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Elminia University
Faculty of Engineering
ngineering Mathematics
Part 1
Dr. li Mohamed Eltamaly
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Preface
Most complex scientific and engineering models of the real world are
Differential equations. Here are some that I know of: heat flow,
electrostatic potential, waves (radio, light, sound, water), metal beam
bending, quantum mechanics, hydrogen bombs, electrons in telegraph
wires, optics, classical mechanics, general relativity, distributions of
organisms, ice sheets, tsunamis, air flow, ocean currents, weather,
auroras, blood flow, plate tectonics, supernovas. For this reason weintroduce this notes for students in faculty of engineer.
By the end of the course the student should:
•
be familiar with the concept of a complex number and be able
perform algebraic operations on complex numbers, both with
numeric and symbolic entries, solve simple equations with
complex roots, and in particular describe geometrically the
roots of unity;
• be familiar with the concept of a matrix and be able to perform
algebraic operations on matrices, both with numeric and
symbolic entries, be able to define a determinant and calculate
one both directly and by using row and column operations,
understand the definition and use of the inverse of a non-
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singular matrix, and be able to solve simple systems both using
inverses and reduction to triangular form, and be able to
compute inverses using Gaussian reduction and explain the
method in terms of elementary matrices;
•
be familiar with many topics in calculus like limits,
differentiations, and all methods of integrations;
• be familiar with first order differential equations (linear and
nonlinear) and their solution by many techniques;
• be familiar with many engineering applications of first order
differential equations like falling bodies, the time rate of change
in temperature of an object varies as the difference in
temperature between the object and surroundings, Chemical
Applications, time required for liquid tanks to get empty, Half
Life Of Nuclear Materials, and Electrical Circuits;
• be familiar with solution of higher order linear differential
equations with constant coefficients and Cuchy differential
equation and their solution by many techniques;
• be familiar with many engineering applications of higher order
differential equations like, Free Oscillation of suspended
bodies, Bending of Beams, and Electrical Circuits;
• be familiar with the benefits of Laplace and inverse Laplace
transforms, using Laplace transform for solving differentialequations, finding Laplace transform of any periodical and non
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periodical waveforms, using Laplace transform to solve many
application problems like Free Oscillation of suspended bodies,
Bending of Beams, and Electrical Circuits;
• be familiar with finding Fourier transform of any waveform by
advanced techniques like jump technique;
• be familiar with curve fitting by using least square technique and
using this technique to fiend Fourier transform for any waveform
numerically;
• be familiar with using power series for solving linear differential
equations of second order, and Bessel function;
•
be familiar with partial differentiation and solving partial
differential equations by many techniques as separation of
variables, Laplace transform, and Fourier transform;
• be familiar with solving differential equations which governs the
conduction of heat in solids;
•
be familiar with eigen values and eigen vectors and using them
for solving simultaneous linear differential equations;
• be familiar with special functions like Gamma and Beta functions;
• be familiar with many topics in numerical analysis like Numerical
solution of equations by many techniques like Simple Iteration,
Bisection, false position, Newton Raphson and secant method; and
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• be familiar with polynomial interpolation and numerical solution of
differential equations by many techniques like Euler’s and Runge-
Kutta’s method.
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Contents
Part 1
Chapter
No.
Title Page
No.
1 Mathematical Numbers 1
2 Matrices 42
3 Calculus 74
4 Ordinary Differential
Equations
101
5 Linear Differential Equations
Of Higher Order
151
6 Laplace Transforms 190
7 Fourier Series238
8 Least Square Technique 259
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References
Contents
Part 2
Chapter
No.
Title Page
No.
9 Power Series Solution Of
Differential Equations
285
10 Partial Differential Equations 346
11 Simultaneous Linear
Differential Equations
389
12 Special Functions 459
13 Numerical Analysis 478
Appendix 543
References 574
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Chapter 1
Mathematical Numbers
1.1 Natural Numbers
Natural numbers known as counting numbers are the numbers
beginning with 1, with each successive number greater than its
predecessor by 1. If the set of natural numbers is denoted by N, then
N = { 1, 2, 3, ......}
1.2 Whole Numbers
Whole numbers are the numbers beginning with 0, with each
successive number greater than its predecessor by 1. It combines the
set of natural numbers and the number 0. If the set of whole
numbers is denoted by W, then
W = { 0, 1, 2, 3, .......}
1.3 Integer Numbers
Integers are the numbers that are in either (1) the set of whole
numbers, or (2) the set of numbers that contain the negatives of the
natural numbers. If the set of integers is denoted by I, then
I = {......, -3, -2, -1, 0, 1, 2, 3, ......}
Positive integers are the numbers in I greater than 0. Negativenumbers are the numbers in I less than 0.
The number zero is neither positive nor negative, i.e., it is both
non-positive and non -negative.
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Mathematical Numbers2
Given the above definitions, the following statements about integers
can be made:
(1) N is the set of positive integers.
(2) W is the union of N and the number 0.
(3) The set of numbers that contain the negatives of the numbers in
N is the set of negative integers.
(4) I is the union of W and the set of negative integers.
1.4 Real Number Line
The set of real numbers can be pictorially represented by the real
number line. It is a straight line, whose "origin" is designated by the
number 0, and continues in both directions. All the positive integers
are ordered, in ascending order from left to right, to the right side of
0; all the negative integers are ordered, in descending order from
right to left, to the left side of 0. Notches are marked to denote the
position of these integers in the following figure (Fig.1).
Fig.1
Every point on the line corresponds to a real number, and every
real number can be paired with a point on this number line. If the
real number is an integer, its point on the number line coincides with
one of the notches for an integer; otherwise, its point lies between
two successive notches. All real numbers represented by points to
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Chapter One 3
the right of the number 0 are positive, while all real numbers
represented by points to the left of the number 0 are negative.
1.5 Absolute Values
The absolute value of a real number is the distance between its
corresponding point on the number line and the number 0. The
absolute value of the real number a is denoted by |a|.
From the diagram shown in Fig.2, it is clear that the absolute
value of non-negative numbers is the number itself, while the
absolute value of negative integers is the negative of the number.
Thus, the absolute value of a real number can be defined as follows:
For all real numbers a,
(1) If a > 0, then aa = .
(2) If a < 0, then aa −= .
Fig.2
Example 1:
| 2 | = 2| -4.5 | = 4.5
| 0 | = 0
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Mathematical Numbers4
1.6 Complex Numbers
1.6.1 Introduction
The solution of a second order equation 02 =++ cbxax can be
obtained by the famous formula,a
acbb x
2
42 −±−=
For example, if 022 =−+ x , then we have:
( )2
31
2
91
2
811 ±−=
±−=
+±−= x
21 −=∴
or x As we see there is no problems with solving the above equation. But
if we solve the equation 0565 2 =+− x in the same way, we get:
( )10
646
10
100366 −±=
−±= x
And the next stage is now to determine the square root of (-64).
Is it (i) 8, (ii) -8, (iii) neither?It is, of course, neither, since + 8 and 8− are the square roots of
64 and not of (-64). In fact, ( )64− cannot be represented by an
ordinary number, for there is no real number whose square is a
negative quantity. However, 64*164 −=−
And therefore we can write:
( ) ( ) 1864*164*164 −=−=−=−
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Chapter One 5
Of course, we are still faced with ( )1− , which cannot be
evaluated as a real number, for the same reason as before, but, if we
can replace 1− with the letter j, then ( ) ( ) 88*164 j=−=− We now have a way of finishing off the quadratic equation we
started before as following:
0565 2 =+−
( )10
86
10
646
10
100366 j x
±=
−±=
−±=
8.06.0 +=∴ or 8.06.0 j x −=
1.6.2 Powers of j
( )
( ) ( ) 11
*1*
1
1
22
24
23
2
=−==
−=−==
−=
−=
j j
j j j j j
j
j
Note especially the last result: 14 = j . Every time a factor 4 j
occurs, it can be replaced by the factor 1, so that the power of j is
reduced to one of the four results above. In the same way we can
replace 12 −= j with –1.
The complex number 41 j x +=
, consists of two separate terms,1, and 4 j These terms cannot be combined any further, since the
second is an imaginary number (due to its having the factor j).
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Mathematical Numbers6
In such an expression as 41 j+=
1 is called the real part of x
4 is called the imaginary part of x
The two together form what is called a complex number .
So, a Complex number = ( Real part ) + j( Imaginary part )
Complex numbers is very important especially in some
engineering application like electrical and mechanical engineering.
So we have to fully understand how to carry out the usual
arithmetical operations.
1.6.3 Addition and Subtraction of Complex Numbers.
Addition and Subtraction are quite easy as shown in the following
example:
Example 2 Find the results of the following arithmetical operations.
( ) ( )2673 j j −++ .
Solution :
( ) ( ) ( ) ( ) 59276326732673 j j j j j j +=−++=−++=−++So, in general, ( ) ( ) ( ) ( )d b jca jd c jba +++=+++
1.6.4 Multiplication of. Complex Numbers
The following example illustrate the multiplication process in
complex numbers.
Example 3 Find the results of the following arithmetical operations.
( )( )7532 J j ++
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Chapter One 7
Solution: These are multiplied together in just the same way as
you would determine the product ( )( )7532 j j ++ . Form the product
terms of
( )( )
2911
212910
21141510
7*37*25*35*27532
2
2
j
j
j j j
j j j j j
+−=
−+=
+++=
+++=++
If the expression contains more than two factors, we multiply the
factors together in stages:
( )( )( ) ( )
( )( )
( )( )
5136585122
58292222
212911
21212910
2121141510217532
2
2
j j
j j j
j j
j j
j j j j j j j
+=++−=
−++−=
−+−=
−−+=
−+++=−++
Example 4 Find the results of the following arithmetical
operations. ( )( )8585 j j −+
Solution:
( )( )
896425
644040258585 2
=+=
−−+=−+ j j j j j
In spite of what we said above, here we have a result containing no
imaginary term. The result is therefore entirely real. This is rather an
exceptional case. If we look at the two complex numbers we can
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Mathematical Numbers8
find that they are identical except for the middle sign in the brackets
are different. These two complex numbers called conjugate complex
numbers and the product of two conjugate complex numbers is
always entirely real. In general we can say:
( )( ) 22 bababa −=−+ difference of two squares and there is no
any imaginary part.
1.6.5 Divison of. Complex Numbers
Division of a complex number by a real number is easy enough.
33.167.13
4
3
5
3
45 j j
j−=−=−
But how do we manage with dividing complex number with other
complex one? If we could, somehow, convert the denominator into a
real number, we could divide out as in the above example. So our
problem is really, how can we convert (4 + j3) into a completely real
denominator and this is explained in the previous item. We know
that we can convert (4 + j3) into a completely real number by
multiplying it by its conjugate (4 - j3). But if we multiply the
denominator by ( )34 j− , we must also multiply the numerator by
the same factor.
( )( )
( )( ) 25
3716
916
123728
3434
3447
34
47 j j
j j
j j
j
j −=
+
−−=
−+
−−=
+
−
48.164.025
37
25
16 j j −=−
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Chapter One 9
Then, to divide one complex number by another, therefore, we
multiply numerator and denominator by the conjugate of the
denominator. This will convert the denominator into a real number
and the final step can then be completed.
Example 5 Simplify the following expression:( )( )
43
2132
j
j j
+
−+
Solution:
( )( )
4.18.025
3520
169
43524
43
43*
43
8
43
8
43
62
43
2132
j j j
j
j
j
j
j
j
j
j
j
j j
−=−
=+
−−=
−
−
+
−=
+
−=
+
+−=
+
−+
Equal Complex Numbers
Now let us see what we can find out about two complex
numbers which we are told are equal.
Let the numbers , jba + and jd c + are equal
jd c jba +=+∴ Rearranging terms, we get ( )bd jca −=−∴
In this last statement the quantity on the left hand side is entirely
real, while that on the right hand side is entirely imaginary, i.e. a
real quantity equals an imaginary quantity. This seems contradictory
and in general it just cannot be true. But there is one special case for
which the statement can be true. That is when each side is zero.
( )bd jca −=−∴ can be true only if caieca ==− .,0
and if d biebd ==− .,0
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Mathematical Numbers10
So we get this important result, If two complex numbers are equal
then,
(i) the two real parts are equal
(ii) the two imaginary parts are equal
For example, if 45 j jy +=+ , then we know 5= and 4= y .
1.6.6 Graphical Representation of a Complex Numbers
Although we cannot evaluate a complex number as a real number,
we can represent it diagrammatically, as we shall now see.
In the usual system of plotting numbers, the number 4 could be
represented by a line from the origin to the point 4 on the scale.
Likewise, a line to represent (-4) would be drawn from the origin to
the point (-4). These two lines are equal in length but are drawn in
opposite directions. Therefore, we put an arrow head on each to
distinguish between them as shown in Fig.3.
0 1 2 3 4
4
-1-2-3-4
-4
Fig.3
A line which represents a magnitude (by its length) and direction
(by the arrow head) is called a vector. We shall be using this word
quite a lot. Any vector therefore must include both magnitude (or
size) and direction. If we multiply (+4) by the factor (-1), we get(-4), i.e. the factor (-1) has the effect of turning the vector through
o180 as shown in Fig.4.
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Chapter One 11
0 1 2 3 4
4
o180
-1-2-3-4
-4
Fig.4
Multiplying by (-1) is equivalent to multiplying by2 j , i.e. by the
factor j twice. Therefore multiply in a single factor j will have half
the effect and rotate the vector through onlyo90 . So, the factor j
always turns a vector througho90 in the positive direction
measuring angles, i.e. anticlockwise. If we now multiply j 4 by a
further factor j, we get 42 j , i.e. (-4) and the following diagram
(Fig.5) agrees with this result. If we multiply (-4) by a further
factor j, sketch showing this new vector ( )4 j− is shown in Fig.6.
0-1-2-3-4
-4 4
o180
1 2 3 4
1
2
3
4
j40-1-2-3-4
-4 4
1 2 3 4
o
1801
2
3
4
j4
-1
-2
-3
-4
-j4
Fig.5 Fig.6
Let us denote the two reference lines by XX, and YY, as usual.
You will see that:
(i) The scale on the X-axis represents real numbers, XX1 is
therefore called the real axis.
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Mathematical Numbers12
(ii) The scale on the Y-axis represents imaginary numbers, YY1 is
therefore called the imaginary axis.
If we now wish to represent 2+ 3 as the sum of two vectors, we
must draw them as a chain, the second vector starting where the first
one finishes as shown in Fig.7.
0 1 2 3 4 5
2 3
5
Fig.7
The two vectors, 2 and 3 are together equivalent to a single vector
drawn from the origin to the end of the final vector (giving naturally
that 2+3=5).
If we wish to represent the complex number (3 + j2), then we add
together the vectors which represent 3 and j2. Notice that the 3 is
now multiplied by a factor j which turns that vector through o90 .
The equivalent single vector to represent (2 + j3) is therefore the
vector from the beginning of the first vector (origin) to the end of
the last one. This graphical representation constitutes an Argand
diagram as shown in Fig.8.
0 1 2 3
1
2
3
2
j3
Fig.8
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Chapter One 13
Example 6 Draw an Argand diagram to represent the following
vectors: 231 j z += , 132 j z +−= , 423 j z −= , and 444 j z −−=
Solution: The Argand diagram of the above vectors are shown in
the following Fig.9.
0-1-2-3-4 1 2 3 4
1
2
3
-1
-2
-3
-4
-j4
3
j2 j1
j4
3
j4
z
1 z
2 z
3 z Fig.9.
1.6.7 Graphical Addition of Complex Numbers
Let us find the sum of 231 j z += and 422 j z −= by Argand
diagram. If we are adding vectors, they must be drawn as a chain.
We therefore draw at the end of 1 z , a vector representing 2 z in
magnitude and direction, and is parallel to it. In the same way, we
therefore draw at the end of 2 z , a vector representing 1 z in
magnitude and direction, and is parallel to it. Therefore we have a
parallelogram. Thus the sum of 1 z and 2 z is given by the vector
joining the starting point to the end of the last vector.
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Mathematical Numbers14
The complex numbers 1 z and 2 z can thus be added together by
drawing the diagonal of the parallelogram formed by 1 z and
2 z .Thus, 25422321 j j j z z −=−++=+ which is clear that thisresults is the same as obtained from Fig10. So the sum of two
vectors on an Argand diagram is given by the diagonal of the
parallelogram of vectors.
0 1 2 3 4
1
2
-1
-2
-3
-4
-j4
3
j2
2 z
21 z z +5
1 z
Fig.10
Regarding to the subtraction it is quite similar to addition but the
only trick is simply this: ( )2121 z z z z −+=−
That is, we draw the vector representing 1 z and the negative
vector of 2 z and add them as before. The negative vector of 2 z issimply a vector with the same magnitude (or length) as 2 z but
pointing in the opposite direction.
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Chapter One 15
Example 7 If 231 j z += and 422 j z −= Find 21 z z −
Solution: It is clear from Argand diagram (Fig.11) that
6121
j z z +=− . We can now check for the above results:
61)42(23)42(2321 j j j j j Z Z +=+−++=−−+=−
0-1-2 1 2 3 4
1
2
3
4
-1
-2
-3
-4
-j4
3
j2
3
2 z
2 z −
u
5
6
21 z z −
1 z
Fig.1
1.6.8 Polar Form of a Complex Numbers
Complex numbers in the form jba + is called rectangular form.
Sometimes, it is convenient to express it in a different form. On an
Argand diagram shown in Fig.12, let OA be a vector jba + . Let r
= length of the vector and θ the angle made with OX. Since
jba z += , this can be written θ θ sincos jr r z += or
( )θ θ sincos jr z += This is called the polar form of the complex
number ba + , where: ( )
=+= −
aband bar 122 tan,, θ
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Mathematical Numbers16
o
Ay
xa
b
r
θ
Fig.12
Example 8 Express 34 j z += in polar form.
Solution: First draw a sketch diagram (that always helps). We can
see that: 591634 22 =+=+=r and o87.364
3tan 1 =
= −θ
( )θ θ sincos jr jba z +=+= oo j z 87.36sin87.36cos5 +=∴
(i) r is called the modulus of the complex number z and is often
abbreviated to ( ) z mod or indicated by z
Thus if 43 j z += , ( ) 516943 22 =+=+=∴ z (ii) θ is called the argument of the complex number and can be
abbreviated to )(arg z . So, if 55 j z += then ( ) o z 45arg ==θ
Warning: In finding θ , there are of course two angles between o0
ando360 , the tangent of which has the value θ . We must be
careful to use the angle in the correct quadrant. Always draw a
sketch of the vector to ensure you have the right one. The follwing
table and Fig.13 show the correct angle range and quadrant.
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Chapter One 17
Value of a Value of b Angle range Quuadrant
+ve +ve oo 900
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Mathematical Numbers18
But from Fig.14. This angle isoo 270180
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Chapter One 19
shorthand versionor θ ∠ to denote the polar form as shown in the
following examples:
Then, if 25 j z +−= ( ) 385.529425 ==+=r and fromabove 2.158=θ . Then, the full polar form is
oo j z 2.158sin2.158cos385.5 += and this can be shortened to
o z 2.158385.5 ∠= .
Example 11 express 34 − in shortened form.
Solution: ( ) 534 22 =+=r 75.0tan = E , o E 87.36=∴ o E 13.323360 =−=∴θ
ooo j z 8.323513.323sin13.323cos5 ∠=+=∴
Of course, given a complex number in polar form, you can convert
it into the basic rectangular form jba + simply by evaluating the
cosine and the sine and multiplying by the value of r .
Example 12 Find the rectangular form of the following: o z 355∠=
Solution:( )
868.3096.4
5736.08192.0535sin35cos5
j
j j z oo
+=
+=+=
1.6.9 Exponential Form of a complex numbers
There is still another way of expressing a complex number which
we must deal with. We shall Expalin it this way:
Many functions can be expressed as series. For example,
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Mathematical Numbers20
......!4!3!2
1!
432
0
+++++== ∑∞
=
x x x x
m
xe
m
m x
(1)
.........!4!2
1)!2(
)1(cos42
0
2
+−+−=−= ∑∞
=
x xm x x
m
mm
(2)
.........!5!3)!12(
)1(sin
53
0
12
+−+−=+
−= ∑
∞
=
+ x x x
m
x x
m
mm
(3)
If we now take the series for x
e and write θ j in place of x, we get
the following series
( ) ( ) ( ) ( )!5!4!3!2
15432 θ θ θ θ
θ θ j j j j
je j +++++= (4)
( ) ( ) ( ) ( )+++++++=∴
!5!4!3!21
55443322 θ θ θ θ θ θ
j j j j je j
( ) ( ) ( ) ( )−−++−−+=∴
!5!4!3!21
5432θ θ θ θ
θ θ j j
je j
( ) ( ) ( ) ( )
−+−+
−+−=∴ ....
!5!3....
!4!21
5342 θ θ θ
θ θ θ j j je j (5)
It is clear from (2),(3) and (5) that the first bracket is in the form of
cosine and the second bracket in the form of sine.
θ θ θ sincos je j +=∴ (6)
Therefore, ( )θ θ sincos jr + can now be written asθ j
re . This is
called the exponential form of the complex number. It can be
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Chapter One 21
obtained from the polar form quite easily since the r value is the
same and the angle θ is the same in both.
The three ways of expressing a complex number are therefore
(i) jba z += (Rectangular form)
(ii) ( )θ θ sincos jr z += (Polar form)
(iii)θ jre z = (Exponential form)
And now a ward about negative angles. We know that:
θ θ θ sincos je j +=
if we replace θ by θ − in this result, we get the following:
( ) ( )θ θ
θ θ θ
sincos
sincos
j
je j
−=
−+−=−
So, θ θ θ sincos je j += And θ θ θ sincos je j −=−
There is one operation that we have been unable to carry out with
complex numbers before this. That is to find the logarithm of a
complex number. The exponential form now makes this possible,
since the exponential form consists only of products and powers.
For, if we haveθ jer z .= we can say: θ jr z += lnln
Example 13 Express4/1 π je − in the rectangular form:
Solution: Well now, we can write
( ) je
je
jeeee j j
−=
−=
−
== −−
122
1
2
1
4sin
4cos4/14/1
π π π π
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Mathematical Numbers22
Since every complex number in polar form is of the same shape,
i.e. ( ) or jr θ θ θ ∠=+ sincos and differs from another complex
number simply by the values of r andθ
, we have a shorthandmethod of quoting the result in polar form.
Example 14 Express 34 j z −= in the polar form.
Solution: The vector has been drawn as shown in Fig. 16.
( ) 534 22 =+=r
Fig.16
From this 5=r , o E E 87.36,7.04
3tan −=∴−=−=
ooo 13.32387.36360 =−=∴θ
Then, the polar form isoo j z 13.323sin13.323cos5 += .
And the polar form iso je z 13.323.5= And the the shortened
form iso z 13.3235∠= . In this last example, we have
oo j z 13.323sin13.323cos5 +=
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Chapter One 23
Fig.17.
But the direction of the vector, measured from OX, could be given
as o87.36− , the minus sign showing that we are measuring the
angle in the opposite direction sense from the usual positive
direction. We could writeoo j z 87.36sin87.36cos5 −+−= .
But you already known as ( ) ( )θ θ coscos =− and
( ) ( )θ θ sinsin −=− .
oo
j z 87.36sin87.36cos5 −=∴
i.e. very much like the polar form but with a minus sign in the
middle. This comes about whenever we use negative angles.
In the same way we can say the following:
oooo j j z 110sin110cos5250sin250cos5 −+−=+=
It is sometimes convenient to use this form when the value of θ is
greater thano180 , i.e. in the 3rd and 4th quadrants. In the same
way we can write the following:
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Mathematical Numbers24
ooooo j j z 1303130sin130cos3230sin230cos3 −∠=−=+=
( ) ( ) ooooo j j z 70470sin70cos4290sin290cos4 −∠=−=+= .
The polar form at first sight seems to be a complicated way ofrepresenting a complex number. However it is very useful as we
shall see.
Suppose we multiply together two complex numbers in this form:
Let ( )1111 sincos θ θ jr z += and ( )2222 sincos θ θ jr z +=
( ) ( )22211121 sincos*sincos* θ θ θ θ jr jr z z ++=∴
()21221
21212121
sinsinsincos
cossincoscos.*
θ θ θ θ
θ θ θ θ
j j
jr r z z
+++=∴
Rearranging the terms and remmembering 12 −= j we get:
( )
( )
++
−=
2121
21212121
sincoscossin
sinsincoscos.*
θ θ θ θ
θ θ θ θ
jr r z z
Now the brackets ( )2121 sinsincoscos θ θ θ θ − and
( )2121 sincoscossin θ θ θ θ + ought to ring the bell. What are they?
( ) ( )212121 cossinsincoscos θ θ θ θ θ θ +=−
( ) ( )212121 sinsincoscossin θ θ θ θ θ θ +=+
( ) ( )[ ]21212121 sincos.* θ θ θ θ +++=∴ jr r z z
Note: This is important result. Then we can say that to multiply
together two complex numbers in the polar form,
(i) Multiply the r 's together,
(ii) Add the angles, θ , together it is just easy as that.
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Chapter One 25
Example 15 Find the result of the following in the polar form:
403*302 ∠∠
Solution: It is easy to do that as we get:
( ) ( ) ooo 70640303*2403*302 ∠=+∠=∠∠ Now let us see if we can discover a similar set of rules for
division. We already know that to simplify 45 j+ we first
obtain a denominator that is entirely real by multiplying top and
bottom by the conjugate of the denominator i.e ( )45 j− Right.
Then let us do the same thing with 1 z and 2 z as following:
( )( )222
111
2
1
sincos
sincos
θ θ
θ θ
jr
jr
z
z
+
+=
( )( )
( )( )22
22
222
111
2
1
sincos
sincos*
sincos
sincos
θ θ
θ θ
θ θ
θ θ
j
j
jr
jr
z
z
−
−
+
+=∴
( )
( )22
22
21212121
2
1
2
1
sincos
sinsinsincoscossincoscos
θ θ
θ θ θ θ θ θ θ θ
+
+−+=∴
j j
r
r
z
z
( ) ( )1
sincoscossinsinsincoscos 21212121
2
1
2
1 θ θ θ θ θ θ θ θ −++=∴ j
r
r
z
z
( ) ( )( ) ( )212
12121
2
1
2
1 sincos θ θ θ θ θ θ −∠=−+−=∴r
r j
r
r
z
z
So, for division the rule is divide the r 's and subtract the angles θ 's.
Example 16 Simplify the following expressions: o
o
442
9510
∠
∠
,
Solution: ( ) oooo
o
51544955442
9510∠=−∠=
∠
∠
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Mathematical Numbers26
1.6.10 DeMoivre’s Theorem
There is very important rule is called DeMoivre’s Theorem. It says
that to raise a complex number in polare form to any power n, we
raise the r to the power n and multiply the angle by n.
( )[ ] ( )θ θ θ θ n jnr jr nn sincossincos +=+∴
Example 17 Use DeMoivre’s Theorem to find the results of the
following expression in polar form: ( )[ ]3110sin110cos3 oo j+ Solution:
( )[ ] ( ) ( )( )oooo j j 110*3sin110*3cos3110sin110cos3 33 +=+
( )[ ] ( ) ( )( ) ooooo j j 33027330sin330cos27110sin110cos3 3 ∠=+=+∴This is where the polar form really comes into its own. For
DeMoivre's theorem also applies when we are raising the complex
number to a fractional power, i.e. when we are finding the roots of a
complex number as shown in the following example.
Example 18 Find the square root ofo z 449∠=
Solution:
We haveo
oo z 223
2
449449 ∠=
∠=∠=
Expansion of θ nsin and θ ncos
By DeMoiver ' s theorem, we know that:
( )n jn jn θ θ θ θ sincossincos +=+ where n is a positive integer.
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Chapter One 27
The method is simply to expand the right hand side as a binomial
series, after which we can equate real and imaginary parts. An
example will soon show you how it is done.
Example 19 To find expansions for θ 3cos and θ 3sin
Solution: We have ( ) ( )33sincos3sin3cos jsc j j +=+=+ θ θ θ θ
Where θ cos=c and θ sin= s just for simplicity.
Now expand this by the binomial series so that:
( ) ( ) ( ) ( )32233 33sincos3sin3cos js jsc jscc j j +++=+=+ θ θ θ θ
3223 333sin3cos jscs sc jc j −−+=+ θ θ
3223 333sin3cos s sc jcsc j −+−=+ θ θ
Now equating real parts and imaginary parts we get:
θ θ θ θ
θ θ θ θ
32
23
sinsincos33sin
sincos3cos3cos
−=
−=
If we wish, we can replace θ θ 22 cos1sin −= and
θ θ 22 sin1cos −= . So that we could write the results above as
following: θ θ θ cos3cos43cos 3 −= , θ θ θ 3sin4sin33sin −=∴ .
While these results are useful, it is really the method that counts. So
now do this one in just the same way as done before, obtain an
expansion for θ 4cos in terms of θ cos .
( ) ( )44sincos4sin4cos jsc j j +=+=+ θ θ θ θ
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Mathematical Numbers28
( ) ( ) ( ) ( )432234 464 js jsc jsc jscc ++++=
432234 464 scs j sc sc jc +−−+=
334224 446 cs sc j s scc −++−=
Equating real parts:
( )( ) ( )22224
4224
116
64cos
cccc
s scc
−+−−=
+−=∴ θ
42424 2166 ccccc +−++−=
188 24 +−= cc
1cos8cos8 24 +−= θ θ
Similarly, ( ) ( )2233 4444sin sccscs sc −=−=θ
( ) scccs scs sc sccs
322
2222
4*414
14
==−=
+−−−=
θ θ θ sin*cos44sin 3=∴
Expansions for θ ncos and θ
nsin in terms of sines and cosines
of muiltiples of θ .
θ θ sincos j z += , θ θ sincos1 1 j z z
−==∴ −
θ cos21
=+∴ z
z and , θ sin21
j
z
z =−
Also by DeMoivre' s theorem θ θ n jn z n sincos +=
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Chapter One 29
And, θ θ n jn z n
sincos1
−=
θ n
z
z
n
n cos21
=+∴ And, θ n j
z
z n
n sin21
=−
Let us collect theses four results together: θ θ sincos j z +=
θ cos21
=+ z
z θ sin21
j z
z =−
θ n z
z n
n cos21
=+ θ n j z
z n
n sin21
=−
Example 20 Expand θ 3cos
Solution: From the previous results,
θ cos21
=+ z Q ( )33
cos21
θ =
+∴
z z
( )
3
3
32
233
1133
113
13cos2
z z z z
z z z
z z z
+++=
+
+
+=∴ θ
Now here is the trick: we rewrite this, collecting the terms up in
pairs from the two extreme ends, thus:
( )
++
+=
z z
z z
13
1cos2
3
33θ
But from the previous results
θ cos21
=+∴ z
z , and θ 3cos213
3 =+ z
z
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Mathematical Numbers30
( ) θ θ θ cos2*33cos2cos2 3 +=∴
( )θ θ θ cos33cos41cos3 +=∴
Example 21 Expand θ 4sin
Solution:
θ sin21
j z
z =−Q , and, θ n j z
z n
n sin21
=−
( )44 1
sin2
−=∴
z z j θ
432
234 1141
61
4 z z
z z
z z
z z +
−
+
−=
61
41
2
2
4
4 +
+−
+=
z z
z z
Now, θ n z
z n
n cos21 =+Q
( )
62cos2*44cos2
61
41
sin22
2
4
44
+−=
+
+−
+=∴
θ θ
θ z
z z
z j
∴ 62cos2*44cos2sin16 4 +−= θ θ θ
( )32cos44cos81sin4 +−=∴ θ θ θ
θ θ θ cos63cos2cos8 3 +=∴
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Chapter One 31
1.6.11 The Roots of Unity
The problem here is to solve the equation 1=n z , where n is usually
a positive whole number.
Write both sides of the equation in polar form. Let z have polar form
( )θ θ sincos jr z += ( )θ θ n jnr z nn sincos +=∴
We know that ( )0sin0cos11 j+= . So our equation becomes:
( ) ( )0sin0cos*1sincos jn jnr z nn +=+=∴ θ θ
Now two complex numbers in standard polar form are equal if and
only if their modulus and arguments are equal. In the case of the
argument this statement has to be handled with care. It means are
equal if reduced to the proper range. So, for exampleo
10 and o370
count as equal from this point of view. So we can say that 1=nr
and that θ n and 0 are equal up to the addition of some multiple of
π 2 radians. π θ k nr
n
201 +== Where k is some wholenumber. Since r is real and positive, the only possibility for r is r = 1.
The other equation gives us:n
k π θ 20 +=
This, in principle, gives us infinitely many answers one for each
possible whole number k . But not all the answers are different.
Remember that changing the angle by π 2 does not change the
number z .
The distinct solutions, of which there are n, are given by 1=r and
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Mathematical Numbers32
1,.........3,2,1,0,2 −== nk n
k π θ and we can write these
solutions as following: k k k j z θ θ sincos +=
Where 1,.........3,2,1,0,2 −== nk n
k π θ
That looks rather complicated. It becomes a lot simpler if you think
in terms of the Argand diagram. All the solutions have modulus 1
and so lie on the circle of radius 1 centered at the origin. The
solution with k = 1 is just z = 1. The other solutions are just 1−n
other points equally spaced round this circle, with angle n/2π
between one and the next. This is shown in Fig.18 for 17=n .
Fig.18 The n roots of 1.
Let's look at some specific examples. The cube roots of unity are the
solutions to 13
= z . There are three of them and they are:,3/2sin3/2cos,1 1 π π j z z o +==
3/4sin3/4cos2 π π j z +=
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Chapter One 33
Fig.19: The three cube roots of 1.
Note that 12 z z = ,212 z z = and 01 21 =++ z z . The roots are
shown in Fig.19.
Similarly the fourth roots of unity are the solutions of 14 = z and
these are: 1= z , j z = , 1−= z , and j z −=
A picture for n = 4 together with those for n = 5 and n= 6 is given
in Fig.20.
Fig.20 The nth roots of 1 for n = 4;5;6.
We can do other equations like this in much the same way.
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Mathematical Numbers34
Example 22 Find the solutions of the equation j z =4 .
Solution:
Put ( )θ θ sincos jr z += . Then ( )θ θ 4sin4cos44
jr z += . We
know that: ( )2/sin2/cos1 π π j j += . So our equation becomes:
( ) ( )2/sin2/cos14sin4cos44 π π θ θ j jr z +=+=
Therefore; 1=r and π π
θ k 22
4 += or 28
π π θ
k +=
There are 4 distinct solutions, given by k = 0;1;2;3. They form a
square on the unit circle.
1.7 Polynomials
We have learned how to manipulate complex numbers, and
suggested that they will prove valuable in engineering calculations.
The original motivation for introducing them was to give the
equation 12 −= x two roots, namely j and j− , rather than it having
no roots. It turns out that this is all we have to do to ensure that
every polynomial has the right number of roots. We now discuss
this, and a number of other basic results about polynomials that are
quite useful to know.
A polynomial in x is a function of the form:
( ) on
nn
n a xa xa xa x p ++++= −
− 11
1 ....
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Chapter One 35
where the a's are (real or complex) numbers and 0≠na . For
example: ( ) ( ) 165,42 3483 −+−=+−= t t t t q x x x p
The highest power in the polynomial is called the degree of the
polynomial. The above examples have degrees 3 and 8.
A number a (real or complex) is said to be a root of the polynomial
( ) x p if ( ) 0=a p . Thus 1= x is a root of 0122 =+− x x
The first important result about polynomials is that a number a (real
or complex) is a root of the polynomial ( ) x p if and only if ( )a x −
is a factor of p( x), in the sense that we can write ( ) x p as:
( ) ( ) ( ) xqa x x p −= . Where ( ) xq is another polynomial. This result
is often called the remainder theorem . For example, 2= x is a root
of ( ) 2723 +−+= x x x x p and it turns out that
( ) ( ) 132 2 −+−= x x x x p
Note that necessarily the polynomial q has degree one less than the
degree of p. It may be the case that you can pull more than one
factor of a x − out of the polynomial. For example, 2 is a root of
( ) 12823 +−−= x x x x p and it turns out that
( ) ( )( )( )322 +−−= x x x x p
In such cases a is said to be a multiple root of ( ) x p . The multiplicity
of the root is the number of factors ( )a x − that you can take out. In
the above example, 2 is a root of multiplicity 2, or a double root . A
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Mathematical Numbers36
root is called a simple root if it produces only one factor. Multiple
roots are a considerable pain in the neck in many applications.
There is a simple test for multiplicity. Suppose a is a root of
( ) x p , so that ( ) 0=a p . If, in addition, ( ) 0=′ a p (derivative) then a
is a multiple root. To take the above example:
( ) 12823 +−−= x x x x pQ ( ) 823 2 −−=′∴ x x x p and ( ) 02 = p
and we have ( ) 02 =′ p , so we know that 2 is a multiple root.
2.9. Theorem (Fundamental Theorem of Algebra).
Let p be any polynomial of degree n. Then p can be factored into a
product of a constant and n factors of the form ( )a x − , where a
may be real or complex.
Also, the factorization is unique; you cannot find two essentially
different factorizations for the same polynomial. The factors need
not all be different because of multiple roots.
The fact that there cannot be more than n such factors is fairly
obvious, since we would have the wrong degree. What is not at all
obvious is that we have all the factors that we want. Note that this
result does not tell you how to find these factors; just that they must
be there!
The result is often stated loosely as: a polynomial of degree nmust have exactly n roots. You have to allow complex roots or the
theorem is not true. For example ( ) 12 += x x p has no real roots at
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Chapter One 37
all. Its roots are j±= and it factorizes as ( ) ( )( ) j x j x x p +−= . In
fact, if 0≠ω then ( ) ω −= n z z p ( )1≥n always has exactly n
distinct roots because we know that it must have n roots in all and it
cannot have any multiple roots because ( ) 1−=′ nnz z p has only 0 as
a root and 0 is not a root of ( ) z p .
There is one other result about roots of polynomials that is worth
knowing. Suppose we have a polynomial with real , as opposed to
complex, coefficients. Suppose that the complex number z is a root
of the polynomial. Then the complex conjugate z is also a root. So
you get two roots for the price of one. You can see this in the
example of the previous paragraph. 12 + x has j as a root, so it
automatically must have j− as a root as well.
Example 23 Let ( ) 201694 234 +−+−= z z z z z p . Given that 2 +
j is a root, express ( ) z p as a product of real quadratic factors and
list all four roots, drawing attention to any conjugate pairs.
Solution:
Since p has real coefficients, and complex roots occur in pairs
consisting of a root and its complex conjugate. Given that j+2 is a
root, it follows that j−2 must also be a root, and so the quadratic:
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Mathematical Numbers38
( )( ) ( )( ) 5422 2 +−=−−+− z z j z j z must be a factor. Dividing
the given polynomial by this factor gives
( ) 45420169422234
++−=+−+−= z z z z z z z z p
The roots of 42 + z are 2 j and its complex conjugate, j2− . Thus
the given polynomial, of degree four, has two pairs of complex
conjugate roots.
Example 24 Express 15 − z as a product of real linear and quadratic
factors.
Solution:
We rely on our knowledge of the nth roots of unity from the
previous section. Let
+
=
=
5
2sin
5
2cos
5
2exp
π π π α j j
Then the roots of 015 =− z are ,,,, 432 α α α α and, 1.
( ) ( )( ) 4322345 1111 α α α α −−−−−=++++−=− z z z z z z z z z z z
For convenience, write2α β = , and note that 3α β = while
4αα = . Our problem is to factorize 1234 ++++ z z z z as a
product of real quadratic factors. We know the roots are
β β α α and ,,, . Now construct the quadratic with roots α and α .
We have: ( )( ) ( ) ( ) 1222 +ℜ−=++−=−− α α α α α α α z z z z
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Chapter One 39
where ( )α ℜ is the real part of α . Since ( )( ) β β −− z z behaves in
the same way, we have:
( ) ( ) ( ) 121211225
+ℜ−+ℜ−−=− β α z z z z
( )
+
−
+
−−=−∴ 1
5
4cos21
5
2cos211 225
π π z z z z
and this is a product of real linear and quadratic factors.
Problems
)
Express the complex number j z 31 −= exactly in
modulus - argument form. Hence find the modulus
and principal argument of4
z .
) Find all solutions w to the equation j273 −=ω and
mark them on an Argand diagram.
) Let 21 j z −= j+= 3ω be complex numbers.
Express each of the following complex numbers in
the rectangular form z z j j z
z −+++
31,2
, ω
ω
) Express the complex number 22 + exactly in
modulus - argument form. Hence find all solutions
w to the equation j22
3
+−=ω and mark them onan Argand diagram.
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Mathematical Numbers40
) Let j z += 3 and j71 −=ω . Express z +ω
ω in a
rectangular form. Find also z z
ω
ω ,,
) Express the complex number j22 +− in polar
form. Hence solve the equation j z 223 +−=
expressing the solutions in polar form and marking
them in the Argand Diagram.
)
Let ( ) 882852345
+−−+−= z z z z z z p
Show that ( ) 02 = p . Show also that 222 +− z z is a factor of ( ) z p .
Hence write p as a product of linear factors.
) Show that ( ) j z +− 1 is a factor of the real
polynomial ( ) 862 23 +−+= z z z z p
Hence write p as a product of linear factors.
) Let ( ) 362753 234 −−+−= z z z z z p Show that
( ) 03 = j p . Hence write p as a product of linear
factors.
) Express in polar 35 j z −−=
) Express in rectangularo1562∠ and o375 −∠
) Ifoo j z 125sin125cos121 += and
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Chapter One 41
oo j z 72sin72cos52 += Then, find (i) 21 * z z and2
1
z
z giving the
results in polar form
) If oo j z 125sin125cos121 += , find3 z and
3
1
z
) If jy x z += , find the equations of the two loci
defined by: (i) 34 =− z and (ii) ( )6
2arg π
=+ z
)
If jy x z += , find the value of x and y when :
j j
z
j
z
−=+
− 3
43
1
3
) Express 32 j+ and 21 j− in polar form and
apply DeMoiver’s theorem to evaluate( )
21
324
j
j
−
+.
Express the result in rectangular and exponential
form.
) Find the fifth roots of 33 +− in polar and
exponential form.
) Express 125 j+ in polar form and hence
evaluate the principle value of ( )3
125 j+ givingthe results in rectangular form.
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Mathematical Numbers42
) Obtain the expansion of θ 7sin in terms of
θ sin .
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Chapter 2
Matrices
2.1 Introduction
A matrix is, by definition, a rectangular array of numeric or
algebraic quantities, which are subject to mathematical operations.
So a real matrix is an arrangement of real numbers into rows and
columns. Matrices can be defined in terms of their dimensions
(number of rows and columns). Let us take a look at a matrix with 4
rows and 3 columns (we denote it as a 4x3 matrix and call it A):
=
0
0
1
1
5
12
8
6
9
2
5
7
A
The dimensions of this matrix are 4 by 3. The dimensions of amatrix tell you the size of the matrix because they tell you the
number of rows and columns in the matrix. By convention, we list
the number of rows before the number of columns.
Definition 1 The dimensions of a matrix are the number of rows
and columns (listed in that order) of the matrix.
Each element of the matrix is named according to its position.
Typically, capital letters represent matrices and small letters with
subscripts represent elements in the matrix. Since vectors can be
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Chapter Two 43
considered to be matrices with only one row or one column, they
could be labeled with capital letters also. However, small letters
usually represents vectors. The element 6 is in the position a12 (read
a one two) because it is in row 1 and column 2. Also by convention,
we list the row number of the element before the column number.
An element in row i and column j would be denoted by ija . This
gives us a compact way to refer to specific elements of a matrix.
Can you represent the same information as before in a 3 by 4
matrix? Yes, you can. It would look like the matrix B which follows.
=
0
5
9
0
12
2
1
8
5
1
6
7
B
Matrix B is the transpose of A, and A is the transpose of B.
Transposing a matrix results in writing the columns as rows and the
rows as columns, but what really happens is that element ija is
placed in the position jib of the new matrix. Therefore, 12a moves
to the position 12b when we form the transpose of A. The transpose
of A is denoted byT
(read A transpose). Therefore, matrix B is
T .
Definition 2 By the transpose of the m by n matrix A, denoted by
T , we mean the n by m matrix, which has ija as its ( )
th ji,
element.
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Matrices44
Definition 3 We say that two m by n matrices, A and B are equal
if their corresponding elements are equal.
In other words, A = B if A and B have the same dimensions and
1111 ba = , 1212 ba = , etc. IsT = ? Usually not, but we have a
special word for a matrix which satisfiesT = .
Definition 4 A matrix is said to be symmetric ifT = .
Observe that the following matrix is symmetric:
=
3681
64058072
1529
A
Notice that jiij aa = for all i and j; as is true for all symmetric
matrices. Symmetric matrices are easy to spot because if you draw a
line down the main diagonal (from 9 to 3 in this matrix), then the
two halves are mirror images of each other. Symmetric matrices
have many special qualities that will be used when you study
matrices in more detail. The matrix A, given above, has another
special property; it is a square matrix because A has the same
number of rows as columns. Notice that A is a 4 by 4 square matrix.
We said that the main diagonal for A runs from 9 to 3. For any
square matrix, the main diagonal runs from the upper left corner to
the lower right corner.
Definition 5 We say that an m by n matrix is square if nm = .
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Chapter Two 45
2.2 Addition and Subtractions of Matrices
Definition 6 Matrices of the same dimensions are added by adding
corresponding elements.
For instance, ija corresponds to ijb because they both lie in the ith
row and jth column of their respective matrices. Therefore, we would
add, ijij ba + to obtain the ( ) ji,th element of + :
Example 1 Find the result of the following:
+
=+
0411
195
069
168
0
0
1
1
5
12
8
6
9
2
5
7
B A
Solution:
=
++
+
+
++
+
+
++
+
+
=+
09201217
11414
21215
0010
01
11
45912
68
66
11952
95
87
B A
Think about the similarities between addition and subtraction. How
do you think matrices are subtracted?
Definition 7 Matrices of the same dimensions are subtracted by
subtracting corresponding elements.
2.3 Multiplication of Matrices
Multiplying a matrix by a scalar value involves multiplying everyelement of the matrix by that value. Here we multiply our 4x3
matrix A by a scalar value k :
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Matrices46
=
=
0*
0*
1*
1*
5*
12*
8*
6*
9*
2*
5*
7*
0
0
1
1
5
12
8
6
9
2
5
7
**
k
k
k
k
k
k
k
k
k
k
k
k
k Ak
The multiplication operation on matrices differs significantly from
its real counterpart. One major difference is that multiplication can
be performed on matrices with different dimensions. The first
restriction is that the first matrix has to have the same amount of
columns as the second has rows. The reason for this will become
clear shortly. Another thing to note is that matrix multiplication is
not commutative i.e, (CD) does not equal ( DC ).
The procedure for matrix multiplication is rather simple. First, we
determine the dimensions of the resultant matrix. All we require is
that there are as many columns in the first matrix as there are rows
in the second. A simple way of determining is to look at the nearest
and farthest dimensions of two matrix symbols written next to eachother, for instance: C [2x3] D[3x2]. The nearest dimensions are both
equal to 3, and so we know that the operation is possible. The
farthest dimensions will give us the dimensions of the product
matrix, so our result will be a 2x2 matrix. The general rule says that
in order to perform the multiplication AB, where A is a mxn matrix
and B a k xl matrix, we must have n=k . The result will be a mxl
matrix.
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Chapter Two 47
Performing the operation product involves multiplying the cells
of a particular rows in the first matrix by the cells of a particular
column in the second matrix, adding the products, and storing the
result in the cell of the resultant matrix whose coordinates
correspond to the row of the first matrix and the column of the
second matrix. For instance, in AB = C , if we want to find the value
of c12, we must multiply the cells of row 1 in the first matrix by the
cells of column 2 in the second matrix and sum the results.
There are several interesting things to notice about matrix
multiplication. We multiplied a 1 by 3 matrix by a 3 by 4 matrix and
got a 1 by 4 matrix. The following picture expresses the
requirements on the dimensions:
Let's also look closely at how we multiply the matrices because we
will multiply matrices with larger dimensions later. This is a hands
on activity. Take your left pointer finger and place it at the
beginning of the first row of the first matrix (the only row we have
in this case). Take your right pointer finger and place it on the first
number of the first column of the second matrix. Multiply the two
numbers to which you are pointing. Each time you move, your left
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Matrices48
hand will go across the row, and your right hand will go down the
column. When you reach the end of the row and column, add the
numbers you have obtained from the multiplications. This number
goes in the first row and first column of your product matrix. This is
the same as taking the inner product of the first row of first matrix
and the first column of the second matrix. Now you can move to the
first row, second column doing the same thing. This number will go
in the first row, second column of your product matrix. In short,
position ij of your product matrix consists of the inner product of the
ith row of your first matrix and the jth column of the second matrix.
This is a lot easier to do than it is to describe! Your left hand will
move across and your right hand will move down. Do this for every
row and column combination to get your product matrix. This
picture depicts the motions necessary to find a product: Inner
product of row i with column j equals position ij
Definition 8 An identity matrix is a square matrix with ones along
the main diagonal and zeros elsewhere.
Example 2
If [ ]341=S And
=
23
1112
0
4
30
4
12
0122
R Find S *
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Matrices50
Example 4 Multiply the following matrices
=
=
14503
167
121548
125063
1204
9013708
51017
23
1112
0430
412
0122
* F R
2.4 Equations
Solving equations is an important part of mathematics. If we are
working with more than one unknown at a time, we need to solvesystems of equations. You may already know how to solve a system
of linear equations, but matrices provide a more compact way to
arrive at the solution. Matrices are also easier to manipulate on a
computer or calculator. Both of these facts will become more
important when you work with larger systems.
Example 5
Solve the following system of equations:
6624
9335
21
21
−=−−
=+
x x
x x
Solution: Let's look at a system of linear equations:
6624
9335
21
21
−=−−
=+
x x
x x
Can be written in matrix form as B X = where
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Chapter Two 51
−−=
24
35 A ;
=
2
1
x
x X , and
−=
66
93 B
When you learned to solve systems of linear equations, you
learned that
(a) You arrive at the same solution no matter which equation you
write first,
(b) The solution doesn't change if you multiply an equation by a
scalar other than zero, and,
(c) You can replace an equation with the sum of that equation and
another equation without changing the solution.
These may not be exactly the words you used when you were
solving a system of linear equations, but you did all these things.
Experiment with the system above to convince yourself that these
statements are true. We can also solve this system entirely in matrix
form. We use the same rules, and we call them Elementary Row
Operations ( EROs). The EROs tell us that we can
(a) Interchange any two rows;
(b) Multiply any row by a non-zero scalar; and
(c) Replace any row by the sum of that row and any other row.
Proper use of EROs will leave us with a system that has the same
solution as our original system, but is much easier to solve. If you
were presented the system
b xa x == 21 ,
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Matrices52
You would be able to solve it instantly because you only have to
read the solution. If this system were written using matrix notation,
it would look like this:
=
b
a
x
x
2
1
10
01 The matrix
10
01is the 2 by 2 identity
matrix. Because you can just read of the solution when a system is
in this form, our first goal is to transform our system into this form.
Let's solve the system above using matrices. We can represent
this entire system with a 2 by 3 matrix, which looks like this:
−−− 66
93
24
35. This is called an augmented matrix because we
combined 2 matrices (a matrix and a vector for this system). In this
case, we combined the 2 by 2 coefficient matrix which is made of
the coefficients for our unknowns and the 2 by 1 matrix from the
right-hand side of the equations into one 2 by 3 matrix. In other
words, we put A to the left of the bar and put b to the right of the
bar. The application of an ERO to the augmented matrix does not
change the solution set of the linear system that the augmented
matrix represents because whatever you do to the left side of an
equation, you also do to the right side. Therefore, we will arrive at
the same solution whether we use augmented matrices or not, and
augmented matrices are more compact to write. Using matrix
notation, our goal is to transform our system into one that looks like
the following form:
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Chapter Two 53
b
a
10
01
In other words, we want the identity matrix to the left of the bar andthe solution to the right of the bar.
Remark 1 The bar is not a formal part of the matrix, so it is not
necessary. It is placed there so that we can refer to the different
parts of the augmented matrix and easily move back and forth
between the augmented matrix and the linear system that it
represents. In this book,1
r represents row 1 and so on.
−−− 66
93
24
35Original augment matrix
−−− 66
6.18
24
6.0151 ÷r
4.8
6.18
4.00
6.01214 r r +
4.021
6.18
10
6.012 ÷
r
12*6.021
6
10
01r r +−
When we convert this from augmented matrix notation back to the
algebraic notation for a system of equations, it looks like this:
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Chapter Two 55
0≠ A because aa /11 =− where 1−a is called the multiplicative
inverse or the reciprocal. There is something analogous to this with
matrices. It is also called the inverse. With scalars, 111
==
−−
aaaa .
Definition 9 The matrix1−
(called A inverse) is the inverse of a
square matrix A if I A == −− 11 where I is the identity matrix.
Once we find A 1; Ax = b can be solved by matrix multiplication
rather than Gauss Jordan elimination. We follow the algebraic steps
below to find an expression for x:
b Ax = b A Ax A 11 −−=∴ b A x I 1*
−=∴
This means that if we find1−; we only need to multiply to solve
systems with the same matrix A for different b vectors. Please
remember that11 −− ≠ Abb A , so you must multiply in the correct
order.
Remark 2 In computational mathematics, the inverse is very seldom found because other methods exist that serve the same
purpose and require fewer steps. However, the inverse will serve
our needs at this level and is important in the theory of matrices.
Example 7 Using the Gauss Jordan elimination method, let's find
1− where A
631
324
420
Solution:
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Matrices56
100
010
001
631
324
420
Original augmented matrix.
Switch 1r and 3r because we cannot have a zero on the main
diagonal, and we would prefer 1 rather 4.
−−−
001
410
100
420
21100
631
214 r r +−
−
001
4.01.00
100
420
1.210
631
( )10/2 −r
−−
− 8.02.014.01.00
100
2.0001.210
631
322 r r +−
−−
−
415
4.01.00
100
100
1.210
631
( )2.0/3 −r
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Chapter 3
Calculus
3.1 Limits
The concept of limits is essential to calculus. A good understanding
of limits will help explain many theories in calculus. So, it is
recommended to start studying calculus from limits.
Consider a function f defined for values of x, as x gets close to a
number a, not necessarily true for a= . If the value of ( ) x f
approaches a number b as x approaches a, then the limit of ( ) x f as
x approaches a is equal to b, denoted as :
b x f a x
=→
)(lim (1)
Example 1 Find the limit of 25)( += x x f as x approaches 3.
Solution: It is clear that as x approaches 3, 5 x approaches 15, and25 + x approaches 17. Thus; 1725lim
3=+
→ x
x
Example 2 Find the limits of102
1)(
−= x f as x approaches 5.
Solution: It is clear as x approach 5, 102 − x approaches zero the
102
1
− x
approaches
0
1 which is undefined. Thus;
)(102
1lim
5undefiend
x∞=
−→
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Calculus76
(or decreases ) indefinitely, then the limit of ( ) x f as x increases (or
decreases ) indefinitely is equal to b, denoted as :
b x f x
=+∞→
)(lim or b x f x
=−∞→
)(lim
A function ( ) x f is continuous at x = a if f is defined at x = a and
either; f is not defined anywhere near a, or f is defined arbitrarily
near a= and, )()(lim x f x f a x
=→
Conversely, A function ( ) x f is discontinuous at a x = if ( ) x f is
defined at a x = and ( ) x f is not continuous at a x = .
3.2 Derivatives
Suppose )( x f y = is shown in Fig.1, the slope of the curve is the
slope of the secant line between point A and another point P on the
graph is shown in the following equation:
( ) ( )( )( )
( ) ( )( )h
x f h x f
xh x
x f h x f m AP
−+=−+
−+=
Notice that h can change and with it the location of point P,
therefore h is the limiting factor of the slope of the curve. As h gets
close to point A, the slope of the curve becomes the tangent of the
graph at point A.
The tangent line of f at point A is:( ) ( )( )
h
x f h x f
h
−+
→0
lim
So, the Differentiation of function f at x is:
( ) ( )( )h
x f h x f
h
−+
→0lim
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Chapter Three 77
If this limit exists, then it is called the derivative of function f
at x, which is denoted bydx
dyor x f )(′ .
So,( ) ( )( )h
x f h x f
dx
dyor x f
h
−+=′
→0lim)(
Fig.1 The Approximate slope of the curve at point A.
Fig.2 The slope of the curve at point A.
So, general rules of differentiation are shown in the appendix of this
book before going in the following example you have to take a look
to the rules of differentiation in the appendix.
Example 3 Find from the first principles( )( ) xe
dx
d tan
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Calculus78
Solution: Put ( ) xu tan= ue y =∴ uedu
dy=∴ and x
dx
du 2sec=
But from chain rule, dx
dy
du
dy
dx
dy
.=
,
( )( ) ( ) xeedx
d x x 2tantan sec*=∴
Example 4 Find
− 3 2 xdx
d
Solution:
−
−
−=−=
− 3
11
3
2
3 2
3
2
3
2 x x x
dx
d
Example 5 Find
−
−
1
3
2 xdx
d
Solution:
( )
−−=
−
− − 21
2
21*3
1
3 x
dx
d
xdx
d ( ) x x 2*1*2
323
2 −−=
( ) ( )222
2221
1*3
11
3
1
3
−
−=
−−=
−
−∴
x
x x
x x
x
xdx
d
Example 6 Find ( )4
75 + xdx
d
Solution: ( ) ( )334 75205*)75(475 +=+=+ x x xdx
d
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Chapter Three 79
Example 7 Find ( )( )65sin + xdx
d
Solution: ( )( ) ( )65cos565sin +=+ x xdx
d
Example 8 Find ( )( )2cos xdx
d
Solution: ( )( ) ( ) ( )222 sin*22*sincos x x x x xdx
d −=−=
Example 9 Find ( )( ) xdx
d cos43ln −
Solution: ( )( ) ( ) x
x x
x x
dx
d
cos43
sin4sin4*
cos43
1cos43ln
−=
−=−
Example 10 Find ( )( )12log10 − xdx
d
Solution: ( )( )( ) ( ) ( ) ( )10ln12
22*
10ln12
112log10
−=
−=−
x x x
dx
d
Example 11 Find ( )( )12ln*5 − xedx
d x
Solution: Assume ( )12ln*5 −= xe y x , ( )12ln5 −== xvand eu x
uv y =∴ ,dx
duv
dx
dvu
dx
dy+=
( ) ( ) x x e x
xe
dx
dy 55 5*12ln2*12
1−+
−=∴
( ) ( )12ln5
12
2 55
−+−
=∴ xe x
e
dx
dy x x
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Calculus80
Example 12 Find ( )( ) x xdx
d sinln3 5
Solution: Assume ( ) x x y sinln3 5= , ( ) xvand xu sinln3 5 ==
uv y =∴ ,dx
duv
dx
dvu
dx
dy+=
( )( ) ( ) 455 5*3*sinlncos*sin
13sinln3 x x x x
x x xdx
d +=∴
( )( ) ( ) x x x x x xdx
d sinln*15cot3sinln3 455 +=∴
Example 13 Find ( )
x
e
dx
d x
3ln
2
Solution: Assume ( ) xe
y x
3ln
2
= , ( ) xvand eu x 3ln2 ==
2v
dx
dvu
dx
duv
dx
dy −
=
( )
( )
( )( )2
222
3ln
3
3*2*3ln
3ln x
xee x
x
e
dx
d x x
x −=
∴
( ) ( )
( )( )
( ) ( )
( )( )2
2
2
2
3ln
)13ln2ln2(
3ln
)1
3ln2ln2(
x x
x x xe
x
x xe x
x
−+=
−+=
Example 14 Find( )
x
x x
dx
d
3cosh
)2sinh(2
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Chapter Three 81
Solution: Assume( ) x x x
y3cosh
)2sinh(2= , 2u = , ( ) xv 2sinh= ,
and ( ) xw 3cosh=
Where
−+=
dx
dw
wdx
dv
vdx
du
uw
uv
w
uv
dx
d 111.
( ) ( ) ( ) ( )
( ) ( )
−
+=
∴
x x
x x
x x x
x x
x
x x
dx
d
3sinh3*3cosh
1
2cosh2*2sinh
12*
1*
3cosh
)2sinh(
3cosh
)2sinh(2
22
( ) ( )( )
( )( )( )
−+=
∴
x x
x x
x x x x
x x x
dxd
3cosh3sinh3
2sinh2cosh22*
3cosh)2sinh(
3cosh)2sinh(
22
( ) ( )( )
( )( )( )
−+=
∴
x
x
x
x
x x
x x
x
x x
dx
d
3cosh
3sinh3
2sinh
2cosh22*
3cosh
)2sinh(
3cosh
)2sinh( 22
( ) ( )( )
( )
( )( ) x x x x
x
x x
x
x x
x
x x
dx
d
3cosh3tanh)2sinh(3
3cosh
2cosh2
3cosh
)2sinh(2
3cosh
)2sinh(
2
22
−
+=
∴
Example 15 Find ( ) x x xdx
d 4cos2sin5
Solution:
Assume x x xuvw y 4cos2sin5== , where 5 xu = , xv 2sin= , and
xw 4cos=
Take the logarithm for both sides we get:
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Calculus82
( ) ( ) ( ) x x x x x x y 4cosln2sinlnln4cos2sinlnln 55 ++==∴
By differentiating both sides of the above equation we get:
( ) ( ) ( ) x
x x
x x xdx
dy y
4sin44cos
12cos22sin
1511 45 −++=∴
x x xdx
dy
y4tan42cot2
51−+=∴
−+=∴ x x
x x x x
dx
dy4tan42cot2
5*4cos2sin5
Example 16 Find
( )
32tan1 x
dx
d +
Solution: ( ) ( ) ( ) xdx
d x x
dx
d 2tan1*2tan132tan1
23 ++=+
( ) ( ) ( ) ( )
++=+∴ x
dx
d x x x
dx
d 22sec0*2tan132tan1 2
23