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1.4 Equivalence classes

Let’s recall the definition of an equivalence relation:

A relation R on a set A is termed an equivalence relation if it is simultaneously reflexive, symmetric and transitive.

Let’s look at more examples:

Example One: Let $A = \{2, 11, 17, 20\}$ be a set with the following relation: $R = \{ (2,2) (11,11) (17,17) (20,20) (2,20) (20,2) (11,17) (17,11) \}.$

The relation described by R is termed “the same parity.” Elements x and y are said to have the same parity if they are both odd or both even. In our case, the elements 11 and 17 are both odd – hence have the same parity. Similarly, 20 and 2 have the same parity because they are both even. An element will always have the same parity as itself.

The elements that share the same parity as 11 can be grouped together to form a set: $O = \{ 11, 17 \}.$ This is the set of all odd elements from A.

Similarly, the even elements can be grouped together to form the set: $E = \{ 2, 20\}.$

The new sets, O and E, form the equivalence classes of the relation R on set A.…

By Yolisa| July 22nd, 2019|Uncategorized|0 Comments

1.3 Relations: Equivalence relations

We know that a relation is called an Equivalence Relation when it is reflexive, symmetric AND transitive on some set A. Let’s look at some examples.

Example One: Let $a,b \in \mathbb{R}.$ Suppose we have that a is related to b (i.e. a ~ b) if $a - b \in \mathbb{Z}.$ We want to show that our relation ~ is an equivalence relation.

First, let’s unpack what the question requires us to prove: It wants us to show that the relation ~ on set A is an equivalence relation. Hence, we need to show that ~ is reflexive, symmetric and transitive.

The relation ~ (in this case) is defined as follows: IF any two real numbers, a and b, are related THEN we know that a – b is some integer.

It’s important to note that the order in which a relation is important! Always write your equations as they’ve been given in the question to avoid confusion and mistakes

Ok, let’s prove this.…

By Yolisa| July 18th, 2019|Uncategorized|1 Comment

1.2 Relations: Properties

Note: Do not confuse binary operations (+, x, -, …) with relations. Recall the definition for a relation as:

A relation R on a set A is a subset R ⊆ A × A. We often abbreviate the statement (x, y) ∈ R as xRy.

For instance, the binary operation “x” has a numeric value: $3 \times 3 = 9 \text{ and } 40 \times \frac{1}{5} = 8.$ Yet a mathematical relation, for example “<“, has a True/False value: $3 < 3 \text{ and } 40 < \frac{1}{5}$ are both False expressions.

We want to look at some properties of relations. We will look at three properties for relation expressions:

Suppose A is a set with relation R, then

Relation R is Reflexive if $\forall x \in A, \text{ } xRx$
Relation R is Symmetric if $\forall x,y \in A, \text{ } xRy \rightarrow yRx$
Relation R is Transitive if $\forall x,y,z \in A, \text{ } xRy \wedge yRz \rightarrow xRz$

The first property tells us that if every element, x, in set A is related to itself, then the relation R acting on the set A is termed “reflective.”

The second property tells us that if “x is related to y” from set A implies that “y is also related to x,” then R is termed “symmetric.”

Lastly, if “x is related to y” and “y is related to z” implies that “x is also related to z,” then the relation R is termed “transitive.”

Let’s look at the following examples: $A = \{ 1, 2, 3, 4\}$ with relation $\le.$ Then:

$\forall x \in A, \text{ } x \le x$

In other words $\le$ is reflexive since every number in set A is equal to itself (i.e.…

By Yolisa| July 18th, 2019|Uncategorized|0 Comments

1.1 Relations: Introduction

What are relations?

In every day life, a person is related to their parents, siblings, cousins, teachers, friends, etc. in some way. Similarly in mathematics, mathematical objects like numbers and sets are related to one another in some way. Many relations (symbols) will be familiar already:

$2 <3$
$\pi \approx 3.14$
$5 \in \mathbb{Z}$
$X \subset Y$
$a \equiv b(modn)$

Consider the following set $A = \{1, 34, 56, 78 \}.$ We can compare the numbers in A using the symbol “<” as follows: $1 < 56, 34 < 78$ etc. We can write this as a set in the following way: $R = \{ (1,34), (1,56), (1,78), (34,56), (34,78), (56,78) \}.$

Each pair in this new set R expresses the relationship x < y (where x and y are numbers from A).

In other words, $1<34, 1<56, 1<78, ...$ So if asked whether $34 < 78$ is true, one only needs to look into our set R to find the pair $(34,78).$ If we didn’t find it, then the relation would be considered false for the given set. The above example is intuitive because we are already comfortable with the relation <. In more abstract cases, thinking of the relationship between mathematical objects in this way may be a little trickier!…

By Yolisa| July 18th, 2019|Uncategorized|1 Comment

The South African Mathematics Olympiad problems

How clear is this post?

…

By Jonathan Shock| July 14th, 2019|Uncategorized|1 Comment

The definite integral

I realise now, in all the excitement of the FTC that I hadn’t written a post about the definite integral…that’s shocking! ok, here we go…the plan for this post:

Look at our Riemann sums and think about taking a limit of them
Define the definite integral
Look at a couple of theorems about the definite integral
Do an example
Look at properties of definite integrals

That’s quite a lot, but we are more or less going to follow along with Stewart. Stewart just has a slightly different style to mine, so I recommend reading his for more detail, and mine for potentially a bit more intuition.

So, let’s begin…

We have seen in previous lectures/sections/semesters/lives that we can approximate the area under a curve by splitting it up into rectangular regions. Here are examples of splitting up one function into rectangles (and, in the last way trapezoids, but you don’t have to worry about this).…

By Jonathan Shock| July 12th, 2019|Courses, First year, MAM1000, Uncategorized, Undergraduate|3 Comments

The Fundamental Theory of Calculus part 2 (part i)

OK, now we come onto the part of the FTC that you are going to use most. We are finally going to show the direct link between the definite integral and the antiderivative. I know that you’ve been holding your breaths until this moment. Get ready to breath a sign of relief:

The Fundamental Theorem of Calculus, Part 2 (also known as the Evaluation Theorem)

If $f$ is continuous on $[a,b]$ then

$\int_a^b f(x) dx=F(b)-F(a)$

where $F$ is any antiderivative of $f$ . Ie any function such that $F'=f$ .

————-

This means that, very excitingly, now to calculate the area under the curve of a continuous function we no longer have to do any ghastly Riemann sums. We just have to find an antiderivative!

OK, let’s prove this one straight away.

We’ll define:

$g(x)=\int_a^x f(t)dt$

and we know from the FTC part 1 how to take derivatives of this. It’s just $g'(x)=f(x)$ . This says that $g$ is an antiderivative of $f$ .…

By Jonathan Shock| July 11th, 2019|Courses, First year, MAM1000, Uncategorized, Undergraduate|0 Comments

The Fundamental Theorem of Calculus part 1 (part iii)

So, we are now ready to prove the FTC part 1. We’re going to follow the proof in Stewart and add in some discussion as we go along to motivate what we are doing. What we are going to prove is that:

$\frac{d}{dx} \int_a^x f(t) dt=f(x)$

for $x\in [a,b]$ when $f$ is continuous on $[a,b]$ .

Proof:

we define $g(x)=\int_a^x f(t)dt$ and we want to find the derivative of $g$ . We will do this by using the fundamental definition of the derivative, so let’s look at calculating this function at $x$ and $x+h$ – ie. how much does it change when we change $x$ by a little bit?

$g(x+h)-g(x)=\int_a^{x+h}f(t) dt-\int_a^x f(t) dt$

But remember that the definite integral is just the area, so this difference is the area between a and x+h minus the area between a and x. Which is just the area between x and x+h. Using the properties of integrals, we can write this formally as:

$g(x+h)-g(x)=\int_a^{x+h}f(t) dt-\int_a^x f(t) dt=\left(\int_a^{x}f(t)+\int_x^{x+h}f(t)\right)-\int_a^{x}f(t)=\int_x^{x+h}f(t)dt$

and we can write, for $h\ne 0$ :

$\frac{g(x+h)-g(x)}{h}=\frac{1}{h}\int_x^{x+h}f(t)dt$

Restated, we can think of this as the area between x and x+h divided by h.…

By Jonathan Shock| July 11th, 2019|Courses, First year, MAM1000, Uncategorized, Undergraduate|0 Comments

The Fundamental Theorem of Calculus part 1 (part ii)

OK, so up to now we can’t actually use the FTC (Fundamental Theorem of Calculus) to calculate any areas. That will come from the FTC part 2.

For now, let’s take some examples and see what the FTC is saying. I’ll restate it here:

The Fundamental Theorem of Calculus, part 1

If $f$ is continuous on $[a,b]$ then the function $g$ defined by:

$g(x)=\int_a^x f(t) dt$ , for $a\le x\le b$

is continuous on $[a,b]$ and differentiable on $(a,b)$ , and $g'(x)=f(x)$ .

——

Let’s look at some examples. We’re going to take an example that we can calculate using a Riemann sum. Let’s choose $f(x)=x^2$ .

If we integrate this from $0$ to some point $x$ – ie. calculate the area under the curve, we get:

$\int_0^x t^2 dt=\frac{x^3}{3}$ .

Make sure that you can indeed get this by calculating the Riemann sum.

So, what does the FTC part 1 tell us? It says that if we take the derivative of this area, with respect to the upper limit, $x$ , then we get back $f(x)$ .…

By Jonathan Shock| July 11th, 2019|Courses, First year, MAM1000, Uncategorized, Undergraduate|0 Comments

The Fundamental Theorem of Calculus, part 1 (part i)

We’ve seen some intriguing things in this course so far, and we’ve developed some clever tricks, from how to find the gradient of just about any function we can throw at you, to proving statements to be true for an infinite number of cases.

To some extent, this is what we have looked at so far (at least in terms of calculus, and building up to calculus):

However, we’re about to see some magic. We’re about to see the most important thing yet on this course, and indeed one of the most important moments in all of mathematical history.

We are going to see…actually, we are going to prove, that there is a relationship between rates of change and the area under a graph. This doesn’t sound that amazing, but its consequences have essentially allowed for the development of much of modern mathematics over the last 350 years.

The link that we are going to prove will allow us to find the area under graphs of functions for which taking the Riemann sum would be really hard.…

By Jonathan Shock| July 9th, 2019|Courses, First year, MAM1000, Uncategorized, Undergraduate|1 Comment

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