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Michael Batty

Essential Engineering Mathematics

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3

Contents

Essential Engineering Mathematics

Contents
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the key skills that top employers demand
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London Business School
Regent’s Park
London NW1 4SA
United Kingdom
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...
edu/mim/
email mim@london
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london
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4

Contents

Essential Engineering Mathematics

30
30
32
34
35

Functions and Limits
Functions
Denition of a Function
Piping Functions Together
Inverse Functions
Limits
Continuity

39
39
39
41
42
43
48

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1
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5
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4
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9

Calculus of One Variable Part 1: Differentiation
Derivatives
The Chain Rule
Some Standard Derivatives
Dierentiating Inverse Functions
Implicit Differentiation
Logarithmic Differentiation
Higher Derivatives
L’Hôpital’s Rule
Taylor Series

51
51
56
58
59
61
63
66
67
69

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Systems of Linear Equations
Row Reduction
Finding the Inverse of a Matrix using Row Reduction
Bases
Eigenvalues and Eigenvectors

3
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3
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3
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3

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6
...
2
6
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6
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6

Calculus of Many Variables
Surfaces and Partial Derivatives
Scalar Fields
Vector Fields
Jacobians and the Chain Rule
Line Integrals
Surface and Volume Integrals

111
111
113
115
117
119
121

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1
7
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3


Second Order Linear Differential Equations with Constant Coefficients:
The Homogeneous Case

126

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Contents

Essential Engineering Mathematics

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Second Order Linear Differential Equations with Constant Coefficients:
The Inhomogeneous Case
Initial Value Problems

131
132

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1
8
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3

Complex Function Theory
Standard Complex Functions
The Cauchy-Riemann Equations
Complex Integrals

134
135
140
142



Index

147

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It is by no means a
comprehensive guide to all the mathematics an engineer might encounter during the course of his or her degree
...
The early sections
on functions and single variable calculus are in this spirit
...
The
necessary linear algebra for multivariate calculus is also outlined
...


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I have had to omit many topics and it is unlikely
that it will cover everything in your course
...

Hopefully there are not too many mistakes in it, but if you find
any, have suggestions to improve the book or feel that I have not
covered something which should be included please send an email
to me at
batty
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com
Michael Batty, Durham, 2010
...
com
9

Essential Engineering Mathematics

Chapter 1

Preliminaries
1
...
You
will probably be used to the idea of real numbers, as numbers on
a “line” and working with graphs of real functions in the product
of two lines, i
...
a plane
...
Here we will mention two important
properties:
• The reals are ordered
...
An example of the properties that the
ordering satisfies is that if x < y and z > 0 then zx < zy
but if x < y and z < 0 then zx > zy
...

• A real number is called rational if it can be written as p for
q
integers p and q (q = 0)
...
This is because they have a property called
completeness which means that if a sequence of real numbers
11

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You
will probably be used to the idea of real numbers, as numbers on
a “line” and working with graphs of real functions in the product
of two lines, i
...
a plane
...
Here we will mention two important
properties:

Preliminaries

• The reals are ordered
...
An example of the properties that the
ordering satisfies is that if x < y and z > 0 then zx < zy
but if x < y and z < 0 then zx > zy
...

• A real number is called rational if it can be written as p for
q
integers p and q (q = 0)
...
This is because they have a property called
completeness which means that if a sequence of real numbers
11

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...
e
...

The rationals do not have this property
...
We also use the notation N for the set of natural
numbers {1, 2, 3,
...
, −2, −1, 0,
√
1, 2,
...
That is, it is an irrational number
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12

Preliminaries

Essential Engineering Mathematics

1
...


1
...
1

Intervals

An interval is a “connected” subset of R and can be bounded i
...

of the form
• [x, y] = {r in R | x

y} (closed),

r

• (x, y) = {r in R | x < r < y} (open),
• [x, y) = {r in R | x

r < y} (half-open) or

• (x, y] = {r in R | x < r

y} (half-open)
...
e
...


1
...
2

Solving Inequalities

When solving inequalities, as opposed to equations, if you
multiply an inequality by a negative number then it reverses
the direction of the inequality
...
People often remember to
do this if they are multiplying by a constant but not when they
multiply by a variable such as x, whose value is not known so it

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So it is very important to split your
working into two cases if you have to multiply an inequality by a
variable
...
2
...


If x + 4 < 0 then 12/(x + 4) < 0 < 2
...
Therefore x + 4 > 0, i
...
x > −4
...
This gives 12
2(x + 4) = 2x + 8
...
The solution set for the inequality is thus
the interval (−4, 2]
...


The portion of the graph above the dotted horizontal line correspons to the correct range of values
...
2
...
The absolute value or modulus of x, written
|x|, is the distance between x and 0 on the number line
...
It is defined by

|x| =

x
−x

if x ≥ 0;
if x < 0
...
com
14

Preliminaries

Essential Engineering Mathematics

The properties of |x| are as follows, for x, y ∈ R:
• | − x| = |x|
• |xy| = |x||y|
•

x
y

=

|x|
|y|

(y = 0)

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Preliminaries

Essential Engineering Mathematics

1
...
4

Inequalities Involving Absolute Value

If y
0 then the statement |x| < y means −y < x < y, that is,
x ∈ (−y, y)
...


Similarly
• |x|

y means −y

x

y
...

• |x|

y means x

y or x

−y
...

Example 1
...
2

1
...


We have |x − 1| 3 if and only if −3 x − 1 3
...
We
can check this by drawing a graph
...
Find all real numbers x ∈ R for which |x + 2|

2
...
e
...
So x satisfies the inequality if and only if it lies in
the set (−∞, −4] ∪ [0, ∞)
...
com
16

Preliminaries

Essential Engineering Mathematics

3
...


x
Suppose that |x2 − 4| > 3
...
If x √
7 or x < − 7, i
...

√
x ∈ (−∞, − 7) ∪ ( 7, ∞)
...
Thus the complete solution set to
the inequality is
√
√
(−∞, − 7) ∪ (−1, 1) ∪ ( 7, ∞),

1
...
3
...
It is the number system formed when we want to include all square roots of the number system inside the system itself
...
Of course there is no
real number with this property as squaring a real number always
results in something greater than or equal to zero
...
Then for some m > 0,
√
√
√
√
√
y = −m = −1 · m = i m
A number of the form yi for y ∈ R is called a purely imaginary
number
...


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3
...

A complex number is an expression of the form x+yi where x
amd y are real numbers and i2 = −1
...


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Preliminaries

Essential Engineering Mathematics

To add, subtract and multiply complex numbers, we use ordinary rules of arithmetic and algebra and whenever it appears, we
substitute i2 with −1
...

If z = a + ib then the complex conjugate of z, written z, is
a − ib
...

• zz = a2 + b2 , which is real
...

• z − z = 2bi, which is purely imaginary
...
For example
1−i 2−i
·
2+i 2−i

=
=
=

1
...
3

22

1
(1 − i)(2 − i)
+ 12

1
(2 − 2i − i − 1)
5
1 3
− i
5 5

Solving Polynomial Equations Using Complex
Numbers

Remember that for quadratic equations
ax + bx + c = 0 implies x =
2

−b ±

√

b2 − 4ac

...
com
19

Preliminaries

Essential Engineering Mathematics

If the discriminant b2 − 4ac < 0 then there are no real roots
...
For example x2 +
√
4x + 5 = 0 gives −2 ± 1 −4 = −2 ± i
...
e
...
We should also mention (but not
prove, as it is difficult) the following
Fundamental Theorem of Algebra: Given a polynomial
equation such as
xn + an−1 xn−1 + · · · + a1 x + a0 ,
the equation has at most n roots, all of which are in C
...


1
...
4

Geometry of Complex Numbers

A complex number z = x + iy can be plotted as (x, y) in the plane
...
f
...
With this geometric interpretation,
• z is the reflection of z in the x-axis
...

• for z = 0, the argument of z, written arg z, is the
angle the same line segment makes with the real axis,
measured anti-clockwise
...
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20

Preliminaries

Essential Engineering Mathematics

x < 0 if and only if it has argument π
...
We can write any complex number in polar form, i
...
as
2
z = r(cos θ + i sin θ) where r is |z| and θ is arg z
...
But it is important not to just
put the relevant values into your calculator and work out inverse
tangent because it will give you the wrong answer half of the time
...
For example,

1+i=
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−1 − i =

√

√

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2 cos

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+ i sin
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4

−3π
4

+ i sin

−3π
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Importantly, 5θ = arg1 does not just mean 5θ = 0 but that 5θ =
2πm for any integer m
...
So there are five fifth roots of 1:
• z1 = cos 0 + i sin 0 = 1
• z2 = cos 2π + i sin 2π
5
5
• z3 = cos 4π + i sin 4π
5
5
• z4 = cos −4π + i sin −4π
5
5
• z5 = cos 2π + i sin 2π
5
5
We can plot them on an Argand diagram as follows: As an exercise,
try to calculate the sixth and seventh roots of 1, and the fifth roots
of −1
...
com
22

Vectors and Matrices

Essential Engineering Mathematics

Chapter 2

Vectors and Matrices
Vectors and matrices can be thought of as generalisations of numbers
...


2
...
More abstractly they are lists of real numbers
...
For
calculations treat them as lists of numbers
...
The vector
a
b

x=

has magniutude, more commonly called modulus,
|x| =

a2 + b2
...
Vectors can have any number of entries but the operations
on them are similar
...

23

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Vectors can be
added and subtracted by adding and subtracting component-wise,
i
...


a
b

c
d

+

=

a+c
b+d

Similarly vectors can be multiplied by a “scalar”, i
...
a real
number, as follows
...
com
24

Vectors and Matrices

Essential Engineering Mathematics

Another useful concept is the dot product or scalar product of
two vectors
...

The set of vectors of length n is denoted by Rn
...
In particular,
• If x and y are unit vectors (vectors whose modulus is 1) then
we have x · y = cos θ, so the dot product tells us the length
of the perpendicular projection of x onto y, and vice versa
...
Moreover, if |x| = 0 and |y| = 0
then x · y = 0 implies that x and y are perpendicular
...
com
25

Vectors and Matrices

Essential Engineering Mathematics

2
...
2
...
If
we have a 2-dimensional vector, for example, we have to say where
each component of the vector goes to, which means 2 lots of 2
...
If we follow it by another transformation
This means that

e f
g h
we get
e(ax + by) + f (cx + dy)
g(cx + dy) + h(ax + by)

=

ea + f c eb + f d
ga + hc gb + hd

x
y

And this shows us how to multiply matrices
...

When multiplying matrices, to get the entry in row p and
column q of the product of two matrices M N , form the dot
product of row p of M with column q of N
...
The product of two matrices is only defined when the
number of columns of the first is equal to the number of rows of
the second
...
Similarly, there is an
opposite operation to multiplying by a matrix
...


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2
...
For 2 × 2
matrices it is
1 0
0 1
and it has the property that for every 2 × 2 matrix A, IA = AI =
A
...

A matrix B is called the inverse of a matrix A if AB =
BA = I
...

There is a test for whether or not a matrix A has an inverse
...
For the 2 × 2 matrix
A=

a b
c d

det A is equal to ad − bc
...

A square matrix A has an inverse if and only if det A is not
zero
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27

Vectors and Matrices

Essential Engineering Mathematics

For a 2 × 2 matrix, if det A is not zero then the inverse is given by
A−1 =

1
detA

d −b
−c a


...


2
...
3

The Cross Product

The cross product or vector product is a construction on vectors
but it has been included here because you need to know about
determinants to calculate them
...

It is only defined for three-dimensional vectors
...
The top row of the matrix contains unit vectors
i,j and k, so it isn’t the determinant of a real matrix but it is
calculated in the same way
...

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The vector product always has
these properties:
• If a and b are not parallel then a × b is perpendicular
to a and b
...


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3
2
...
1

Systems of Linear Equations and Row
Reduction
Systems of Linear Equations

What we call “systems of linear equations” are usually called “simultaneous equations” at school
...
For example,
x+y+z = 2
x − 2y + 2z = 5
2x + y + z = 3

can be written using a 3 × 3 matrix:


   
1 1 1
x
2
 1 −2 2   y  =  5 
2 1 1
z
3

2
...
2

Row Reduction

There is a systematic way to solve the system of equations called
row reduction (sometimes it is also called Gaussian elimination)
...
To do this

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• And so on, if the matrix is larger than 3 × 3
...

A solution to a linear system of three equations in three unknowns
can be either
• a point, i
...
a unique solution
• a line of solutions, if one of the variables is unconstrained
• a plane of solutions, if two of the variables are unconstrained
• the entire three dimensional space, if all of the variables are
unconstrained
For row reduction it is common to omit the x, y and z and use a
bar to separate the left hand side from the right hand side
...



1 1 1 2
 1 −2 2 5 
2 1 1 3


1 1
1
2
R2 −R1 ,R3 −2R1
 0 −3 1
3 
→
0 −1 −1 −1


1 1
1
2
3R3 −R2
 0 −3 1
3 
→
0 0 −4 −6

The third row tells us that −4z = −6, i
...
z = 3
...
Finally, row one gives
2
2
x + 1 = 2, so x = 1 and we have a unique solution
...
com
31

Vectors and Matrices

Essential Engineering Mathematics

2
...
3

Finding The Inverse of a Matrix Using Row
Reduction
...
To do this we make another type of augmented matrix
with the identity matrix on the right hand side
...
If it is not possible to reduce the
left hand side to the identity then that is because the left hand side
is not an invertible matrix
...
3
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Bases

A basis is like a linear co-ordinate system
...
The
pair (0, 1) and (1, 0) are called the standard basis of R2
...
For example, (1, 2) and (3, 4) will do
...

2
2

Moreover, for any (x, y) the required a and b is unique
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Of course, we can speak similarly about a basis of Rn for any n, the
standard basis in Rn , which has the obvious definition, and it turns
out that any basis for Rn has to consist of precisely n vectors
...
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34

Vectors and Matrices

Essential Engineering Mathematics

2
...
Then the matrix of T
with respect to B is
a b
c d
In particular, if T sends B to {λ1 v1 , λ2 v2 } then B is called an
eigenbasis of T , λ1 and λ2 are called eigenvalues of T and v1 and
v2 are called eigenvectors of T
...
This is useful because we
can more easily find powers and limits of a matrix representing
T
...

For example,
λ1 0
A=
0 λ2
is diagonal, and An is just
λn 0
1
0 λn
2
There is a process for calculating eigenvalues and eigenvectors which
relies on (a) calculating determinants (b) row reduction
...
This is a system of linear equations with zeros in the right hand side, and has
a solution if and only if that solution is non-unique
...
This could potentially have repeated roots or even complex
roots
...
This entails the following
(usually four) computational tasks:

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• For each eigenvalue, substitute it into the equation and
solve the resulting linear system of equations to find
the eigenvectors
...
5
...
Now for example, to
find an eigenvector for λ = 3 we solve

0
1−λ
3
0
 3
−2 − λ −1 0 
0
−1
1−λ 0

for λ = 3
...
com
36

Vectors and Matrices

Essential Engineering Mathematics

We use row reduction to do this
...
We saw
in that section an example where the solution was a point (i
...
a
unique solution)
...
This is because
the bottom row says 0z = 0, which tells us nothing about z, meaning that z is unconstrained
...
This is because
any scalar multiple of an eigenvector is also an eigenvector, by the
definition of an eigenvector
...

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the key skills that top employers demand
...


London Business School
Regent’s Park
London NW1 4SA
United Kingdom
Tel +44 (0)20 7000 7573
Email mim@london
...


For more information visit www
...
edu/mim/
email mim@london
...
london
...
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37

Vectors and Matrices

Essential Engineering Mathematics

Continuing with the solution, the second row tells us that y =
−2z and the first then tells us that x = −3z
...


You should continue this example and show that the eigenvectors
for the eigenvalue −4 are all of the form


−3
α 5 
1
and find those for the eigenvalue 1, verifying your answer
...
com
38

Functions and Limits

Essential Engineering Mathematics

Chapter 3

Functions and Limits
3
...
1
...


Let X and Y be sets
...
So technically, f is actually a certain type
of set of pairs of points (x, y) with x ∈ X and y ∈ Y
...

X is called the domain of f and Y is called the range of f
...
1
...
Suppose that we define f : X → X by the rule: f (x) is the
father of x
...
On the other hand, we may define
c : X → X by defining c(x) to be the child of x
...

Example 3
...
2 Let X = Y = R
...

We usually write f for a function and f (x) for the value of f at
x
...
Often whether
39

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And that’s
just what we are looking for
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...

1
For example, f (x) = x is not a function from R to R because f (0)
is not defined
...


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1
...

One of these is called composition
...

For example, if f (x) = x2 and g(x) = x + 1 then g ◦ f (x) = x2 + 1
whereas f ◦ g(x) = (x + 1)2
...
f + g : R → R by (f + g)(x) = f (x) + g(x),
2
...
f g : R → R by (f g)(x) = f (x)g(x),
4
...
kf : R → R by (kf )(x) = kf (x) (k ∈ R) and
6
...

Later, as we meet rules for manipulating limits, derivatives and
integrals, you will see that they are usually concerned with building
functions up in the ways above
...
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41

Functions and Limits

Essential Engineering Mathematics

3
...
3

Inverse Functions

A function f : R → R is injective if f (x1 ) = f (x2 ) always implies
that x1 = x2
...
If we draw a horizontal line across the graph of an
injective function the line will never meet more than one point of
the graph
...
For
example, g(−1) = g(1)
...
The function f (x) = x3 is surjective,
1
since given y ∈ R the number y 3 maps onto y (every real number
has a unique cube root, unlike the case for square roots)
...
The function
f (x) = sin(x) is an example of a function from R to R which is
not surjective
...

If f is both injective and surjective then we say it is bijective, or
that it is a bijection
...
In order for us to be able to define an inverse
for f (i
...
a function f −1 such that f −1 ◦ f (x) = x for all x), f
must be a bijection
...


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...
2

Limits

The formal definition of a limit of a function is technical and has
been omitted
...
It is perhaps easier to
see the need for the first two cases
...

2

Example 3
...
1
1
...
This also occurs as x becomes arbitrarily negative
...


1
2
...
com
43

Functions and Limits

Essential Engineering Mathematics

As x becomes arbitrarily close to 0, either from the left or
from the right, y does not become arbitrarily close to any
value
...


3
...


As x becomes arbitrarily large, so does f (x)
...
Similarly,
limx→

−∞ f (x)

does not exist
...
com
44

Functions and Limits

Essential Engineering Mathematics

Rules For Calculating Limits Part 1
Let k, L and M and a be real numbers and let f and g be real
functions
...
Then
• Sum Rule: limx→a (f + g)(x) = L + M
• Difference Rule: limx→a (f − g)(x) = L − M
• Product Rule: limx→a (f g)(x) = LM
• Quotient Rule: if g(x) is never 0 and M = 0 then
limx→a f (x) = L / M
g
• Constant Rule: limx→a (kf )(x) = kL
• Power Rule limx→a (f k )(x) = Lk (where defined)

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These also hold if we replace x → a by x → ∞ or x → −∞
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45

Functions and Limits

Essential Engineering Mathematics

Example 3
...
2

1
...

h

Then
f (h) =

h2 + 2hx
x2 + 2hx + h2 − x2
=

...


2
...


Let x > 0
...
Similarly if h(x) = 3x2 +x+1,
h
g (x) → 3 as x → ∞
...

3

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...


|g(x)|

– If g(x) → 0 as x → 0 then f (x) → 0 as x → 0
...

• Exponential v
...

– If b > 1 and p is fixed then bx xp → ∞ as x → ∞
...
Polynomial: Let p < 0 be fixed
...

Example 3
...
3

1
...
Since x2 → 0 as x → ∞, so does
the sandwich theorems
...
What is
lim e−x x2010 ?

x→∞

In the “exponential v
...
Then e−x x2010 = ax xp → 0 as x → ∞
...
com
47

Functions and Limits

Essential Engineering Mathematics

3
...
That is, a very
small change in x will only produce a very small change in f (x)
...


Example 3
...
1 The function
f (x) =

x+1
−x

if x 0
if x < 0

is continuous everywhere except at 0
...
Join us
...
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...
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48

Functions and Limits

Essential Engineering Mathematics

Rules of Continuity Let f : R → R and g : R → R be
functions which are continuous at x0 ∈ R
...
f + g and f − g are continuous at x0
2
...
kf is continuous at x0 for any k ∈ R
...
If g(x) is never 0 then

f
g

is continuous at x0
...
f p is continuous at x0 for any p ∈ R where defined
6
...

Most functions that you are familiar with are continuous
...
They
are not defined at the points x where bm xm + · · · b1 x + b0 = 0, so
it doesn’t make sense to ask whether or not they are continuous
there
...
However if we let L be any real value
then the function f : R → R defined by
f (x) =

1
x

L

if x = 0
if x = 0

is not
...
Then for all y ∈ R between f (a) and
f (b) there exists a value x ∈ [a, b] such that f (x) = y
...
com
49

Functions and Limits

Essential Engineering Mathematics

This theorem is used for showing that roots of equations exist
...
3
...
f (1) = −1 and f (2) = 5
...
e
...
In fact the root is
about 1
...

The conclusion of the intermediate value theorem is only true for
continuous functions
...


If we choose any value y ∈ (−1, 1) then there is no x ∈ [−1, 1] with
f (x) = y
...
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50

Calculus of One Variable

Essential Engineering Mathematics

Chapter 4

Calculus of One Variable
Part 1: Differentiation
4
...


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51

Calculus of One Variable

Essential Engineering Mathematics

For example, what is the gradient of the graph of f (x) = |x|
at x = 0? Is it 1? -1? Or should we take an average of these and
say that the gradient is 0? None of those make sense, particularly,
so we just say it is not defined
...

Recall how the gradient to a curve at a point is defined
...
The gradient of this
approximation is
f (x + h) − f (x)

...

f : R → R is differentiable at x ∈ R if this limit exists:
f (x + h) − f (x)
h→0
h
lim

If the limit does exist, then we define the derivative of f
at x, written f (x), to be the value of the limit
...

dx
If f is differentiable at all points x ∈ R then we say that it
is differentiable and define the derivative of f to be the
function f given by
f (x + h) − f (x)

...
com
52

Calculus of One Variable

Essential Engineering Mathematics

Example 4
...
1
Then

1
...

f (0 + h) − f (0)
|h|
=
h
h

The graph of g : (−∞, 0) ∪ (0, ∞) → R given by g(h) = |h|/h
is as follows
...

Hence limh→0 g(h) doesn’t exist, or these one sided limits
would both give the same value
...

2
...
Then
f (x + h) − f (x)
(x + h)2 − x2
=
h
h
We saw in section 1
...
Thus the derivative of f is the function f given by
f (x) = 2x
...

Differentiability at a point implies continuity at that point
...
However, continuous functions need not neccessarily be differentiable, as the
example of the function f given by f (x) = |x| shows
...
com
53

Calculus of One Variable

Essential Engineering Mathematics

Rules of Differentiability Let f : R → R and g : R → R
be differentiable functions
...
Sum Rule: f +g is differentiable and (f +g) = f +g
...
Difference Rule: f −g is differentiable and (f −g) =
f −g
...
Product Rule: f g is differentiable and (f g) = f g +
gf
...
Quotient Rule: f /g is differentiable (where defined)
and
gf − f g

...
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1
...
Differentiate f (x) = x3 from first principles
(i
...
Don’t just write
down “the answer is 3x2 ”
...
Use the above rules to differentiate
x1999 + x2 + 1

...
com
55

Calculus of One Variable

Essential Engineering Mathematics

4
...

The Chain Rule: If f and g are functions from R to R, f
is differentiable at x ∈ R and g is differentiable at f (x) then
g o f is differentiable at x and
(g ◦ f ) (x) = g (f (x))f (x)
This theorem is also true for functions only defined on subsets of
R, provided that we can compose them
...

We can now prove the power rule for all rational indices p
...

dx
We did this for x2 from first principles
...
xp−1 )
dx
dx
= (p − 1)xp−2
...
xp−1 (by the product rule)
= (p − 1 + 1)xp−1

= pxp−1
...


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...
Let q = −p
...

d
dx

1
xq

xq
...
1
x2q
−q
=
xq+1
= pxp−1
...
To extend it to rational
n
indices p = m we can use the chain rule
...
m x m
dx

m−1

by the chain rule
...
n
m (x m )m−1
n n−1−n+ n
m

...


=

Thus the power rule holds for all rational indices
...


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...
3

Some Standard Derivatives

The following derivatives should be learnt
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4

Differentiating Inverse Functions

If f is differentiable and bijective, and hence has an inverse function
f −1 , then we can show that f −1 is also differentiable at points
where f (x) = 0
...

dx
dx
and by the chain rule, if we write y = f (x) then
(f −1 ) (y) =

1
f (x)

provided f (x) = 0, i
...
we have
(f −1 ) =

1
f

For example, if we let y = sin(x) then we can differentiate sin−1 (x)
using this method
...


We have

d
sin−1 (y) =
dy
=
=
=

1
d
dx (sin(x))

1
cos(x)

1
1−
1

sin2 (x)

1 − y2

(since cos2 x + sin2 x = 1)


...
The following graphs may
help to understand this
...
com
59

Calculus of One Variable

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Essential Engineering Mathematics

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60

Calculus of One Variable

Essential Engineering Mathematics

4
...
Differentiate the entire equation, possibly using the
chain rule
...
Rearrange the result to get the desired derivative
...
For instance
in the (x, y)-plane, the circle of radius 1 is the set of all points
(x, y) such that x2 + y 2 = 1
...
Note that the graph of the above curve is not the graph
of a function
...
In the graph of a function
there can only be one such point
...
For instance, above we have
F (x, y) = x2 + y 2 − 1
...

Example 4
...
1 Differentiating the formula
x2 + y 2 = 1

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...

dx
y
This is only valid for y = 0
...
In both of these places the graph of the circle
has a vertical tangent and so its gradient at these points is not
defined
...
This allows us to define two functions (called branches)
√
by
f+ : [−1, 1] → R defined√ f+ (x) = + 1 − x2 and f− : [−1, 1] →
R defined by f− (x) = − 1 − x2
...

For most implicit formulae there are many different branches and
many different values of the derivative for a given x
...
5
...

We have
5(y + x)4 (

dy
dy
+ 1) = 7x6 cos(x7 )e3y + sin(x7 )
...

dx
dx

Therefore
dy
(5(y + x)4 − 3e3y sin(x7 )) = 7x6 cos(x7 )e3y − 5(y + x)4
dx
which gives
7x6 cos(x7 )e3y − 5(y + x)4
dy
=

...
com
62

Calculus of One Variable

Essential Engineering Mathematics

4
...
Take logs of the whole equation and apply the logarithmic manipulation rules
...
Differentiate implicitly
...
Rearrange to get the desired derivative
...
You can’t treat this as a polynomial because the power is not constant
...
Let y = xx
...
com
63

Calculus of One Variable

Essential Engineering Mathematics

hence

dy
= y(1 + log(x)) = xx (1 + log(x))
dx
This illustrates the principle of logarithmic differentiation, really
just a special case of implicit differentiation
...

Example 4
...
1 Suppose that we want to differentiate
√
x2 + 1(x + 5)3
y=
(x + 1)2
First take logs and use logarithm manipulation rules:
log(y) =

1
log(x2 + 1) + 3 log(x + 5) − 2 log(x + 1)
2

Then differentiate implicitly:
1 dy
1 2x
3
2
=
...

( 2
2
(x + 1)
x +1 x+5 x+1

Of course this is not a particularly elegant expression, but using
the quotient rule would require similar simplification
...
6
...
Differentiate implicitly
cos(y 2 ) = sin(y 3 ) 1 + x2

2
...
com
64

Calculus of One Variable

Essential Engineering Mathematics

(b) f (x) = (xx )x
(c) f (x) = x(x

x)

Note that the functions in (b) and (c) are not the same
...
However, 33 = 27
and 327 is far larger than 273
...
com
65

Calculus of One Variable

Essential Engineering Mathematics

4
...
The derivative of f is denoted by f and
is called the second derivative of f
...
The n should be in brackets to
distinguish from the nth power of f
...

dxn
A function which is n-times differentiable for all n is called smooth
...


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66

Calculus of One Variable

Essential Engineering Mathematics

4
...

Quite often when we try to find a limit using the quotient rule we
obtain an “answer” like
0 ∞
, , 0 × ∞, ∞ − ∞, 1∞ , 00 , ∞0 , or 0∞
...
If
you obtain an indeterminate form you have to resort to some other
means of finding the limit
...
8
...

x
We see that as x → 0, both sin(x) and x tend to 0, leading to the
indeterminate form 0
...

When the relevant limits exist, are defined and are nonzero
if in the denominator, and the relevant functions are differentiable,
1
...

lim

x→x0

f (x)
f (x0 )
=

...

x→x0 g (x)
g(x)

3
...
If

f (x)
g (x)

→ −∞ as x → x0 then

→ ∞ as x → x0
...


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...

Example 4
...
2
1
...
Both
functions are differentiable
...
Thus
sin(x)
cos(0)
=
= 1
...
Find

cos(x) − 1
x→0
x
Let f (x) = cos(x) − 1 and let g(x) = x
...
Alternatively, if we assume the above limits we can differentiate sin(x) from first principles
...
Then
sin(x + h) − sin(x)
h
sin(x) cos(h) − cos(x)sin(h) − sin(x)
= lim
h→0
h
sin(x)(cos(h) − 1) + cos(x) sin(h)
= lim
h→0
h
cos(h) − 1
sin(h)
= lim sin(x)
...

h→0
h→0
h
(h)
(by the sum rule for limits)

f (x) =

lim

h→0

= sin(x)
...
1 (by the product rule for limits)
= cos(x)
...


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...
9

Taylor Series

Polynomials are easy to differentiate and integrate
...
We would
therefore like to be able to approximate functions by polynomials
...
, an are constants
...
e
...


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...
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69

Calculus of One Variable

Essential Engineering Mathematics

Clearly we have not taken much of the behaviour of f into
account
...


The constant function T0 is a degree 0 polynomial whose graph
passes through f (x0 )
...
The idea of the nth Taylor polynomial Tn is that
it continues this sequence of approximations
...
9
...
Suppose that f (x) = ex and we want a degree 2 polynomial (quadratic) T2 (x) satisfying
T2 (0) = f (0) = 1
T2 (0) = f (0) = 1
T2 (0) = f (0) = 1

Let
T2 (x) = a2 x2 + a1 x + a0
...
com
70

Calculus of One Variable

Essential Engineering Mathematics

These equations hold for all x, in particular for x = 0
...
We therefore have
2
T2 (x) = 1 + x +

x2
2

If you like think of this as the “tangent quadratic” to f at
0, although “tangent” now means that the second derivative of the quadratic also matches that of ex
...


2
...
e
...
Substituting x = 2 will not have this effect
...


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...
Now, as in the
previous example we have
T2 (y) = a2 y 2 + a1 y + a0
T2 (y) = 2a2 y + a1
T2 (y) = 2a2
When x = 2, y = 0 so we have 2a2 = T2 when y = 0
...
e
...
Thus
a2 = e2 /2
...
The second Taylor
polynomial of f at x = 2 is hence given by
T2 (y) = e2 + e2 y +

e2 y 2
2

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72

Calculus of One Variable

Essential Engineering Mathematics

We now have to write it in terms of x as
T2 (x) = e2 + e2 (x − 2) +
= e2 − e2 x +

e2 (x − 2)2
2

e2 x2
2

Let f be a function which is n-times differentiable
...
, Tn (a) = f (n) (a)

Tn is given by the following formula
Tn (x) = f (a)+f (a)(x−a)+

f (a)
f (n) (a)
(x−a)2 +· · ·+
(x−a)n
2!
n!

The proof is just an extension of the examples seen above
...
9
...
The nth Taylor polynomial of ex at 0 is
1+x+

xn
x2
+ ··· +
2!
n!

The graphs of the first few of these polynomials are as shown
...


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...
Suppose that we wish to calculate the nth Taylor polynomial
of f (x) = cosx about 0
...
So if k is odd
then f (k) (0) = 0 and if k is even, say k = 2m, we have
f (k) (0) = (−1)m
...
com
74

Calculus of One Variable

Essential Engineering Mathematics

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Exercise 4
...
3 For both of the following functions f find the Taylor polynomials T0 , T1 , T2 and T3 about the given point a
...


(a) f (x) =

√

x + 4, a = 0 (b) f (x) = sinx, a =

π
4

We have now seen how to construct polynomials of arbitrarily high
degree n, whose mth derivatives (0 m n) all match those of f
at a given point a
...


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...
9
...
Then the
Taylor series of f at a is
∞
k=0

f (k) (a)
(x − a)k = f (a) + f (a)(x − a) +
k!
=

f (a)
(x − a)2 + · · ·
2!
f (k) (a)
(x − a)k + · · ·
··· +
k!

If a = 0 then the above series is called a MacLaurin series
...
It defines such a function precisely
when it converges everywhere on I
...
9
...
However, we can calculate a Taylor
series for f at a = 1
...
k!
...
Hence f (k) (1) =
(−1)k
...
The Taylor series is then given by
f (1) + f (1)(x − 1) +
But we have

f (1)
f (k) (1)
(x − 1)2 + · · · +
(x − 1)k + · · ·
2!
k!
f (k) (1)
= (−1)k
k!

So the Taylor series is
1 − (x − 1) + (x − 1)2 − · · · + (−1)k (x − 1)k + · · ·
Notes

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...
It is
also important where the series are defined (i
...
converge to
a limit)
...


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...
1

Summing Series

Let an be a real number for all n ∈ {0, 1, 2,
...
We have the following formal
rules for manipulating finite series
...

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80

Calculus of One Variable

Essential Engineering Mathematics

5
...
e
...
)

Let f be a real function which is bounded on any closed interval,
e
...
a continuous function
...
Let [a, b]max denote the maximum value of f on [a, b] and
let [a, b]min denote the minimum value
...
Let n be a positive integer
and divide [a, b] into n equal intervals
...


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...
e
...

n
Then ck = a + k∆ and the shaded area, which we call Un , is given
by
n

n

∆[ck−1 , ck ]max =
k=1

k=1

∆[a + (k − 1)∆, a + k∆]max

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This is called the nth upper Riemann sum
...
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82

Calculus of One Variable

Essential Engineering Mathematics

Secondly, take the following shaded area
...

Note that the required definition of limit is that of the limit of a
sequence
...

Example 5
...
1 Evaluate, from first principles,
b

x2 dx

0

f (x) =
is continuous so it is integrable
...
com
83

Calculus of One Variable

Essential Engineering Mathematics

upper sum
...

n

Un =

k=1
n
b3

=

n3

kb
n

2

k 2 (since b and n are constants)

k=1

n(n + 1)(2n + 1)
(by the last section)
n3
6
3
1
b3
2+ + 2
6
n n
b3

=
=
Hence

b

n

x dx =
2

0

=
=
=

lim

n→∞

k=1

b
n

kb
n

2

3
b3
1
lim 2 + + 2
6 n→∞
n n
3
b

...
com
84

Calculus of One Variable

Essential Engineering Mathematics

5
...
A differentiable real function F is an antiderivative or indefinite integral of f if for all x, F (x) = f (x)
...

We write
f (x)dx = F (x) + C
to mean that F (x) is the unique derivative of x, up to possibly
adding a constant C
...
This is because of the fundamental theorem
of calculus which states formally that integration (finding areas
under curves) and differentiation (finding gradients) are inverse
operations of each other
...
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...
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...
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85

Calculus of One Variable

Essential Engineering Mathematics

To find antiderivatives there are various techniques
...

Example 5
...
1

xn+1
+C
n+1

xn dx =
since

d
dx

xn+1
n+1

=

n+1 n
x = xn
...


kf (x) = k
(f (x) + g(x))dx =

f (x)dx (for a constant k)
f (x)dx +

g(x)dx

From these we can deduce that
−f (x)dx = −
(f (x) − g(x))dx =

f (x)dx and
f (x)dx −

g(x)dx

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...
Similarly,
f (x)
dx
g(x)
does not have a simple formula
...

Examples 5
...
2
backwards,

1
...
Using the chain rule for differentation backwards,
2x cos(x2 )dx = sin(x2 ) + C

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...
4

Integration by Substitution

We saw in the last section that the chain rule may be applied in
reverse to find antiderivatives
...
A good way of keeping track is to use substitution
...

So we substitute u = x9 + 1 in the integral
...
P
...


www
...
com/mitas
Download free ebooks at bookboon
...
So we calculate
du
du
= 9x8 =⇒dx = 8
dx
9x
Which gives
x8 (x9 + 1)2 dx =
=
=
=
substituting back,

=

x8 u2

du
9x8

1
u2 du
9
1 u3

...

d
dx

1 9
(x + 1)3 + C
27

=

1

...
9x8 + 0,
27

which is correct
...


f (g(x))g (x)dx =

f (u)du where u = g(x)
...

Example 5
...
1

1
...
com
89

Calculus of One Variable

Essential Engineering Mathematics

Let u = cosx
...
−

= −

du
sinx

eu du

= −eu + C

= −ecosx + C
2
...
g
...
com
90

Calculus of One Variable

Essential Engineering Mathematics

3
...
g
...
Then dx = du which gives
dx
x (x + 1)

=

du
√
(u − 1) u

Then let u = v 2 to get rid of the square root
...
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v
2dv
(v + 1)(v − 1)
−1
1
+
dv
v+1 v−1
− log(v + 1) + log(v − 1) + C
v−1
log
+ C
...


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...
5

Partial Fractions

The method of partial fractions is an algebraic technique which can
be thought of as “finding a common denominator in reverse”
...

Example 5
...
1
1
...
Let x = 7 to make the coefficient of A equal to
0
...
Let x = 1 to make
3
the coefficient of B equal to 0
...
Thus
A = − 1
...


2
...
com
93

Calculus of One Variable

Essential Engineering Mathematics

We have 2 − x = A(x − 3) + B(x + 1)
...
Thus
4
2−x
−3
−1
=
+
(x + 1)(x − 3)
4(x + 1) 4(x − 3)

The main application of partial fractions is to integration of rational functions
...
5
...
If x can take values
less than 0 then
dx
= log|x| + C
...

Example 5
...
3

1
...
com
94

Calculus of One Variable

Essential Engineering Mathematics

This gives 6x + 7 = A(x + 2) + B = Ax + 2A + B
...
Coefficients of 1 give B = −5
...
(Improper fractions) How do we express
5x3 − 10x2 − 16x + 2
x2 − 2x − 3

as partial fractions? First we use long division
...

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Now we can integrate term-by-term as before
...
(Irreducible quadratics in the denominator) Integrate
(x2

(x + 3)
dx
+ 1)(x − 1)

We have to write the partial fractions expression as
(x2

x+3
Ax + B
C
= 2
+
+ 1)(x − 1)
x +1
x−1

This gives (Ax + B)(x − 1) + C(x2 + 1) = x + 3, i
...

(A + C)x2 + (B − A)x + C − B = x + 3
Comparing coefficients of x2 , x and 1 we obtain the following
system of linear equations
...
Hence
(x2

x+3
−2x − 1
2
= 2
+
+ 1)(x − 1)
x +1
x−1

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6

Integration by Parts

Integration by parts is a substitute for having no product rule for
integration
...
The
product rule for differentiation is
(uv) = uv + u v
Which we can integrate to get
uv =

uv dx +

u vdx

u vdx = uv −

uv dx

or

Example 5
...
1 Find
xex dx
Let u(x) = ex and v(x) = x
...
1dx = xex − ex + C = (x − 1)ex + C

If we had done it the other way, i
...
u(x) = x and v(x) = ex we
would have got
x2 x 1
e −
2
2

x2 ex dx

which is worse!
Sometimes we have to apply the formula more than once
...
com
98

Calculus of One Variable

Essential Engineering Mathematics

Example 5
...
2
π
2

x cosxdx =
2

x sinx
2

0

=
=
=
=
=

π2
−2
4
π2
−
4

π
2

0

−
π
2

π
2

2xsinxdx

0

xsinxdx

0
π
2

π
2

2 [−xcosx]0 − 2

π
π2
+ 2(
...
1) − 2
4
2
2
π
π
2
− 2 [sinx]0
4
π2
−2
4

(−cosx)
...

Example 5
...
3
ex sinxdx = ex sinx −

ex cosxdx

= ex sinx − ex cosx +
= ex sinx − ex cosx −

ex sinxdx
ex sinxdx

Rearranging,
2

ex sinxdx = ex sinx − ex cosx + K

which gives
ex sinxdx =

1 x
(e sinx − ex cosx) + C
2

Using the “trick” of writing logx = 1
...
sin−1 x etc
...


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...
6
...

logxdx =

1
...
dx
x

= xlogx −

dx

= xlogx − x + C

2
...
sin−1 xdx

= xsin−1 x −

(1 − x2 )

dx

(1 − x2 ) + C

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= xsin−1 x +

x

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...
7

Reduction Formulae

Integration by parts becomes tedious for integrals with large powers
e
...

x259 ex dx
x6 cosxdx
For such integrals it is better to use a reduction formula
...
Suppose that we want an expression
for
In = xn ex dx
in terms of
In−1 =

xn−1 ex dx
...
e
...
Hence

I3 = x3 ex − 3x2 ex + 6xex − 6ex + C

We can also derive reduction formulae for definite integrals
...
com
101

Calculus of One Variable

Essential Engineering Mathematics

Example 5
...
1 Find a reduction formula for
1

In =

xn e−x dx

0

We have
In =

−e−x xn

1
0

1

−

nxn−1
...
In should always be positive
because xn and ex are both positive between 0 and 1
...
com
102

Calculus of One Variable

Essential Engineering Mathematics

Sometimes a reduction formula does not reduce to In−1 but to
In−2
...

Example 5
...
2 Find a reduction formula for
xn cosx dx
Let In =

xn cosx dx
...
nxn−1 dx

= xn sinx − nxn−1
...
(n − 1)xn−2 dx
xn−2 cosx dx

= xn sinx + nxn−1 cosx − n(n − 1)In−2

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103

Calculus of One Variable

Essential Engineering Mathematics

Thus, for example,
I4 = x4 sinx + 4x3 cosx − 4
...
com
104

Calculus of One Variable

Essential Engineering Mathematics

5
...
All of the area under the
curve is above the x-axis so we should get a positive answer
...
One is an integral in which one or both of the
limits of integration are infinite, e
...

∞ 1
1 x2

or

∞ −x2
dx
−∞ e

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...

The other type is where we integrate over a bounded interval but
the function which we are integrating is unbounded on this interval,
e
...

1
1
dx
1/2
0 x

These areas may or may not be finite
...
8
...
We
define
∞

f (x) dx = lim

b

b→∞ a

a

f (x) dx

if this limit exists
...

If the limit doesn’t exist, we say that the integral diverges
...
8
...

∞
1

1
dx
x2

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...

∞
1

1
dx =
x2

1
dx
b→∞ 1 x2
1
= lim 1 −
b→∞
b
= 1

2
...


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...
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...

Now we consider the second type of improper integral
...
8
...
If f is integrable on [a, b) (possibly undefined
at b) then we define
b

c

f (x) dx = lim

c→b−

a

f (x) dx

a

whenever it exists
...
8
...


1
...
com
108

Calculus of One Variable

Essential Engineering Mathematics

2
...

1
0

1
dx
x2

We have
1
c

1
dx =
x2

−

1
x

1
c

1
= −1 −
c
→ − ∞ as c → 0+

We can also integrate from −∞ to ∞
...
8
...
H
...
exist
...
We define the value of the integral
in this case to be
c
a

b

f (x) dx +

f (x) dx

c

But both integrals must converge
...
We were using an antiderivative which we weren’t
allowed to, because it wasn’t defined at 0
...


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...
We define the value of the integral
in this case to be
Essential Engineering Mathematics
c

b

f (x) dx +

a

Calculus of One Variable

f (x) dx

c

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But both integrals must converge
...
We were using an antiderivative which we weren’t
allowed to, because it wasn’t defined at 0
...


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Two important types of functions which fall
into this category are
• scalar fields which are functions f : Rn → R
• vector fields which are functions f : Rn → Rn
This section generalises the single-variable calculus of functions
f : R → R and involves a mixture of techniques from single-variable
calculus and from linear algebra, the theory of vectors and matrices
...


6
...
Geometrically, this
can be viewed as a surface z = f (x, y)
...
com
111

Calculus of Many Variables

Essential Engineering Mathematics

the plane y = y0
...
Partial differentiation allows us to write down a function
of x and y such that this gradient is given at any point (x, y)
...
Similarly we
can define ∂f = fy (x, y)
...
1
...
com
112

Calculus of Many Variables

Essential Engineering Mathematics

6
...
e
...
Suppose that a scalar field be given by f (x1 ,
...

Then we can define the gradient of f , which is denoted by f and
is defined by

f =(

∂f
∂f
,
...

x1
xn

Or it may just be written as f , but it is underlined here to remind
us that it is a vector operator
...
It has a geometric interpretation for n = 2:
the value of f (x, y) at a point (x, y) is a vector pointing
in the direction where the surface slopes the most steeply
upwards
Example 6
...
1 If we take the function f (x, y) = x2 y + 2y 2 x3 as
in the last section then
f (x, y) = (2xy + 6y 2 x2 , x2 + 4yx3 )
You might like to try to visualise this surface
...
You might also like to consifer
what happens at (0, 0)
...
Suppose that u is a unit vector (i
...
|ˆ | = 1)
...
We write it as
ˆ
f (x, y) and it is given by
u
ˆ
u f (x, y)
ˆ

=

f (x, y) · u
...
com
113

Calculus of Many Variables

Essential Engineering Mathematics

Example 6
...
2 if we take u =
ˆ
ˆ
u f (x, y)

1
1
√ ,√
2
2

then

= (2xy + 6y 2 x2 , x2 + 4yx3 ) ·
=

1 1
√ ,√
2 2

1
√ (2xy + 6y 2 x2 + x2 + 4yx3 )
2

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3

Vector Fields

Recall that a vector field is a function f : Rn → Rn
...
, xn ),
...
xn ))
...

If a vector field v can be obtained as v = f for some scalar field
f then it is called a conservative field and f is called a (scalar)
potential for f
...

We can view vector fields by showing the vector values on a grid
There are two commonly used differential operators on vector fields,
the divergence and the curl of a vector field
...

The divergence of a vector field v is a scalar field given by
div
div v =

·v =

∂v1
∂vn
+ ··· +
x1
xn

The curl of a vector field v : R3 → R3 , which will be written as
(vx (x, y, z), vy (x, y, z), vz (x, y, z))
is a vector field given by

curl v =

×v =

∂vy ∂vx ∂vz ∂vy
∂vx
∂vz
−
,
−
,
−
∂y
∂z ∂z
∂x ∂x
∂y

Curl has a nice interpretation
...


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...
3
...
Then we have
∂
∂ 2
∂
(xy) +
(xyz) +
(y z)
∂x
∂y
∂z
= y + xz + y 2

div v =

and

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∂
∂ 2
∂ 2
∂
(y z) −
(xyz), (xy) −
(y z),
∂y
∂z
∂z
∂x
∂
∂
(xyz) −
(xy)
∂x
∂y
= (2yz − xy, 0, yz − x)

You’re full of energy
and ideas
...


© UBS 2010
...


curl v =

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4

Jacobians and The Chain Rule

This section concerns functions f : Rn → Rm for arbitrary m and
n
...
, xn ),
...
, xn ))
Given such a (differentiable) function we can write down an m × n
matrix of partial derivatives called the Jacobian matrix of f :



Jf = 

∂f1
∂x1


...


...


...


...


∂fm
∂xn





Recall the chain rule for functions f : R → R and g : R → R, that
d
dg
df
(g(f (x)) =
(f (x)) ·
(x)
dx
dx
dx
In a very aesthetically pleasing way, this translates into something
about Jacobian matrices for functions f : Rn → Rm where the
product in the ordinary chain rule is replaced by a matrix product
...
Let a be a point
in Rn
...
4
...

Example 6
...
2 Let f (x, y) = (x + y, xy), f : R2 → R2 , and let
g(x, y) = (x + y)2 , g : R2 → R
...
com
117

Calculus of Many Variables

Essential Engineering Mathematics

Note here that AT is the transpose of a matrix, where the (i, j)th
entry of AT is the (j, i)th entry of A
...


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...
5

Line Integrals

A line integral is an integral done on a vector field
...
Line integrals are of the
form
b

a

v · dr

where v is a vector field, and r is a curve from a to b
...
We can then write
as r(λ) = (x(λ), y(λ), z(λ)) for λa
down a function f : R → R
f (λ) = v(x(λ), y(λ), z(λ)) · (x(λ), y(λ), z(λ))

and treat it as a normal definite integral
λb

f (λ)dλ

λa

Example 6
...
1 Let v(x, y) = (y, x) and the curve C be paramπ
λ
eterized by r(λ) = (cos λ, sin λ) for 0
4
...
com
119

Calculus of Many Variables

Essential Engineering Mathematics

A special case is where a = b and the integral is over a closed curve
C it is called a circulation integral, and written as

C

v · dr

In the case where v is a conservative field, the value of the line
integral only depends on the endpoints, not the curve r
...

More generally we have:
Green’s Theorem: If C is a smooth closed curve in R2
and v : R2 → R2 is a smooth vector field (v = (vx , vy )) then
C

∂vy
∂vx
−
∂x
∂y

v · dr =

dx dy

R

where R is the region enclosed by C
...
com
120

Calculus of Many Variables

Essential Engineering Mathematics

6
...
A surface integral looks like

S

v · dS

where v is a vector field and S is a smooth surface
...
A normal
vector to S at a point p on S is a vector perpendicular to the
tangent plane to S at p
...
A parameterisation of S takes the
form
s(λ, µ) = (x(λ, µ), y(λ, µ), z(λ, µ))
where λ and µ are real numbers
...
Thus, as with a line integral we can rephrase the surface
integral as some more easily understood real integrals
λb
λa

µb

f (λ, µ)dµdλ

µa

where
f (λ, µ) = v(s(λ, µ)) · n(λ, µ)

The following theorem, Stokes’ theorem, allows us to evaluate a
surface integral of the curl of a vector field in terms of a line integral

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One word of warning
...
A choice which varies smoothly
over the surface S is called an orientation of S
...
If you are interested then google “mobius
band”
...

Theorem 6
...
1 (Stokes’ Theorem) If S is a smooth orientable surface with boundary C and v is a smooth vector
field in R3 , then

C

v · dr =

(

× v) · dS

S

A volume integral is simply a triple integral of some function
f : R 3 → R3
xb

f (x, y, z)dV =
V

xa

yb

zb

ya

f (x, y, z)dzdydx

za

The following theorem is usually used to evaluate a surface
integral of a vector field v in terms of a volume integral of the
divergence of v
...

Theorem 6
...
2 (The Divergence Theorem) If S is a
smooth closed surface which encloses a solid region R, and
v is a smooth vector field in R3 then

S

v · dS =

(
R

· v) dV

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...
The
solution to such an equation is not a number but a function
...
There are many
kinds which can be solved algebraically but it is often not possible
...
There are also ordinary differential equations (ODE) and
partial differential equations (PDE)
...
The latter contain partial derivatives
and their solution is a function of more than one variable
...


7
...
com
123

Ordinary Differential Equations

Essential Engineering Mathematics

can be solved for y = f (x) by using what is called an integrating
factor
...

1
...

2
...

3
...
Rearrange to get y(x) =
grate
...

Example 7
...
1 Solve
dy
+ y = ex
...
The integrating factor is
R

e

1dx

= ex
...
) Multiplying the differential equation by the integrating

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2

First Order Separable Differential Equations

A first order separable differential equation is one of the form
dy
= f (x)g(y)
dx
To solve we simply “rearrange” it and integrate both sides
...
2
...


7
...
Their
solutions are an interesting combination of what happens with first
order differential equations with constant coefficients (where the solutions are typically exponential functions) and what happens with

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...
As a quadratic equation has two roots, the
equations in this subsection have two solutions, called particular
solutions
...

First we form what is called the characteristic equation which is
the quadratic equation
p2 + ap + b = 0
We solve this to get the roots p1 and p2
...

1
...
P
...


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...
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...

2
...

3
...
The particular solutions are

y1 (x) = erx cos(sx)
y2 (x) = erx sin(sx)
(these follow from Euler’s formula eiθ = cos θ + i sin θ) and
the general solution is

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...

Example 7
...
1

1
...

2
...
So
the differential equation has general solution
y(x) = Ae−3x + Bxe−3x
...


d2 y
+ 4y = 0
dx2
(Undamped simple harmonic motion)
...
So
the differential equation has the general solution
y(x) = A cos 2x + B sin 2x
...
com
129

Ordinary Differential Equations

Essential Engineering Mathematics

4
...
The characteristic equation is
p2 + 2p + 2 = 0
which has complex roots −1 + i and −1 − i
...


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130

Ordinary Differential Equations

Essential Engineering Mathematics

7
...
Solve the corresponding homogenous equation, that is when
f (x) = 0, as in the previous section
...
Find a particular integral
...

3
...
Depending on the function f there are different trial functions you
can use
...

• kx, for which the trial solution is Kx + L
• kx2 , for which the trial solution is Kx2 + Lx + M
Example 7
...
1

d2 y
dy
− 2y = x2
+
dx2 dx

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...

A trial solution should be Kx2 + Lx + M
...

2
2
4
giving the solution to the differential equation as
3
1
1
y(x) = Ae−2x + Bex − x2 − x −
...
5

Initial Value Problems

An initial value problem or IVP for short, is where we are also
supplied with enough information to determine the constants in
the solution (e
...
A and B in the previous section)
...
Then

y(x) = Ae−2x + Bex
y (x) = −2Ae−2x + Bex

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...
So the solution to the
IVP is
y(x) = −e−2x + 3ex
...

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Copenhagen, Denmark
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– make new friends am
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This calculus has more
structure than real calculus and as a result has some surprising and
strong properties
...
1

Standard Complex Functions

Recall that complex numbers are written z = x+iy where i2 = −1
...
Then in general we can write w = u(x, y) + iv(x, y)
...
For example,

z 2 = (x + iy)2 = x2 + 2xyi + i2 y 2 = (x2 − y 2 ) + i(2xy)
So in this case, u(x, y) = x2 − y 2 and v(x, y) = 2xy
...
com
134

cerns the calculus of the complex numbers
...

Complex Function Theory

Essential Engineering Mathematics

8
...

Suppose that we are given a complex function f : C → C, so that
w = f (z)
...
u
is the real part of f and v is the imaginary part of f
...
We use power
series to define some complex functions, for example

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135

Complex Function Theory

Essential Engineering Mathematics

z2
+ ···
2!
z5
+
+ ···
5!
z4
+
+ ···
4!

ez = 1 + z +
z3
3!
z2
cos z = 1 −
2!
sin z = z −

By manipulating power series we can prove that certain identities
from the real case carry through to the complex case, such as:

ez ew = ez+w
sin(−z) = − sin z

cos(−z) = cos z

sin(z + w) = sin z cos w + cos w sin z
cos(x + w) = cos z cos w − sin z sin w
Most importantly we have Euler’s formula
eiz = cos z + i sin z
In particular for θ ∈ R, eiθ = cos θ + i sin θ giving the famous
identity eiπ = −1, and de Moivre’s theorem
(cos θ + i sin θ)n = cos nθ + i sin nθ
Also, ez = ex+iy = ex eiy gives
ez = ex (cos y + i sin y)
Thus |ez | = ex and Arg(ez ) = y
...
Recall that the real function

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...
This means that there is a number 0 = a ∈ R with
sin(x + a) = sin x
...
Since
ez+2πi = ez e2πi = ez
we have the following fact
ez is periodic, with period 2πi
...

We could also define the complex logarithm using a power series
...
Suppose that w ∈ C satisfies ew = z
...
Then

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ew = eu+iv = eu eiv = reiθ

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...
That is,
log z = log r + i(θ + 2nπ), n ∈ Z
Note that the complex logarithm is many-valued
...

The complex logarithm allows us to define complex powers of comw
plex numbers
...

There is also a simple ezpression for the complex sin and cos functions
...
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138

Complex Function Theory

Essential Engineering Mathematics

sin z = sin x cosh y + i cos x sinh y
cos z = cos x cosh y − i sin x sinh y

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139

Complex Function Theory

Essential Engineering Mathematics

8
...
f : C → C is differentiable at z ∈ C
if the limit
f (z + h) − f (z)
lim
h→0
h
df
exists; if it does them the limit is called the derivative f (z) = dz
of f at z
...
e
...
By taking two
different limits, in the direction of the real axis, and in the direction
of the imaginary axis, we obtain two different expressions for the
derivative:

Let f (z) = u(x, y) + iv(x, y)
...

The Cauchy-Riemann Equations
ux = vy , vx = −uy
Examples 8
...
1
1
...
Then f (z) = x − iy which means that u(x, y) = x and
v(x, y) = −y
...

2
...
Then we can
differentiate f to get
f (z) = ux (x, y) + ivx (x, y)
= cos x cosh y − sin x sinh y

= cos z

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...

As an exercise, try to show that
1
d
and dz (log z) = z
...
If f is a differentiable complex function (whose components are sufficently differentiable) then by the
Cauchy-Riemann equations,
uxx + uyy = vyx − vxy = 0
Because by the theorem of mixed partial derivatives vyx = vxy
...
If two harmonic functions are the
real and imaginary parts of a complex function then they are called
harmonic conjugates
...
2
...
Then ux x = 6x and
uy y = −6x so u is harmonic
...
We have vy = ux =
3x2 − 3y 2 and vx = −uy = 6xy
...


Differentiating with respect to y, we obtain vy = 3x2 +g (y), giving
g (y) = −3y 2
...
Taking c = 0
gives v(x, y) = 3x2 y − y 3
...
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141

Complex Function Theory

Essential Engineering Mathematics

8
...
Complex integrals are similar to line integrals, that
is they are performed on a curve in the complex plane
...
It is often convenient
to parameterize using the complex exponential
...
3
...
Let C be the circle |z| = r, with centre
a ∈ C parameterized in an anticlockwise direction
...
We wish to evaluate
1
iθ dθ and we have z − a = reiθ
...
Now, dz = ire
Hence

C

1
dz =
z−a

2π
0

ireiθ dθ
reiθ

2π

= i

dθ = 2πi

0

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142

Complex Function Theory

Essential Engineering Mathematics

2
...
Then we have
C

2π

(z − a)n dz =

(reiθ )n ireiθ dθ

0
2π

= irn+1

ei(n+1)θ dθ

0
2π
irn+1
=
ei(n+1)θ
i(n + 1)
0
n+1
r
e2π(n+1)i − 1
=
n+1
= 0

Since e2πim = 1 for any integer m
...
Note
• The value of the integral did not depend on a or r
• There is something special about the power −1
...
Hence this is the value of the integral
...
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143

Complex Function Theory

Essential Engineering Mathematics

Theorem 8
...
2 (Cauchy’s Integral Formula) Let f :
C → C be any function which is differentiable on and inside
any simple closed contour C which encloses (but doesn’t pass
through) the point a ∈ C
...
But you might ask
• What happens for a function which can be expressed by a
Taylor series (without having to divide by z)?
• What about functions divided by other powers of z?
In the first case we simply get 0
...

z
Firstly the computation above with, say, ez instead of ez will clearly
give 0
...
In the analytic case, they correspond to
conservative fields
...

Theorem 8
...
3 Let C and f be as in the statement of
Cauchy’s integral formula
...


All of these ideas lead up to the most powerful theorem for computing complex integrals, called the Theorem of Residues
...
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144

Complex Function Theory

Essential Engineering Mathematics

properties at only finitely many points, the singularities or poles of
the function we are integrating
...
Then
• We say that f has a pole of order k at a
...


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...
3
...
, an inside C
...
3
...
Let
f (z) =

1

...

2
Similarly, res(f, −1) = 1 and res(f, 2) = 1
...
com
146

Index

Essential Engineering Mathematics

Index
complex sine, 136
component of a vector, 24
absolute value, 14
composition of functions, 41
analytic function, 144
conservative field, 115
angle between vectors, 25
continuous function, 48
antiderivative, 85
contour, 142
area under a curve, 81
convergence of an integral, 106
Argand diagram, 20
cross product, 28
argument of a complex number,
curl of a vector field, 115
20

In , 27

basis, 34
bijection, 42

de Moivre’s Theorem, 22
derivative, 52
determinant, 27
Cauchy’s Integral Formula, 143 diagonal matrix, 35
Cauchy-Riemann Equations, 140 differentiability, 52
chain rule, one-variable, 56
differentiability of complex funccharacteristic equation (of an ODE),
tions, 140
127
differentiating inverse functions,
circulation integral, 120
59
complementary function, 131
directional derivative, 113
complex conjugate, 19
divergence of a vector field, 115
complex cosine, 136
divergence of an integral, 106
complex exponential, 136
domain of a function, 39
complex function, 135
dot product, 25
complex logarithm, 138
complex number, 18
eigenbasis, 35
complex plane, 20
complex roots of quadratic equa- eigenvalues, 35
tions, 20
eigenvectors, 35
147

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148

Index

Essential Engineering Mathematics

product rule for differentiation, Taylor series, 77
54
Taylor’s theorem, 78
purely imaginary number, 17
The Divergence Theorem, 122
theorem of residues, 146
range of a function, 39
trial solution, 131
rational number, 11
undamped simple harmonic moreal numbers, 11
tion, 129
reduction formulae, 101
residue, 145
vector field, 111
Riemann sum, 82
vector product, 28
roots of complex numbers, 22
vectors, 23
row reduction, 30
volume integral, 122
row-echelon form, 30
sandwich theorems, 46
scalar field, 111, 113
scalar product, 25
second order differential equation,
126
separable differential equation, 126
series, 79
signed area, 81
simple harmonic motion (undamped),
129
smooth function, 66
square matrix, 27
standard basis, 34
standard derivatives, 58
standard integrals, 86
Stokes’ Theorem, 122
summation of series, 79
surface, 111
surface integral, 121
surjective function, 42
system of linear equations, 30
tangent plane to a surface, 121
Taylor polynomial, 70

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