Mathematics Education Project
Shaun Mooney
April 5, 2013
1
Contents
1 Teaching Experience
1.1 Alexandra College . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Trinity College Dublin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
4
4
2 Topic
6
3 History of Differentiation
3.1 Newton-Leibniz Controversy . .
3.2 Early Work . . . . . . . . . . .
3.3 16th Century . . . . . . . . . .
3.4 Newton & Leibniz . . . . . . . .
3.5 Controversy Over Infinitesimals
3.6 Cauchy & Weierstrass . . . . .
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4 Differential Calculus in the Curriculum
4.1 History of Differential Calculus in the Leaving Certificate . . . . . . . . . . .
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5 Teaching Plan
5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Lesson Plan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3 Unit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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References
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Acknowledgments
I would like to thank Elizabeth Oldham for putting on a such an interesting course
and also for being so patient and understanding. I learnt a lot of things in this class
that I believe will serve me well in the years to come.
I must also think Professor Donal O’Donovan for organising the schools for us and
Emma Howard for all of her help over the course of this project.
3
Part 1
1
1.1
Teaching Experience
Alexandra College
For part of my teaching experience I spent one class in Alexandra College. The class was a
third year one and the students were borderline pass according to the teacher. They were in
the process of learning about the volume of three dimensional shapes, mostly cylinders, and
it was a purely behaviourist class with little cognitive or constructivist learning. The whole
class was taken up with correcting homework which most of the students had not seemed to
have done.
Unfortunately I did not have an opportunity to witness any part of the project maths
course due to the short amount of time that I spent there. I was treated to a class of
behaviourist learning. The students were attempting exam style questions and they clearly
did not understand what they were doing. Most of them had to turn back pages in order
to try and link aspects of certain questions together and apparently try and “force” the
question to work out (at least that is what I gathered from my perspective. Of course, this
is my own opinion from what I observed and I may be, and hope I am, completely wrong.
However, the way that the students approached the class leads me to believe that I am right.
There were many feeble excuses for why the homework was not done, all of which I have
heard from my own schooling days).
I must say that I was very impressed with the teacher in the way that she handled the
class. I felt that she was rather handicapped by the nature of the class, but I feel that that
was not her fault. The maths syllabus, and the Irish education system in general, does not
lend itself to the type of learning that is required to properly develop the minds of the young
generation. There seemed to be far too much teaching to the exam rather than any actual
learning of maths. As I say, this is a shortcoming of the system rather than the teacher. I
can not help but feel that they are as much victims of this system as the students.
As I have already stated, I was only present for one class. It may be that I happened to
be present for the one and only such class. However, I fear that this is not the case.
1.2
Trinity College Dublin
As well as the above teaching experience I also spent some time helping mature BESS
students in Trinity College Dublin. The major problem with BESS students is that many of
them are more interested in their other subjects and do not like maths.
As the maths and statistics is based on economics, there is a lot of calculus and probability
on the course. I was mainly helping with the maths side of things. There seemed to be a
severe lack of understanding of fundamental concepts in maths. The most worrying of these
was that they struggled with the ideas of what functions are and how to deal with them
4
properly. When dealing with differentiation some people had trouble understanding that y is
a function of x and, while I attempted to convince them, I’m not sure that they understood
it properly. I also had to explain how to differentiate a function as well as using the product
rule, quotient rule and chain rule.
While these issues are quite alarming, they are also predictable. Since it is a mix of
multiple different subjects, BESS is a subject that attracts many different types of people
and that means that there will be quite a high proportion of people who do not like maths.
5
Part 2
2
Topic
The topic I have chosen to look at for this project is differentiation. I chose this topic for
many reasons. Firstly, it has always been an area that I am very interested in. Ever since
my secondary school teacher first went through the mechanics of calculus I have found it
very interesting. I am also extremely interested in the research options related to calculus.
For my final year project I looked at interactions of partial differential equations which are
completely based on calculus.
The idea of calculus is, in my opinion, one of the greatest formulated in modern maths.
It is not a coincidence that it is used in nearly all areas of maths. It is the cornerstone of
whole areas of the subject and yet nobody still really knows who first discovered it.
6
3
3.1
History of Differentiation
Newton-Leibniz Controversy
Depending on who you were to ask you would get different answers on who discovered
calculus. In this area of the world Isaac Newton would be considered to be the first person
to have formulated calculus. However, a German, say, might disagree and claim it was in
fact Gottfried Leibniz who must take the lion’s share of the credit. This controversy has
come to be known, rather unimaginatively I might add, as the Newton-Leibniz Controversy.
Nowadays, since science is not as competitive as once it was, credit is generally given equally
to both men. Back in the day, however, there was great debate over which man deserved
greater recognition.
3.2
Early Work
When Newton and Leibniz formulated their respective theories they did not think about
limits as we do today; they worked with infinitesimals. Infinitesimals are quantities that
are infinitely small. They are as close to zero as it is possible to get without actually being
zero. To see an example of infinitesimal numbers, consider the graph f (x) = x1 . It can been
seen from the graph that, as x → ∞, f (x) → 0. However, f (x) will never be 0. The curve
makes an asymptote with f (x) = 0, i.e. it will never touch the line. This idea had been in
circulation in the maths world well before the late 1600’s.
Infinitesimals and the idea of infinity can be traced back to Ancient Greece and the
mathematicians and philosophers of antiquity. Zeno of Elea(490BC-430BC) famously posed
paradoxes, many pf which are about the infinite[17].
One of the most famous of these is ’Achilles and the Tortoise’ which states that, in a race,
if the slowest runner (the tortoise) will never be over-taken by the faster runner (Achilles).
We allow the slow runner to be slightly ahead with a speed much less than the fast runner.
However, in every moment of time the slower runner moves a non-zero distance. Therefore,
in each of these moments the fast runner must first reach where the slow runner was in the
previous moment. This problem looks at time a discrete, disjoint quantity and to anybody
today the flaws are obvious.
Another famous and similar problem is that of ’The Arrow’, which says that an arrow
will never reach it’s target and, hence, that movement is infinite. Again, Zeno splits time
into moments and claims that in order to traverse the full distance, the arrow must first fly
half the distance, then half that again, etc. Therefore, it will never reach it’s target; the
distance just keep being halved.
These problems perplexed most Greeks, as was noted by Aristotle: “Zeno’s arguments
of motion, which cause so much disquietude to those who try to solve the problems”[8].
7
However, Aristotle, himself, dismissed the paradoxes, saying that reducing time to moments
“is not granted”[8].
There were many reasons why the Ancient Greeks found Zeno’s ideas hard to comprehend,
not least of which being their number system. They were not as fortunate as we are today
with our decimal number system[18]. This meant that they could only really work in whole
√
numbers, or, to be more precise, finite whole numbers. This is why the proof that 2 is
irrational caused such furore.
This inability to deal with infinity lead to something called the method of exhaustion
which was effectively an early form of calculus[4].
The person most famous for implementing the method of exhaustion is Archimedes of
Syracuse (287BC-212BC) in his book entitled “The Method of Mechanical Theorems” or,
more simply, “The Method”. The aforementioned method is a pre-calculus way of computing
areas. We do this today with integration which uses limits, but before there were limits,
the Greeks would define area by saying that the area of one object is similar to the area
of another. For example to get the area of a circle of radius r it could be compared to a
triangle whose height is the same as the radius of the circle and whose base is the same as
the circumference of the circle[5]. So,
1
(r) (2πr)
2
= πr2
Area of triangle =
(1)
= Area of circle
Archimedes’ use of the method of exhaustion to find the area of a circle was to construct
a circumscribed and inscribed n-sided polygon (see figure 1) and to estimate the area this
way.
Figure 1: Archimedes’ method of exhaustion for finding the area of a circle[6]
8
The Greeks also had no concrete definition of length. In fact, today we define length in
terms of limits and, so, Archimedes and his peers had to limit themselves to relatively simple
curves. However, they were not limited to two dimensions. We also know that they could
find the volume of shapes such as cones, cylinders and spheres[11].
3.3
16th Century
For over a thousand years the development of calculus remained relatively static. In the 16th
century Johannes Kepler(1571-1630), on realising that the orbit of the planets could not be
circular but elliptical, cut an ellipse into segments and summed them to find the area of
it. Similarly, Bonaventura Cavalieri (1598-1647) and Pierre de Fermat(1602-1665), among
many others, used this method of indivisibles in order to sum the area under curves and
surfaces[10]. For example, Cavalieri showed that
Z
a
xn =
0
an+1
n+1
(2)
and Fermat showed that the derivative at the maxima and minima (or turning points) are
zero.
These were clearly early attempts at integration[6]. These mathematicians managed to,
unknowingly, evaluate many integrals and set up the basis for integral calculus, which would
be expanded on by Newton and Leibniz.
3.4
Newton & Leibniz
Sir Isaac Newton (1642-1727) was primarily concerned with Physics. For this reason, when
he developed calculus, he did it in terms of time, that is, he considered how variables changed
with time.
While Newton considered how variables changed with time, Gottfried Wilhelm von Leibniz (1646-1716) was from a more pure side of maths and he developed calculus by thinking
about how variables ranged over sequences that are infinitely close[2]. While Newton was
quite flippant about the notation he used, Leibniz was very precise in the way he did everything, even down to his notation. It was himself that introduced the symbols dx and
dy
today is precisely because
dy that are still commonly used today. The reason we use dx
of Leibniz’ pedantry. This notation was carefully thought through, representing differences
between successive sequences.
As with those that came before them, Newton and Leibniz used infinitesimals. Since
limits had not yet been introduced these were the most convenient for their calculations.
There was, however, controversy about the use of infinitesimals.
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3.5
Controversy Over Infinitesimals
The problem with infinitesimals is that are not good for rigorously defining anything. This
is because they do not really exist, in so much as that they can be manipulated easily into
what you want.
To illustrate the controversy, I will give an example. The definition of a derivative, as we
now know, is
f (x + h) − f (x)
(3)
f 0 (x) = lim
h→0
h
where f is a function, x is a variable and h is an infinitesimal (by which I mean that I am
simulating it as an infinitesimal by sending it to zero). It is important to note that neither
Newton nor Leibniz, or anyone else alive at that time for that matter, had this formula.
For this simple example I will compute the derivative of f = x3
(x + h)3 − x3
h→0
h
3
x + 3x2 h + 3xh2 + h3 − x3
= lim
h→0
h
3x2 h + 3xh2 + h3
= lim
h→0
h
2
= lim 3x + 3xh + h2
f 0 (x) = lim
(4)
h→0
= 3x2
d 3
x = 3x2 ). What many people had a
As we can clearly see this gives the correct result ( dx
problem with however is the rôle that h plays. Up until the second to last line h is considered
to be an ordinary, non-zero variable. However, on the second last line it supposed so small
that it makes all terms containing it negligible. The use of limits redeems this inconsistency,
however, before the introduction of limits, this was seen as a major problem in the work
of Newton, Leibniz and anyone else working with them by many people, one of the most
notable of whom was Berkeley.
Lord Bishop George Berkeley (1685-1753) was a philosopher with an interest in maths
who attacked the propagators of calculus [1] in his book “The Analyst: A Discourse Addressed to an Infidel Mathematician”. In it, he makes his reservations clear about infinitesimals or “ghosts of departed quantities”[9]. In fact, Berkeley’s criticisms were so well founded
that it was not until very recently that infinitesimals were put on a rigorous foothold, namely
in a book by the name of ’Non-Standard Analysis’ by Abraham Robinson[16]. In his book,
Robinson introduces hyperreal numbers, ∗ R, whose main application is in non-standard analysis. An immediate consequence of hyperreal numbers is that the infinitesimal quantity can
be rigorously defined.
These criticisms about the foundations of calculus meant that it took a long time for it
to be rigorously developed and widely accepted.
10
3.6
Cauchy & Weierstrass
Augustin Louis Cauchy(1789 - 1857) had a profound impact on differential calculus. In his
book “Course D’Analyse”, he gave what is believed to be the first appearance of the − δ
definition[15]. This powerful definition provides a way to prove continuity of a curve which
is very important for calculus since it is not possible differentiate a curve thatis not smooth.
The definition states that, for any > 0, there exists some δ > 0 such that
|x − x0 | < δ ⇒ |f (x) − f (x0 )| < (5)
Another very important development in calculus was Cauchy’s proof of Taylor’s Theorem.
This theorem allows one to express a function at a particular point in terms of it’s derivative
and is an extremely powerful tool.
One thing I feel I should point out here is the major contribution that Cauchy made
to analysis with the Fundamental Theorem of Calculus[14]. As this is primarily about
differentiation I will not discuss this in detail, however, it is one of the most fundamental
and brilliant concepts in calculus and I feel I would be remiss without mentioning it.
The reason that it is so important is because it relates differentials to integrals and,
therefore, makes a link between differential and integral calculus. The Fundamental Theorem
of Calculus states that for a continuous f on a closed interval [a, b] and with integral F , then
Z
b
f (x)dx = F (b) − F (a)
(6)
a
We know that Newton and Leibniz at least understood the concept of this theorem, however, they were not able to prove it. What the above theorem effectively states is that
differentiation is the inverse of integration.
Just as Cauchy did, Karl Weierstrass (1815-1897) attempted to put the work of Newton
and Leibniz on a more rigorous footing. While it is believed that Cauchy first introduced the
− δ definition formally to the world of maths, the proof he gave was not entirely correct.
Cauchy proved it for pointwise convergence when he should have proved it for uniform
convergence. Weierstrass realised the importance of this definition and proved it rigorously
himself. The form of this definition given in equation (5) was presented by Weierstrass[15].
11
4
4.1
Differential Calculus in the Curriculum
History of Differential Calculus in the Leaving Certificate
Between 1920 and 1960 there were very few changes in the Irish maths curriculum[21]. In fact,
the syllabus remained quite stable after the free state was set up in 1922. The papers for the
leaving certificate were divided in three sections. These were arithmetic, algebra & calculus
and geometry & trigonometry. The calculus portion of the course included derivatives and
integrals of elementary functions (Algebraic and Trigonometric) with easy applications to
problems on rates, tangents, areas, volumes and to problems of motion[7].
It was during the 1960s that changes began to take place, with a new phase in the teaching
of maths being implemented in 1964[20]. This new phase was put into motion in 1959 ”when
the Department of Education sent a representative to the OEEC conference ’New Thinking
In School of Mathematics’”[19].
For over 60 years the scope of leaving certificate maths was relatively limited. In 1964
there were many subjects added to the syllabus in order to give students a better grounding
in maths education. Subjects added include sets, statistics & probability and vectors, among
other things. To accommodate these new additions there were many aspects of the old course
dropped, including certain theorems and troublesome calculations. This revision also meant
that part of the calculus course began to be taught at lower levels[19].
During this revision it was decided that regular updates to the curriculum would be
necessary and, accordingly, the next revision happened in the 1970s. During this decade
matrices and linear transformations were added to the course whereas the areas of Taylor
series and polar co-ordinates were reduced and moved to the applied maths course[21].
The maths course has always been a very long one. In many schools today students who
are capable for the higher level opt to do pass instead because of the workload and this was
no different back in the 1970s[21]. A proposed fix for this was to count a suitable higher
level grade as the equivalent of two other courses. This was also done because universities
wanted students who were better at maths.
However, it was not just the universities. Employers were complaining that students
failed were not able to perform basic arithmetic and blamed the course[21]. This led to
a greater emphasis being put on algebraic manipulation and arithmetic calculation and a
reduction in applications.
The next, and outgoing, curriculum was the 1992 one. This course was still examined
on two papers and had the following subjects: algebra, geometry, trigonometry, sequences
& series, functions & calculus and discrete maths & statistics[3]. All of these were examined
on paper 1 and section A of paper 2. There were also four options to choose from (further
calculus & series, further probability & statistics, groups and further geometry) which were
examined in section B of paper 2. With regard to calculus, the main change from the previous
course was the introduction of the option, further calculus and series.
Project Maths, which is quickly being introduced at the moment in the Irish curriculum,
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is set to completely change the way the maths is taught and thought about in Irish schools.
It is set to be completely functional in the junior certificate curriculum by 2015 and the
leaving certificate curriculum by 2014. It is made up of five strands
• Strand 1: Statistics and Probability
• Strand 2: Geometry and Trigonometry
• Strand 3: Numbers
• Strand 4: Algebra
• Strand 5: Functions
with a large proportion of the weight stacked on the shoulders of Probability and Statistics.
On first viewing the list of strands, none the words calculus, differentiation or integration
appear. This led to a number of people believing that they went the were axed. Fortunately,
this is not the case, as, if it were, it would be a tragedy of the highest proportions. In fact,
calculus is tucked neatly into the fifth strand, the ambiguously named functions.
The amount of calculus being taught, however, has reduced. The changes from the
previous curriculum include
• a reduction in first principle differentiation. Only linear and quadratic are to be examined,
• proof of induction that
d
(xn )
dx
= nxn−1 has been removed,
• differentiation of implicit and parametric functions appears to have been removes also.
As I have noted earlier the time that I spent in Alexandra College I observed the students
struggling through exam-style questions. The life of a leaving certificate student (not only
maths) is usually spent grappling with these type of questions, practicing for hours on end so
when the right questions finally appear in the exam the student has three hours to regurgitate
the numerous answers they have so painstakingly learned off. This is not what I consider
learning and it is the type of approach that project maths is trying to eliminate. However,
could they be going too far too fast? The authors of [12] claim that a complete reform of the
system may be going too far. They claim that “there seemed to be far too much teaching
to the exam rather than any actual learning of maths.” They suggest that it may be more
efficient to make the teachers more skilled and reform the examination system.
Personally, I tend to agree with the above view. I feel that project maths is an overreaction, a solution to the wrong problem, if you will, a mis-diagnosis.
13
5
Teaching Plan
5.1
Introduction
In my lesson plan I am choosing to focus on differential calculus and I hope to cover most
of the course here. I plan to teach a high level honours class differentiation. The way I
am choosing to teach it is with rates of change by way of limits. I try not to go into too
much detail where at all possible and I plan to cover as much relevant material as possible,
including a review of functions at the beginning as they are very important for a course such
as this.
There is certain prior knowledge assumed for this course, namely that the students have
covered all of the material in the junior cycle as well as geometry, trigonometry and functions
in the senior cycle.
My main resources for the teaching of this class will be the prescribed textbook and the
past examination papers, however, how close I will stick to the textbook is still up for debate.
As I have stated before I look to justify the calculus by the concepts of rates of change.
Most of the class will be taught using the blackboard and the students are assumed to have
the following attributes:
Year
6th
5.2
Subject
Mathematics: Higher Level
Topic
Differential Calculus
Number of students
20
Characteristic of Class
Able
Class Length
45 minutes
Lesson Plan1
Since strand 5 of project maths is not due to be implemented until the summer of 2014 I
shall outline my lesson plan based on the syllabus of the previous incarnation, that is, the
1992 syllabus.
1
This lesson plan is based on notes given by [13]
14
Lesson 1
Review of linear, quadratic and cubic
functions
Lesson 2
Review of trigonometric, exponential
and logarithmic functions
Lesson 3
Limits
Lesson 4
Limits at Infinity
Lesson 5
Differentiation
Lesson 6
Rate of Change and turning points
Lesson 7
Maxima, minima and points of inflection
Lesson 8
Derivatives of trigonometric, exponential and logarithmic functions
Lesson 9
Product, quotient and chain rules
Lesson 10
Implicit and parametric differentiation
Lesson 11
In-class test
5.3
5.3.1
Unit
Lesson 1
Prior Knowledge A basic knowledge of functions is assumed, although there will be a
review so it does not have to be very in-depth.
Content & Skills Graphs of constant, linear, quadratic and cubic functions.
Principal Aim Student will understand the difference between a constant, linear, quadratic
and cubic function. They will understand how each of these work and how to find points of
intersection of functions like these.
Behavioural Objectives At the end of the class, students should also be able to
• draw a graph using a given function,
• come up with their own conclusions from looking at a graph of a curve.
Introduction The first thing to do is to make sure that the students understand the
difference between a variable and a constant. I will begin by writing the functions f (x) = 2
and f (x) = x. In the interest of class participation I will open the floor to the meaning of
these functions. I will explain that while the first function remains the same everywhere,
the second one will change depending on the value of x. While this seems very basic, I view
it as a necessary step since a lack of comprehension of what variables are can lead to many
problems.
15
After giving an explanation I will draw each of the functions on the board so the students
can get a visual representation of what is happening. From this it should be clear that
f (x) = 2 will forever remain the same, while f (x) = x rises steadily and uniformly. In order
to further illustrate the latter’s multi-valued nature I will plug in a few values of x to show
how it changes.
Development
Presentation 1 On the blackboard write down f (x) = 2x.
Question: What can we say about this curve?
Application 1 Ask the students to consider a few different values of x and, hence, to
draw the curve on graph paper. Ask them to compare their graph to the one drawn on the
blackboard for f (x) = x. What do they notice?
Presentation 2 On the blackboard write down f (x) = x2 . Again we find values of
f (x) and plot these points. Do the same for f (x) = x2 + 1.
Question: What are the similarities with these graphs?
Application 2 Ask the students to do the same for the graphs f (x) = 2x2 and f (x) =
3x2 + 2.
Explain to the students that these are called quadratic curves and that they always look
like a parabola.
Application 3 By now functions should be coming back. Ask students to look at cubic
curves. The remainder of the class will be spent considering compositions of functions and
intersection of functions.
5.3.2
Lesson 2
Prior Knowledge Basic trigonometry. Students should also know basic facts about the
exponential and logarithmic functions.
Content & Skills Graphs of trigonometric, exponential and logarithmic functions.
Principal Aim The aim of this class is to review trigonometric, exponential and logarithmic functions.
16
Behavioural Objectives At the end of the class, the students should be able to:
• recognise the graphs of the above functions,
• draw and infer conclusions from these graphs.
Introduction Today we want to review trigonometric functions, such as sin, cos and tan
and the exponential and logarithmic functions.
Development
Presentation 1 I begin by drawing the graphs of sin x and cos x on the blackboard.
We note that they both oscillate with the same period.
I will quickly go over how sin and cos are defined using the unit circle.
Application 1 I will ask the students to tell me specific values of sin x and cos x for
certain values of x. We will also consider the C.A.S.T. diagram, that is, the one that gives
the sign of sin, cos and tan in different quadrants.
Presentation 2 Next we wish to look at tan. Again, I will go over the definition of
tan and then proceed to graph it. If asymptotes have not been discussed before, this is the
time to do it. I will show how tan x has an asymptote at x = nπ
.
2
Application 2 As with sin and cos we will now look at different values of tan and how
it behaves.
Presentation 3 The final part of this class will be spent looking at the natural exponential and logarithm. Again, we graph the functions and we note that they are inverses of
each other.
The exponential has special properties, such as e0 = 1, ex > 0 and it is always increasing.
We also know that it approaches zero as x approaches −∞ and it approaches ∞ as x
approaches ∞.
Log also has special properties like log 1 = 0. There is no need to discuss bases as we are
only using the natural logarithm.
Since ex = (log x)−1 , we have that elog x = x = log ex .
5.3.3
Lesson 3
Prior Knowledge Students should understand functions and be confident graphing them
and inferring conclusions from graphs.
17
Content & Skills Limits
Principal Aim In this lesson I want to successfully introduce the idea of limits and give
some properties of them.
Behavioural Objectives At the end of the class, the students should be able to:
• compute limits of simple functions.
Introduction In this class we want to consider the following question: If we have a function
f (x), what happens when x gets close to some point a? Let’s look at this question with
some examples.
Development
Presentation 1 Let’s take
f (x) =
2x2 − x + 5
x+1
(7)
and a = 1. As x approaches a, f (x) approaches 3. This value, f (a) = 3, is called the limit
of the function f (x) at the point a. In other words, the limit of a function is the value it
approaches at a certain point. We call this process taking a limit and, if our function is a
polynomial, we can simply substitute this value into the function in order to find the limit
at that point. In mathematical notation this looks like
lim f (x) = f (a)
x→a
(8)
Application 1 Evaluate the following limits:
•
f (x) = (x − 3)3
(9)
f (x) = cos(x)
(10)
as x approaches a = 4.
•
as x approaches a = π.
18
Presentation 2 We now look to estimate
x2 + 4x − 12
x→2
x2 − 2x
lim f (x) = lim
x→2
(11)
Notice I say estimate. Solving this normally would yield zero in the denominator. Therefore,
we look at what happens in the limit.
x
f (x)
x
f (x)
2.5
2.1
2.01
2.001
3.4
3.857
3.985
3.999
1.5
1.9
1.99
1.999
5.0
4.158
4.015
4.000
Subbing x = 2 straight into the equation is clearly not possible. Therefore, we have
looked at values close to this point on either side of our curve. From this it should be clear
that
x2 + 4x − 12
lim
=4
(12)
x→2
x2 − 2x
Explain that the function does not actually exist at x = 2. What we are doing above is
considering what happens around this point. A limit does not care what happens at the
point itself.
At this stage it should be useful to look at some of the properties of limits. As with all
operations in maths, limits have many properties. Firstly, limx→a c = c if c is a constant.
We also have limx→a x = a and limx→a xn = an .
Limits are linear. This means that limx→a [cf (x)] = c limx→a f (x) and limx→a [f (x) +
g(x)] = limx→a f (x) + limx→a g(x). Hopefully this is straightforward enough as the students
should have seen something similar before.
We can do something similar for hproducts
and quotients in this way: limx→a [f (x)g(x)] =
i
limx→a f (x)
f (x)
limx→a f (x) limx→a g(x) and limx→a g(x) = limx→a g(x) . Again, it is my hope that the students will be able to pick up on these without too much effort.
It is probably necessary to give examples of all of these with figures after I state each
property.
5.3.4
Lesson 4
Prior Knowledge Students should understand the basics of taking limits.
Content & Skills Limits at infinity
Principal Aim The aim of this lesson is to introduce limits at infinity.
Behavioural Objectives At the end of the class, the students should be able to:
• compute limits near zero and infinity.
19
Introduction As limits have been introduced, the next step is to consider limits at infinity.
Development
Presentation 1 To begin this class we look at functions that go to plus or minus
infinity. These are called limits at infinity and they are very important.
A function that takes the value of infinity at a is denoted
lim f (x) = ∞
x→a
(13)
and a function that takes the value of minus infinity at a is denoted
lim f (x) = −∞
x→a
(14)
To demonstrate this, look at the example
1
=∞
x→0 x
lim
(15)
To see this, graph the function on the blackboard. We can see that as x gets smaller and
smaller, the curve appears to get closer and closer to the y-axis. In fact it will never touch
it. The y-axis here is known as the asymptote of the curve. Since the function never touches
it, the value of f will approach infinity as x approaches 0.
To further see this, look at f (x) for specific values of x:
x
f (x)
1
0.1
0.01
0.001
1
10
100
1000
We can see from that table that, as x decreases, f (x) increases.
Now we shall consider what happens when x get bigger. Again, from the graph, it is
clear that as x increases, f decreases. As before, we look at some specific points:
x
f (x)
1
10
100
1000
1
0.1
0.01
0.001
So, f gets smaller and smaller without touching the x-axis. We say that this axis is an
asymptote to the curve. This shows us that limx→0 x1 = ∞.
20
Application 1
• Plot the function
f (x) =
2
x−1
(16)
• Evaluate the following limit
2
x→1 x − 1
lim
(17)
Does your graph correspond to the limit?
Presentation 2 We now wish to look at more complex limits at infinity. For brevity
I will do it here in variables rather than numbers. Let f (x) = a0 + · · · + an xn and g(x) =
b1 + · · · + bn x n .
What happens in the following limit? limx→∞ f (x). It would be tempting to simply plug
in x = ∞. However, we can not do this. It leads to too much ambiguity. It will probably
be important to explain this properly. The way we fix this is to factor out xn . This gives us
f (x) = xn xan0 + · · · + an . Evaluating this limit yields limx→∞ f (x) = limx→∞ an xn , where
f is defined as before.
n
(x)
nx
= limx→∞ ab00 +···a
. We look to do the same as
Next we consider limx→∞ fg(x)
+···bn xn
before, that is, factoring out xn . This leads to
lim
x→∞
f (x)
g(x)
xn
= lim n
x→∞ x
= lim
x→∞
=
Similarly, limx→−∞
f (x)
g(x)
=
a0
xn
b0
xn
· · · an
· · · bn
!
a0
· · · an
xn
b0
· · · bn
xn
!
an
bn
an
.
bn
Conclusion Homework:
• Evaluate limx→0 log x,
• Evaluate limx→∞
x4 +x2 −2
,
x−1
• Evaluate limx→−∞
x2 +1
.
x3 −2x+1
Hint: To verify answers it is usually useful to graph functions.
21
(18)
5.3.5
Lesson 5
Prior Knowledge Limits
Content & Skills Basic differentiation.
Principal Aim I wish to introduce differentiation from first principles
Behavioural Objectives At the end of the class, the students should be able to:
• differentiate polynomials from first principles,
• differentiate quotients from first principles.
Introduction Today I will introduce the following limit
f (x) − f (a)
x→a
x−a
lim
(19)
This is called the derivative of a function f at the point a. The reason why we are looking at
this limit in particular will be discussed in the next class. For now, however, it is important
to see if one understands the idea of limits as they were introduced.
We can rewrite the above equation in a more computationally friendly form. If we make
the substitution x − a = h, we find that
f (a + h) − f (a)
h→0
h
lim
(20)
Now, we have the following definition:
The derivative of a function f at a point x is denoted f 0 (x) and is given by
f (x + h) − f (x)
h→0
h
f 0 (x) = lim
(21)
This is called differentiating a function from first principles.
Development
Presentation 1 To see how this definition works we shall compute a few examples.
Find the derivative of
f (x) = 2x2 + 3x − 5
(22)
22
By the above definition, we have
f (x + h) − f (x)
h→0
h
2(x + h)2 + 3(x + h) − 5 − (2x2 + 3x − 5)
= lim
h→0
h
2x2 + 4xh + 2h2 + 3x + 3h − 5 − 2x2 − 3x + 5
= lim
h→0
h
2
4xh + 2h + 3h
= lim
h→0
h
= lim 4x + 2h + 3
f 0 (x) = lim
(23)
h→0
= 4x + 3
Therefore, the derivative of f (x) = 2x2 + 3x − 5 with respect to x is f 0 (x) = 4x + 3. This
computation is the same for all polynomials.
I would also like to introduce alternative notation. Recall that we can also write a
function with y instead of f (x). The above equation would look like y = 2x2 + 3x − 5 in this
dy
form. Taking a derivative with respect to x is now denoted dx
. It should be noted that this
is the exact same as f 0 (x), just the notation is different.
Application Find the derivative of the following polynomials:
• f (t) = t2 − 5t − 2,
• y = (x − 3)3 .
Presentation 2 We will now consider a more complicated example. Recall from a
previous class we considered the function f (x) = x1 . What is the derivative of this?
1
x+h
− x1
f (x) = lim
h→0
h
1 x − (x + h)
= lim
h→0 h x(x + h)
1 −h
= lim
h→0 h x(x + h)
−1
= lim
h→0 x(x + h)
1
=− 2
x
0
It is very important that the students understand all of the above steps.
Application 2 Find the derivative of
• g(x) =
x
.
x+1
23
(24)
Presentation 3 Recall the first few examples from this class. Is there any relation
between the function and it’s derivative? Are there any conclusions we can make? If noone knows, do another example. We should eventually come to the conclusion that we can
multiply by the power and then subtract one from the power. Explain that this only works
for polynomials. Formally, if f is of the form xn , then it’s derivative is f 0 (x) = nxn−1 .
Therefore, in the first example we can easily compute the derivative in this way. We have
f (x) = 2x2 + 3x − 5. Looking at the first term in this function we get 2(2x2−1 ), by using the
above rule, or 4x. The second term becomes 3x1−1 = 3. Note that constants become zero.
Putting this together yields f 0 (x) = 4x + 3.
Presentation 4 Finally I will differentiate f (x) =
√
x:
f (x + h) − f (x)
h→0
h
√
√
x+h− x
= lim
h→0
h
√
√ √
√
x+h− x x+h+ x
√
= lim
√
h→0
h
x+h+ x
x+h−x
= lim √
√
h→0 h( x + h +
x)
1
= √
2 x
lim f (x) = lim
h→0
(25)
Note that this last calculation may overflow into the next class as it may need a more lengthly
explanation.
Conclusion Homework:
Differentiate the following and confirm your answer by differentiating from first principles
• f (x) = x3 − 2x2 + 3x − 4,
• f (x) = (2x + 3)4 .
Differentiate the following from first principles
• g(x) =
x+1
.
x
Remind the class about the rules of indices
5.3.6
Lesson 6
Prior Knowledge Slope from geometry
Content & Skills Slope, rate of change and turning points.
24
Principal Aim The aim here is to consider practical uses of differentiation. We want to
look at what differentiation means in maths.
Behavioural Objectives At the end of the class, the students should be able to:
• find the slope of a curve using differentiation,
• compute turning points of a function.
Introduction Differentiation has many applications in many areas of life and the idea of
rates of change is one of the most important.
First, we recall from geometry the formula for slope:
m=
f (x) − f (x1 )
x − x1
(26)
where m is the slope and (x1 , f (x1 )), (x, f (x)) are points on the curve.
To demonstrate this formula, I draw the graph of f (x) = x on the blackboard. I mark
two arbitrary points and call them (x1 , f (x1 )), (x, f (x)). It will also be necessary to show
where each point intersects the x- and y- axes.
As we can see, the slope of this curve is the vertical distance divided by the horizontal
distance. In other words, the slope can be described as the change in f (x) divided by the
change in x. Using the above formula we find that, for f (x) = x, we have
m=
x − x1
x − x1
(27)
=1
This tells us that our function changes at the same rate everywhere, which is the same with
all linear functions. This is not an easy concept to grasp and it is, therefore, necessary to
explain it fully.
Development
Presentation 1 We can use the above formula for slope, equation (26), to find how
a function changes between two given points. Let’s look at this for a quadratic function,
f (x) = 2x2 , say. We will try to find the slope of this curve at x1 = 3.
By equation (26),
2x2 − 2x21
m=
x − x1
(28)
2x2 − 18
=
x−3
25
We can use this formula in order to find how f (x) changes between x = 3 and some other
point on the line.
What if we want to know how it changes at x = 3? This formula clearly will not work
because, if we substitute x = 3 into (28), we find that m = 0. Drawing a rough diagram on
the blackboard we can see that this is not the case. I mark the point x = 3 and we see that
the slope is non-zero. If it was the curve would be constant at that point.
What are we to do? Does anyone offer a suggestion? The formula for slope gives us the
slope between two points. How can we incorporate this? What if we look for our second
point to be as close to x = 3 as possible? Remind the class about the lesson on limits.
What happens if we take a limit as x approaches 3? To do this we consider the slope
between our point x and a point infinitesimally close to it. We call this point x + h, where
h is an infinitesimal.
Applying this gives us
2x2 − 2x21
m=
x − x1
(29)
2(x + h)2 − 2x2
=
x+h−x
Since h is an infinitesimal we need to take the limit as h approaches 0, therefore
2(x + h)2 − 2x2
h→0
h
f (x + h) − f (x)
= lim
h→0
h
m = lim
(30)
Hopefully, the students will see that this is the same as the definition we have for the
derivative. Therefore, the value of the derivative of a function at a point is the same as the
slope at that point.
Presentation 2 We can see from the graph of the function that before the y-axis it
appears to be coming down while after the y-axis it appears to be going up, or, in other
words, the function seems to be decreasing and then increasing. Recall from geometry that
the slope can be found using trigonometry. We find it by taking the tan of the angle the
function makes at a certain point with the x-axis.
Therefore, if we take f (x) = x, we know that m = tan(45) = 1. If we take f (x) = −x, we
know that m = tan(135) = −1. Note that the first function is always increasing, while the
second function is always decreasing. Also note that the slope of the first function is positive
and the slope of the second function is negative. This relation between the behaviour of a
function and the sign of the slope, or derivative, always holds. That is, a function with a
positive slope at a certain point is said to be increasing, while a function with a negative
slope at a certain point is said to be decreasing.
Returning to f (x) = 2x2 , we know that m = f 0 (x) = 4x. Taking x = 3, we find that
f 0 (3) = 12, which is positive, therefore, the function is increasing at this point. Similarly,
26
if we take x = −2, we find that f 0 (x) = −8, which is negative, therefore, the function is
decreasing at this point.
Presentation 3 This is all very good but we must now consider what happens at
the origin. Note that as the graph cuts the y-axis the function stops decreasing and starts
increasing. This is called a turning point. It is important because it is a transition point.
What is interesting is that at x = 0 the function becomes constant and, therefore, the
derivative is zero. It is not difficult to understand why the function is constant at a turning
point. To get from the decreasing part to the increasing part it must flatten out somewhere.
In our example this happens at x = 0.
To see this we want to find the point where the slope is equal to 0. Since the slope is the
same as the derivative we differentiate our function and set it equal to 0. Doing this we find
that 4x = 0, which implies x = 0. In order to find the exact turning point we need a second
co-ordinate. Subbing x = 0 into f (x) = 2x2 gives us that co-ordinate. Therefore, we now
know that we have a turning point at (0, 0).
Application The rest of the class will be spent looking at turning points of functions.
5.3.7
Lesson 7
Prior Knowledge Turning points and differentiation.
Content & Skills Maxima and minima and points of inflection.
Principal Aim In this lesson I want to introduce the idea of maxima and minima of
a curve and points of inflection. I also want to discuss how to sketch a graph using this
information.
Behavioural Objectives At the end of the class, the students should be able to:
• find the local maxima and minima of a function,
• find the point of inflection of a function,
• determine where a function is increasing and decreasing
• sketch functions using the above information.
27
Introduction Recall from the last lesson we saw how to find the turning points of a
function. In this lesson we will see a practical use for these turning points, namely we will
see how to find local maxima and minima of a function.
What do I mean by maximum and minimum. A function has a maximum at the point
where it has the greatest value on the y-axis. Similarly, a function has a minimum at the
point where it has the lowest value on the y-axis.
Development
Presentation 1 Recall that a turning point is found by finding where the derivative
is zero. Let’s take f (x) = 2x3 + 3x2 − 12x + 4. Therefore, f 0 (x) = 6x2 + 6x − 12. This
function has two turning points, namely (1, −3) and (−2, 24). This function has a maximum
at (−2, 24) and a minimum at (1, −3).
How do we know that this is true? Do we not have to check all points? We know that
these are right because maxima and minima can only occur at turning points.
What are the maxima and minima of f (x) = 1/4x4 + 2/3x3 − 1/2x2 − 2x − 1/12? We
have three turning points: (−2, 7/12), (−1, 1) and (1, −5/3). From this we see that we have
a minimum at the point (1, −5/3). However, what is our maximum? It is the one that takes
the highest value on the y-axis, namely (−1, 1).
Application 1 Find the maxima and minima of the following functions:
• f (x) = x3 − 2x2 + 1,
• f (t) = t5 + t4 − 3t3 + 1.
Presentation 2 First derivatives are very useful. They can give us lots of information
about out function. We can use the first derivative, for example, to tell if our function is
increasing, decreasing, or neither. What do these words mean in maths? If x1 < x2 , we
say f is increasing if f (x1 ) < f (x2 ) and we say f is decreasing if f (x1 ) > f (x2 ). If we take
the above example f (t) = t5 + t4 − 3t3 + 1, we find that the turning points are (0, 1), (1, 0)
and (− 59 , 10). This tell us that it is increasing on (−∞, − 95 ) and (1, ∞) and decreasing on
(− 95 , 1).
Application 2 Find where f (x) = x3 − 2x2 + 1 is increasing and decreasing.
Presentation 3 We are now able to sketch graphs. I will demonstrate this by sketching
f (x) = x3 − 2x2 + 1 using turning points and its increasing and decreasing properties.
Application 3 Sketch the graph of f (t) = t5 + t4 − 3t3 + 1.
28
Presentation 4 We now wish to look at points of inflection. These are where the
direction of the curve changes. At these points the second derivative is zero. To demonstrate
this consider f (x) = x3 . We find that f 00 (x) = 6x. This is zero when x = 0. Looking at the
graph of the curve we see that at the origin it moves from being concave shape to a convex
shape.
This leads us to the following. A function, f , is called concave at a point a if f 00 (a) < 0
and it is called convex at a if f 00 (a) > 0.
Application 4 Determine where f (x) = x4 − x2 + 1 is concave and convex.
The rest of the class will be taken up doing real life examples such as: a population is
modelled by the function f (t) = t2 log(3t) + 6. After how long will it start to increase?
5.3.8
Lesson 8
Prior Knowledge Trigonometric, exponential and logarithmic functions, limits and the
definition of the derivative.
Content & Skills Trigonometric, exponential and logarithmic functions.
Principal Aim We wish to look at the derivatives of trigonometric, exponential and logarithmic functions.
Behavioural Objectives At the end of the class, the students should be able to:
• differentiate trigonometric, exponential and logarithmic functions.
Introduction We wish to use the definition of the derivative to differentiate certain special
functions.
Development
Presentation 1 We will derive the derivatives of sin(x) and cos(x). We first take
f (x) = sin(x)
f (x + h) − f (x)
h→0
h
sin(x + h) − sin(x)
= lim
h→0
h
(31)
sin(x) cos(h) + cos(x) sin(h) − sin(x)
= lim
h→0
h
sin(x) cos(h) − sin(x) cos(x) sin(h)
= lim
+
h→0
h
h
At this point I will tell the class to note that the first fraction can be written as follows
f 0 (x) = lim
29
sin(x)(cos(h) − 1)
sin(x) cos(h) − sin(x)
= lim
h→0
h→0
h
h
=0
lim
(32)
We now wish to consider sin(h)
for small h. The best way to understand this is to plot
h
it. It should be clear from the graph that the function reaches a maximum at f (0) = 1 and
then goes to zero at ±∞. Therefore, we can conclude that limh→0 cos(x)hsin(h) = cos(x). So,
from first principles, we can clearly see that the derivative of sin(x) is cos(x).
We will proceed to compute the derivative of cos(x) and show that it is − sin(x).
sin(x)
. We should find that
We now look to compute the derivative of f (x) = tan(x) = cos(x)
f 0 (x) = cos12 (x)
Presentation 2 We look at f (x) = ax . Working this down, we find that
ah − 1
f (x) = a lim
h→0
h
0
x
= f (0)a
0
x
(33)
This gives us a problem. In order to compute the derivative we need the derivative. Remind
the students about the exponential function. I give the following definition that they will
not have seen before:
eh − 1
=1
(34)
lim
h→0
h
This tells us that the derivative of the exponential is itself.
Now, we recall the natural logarithm, log x, the derivative of which is x1 .
Application The rest of the class will be spent computing derivatives of the above
types of functions.
5.3.9
Lesson 9
Prior Knowledge The students should know how to compute derivatives.
Content & Skills Product, quotient and chain rule.
Principal Aim In this class we wish to look at how to compute derivatives of more complex
functions.
Behavioural Objectives At the end of the class, the students should be able to:
• Implement the product rule,
30
• Implement the quotient rule,
• Implement the chain rule.
Introduction We now have a few ways to compute derivatives of functions. We can do
it by first principles or we can use the formula for polynomials. However, this leaves many
functions very difficult to differentiate. First principles can get very tricky and the latter
can only be used for polynomials. Today we will consider how to differentiate more complex
functions.
Development
Presentation 1 Today we will look at differentiation a little closer. Consider the
functions f (x) = x2 and g(x) = x3 . We can multiply these to find that (f g)(x) = x5 .
Differentiating we get (f g)0 (x) = 5x4 . Now, we try f 0 (x)g 0 (x) = (2x)(3x2 ) = 6x3 . We can
clearly see that (f g)0 (x) 6= f 0 (x)g 0 (x), that is, the derivative of a product is not the product
of a derivative.
0
3
2
= xx2 . We now find that fg 0(x)
= 3x
6= 1 =
The same is true for quotients. Taking fg(x)
(x)
(x)
2x
f 0
( g ) (x).
In order to differentiate these properly we use the product and quotient rules.
The product rule for two functions, f (x) and g(x), is
(f g)0 (x) = f 0 (x)g(x) + f (x)g 0 (x)
(35)
The quotient rule for two functions, f (x) and g(x) 6= 0, is
0
f
f 0 (x)g(x) − f (x)g 0 (x)
(x) =
g
g(x)2
(36)
We now look at some examples. Take f (x) = x and g(x) = ex . What is (f g)0 (x)? We
use the product rule to get ex (1 + x).
What if we want
f
g
0
(x)? In this case we use the quotient rule and get
ex (1−x)
e2x
=
1−x
.
ex
Application 1 This is an important concept and also one that students tend to have
trouble with. Therefore, it is necessary to do many examples.
Presentation 2 Recall from a few classes ago we looked at the example f (x) = (x−3)3 .
This was computed from first principles, however, it is not a very nice calculation as there
are many terms. For higher powers computing a derivative this way becomes ridiculous. For
31
this reason there is a formula to deal with functions such as these. It is called the chain rule.
Explain how the chain rule works and what functions it is useful to use it for.
Differentiate f (x) = (x − 3)3 by the chain rule.
f 0 (x) = 3(x − 3)3−1 = 3(x − 3)2
(37)
Does this correspond with what we got earlier?
Application 2 As before, this is a very important concept and plenty of examples are
necessary.
5.3.10
Lesson 10
Prior Knowledge Differentiation, product rule
Content & Skills Implicit and parametric differentiation.
Principal Aim In this class we want to look at how to differentiate implicitly and parametrically.
Behavioural Objectives At the end of the class, the students should be able to:
• differentiate implicitly,
• differentiate parametrically.
Introduction The first thing to do is to remind the class about the notation we used
previously, that is, using y instead of f (x) as our function. Up until now we looked at
functions of the form y = f (x). Now we want to look what happens when out x’s and y’s
are not split up.
Development
Presentation 1 We will begin with an example: x + xy = 1. We want to differentiate
this with respect to x. There are two ways to do this. The first way is to rewrite the equation
dy
.
in a form we already know, that is, y = 1−x
. Therefore, dx
= −1
x
x2
The second method is to use implicit differentiation. We look to differentiate term by
d
term. The first term we have seen before, dx
x = 1. The second term is more difficult. We
need to use the product rule where f (x) = x and g(x) = y. By the product rule, we get
dy
dy
dy
dy
x dx
+ y. Putting this together gives 1 + x dx
+ y = 0. Solving for dx
gives dx
= y−1
. Subbing
x
1−x
in y = x gives
dy
1
=− 2
(38)
dx
x
32
which is the same as we got before.
Although it is possible to avoid using implicit differentiation here, there are many situation where it is necessary.
Application 1 We want to look at some examples. Differentiate the following implicitly:
• 2x2 y 2 − xy 5 = x,
• x sin y + y cos x = 0.
dy
d (n)
y (x) = ny (n−1) (x) dx
, where y (n) (x) is the nth derivative of y with respect
We note that dx
to x, which is the chain rule.
Presentation 2 In the second half of this class we will look at parametric differentiation. Firstly, what does the word parametric mean. Until now x has not depended on
anything else. This means that we can not write x in terms of any other variable. We also
had, until this class, that y has not depended on anything. What if x and y depend on the
same variable, t, say. By this I mean x = x(t) and y = y(t). This is best illustrated by an
example.
dy
. We do this by finding dx
= 2t and
Take x = t2 − 2 and y = t + 1. The idea is to find dx
dt
dy
dy
2t
= 1. Expressing the first as a quotient of the second gives dx = 1 = 2t.
dt
Application 2 Look at examples
• x = cos t and y = sin t,
• x = tet and y = t2 + t.
33
References
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[2] History@ONLINE.
[3] Leaving cert syllabus.
[4] Method of Exhaustion.
[5] Method of Exhaustion@ONLINE.
[6] The Rise of Calculus@ONLINE.
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[8] Aristotle. Physics.
[9] George Berkeley. The Analyst: A Discourse Addressed to an Infidel Mathematician.
[10] Carl Boyer. History of Mathematics.
[11] Scott E. Brodie. An ancient extra-geometric proof.
[12] J. J. Grannell, P. D. Barry, M. Cronin, F. Holland, D. Hurley. Interim report on project
maths.
[13] Paul Dawkins. Calculus i - notes.
[14] Karen E. Donnelly. Cauchy and the rigorous development of calculus@ONLINE.
[15] Judith V. Grabiner. Who gave you the epsilon? cauchy and the origins of rigorous
calculus.
[16] H. Jerome Keisler. Elementary Calculus: An Infinitesimal Approach.
[17] Jim Loy. Zeno’s paradoxes.
[18] The MacTutor History of Mathematics archive. Greek number systems.
[19] Elizabeth Oldham. Irish curriculum changes 1959-1979.
[20] Elizabeth Oldham. Second iea mathematics survey. 1964.
[21] Elizabeth Oldham. Second iea mathematics survey. 1976.
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