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Engineering Analysis 2
P.D. Ledger
Civil and Computational Engineering Centre
Swansea University
Singleton Park
Swansea SA2 8PP
Wales. U.K
Script to accompany the lecture Engineering Analysis 2 running in the summer term 2007
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Contents
1 Introduction 3
2 Linear Algebra 5
2.1 Simultaneous Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2.2 Gauss Elimination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2.2.1 Schematic representation . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2.2.2 Algorithm for m = n . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.2.3 Linear systems of equations with multiple right hand sides . . . . . . . . 13
2.2.4 The algorithm for m equations and n unknowns . . . . . . . . . . . . . . 132.2.5 Rank of a linear system of equations . . . . . . . . . . . . . . . . . . . . 14
2.3 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.3.1 Matrix definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.3.2 Computations with matrices . . . . . . . . . . . . . . . . . . . . . . . . 17
2.3.3 Matrix notation for linear systems of equations . . . . . . . . . . . . . . 19
2.3.4 Matrix inverse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.3.5 Rank of a matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.3.6 Linear independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.4 Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.4.1 Definition and properties . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.4.2 Efficient calculation of determinants . . . . . . . . . . . . . . . . . . . . 24
2.4.3 Determinants and linear equation systems . . . . . . . . . . . . . . . . . 24
2.5 Eigenvalue Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
2.5.1 Eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
2.5.2 Eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
2.5.3 Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
3 Functions of more than one variable 32
3.1 Visualisation of Functions of Two and Three variables . . . . . . . . . . . . . . . 32
3.2 Partial Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
3.2.1 Chain rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
3.2.2 Higher order partial derivatives . . . . . . . . . . . . . . . . . . . . . . . 37
3.2.3 Total differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
3.3 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
3.3.1 Line integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
3.3.2 Surface integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
3.3.3 Volume integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
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4 Sequences and Series 48
4.1 Sequences and Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
4.1.1 Graphical representation of sequences . . . . . . . . . . . . . . . . . . . 49
4.2 Finite sequences and series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
4.2.1 Arithmetical sequences and series . . . . . . . . . . . . . . . . . . . . . 50
4.2.2 Geometric sequences and series . . . . . . . . . . . . . . . . . . . . . . 51
4.2.3 Other finite series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
4.3 Limit of a sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
4.3.1 Convergent sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
4.3.2 Proprieties of convergent sequence . . . . . . . . . . . . . . . . . . . . . 54
4.3.3 Divergent sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
4.3.4 Cauchys test for convergence . . . . . . . . . . . . . . . . . . . . . . . 55
4.4 Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
4.4.1 Convergence of an infinite series . . . . . . . . . . . . . . . . . . . . . . 55
4.4.2 Tests of convergence of positive series . . . . . . . . . . . . . . . . . . . 56
4.4.3 Absolute convergence of a general series . . . . . . . . . . . . . . . . . 57
4.5 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 584.5.1 Convergence of power series . . . . . . . . . . . . . . . . . . . . . . . . 58
4.5.2 Binomial Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
4.5.3 Maclaurin Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
4.5.4 Taylor series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
5 Differential Equations 62
5.1 Classification of differential equations . . . . . . . . . . . . . . . . . . . . . . . 62
5.1.1 Ordinary and partial differential equations . . . . . . . . . . . . . . . . . 62
5.1.2 Independent and dependent variables . . . . . . . . . . . . . . . . . . . 62
5.1.3 Order of a differential equation . . . . . . . . . . . . . . . . . . . . . . . 635.1.4 Linear and nonlinear equations . . . . . . . . . . . . . . . . . . . . . . 63
5.1.5 Homogeneous and nonhomogeneous equations . . . . . . . . . . . . . 64
5.2 First order differential equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
5.2.1 Implicit and explicit solutions . . . . . . . . . . . . . . . . . . . . . . . 64
5.2.2 General and Particular Solutions . . . . . . . . . . . . . . . . . . . . . . 65
5.2.3 Boundary and initial conditions . . . . . . . . . . . . . . . . . . . . . . 65
5.2.4 Variable Separable Type . . . . . . . . . . . . . . . . . . . . . . . . . . 65
5.2.5 Separable after substitution type . . . . . . . . . . . . . . . . . . . . . . 67
5.2.6 Linear Type . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
5.2.7 More specialised types . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
5.3 Second Order ODEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
5.3.1 Homogeneous equations . . . . . . . . . . . . . . . . . . . . . . . . . . 74
5.3.2 Linear equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
5.3.3 Linear equations with constant coefficients . . . . . . . . . . . . . . . . 74
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Chapter 1
Introduction
This script is intended to accompany the lecture course Engineering Analysis 2 which will run
during the summer term 2007 and is common for Aerospace, Civil, Chemical, Electrical and
Mechanical first year undergraduate engineering students. The material contained in the script is
intended to provide the basis of what students should learn during the course, however, there willalso be occasions when additional material (in particular applications of the theory) which will
be discussed during the course and it is the students responsibility to make their own notes where
necessary.
This lecture course, aims to build on the material taught in Engineering Analysis 1 so that
students are provided with further mathematical skills that are required to solve real world en-
gineering problems. Unlike, Engineering Analysis 1, the mathematical concepts that we shall
discuss will be new to nearly all students. However, if you are already familiar with a particular
topic, then please treat it as valuable revision, as you will certainly use it again in your other
engineering courses. If you are having difficulty to understand a particular aspect of the course
please feel free to ask questions.The course itself will comprise of lecturers and handson MATLAB classes. In the lecture
classes we will discuss the theory and examples, while in the handson classes you get chance to
experiment with a computer based mathematics software called MATLAB. This software is very
useful in solving mathematical problems which would otherwise be difficult, time consuming or
impossible to solve by hand.
As well as the lectures and handson classes you are expected to conduct your own private
study. You should use the private study for reading the lecture notes and text books, attempting
exercises, completing coursework and preparing for the examination.
The course will be assessed in terms of both continual assessment and an end of semester
examination. The continual assessment will be in the form of exercises which students should
complete (alone) and hand in on a specified date. The weighting is 20% for continual assessment
and 80% for the examination. Exercise sheets will be issued each week and will state if and
when exercises are to be submitted. Late submission will normally be awarded zero marks. Late
submissions due to certified illness will be dealt with according to standard procedures. Further
details about the exercises will follow later.
The university has its own set of calculators which are available for use within examinations.
Only the university owned calculators will be permitted for use within the examination. The
model numbers for the calculators owned by the university can be found on the link
http://www.swan.ac.uk/registry/A-ZGuide/E/Examinations/
There are a range of text books that also discuss the material we shall cover in the course.
The recommended textbooks for the course are
G. James, Modern Engineering Mathematics, Third Edition, Prentice Hall, ISBN 0-13-
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018319-9
A. Croft & R. Davison Mathematics for Engineers: An Interactive Approach, Second Edi-tion, Prentice Hall, ISBN 0-13-120193-X
Copies of these books have been ordered at the university branch of Waterstones and may also
be found in the university library. If these textbooks are not to your liking or if you are havingdifficulty obtaining a copy, then you may also wish to consult any other book with Engineering
Mathematics in the title.
MATLAB is installed on the computers in rooms 42, 43, 43a, 942 and 943 of the Talbot
Building, but not those in the LIS. If you wish to use MATLAB on your own machine, a student
version may be purchased from Mathworks
http://www.mathworks.co.uk/academia/student_version/
or from Waterstones. There are many textbooks that have been written about MATLAB some
examples are
H. Moore, MATLAB for Engineers, Pentice Hall 2007
A. Biran, M. Breiner, MATLAB 6 for engineers, Prentice Hall 2002 C. VanLoan, Introduction to scientific computing : a matrix-vector approach using MAT-
LAB, Prentice Hall 2000
S.R. Otto and J.P. Denier, An introduction to programming and numerical methods in MAT-LAB, Springer 2005.
Typing MATLAB as a keyword in voyager will reveal many others. A free book published on-
line written by one of the founders of MATLAB can be found at
http://www.mathworks.com/company/aboutus/founders/clevemoler.htmlLecture notes and exercise sheets will be issued in PDF format over Blackboard. Blackboard is a
piece of software that lets you access learning material related to lectures at Swansea University.
It accessed over the Internet (either on or offcampus) at the following link
https://blackboard.swan.ac.uk/
instructions on how to logon can also be found on this site. Once logged on you will be able
to access details relating to the specific courses that you are attending. For this course, the latest
version of the lecture notes and exercise sheets will be available on Blackboard as the course
progresses.
My office is room 150 in the Talbot building. There will be office hours available for this
course, please try to restrict your enquires to the office hours, for times see the announcement on
blackboard.
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Chapter 2
Linear Algebra
In this chapter we will introduce the concept of Linear Algebra. No doubt you have come across
set of two or three simultaneous equations. We shall extend the idea of simultaneous equations in
to the wider field of linear algebra. In doing so, we shall introduce a new notation for representing
the equations, called matrices and a solution process called Gauss elimination.When exploring solutions to sets of simultaneous equations, we need to know what solutions
we should expect, a single set of values or no solution, infinitely many solutions and this is where
are discussions begin.
2.1 Simultaneous Equations
Let us start with an example.
Example
x1 + 2x2 = 5
2x1 + 3x2 = 8
Solution
This is an example for linear equation system with two equations and two unknowns. We wish to
find x1 and x2 so that both equations are fulfilled. The values x1 = 1 and x2 = 2, which are foundby substituting one equation in to another, satisfy both equations. They represent the solution of
the linear equation system.
Note that in general a linear equation system may have m equations and n unknowns. Letslook at some more examples.
Example
x1 + x2 = 4
2x1 + 2x2 = 5
Here m = 2 and n = 2.Solution
This is linear equation system has no solution. If one multiplies the first equation by 2 one obtains2x1 + 2x2 = 8 which disagrees with the second equation.
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Example
x1 x2 + x3 = 22x1 + x2 x3 = 4
Here m = 2 and n = 3.Solution
One possible solution is x1 = 2, x2 = 0 and x3 = 0 another is x1 = 2, x2 = 1 and x3 = 1. Ingeneral there are infinitely many solutions, namely x1 = 2, x2 = and x3 = where is anyreal number.
Example
x1 + x2 = 2
x1 x2 = 1x1 = 4
Here m = 3 and n = 2.Solution
This is linear equation system has no solution. Through addition of the first two equations, one
has 2x1 = 3 which disagrees with the last equation.
The set of all solutions of a linear equation system is called the solution set of the linear
equation system.
2.2 Gauss Elimination
The aim of this section is to describe the development of an efficient strategy for determining
the solution set of a linear equation system. By efficient, we mean that with the fewest possible
calculations. The procedure that we describe is called Gauss elimination. The idea of Gauss elim-
ination is to transform the linear system so that it is easier to solve. Note that the transformation
is performed so that the solution set is not changed during the process.
The following two operations transform a linear equation system in to an equivalent system.
Exchanging equations
x1 + 2x2 = 52x1 + 3x2 = 8
is equivalent to2x1 + 3x2 = 8
x1 + 2x2 = 5
It is clear that both equation systems posses the same solution set.
Addition of factored equation to another equation
x1 + 2x2 = 52x1 + 3x2 = 8
is equivalent tox1 + 2x2 = 5
x2 =
2
Here, the first equation remains the same. To obtain the revised second equation, we must multi-
plied the first equation by 2 and then subtracted it from the second equation in the original system.
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The solution set of the left equation system is the same as solution set of the equation system on
the right.
A linear equation with n equations and n unknowns is easier to solve when it is has a trian-gular form. In this case, one can easily obtain the solution set by back substitution.
Example
3x1 + 2x2 + x3 = 1
x2 x3 = 22x3 = 4
Here m = n = 3.Solution
From the third equation we have x3 = 2. Substituting this in to the second equation gives x2 = 4and finally using both values in the first equation gives x1 = 3.
Lets now attempt to solve a linear equation system using these ideas.
Example
2x2 + 2x3 = 1
2x1 + 4x2 + 5x3 = 9
x1 x2 + 2x3 = 3
Here m = n = 3.Solution
First we exchange the first and second equations
2x1 + 4x2 + 5x3 = 9
2x2 + 2x3 = 1
x1 x2 + 2x3 = 3
Next we multiply the first equation by 12
and subtract it from the third equation
2x1 + 4x2 + 5x3 = 9
2x2 + 2x3 = 1
3x2 12
x3 = 32
Finally we multiply the second equation by 32
and add it from the third equation, giving
2x1 + 4x2 + 5x3 = 9
2x2 + 2x3 = 15
2x3 = 0
Then by back substitution we find the solution x1 =72
, x2 =12
and x3 = 0. It therefore followsthat this is the only solution to the linear system, i.e. the linear set contains only this solution.
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What we have just performed is the Gauss elimination algorithm, to explain it in more details
let us consider a system where m = n = 3
a11x1 + a12x2 + a13x3 = b1
a21x1 + a22x2 + a23x3 = b2
a31x1 + a32x2 + a13x3 = b3
here the values aij , i = 1, , 3 and j = 1, , 3 and bi, i = 1, , 3 are known real numbers.The values aij are the coefficients of the unknown xj in the ith equation of the system. Thenumber bi is the right hand side of the ith equation.
2.2.1 Schematic representation
To aid the application of the Gauss elimination algorithm, we express the system in the form
x1 x2 x3 1
a11 a12 a13 b1a21 a22 a23 b2a31 a32 a33 b3
The coefficients are written in the main part of the schematic, in the headerrow stand the
respective unknowns x1, x2 and x3. The righthand side is labelled the 1column. The schematicis just another way of writing the linear system. When we wish to perform operations on the
equations, we just perform the analogue on the schematic.
Step 1
Assumption: One ofai1, i = 1, , 3 is not zero, in other words at least one values in the firstcolumn isnt zero.
a) If a11 = 0 then swap the first row with a row whose first element isnt zero. Next were-name the coefficients according to the position in which they are now lying in. Continue
to case b).
b) If a11 = 0 we create a new equivalent scheme in which the coefficients in the secondrow are given by their original value minus the value in the first row factored by a21/a11.Similarly, the values in the third row are given by their original value minus the value in
the first row factored by a31/a11.
As a result of this operation the first number in the second and third row, will now be zero. To
highlight the fact that the values in the second and third column have changed we label them
with the superscript (2). We have now eliminated the unknown x1 from the second and thirdequations:
x1 x2 x3 1
a11 a12 a13 b10 a
(2)22 a
(2)23 b
(2)2
0 a(2)32 a
(2)33 b
(2)3
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Step 2
Assumption:One ofa(2)i2 , i = 2, 3 is non zero.
a) Ifa(2)22 = 0 we swap rows and then proceed to b).
b) Ifa
(2)
22 = 0 we create a new equivalent system whose entries in the third row are given bytheir original values minus the values in the second row factored by a
(2)32 /a
(2)22 .
We have now eliminated the unknown x2 from the third equation, resulting in a scheme withtriangular form
x1 x2 x3 1
a11 a12 a13 b10 a
(2)22 a
(2)23 b
(2)2
0 0 a(3)33 b
(3)3
Step3
The equivalent system is
a11x1 + a12x2 + a13x3 = b1
a(2)22 x2 + a(2)23 x3 = b
(2)2
a(3)33 x3 = b
(3)3
The solution set can be found by back substitution.
2.2.2 Algorithm for m = n
What we have just performed can be expressed in terms of an algorithm. In the algorithm we use
the construction For j = 1, , n which means that all the lines which follow this statement areevaluated with j = 1 until the statement End is reached. At this point the lines are then executedagain, in turn, but this time with j = 2 until End is reached. The process is repeated until j = n.This enables us to write the Gauss elimination algorithm in a concise way and will turn out to be
useful for The algorithm for the case where we have n equations and n unknowns is as follows:Perform the elimination process:
For j = 1, , n 1:
Determine the row index p j, , n for which a(j)pj = 0Ifp = j exchange rows and renumber coefficients according to their new locations.For k = j + 1, , n :
Compute lkj = a(j)kj /a
(j)jj
For p = j, , n:Set a
(j+1)kp = a
(j)kp lkja(j)jp
End
Set b(j+1)k = b(j)k lkjb(j)jEnd
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End
Determine the solution by back substitution.
Set xn = bn/annFor j = n 1, , n
Set v = bj
For k = j + 1, , nCalculate v = v ajkxk
End
Set xj = v/ajj
End
Example
x1 + 2x2 + 3x3 + x4 = 5
2x1 + x2 + x3 + x4 = 3
x1 + 2x2 + x3 = 4
x2 + x3 + 2x4 = 0
Solution
First we write the equation in schematic representationx1 x2 x3 x4 1
1 2 3 1 52 1 1 1 3
1 2 1 0 4
0 1 1 2 0
We now proceed with the Gauss elimination algorithm:
x1 x2 x3 x4 1
1 2 3 1 50 -3 -5 -1 -7
0 0 -2 -1 -1
0 1 1 2 0
x1 x2 x3 x4 1
1 2 3 1 50 -3 -5 -1 -7
0 0 -2 -1 -1
0 0
23
53
73
x1 x2 x3 x4 1
1 2 3 1 50 -3 -5 -1 -7
0 0 -2 -1 -1
0 0 0 63
63
We then determine the solution through back substitution giving x4 = 1, x3 = 1, x2 = 1 andx1 = 1
What happens when we apply the Gauss elimination algorithm to a system that has no solu-
tion? Lets consider the following example
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Example
x1 + x2 = 4
2x1 + 2x2 = 5
Solution
First we write the equation in schematic representationx1 x2 1
1 1 42 2 5
Then we proceed with the Gauss elimination algorithm, givingx1 x2 1
1 1 40 0 -3
Clearly 0x1 + 0x2=
3 so the linear system has no solution.
Let us now look at an example in which the right hand side vector contains a unknown pa-
rameter.
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End
Set b(j+1)k = b
(j)k lkib(j)i
End
Set i = i + 1 and j = j + 1 , goto *.
End
Compute the solution set by back substitution
2.2.5 Rank of a linear system of equations
The rank of a linear equation system is defined as the number, r, of nonzero rows after perform-ing the Gauss elimination algorithm. Using the rank, one can immediately describe the solution
set of the linear system of equations: A linear equation system has at least one solution if
r = m, or
r < m and ci = 0, i = r + 1, , m where c is the right hand side vector after Gausselimination.
Example
x1 x2 + x3 = 22x1 + x2 x3 = 4
Solution
First we write the equation in schematic representationx1 x2 x3 1
1 1 1 22 1 -1 4
Then we proceed with the Gauss elimination algorithm, givingx1 x2 x3 1
1 1 1 20 3 -3 0
The rank of this system is therefore r = 2 and we have at least one solution, explicitly
x2 = x3
x1 = 2 + x2 x3 = 2
which means that x3 is a free parameter. Thus we have infinitely many solutions, x1 = 2,x2 = x3 = where is any real number.
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Example
x1 + x2 = 2
x1 x2 = 1
x1 = 4
Solution
First we write the equation in schematic representationx1 x2 1
1 1 21 -1 1
1 0 4
Then we proceed with the Gauss elimination algorithm, givingx1 x2 1
1 1 20 -2 -10 -1 2
x1 x2 1
1 1 20 -2 -10 0 5
2
The rank of this system is r = 2 < m = 3 and c3 = 0 so that the system has no solution.
2.3 Matrices
In the last section we introduced the Gauss elimination method for solving linear equations. To
simplify the way we write large systems of equations a new notation is introduced called matrix
notation. Matrices can be used not only in connection with linear systems of equations, but also
in mappings and in the solution of systems of differential equations.
2.3.1 Matrix definitions
A m n matrix is a schematic of mn numbers ordered in to m rows and n columns. The mnnumbers are called elements of the m n matrix.
In these lecture notes we shall use capital letters to represent matrices. The element of a
matrix A which lies on the ith row and jth column is denoted by aij or (A)ij. We write a matrixas follows
A =
a11 a12
a1n
a21 a22 a2n...
...
am1 am2 amn
For example
A =
2 3 15 1 2
is a 2 3 matrix. In the first row is the second element (A)12 = a12 = 3.
A n n matrix has an equal number of rows and columns and is called a square matrix.The two matrices A and B are said to be equal when they have the same number of rows and
columns and when the respective elements of the two matrices are the same
Aij = Bij for all i, j
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For example the following two matrices are equal5 12 4
=
10/2 13 1 22
Below we introduce some common matrices which are used in engineering:
A m n matrix is called the null matrix (or zero matrix) if every element of the matrix iszero. For example, the matrix
0 0 00 0 0
is 2 3 null matrix.
A square matrix U is called a upper triangular matrix if(U)ij = 0 for i > j. For example
U =
1 3 10 2 40 0 3
A square matrix L is called a lower triangular matrix if(L)ij = 0 for i < j. For example
L =
2 0 0 03 4 0 01 2 2 01 0 0 3
A nn matrix is called a diagonal matrix if(D)ij = 0 for i = j. The elements (D)ii = diiare called the diagonal elements. For a diagonal matrix with given diagonal elements,
d11, d22, , dnn we write D = diag(d11, d22, , dnn). For example 5 0 00 2 0
0 0 3
= diag(5, 2, 3)
The n n matrix In = diag(1, 1, , 1) is called the identity matrix. For example
I3 =
1 0 00 1 0
0 0 1
A further class of matrices are the 1-column or n
1 matrices. The n
1 matrix are
commonly known as column vectors. We write column vectors using lower case letters.The elements of column vectors are called components. Components are only identified
with a single index. For example, the 4 1 matrix
b =
24
70
is a column vector. We also have that b =
b1b2
b3b4
with b1 = 2, b2 = 4, b3 = 7 and
b4 = 0.
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2.3.2 Computations with matrices
Addition
Consider two m n matrices A and B. To add the matrices A and B together, we add therespective elements ofA and B together. Written more precisely: the m n matrix A + B with(A)ij + (B)ij is called the sum of matrices A and BExample
A =
3 1 02 2 1
B =
1 2 00 1 1
Find A + B.Solution
A + B =
3 1 02 2 1
+
1 2 00 1 1
=
4 3 02 1 2
Multiplication by a scaler
If a m n matrix is multiplied by scaler number , this means that every element of the matrix ismultiplied by . The matrix A with (A)ij = (A)ij is called the multiple of the matrix A.
Multiplication of two matrices
Let A be an m n matrix and B a n p matrix. The m p matrix AB, with (AB)ij =
nk=1(A)ik(B)kj is called the matrix product of matrices A and B.
Note that the matrix product AB can only be computed when the number of columns of matrix
A is exactly the same as the number of rows of matrix B. An illustration of matrix multiplicationis shown in Figure 2.1. In this figure, we observe how row i of matrix A is multiplied by column
m
n
n
p
p
m
B ABA
x =
ith row
jth column
(AB) ith row
jth column
ij
Figure 2.1: Illustration of matrix multiplication
j of matrix B to obtain the element (AB)ij of matrix AB. Explicitly this given as
(AB)ij = ai1b1j + ai2b2j + + ainbnj
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Example
A =
3 1 02 2 1
is a 2 3 matrix and B =
1 1 0 01 2 2 1
2 1 1 2
is a 3 4 matrix
Find AB.Solution
AB =
4 5 2 12 3 5 0
. The two elements in the first column were computed as follows
(AB)11 = a11b11 + a12b21 + a13b31 = 3 1 + 1 1 + 0 2 = 4(AB)21 = a21b11 + a22b21 + a23b31 = 2 1 + (2) 1 + 1 2 = 2
Rules
When computing with matrices, the following rules should be obeyed: For m n matrices A and B, the commutative law of addition holds
A + B = B + A
For m n matrices A, B and C the associative law of addition holds(A + B) + C = A + (B + C)
For every m n matrix A, n p matrix B and p q matrix C, the associative law ofmultiplication holds
(AB)C = A(BC) For m n matrices A and B and n p matrices C and D, the distributive law of multipli-
cation holds
(A + B)C = AC+ BC
A(C+ D) = AC+ AD
Note, however that the commutative law of multiplication does NOT hold for matrices.That is to say that in general for two matrices A and B
AB
= BA
Example
A =
2 61 3
B =
1 45 2
Find AB and BASolution
AB =
32 2016 10
= BA =
6 18
12 36
(2.1)
For every m n matrix A, it holds that ImA = AIn = A. Thus giving the name for theidentity matrix Im and In.
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Transpose
Let A be a m n matrix. Then the n m matrix AT with (AT)ij = (A)ji is called the transposeofA. A matrix is called symmetric if AT = A holds.
Example
Determine the transpose of the following matrices
A =
1 2 3 45 6 7 8
B =
2 3 53 1 2
5 2 7
Solution
AT =
1 52 63 74 8
= A is NOT symmetric BT =
2 3 53 1 2
5 2 7
= B is symmetric
The matrix transpose obeys the following rules
For general m n matrices A and B, (A + B)T = AT + BT holds For every m n matrix A and every n p matrix B, (AB)T = BTAT holds.
2.3.3 Matrix notation for linear systems of equations
We are now in a position to write the linear equation system
a11x1 + a12x2 + + a1nxn = b1... ...am1x1 + am2x2 + + amnxn = bm
(2.2)
in a much shorter way. To enable us to do this, we define the matrix
A =
a11 a1n...
...
am1 amn
and the column vectors
x =
x1...xn
b = b1...
bm
(2.3)The matrix A is called the coefficient matrix and b is called the right hand side of the linearequation system. The equation system( 2.2) is equivalent to the matrix equation
Ax = b (2.4)
To solve this linear system of equations we use the Gauss elimination method discussed ear-
lier.
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2.3.4 Matrix inverse
The matrix inverse only makes sense for square matrices. It is defined as follows: The n nmatrix X is called the inverse of matrix A ifAX = In. If the matrix A has an inverse, the matrixA is called invertible or regular, if the matrix has no inverse it is called singular. For a regularn
n matrix A, we denote its inverse by A1.
Let A and B be invertible n n matrices, then A1A = In A1 is invertible and (A1)1 = A AB is invertible and (AB)1 = B1A1
AT is invertible and (AT)1 = (A1)T
The following statements are equivalent:
A is invertible The linear equation system Ax = b is solvable for every b The linear equation system Ax = 0 has only the trivial solution x = 0The matrix inverse is very rarely computed as it is an expensive computation. In theory, one
could compute the solution to a n n linear equation system Ax = b using the matrix inverse,since A1Ax = A1b and A1A = In so that A1Ax = Inx = x = A1b. However, this is notrecommended and Gauss elimination should be used.
To compute the matrix inverse for a regular nn matrix A we proceed as follows: We denotethe matrix inverse by X and note that
AX =
a(1) a(n) x(1) x(n) = In = b(1) b(n) where a(1), , a(n) are column vectors which make up the columns of matrix A and x(1), , x(n)are column vectors which make up the inverse of A. To determine x(1), , x(n), we can solvelinear systems Ax(1) = b(1), , Ax(n) = b(n) for x(1), , x(n) where b(1), , b(n) are columnsof the identity matrix In. Then, the inverse of A is given by the matrix whose columns arex(1), , x(n).
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In practise, we perform Gauss elimination on the matrix A to decide whether the vectors arelinearly independent or not. In particular for a m n matrix we have
Ifr = n the vectors are linearly independent. Ifr < n the vectors are linearly dependent. Ifr = m the vectors are called generating. Ifr = n = m the vectors are generating and linearly independent and form a basis.
Example
Determine whether the following vectors are linearly dependent or not
111
and
000
SolutionWe form the matrix whose columns are the two vectors
A =
1 01 0
1 0
Next, we perform Gauss elimination on the system Ax = 0x1 x2 111 0 0
1 0 01 0 0
x1 x2 111 0 0
0 0 00 0 0
We observe that r = 1, m = 3 and n = 2. This means that r < n so that the system is linearlydependent and not generating.
2.4 Determinants
The determinate of a square matrix is an important aspect of linear algebra. It enables one to
characterise whether a matrix is regular or singular. With help of determinants one can discuss
linear equation systems. There also lies a connection between determinants and volumes. Further
topics of linear algebra such as eigenvalues and eigenvectors require the use of determinants.
2.4.1 Definition and properties
A determinate is a number which can be computed from each square matrix A. The number iswritten as det A or |A|. Below, we illustrate some simple examples a11 a12a21 a22
= a11a22 a12a21 (2.5)
a11 a12 a13a21
a22
a23a31 a32 a33 = a11
a22 a23
a32 a33 a12
a21 a23
a31 a33
+ a13 a21 a22
a31 a32 (2.6)
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To determine the explicit value for the case given in equation (2.6) we use the result from equa-
tion (2.5).
The definition of a determinate is a follows
For a 1 1 matrix A = (a)det A = |A| = a
Set
A =
a11 a12 a1na21 a22 a2n
......
an1 an2 ann
to be a n n matrix with n 2. For i = 1, 2, , n set A1i to be the (n 1) (n 1) matrixthat one obtains when the first row and the ith column has been deleted. Then the number
det A = |A| = a11det A11 a12det A12 + a13det A13 + (1)n+1a1ndet A1n (2.7)
is called the determinate ofA.ExampleDetermine the determinants of the following matrices
A =
3 21 2
B =
1 2 12 3 2
4 1 2
C =
1 0 1 00 4 1 22 0 2 11 0 2 2
Solution
det A = 3 21 2 = 3 2 2 1 = 4
det B =
1 2 12 3 24 1 2
= 1
3 21 2 2
2 24 2 + 1
2 34 1
= 1 4 2 (4) + 1 (10) = 2
det C =
1 0 1 02 4 1 22 0 2 11 0 2 2
= 1
4 1 20 2 10 2 2
0
2 1 22 2 11 2 2
+ 1
2 4 22 0 11 0 2
0
2 4 12 0 21 0 2
= 1 4 2 1
2 2 1
0 1
0 2 + 2 0 2
0 2+
+1
2
0 10 2 4
2 11 2 + 2
2 02 0
= 1[4 2 1 0 + 2 0] + 1[2 0 4 3 + 2 0]= 4
Some important proprieties of determinants are listed below
If two row of a square matrix are interchanged, the determinate changes sign
aj bj cj
ai bi ci
=
ai bi ci
aj bj cj
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If a one row is multiplied by a constant factor and added to another row, the determinateremains unaltered
ai + aj bi + bj ci + cj
aj bj cj
=
ai bi ci aj bj cj
If a row of a matrix is multiplied by a factor , the determinate also becomes multiplied bythat factor.
ai bi ci
=
ai bi ci
The determinate of a triangular matrix is equal to the product of the diagonal terms.
For every n n matrix A, it holds that det A = det AT If the n n matrix A is invertible then det A = 0 and det A1 = 1
detA
2.4.2 Efficient calculation of determinants
We have seen that the multiples of other rows of a matrix can be added to other rows without
altering the determinate of a matrix. Also, we noted that when the matrix is in triangular form,
the determinate is just simply the product of the diagonal terms. This means that we can use
Gauss elimination to achieve efficient calculation of the determinate. After performing Gauss
elimination, the determinate is just simply the product of the diagonal entries. Note that if rowsare exchanged during Gauss elimination, the determinate changes sign.
Example
Determine the determinant of the following matrix using Gauss elimination
A =
0 3 24 2 1
2 1 1
(2.8)
Solution
det A =
0 3 24 2 12 1 1
=
2 1 14 2 10 3 2
=
2 1 10 0 10 3 2
=
2 1 10 3 20 0 1
The determinate is then the product of the diagonal terms det A = 2 3 (1) = 6
2.4.3 Determinants and linear equation systems
The first thing to note is that when A is a n n matrix, then performing the Gauss eliminationprocedure leads one to the conclusion that det A = 0 exactly when a matrix has full rank (r = n).
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Furthermore the following statements are equivalent:
The matrix A is invertible det A = 0
After performing Gauss elimination r = n The linear equation system is solvable for every b. The solution of the linear equation system Ax = b is unique. The linear equation system Ax = 0 has only the trivial solution x = 0.If we set A to be a n n matrix, then the following holds The homogeneous linear equation system Ax = 0 has only the trivial solution when
det A = 0. The linear equation system Ax = b is solvable for any right hand sides when det A = 0. The solution of the linear equation system Ax = b is unique when det A = 0.Now if we consider the solution set of a linear equation system with n equations and n un-
knowns we have the following
If det A = 0 the homogeneous linear equation system Ax = 0 has only the trivial solution. If det A = 0 the homogeneous linear equation system Ax = 0 has infinitely many solutions.
If det A = 0 the linear equation system Ax = b has for a general right hand side vectorexactly one solution. If det A = 0 the linear equation system Ax = b has no solution or infinitely many solutions,
depending on the right hand side vector.
2.5 Eigenvalue Problems
The eigenvalue problem is one of the most important exercises in linear algebra. In what follows
we shall describe how eigenvalues and eigenvectors may be computed by hand for small matrices.
2.5.1 Eigenvalues
Let us consider a n n matrix A. We now ask the question, does a a (column) vector exist thatallows us to write the matrix vector product Ax in a particularly simple way? In other words, isthere a vector such that we can write Ax as a number multiplied by x?
Eigenvalues and eigenvectors are defined as follows
The number is called an eigenvalue of matrix A, if there exists a vector x such thatAx = x holds.
If is an eigenvalue of the matrix A, then the vector x, for which Ax = x holds, is calledthe eigenvector of matrix A corresponding to eigenvalue .
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For the moment, we just want to characterise the eigenvalue of matrix A. As we describedabove, the value is an eigenvalue of A when there is a vector x = 0 such that Ax x = 0holds. The equation can also be written as Ax Inx = 0 or (A In)x = 0. The number istherefore the eigenvalue of the matrix A when the homogeneous equation system (AIn)x = 0has a nontrivial solution. It follows from the previous section, that this is exactly the case when
det (A
In
) = 0.
Example
Determine the eigenvalues of the following matrix
A =
2 1 01 2 1
0 1 2
Solution
A
In =
2 1 01
2
1
0 1 2 Next, we compute the determinate of this matrix
det (A I) = (2 ) (2 )2 1 1 [(2 ) 0]= (2 ) (2 )2 2 = (2 + )(2 + 4 + 2)
The cubic equation det (A I) = 0 has the following roots, 1 = 2, 2 = 2 +
2 and3 = 2
2 which are also in turn the eigenvalues of matrix A.
In general for a nn matrix A we observe that det (AIn) is a polynomial ofnth degree in. We call the polynomial det (A
In) the characteristic polynomial of matrix A and denote
it by PA(). If the polynomial PA() has a root which is repeated k times, we call k thealgebraic multiplicity of eigenvalue .
We have the following properties
Every n n matrix has at least one eigenvalue. Every n n matrix has at most n eigenvalues. The algebraic multiplicity of every eigenvalue is greater or equal to 1 and less than or equal
to n.
Every n n matrix has exactly n eigenvalues when the algebraic multiplicity of eacheigenvalue is taken in to account.
For every real matrix the coefficients of the characteristic polynomial are real. In this case,the eigenvalues are either real or appear in complex conjugate pairs.
The following holds for the characteristic polynomial PA() = cnn + cn1n1 + +c1 + c0
cn = (1)ncn1 = (
1)n1(a11 + a22 +
+ ann)
c0 = det A
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Example
Determine the eigenvalues of the following matrix
A =
2 1 11 2 11 1 2
Solution
This time we use Gauss elimination to compute det (A In)
det (A In) =
2 1 11 2 11 1 2
=
1 1 2 1 2 1
2 1 1
=
1 1 2 0 1 1 + 0 1 + 1 (2 )2
=
1 1 2 0 1 1 + 0 0 4 + 5 2
= (1 )(4 5 + 2
)
Therefore we have PA() = ( 1)(4 5 + 2) = ( 1)2( 4). Therefore there aretwo eigenvalues 1 and 4.
The eigenvalue = 1 has algebraic multiplicity 2.The eigenvalue = 4 has algebraic multiplicity 1.
2.5.2 Eigenvectors
Let us set A to be an n n matrix and to be an eigenvalue of this matrix. We have seen, thatwhen the determinate of the matrix (A
In) is equal to zero, there exists an eigenvector x
= 0
to matrix A corresponding to the eigenvalue , if
(A In)x = 0 (2.9)
The set of eigenvectors corresponding to eigenvalue is equal to set the set of nontrivial so-lutions to the equation system (2.9). We call this set of nontrivial solutions the eigenspace of
A corresponding to eigenvalue and is given the symbol E. The dimension ofE is calledthe geometric multiplicity of the eigenvalue . The geometric multiplicity is always greater orequal to 1.
The span is the set of all linear combinations of a set of vectors which make up the eigenspace.
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Example
Given the following matrix
A =
2 1 11 2 11 1 2
For which we have already found that its eigenvalues are 1 = 2 = 1, and 3 = 4, now computethe corresponding eigenspaces
Solution
Eigenspace for = 1. With help of Gauss elimination we havex1 x2 x3 1
1 1 1 01 1 1 01 1 1 0
x1 x2 x3 1
1 1 1 00 0 0 00 0 0 0
The solution set is {x3 = , x2 = , x1 = |, R}, thus
E1 =
|, R
=
10
1
+
11
0
|, R
= span
10
1
,
11
0
This solution set has two free parameters. The dimension of E1 is therefore 2, we say that thegeometric multiplicity of eigenvalue = 1 is 2.Eigenspace for = 4. With help of Gauss elimination we have
x1 x2 x3 1-2 1 1 01 2 1 01 1 2 0
x1 x2 x3 11 2 1 00 3 3 00 3 3 0
x1 x2 x3 11 2 1 00 3 3 00 0 0 0
The solution has the form x3 = , x2 = , x1 = , R
E4 =
11
1
| R
= span
11
1
(2.10)
The geometric multiplicity of = 4 is 1.
2.5.3 Application
We now consider the application of eigenvalues to the vibration of a simple 2 degree of freedom
system. Consider a system in which two particles are joined by 3 springs shown in Figure 2.2.
The equations of motion for particles 1 and 2 are
m1u1 = k1u1 + k2(u2 u1)m2u2 = k3u2 k2(u2 u1)
which we can rearrange as
m1u1 + (k1 + k2)u1 k2u2 = 0m2u2 k2u1 + (k2 + k3)u2 = 0
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Thus the solution is of the form x2 = , x1 = , R so that
Ekm
=
11
| R
= span
11
The final solution to the problem is given by superposition of the different modes
u1u2
=
2j=1
Cju(j)
where u(j) = x(j) sin(jt + j) and Cj is a constant and x(j) is thejth eigenvector. Thus we have
u1u2
= C1
11
sin
k
mt + 1
+ C2
1
1
sin
3k
mt + 2
where C1, C2, 1 and 2 are found from initial conditions.
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Chapter 3
Functions of more than one variable
Last semester we looked at differentiation and integration for functions of a single variable. We
saw how we could differentiate and integrate a variety of functions and looked at their importance
in engineering. However, many of the functions that we come across in engineering depend on
more than one variable, for example the area of a rectangular plate of width x and breadth y isgiven by
A = xy (3.1)
The variables x and y are clearly independent of each other, so we say that the dependent variableA is a function of the two independent variables x and y. This is expressed by writing A = f(x, y)or A(x, y). Let us now consider the volume of a plate given by
V = xyz (3.2)
where the thickness of the plate is z. In this case V is the dependent variable and x , y and z are
independent variables. We write V = f(x,y,z) or V(x,y,z).In general if we have a variable t which is a function ofn independent variables x1, x2, x3, , xn
we can express this as
t = f(x1, x2, x3, , xn) (3.3)As for functions of one variable f(x) which we discussed last semester, the function ofn variableshas an associated domain in ndimensional space, a range and a rule that assigns each ntupleof real numbers (x1, x2, x3, , xn) in the ndimensional domain with a real number z in therange.
We do not wish to pursue deeper in to these issues as our interest here lies with the differenti-
ation and integration of functions of more than one variable. We begin this chapter with looking
at how we visualise functions of more than one variable, then we move on to the topic of partialdifferentiation. We finish the chapter by considering integrals of surfaces and volumes.
3.1 Visualisation of Functions of Two and Three variables
For purposes of illustration we restrict ourselves to functions of two or three independent vari-
ables. Let us consider the function
z = f(x, y) (3.4)
which is a function of two independent variables x and y. We have two ways of visualising
such a function: The first way uses level curves which curves in the x, y domain on which thefunction f(x, y) has a constant value. Level curves follow the same ideas as contours which areused to show elevation on a ordnance survey map. The second alternative is to plot the points
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corresponding to (x,y,z) with z = f(x, y) in a rectangular coordinate system with axis x,y,z.By doing this we end up with function being represented as a surface.
Example
We wish to visualise the function
z = x2 + y2
SolutionBy using MATLAB, we can make level surface plots and surface plots of this functions. Illustra-
tions of both are shown below
10 8 6 4 2 0 2 4 6 8 1010
8
6
4
2
0
2
4
6
8
10
64
20
24
6
5
0
5
0
20
40
60
80
x
x2+y
2
y
Note that for functions of three independent variables, eg w = f(x , y, z) we cannot plotsurfaces like we did for functions of two variables. We can, however, plot level surfaces. Level
surfaces are like level curves, they represent a surface on which w is constant.
3.2 Partial Differentiation
We recall from last semester that the derivative of a function f(x) of one variable measures theslope of the tangent to the graph of the function. If we now consider a function z = f(x, y)of two variables, slope no longer makes sense because z = f(x, y) defines a surfaces in threedimension. Consider the following two cases:
Lets start with the simplest surface z = 0 ie, a surface which is flat in both the x and ydirections, as shown in Figure 3.1 (a). If we move along a line for which y is fixed and x
is increasing, the slope of this line will be 0. Similarly if we move along a line for which xis fixed and y is increasing this line will also have zero slope.
Next we consider the surface z = x + 2y, as shown in Figure 3.1 (b). For this example,the slope is equal to 1 if we move along a line of fixed y and increasing x. If, however,we move along a line for which x is fixed and y is increasing then we find that the slope isequal to 2.
It turns out that for a general surface the slope will be different depending on which direction
we move in. To measure this a new kind of derivative is introduced called the partial derivative.
Formally the partial derivative off(x, y) with respect to x is defined as
limx0
f(x + x, y) f(x, y)x
(3.5)
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5
0
5
5
0
5
1
0.5
0
0.5
1
0
xy5
0
5
5
0
5
20
10
0
10
20
x
x+2 y
y
(a) (b)
Figure 3.1: Visualisation surfaces for z = 0 and z = x + 2y
This means that we differentiate f(x, y) with respect to x while keeping y constant (fixed). Thepartial derivative off(x, y) with respect to x is the same as measuring the slope in the x direction.We denote this partial derivative by
f
xor f/x
Note the use of curly dees to distinguish between partial differentiation and normal differ-
entiation. In writing care must be taken to distinguish between
df
dx ,
f
x and
f
x (3.6)
In a similar way to the partial derivative of f(x, y) with respect to x, we define the partialderivative off(x, y) with respect to y as
f
y= lim
x0f(x, y + y) f(x, y)
y(3.7)
which we determine by differentiating f(x, y) with respect to y by keeping x constant. Thispartial derivative is the same as measuring the slope in the y direction.
If we know both fx
and fy
we can work out slope of the surface for any direction. If we
consider a direction at an angle to the x axis the slope is given by
f
xcos +
f
ysin (3.8)
we call this the directional derivative.
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Example
Given the function f(x, y) = x2y3 + 3y + x, determine its partial derivative with respect to x andy. Hence determine its directional derivative for a direction at angle to the x axis.Solution
To find the partial derivative of f(x, y) with respect to x, we differentiate f(x, y) and keep yconstant. Thus
fx
= 2xy3 + 1
Similarly, we obtain the partial derivative of f(x, y) with respect to y, by differentiating f(x, y)while keeping x constant
f
y= 3x2y2 + 3
We obtain the directional derivative by applying formula (3.8), giving
(2xy3 + 1) cos + (3x2y2 + 3) sin
Here are some more examples
Example
Determine f/x and f/y when f(x, y) is
a) x2y2 + 3xy x + 2 b) sin(x2 3y)
Solution
a) For f(x, y) = x2y2 + 3xy x + 2 we havef
x = 2xy2
+ 3y 1f
y = 2x2
y + 3x
b) For f(x, y) = sin(x2 3y) we havef
x= cos(x2 3y)
x(x2 3y) = 2x cos(x2 3y) f
y= 3cos(x2 3y)
In the examples we have considered so far we have used partial differentiation in the context
of function of two variables. However, the concept may be extended to functions of as many
variables as we please. For a function f(x1, x2, xn) ofn variables, the partial derivative withrespect to xi is given by
f
xi= lim
xi0f(x1, x2, , xi + xi, xi+1, , xn) f(x1, x2, , xn)
xi
in practise we obtain this by differentiating the function with respect to xi while keeping all othern 1 variables constant.
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Example
Determine f/x, f/y and f/z when
f(x , y, z) = xyz2 + 3xy z
Solution
We obtain that
f
x= yz2 + 3y
f
y= xz2 + 3x
f
x= 2xyz 1
3.2.1 Chain ruleWe already came across the chain rule when we performing standard differentiation for functions
of a single variable. We now wish to extend these ideas to functions of more than one variable.
Lets consider the case where z = f(x, y) and x and y are themselves functions of two indepen-dent variables s and t. This means that we can also write z as a function of s and t, say F(s, t).If we want to differentiate z with respect to s or t we have
z
s=
z
x
x
s+
z
y
y
s
z
t=
z
x
x
t+
z
y
y
t(3.9)
We can write this in matrix notation as followszszt
=
xs
ys
xt
yt
zxzy
(3.10)
This result is called the chain rule.
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Example
Find T/r and T/ when
T(x, y) = x2 + 2xy + y3x2
and x = r cos and y = r sin
SolutionBy the chain rule
T
r=
T
x
x
r+
T
y
y
r
In this exampleT
x= 2x + 2y + 2xy3
T
y= 2x + 3x2y2
andx
r= cos
y
r= sin
so that
T
r= (2x + 2y + 2xy3)cos + (2x + 3x2y2)sin
= (2r cos + 2r sin + 2r4 cos sin3 )cos + (2r cos + 3r4 cos2 sin2 )sin
Similarly
T
= (2x + 2y + 2xy3)r sin + (2x + 3x2y2)r cos =
(2r cos + 2r sin + 2r4 cos sin3 )r sin + (2r cos + 3r4 cos2 sin2 )r cos
Example
Find dR/ds whenR(s) = cosh(x2 + 3y)
and x = s2 + 3s and y = sin sSolution
For this example, x and y are functions ofs only so
dR
ds=
R
x
dx
ds+
R
y
dy
ds
which gives
dR
ds= 2x(2s + 3) sinh(x2 + 3y) + 3 cos s sinh(x2 + 3y)
= 2(s2 + 3s)(2s + 3) sinh((s2 + 3s)2 + 3 sin s) + 3 cos s sinh((s2 + 3s)2 + 3 sin s)
= 2(2s3 + 9s2 + 9s) sinh((s2 + 3s)2 + 3 sin s) + 3 cos s sinh((s2 + 3s)2 + 3 sin s)
3.2.2 Higher order partial derivatives
So far we have considered functions like f(x, y) and found its partial derivatives fx and fy . Ifthe partial derivatives are also functions of x and y, they can also be differentiated with respect
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to x an y. We define higher order partial derivatives as follows
2f
x2=
x
f
x
2f
y2 =
yf
y
2f
xy=
x
f
y
2f
yx=
y
f
x
If fx
, fy
, 2f
xyand
2fyx
exist and are continuous, then it follows that
2f
xy
=2f
yx
(3.11)
Note, however, that if the conditions are not fulfilled these so called mixed partial deriva-
tives are not equal.
Example
For the function
f(x, y) = sin x cos y + x3ey (3.12)
find all the second order partial derivatives
Example
First we find the first order partial derivatives
fx
= cos x cos y + 3x2eyfy
= sin x sin y + x3ey
Then by differentiating these expressions again we can find the second order derivatives
2f
x2= sin x cos y + 6xey
2f
y2= sin x cos y + x3ey
2f
xy
=
cos x sin y + 3x2ey
2f
yx= cos x sin y + 3x2ey
In this case, we have that 2f
xy=
2fyx
3.2.3 Total differentiation
Let us consider the function z = f(x, y) which is a function of two variables x and y. Now letx represent a small change in x, y a small change in y and z a small change in z.
It follows thatz = f(x + x, y + y) f(x, y)
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we can rewrite this as the sum of two terms, the first of which shows the change in z due to achange in x and the second which shows the change in z due to a change in y
z = [f(x + x, y + y) f(x, y + y)] + [f(x, y + y) f(x, y)]
Next, we multiply the first term by x/x = 1 and the second term by y/y = 1
z =[f(x + x, y + y) f(x, y + y)]
xx +
[f(x, y + y) f(x, y)]y
y
By letting x, y and z tend to zero we get
dz =f
xdx +
f
ydy (3.13)
In this expression dx, dy and dz are called differentials. If z = f(x) so that it is a function ofone variable, the formula takes the form
dz =df
dxdx (3.14)
Ifw = f(x,y,z) is a function of three variables we have
dw =f
xdx +
f
ydy +
f
zdz (3.15)
We can use the idea of differentials to calculate errors. If z = f(x, y) and x and y areerrors in x and y, then the error in z is approximately given by
z fx
x + fy
y (3.16)
Example
We want to estimate
(3.01)2 + (3.97)2
Solution
Let z = f(x, y) =
x2 + y2. If we set x = 3 and y = 4 we can easily compute z =
32 + 42 =
5. Now
(3.01)2 + (3.97)2 is z when x is increased by x = 0.01 and when y is decreased by0.03, ie y = 0.03
z f
x x +
f
y y
=1
22x(x2 + y2)1/2x +
1
22y(x2 + y2)1/2y
=x
x2 + y2x +
yx2 + y2
y
=
3
5 0.01
+
4
5 (0.03) = 0.018
So
(3.01)2 + (3.97)2 5 + z = 5 0.018 4.98
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Example
The height of a cylinder is under measured by 3% and the raduis is over measured by 2% we wishto estimate the percentage error in the volume.
Solution
The volume of a cylinder is given by V = r2h so
Vr
= 2rh Vh
= r2
So that the error in the volume may be written as
V Vr
r +V
hh
= 2rhr + r2h
As we are interested in the percentage error, we divide this equation by V
V 2rhr2h
r + r2
r2hh
=2r
r+
h
h
From the question we know that rr
= 2100
and hh
= 3100
giving VV
= 1100
. This means that the
volume is overestimated by 1%.
3.3 Integration
As well as being able to differentiate multivariate functions we also need to be able to integratethem. In engineering, three types of integrals commonly occur: line integrals, surface integrals
and volume integrals. In this section we shall look at how these may be performed.
3.3.1 Line integrals
Let us consider the integral ba
f(x, y)dx where y = g(x) (3.17)
we can perform the integration in the usual way, once we have substituted y for g(x)ba
f(x, g(x))dx (3.18)
Clearly the value of the integral depends on the function y = f(x). We can interpret it as
evaluatingba
f(x, y)dx along the curve y = g(x), as shown in Figure 3.2. The result of thisintegral is no longer the area under the curve and to distinguish it from our earlier integrals we
call it a line integral.
This isnt the only type of line integral, other examples are
C
f(x, y)dxC
f(x, y)dsC
f(x, y)dtC
[f1(x, y)dx + f2(x, y)dy]
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a b
AB
C
x
y
Figure 3.2: Illustration of a line integral
Note that in the above the symbol C, this means that the integral is evaluated along the curve orpath C. The path is not restricted to two dimensions and may be in as many dimensions as weplease. It is generally preferred to use C instead of the usual limits a and b when talking aboutline integrals, as the limits of integration are usually clear from how C is defined.
ExampleEvaluate
C
xydx from (1, 0) to (0, 1) along the curve C that is the portion ofx2 + y2 = 1 in thefirst quadrant.
(1,0)
(0,1)
x
y
C
SolutionOn this curve y =
1 x2 so that
C
xydx =
01
x
1 x2dx =1
2
2
3(1 x2)3/2
01
= 13
Example
Evaluate the integral
I =
C
[(x2 + 2y)dx + (x + y2)dy]
from (0, 1) to (2, 3) along the curve C defined by y = x + 1Solution
Since y = x + 1 then dy = dx and
I =
20
[(x2 + 2(x + 1)) + (x + (x + 1)2)]dx
=
20
(2x2 + 5x + 3)dx =
2
3x3 +
5
2x2 + 3x
20
=64
3
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Example
Evaluate C
(zdx + x2dy 2ydz)
along the curve C which is specified parametrically as x = t, y = t2 and z = t3 from (0, 0, 0) to
(1, 1, 1).SolutionOn the curve C, dx = dt, dy = 2tdt and dz = 3t2dt. Also at the point (0, 0, 0) t = 0 and at thepoint (1, 1, 1) t = 1 so that
C
(zdx + x2dy 2ydz) =10
(t3dt + 2t3dt 6t4dt)
=
10
(3t3 6t4)dt
= 34
t4
6
5
t51
0
=
9
20
As we mentioned earlier, some line integrals may be given in the formC
f(x, y)ds where sindicates the arc length along the curve defined by y = g(x). One of the simplest examples ofsuch integrals is
C
ds which is equal to the length of the curve C. To evaluate this kind integralswe note that ds is given by
ds =
1 +dydx
2dx in Cartesian form
ds = dxdt
2+ dydt
2dt in parametric form
ds =
r2 +drd
2d in polar form
Furthermore, if a line integral is such that the integration is performed around a closed (sim-
ple)curve, then we denote this type of integral byC
ds with the convention that the integral isevaluated by travelling around C in an anticlockwise direction.
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Example
Evaluate the integral C
dsx2 + y2
where C is the unit square with vertices (1, 1), (1, 1), (1, 1), (1, 1).
x
y
(1,1)(1,1)
(1,1)(1,1)
AB
D E
Solution
We can break the integral into four partsC
=
BA
+
DB
+
ED
+
AE
Along AB y = 1 and ds = dx Along BD x = 1 and ds = dy Along DE y = 1 and ds = dx
Along EA x = 1 and ds = dyThus the integral becomes
C
dsx2 + y2
=
11
dx1 + x2
+
11
dy1 + y2
+
11
dx1 + x2
+
11
dy1 + y2
= 4
11
dt1 + t2
= 4[sinh1 t]11 = 8 sinh1 1
3.3.2 Surface integralsWe recall the definition of an integral of a function f(x) from Engineering Analysis 1,
ba
f(x)dx = limn
xi 0
ni=1
f(xixi
where a = x0 < x1 < < xn = b, xi = xi xi1 and xi1 xi xi. We remember thatthis integrals is equal to the area under the curve f(x) between x = a and x = b, as shown inFigure 3.3.
We now wish to extend this to integrals of functions of more than one variable. Next we
consider z = f(x, y) and a region R of the xy plane, as illustrated in Figure 3.4. We define the
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*r
n10
ba
f(x )
rx
*
rx
r1x xxx0 x
f(x)
Figure 3.3: Integral of a function of a single variable
integral off(x, y) over R by
R
f(x, y)dA = limn
Ai 0
ni=1
f(xi, yi)Ai (3.19)
where Ai is an elemental area of R and (xi, yi) is a point in Ai. As we have already seenf(x, y) represents a surface and so f(xi, yi)Ai = ziAi is the volume between the z = 0 andz = zi whose base cross section is Ai. The integral is the limit of the sum of all such volumesand so it is the volume under the surface of z = f(x, y) above R.
A
x
y
z
i
Figure 3.4: Integral of a function of two variables
If we introduce a series of lines which are parallel to the x and y axis, as shown on Figure 3.5,we can write Ai = xiyi, giving
R
f(x, y)dA =
R
f(x, y)dxdy = limn
ni=1
f(xi, yi)xiyi (3.20)
Note that we can evaluate integrals of the type R f(x, y)dxdy as repeated single integrals in xand y and consequently they are usually called double integrals. For the particular case of theintegral
R
dA we note that this equal to the area of region R.
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Example
Evaluate the integral 10
31
(x2 + y2)dxdy
Solution
If we integrate with respect to x first, then we obtain10
31
(x2 + y2)dxdy =
10
1
3x3 + y2x
x=3x=1
dy
=
10
26
3+ 2y2
dy =
26
3y +
2
3y310
=28
3
Alternatively with respect to y first
1
0
3
1
(x2 + y2)dxdy = 3
1x2y +
1
3
y3y=1
y=0
dx
=
31
x2 +
1
3
dx =
28
3
Example
Evaluate
R(x2 + y2)dA over a triangle with vertices (0, 0), (2, 0) and (1, 1).
Solution
x
y
1 2
y=2xy=x1
x
y
1 2
1
x
y
1 2
1
First, integrating with respect to x first gives R
(x2 + y2)dA =
10
x=2yx=y
(x2 + y2)dxdy
=
10
1
3x3 + y2x
x=2yx=y
dy =
10
8
3 4y + 4y2 8
3y3
dy =4
3
Next integrating with respect to y first R
(x2 + y2)dA =
10
y=xy=0
(x2 + y2)dydx +
21
y=2xy=0
(x2 + y2)dydx
Here the integrals are10
y=xy=0
(x2 + y2)dydx =
10
x2y +
1
3y3y=xy=0
dx =
10
4
3x3dx =
1
3
2
1
y=2x
y=0
(x2 + y2)dydx = 2
1x2y +
1
3
y3y=2x
y=0
dx = 2
1
8
3 4x + 4x2
4
3
x3dx = 1So
R
(x2 + y2)dA = 1 + 13
= 43
.
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3.3.3 Volume integrals
Volume integrals are evaluated by carrying out three successive integrals. Volume integrals are
of the form V
dV (3.22)
and are called triple integrals. They are evaluated in the same way as double integrals, westart by evaluating the inner integral and work outwards. The main difficulty is associated with
determining the correct limits for the integration. To aid, this one may make a sketch of the region
to be integrated. Also useful to note that if integrals are evaluated in the order x, y, z then thelimits on the y integral may depend on z but not on x.Example
A cube 0 x,y,z 1 has a variable density given by = 1 + x + y + z, what is the total massof the cube
Solution
The total mass is given by
M =
V
dV
=
10
10
10
(1 + x + y + z)dxdydz =
10
10
x +
x2
2+ xy + xz
10
dydz
=
10
10
3
2+ y + z
dydz =
10
3y
2+
y2
2+ yz
10
dz
=
10
(2 + z)dz =
2z +
z2
2
10
=5
2
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Chapter 4
Sequences and Series
This chapter investigates sequences and series and there importance in engineering. Sequences
are important and arise if a continuous function is measured or sampled at periodic intervals.
They also arise when attempts are made to find approximate solutions of equations that model
physical phenomena. Closely related to sequences are series. They are important as certainmathematical problems can expressed as series. Two well known series that we shall consider
are the Taylor and Maclaurin series.
4.1 Sequences and Series
A sequence is a set of numbers which are written down in a specific order. Examples of se-
quences are 2, 4, 6, 8 and 7, 9, 11, 13. We call each number a term of the sequence. Thecontinuation dots are sometimes used to illustrate that the sequence continues.
Often sequences arise from the evaluation of a function, for example if we consider the set
of whole number {0, 1, 2, 3, }, the set of values {f(0), f(1), f(2), f(3), } which arise fromevaluating the function on the set of whole numbers is also called a sequence. In this case, we give
the identify terms in the sequence as follows f0 = f(0), f1 = f(1) and so on. Thus the first termin the sequence is f0, the second term in the sequence is f1. If the sequence has a given numberof terms such as {f0, f1, , fn} we call it a finite sequence. Sequences like {f0, f1, , f}which extend to infinity are called infinite sequences.
If the next term in a sequence can be generated from some combination of previously com-
puted terms, the formula which gives the next term is called a recurrence relation.
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Example
One way to compute square roots is Newton formula. This states that if x is an approximationto the square root ofa, then a/x is also an approximation to
a. A better approximation can be
obtained by taking an average of the two values. Thus ifx0 is an approximation to
a then
x1 =
1
2
x0 +
a
x0
similarly
x2 =1
2
x1 +
a
x1
is a better approximation than x0. In general xn+1 given by
xn+1 =1
2
xn +
a
xn
is better approximation than xn, this is an example of recurrence relation. If we wish to compute2 then starting with x0 = 1 gives the sequence
x0 = 1 x1 =3
2= 1.5 x2 =
17
12= 1.416666(6dp) x3 =
577
408= 1.414216(6dp)
A series is obtained when terms of a sequence are added. For example, if a sequence contains
2, 4, 6, 8, 10, then by adding the terms we obtain the series
2 + 4 + 6 + 8 + 10
We can use sigma notation to write a series more concisely. For example, if a sequence contains
the integers 0, 1, 2, , n a series is given by
Sn = 0 + 1 + 2 + + n =n
k=0
k
Example
Use summation notation to write the series consisting of a) the first six odd numbers and b) the
first seven even numbers.
Solution
a) A series which sums the first six odd numbers is given by
6k=1
(2k 1) = 1 + 3 + 5 + 7 + 9 + 11
b) A series which sums the first seven even numbers is given by
7k=1
2k = 2 + 4 + 6 + 8 + 10 + 12 + 14
4.1.1 Graphical representation of sequences
Sometimes it helpful to display a sequence graphically. We can do this by plotting each term in
the sequence on a standard x, y graph. For example, terms in a particular sequence are defined
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by xn = 1 + (1)n/n, starting with n = 1 and considering terms up to n = 10 gives to 2dp0, 1.50, 0.67, 1.25, 0.80, 1.17, 0.86, 1.12, 0.89, 1.10
By plotting each term of the sequence as a graph, where the terms index is used as the x coor-dinate and the terms value is used as the y coordinate, gives the plot shown in Figure 4.1. From
1 2 3 4 5 6 7 8 9 10
0
0.5
1
1.5
x
xn
Figure 4.1: Graph of the sequence xn = 1 + (1)n/n
this figure, we can observe that values of the sequence oscillate around 1 and become closer to 1as n increases. Thus plotting a sequence can often give us valuable insights in to its behaviour.
4.2 Finite sequences and series
We now wish to look at finite sequences and series in more detail.
4.2.1 Arithmetical sequences and series
An arithmetical sequence is a sequence in which the difference between successive terms is a
constant number. Examples of arithmetical sequences are {0, 3, 6, 9, 12, 15} and {1, 0, 1, 2, 3}.Traditionally arithmetical series were called arithmetical progressions, however the former
name is now preferred. We can write arithmetical sequences as {a + kd}n1k=0 where a is thefirst term, d is the difference between the terms and n is the number of terms in the sequence. So,for the first example a = 0, d = 3 and n = 6, for the second example a = 1, d =
1 and n = 5.
The sum of terms in an arithmetical sequence is an arithmetical series. In general this canbe written as
Sn = a + (a + d) + (a + 2d) + + [a + (n 1)d] =n1k=0
(a + kd) (4.1)
We can obtain an expression for the sum ofn terms in the series. If we expand the summationand then write it in reverse order we have
Sn = a + (a + d) + (a + 2d) + + [a + (n 1)d]Sn = [a + (n 1)d] + [a + (n 2)d] + [a + (n 3)d] + a
Now if we add these expressions we obtain
2Sn = [2a + (n 1)d] + [2a + (n 1)d] + [2a + (n 1)d] + [2a + (n 1)d]
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Thus giving
Sn =1
2n[2a + (n 1)d] (4.2)
as the sum ofn terms of an arithmetical series.Example
How many terms of the arithmetical sequence 2, 4, 6, 8,
will give rise to 420?
SolutionFor this example, a = 2, d = 2, Sn = 420, we need to find n
Sn = 420 =1
2n[4 + 2(n 1)] = 2n + n(n 1)
Thus
n2 + n 420 = 0 n = 1
1 4(420)2
Hence n = 20 or n = 22, since n must be a positive number, n = 20Example
A building company offers to place a foundation pile at a cost of 100 pounds for the first metre,110 pounds for the second metre and increasing at a cost of 10 pounds per metre thereafter. It is
decided to set piles at 5 metres.
a) What is the the total cost of the piling?
b) What is the cost of piling the last metre?
Solution
a) The cost is the sum of the arithmetic series where a = 100, b = 10 and n = 40
Sn = 52
[2(100) + (5 1)10] = 600 pounds
b) The cost of piling the last metre is given by the fifth term in the sequence. This 100 + (5 1)10 = 140 pounds.
4.2.2 Geometric sequences and series
A geometric sequence is one in which the ratio of successive terms is a constant number. Ex-
amples of geometric sequences are {3, 6, 12, 24, 48} and {1, 12
, 14
, 116
, 132
}. A geometricsequence always takes the form
{ark
}n1k=0 where a is the first term in the sequence, r is the ratio
between the terms and n is the number of terms in the sequence. Thus in the first example a = 3,r = 2 and n = 5, for the second example a = 1, r = 1
2and n = 5. Geometric sequences are
sometimes still called geometric progressions. The sum of a geometric sequence is a geometric
series. The general geometric series has the form
Sn = a + ar + ar2 + ar3 + + arn1 =
n1k=0
ark (4.3)
To obtain the sum Sn, we first multiply the equation by r
rSn = ar + ar2 + ar3 + ar4 + + arn
then if we subtract this from Sn we obtain
(1 r)Sn = a arn
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so that
Sn =n1k=0
ark = a1 rn1 r (4.4)
Example
An insurance company garantees that, for a fixed annual premium payable at the beginning of
each year, for a period of 25 years, the return will be equal to the premium paid together with 3%compound interest. For an annual premuim of 250 pounds what is the guaranteed sum at the end
of25 years?Solution
The first year premium earns interest for 25 years and so grantees 250(1 + 0.03)25
The second year premium earns interest for 24 years and so grantees 250(1 + 0.03)24
...
The final year premium earns interest for 1 year and so grantees 250(1 + 0.03)The total sum is therefore
250[(1.03) + (1.03)2 + + (1.03)25]the term inside the square brackets is a geometric sequence. Taking a = 1.03, r = 1.03 andn = 25 gives the total cost as
250
1.03
(1 1.0325)(1 1.03)
= 9388 pounds
4.2.3 Other finite series
Sometimes engineers are required to use finite series other than arithmetical and geometrical
sequences. We investigate a method that can be generalised to finding the sums of different finite
series and apply it to the case of finding the sum of squares.
We wish to find the summation of squares
Sn = 12 + 22 + 32 + + n2 =
nk=1
k2
To do this we use the identity (k + 1)3 k3 = 3k2 + 3k + 1. This means we can writen
k=1[(k + 1)
3
k3
] =
nk=1(3k
2
+ 3k + 1)
In this expression, we can expand the left hand side to find that
23 13 + 33 23 + 43 33 + + (n + 1)3 n3 = (n + 1)3 1and the right hand side is equal to
3n
k=1
k2 + 3n
k=1
k +n
k=1
1
We already know that nk=1 k = 12n(n + 1) and nk=1 1 = n so this means that(n + 1)3 1 = 3
nk=1
k2 +3n
2(n + 1) + n
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a
a+
a
nN
an
Figure 4.2: Convergence of a sequence
which finally gives
Sn =n
k=1
k2 = 16
n(n + 1)(2n + 1) (4.5)
4.3 Limit of a sequence
4.3.1 Convergent sequences
We previously saw how we could use Newtons formula to gain an ever improving approximation
to the square root of a value. Starting with 1 we obtained the following improving approximationsto
2
x0 = 1x1 = 1.50x2 = 1.42
if the process is continued we would obtain
x22 = 1.41x23 = 1.41
indeed for n 22 we have xn = 1.41 to 2dp. We observe that the difference between x22 andx23 is indistinguishable when the numbers are expressed to two decimal places, in other wordsthe difference is less than the rounding error. When this happens, we say that the sequence tends
to a limit or has a limiting value or converges or that it is convergent.Given a general sequence {ak}k=0 we say it has the limiting value a as n becomes large, if
given a small positive number , an differs from a by less than for all sufficiently large n, ie
an a as n if, given any > 0 , there is a numberN such that |a an| < for alln > N
We remark that stands for tend to the value or converges to the limit. An alternativenotation would be to write
limn
an = a (4.6)
We illustrate this process graphically in Figure 4.2.Note that the limit of a sequence need not actually be an element of the sequence. For example
{n1}n=1 has the limit of0, but 0 is not an element of the sequence.
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4.3.2 Proprieties of convergent sequence
It turns out that a convergent sequence satisfies a number of properties which are given below
Every convergent sequence is bounded; that is, if {an}n=0 is convergent then there is apositive number M such that |an| < M for all n.
If{an} has limit a and {bn} has limit b then {an + bn} has limit a + b {an bn} has limit a b {anbn} has limit ab {an/bn} has limit a/b for bn = 0 and b = 0.
Example
Find the limits of the sequence {xn}n=0 when xn is given by
a) xn =n
n + 1b) xn =
2n2 + 3n + 1
5n2 + 6n + 2
Solution
a) xn = n/(n + 1) leads to the sequence {0, 12 , 34 , 45 , }. Already from these values it seams thatxn 1 as n . We can prove this by rewriting xn as
xn = 1 1n + 1
As n increases 1/(n + 1) becomes smaller and smaller, thus we have
limn
n
n + 1= 1
b) Now considering xn =2n2+3n+15n2+6n+2
it is easiest to divide the numerator and denominator by the
highest power ofn, giving
xn =2 + 3
n+ 1
n2
5 + 6n
+ 2n2
We have that limn 2 + 3n +1n2
= 2 and limn 5 + 6n +2n2
= 5. Hence we have that
limn 2n
2
+ 3n + 15n2 + 6n + 2 = 25
4.3.3 Divergent sequences
To illustrate the fact that not all sequences converge we consider the following geometric se-
quence
an = rn r constant (4.7)
For this sequence we have
limn an = 0 (1 < r < 1)1 (r = 1)
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ifr > 1 the sequence increases without bound as n and we say that it diverges. Ifr = 1the sequence has takes the values of alternating 1, and has no limiting value. If r < 1 thesequence is unbounded and the terms alternate in sign.
4.3.4 Cauchys test for convergence
The following test for convergence is used in a computational context. If we do not know the limit
a to which a sequence {an} converges we cannot measure |a an|. However, in a computationalcontext where we often use a recurrence relationship to compute the sequence {an}, we say thatis has converged when all subsequent terms yield the same level of approximation required. We
say that a sequence of finite terms is convergent if for any n and m > N
|an am| < (4.8)
where is specified. This means that a sequence terns to a limit if all the terms of the sequencefor n > N are restricted to an interval that can be made arbitrarily small by making N arbitrarily
large. This is called Cauchys test for convergence.
4.4 Infi nite Series
We must exercise care when dealing with infinite series as mistakes can be made if they are not
dealt with correctly. If we consider the series
S = 1 2 + 4 8 + 16 32 +
then by multiplying it by 2 we obtain
2S = 2 4 + 8 16 + 32 64 +
if we add these equations we might come to the conclusion that 3S = 1 or S = 13
, however this
result is clearly incorrect. To avoid making such mistakes we have introduce methods for dealing
with infinite series correctly.
4.4.1 Convergence of an infinite series
We have already seen that series and sequences are closely related. When the sum Sn of a seriesofn terms tends to a limit as n
we say it is convergent. Provided that we can express Sn
is a simple form it is usually easy to say whether or not the series converges. When considering
infinite series, the sequence of partial terms is taken to the limit.
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Example
We wish to examine the following series for convergence
a) 1 + 3 + 5 + 7 + 9 + b) 12 + 22 + 32 + 42 +
c) 1 +1
2+
1
4+
1
8+
1
16+
d)1
1 2 +1
2 3 +1
3 4 +1
4 5 +
a) The first case is an arithmetic sequence which we can write as
Sn =n1k=0
(2k + 1) = 1 + 3 + 5 + + (2n 1) = n2
we can see that Sn as n and the series does not converge to a limits. It is an exampleof a divergent series.
b) The second case can be written as
Sn = 12 + 22 + 32 + + n2 = 1
6n(n + 1)(2n + 1)
This is another example where Sn as n , ie the series is divergent.c) For the third example
Sn = 1 +1
2+
1
4+ + 1
2n1
we have a geometric sequence, the sum can be written as
Sn =1 1
2n
1 12
= 2
1 1
2n
we have that as n , Sn 2, hence the sum converges to 2.d) In the final example we have
Sn =1
2+
1
6+
1
12+
1
20+ + 1
n(n + 1)=
nk=1
1
k(k + 1)=
nk=1
1
k
nk=1
1
k + 1
Expanding we have
Sn = 1 12
+1
2 1
3+
1
3 1
4+ + 1
n 1
n + 1= 1 1
n + 1
thus as n , Sn 1, hence the sum converges to 1.
4.4.2 Tests of convergence of positive series
Unfortunately the sum of a series cant always be expressed in a closed form expression. For
such cases we use a series of tests to examine the convergence of a series.
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Comparison Test
Given a sequence
k=0 ck which consists of positive terms (ck 0 for all k) which is convergent,then if we have a different
