 2 and 𝑧 = − 2.
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Author: Harold Jan R. Teran
CHAPTER 6
MATRICES AND DETERMINANTS
Exercise 6.6
Solve the following systems of linear equations.
1. {
2𝑥 − 6𝑦 = −3
3𝑥 + 12𝑦 = −8
2. {
2𝑥 + 𝑦 = 4
−5𝑥 − 6𝑦 = 4
3. {
7𝑥 − 3𝑦 = 6
4𝑥 − 8𝑦 = −2
4𝑥 − 3𝑦 − 6𝑧 = −7
4. { −𝑥 + 𝑦 + 3𝑧 = 7
6𝑥 − 5𝑦 − 2𝑧 = 19
3𝑥 − 4𝑦 − 10𝑧 = −3
{
5. 6𝑥 + 8𝑦 − 15𝑧 = 3
3𝑥 + 10𝑦 + 5𝑧 = 7
6.8
Cramer’s Rule
For a system of two linear equations in two unknowns, it may be solve by using
the second-order determinants. Given the system of equations
𝑎1 𝑥 + 𝑏1 𝑦 = 𝑐1
𝑎2 𝑥 + 𝑏2 𝑦 = 𝑐2
The values of 𝑥 and 𝑦 are given by
𝑫
𝑫𝒚
𝒙 = 𝑫𝒙 and 𝒚 = 𝑫
where
𝑎
𝐷=| 1
𝑎2
𝑏1
|
𝑏2
and
𝑐 𝑏1
|
𝐷𝑥 = | 1
𝑐2 𝑏2
𝑎1 𝑐1
𝐷𝑦 = |𝑎 𝑐 |
2
2
Example 1:
𝑥 + 2𝑦 = −5
Solve the system {
.
𝑥 − 2𝑦 = 3
Solution:
The denominator 𝐷 is,
𝐷=|
1
2
| = (1)(−2) − (1)(2) = −4
1 −2
For 𝐷𝑥 ,
−5
2
| = (−5)(−2) − (3)(2) = 4
𝐷𝑥 = |
3 −2
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Author: Harold Jan R. Terano, ECE, MET
MATRICES AND DETERMINANTS
CHAPTER 6
For 𝐷𝑦 ,
1 −5
| = (1)(3) − (1)(−5) = 8
𝐷𝑦 = |
1
3
Then,
𝐷
4
𝑥 = 𝐷𝑥 = −4 = −1
𝑦=
𝐷𝑦
𝐷
=
8
−4
= −2
Example 2:
2𝑥 − 3𝑦 = 5
Solve the system {
.
𝑥 + 2𝑦 = 6
Solution:
The denominator 𝐷 is,
𝐷=|
2 −3
| = (2)(2) − (1)(−3) = 7
1
2
For 𝐷𝑥 ,
5 −3
| = (5)(2) − (6)(−3) = 28
𝐷𝑥 = |
6
2
For 𝐷𝑦 ,
2 5
| = (2)(6) − (1)(5) = 7
𝐷𝑦 = |
1 6
Then,
𝐷
28
𝐷𝑦
7
𝑥 = 𝐷𝑥 = 7 = 4
𝑦= 𝐷 =7=1
Example 3:
3𝑥 − 4𝑦 = −3
Solve the system {
.
−5𝑥 + 2𝑦 = −4
Solution:
The denominator 𝐷 is,
3 −4
| = (3)(2) − (−5)(−4) = −14
𝐷=|
−5
2
For 𝐷𝑥 ,
−3 −4
| = (−3)(2) − (−4)(−4) = −22
𝐷𝑥 = |
−4
2
For 𝐷𝑦 ,
3 −3
| = (3)(−4) − (−5)(−3) = −27
𝐷𝑦 = |
−5 −4
Then,
𝐷
−22
11
𝐷𝑦
−27
27
𝑥 = 𝐷𝑥 = −14 = 7
𝑦 = 𝐷 = −14 = 14
96
Author: Harold Jan R. Teran
CHAPTER 6
6.9
MATRICES AND DETERMINANTS
Cramer’s Rule for a System of Three Linear Equations
For a system of three linear equations in three unknowns, it may be solved by
using the third-order determinants. Given the system of equations
𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 𝑧 = 𝑑1
𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐2 𝑧 = 𝑑2
𝑎3 𝑥 + 𝑏3 𝑦 + 𝑐3 𝑧 = 𝑑3
The values of 𝑥, 𝑦 and 𝑧 are given by
𝑫𝒚
𝑫
𝑫
𝒙 = 𝑫𝒙 , 𝒚 = 𝑫 and 𝒛 = 𝑫𝒛
where
𝑎1
𝐷 = |𝑎2
𝑎3
𝑏1
𝑏2
𝑏3
𝑐1
𝑐2 |
𝑐3
𝑑1
𝐷𝑥 = |𝑑2
𝑑3
𝑏1
𝑏2
𝑏3
𝑐1
𝑐2 |
𝑐3
𝑎1
𝐷𝑦 = |𝑎2
𝑎3
𝑑1
𝑑2
𝑑3
𝑐1
𝑐2 |
𝑐3
𝑎1
𝐷𝑧 = |𝑎2
𝑎3
𝑏1
𝑏2
𝑏3
𝑑1
𝑑2 |
𝑑3
and
Example 1:
2𝑥 − 3𝑦 − 𝑧 = 9
Solve the system { 𝑥 + 3𝑦 + 3𝑧 = 5 .
−2𝑥 − 𝑦 + 𝑧 = 3
Solution:
The denominator 𝐷 is,
2 −3 −1
2 −3 −1 2 −3
|
|
𝐷=| 1
=
3
3
1
3
3| 1
3|
−2 −1
1
−2 −1
1 −2 −1
= (2)(3)(1) + (−3)(3)(−2) + (−1)(1)(−1) − (−2)(3)(−1) −
(−1)(3)(2) − (1)(1)(−3)
𝐷 = 28
For 𝐷𝑥 ,
9 −3 −1
9 −3 −1 9 −3
𝐷𝑥 = |5
3
3 | = |5
3
3| 5
3|
3 −1
1
3 −1
1 3 −1
= (9)(3)(1) + (−3)(3)(3) + (−1)(5)(−1) − (3)(3)(−1) −
(−1)(3)(9) − (1)(5)(−3)
𝐷𝑥 = 56
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Author: Harold Jan R. Terano, ECE, MET
MATRICES AND DETERMINANTS
CHAPTER 6
For 𝐷𝑦 ,
2 9 −1
2 9 −1 2 9
|
|
𝐷𝑦 = | 1 5
=
3
1 5
3 | 1 5|
−2 3
1
−2 3
1 −2 3
= (2)(5)(1) + (9)(3)(−2) + (−1)(1)(3) − (−2)(5)(−1) −
(3)(3)(2) − (1)(1)(9)
𝐷𝑦 = −84
For 𝐷𝑧 ,
2 −3 9
2 −3 9 2
𝐷𝑧 = | 1
3 5| = | 1
3 5| 1
−2 −1 3
−2 −1 3 −2
−3
3|
−1
= (2)(3)(3) + (−3)(5)(−2) + (9)(1)(−1) − (−2)(3)(9) −
(−1)(5)(2) − (3)(1)(−3)
𝐷𝑧 = 112
Then,
𝐷
56
𝐷𝑦
−84
𝐷
112
𝑥 = 𝐷𝑥 = 28 = 2
𝑦 = 𝐷 = 28 = −3
𝑧 = 𝐷𝑧 = 28 = 4
Example 2:
−2𝑥 − 3𝑦 + 5𝑧 = −5
Solve the system { 2𝑥 − 6𝑦 − 2𝑧 = −4 .
−𝑥 + 6𝑦 − 2𝑧 = 8
Solution:
The denominator 𝐷 is,
−2 −3
5
−2 −3
5 −2 −3
𝐷 = | 2 −6 −2| = | 2 −6 −2| 2 −6|
6
−1
6 −2
−1
6 −2 −1
= (−2)(−6)(−2) + (−3)(−2)(−1) + (5)(2)(6) −
(−1)(−6)(5) − (6)(−2)(−2) − (−2)(2)(−3)
𝐷 = −36
For 𝐷𝑥 ,
−5 −3
5
−5
𝐷𝑥 = |−4 −6 −2| = |−4
8
6 −2
8
−3
−6
6
5 −5 −3
−2 | −4 −6|
−2 8
6
= (−5)(−6)(−2) + (−3)(−2)(8) + (5)(−4)(6) −
(8)(−6)(5) − (6)(−2)(−5) − (−2)(−4)(−3)
𝐷𝑥 = 72
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CHAPTER 6
MATRICES AND DETERMINANTS
For 𝐷𝑦 ,
−2 −5
5
−2 −5
5 −2 −5
𝐷𝑦 = | 2 −4 −2| = | 2 −4 −2| 2 −4|
−1
8 −2
−1
8 −2 −1
8
= (−2)(−4)(−2) + (−5)(−2)(−1) + (5)(2)(8) −
(−1)(−4)(5) − (8)(−2)(−2) − (−2)(2)(−5)
𝐷𝑦 = −18
For 𝐷𝑧 ,
−2 −3 −5
−2 −3 −5 −2 −3
𝐷𝑧 = | 2 −6 −4| = | 2 −6 −4| 2 −6|
6
−1
6
8
−1
6
8 −1
= (−2)(−6)(8) + (−3)(−4)(−1) + (−5)(2)(6) −
(−1)(−6)(−5) − (6)(−4)(−2) − (8)(2)(−3)
𝐷𝑧 = 54
Then,
𝐷
72
𝐷𝑦
−18
𝐷
54
𝑥 = 𝐷𝑥 = −36 = −2
1
𝑦 = 𝐷 = −36 = 2
3
𝑧 = 𝐷𝑧 = −36 = − 2
Exercise 6.7
Solve the systems of linear equations using Cramer’s rule.
1. {
𝑥 − 2𝑦 = −5
−2𝑥 + 3𝑦 = 7
2. {
−3𝑥 − 5𝑦 = 8
4𝑥 − 2𝑦 = −7
3. {
2𝑥 − 3𝑦 = 6
−6𝑥 + 2𝑦 = 3
5𝑥 − 3𝑦 − 2𝑧 = 14
4. { 2𝑥 + 6𝑦 + 3𝑧 = 0
3𝑥 − 𝑦 − 𝑧 = 6
2𝑥 − 4𝑦 − 3𝑧 = 6
5. { 2𝑥 + 2𝑦 + 𝑧 = −2
−6𝑥 + 9𝑦 − 2𝑧 = 13
.
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Author: Harold Jan R. Terano, ECE, MET
MATRICES AND DETERMINANTS
CHAPTER 6
100
Author: Harold Jan R. Teran
CHAPTER 7
VECTOR ANALYSIS
Chapter 7
VECTOR ANALYSIS
Chapter Outline:
7.1
7.2
7.3
7.4
7.5
7.6
Vector and Scalar
Properties of Vectors
Laws of Vector Algebra
Components of a Vector
Dot Product
Cross Product
Learning Objectives:
At the end of the lesson, the students will be able to:
1. Differentiate between vector and scalar quantities.
2. Define the properties of vectors.
3. Define the laws of vector algebra.
4. Solve dot product of vectors.
5. Solve cross product of vectors.
Overview:
This chapter will discuss vectors and its properties. This includes algebraic
operations of vectors, dot product and cross product of vectors.
101
Author: Harold Jan R. Terano, ECE, MET
VECTOR ANALYSIS
CHAPTER 7
7.1 Vector and Scalar
A vector is a quantity having both magnitude and direction. Examples include
displacement, force, velocity, acceleration, etc.
A vector is represented by an arrow which defines the direction and the length
which indicates the magnitude.
𝐏
̅
𝐀 or 𝐀
𝐎
̅ (other designation for vector). The tail end 𝐎
The figure above is a vector 𝐀 or 𝐀
of the arrow is the origin (initial point) of the vector and the head 𝐏 is the terminal point
of the vector.
A scalar is a quantity having magnitude but no direction. Examples include
mass, length, distance, speed, temperature, etc.
7.2 Properties of Vectors
1. Consider vectors 𝐀 and 𝐁. These vectors are equal if they have the same magnitude
and direction regardless of the positions of their initial points.
𝐀
𝐀 = 𝐁.
𝐁
2. A vector with a direction opposite to that of a vector 𝐀 but having the same magnitude
is denoted by – 𝐀.
𝐀
𝐀 = −𝐀
−𝐀
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CHAPTER 7
VECTOR ANALYSIS
3. Vectors 𝐀 and 𝐁 has a sum or resultant 𝐂 formed by placing the initial point of 𝐁 on
the terminal point of 𝐀 and then joining the initial point of 𝐀 to the terminal point of 𝐁.
The sum is denoted as 𝐂 = 𝐀 + 𝐁.
𝐁
𝐀
𝐀
𝐁
𝐂= 𝐀+𝐁
4. The difference of vectors 𝐀 and 𝐁 is denoted by 𝐀 − 𝐁. It can also be denoted as
𝐀 + (−𝐁). If 𝐀 = 𝐁, then 𝐀 − 𝐁 is defined as a null vector (zero vector).
5. The product of a vector 𝐀 by any scalar 𝐦 is a vector 𝐦𝐀.
7.3 Laws of Vector Algebra
1. 𝐀 + 𝐁 = 𝐁 + 𝐀 (Commutative Law for Addition)
2. 𝐀 + (𝐁 + 𝐂) = (𝐀 + 𝐁) + 𝐂 (Associative Law for Addition)
3. 𝐦𝐀 = 𝐀𝐦 (Commutative Law for Multiplication)
4. 𝐦(𝐧𝐀) = (𝐦𝐧)𝐀 (Associative Law for Multiplication)
5. (𝐦 + 𝐧)𝐀 = 𝐦𝐀 + 𝐧𝐀 (Distributive Law)
6. 𝐦(𝐀 + 𝐁) = 𝐦𝐀 + 𝐦𝐁 (Distributive Law)
Rectangular unit vectors 𝒂𝒙 , 𝒂𝒚 and 𝒂𝒛 are important set of vectors having the
same directions of the positive
𝑥, 𝑦
and
𝑧
of a three-dimensional rectangular
coordinate system.
𝒛
𝒂𝒛
𝒂𝒚
𝒚
𝒂𝒙
𝒙
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Author: Harold Jan R. Terano, ECE, MET
VECTOR ANALYSIS
CHAPTER 7
Unit Vector is a vector having a unit magnitude. Consider 𝐀 as vector and |𝐀|
as the magnitude of the vector, thus,
A
|𝐀|
is a unit vector having the same direction as
𝐀.
Example 1:
Find the unit vector of 𝐀 = 𝟑𝒂𝒙 + 𝟐𝒂𝒚 − 𝟑𝒂𝒛
Solution:
𝐀 = 𝟑𝒂𝒙 + 𝟐𝒂𝒚 − 𝟑𝒂𝒛
|𝐀| = √(𝟑)𝟐 + (𝟐)𝟐 + (−𝟑)𝟐 = √𝟐𝟐
𝟑𝒂𝒙 +𝟐𝒂𝒚 −𝟑𝒂𝒛
𝐀
𝟑√𝟐𝟐
𝟑√𝟐𝟐
√𝟐𝟐
=
= 𝟐𝟐 𝒂𝒙 + 𝟏𝟏 𝒂𝒚 − 𝟐𝟐 𝒂𝒛
|𝐀|
√𝟐𝟐
To check:
𝟑√𝟐𝟐
|
𝟐𝟐
𝒂𝒙 +
𝟐
𝟐
𝟐
𝟑√𝟐𝟐
𝟑√𝟐𝟐
𝟑√𝟐𝟐
√𝟐𝟐
√𝟐𝟐
𝒂𝒚 −
𝒂𝒛 | = √(
) + ( ) + (−
) =𝟏
𝟏𝟏
𝟐𝟐
𝟐𝟐
𝟏𝟏
𝟐𝟐
7.4 Components of a Vector
A vector 𝐀 in a three-dimension can be represented as
𝐀 = 𝐀 𝟏 𝒂𝒙 + 𝐀 𝟐 𝒂𝒚 + 𝐀 𝟑 𝒂𝒛. The vectors 𝐀 𝟏 𝒂𝒙 , 𝐀 𝟐 𝒂𝒚 and 𝐀 𝟑𝒂𝒛 are called rectangular
components of vectors.
The sum or resultant of 𝐀 𝟏 𝒂𝒙 , 𝐀 𝟐 𝒂𝒚 and 𝐀 𝟑𝒂𝒛 is the vector 𝐀 = 𝐀 𝟏 𝒂𝒙 + 𝐀 𝟐 𝒂𝒚 +
𝐀 𝟑 𝒂𝒛. The magnitude of 𝐀 is
𝐴 = |𝐀 | = √ ( 𝐀 𝟏 ) 𝟐 + ( 𝐀 𝟐 ) 𝟐 + ( 𝐀 𝟑 ) 𝟐
Example 1:
Given 𝐀 = 𝟐𝒂𝒙 − 𝟑𝒂𝒚 + 𝟒𝒂𝒛 , 𝐁 = 𝟓𝒂𝒙 − 𝟐𝒂𝒚 + 𝟑𝒂𝒛 and
𝐂 = −𝟑𝒂𝒙 + 𝟓𝒂𝒚 + 𝟒𝒂𝒛 . Find:
a. 𝐀 + 𝟐𝐁 − 𝐂
b. |𝐀|
c. |𝟐𝐀 − 𝟑𝐁 + 𝐂|
Solution:
a. 𝐀 + 𝟐𝐁 − 𝐂 = (𝟐𝒂𝒙 − 𝟑𝒂𝒚 + 𝟒𝒂𝒛 ) + 𝟐(𝟓𝒂𝒙 − 𝟐𝒂𝒚 + 𝟑𝒂𝒛 ) − (−𝟑𝒂𝒙 + 𝟓𝒂𝒚 +
𝟒𝒂𝒛 )
= 𝟐𝒂𝒙 − 𝟑𝒂𝒚 + 𝟒𝒂𝒛 + 𝟏𝟎𝒂𝒙 − 𝟒𝒂𝒚 + 𝟔𝒂𝒛 + 𝟑𝒂𝒙 − 𝟓𝒂𝒚 − 𝟒𝒂𝒛
= 𝟏𝟓𝒂𝒙 − 𝟏𝟐𝒂𝒚 + 𝟔𝒂𝒛
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b. |𝐀| = |𝟐𝒂𝒙 − 𝟑𝒂𝒚 + 𝟒𝒂𝒛 | = √(𝟐)𝟐 + (−𝟑)𝟐 + (𝟒)𝟐 = √𝟐𝟗
c. |𝟐𝐀 − 𝟑𝐁 + 𝐂| = |𝟐(𝟐𝒂𝒙 − 𝟑𝒂𝒚 + 𝟒𝒂𝒛) − 𝟑(𝟓𝒂𝒙 − 𝟐𝒂𝒚 + 𝟑𝒂𝒛 ) + −𝟑𝒂𝒙 + 𝟓𝒂𝒚 +
𝟒𝒂𝒛 |
= |𝟒𝒂𝒙 − 𝟔𝒂𝒚 + 𝟖𝒂𝒛 − 𝟏𝟓𝒂𝒙 + 𝟔𝒂𝒚 − 𝟗𝒂𝒛 − 𝟑𝒂𝒙 + 𝟓𝒂𝒚 + 𝟒𝒂𝒛 |
= |−𝟏𝟒𝒂𝒙 + 𝟓𝒂𝒚 + 𝟑𝒂𝒛 |
= √(−𝟏𝟒)𝟐 + (𝟓)𝟐 + (𝟑)𝟐
= √𝟐𝟑𝟎
Example 2:
If 𝐀 = 𝟑𝒂𝒙 − 𝟐𝒂𝒚 + 𝟑𝒂𝒛, 𝐁 = −𝒂𝒙 + 𝟐𝒂𝒚 + 𝟓𝒂𝒛, 𝐂 = 𝟑𝒂𝒙 + 𝟐𝒂𝒚 + 𝟐𝒂𝒛 and
𝐃=
𝟑𝒂𝒙 − 𝒂𝒚 − 𝟐𝒂𝒛 , find scalars 𝑎, 𝑏 and 𝑐 such that 𝐃 = 𝐀𝒂 + 𝐁𝒃 + 𝐂𝒄.
Solution:
𝟑𝒂𝒙 − 𝒂𝒚 − 𝟐𝒂𝒛 = 𝒂(𝟑𝒂𝒙 − 𝟐𝒂𝒚 + 𝟑𝒂𝒛 ) + 𝒃(−𝒂𝒙 + 𝟐𝒂𝒚 + 𝟓𝒂𝒛 )
+𝒄(𝟑𝒂𝒙 + 𝟐𝒂𝒚 + 𝟐𝒂𝒛 )
𝟑𝒂𝒙 − 𝒂𝒚 − 𝟐𝒂𝒛 = (𝟑𝒂 − 𝒃 + 𝟑𝒄)𝒂𝒙 + (−𝟐𝒂 + 𝟐𝒃 + 𝟐𝒄)𝒂𝒚
+(𝟑𝒂 + 𝟓𝒃 + 𝟐𝒄)𝒂𝒛
Then,
𝟑𝒂𝒙 = (𝟑𝒂 − 𝒃 + 𝟑𝒄)𝒂𝒙
𝟑𝒂 − 𝒃 + 𝟑𝒄 = 𝟑
;
−𝒂𝒚 = (−𝟐𝒂 + 𝟐𝒃 + 𝟐𝒄)𝒂𝒚 ;
−𝟐𝒂 + 𝟐𝒃 + 𝟐𝒄 = −𝟏
−𝟐𝒂𝒛 = (𝟑𝒂 + 𝟓𝒃 + 𝟐𝒄)𝒂𝒛 ;
𝟑𝒂 + 𝟓𝒃 + 𝟐𝒄 = −𝟐
Solving the equations lead,
𝟏
𝟑
𝟏
𝒂 = 𝟒,𝒃 = −𝟒,𝒄 = 𝟐
Exercise 7.1
Solve the following problems.
1. Given 𝐀 = 𝟐𝒂𝒙 − 𝟑𝒂𝒚 + 𝟒𝒂𝒛, 𝐁 = −𝟑𝒂𝒙 − 𝒂𝒚 + 𝟓𝒂𝒛 and 𝐂 = 𝟔𝒂𝒙 + 𝟐𝒂𝒚 − 𝒂𝒛, find:
a.) 𝟐𝐀 + 𝟒𝐁 − 𝟐𝐂
f.) |𝟑𝐀 − 𝟐𝐁 + 𝐂|
b.) −𝟑(𝐀 + 𝐂)
g.) 𝟐|𝐀| + 𝟑|𝐁|
c.) 𝟐(−𝟑𝐀 + 𝟔𝐁)
h.)
d.) −𝟑(𝐀 − 𝐁) + (𝟑𝐁 − 𝐂)
i.) unit vector 𝐀
e.) |𝐀|
j.) unit vector (𝐀 − 𝟐𝐁)
𝐀−𝟐𝐁+𝐂
|𝐀+𝐁|
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2. Given 𝐀 = −𝟑𝐚𝐱 + 𝟒𝐚𝐲 + 𝐚𝐳 , 𝐁 = 𝐚𝐱 + 𝐚𝐲 − 𝟐𝐳, 𝐂 = −𝟐𝐚𝐱 + 𝟔𝐚𝐲 − 𝟑𝐚𝐳 and 𝐃 = 𝟐𝐚𝐱 +
𝟑𝐚𝐲 − 𝟔𝐚𝐳 , find:
a.) scalars 𝒂, 𝒃 and 𝒄 such that 𝐃 = 𝒂𝐀 + 𝒃𝐁 + 𝐜𝐁
b.) scalars 𝑥, 𝑦 and 𝑧 such that 𝐀 = 𝟑𝐁𝒙 + 𝟐𝐂𝒚 − 𝐃𝒛
7.5 Dot Product
Dot Product or Scalar Product of two vectors 𝐀 and 𝐁 is denoted by 𝐀 ∙ 𝐁
(read as 𝐀 dot 𝐁) is defined as the product of the magnitudes of 𝐀 and 𝐁 and the
cosine of the angle 𝛉 between them.
𝐀 ∙ 𝐁 = |𝐀||𝐁|𝐜𝐨𝐬 𝛉 ;
𝐀 ∙ 𝐁 is a scalar
Laws:
1. 𝐀 ∙ 𝐁 = 𝐁 ∙ 𝐀(Commutative Law for Dot Product)
2. 𝐀 ∙ (𝐁 + 𝐂) = 𝐀 ∙ 𝐁 + 𝐀 ∙ 𝐂(Distributive Law)
3. 𝐦(𝐀 ∙ 𝐁) = (𝐦𝐀) ∙ 𝐁 = 𝐀 ∙ (𝐦𝐁) = (𝐀 ∙ 𝐁)𝐦(where 𝐦 is a scalar)
4. 𝒂𝒙 ∙ 𝒂𝒙 = 𝒂𝒚 ∙ 𝒂𝒚 = 𝒂𝒛 ∙ 𝒂𝒛 = 𝟏, 𝒂𝒙 ∙ 𝒂𝒚 = 𝒂𝒚 ∙ 𝒂𝒛 = 𝒂𝒛 ∙ 𝒂𝒙 = 𝟎
5. If 𝐀 = 𝐀 𝟏 𝒂𝒙 + 𝐀 𝟐 𝒂𝒚 + 𝐀 𝟑 𝒂𝒛 and 𝐁 = 𝐁𝟏 𝒂𝒙 + 𝐁𝟐 𝒂𝒚 + 𝐁𝟑 𝒂𝒛 , then,
𝐀 ∙ 𝐁 = 𝐀 𝟏 𝐁𝟏 + 𝐀 𝟐 𝐁𝟐 + 𝐀 𝟑 + 𝐁𝟑
𝐀 ∙ 𝐀 = 𝐀𝟐 = (𝐀 𝟏)𝟐 + (𝐀 𝟐 )𝟐 + (𝐀 𝟑 )𝟐
𝐁 ∙ 𝐁 = 𝐁 𝟐 = (𝐁𝟏 )𝟐 + (𝐁𝟐 )𝟐 + (𝐁𝟑 )𝟐
6. If 𝐀 ∙ 𝐁 = 𝟎 and 𝐀 and 𝐁 are null vectors, then 𝐀 and 𝐁 are perpendicular.
Example 1:
Find the angle between 𝐀 = 𝟑𝒂𝒙 + 𝟐𝒂𝒚 − 𝟓𝒂𝒛 and 𝐁 = 𝟓𝒂𝒙 − 𝟒𝒂𝒚 + 𝟐𝒂𝒛 .
Solution:
𝐀 ∙ 𝐁 = (𝟑)(𝟓) + (𝟐)(−𝟒) + (−𝟓)(𝟐) = −𝟑
|𝐀| = √(𝟑)𝟐 + (𝟐)𝟐 + (−𝟓)𝟐 = √𝟑𝟖
|𝐁| = √(𝟓)𝟐 + (−𝟒)𝟐 + (𝟐)𝟐 = 𝟑√𝟓
Since,
𝐀 ∙ 𝐁 = |𝐀||𝐁|𝐜𝐨𝐬 𝛉
−𝟑 = (√𝟑𝟖 )(𝟑√𝟓 ) 𝐜𝐨𝐬 𝛉
𝛉 = 𝟗𝟒. 𝟏𝟔𝟎
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Example 2:
Evaluate the following:
a.) (𝟑𝒂𝒙 + 𝒂𝒚 ) ∙ (−𝟐𝒂𝒙 + 𝟐𝒂𝒚 − 𝒂𝒛 )
b.) 𝒂𝒛 ∙ (𝟑𝒂𝒙 − 𝟐𝒂𝒚 )
c.) (𝟐𝒂𝒙 + 𝟑𝒂𝒚 − 𝟓𝒂𝒛 ) ∙ (−𝟒𝒂𝒙 + 𝟐𝒂𝒚 − 𝒂𝒛 )
Solution:
a.) (𝟑𝒂𝒙 + 𝒂𝒚 ) ∙ (−𝟐𝒂𝒙 + 𝟐𝒂𝒚 − 𝒂𝒛 ) = (𝟑)(−𝟐)𝒂𝒙 ∙ 𝒂𝒙 + (𝟑)(𝟐)𝒂𝒙 ∙ 𝒂𝒚 +
(𝟑)(−𝟏)𝒂𝒙 ∙ 𝒂𝒛 + (𝟏)(−𝟐)𝒂𝒚 ∙ 𝒂𝒙 + (𝟏)(𝟐)𝒂𝒚 ∙ 𝒂𝒚 + (𝟏)(−𝟏)𝒂𝒚 ∙ 𝒂𝒛
Since 𝒂𝒙 ∙ 𝒂𝒙 = 𝒂𝒚 ∙ 𝒂𝒚 = 𝒂𝒛 ∙ 𝒂𝒛 = 𝟏 and other dot products are zero, therefore,
(𝟑𝒂𝒙 + 𝒂𝒚 ) ∙ (−𝟐𝒂𝒙 + 𝟐𝒂𝒚 − 𝒂𝒛 ) = −𝟔 + 𝟐 = −𝟒
b.) 𝒂𝒛 ∙ (𝟑𝒂𝒙 − 𝟐𝒂𝒚 ) = 𝟑𝒂𝒛 ∙ 𝒂𝒙 − 𝟐𝒂𝒛 ∙ 𝒂𝒚 = 𝟎
c.) (𝟐𝒂𝒙 + 𝟑𝒂𝒚 − 𝟓𝒂𝒛 ) ∙ (−𝟒𝒂𝒙 + 𝟐𝒂𝒚 − 𝒂𝒛 ) = (𝟐)(−𝟒)𝒂𝒙 ∙ 𝒂𝒙 + (𝟐)(𝟐)𝒂𝒙 ∙
𝒂𝒚 + (𝟐)(−𝟏)𝒂𝒙 ∙ 𝒂𝒛 + (𝟑)(−𝟒)𝒂𝒚 ∙ 𝒂𝒙 + (𝟑)(𝟐)𝒂𝒚 ∙ 𝒂𝒚 + (𝟑)(−𝟏)𝒂𝒚 ∙ 𝒂𝒛 + (−𝟓)(−𝟒)𝒂𝒛 ∙
𝒂𝒙 + (−𝟓)(𝟐)𝒂𝒛 ∙ 𝒂𝒚 + (−𝟓)(−𝟏)𝒂𝒛 ∙ 𝒂𝒛
= −𝟖 + 𝟔 + 𝟓 = 𝟑
Example 3:
Given 𝐀 = −𝟒𝒂𝒙 + 𝟑𝒂𝒚 + 𝟐𝒂𝒛, 𝐁 = 𝟐𝒂𝒙 + 𝟐𝒂𝒚 − 𝒂𝒛 and 𝐂 = −𝟑𝒂𝒙 − 𝟐𝒂𝒚 + 𝒂𝒛 ,
find:
a.) 𝐀 ∙ 𝐁
b.) (𝐀 + 𝐁) ∙ (𝐁 + 𝐂)
c.) 𝟐𝐀 ∙ (𝟑𝐁 − 𝐂)
Solution:
a.) 𝐀 ∙ 𝐁 = (−𝟒)(𝟐) + (𝟑)(𝟐) + (𝟐)(−𝟏) = −𝟒
b.) (𝐀 + 𝐁) ∙ (𝐁 + 𝐂) = [( −𝟒𝒂𝒙 + 𝟑𝒂𝒚 + 𝟐𝒂𝒛 ) + (𝟐𝒂𝒙 + 𝟐𝒂𝒚 − 𝒂𝒛 )] ∙
[(𝟐𝒂𝒙 + 𝟐𝒂𝒚 − 𝒂𝒛 ) + (−𝟑𝒂𝒙 − 𝟐𝒂𝒚 + 𝒂𝒛 )]
= (−𝟐𝒂𝒙 + 𝟓𝒂𝒚 + 𝒂𝒛 ) ∙ (−𝒂𝒙 )
= (−𝟐)(−𝟏) + (𝟓)(𝟎) + (𝟏)(𝟎)
=𝟐
c.) 𝟐𝐀 ∙ (𝟑𝐁 − 𝐂) = 𝟐(−𝟒𝒂𝒙 + 𝟑𝒂𝒚 + 𝟐𝒂𝒛 ) ∙
[𝟑(𝟐𝒂𝒙 + 𝟐𝒂𝒚 − 𝒂𝒛 ) − (−𝟑𝒂𝒙 − 𝟐𝒂𝒚 + 𝒂𝒛 )]
= (−𝟖𝒂𝒙 + 𝟔𝒂𝒚 + 𝟒𝒂𝒛) ∙ [(𝟔𝒂𝒙 + 𝟔𝒂𝒚 − 𝟑𝒂𝒛 ) − (−𝟑𝒂𝒙 − 𝟐𝒂𝒚 + 𝒂𝒛 )]
= (−𝟖𝒂𝒙 + 𝟔𝒂𝒚 + 𝟒𝒂𝒛) ∙ (𝟗𝒂𝒙 + 𝟖𝒂𝒚 − 𝟒𝒂𝒛 )
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= (−𝟖)(𝟗) + (𝟔)(𝟖) + (𝟒)(−𝟒)
= −𝟒𝟎
Example 4:
(𝐀+𝐁)∙(𝐀−𝐁)
Given 𝐀 = −𝟓𝒂𝒙 − 𝟐𝒂𝒚 + 𝟔𝒂𝒛 and 𝐁 = 𝟒𝒂𝒙 + 𝟑𝒂𝒚 − 𝟐𝒂𝒛 , find ||𝐀+𝐁|+|𝐀−𝐁||.
Solution:
𝐀 + 𝐁 = (−𝟓𝒂𝒙 − 𝟐𝒂𝒚 + 𝟔𝒂𝒛 ) + (𝟒𝒂𝒙 + 𝟑𝒂𝒚 − 𝟐𝒂𝒛 ) = −𝒂𝒙 + 𝒂𝒚 + 𝟒𝒂𝒛
𝐀 − 𝐁 = (−𝟓𝒂𝒙 − 𝟐𝒂𝒚 + 𝟔𝒂𝒛 ) − (𝟒𝒂𝒙 + 𝟑𝒂𝒚 − 𝟐𝒂𝒛 ) = −𝟗𝒂𝒙 − 𝟓𝒂𝒚 + 𝟖𝒂𝒛
(−𝒂 +𝒂𝒚+𝟒𝒂𝒛 )∙(−𝟗𝒂𝒙 −𝟓𝒂𝒚 +𝟖𝒂𝒛 )
(𝐀+𝐁)∙(𝐀−𝐁)
||𝐀+𝐁|+|𝐀−𝐁|| = ||−𝒂 𝒙
𝒙 +𝒂𝒚 +𝟒𝒂𝒛 |+|−𝟗𝒂𝒙 −𝟓𝒂𝒚 +𝟖𝒂𝒛 |
𝟑𝟔
|=|
(−𝟏)(−𝟗)+(𝟏)(−𝟓)+(𝟒)(𝟖)
√(−𝟏)𝟐 +(𝟏)𝟐+(𝟒)𝟐 +√(−𝟗)𝟐 +(−𝟓)𝟐 +(𝟖)𝟐
|
𝟑𝟔
= |𝟑√𝟐 +√𝟏𝟕𝟎 | = |𝟏𝟕.𝟐𝟖| = 𝟐. 𝟎𝟖
Example 5:
𝐀 = 𝐀 𝟏 𝒂𝒙 + 𝐀 𝟐 𝒂𝒚 + 𝐀 𝟑 𝒂𝒛 and 𝐁 = 𝐁𝟏 𝒂𝒙 + 𝐁𝟐 𝒂𝒚 + 𝐁𝟑 𝒂𝒛 , prove that
Given that
𝐀 ∙ 𝐁 = 𝐀 𝟏 𝐁𝟏 + 𝐀 𝟐 𝐁𝟐 + 𝐀 𝟑 + 𝐁𝟑 .
Solution:
𝐀 ∙ 𝐁 = (𝐀 𝟏 𝒂𝒙 + 𝐀 𝟐 𝒂𝒚 + 𝐀 𝟑 𝒂𝒛 ) ∙ (𝐁𝟏 𝒂𝒙 + 𝐁𝟐 𝒂𝒚 + 𝐁𝟑 𝒂𝒛 )
= 𝐀 𝟏 𝒂𝒙 ∙ (𝐁𝟏 𝒂𝒙 + 𝐁𝟐 𝒂𝒚 + 𝐁𝟑 𝒂𝒛 ) + 𝐀 𝟐 𝒂𝒚 ∙ (𝐁𝟏 𝒂𝒙 + 𝐁𝟐 𝒂𝒚 + 𝐁𝟑 𝒂𝒛)
+𝐀 𝟑 𝒂𝒛 ∙ (𝐁𝟏 𝒂𝒙 + 𝐁𝟐 𝒂𝒚 + 𝐁𝟑 𝒂𝒛)
= 𝐀 𝟏 𝐁𝟏 𝒂𝒙 ∙ 𝒂𝒙 + 𝐀 𝟏 𝐁𝟐 𝒂𝒙 ∙ 𝒂𝒚 + 𝐀 𝟏 𝐁𝟑 𝒂𝒙 ∙ 𝒂𝒛 + 𝐀 𝟐 𝐁𝟏 𝒂𝒚 ∙ 𝒂𝒙
+𝐀 𝟐 𝐁𝟐 𝒂𝒚 ∙ 𝒂𝒚 + 𝐀 𝟐 𝐁𝟑 𝒂𝒚 ∙ 𝒂𝒛 + 𝐀 𝟑 𝐁𝟏 𝒂𝒛 ∙ 𝒂𝒙 + 𝐀 𝟑 𝐁𝟐𝒂𝒛 ∙ 𝒂𝒚
+𝐀 𝟑 𝐁𝟑 𝒂𝒛 ∙ 𝒂𝒛
Since 𝒂𝒙 ∙ 𝒂𝒙 = 𝒂𝒚 ∙ 𝒂𝒚 = 𝒂𝒛 ∙ 𝒂𝒛 = 𝟏 and other dot products are zero, therefore,
𝐀 ∙ 𝐁 = 𝐀 𝟏 𝐁𝟏 + 𝐀 𝟐 𝐁𝟐 + 𝐀 𝟑 𝐁𝟑
Exercise 7.2
1. Evaluate the following:
a.) 𝒂𝒙 ∙ (𝟐𝒂𝒙 − 𝟑𝒂𝒚 + 𝟓𝒂𝒛)
b.) (𝟐𝒂𝒙 − 𝒂𝒚 ) ∙ (𝟑𝒂𝒙 + 𝒂𝒚 − 𝟐𝒂𝒛 )
c.) (−𝟓𝒂𝒙 + 𝟑𝒂𝒚 ) ∙ (𝟔𝒂𝒙 + 𝟕𝒂𝒛 )
d.) (−𝟐𝒂𝒙 + 𝟓𝒂𝒚 + 𝒂𝒛) ∙ (𝟑𝒂𝒙 + 𝟓𝒂𝒚 − 𝟑𝒂𝒛 )
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2. Given 𝐀 = −𝟕𝒂𝒙 + 𝟒𝒂𝒚 − 𝟔𝒂𝒛 and 𝐁 = −𝟒𝒂𝒙 − 𝟑𝒂𝒚 + 𝟓𝒂𝒛 , find:
a.) the angle between them
e.) 𝟑𝐀 ∙ (𝐀 + 𝐁)
b.) 𝐀 ∙ 𝐁
f.) (𝐀 + 𝟐𝐁) ∙ (𝐁 − 𝐀)
c.) 𝐀𝟐
g.) (𝟐𝐀 − 𝟑𝐁) ∙ (𝟒𝐀 + 𝟓𝐁)
d.) 𝐁 𝟐 − 𝐀𝟐
h.) (𝟓𝐀 ∙ 𝐁) + (𝟑𝐀 ∙ 𝟐𝐁)
3. Given 𝐀 = 𝟓𝒂𝒙 − 𝟐𝒂𝒚 , 𝐁 = −𝒂𝒙 − 𝟔𝒂𝒚 + 𝟐𝒂𝒛 and 𝐂 = 𝟐𝒂𝒙 − 𝟒𝒂𝒚 + 𝟒𝒂𝒛, find:
|𝐀+𝐁|
a.) (𝐀 − 𝟐𝐁) ∙ (𝐁 + 𝟑𝐂)
b.)
c.)
e.) (𝐀+𝐁)∙(𝐁+𝐂)
𝐀+𝟐𝐁−𝐂
f.)
(𝐀∙𝐁)
𝟐𝐀𝟐 +𝟑𝐁𝟐 −𝟒𝐂𝟐
|𝐀+𝐁|
(𝟐𝐁+𝟑𝐂)∙(𝐀+𝟐𝐁)
−𝟐𝐀+𝟑𝐁+𝐂
g.) | (𝐀+𝟐𝐁)∙(𝐁+𝟐𝐂) |
(𝟐𝐁∙𝐂)
4. Given 𝐀 = 𝐀 𝟏 𝒂𝒙 + 𝐀 𝟐 𝒂𝒚 + 𝐀 𝟑 𝒂𝒛, prove that 𝐀𝟐 = (𝐀 𝟏 )𝟐 + (𝐀 𝟐)𝟐 + (𝐀 𝟑 )𝟑.
7.6 Cross Product
Cross Product or Vector Product of two vectors 𝐀 and 𝐁, denoted by 𝐀 × 𝐁
(read 𝐀 cross 𝐁). It is defined as the product of the magnitudes of 𝐀 and 𝐁, the sine
of the angle 𝛉 between them and 𝐮 (a unit vector indicating the direction of 𝐀 × 𝐁.
𝐀 × 𝐁 = |𝐀||𝐁| 𝐬𝐢𝐧 𝛉 𝐮
If 𝐀 = 𝐁, that is: 𝐀 is parallel to 𝐁, then 𝐬𝐢𝐧 𝛉 = 𝟎, therefore 𝐀 × 𝐁 = 𝟎.
Laws:
1. 𝐀 × 𝐁 = −𝐁 × 𝐀(Commutative Law for Cross Product fails)
2. 𝐀 × (𝐁 + 𝐂) = 𝐀 × 𝐁 + 𝐀 × 𝐂(Distributive Law)
3. 𝐦(𝐀 × 𝐁) = (𝐦𝐀) × 𝐁 = 𝐀 × (𝐦𝐁) = (𝐀 × 𝐁)𝐦(where 𝐦 is a scalar)
4. 𝒂𝒙 × 𝒂𝒙 = 𝒂𝒚 × 𝒂𝒚 = 𝒂𝒛 × 𝒂𝒛 = 𝟎, 𝒂𝒙 × 𝒂𝒚 = 𝒂𝒛 ; 𝒂𝒚 × 𝒂𝒛 = 𝒂𝒙 ; 𝒂𝒛 × 𝒂𝒙 = 𝒂𝒚
5. If 𝐀 = 𝐀 𝟏 𝒂𝒙 + 𝐀 𝟐 𝒂𝒚 + 𝐀 𝟑 𝒂𝒛 and 𝐁 = 𝐁𝟏 𝒂𝒙 + 𝐁𝟐 𝒂𝒚 + 𝐁𝟑 𝒂𝒛, then,
𝒂𝒙
𝐀 × 𝐁 = |𝐀 𝟏
𝐁𝟏
𝒂𝒚
𝐀𝟐
𝐁𝟐
𝒂𝒛
𝐀𝟑|
𝐁𝟑
6. The magnitude of 𝐀 × 𝐁 is the same as the area of a parallelogram with sides 𝐀
and 𝐁.
7. If 𝐀 = 𝐁, that is: 𝐀 is parallel to 𝐁, then, 𝐬𝐢𝐧 𝛉 = 𝟎, therefore 𝐀 × 𝐁 = 𝟎.
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Example 1:
Evaluate the following:
a.) (−𝟑𝒂𝒙 ) × (𝟒𝒂𝒚 )
b.) 𝟐𝒂𝒙 × (𝟑𝒂𝒙 + 𝟒𝒂𝒚 − 𝟑𝒂𝒛 )
c.) (𝟑𝒂𝒙 + 𝟐𝒂𝒚 ) × (−𝟐𝒂𝒙 + 𝟒𝒂𝒚 )
Solution:
a.) (−𝟑𝒂𝒙 ) × (𝟒𝒂𝒚 ) = (−𝟑)(𝟒)𝒂𝒙 × 𝒂𝒚 = −𝟏𝟐𝒂𝒛
b.) 𝟐𝒂𝒙 × (𝟑𝒂𝒙 + 𝟒𝒂𝒚 − 𝟑𝒂𝒛 ) = (𝟐)(𝟑)𝒂𝒙 × 𝒂𝒙 + (𝟐)(𝟒)𝒂𝒙 × 𝒂𝒚
+(𝟐)(−𝟑)𝒂𝒙 × 𝒂𝒛
= 𝟔(𝟎) + 𝟖𝒂𝒛 − 𝟔(−𝒂𝒚 ) = 𝟔𝒂𝒚 + 𝟖𝒂𝒛
c.) (𝟑𝒂𝒙 + 𝟐𝒂𝒚 ) × (−𝟐𝒂𝒙 + 𝟒𝒂𝒚 ) = (𝟑)(−𝟐)𝒂𝒙 × 𝒂𝒙 + (𝟑)(𝟒)𝒂𝒙 × 𝒂𝒚
+(𝟐)(−𝟐)𝒂𝒚 × 𝒂𝒙 + (𝟐)(𝟒)𝒂𝒚 × 𝒂𝒚
= −𝟔(𝟎) + 𝟏𝟐𝒂𝒛 − 𝟒(−𝒂𝒛 ) + 𝟖(𝟎)
= 𝟏𝟔𝒂𝒛
Example 2:
If 𝐀 = 𝟑𝒂𝒙 + 𝟑𝒂𝒚 − 𝟐𝒂𝒛 and 𝐁 = 𝟒𝒂𝒙 − 𝟐𝒂𝒚 − 𝒂𝒛 , find:
a.) 𝐀 × 𝐁
b.) 𝐁 × 𝐀
c.) 𝐀 × 𝐀
Solution:
𝒂𝒙
a.) 𝐀 × 𝐁 = | 𝟑
𝟒
𝒂𝒚
𝟑
−𝟐
𝒂𝒛
𝒂𝒙
−𝟐 | = | 𝟑
−𝟏
𝟒
𝒂𝒚
𝟑
−𝟐
𝒂𝒛 𝒂𝒙
−𝟐 | 𝟑
−𝟏 𝟒
𝒂𝒚
𝟑|
−𝟐
= (−𝟑𝒂𝒙 − 𝟖𝒂𝒚 − 𝟔𝒂𝒛 ) − (𝟏𝟐𝒂𝒛 + 𝟒𝒂𝒙 − 𝟑𝒂𝒚 )
= −𝟑𝒂𝒙 − 𝟖𝒂𝒚 − 𝟔𝒂𝒛 − 𝟏𝟐𝒂𝒛 − 𝟒𝒂𝒙 + 𝟑𝒂𝒚
= −𝟕𝒂𝒙 − 𝟓𝒂𝒚 − 𝟏𝟖𝒂𝒛
𝒂𝒙
b.) 𝐁 × 𝐀 = | 𝟒
𝟑
𝒂𝒚
−𝟐
𝟑
𝒂𝒛
𝒂𝒙
−𝟏 | = | 𝟒
−𝟐
𝟑
𝒂𝒚
−𝟐
𝟑
𝒂𝒛 𝒂𝒙
−𝟏 | 𝟒
−𝟐 𝟑
= (𝟒𝒂𝒙 − 𝟑𝒂𝒚 + 𝟏𝟐𝒂𝒛) − (−𝟔𝒂𝒛 − 𝟑𝒂𝒙 − 𝟖𝒂𝒚 )
= 𝟒𝒂𝒙 − 𝟑𝒂𝒚 + 𝟏𝟐𝒂𝒛 + 𝟔𝒂𝒛 + 𝟑𝒂𝒙 + 𝟖𝒂𝒚
= 𝟕𝒂𝒙 + 𝟓𝒂𝒚 + 𝟏𝟖𝒂𝒛
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𝒂𝒚
−𝟐 |
𝟑
CHAPTER 7
VECTOR ANALYSIS
𝒂𝒙
c.) 𝐀 × 𝐀 = | 𝟑
𝟑
𝒂𝒚
𝟑
𝟑
𝒂𝒛
𝒂𝒙
−𝟐 | = | 𝟑
−𝟐
𝟑
𝒂𝒚
𝟑
𝟑
𝒂𝒛 𝒂𝐱
−𝟐 | 𝟑
−𝟐 𝟑
𝒂𝒚
𝟑|
𝟑
= (−𝟔𝒂𝒙 − 𝟔𝒂𝒚 + 𝟗𝒂𝒛 ) − (𝟗𝒂𝒛 − 𝟔𝒂𝒙 − 𝟔𝒂𝒚 )
=𝟎
This explains law 7 where two vectors are equal.
Example 3:
Given 𝐀 = −𝟑𝒂𝒙 + 𝟐𝒂𝒚 − 𝟓𝒂𝒛 , 𝐁 = −𝟐𝒂𝒙 − 𝟑𝒂𝒚 and 𝐂 = 𝟑𝒂𝒙 + 𝟒𝒂𝒚 − 𝟒𝒂𝒛, find:
a.) (𝐀 + 𝐁) × (𝐀 − 𝐁)
b.) 𝐀 ∙ (𝐁 × 𝐂)
c.) |𝟐𝐀 × 𝐁|
d.) (𝐀 × 𝐁) × (𝐁 × 𝐂)
Solution:
a.) 𝐀 + 𝐁 = −𝟑𝒂𝒙 + 𝟐𝒂𝒚 − 𝟓𝒂𝒛 + (−𝟐𝒂𝒙 − 𝟑𝒂𝒚 ) = −𝟓𝒂𝒙 − 𝒂𝒚 − 𝟓𝒂𝒛
𝐀 − 𝐁 = −𝟑𝒂𝒙 + 𝟐𝒂𝒚 − 𝟓𝒂𝒛 − (−𝟐𝒂𝒙 − 𝟑𝒂𝒚 ) = −𝒂𝒙 + 𝟓𝒂𝒚 − 𝟓𝒂𝒛
𝒂𝒙
(𝐀 + 𝐁) × (𝐀 − 𝐁) = | −𝟓
−𝟏
𝒂𝒚
−𝟏
𝟓
𝒂𝒛
𝒂𝒙
−𝟓| = | −𝟓
−𝟓
−𝟏
𝒂𝒚
−𝟏
𝟓
𝒂𝒛 𝒂𝒙
−𝟓| −𝟓
−𝟓 −𝟏
𝒂𝒚
−𝟏 |
𝟓
= (𝟓𝒂𝒙 + 𝟓𝒂𝒚 − 𝟐𝟓𝒂𝒛) − (𝒂𝒛 − 𝟐𝟓𝒂𝒙 + 𝟐𝟓𝒂𝒚 )
= 𝟑𝟎𝒂𝒙 − 𝟐𝟎𝒂𝒚 − 𝟐𝟔𝒂𝒛
𝒂𝒙
b.) 𝐁 × 𝐂 = | −𝟐
𝟑
𝒂𝒚
−𝟑
𝟒
𝒂𝒛
𝒂𝒙
𝟎 | = | −𝟐
−𝟒
𝟑
𝒂𝒚
−𝟑
𝟒
𝒂𝒛 𝒂𝒙
𝟎 | −𝟐
−𝟒 𝟑
𝒂𝒚
−𝟑 |
𝟒
= (𝟏𝟐𝒂𝒙 + 𝟎 − 𝟖𝒂𝒛 ) − (−𝟗𝒂𝒛 + 𝟎 + 𝟖𝒂𝒚 )
= 𝟏𝟐𝒂𝒙 − 𝟖𝒂𝒚 +𝒂𝒛
𝐀 ∙ (𝐁 × 𝐂) = (−𝟑𝒂𝒙 + 𝟐𝒂𝒚 − 𝟓𝒂𝒛) ∙ (𝟏𝟐𝒂𝒙 − 𝟖𝒂𝒚 +𝒂𝒛 )
= (−𝟑)(𝟏𝟐) + (𝟐)(−𝟖) + (−𝟓)(𝟏)
= −𝟓𝟕
𝒂𝒙 𝒂𝒚
c.) 𝟐𝐀 × 𝐁 = 𝟐 | −𝟑
𝟐
−𝟐 −𝟑
𝒂𝒛
𝒂𝒙
−𝟓 | = 𝟐 | −𝟑
𝟎
−𝟐
𝒂𝒚
𝟐
−𝟑
𝒂𝒛 𝒂𝒙
−𝟓 | −𝟑
𝟎 −𝟐
𝒂𝒚
𝟐|
−𝟑
= 𝟐[(𝟎 + 𝟏𝟎𝒂𝒚 + 𝟗𝒂𝒛 ) − (−𝟒𝒂𝒛 + 𝟏𝟓𝒂𝒙 + 𝟎)]
= 𝟐(−𝟏𝟓𝒂𝒙 + 𝟏𝟎𝒂𝒚 + 𝟏𝟑𝒂𝒛 )
= −𝟑𝟎𝒂𝒙 + 𝟐𝟎𝒂𝒚 + 𝟐𝟔𝒂𝒛
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𝒂𝒙
d.) 𝐀 × 𝐁 = | −𝟑
−𝟐
𝒂𝒚
𝟐
−𝟑
𝒂𝒛
𝒂𝒙
−𝟓 | = | −𝟑
𝟎
−𝟐
𝒂𝒚
𝟐
−𝟑
𝒂𝒛 𝒂𝒙
−𝟓 | −𝟑
𝟎 −𝟐
𝒂𝒚
𝟐|
−𝟑
= (𝟎 + 𝟏𝟎𝒂𝒚 + 𝟗𝒂𝒛 )— 𝟒𝒂𝒛 + 𝟏𝟓𝒂𝒙 + 𝟎
= −𝟏𝟓𝒂𝒙 + 𝟏𝟎𝒂𝒚 + 𝟏𝟑𝒂𝒛
𝒂𝒙
𝐁 × 𝐂 = | −𝟐
𝟑
𝒂𝒚
−𝟑
𝟒
𝒂𝒛
𝒂𝒙
𝟎 | = | −𝟐
−𝟒
𝟑
𝒂𝒚
−𝟑
𝟒
𝒂𝒛 𝒂𝒙
𝟎 | −𝟐
−𝟒 𝟑
𝒂𝒚
−𝟑 |
𝟒
= (𝟏𝟐𝒂𝒙 + 𝟎 − 𝟖𝒂𝒛 ) − (−𝟗𝒂𝒛 + 𝟎 + 𝟖𝒂𝒚 )
= 𝟏𝟐𝒂𝒙 − 𝟖𝒂𝒚 +𝒂𝒛
𝒂𝒙
(𝐀 × 𝐁) × (𝐁 × 𝐂) = |−𝟏𝟓
𝟏𝟐
𝒂𝒚
𝟏𝟎
−𝟖
𝒂𝒛
𝒂𝒙
𝟏𝟑| = |−𝟏𝟓
𝟏
𝟏𝟐
𝒂𝒚
𝟏𝟎
−𝟖
𝒂𝒛 𝒂𝒙
𝟏𝟑| −𝟏𝟓
𝟏 𝟏𝟐
𝒂𝒚
𝟏𝟎 |
−𝟖
= (𝟏𝟎𝒂𝒙 + 𝟏𝟓𝟔𝒂𝒚 + 𝟏𝟐𝟎𝒂𝒛 ) − (𝟏𝟐𝟎𝒂𝒛 − 𝟏𝟎𝟒𝒂𝒙 − 𝟏𝟓𝒂𝒚 )
= 𝟏𝟏𝟒𝒂𝒙 + 𝟏𝟕𝟏𝒂𝒚
Example 4:
Given 𝐀 = 𝐀 𝟏 𝒂𝒙 + 𝐀 𝟐 𝒂𝒚 + 𝐀 𝟑 𝒂𝒛, 𝐁 = 𝐁𝟏 𝒂𝒙 + 𝐁𝟐 𝒂𝒚 + 𝐁𝟑 𝒂𝒛 and
𝐀𝟏
𝐂 = 𝐂𝟏 𝒂𝒙 + 𝐂𝟐 𝒂𝒚 + 𝐂𝟑 𝒂𝒛 , show that 𝐀 ∙ (𝐁 × 𝐂) = |𝐁𝟏
𝐂𝟏
𝐀𝟐
𝐁𝟐
𝐂𝟐
𝐀𝟑
𝐁𝟑 |.
𝐂𝟑
Solution:
𝒂𝒙
𝐁 × 𝐂 = | 𝐁𝟏
𝐂𝟏
𝒂𝒚
𝐁𝟐
𝐂𝟐
𝒂𝒛
𝒂𝒙
𝐁𝟑 | = | 𝐁𝟏
𝐂𝟑
𝐂𝟏
𝒂𝒚
𝐁𝟐
𝐂𝟐
𝒂𝒛 𝒂𝒙
𝐁𝟑 | 𝐁𝟏
𝐂𝟑 𝐂𝟏
𝒂𝒚
𝐁𝟐 |
𝐂𝟐
= (𝐁𝟐 𝐂𝟑 𝒂𝒙 + 𝐁𝟑 𝐂𝟏 𝒂𝒚 + 𝐁𝟏 𝐂𝟐 𝒂𝒛 ) − (𝐁𝟐 𝐂𝟏 𝒂𝒛 + 𝐁𝟑 𝐂𝟐 𝒂𝒙 + 𝐁𝟏 𝐂𝟑 𝒂𝒚 )
= (𝐁𝟐 𝐂𝟑 − 𝐁𝟑 𝐂𝟐 )𝒂𝒙 + (𝐁𝟑 𝐂𝟏 − 𝐁𝟏 𝐂𝟑 )𝒂𝒚 + (𝐁𝟏 𝐂𝟐 − 𝐁𝟐 𝐂𝟏 )𝒂𝒛
𝐀 ∙ (𝐁 × 𝐂) = 𝐀 𝟏 (𝐁𝟐 𝐂𝟑 − 𝐁𝟑 𝐂𝟐 ) + 𝐀 𝟐 (𝐁𝟑 𝐂𝟏 − 𝐁𝟏 𝐂𝟑 ) + 𝐀 𝟑(𝐁𝟏 𝐂𝟐 − 𝐁𝟐 𝐂𝟏 )
= 𝐀 𝟏 𝐁𝟐 𝐂𝟑 − 𝐀 𝟏 𝐁𝟑 𝐂𝟐 + 𝐀 𝟐 𝐁𝟑 𝐂𝟏 − 𝐀 𝟐 𝐁𝟏 𝐂𝟑 + 𝐀 𝟑 𝐁𝟏 𝐂𝟐 − 𝐀 𝟑 𝐁𝟐 𝐂𝟏
= (𝐀 𝟏 𝐁𝟐 𝐂𝟑 + 𝐀 𝟐 𝐁𝟑 𝐂𝟏 + 𝐀 𝟑 𝐁𝟏 𝐂𝟐 ) − (𝐀 𝟏 𝐁𝟑𝐂𝟐 + 𝐀 𝟐 𝐁𝟏𝐂𝟑 + 𝐀 𝟑 𝐁𝟐𝐂𝟏 )
Since,
𝐀𝟏
| 𝐁𝟏
𝐂𝟏
𝐀𝟐
𝐁𝟐
𝐂𝟐
𝐀𝟑
𝐀𝟏
𝐁𝟑 | = | 𝐁𝟏
𝐂𝟑
𝐂𝟏
𝐀𝟐
𝐁𝟐
𝐂𝟐
𝐀𝟑 𝐀𝟏
𝐁𝟑 | 𝐁 𝟏
𝐂𝟑 𝐂𝟏
𝐀𝟐
𝐁𝟐 |
𝐂𝟐
= (𝐀 𝟏 𝐁𝟐 𝐂𝟑 + 𝐀 𝟐 𝐁𝟑 𝐂𝟏 + 𝐀 𝟑 𝐁𝟏 𝐂𝟐 ) − (𝐀 𝟑 𝐁𝟐𝐂𝟏 + 𝐀 𝟏 𝐁𝟑𝐂𝟐 + 𝐀 𝟐 𝐁𝟏𝐂𝟑 )
= (𝐀 𝟏𝐁𝟐 𝐂𝟑 + 𝐀 𝟐 𝐁𝟑 𝐂𝟏 + 𝐀 𝟑 𝐁𝟏 𝐂𝟐 ) − (𝐀 𝟏 𝐁𝟑 𝐂𝟐 + 𝐀 𝟐 𝐁𝟏 𝐂𝟑 + 𝐀 𝟑 𝐁𝟐 𝐂𝟏 )
Therefore,
𝐀𝟏
(
)
|
𝐀 ∙ 𝐁 × 𝐂 = 𝐁𝟏
𝐂𝟏
𝐀𝟐
𝐁𝟐
𝐂𝟐
𝐀𝟑
𝐁𝟑 |
𝐂𝟑
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CHAPTER 7
VECTOR ANALYSIS
Exercise 7.3
1. Evaluate the following:
a.) (𝟐𝒂𝒙 ) × (−𝟑𝒂𝒙 )
b.) (−𝟓𝒂𝒛) × (−𝒂𝒚 )
c.) (−𝟑𝒂𝒚 + 𝟐𝒂𝒛) × (𝟐𝒂𝒙 + 𝟒𝒂𝒚 )
d.) (𝟔𝒂𝒙 + 𝟓𝒂𝒚 − 𝟑𝒂𝒛 ) × (−𝟐𝒂𝒙 + 𝟓𝒂𝒛 )
2. Given 𝐀 = 𝟐𝒂𝒙 − 𝟑𝒂𝒚 + 𝟐𝒂𝒛, 𝐁 = −𝒂𝒙 + 𝟐𝒂𝒚 + 𝟓𝒂𝒛 and 𝐂 = −𝟐𝒂𝒙 + 𝟑𝒂𝒚 − 𝒂𝒛, find:
a.) 𝐀 × 𝐁
e.) (𝐀 × 𝐁) + (𝐁 × 𝐂)
b.) 𝐁 × 𝐂
f.) (𝐀 + 𝐁) × (𝐁 × 𝐂)
c.) (𝟑𝐁 × 𝟐𝐂)
g.) (𝟑𝐀 × 𝐁) ∙ (𝟐𝐁 × 𝐂)
d.) (𝐀 × 𝐁) ∙ 𝐂
h.)
(𝐀×𝐁)−(𝐂×𝐁)
𝐀∙𝐁
3. Given 𝐀 = 𝟑𝒂𝒙 − 𝟒𝒂𝒛 , 𝐁 = 𝟔𝒂𝒙 + 𝟔𝒂𝒚 − 𝟑𝒂𝒛 and 𝐂 = 𝟒𝒂𝒚 , find:
a.) 𝐀 × 𝐂
e.) (𝟐𝐀 + 𝟑𝐁) × (𝐂 − 𝐁)
b.) 𝐂 × 𝐁
f.) 𝐀 × (𝐀 × 𝐁)
c.) (𝐀 × 𝐁) × 𝐂
g.)
d.) |𝐀 × 𝐁|
h.)
(𝐀×𝐁)∙(𝐁×𝐂)
|𝐀×𝐁|
(𝟐𝐀+𝟑𝐁−𝐂)×(𝐁−𝟐𝐂)
𝐀∙𝐂
4. Given 𝐀 = 𝐀 𝟏 𝒂𝒙 + 𝐀 𝟐 𝒂𝒚 + 𝐀 𝟑 𝒂𝒛, 𝐁 = 𝐁𝟏 𝒂𝒙 + 𝐁𝟐 𝒂𝒚 + 𝐁𝟑 𝒂𝒛 and
𝐂 = 𝐂𝟏 𝒂𝒙 + 𝐂𝟐 𝒂𝒚 + 𝐂𝟑 𝒂𝒛 , show that 𝐀 ∙ (𝐁 × 𝐂) = (𝐀 × 𝐁) ∙ 𝐂.
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VECTOR ANALYSIS
CHAPTER 7
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CHAPTER 8
NUMERICAL METHODS
Chapter 8
NUMERICAL METHODS
Chapter Outline:
8.1
8.2
8.3
8.4
8.5
Numerical Methods
Solutions of Equations
Numerical Solution of Linear System of Equations
Interpolation
Numerical Integration
Learning Objectives:
At the end of the lesson, the students will be able to:
1. Define numerical method.
2. Solve roots of equation using numerical methods.
3. Solve system of linear equations using numerical methods.
4. Interpolate functions.
5. Solve integral using numerical integration.
Overview:
This chapter will discuss numerical methods and its applications in solving
solutions of equations, systems of linear equations, interpolation and numerical
integration.
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NUMERICAL METHODS
CHAPTER 8
8.1 Numerical Methods
Numerical methods are used in solving complex numerical problems using only
the simple operations of arithmetic. This also allows evaluation methods for computing
numerical results from given data.
8.2 Solution of Equations
I. Newton’s Method
Linear equations can be solved using the simple algebraic operations. Some
equations such as quadratic equations can be solved using the quadratic formula. In
case an equation, containing polynomials of higher degree or an equation involving
transcendental functions, it is difficult and often impossible to obtain an exact solution.
Such equations are,
𝑥 3 + 𝑥 = 4 and 𝑥 3 − 3𝑥 + 1 = 0
There are techniques for approximating solutions and one of them is the
Newton’s Method, also known as Newton-Raphson Method.
Consider a function 𝑦 = 𝑓 (𝑥 ) and its graph as shown below,
𝒚
𝒚 = 𝒇(𝒙)
(𝒙, 𝒚)
𝒙
𝑥5
𝑥4
𝑥3
𝑥2
𝑥1
where 𝑥1 is the first approximation, 𝑥2 is the second, 𝑥3 is the third and so on.
Newton’s Method are based on,
1. If 𝑓 is differentiable at 𝑥, then the tangent line at (𝑥, 𝑦), can be used to
approximate the values of 𝑓 near 𝑥.
2. The point where the tangent line crosses the 𝑥 − axis is easy to determine.
Consider the first approximation 𝑥1 . The tangent line at (𝑥1 , 𝑓 (𝑥1 )) is,
𝑦 − 𝑓 (𝑥1 ) = 𝑓′(𝑥1 )(𝑥 − 𝑥1 )
Consider 𝑥2 the point where the tangent line crosses the 𝑥 − axis, when 𝑥 = 𝑥2
and 𝑦 = 0, thus,
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CHAPTER 8
NUMERICAL METHODS
0 − 𝑓(𝑥1 ) = 𝑓′(𝑥1 )(𝑥2 − 𝑥1 )
𝑓(𝑥 )
𝑥2 = 𝑥1 − 𝑓′(𝑥1 )
1
Now, let 𝑥2 be the second approximation and repeat the process, we find,
𝑓(𝑥 )
𝑥3 = 𝑥2 − 𝑓′(𝑥2 )
2
For the fourth approximation, fifth approximation and so on, the process are the
same, therefore, for 𝑛𝑡ℎ approximation, we have,
𝒇(𝒙 )
𝒙𝒏+𝟏 = 𝒙𝒏 − 𝒇′(𝒙𝒏 ), provided 𝑓′(𝑥𝑛 ) ≠ 0 for 𝑛 = 0,1,2, …
𝒏
Example 1:
Approximate a solution of 𝑥 3 + 𝑥 = 4.
Solution:
Set 𝑥 3 + 𝑥 = 4 and 𝑓 (𝑥 ) = 𝑥 3 + 𝑥 − 4
Graph the function,
𝒚
(𝟐, 𝟔)
𝒚 = 𝒙𝟑 + 𝒙 − 𝟒
𝟔
𝟓
𝟒
𝟑
Tangent Line 𝐿
𝟐
𝟏
First Approximation
𝒙
−𝟏
𝟏
𝟐
Second Approximation
−𝟐
−𝟑
−𝟒
(𝟎, −𝟒)
From the figure, we estimate where the 𝑓 crosses the axis. Let us approximate
the value between 1 and 2. Choose the number near the zero as a first approximation,
consider 2.
Setting 𝑥0 = 2, thus,
𝑓(𝑥 )
𝑥1 = 𝑥0 − 𝑓′ (𝑥0 )
0
(2)3+2−4
𝑥1 = 2 − 3(2)2+1 = 1.538462
Continue the process until we found two consecutive answers that are the same.
The result for iterations is shown in the table below.
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NUMERICAL METHODS
𝑛
0
1
2
3
4
CHAPTER 8
𝑥𝑛
2.000000
1.538462
1.392819
1.378917
1.378797
𝑁𝑛
6.000000
1.179791
0.094811
0.000805
0.000000
𝐷𝑛
13.000000
8.100592
6.819834
6.704234
6.703241
𝑥𝑛+1
1.538462
1.392819
1.378917
1.378797
1.378797
𝑥4 and 𝑥5 found that the value is 1.378797 and is exact to six decimal places.
Thus, the positive solution of 𝑥 3 + 𝑥 = 4 is approximately 1.378797.
Example 2:
Find to four decimal places the smaller positive root of the equation 𝑥 3 − 2𝑥 −
1 = 0. Use 𝑥0 = 1.5.
Solution:
𝑓 (𝑥 ) = 𝑥 3 − 2𝑥 − 1, 𝑓′(𝑥 ) = 3𝑥 2 − 2.
Put 𝑦 = 𝑥 3 − 2𝑥 − 1.
When 𝑥 = 0, 𝑦 = −1.
When 𝑥 = 1, 𝑦 = −2.
When 𝑥 = 2, 𝑦 = 3.
The root lies between 𝑥 = 1 and 𝑥 = 2, try 𝑥0 = 1.5 and by direct substitution,
we find,
𝑥1 = 1.5 −
(1.5)3−2(1.5)−1
3(1.5)2 −2
𝑥1 ≈ 1.631579
𝑥2 = 1.631579 −
(1.631579)3−2(1.631579)−1
3(1.631579)2 −2
𝑥2 ≈ 1.618184
𝑥3 = 1.618184 −
(1.618184)3 −2(1.618184)−1
3(1.618184)2 −2
𝑥3 ≈ 1.618034
𝑥4 = 1.618034 −
(1.618034)3 −2(1.618034)−1
3(1.618034)2 −2
𝑥4 ≈ 1.618034
The smaller positive value of the equation is approximately 1.6180.
Example 3:
Find to three decimal places the cube root of 6.
Solution:
Cube root of 6 can be denoted into an equation 𝑥 3 = 6.
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𝑓 (𝑥 ) = 𝑥 3 − 6, 𝑓′(𝑥 ) = 3𝑥 2 .
Put 𝑦 = 𝑥 3 − 6.
When 𝑥 = 0, 𝑦 = −6.
When 𝑥 = 1, 𝑦 = −5.
When 𝑥 = 2, 𝑦 = 2.
The root lies between 𝑥 = 1 and 𝑥 = 2, try 𝑥0 = 1.5 and by direct substitution,
we find,
(1.5)3−6
𝑥1 = 1.5 − 3(1.5)2
𝑥1 ≈ 1.888889
𝑥2 = 1.888889 −
(1.888889)3−6
3(1.888889)2
𝑥2 ≈ 1.819813
(1.819813)3−6
𝑥3 = 1.819813 − 3(1.819813)2
𝑥3 ≈ 1.817125
𝑥4 = 1.817125 −
(1.817125)3 −6
3(1.817125)2
𝑥4 = 1.817121
The cube root of 6 is approximately 1.817.
Exercise 8.1
Solve the following equations using Newton’s Method.
1. 𝑥 4 + 𝑥 − 3 = 0 on [0,2]
2. 𝑥 2 + 𝑥 − 1 = 0 on [−2, −1]
3. 𝑥 3 − 2𝑥 − 5 = 0 on [2,3]
4. 𝑥𝑒 −𝑥 + 𝑥 2 = 5 on [1,2]
5. 3𝑒 𝑥 + sin 𝑥 − 6 = 0 on [0,1]
II. Secant Method
Newton’s method has a disadvantage that 𝑓′(𝑥) may sometimes difficult to
evaluate, thus, replacing the the derivative with the difference quotient,
𝑓 ′(𝑥𝑛 ) =
𝑓(𝑥𝑛 )−𝑓(𝑥𝑛−1 )
𝑥𝑛 −𝑥𝑛−1
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NUMERICAL METHODS
CHAPTER 8
Thus, the formula for Secant method is,
𝑓(𝑥 )
𝑛
𝑥𝑛+1 = 𝑥𝑛 − 𝑓(𝑥𝑛)−𝑓(𝑥
𝑛−1 )
or
𝑥𝑛 −𝑥𝑛−1
𝒇(𝒙 ) (𝒙 −𝒙
)
𝑵
𝒏
𝒏−𝟏
𝒙𝒏+𝟏 = 𝒙𝒏 − 𝒇(𝒙𝒏 )−𝒇(𝒙
= 𝒙𝒏 − 𝑫 𝒏
)
𝒏
𝒏−𝟏
𝒏
where, 𝑁𝑛 = 𝑓(𝑥𝑛 ) (𝑥𝑛 − 𝑥𝑛−1 )
𝐷𝑛 = 𝑓 (𝑥𝑛 ) − 𝑓 (𝑥𝑛−1 )
Example 1:
Find to four decimal places the smaller positive root of the equation 𝑥 3 − 2𝑥 −
1 = 0.
Solution:
𝑓 (𝑥 ) = 𝑥 3 − 2𝑥 − 1
Put 𝑦 = 𝑥 3 − 2𝑥 − 1.
When 𝑥 = 0, 𝑦 = −1.
When 𝑥 = 1, 𝑦 = −2.
When 𝑥 = 2, 𝑦 = 3.
The root lies between 𝑥 = 1 and 𝑥 = 2, set 𝑥0 = 1 and 𝑥1 = 2, then,
𝑥 −𝑥
0
𝑥2 = 𝑥1 − 𝑓 (𝑥1 ) 𝑓(𝑥 1)−𝑓(𝑥
)
1
0
[(2)3 −2(2)−1](2−1)
3
𝑥2 = 2 − [(2)3 −2(2)−1]−[(1)3 −2(1)−1] = 2 − 5 = 1.400000
Continue the process until we found two consecutive answers that are the same.
The result for iterations is shown in the table below.
𝑛
1
2
3
4
5
6
7
𝑥𝑛−1
1.000000
2.000000
1.400000
1.556213
1.631554
1.617320
1.618026
𝑥𝑛
2.000000
1.400000
1.556213
1.631554
1.617320
1.618026
1.618034
𝑁𝑛
3.000000
0.633600
-0.053673
0.006030
0.000059
0.000000
0.000000
𝐷𝑛
5.000000
-4.056000
0.712409
0.423626
-0.084215
0.004133
0.000047
𝑥𝑛+1
1.400000
1.556213
1.631554
1.617320
1.618026
1.618034
1.618034
𝑥7 and 𝑥8 found that the value is 1.618034 and is exact to six decimal places.
Example 2:
Find the root of 𝑓 (𝑥 ) = 2𝑒 𝑥 + 𝑥 − 3 on [0,1].
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NUMERICAL METHODS
Solution:
The root lies between 𝑥 = 0 and 𝑥 = 1, set 𝑥0 = 0 and 𝑥1 = 1, then,
𝑥 −𝑥
0
𝑥2 = 𝑥1 − 𝑓 (𝑥1 ) 𝑓(𝑥 1)−𝑓(𝑥
)
1
0
(2𝑒 1 +1−3)(1−0)
𝑥2 = 1 − (2𝑒1 +1−3)−(2𝑒0 +0−3) = 0.225400
Continue the process until we found two consecutive answers that are the same.
The result for iterations is shown in the table below.
𝑛
1
2
3
4
5
6
𝑥𝑛−1
0.000000
1.000000
0.225400
0.281622
0.300581
0.300073
𝑥𝑛
1.000000
0.225400
0.281622
0.300581
0.300073
0.300076
𝑁𝑛
3.436564
0.208332
-0.003813
0.000035
0.000000
0.000000
𝐷𝑛
4.436564
-3.705517
0.201130
0.069690
-0.001878
0.000013
𝑥𝑛+1
0.225400
0.281622
0.300581
0.300073
0.300076
0.300076
𝑥6 and 𝑥7 found that the value is 0.300076 and is exact to six decimal places.
Exercise 8.2
Solve the following equations using Secant Method.
1. 𝑥 3 = 3 − 5𝑥 on [0,2]
2. 𝑥 5 = 1 − 𝑥 on [0,1]
3. 𝑥 2 + 3𝑥 − 2 = 0 on [−4, −3]
4. 𝑒 𝑥 + 2𝑥 − 6 = 0 on [1,2]
5. sin 𝑥 + 𝑒 3𝑥 = 2 on [0,1]
8.3 Numerical Solution of Linear Systems of Equations
I. Gaussian Elimination
Gaussian elimination is a process of reducing a matrix into Gaussian. Gaussian
is a resultant matrix in which all the elements below the diagonal are all zeros.
The first step of the elimination process involves assuming the first element on
the first row equal to zero and the process continues until all the elements below it are
all zeros. Row operations will be applied continuously so that the resulting matrix is a
Gaussian. Then, the next process is called back substitution.
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Example 1:
𝑥 + 2𝑦 − 𝑧 = −5
Solve { 2𝑥 − 2𝑦 + 𝑧 = 8 .
5𝑥 + 3𝑦 + 4𝑧 = 7
Solution:
1
2 −1 −5
The matrix is [2 −2
1| 8]
5
3
4 7
Consider the second row: Multiply row 1 by (-2) and add it to row 2.
For the third row: Multiply row 1 by (-5) and add it to row 3. The result is,
1
2 −1 −5
[0 −6
3 | 18]
0 −7
9 32
Lastly, multiply row 2 by (-7/6) and add it to row 3.
1
2
[0 −6
0
0
−1 −5
3 | 18]
11/2 11
Then, by back substitution,
11
2
𝑧 = 11
𝑧=2
On the second row,
−6𝑦 + 3𝑧 = 18
−6𝑦 + 3(2) = 18
𝑦 = −2
On the first row,
𝑥 + 2𝑦 − 𝑧 = −5
𝑥 + 2(−2) − 2 = −5
𝑥=1
Example 2:
2𝑥 − 4𝑦 + 𝑧 = −11
Solve { 𝑥 + 3𝑦 + 2𝑧 = −4 .
3𝑥 + 5𝑦 + 2𝑧 = −2
Solution:
2 −4 1 −11
[1
3 2| −4 ] 𝑅1 (−1/2) + 𝑅2
3
5 2 −2 𝑅1 (−3/2) + 𝑅3
2
= [0
0
−4
1 −11
5 3/2| 3/2 ]
11 1/2 29/2 𝑅2 (−11/5) + 𝑅3
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2
= [0
0
−4
1
−11
5
3/2 | 3/2 ]
0 −14/5 56/5
By back substitution,
14
56
−5𝑧= 5
𝑧 = −4
On the second row,
3
3
5𝑦 + 2 𝑧 = 2
3
3
5𝑦 + 2 (−4) = 2
3
𝑦=2
On the first row,
2𝑥 − 4𝑦 + 𝑧 = −11
3
2𝑥 − 4 (2) − 4 = −11
1
𝑥 = −2
Example 3:
𝑤 + 6𝑥 − 3𝑦 − 𝑧 = 4
−2𝑤 + 6𝑥 − 3𝑦 − 𝑧 = −5
Solve {
.
𝑤 − 3𝑥 + 𝑦 − 2𝑧 = −9
5𝑤 + 3𝑥 − 𝑦 − 2𝑧 = 7
Solution:
1
[−2
1
5
6 −3 −1 4
6 −3 −1| −5] 𝑅1 (2) + 𝑅2
−3
1 −2 −9 𝑅1 − 𝑅3
3 −1 −2 7 𝑅1 (−5) + 𝑅4
1
6 −3
= [0 18 −9
0
9 −4
0 −27 14
−1
4
−3| 3 ]
1 13 𝑅2 + 𝑅3 (−2)
3 −13 𝑅2 (3/2) + 𝑅4
1
0
=[
0
0
−1
4
−3
3
−5 | −23 ]
−3/2 −17/2 𝑅3 (1/2) + 𝑅4
6 −3
18 −9
0 −1
0 1/2
1 6
= [0 18
0 0
0 0
−3
−9
−1
0
−1
4
−3| 3]
−5 −23
−4 −20
Then, by back substitution,
−4𝑧 = −20
𝑧=5
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On the third row,
−𝑦 − 5𝑧 = −23
−𝑦 − 5(5) = −23
𝑦 = −2
On the second row,
18𝑥 − 9𝑦 − 3𝑧 = 3
18𝑥 − 9(−2) − 3(5) = 3
𝑥=0
On the first row,
𝑤 + 6𝑥 − 3𝑦 − 𝑧 = 4
𝑤 + 6(0) − 3(−2) − 5 = 4
𝑤=3
Exercise 8.3
Solve the following systems of linear equations using Gaussian Elimination.
𝑥 + 3𝑦 + 6𝑧 = −3
1. {−2𝑥 − 5𝑦 + 3𝑧 = 3
3𝑥 + 4𝑦 − 6𝑧 = 6
2𝑥 + 3𝑦 = −5
2. {2𝑥 + 3𝑦 − 𝑧 = −6
𝑦 − 4𝑧 = −7
3𝑎 − 2𝑏 − 5𝑐 = 9
3. {−𝑎 + 𝑏 − 3𝑐 = −7
2𝑎 + 𝑏 − 6𝑐 = 8
2𝑤 + 3𝑥 − 5𝑦 + 6𝑧 = 4
4𝑤 + 6𝑥 − 5𝑦 = −6
4. {
2𝑤 − 6𝑥 − 6𝑧 = −13
−6𝑤 + 3𝑥 + 10𝑦 − 3𝑧 = 10
−𝑎 − 3𝑏 + 5𝑐 − 6𝑑 = 14
4𝑎 + 3𝑏 − 6𝑑 = 3
5. {
−2𝑎 − 3𝑏 + 15𝑐 + 6𝑑 = 8
𝑏 − 5𝑐 + 3𝑑 = −7
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II. Gauss-Seidel Method
𝐴1 𝑥 + 𝐵1 𝑦 + 𝐶1 𝑧 = 𝐷1
Given {𝐴2 𝑥 + 𝐵2 𝑦 + 𝐶2 𝑧 = 𝐷2 .
𝐴3 𝑥 + 𝐵3 𝑦 + 𝐶3 𝑧 = 𝐷3
The matrix of the coefficients is,
𝐴1
𝐴
[ 2
𝐴3
𝐵1
𝐵2
𝐵3
𝐶1
𝐶2 ]
𝐶3
The system should be diagonally dominant, that is, in each row of the coefficient
of the matrix, the absolute value of the element lying on the leading diagonal should
exceeds the sum of the absolute value of all the other elements in the row. Thus,
|𝐴1 | > |𝐵1 | + |𝐶1 |
|𝐵2 | > |𝐴2 | + |𝐶2 |
|𝐶3 | > |𝐴3 | + |𝐵3 |
Then, the Gauss-Seidel method can be applied and the process can be defined
as,
𝑥 (𝑛) =
1
𝐴1
(𝐷1 − 𝐵1 𝑦 (𝑛−1) − 𝐶1 𝑧 (𝑛−1) )
1
𝑦 (𝑛) = 𝐵 (𝐷2 − 𝐴2 𝑥 (𝑛) − 𝐶2 𝑧 (𝑛−1) )
2
1
𝑧 (𝑛) = 𝐶 (𝐷3 − 𝐴3 𝑥 (𝑛) − 𝐵3 𝑦 (𝑛) )
3
Example 1:
Solve
4𝑥 + 2𝑦 − 𝑧 = 1
{2𝑥 + 5𝑦 − 𝑧 = −12 using Gauss-Seidel method. Assume 𝑥 (0) = 𝑦 (0) =
𝑥 + 3𝑦 + 6𝑧 = −1
𝑧 (0) = 0.
Solution:
The system is diagonally dominant, such that
|4| > |2| + |−1|
|5| > |2| + |−1|
|6| > |1| + |3|
By Gauss-Seidel method, the system becomes,
1
𝑥 (𝑛) = 4 (1 − 2𝑦 (𝑛−1) + 𝑧 (𝑛−1) )
1
𝑦 (𝑛) = 5 (−12 − 2𝑥 (𝑛) + 𝑧 (𝑛−1) )
1
𝑧 (𝑛) = 6 (−1 − 𝑥 (𝑛) − 3𝑦 (𝑛))
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If initially, 𝑥 (0) = 𝑦 (0) = 𝑧 (0) = 0, then, when 𝑛 = 1,
1
𝑥 (1) = 4 [1 − 2(0) + (0)] = 0.250000
1
1
𝑦 (1) = 5 [−12 − 2 (4) + 0] = −2.500000
1
1
5
𝑧 (1) = 6 [−1 − 4 − 3 (− 2)] = 1.041667
The result for succeeding iterations is shown in the table below.
Iteration
Numbers
0
1
2
3
4
𝑥
𝑦
𝑧
0
0.250000
1.760417
1.944878
1.990035
0
-2.500000
-2.895833
-2.980382
-2.996139
0
1.041667
0.987847
0.999378
0.999730
For the 4th iteration, it can be shown that the values are 𝑥 = 2, 𝑦 = −3 and 𝑧 =
1.
Example 2:
Solve
2𝑥 + 5𝑦 − 2𝑧 = 14
{ 6𝑥 − 𝑦 + 8𝑧 = −6 using Gauss-Seidel method. Assume 𝑥 (0) = 𝑦 (0) =
6𝑥 + 4𝑦 − 𝑧 = 9
𝑧 (0) = 2.
Solution:
The system is not diagonally dominant. Arranging the system such that,
6𝑥 + 4𝑦 − 𝑧 = 9
{2𝑥 + 5𝑦 − 2𝑧 = 14
6𝑥 − 𝑦 + 8𝑧 = −6
becomes diagonally dominant, hence,
|6| > |4| + |−1|
|5| > |2| + |−2|
|8| > |6| + |−1|
By Gauss-Seidel method, the system becomes,
1
𝑥 (𝑛) = 6 (9 − 4𝑦 (𝑛−1) + 𝑧 (𝑛−1) )
1
𝑦 (𝑛) = 5 (14 − 2𝑥 (𝑛) + 2𝑧 (𝑛−1) )
1
𝑧 (𝑛) = 8 (−6 − 6𝑥 (𝑛) + 𝑦 (𝑛))
If initially, 𝑥 (0) = 𝑦 (0) = 𝑧 (0) = 2, then, when 𝑛 = 1,
1
𝑥 (1) = 6 [9 − 4(2) + 2] = 0.500000
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1
𝑦 (1) = 5 [14 − 2(0.500000) + 2(2)] = 3.400000
1
𝑧 (1) = 8 [−6 − 6(0.500000) + 3.400000] = −0.700000
The result for succeeding iterations is shown in the table below.
Iteration
Numbers
0
1
2
3
4
5
6
7
8
9
10
𝑥
𝑦
𝑧
2
0.500000
-0.883333
-0.370278
-0.552884
-0.483751
-0.507598
-0.498143
-0.501179
-0.499845
-0.500207
2
3.400000
2.873333
3.056778
2.985076
3.008619
2.998595
3.001466
2.999988
3.000291
3.000051
2
-0.700000
0.271667
-0.090194
0.037798
-0.011109
0.005523
-0.001209
0.000883
-0.000080
0.000162
For the 10th iteration, it can be shown that the values are 𝑥 = −0.5, 𝑦 = 3 and
𝑧 = 0.
Example 3:
10𝑤 − 2𝑥 + 3𝑦 − 2𝑧 = −14
−𝑤 + 5𝑥 − 𝑦 + 2𝑧 = 22
Solve {
using Gauss-Seidel method. Assume 𝑥 (0) =
3𝑤 + 2𝑥 − 8𝑦 + 2𝑧 = −4
𝑤 − 𝑥 + 2𝑦 − 6𝑧 = 5
(0)
(0)
𝑦 = 𝑧 = 0.
Solution:
The system is diagonally dominant. Thus,
1
𝑤 (𝑛) = 10 (−14 + 2𝑥 (𝑛−1) − 3𝑦 (𝑛−1) + 2𝑧 (𝑛−1) )
1
𝑥 (𝑛) = 5 (22 + 𝑤 (𝑛) + 𝑦 (𝑛−1) − 2𝑧 (𝑛−1) )
1
𝑦 (𝑛) = − 8 (−4 − 3𝑤 (𝑛) − 2𝑥 (𝑛) − 2𝑧 (𝑛−1) )
1
𝑧 (𝑛) = − 6 (5 − 𝑤 (𝑛) + 𝑥 (𝑛) − 2𝑦 (𝑛) )
If initially, 𝑥 (0) = 𝑦 (0) = 𝑧 (0) = 0, then, when 𝑛 = 1,
1
𝑤 (1) = 10 [−14 + 2(0) − 3(0) + 2(0)] = −1.400000
1
𝑥 (1) = 5 [22 + (−1.400000) + 0 − 2(0)] = 4.120000
1
𝑦 (1) = − 8 [−4 − 3(−1.400000) − 2(4.120000) − 2(0)] = 1.005000
1
𝑧 (1) = − 6 [5 − (−1.400000) + 4.120000 − 2(1.005000)] = −1.418333
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CHAPTER 8
The result for succeeding iterations is shown in the table below.
Iteration
Numbers
0
1
2
3
4
5
6
𝑤
𝑥
𝑦
𝑧
0
-1.400000
-1.161167
-1.002957
-0.997571
-0.999879
-1.000041
0
4.120000
4.936100
5.002160
5.000208
4.999875
4.999982
0
1.005000
0.944004
0.990712
0.999975
1.000105
0.999999
0
-1.418333
-1.534876
-1.503949
-1.499638
-1.499924
-1.500004
For the 6th iteration, it can be shown that the values are 𝑤 = −1, 𝑥 = 5, 𝑦 = 1
3
and 𝑧 = − 2.
Exercise 8.4
Solve the following systems of linear equations using Gauss-Seidel method.
−6𝑥 − 3𝑦 + 2𝑧 = −1
1. { 𝑥 + 5𝑦 − 3𝑧 = −16
−2𝑥 + 3𝑦 + 7𝑧 = −6
4𝑎 − 8𝑏 − 3𝑐 = 14
2. { 6𝑎 + 2𝑏 − 𝑐 = −7
−2𝑎 + 𝑏 + 5𝑐 = 16
3𝑥 − 8𝑦 + 4𝑧 = −11
3. {−6𝑥 − 2𝑦 + 10𝑧 = 13
6𝑥 + 2𝑦 − 𝑧 = 5
8𝑤 + 𝑥 − 2𝑦 + 3𝑧 = 10
−2𝑤 − 9𝑥 − 𝑦 + 5𝑧 = 18
4. {
−𝑤 − 𝑥 − 6𝑦 + 3𝑧 = −2
𝑥 − 2𝑦 + 5𝑧 = −8
−𝑎 + 6𝑏 + 2𝑐 + 𝑑 = −25
5. { 2𝑎 + 𝑏 + 7𝑐 − 𝑑 = 9
2𝑎 + 3𝑏 + 6𝑑 = −6
5𝑎 + 3𝑏 − 𝑐 = 2
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8.4 Interpolation
The general formula for an (𝑛 − 1)th order polynomial is,
𝒇(𝒙) = 𝒑𝟏 𝒙𝒏−𝟏 + 𝒑𝟐 𝒙𝒏−𝟐 + ⋯ + 𝒑𝒏−𝟏 𝒙 + 𝒑𝒏
Common interpolating polynomials include:
1. First-order (linear) interpolating polynomial. This consider connecting two points.
𝑓1 (𝑥 ) = 𝑝1 𝑥 + 𝑝2
2. Second-order (quadratic or parabolic) interpolating polynomial. This consider
connecting three points.
𝑓2 (𝑥 ) = 𝑝1 𝑥 2 + 𝑝2 𝑥 + 𝑝3
3. Third-order (cubic) interpolating polynomial. This consider connecting four points.
𝑓3 (𝑥 ) = 𝑝1 𝑥 3 + 𝑝2 𝑥 2 + 𝑝3 𝑥 + 𝑝4
Example 1:
Consider the data which represent values of current at given voltages.
V (volts)
20
23
25
I (ampere)
0.0524
0.2512
0.3204
Determine the value of the current when the voltage is 22 V. Use linear and
parabolic interpolation.
Solution:
For linear interpolation:
From the table:
𝑥1 = 20
𝑓 (𝑥1 ) = 0.0524
𝑥2 = 23
𝑓 (𝑥2 ) = 0.2512
The interpolating polynomial is,
𝑓1 (𝑥 ) = 𝑝1 𝑥 + 𝑝2
Solve for 𝑝1 and 𝑝2 , thus,
0.0524 = 𝑝1 (20) + 𝑝2
;
20𝑝1 + 𝑝2 = 0.0524
0.2512 = 𝑝1 (23) + 𝑝2
;
23𝑝1 + 𝑝2 = 0.2512
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Solving the systems of equations,
𝑝1 = 0.066267
𝑝2 = −1.272933
The first-order interpolating polynomial is,
𝑓1 (𝑥 ) = 0.066267𝑥 − 1.272933
At 𝑥 = 22𝑉,
𝑓1 (22) = 0.066267 (22) − 1.272933
𝑓1 (22) = 0.1849
Therefore, using linear interpolation, the current at 22 volts is 0.1849 amperes.
For quadratic/parabolic interpolation:
From the table:
𝑥1 = 20
𝑓 (𝑥1 ) = 0.0524
𝑥2 = 23
𝑓 (𝑥2 ) = 0.2512
𝑥3 = 25
𝑓 (𝑥3 ) = 0.3204
The interpolating polynomial is,
𝑓2 (𝑥 ) = 𝑝1 𝑥 2 + 𝑝2 𝑥 + 𝑝3
Solve for 𝑝1 , 𝑝2 and 𝑝3 , thus,
0.0524 = 𝑝1 (20)2 + 𝑝2 (20) + 𝑝3
;
400𝑝1 + 20𝑝2 + 𝑝3 = 0.0524
0.2512 = 𝑝1 (23)2 + 𝑝2 (23) + 𝑝3
;
529𝑝1 + 23𝑝2 + 𝑝3 = 0.2512
0.3204 = 𝑝1 (25)2 + 𝑝2 (25) + 𝑝3
;
625𝑝1 + 25𝑝2 + 𝑝3 = 0.3204
Solving the systems of equations,
𝑝1 = −0.006333
𝑝2 = 0.3386
𝑝3 = −4.186267
The second-order interpolating polynomial is,
𝑓2 (𝑥 ) = −0.006333𝑥 2 + 0.3386𝑥 − 4.186267
At 𝑥 = 22𝑉,
𝑓2 (22) = −0.006333(22)2 + 0.3386(22) − 4.186267
𝑓2 (22) = 0.1978
Therefore, using quadratic interpolation, the current at 22 volts is 0.1978
amperes.
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Example 2:
Estimate the natural logarithm of 2 given ln 1 = 0, ln 4 = 1.386294 and ln 6 =
1.791759. Use linear and quadratic interpolation.
Solution:
The data is given as,
𝑥
1
4
6
𝑓(𝑥)
0
1.386294
1.791759
For linear interpolation:
𝑥1 = 1
𝑓 (𝑥1 ) = 0
𝑥2 = 4
𝑓 (𝑥2 ) = 1.386294
The interpolating polynomial is,
𝑓1 (𝑥 ) = 𝑝1 𝑥 + 𝑝2
Solve for 𝑝1 and 𝑝2 , thus,
0 = 𝑝1 (1) + 𝑝2
;
𝑝1 + 𝑝2 = 0
1.386294 = 𝑝1 (4) + 𝑝2
;
4𝑝1 + 𝑝2 = 1.386294
Solving the systems of equations,
𝑝1 = 0.462098
𝑝2 = −0.462098
The first-order interpolating polynomial is,
𝑓1 (𝑥 ) = 0.462098𝑥 − 0.462098
When 𝑥 = 2,
𝑓1 (2) = 0.462098
By linear interpolation, ln 2 ≈ 0.462098.
For quadratic interpolation:
𝑥1 = 1
𝑓 (𝑥1 ) = 0
𝑥2 = 4
𝑓 (𝑥2 ) = 1.386294
𝑥3 = 6
𝑓 (𝑥3 ) = 1.791759
The interpolating polynomial is,
𝑓2 (𝑥 ) = 𝑝1 𝑥 2 + 𝑝2 𝑥 + 𝑝3
Solve for 𝑝1 , 𝑝2 and 𝑝3 , thus,
0 = 𝑝1 (1)2 + 𝑝2 (1) + 𝑝3
;
𝑝1 + 𝑝2 + 𝑝3 = 0
1.386294 = 𝑝1 (4)2 + 𝑝2 (4) + 𝑝3
;
16𝑝1 + 4𝑝2 + 𝑝3 = 1.386294
1.791759 = 𝑝1 (6)2 + 𝑝2 (6) + 𝑝3
;
36𝑝1 + 6𝑝2 + 𝑝3 = 1.791759
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CHAPTER 8
Solving the systems of equations,
𝑝1 = −0.051873
𝑝2 = 0.721464
𝑝3 = −0.669590
The second-order interpolating polynomial is,
𝑓2 (𝑥 ) = −0.051873𝑥 2 + 0.721464𝑥 − 0.669590
When 𝑥 = 2,
𝑓2 (2) = 0.565846
Exercise 8.5
Solve the following problems.
1. Given the table representing the density of air at a given temperature.
𝑇, (0 𝐶)
0
20
50
𝜌, kg/m3
1.29
1.20
1.09
Find the density at 𝑇 = 250 𝐶 using linear and quadratic interpolation.
2. Estimate the value of √5 using linear and quadratic interpolation
given √2 = 1.414214, √3 = 1.732051 and √7 = 2.645751.
3. Given the table representing the current at a given resistance.
𝑅 (Ω )
1.45
3.24
4.65
𝐼 (𝑚𝐴)
122.45
194.23
223.16
Find the current at 𝑅 = 1.76Ω using linear and quadratic interpolation.
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CHAPTER 8
NUMERICAL METHODS
8.5 Numerical Integration
In evaluating definite integrals using the Fundamental Theorem, we must first
find the antiderivative of the integral. It is not always easy to find the antiderivative of
some functions.
2
Consider the ∫ 𝑒 𝑥 𝑑𝑥. The integral does not exist and no one from the methods
of integration can find for the antiderivative of that function.
Suppose that a function 𝑓(𝑥) is continuous on [𝑎, 𝑏]. If we don’t know the
𝑏
antiderivative of the function, how can we approximate for the value of ∫𝑎 𝑓 (𝑥 )𝑑𝑥. Aside
from the Reimann sums in approximating definite integrals, there are other methods of
approximations that are available. These methods are the Trapezoidal Rule, the
Simpson’s Rule and the Midpoint Rule.
I. Trapezoidal Rule
Let 𝑓(𝑥) be continuous on the interval [𝑎, 𝑏]. Consider the figure below,
𝒇(𝒙)
𝒇(𝒙𝒊 )
𝒇(𝒙𝒊−𝟏 )
𝒙𝒊−𝟏
𝒂
𝒙𝒊
∆𝒙𝒊 =
𝒃
𝒃−𝒂
𝒏
where 𝑛 are the subintervals of [𝑎, 𝑏] and each one has a length of
𝑏−𝑎
𝑛
.
Taking the two Riemann sums of the two endpoints 𝑥𝑖−1 and 𝑥𝑖 ,
∑𝑛𝑖=1 𝑓 (𝑥𝑖−1 )∆𝑥𝑖 and ∑𝑛𝑖=1 𝑓(𝑥𝑖 )∆𝑥𝑖 ,
and their average,
1
2
1
[∑𝑛𝑖=1 𝑓 (𝑥𝑖−1 )∆𝑥𝑖 + ∑𝑛𝑖=1 𝑓 (𝑥𝑖 )∆𝑥𝑖 ] = ∑𝑛𝑖=1 [𝑓 (𝑥𝑖−1 ) + 𝑓(𝑥𝑖 )] ∆𝑥𝑖
2
𝑏−𝑎
= 2𝑛 ∑𝑛𝑖=1[𝑓 (𝑥𝑖−1 ) + 𝑓(𝑥𝑖 )]
=
𝑏−𝑎
2𝑛
{[𝑓 (𝑥0 ) + 𝑓 (𝑥1 )] + [𝑓 (𝑥1 ) + 𝑓(𝑥2 )] + ⋯ + [𝑓(𝑥𝑛−1 ) + 𝑓 (𝑥𝑛 )]}
𝑏−𝑎
= 2𝑛 [𝑓(𝑎) + 2𝑓(𝑥𝑖 ) + ⋯ + 2𝑓(𝑥𝑛−1 ) + 𝑓(𝑏)]
The average is the area of the trapezoid which is approximately the area under
the curve from 𝑥𝑖−1 to 𝑥𝑖 .
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Author: Harold Jan R. Terano, ECE, MET
NUMERICAL METHODS
CHAPTER 8
Therefore,
𝒃
𝒃−𝒂
∫𝒂 𝒇(𝒙)𝒅𝒙 ≈ 𝟐𝒏 [𝒇(𝒂) + 𝟐𝒇(𝒙𝟏 ) + ⋯ + 𝟐𝒇(𝒙𝒏−𝟏) + 𝒇(𝒃)]
II. Simpson’s Rule
Simpson’s rule uses parabolas to approximate the area under a curve.
Simpson’s rule is more accurate than the trapezoidal rule.
Consider the figure below,
𝒇(𝒙)
Parabola
(𝟎, 𝒇(𝒙𝒊 ))
(−∆𝒙𝒊 , 𝒇(𝒙𝒊−𝟏))
(∆𝒙𝒊 , 𝒇(𝒙𝒊+𝟏 ))
𝒂
−∆𝒙𝒊
𝟎
∆𝒙𝒊 =
∆𝒙𝒊
𝒃
𝒃−𝒂
𝒏
Let 𝑦 = 𝐴𝑥 2 + 𝐵𝑥 + 𝐶 be the equation of the parabola passing through the three
∆𝑥
points as indicated in the illustration. The area under the curve is ∫−∆𝑥𝑖 (𝐴𝑥 2 + 𝐵𝑥 +
𝑖
𝐶 ) 𝑑𝑥.
Then,
∆𝑥
𝐴 = ∫−∆𝑥𝑖 (𝐴𝑥 2 + 𝐵𝑥 + 𝐶 ) 𝑑𝑥
𝑖
𝐴
𝐵
= [ 3 𝑥 3 + 2 𝑥 2 + 𝐶𝑥]
=
2𝐴
3
∆𝑥𝑖
−∆𝑥𝑖
∆𝑥𝑖3 + 2𝐶∆𝑥𝑖
Since the coordinates of the three points satisfy the equation of the curve 𝑦 =
𝐴𝑥 2 + 𝐵𝑥 + 𝐶, therefore,
𝑓 (𝑥𝑖−1 ) = 𝐴∆𝑥𝑖2 − 𝐵∆𝑥𝑖 + 𝐶
𝑓(𝑥𝑖 ) = 𝐶
𝑓 (𝑥𝑖+1 ) = 𝐴∆𝑥𝑖2 + 𝐵∆𝑥𝑖 + 𝐶
Equating the equations, we can show,
2𝐴∆𝑥𝑖2 = 𝑓 (𝑥𝑖−1 ) + 𝑓(𝑥𝑖+1 ) − 2𝑓 (𝑥𝑖 )
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Author: Harold Jan R. Teran
CHAPTER 8
NUMERICAL METHODS
Substituting 2𝐴∆𝑥𝑖2 = 𝑓 (𝑥𝑖−1 ) + 𝑓 (𝑥𝑖+1 ) − 2𝑓 (𝑥𝑖 )
to the equation of the area
obtained, we have,
2𝐴
𝐴 = 3 ∆𝑥𝑖3 + 2𝐶∆𝑥𝑖
∆𝑥
= 3 𝑖 [𝑓(𝑥𝑖−1 ) + 𝑓 (𝑥𝑖+1 ) − 2𝑓 (𝑥𝑖 )] + 2𝑓 (𝑥𝑖 )∆𝑥𝑖
1
1
4
3
3
3
= ∆𝑥𝑖 [ 𝑓(𝑥𝑖−1 ) + 𝑓 (𝑥𝑖+1 ) + 𝑓 (𝑥𝑖 )]
∆𝑥𝑖
= 3 [𝑓(𝑥𝑖−1 ) + 𝑓 (𝑥𝑖+1 ) + 4𝑓 (𝑥𝑖 )]
Since ∆𝑥𝑖 =
𝐴=
𝑏−𝑎
𝑛
, then,
𝑏−𝑎
[𝑓(𝑥𝑖−1 ) + 𝑓(𝑥𝑖+1 ) + 4𝑓(𝑥𝑖 )]
3𝑛
The Simpson’s rule is,
𝒃
𝒃−𝒂
∫𝒂 𝒇(𝒙)𝒅𝒙 ≈ 𝟑𝒏 [𝒇(𝒂) + 𝟒𝒇(𝒙𝟏 ) + 𝟐𝒇(𝒙𝟐 ) + 𝟒𝒇(𝒙𝟑 ) + ⋯
+𝟐𝒇(𝒙𝒏−𝟐) + 𝟒𝒇(𝒙𝒏−𝟏) + 𝒇(𝒃)]
where 𝑃 = {𝑥0 , 𝑥1 , … , 𝑥𝑛 }, is a regular partition and 𝑛 is an even integer.
III. Midpoint Rule
Let 𝑓(𝑥) be continuous on the interval [𝑎, 𝑏]. Consider the figure below,
𝒇(𝒙)
𝒇(𝒙𝒊 )
𝒇(𝒙𝒊−𝟏 )
𝒙𝒊−𝟏 + 𝒙𝒊
𝟐
𝒙𝒊−𝟏
𝒂
∆𝒙𝒊 =
𝒙𝒊
𝒃
𝒃−𝒂
𝒏
𝑏−𝑎
where 𝑛 are the subintervals of [𝑎, 𝑏] and each one has a length of 𝑛 .
Then,
𝑥 +𝑥
𝑏−𝑎
𝑥 +𝑥
𝑥 +𝑥
𝑥
+𝑥
∑𝑛𝑖=1 𝑓 ( 𝑖−1 𝑖 ) ∆𝑥𝑖 =
[𝑓 ( 0 1 ) + 𝑓 ( 1 2 ) + ⋯ + 𝑓 ( 𝑛−1 𝑛 )]
2
𝑛
2
2
2
Thus,
𝒃
∫𝒂 𝒇(𝒙)𝒅𝒙 ≈
𝒃−𝒂
𝒏
𝒙 +𝒙𝟏
[𝒇 ( 𝟎
𝟐
𝒙 +𝒙𝟐
)+𝒇( 𝟏
𝟐
𝒙
) + ⋯ + 𝒇 ( 𝒏−𝟏
+𝒙𝒏
𝟐
)]
Example 1:
2
Approximate the ∫0 √4 + 𝑥 2 𝑑𝑥 using trapezoidal rule. Use 𝑛 = 6.
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Author: Harold Jan R. Terano, ECE, MET
NUMERICAL METHODS
CHAPTER 8
Solution:
First, find the points of the partition. Using 𝑛 = 6, the length of each subinterval
is
𝑏−𝑎
𝑛
=
2−0
6
1
= 3. Then,
1
2
4
5
𝑥0 = 𝑎 = 0, 𝑥1 = 3 , 𝑥2 = 3 , 𝑥3 = 1, 𝑥4 = 3 , 𝑥5 = 3 , 𝑥6 = 𝑏 = 2
Compute 𝑓 (𝑎), 𝑓(𝑏) and 2𝑓 (𝑥𝑖 ),
𝑓 (𝑎) = 𝑓 (0) = 2.0000
1
2𝑓 (𝑥1 ) = 2𝑓 (3) = 4.0552
2
2𝑓 (𝑥2 ) = 2𝑓 (3) = 4.2164
2𝑓 (𝑥3 ) = 2𝑓 (1) = 4.4721
4
2𝑓 (𝑥4 ) = 2𝑓 ( ) = 4.8074
3
5
2𝑓 (𝑥4 ) = 2𝑓 ( ) = 5.2068
3
𝑓 (𝑏) = 𝑓 (2) = 2.8284
Add the results and multiply by
𝑏−𝑎
2𝑛
. The sum is 27.5863 and
2
𝑏−𝑎
2𝑛
1
= 6. Thus,
1
∫0 √4 + 𝑥 2 𝑑𝑥 ≈ 27.5863 (6) ≈ 4.5977
2
Using the fundamental theorem, the ∫0 √4 + 𝑥 2 𝑑𝑥 is 4.5912. The relative error in
the answer obtained using the trapezoidal rule is,
4.5977−4.5912
4.5912
≈ 0.0014 ≈ 0.14%
Example 2:
2
Approximate the ∫0 √4 + 𝑥 2 𝑑𝑥 using simpson’s rule. Use 𝑛 = 6.
Solution:
First, find the points of partition using 𝑛 = 6. The length of each subinterval is
𝑏−𝑎
𝑛
=
2−0
6
1
= 3. Then,
1
2
4
5
𝑥0 = 𝑎 = 0, 𝑥1 = 3 , 𝑥2 = 3 , 𝑥3 = 1, 𝑥4 = 3 , 𝑥5 = 3 , 𝑥6 = 𝑏 = 2
Compute 𝑓 (𝑎), 𝑓(𝑏), 4𝑓 (𝑥𝑖 ) and 2𝑓 (𝑥𝑖 ). For 4𝑓 (𝑥𝑖 ), 𝑖 = 1,3, …, and for 2𝑓 (𝑥𝑖 ),
𝑖 = 2,4, …,
𝑓 (𝑎) = 𝑓 (0) = 2.0000
1
4𝑓 (𝑥1 ) = 4𝑓 (3) = 8.1104
2
2𝑓 (𝑥2 ) = 2𝑓 (3) = 4.2164
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CHAPTER 8
NUMERICAL METHODS
4𝑓 (𝑥3 ) = 4𝑓 (1) = 8.9443
4
2𝑓 (𝑥4 ) = 2𝑓 (3) = 4.8074
5
4𝑓 (𝑥5 ) = 4𝑓 (3) = 10.4137
𝑓 (𝑏) = 𝑓 (2) = 2.8284
Add the results and multiply by
𝑏−𝑎
3𝑛
. The sum is 41.3206 and
2
𝑏−𝑎
3𝑛
1
= 9. Thus,
1
∫0 √4 + 𝑥 2 𝑑𝑥 ≈ 41.3206 (9) ≈ 4.5912
2
Using the fundamental theorem, the ∫0 √4 + 𝑥 2 𝑑𝑥 is 4.5912. Approximately the
same answers were obtained. Therefore, Simpson’s rule is more accurate than
trapezoidal rule.
Example 3:
2
Approximate the ∫0 √4 + 𝑥 2 𝑑𝑥 using midpoint rule. Use 𝑛 = 6.
Solution:
Find the points of partition using 𝑛 = 6. The length of each subinterval is
2−0
6
𝑏−𝑎
𝑛
=
1
= 3. Then,
1
2
4
5
𝑥0 = 𝑎 = 0, 𝑥1 = 3 , 𝑥2 = 3 , 𝑥3 = 1, 𝑥4 = 3 , 𝑥5 = 3 , 𝑥6 = 𝑏 = 2
𝑥 +𝑥
𝑥 +𝑥
𝑥
+𝑥𝑛
Compute, 𝑓 ( 0 2 1 ) , 𝑓 ( 1 2 2 ) , … , 𝑓 ( 𝑛−12
𝑥 +𝑥
),
1
𝑓 ( 0 2 1 ) = 𝑓 (6) = 2.0069
𝑥 +𝑥2
𝑓( 1
2
1
) = 𝑓 ( ) = 2.0616
2
𝑥 +𝑥
5
𝑥 +𝑥
7
𝑓 ( 2 2 3 ) = 𝑓 (6) = 2.1667
𝑓 ( 3 2 4 ) = 𝑓 (6) = 2.3154
𝑥 +𝑥5
𝑓( 4
2
3
) = 𝑓 ( ) = 2.5000
𝑥 +𝑥
2
11
𝑓 ( 5 2 6 ) = 𝑓 ( 6 ) = 2.7131
Add the results and multiply by,
𝑏−𝑎
𝑛
. The sum is, 13.7637 and
𝑏−𝑎
𝑛
2
1
= 6 = 3.
Thus,
2
1
∫0 √4 + 𝑥 2 𝑑𝑥 ≈ 13.7637 (3) ≈ 4.5879
2
Using the fundamental theorem, the ∫0 √4 + 𝑥 2 𝑑𝑥 is 4.5912. The relative error in
the answer obtained using the Midpoint rule is,
4.5879−4.5912
4.5912
≈ −0.00072 ≈ −0.072%
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Author: Harold Jan R. Terano, ECE, MET
NUMERICAL METHODS
CHAPTER 8
Example 4:
1
2
Approximate the ∫0 𝑒 𝑥 𝑑𝑥 using Simpson’s rule. Use 𝑛 = 6.
Solution:
Find the points of partition. Using 𝑛 = 6, the length of each subinterval is
1−0
6
1
= 6. Then,
1
1
1
2
𝑏−𝑎
𝑛
=
5
𝑥0 = 𝑎 = 0, 𝑥1 = 6 , 𝑥2 = 3 , 𝑥3 = 2 , 𝑥4 = 3 , 𝑥5 = 6 , 𝑥6 = 𝑏 = 1
Compute 𝑓 (𝑎), 𝑓(𝑏), 4𝑓 (𝑥𝑖 ) and 2𝑓 (𝑥𝑖 ). For 4𝑓 (𝑥𝑖 ), 𝑖 = 1,3, …, and for 2𝑓 (𝑥𝑖 ),
𝑖 = 2,4, …,
𝑓 (𝑎) = 𝑓(0) = 1.0000
1
4𝑓 (𝑥1 ) = 4𝑓 (6) = 4.1127
1
2𝑓 (𝑥2 ) = 2𝑓 ( ) = 2.2350
3
1
4𝑓 (𝑥3 ) = 4𝑓 (2) = 5.1361
2
2𝑓 (𝑥4 ) = 2𝑓 ( ) = 3.1192
3
5
4𝑓 (𝑥5 ) = 4𝑓 ( ) = 8.0104
6
𝑓 (𝑏) = 𝑓 (1) = 2.7183
Add the results and multiply by
1
𝑏−𝑎
3𝑛
. The sum is 26.3317 and
𝑏−𝑎
3𝑛
1
= 18. Thus,
1
2
∫0 𝑒 𝑥 𝑑𝑥 ≈ 26.3317 (18) ≈ 1.4629
The value of the
1
2
∫0 𝑒 𝑥 𝑑𝑥
using the fundamental theorem is
relative error in the answer obtain using the Simpson’s rule is,
1.4629−1.4627
1.4627
≈ 0.00014 ≈ 0.014%
Exercise 8.6
Solve the following.
4
3
1. ∫0 (1 − 𝑥 ) 𝑑𝑥, use trapezoidal method, n=5
2
2. ∫−1 𝑥(2𝑥 2 − 4) 𝑑𝑥, use midpoint rule, n=7
2
𝑥2
3. ∫1 √2𝑥3
+1
4 √3𝑥 2 −9
4. ∫2
𝑥
𝑑𝑥 , use Simpson’s rule, n=6
𝑑𝑥, use midpoint rule, n=6
5 𝑥+1
5. ∫0 𝑥2 +3 𝑑𝑥 , use Simpson’s rule, n=10
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1.4627. The
CHAPTER 8
1
NUMERICAL METHODS
𝑑𝑥
6. ∫−1 √𝑒2𝑥
3
+4
, use trapezoidal rule, n=8
𝑑𝑥
7. ∫1 9𝑥3 +6𝑥2 −5𝑥, use midpoint rule, n=5
3
𝑑𝑦
8. ∫1 𝑦(ln2 𝑦−2 ln 𝑦+5), use trapezoidal rule, n=6
3
9. ∫0 sin 𝑥 cos 𝑥 𝑑𝑥, use Simpson’s rule, n=10
0
10. ∫−1 𝑥 𝑒 2𝑥 𝑑𝑥, use midpoint rule, n=7
139
Author: Harold Jan R. Terano, ECE, MET
NUMERICAL METHODS
CHAPTER 8
140
Author: Harold Jan R. Teran
CHAPTER 9
INTRODUCTION TO
PARTIAL DIFFERENTIAL EQUATIONS
Chapter 9
INTRODUCTION TO PARTIAL
DIFFERENTIAL EQUATIONS
Chapter Outline:
9.1
9.2
Partial Differential Equations
Solutions by Direct Partial Integration
Learning Objectives:
At the end of the lesson, the students will be able to:
1. Define partial differential equations.
2. Solve partial differential equations.
Overview:
This chapter will discuss introduction to partial differential equations and
solutions by direct partial integration.
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Author: Harold Jan R. Terano, ECE, MET
INTRODUCTION TO
PARTIAL DIFFERENTIAL EQUATIONS
CHAPTER 9
9.1 Partial Differential Equations
A partial differential equation is an equation containing one or more partial
derivatives.
Examples include:
(1)
(2)
𝜕𝑢
𝜕𝑢
4 𝜕𝑥 + 5 𝜕𝑦 = 𝑐
𝜕2 𝑢
𝜕𝑥 2
𝜕2 𝑢
+ 𝜕𝑦2 = 0 (this is Laplace’s equation)
Equation (1) is a first-order partial differential equation since the derivative is 1
whereas equation (2) is a second-order partial differential equation since the highest
derivative is 2.
Example 1:
Verify that 𝑢 =
𝜕2 𝑢
𝜕𝑧 2
1
√𝑥 2 +𝑦2 +𝑧2
satisfies the partial differential equation
𝜕2 𝑢
𝜕𝑥 2
𝜕2 𝑢
+ 𝜕𝑦2 +
= 0.
Solution:
The partial differential equation
𝜕2 𝑢
𝜕2 𝑢
𝜕2 𝑢
+ 𝜕𝑦2 + 𝜕𝑧 2 = 0
𝜕𝑥 2
is called the Laplace’s
equation.
𝑢=
1
1
√𝑥 2 +𝑦 2+𝑧
= (𝑥 2 + 𝑦 2 + 𝑧 2 )−2
2
Then,
𝜕𝑢
𝜕𝑥
3
1
3
= − (𝑥 2 + 𝑦 2 + 𝑧 2 )−2 (2𝑥 ) = −𝑥 (𝑥 2 + 𝑦 2 + 𝑧 2 )−2
𝜕2 𝑢
𝜕𝑥 2
2
5
3
3
= −𝑥 [− 2 (𝑥 2 + 𝑦 2 + 𝑧 2 )−2 ] (2𝑥 ) + (𝑥 2 + 𝑦 2 + 𝑧 2 )−2 (−1)
5
𝜕2 𝑢
3
= 3𝑥 2 (𝑥 2 + 𝑦 2 + 𝑧 2 )−2 − (𝑥 2 + 𝑦 2 + 𝑧 2 )−2
𝜕𝑥 2
𝜕2 𝑢
𝜕𝑥 2
5
= (𝑥 2 + 𝑦 2 + 𝑧 2 )−2 [3𝑥 2 − (𝑥 2 + 𝑦 2 + 𝑧 2 )]
𝜕2 𝑢
5
= (𝑥 2 + 𝑦 2 + 𝑧 2 )−2 (2𝑥 2 − 𝑦 2 − 𝑧 2 )
𝜕𝑥 2
Similarly,
𝜕2 𝑢
5
𝜕2 𝑢
5
= (𝑥 2 + 𝑦 2 + 𝑧 2 )−2 (2𝑦 2 − 𝑥 2 − 𝑧 2 )
𝜕𝑦 2
And,
= (𝑥 2 + 𝑦 2 + 𝑧 2 )−2 (2𝑧 2 − 𝑥 2 − 𝑦 2 )
𝜕𝑧 2
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CHAPTER 9
INTRODUCTION TO
PARTIAL DIFFERENTIAL EQUATIONS
Thus,
𝜕2 𝑢
𝜕𝑥 2
𝜕2 𝑢
𝜕2 𝑢
+ 𝜕𝑦2 + 𝜕𝑧 2 = 0
Example 2:
𝜕2 𝑢
𝜕2 𝑢
Verify that 𝑢 = 𝑒 𝑥 sin 𝑦 is a solution of 𝜕𝑥2 + 𝜕𝑦2 = 0.
Solution:
𝑢 = 𝑒 𝑥 sin 𝑦
𝜕𝑢
= 𝑒 𝑥 sin 𝑦
𝜕𝑥
𝜕𝑢
= 𝑒 𝑥 cos 𝑦
𝜕𝑦
𝜕2 𝑢
𝜕𝑥 2
𝜕2 𝑢
𝜕𝑦 2
= 𝑒 𝑥 sin 𝑦
= −𝑒 𝑥 sin 𝑦
Therefore,
𝜕2 𝑢
𝜕2 𝑢
+ 𝜕𝑦2 = 0
𝜕𝑥 2
Exercise 9.1
Solve the following problems.
𝜕𝑢
𝜕𝑢
1. Verify that 𝑢 = 𝑥 2 − 𝑥𝑦 is a solution of 𝜕𝑥 + 2 𝜕𝑦 + 𝑦 = 0.
𝜕2 𝑢
𝜕2 𝑢
2. Verify that 𝑢 = 𝑥 sin 𝑦 − 𝑒 𝑥 cos 𝑦 is a solution of 𝜕𝑥2 + 𝜕𝑦 2 + 𝑥 sin 𝑦 = 0.
𝑥
𝑥 𝜕2 𝑧
1 𝜕2 𝑢
3. Verify that 𝑧 = 𝑥 2 𝑦 2 + 𝑦 is a solution of 𝑦 𝜕𝑥𝜕𝑦 + 2 𝜕𝑦2 = 5𝑥 2 .
9.2 Solution by Direct Partial Integration
For simple partial differential equation, direct partial integration can be done to
obtain the solution of the differential equation.
Example 1:
𝜕𝑢
Solve 𝜕𝑥 = 4𝑥 sin 𝑡.
Solution:
𝜕𝑢
𝜕𝑥
= 4𝑥 sin 𝑡
Arranging the equation, we have,
𝜕𝑢 = 4𝑥 sin 𝑡 𝜕𝑥
Integrating partially, then the term (4sin 𝑡) is considered to be constant,
∫ 𝜕𝑢 = 4sin 𝑡 ∫ 𝑥 𝜕𝑥
𝑥2
𝑢 = 4 sin 𝑡 ( 2 ) + 𝑓 (𝑡)
𝑢 = 2𝑥 2 sin 𝑡 + 𝑓(𝑡)
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Example 2:
𝜕2 𝑢
Solve 𝜕𝑥𝜕𝑦 = 2𝑥 sin 𝑦.
Solution:
𝜕2 𝑢
𝜕𝑥𝜕𝑦
𝜕
= 2𝑥 sin 𝑦
𝜕𝑢
( ) = 2𝑥 sin 𝑦
𝜕𝑦 𝜕𝑥
𝜕𝑢
𝜕 (𝑑𝑥 ) = 2𝑥 sin 𝑦 𝜕𝑦
Integrating partially with respect to 𝑦, we have,
𝜕𝑢
∫ 𝜕 (𝑑𝑥 ) = ∫ 2𝑥 sin 𝑦 𝜕𝑦
𝜕𝑢
∫ 𝜕 (𝑑𝑥 ) = 2𝑥 ∫ sin 𝑦 𝜕𝑦
𝜕𝑢
𝜕𝑥
= −2𝑥 cos 𝑦 + 𝑓(𝑥)
Then, integrate again with respect to 𝑥,
∫ 𝜕𝑢 = ∫(−2𝑥 cos 𝑦 + 𝑓(𝑥))𝜕𝑥
𝑢 = −𝑥 2 cos 𝑦 + 𝑥𝑓 (𝑥 ) + 𝑔(𝑦)
Functions 𝑓(𝑥) and 𝑔(𝑦) may be determined if there are given initial conditions
or boundary conditions.
Example 3:
Solve
𝜕𝑢
𝜕𝑥
𝜕2 𝑢
𝜕𝑥 2
= 3𝑥(𝑦 2 − 2) given the boundary conditions that at 𝑥 = 0,
= sin 𝑦 and 𝑢 = cos 𝑦.
Solution:
𝜕2 𝑢
𝜕𝑥 2
= 3𝑥(𝑦 2 − 2)
Integrating partially with respect to 𝑥,
𝜕
𝜕𝑢
( ) = 3𝑥 (𝑦 2 − 2)
𝜕𝑥 𝜕𝑥
𝜕𝑢
𝜕 (𝜕𝑥 ) = 3𝑥 (𝑦 2 − 2)𝜕𝑥
𝜕𝑢
∫ 𝜕 (𝜕𝑥 ) = ∫ 3𝑥 (𝑦 2 − 2)𝜕𝑥
𝑥2
𝜕𝑢
6𝑥 2
= 3𝑦 2 ( 2 ) − 2 + 𝑓 (𝑦)
𝜕𝑥
𝜕𝑢
𝜕𝑥
3
= 2 𝑥 2 𝑦 2 − 3𝑥 2 + 𝑓(𝑦)
𝜕𝑢
At 𝑥 = 0, 𝜕𝑥 = sin 𝑦, then,
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3
sin 𝑦 = 2 (0)2 𝑦 2 − 3(0)2 + 𝑓 (𝑦)
𝑓 (𝑦) = sin 𝑦
Thus,
𝜕𝑢
𝜕𝑥
3
= 2 𝑥 2 𝑦 2 − 3𝑥 2 + sin 𝑦
Integrating partially with respect to 𝑥,
3
𝜕𝑢 = (2 𝑥 2 𝑦 2 − 3𝑥 2 + sin 𝑦) 𝜕𝑥
3
∫ 𝜕𝑢 = ∫ (2 𝑥 2 𝑦 2 − 3𝑥 2 + sin 𝑦) 𝜕𝑥
3 𝑥3
𝑥3
𝑢 = 2 ( 3 ) 𝑦 2 − 3 ( 3 ) + 𝑥 sin 𝑦 + 𝑔(𝑦)
1
𝑢 = 𝑥 3 𝑦 2 − 𝑥 3 + 𝑥 sin 𝑦 + 𝑔(𝑦)
2
When 𝑥 = 0, 𝑢 = cos 𝑦,
1
cos 𝑦 = 2 (0)3 𝑦 2 − (0)3 + (0) sin 𝑦 + 𝑔(𝑦)
𝑔(𝑦) = cos 𝑦
Therefore,
1
𝑢 = 2 𝑥 3 𝑦 2 − 𝑥 3 + 𝑥 sin 𝑦 + cos 𝑦
Example 4:
𝜕2 𝑢
𝜕𝑢
Solve 𝜕𝑥𝜕𝑦 = 𝑒 𝑥+𝑦 , given that 𝜕𝑥 = 1 when 𝑦 = 0, and 𝑢 = 𝑦 when 𝑥 = 0.
Solution:
𝜕2 𝑢
𝜕𝑥𝜕𝑦
= 𝑒 𝑥+𝑦
Integrating partially with respect to 𝑦,
𝜕
𝜕𝑢
( ) = 𝑒 𝑥+𝑦
𝜕𝑦 𝜕𝑥
𝜕𝑢
𝜕 (𝜕𝑥 ) = 𝑒 𝑥+𝑦 𝜕𝑦
𝜕𝑢
∫ 𝜕 (𝜕𝑥 ) = ∫ 𝑒 𝑥+𝑦 𝜕𝑦
𝜕𝑢
𝜕𝑥
= 𝑒 𝑥+𝑦 + 𝑓(𝑥)
𝜕𝑢
When 𝑦 = 0, 𝜕𝑥 = 1, then,
1 = 𝑒 𝑥 + 𝑓(𝑥)
𝑓 (𝑥 ) = 1 − 𝑒 𝑥
Thus,
𝜕𝑢
𝜕𝑥
= 𝑒 𝑥+𝑦 + 1 − 𝑒 𝑥
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CHAPTER 9
Integrating partially with respect to 𝑥,
𝜕𝑢 = (𝑒 𝑥+𝑦 + 1 − 𝑒 𝑥 )𝜕𝑥
∫ 𝜕𝑢 = ∫(𝑒 𝑥+𝑦 + 1 − 𝑒 𝑥 )𝜕𝑥
𝑢 = 𝑒 𝑥+𝑦 + 𝑥 − 𝑒 𝑥 + 𝑓(𝑦)
When 𝑥 = 0, 𝑢 = 𝑦,
𝑦 = 𝑒 𝑦 − 1 + 𝑓 (𝑦 )
𝑓 (𝑦 ) = 𝑦 − 𝑒 𝑦 + 1
Therefore,
𝑢 = 𝑒 𝑥+𝑦 + 𝑥 − 𝑒 𝑥 + 𝑦 − 𝑒 𝑦 + 1
Exercise 9.2
Solve the following partial differential equations.
𝜕𝑢
1. 𝜕𝑥 = 5𝑥𝑦
𝜕𝑢
2. 𝜕𝑦 = 3𝑥 2 sin 𝑦, given 𝑢 = 4𝑥 when 𝑥 = 0.
𝜕2 𝑢
𝜕𝑢
3. 𝜕𝑥𝜕𝑦 = 2𝑥 𝑒 3𝑦 , given at 𝑦 = 0, 𝜕𝑥 = 4𝑥 2 and at 𝑥 = 2, 𝑢 = 3𝑦.
𝜕2 𝑢
𝜕𝑢
4. 𝜕𝑥2 = 3𝑥 sin 2𝑦, given that at 𝑥 = 0, 𝑢 = cos 2𝑦 and 𝜕𝑥 = 𝑦 2.
𝜕2 𝑢
𝜕𝑢
5. 𝜕𝑥𝜕𝑦 = 𝑦 𝑒 𝑥𝑦 , given that at 𝑥 = 1, 𝜕𝑦 = sin 𝑦 and at 𝑦 = 0, 𝑢 = 𝑥 3
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REFERENCES
References
John Bird, Higher Engineering Mathematics, 4th ed., (Great Britain: Elsevier
Ltd., 2006).
Alan Jeffrey, Advanced Engineering Mathematics, (USA: Harcourt/Academic
Press, 2002).
Erwin Kreyszig, Advanced Engineering Mathematics, (USA: John Wiley &
Sons, Inc., 2006).
C.F. Chan Man Fong, D. De Kee, P.N. Kaloni, Advanced Mathematics for
Engineering and Science, (USA: World Scientific Publishing Co. Pte. Ltd., 2003).
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