First year – Page 5 – Mathemafrica

Logical implications and the structure of if and only if statements

We had a homework assignment a couple of weeks back. It was looking at mathematics in a very different way from how many had seen it before, and it caused a lot of confusion. I would like to try and add some clarity to what we were doing. My thought was, rather than going through the questions themselves, I would like to add annotations to the proof itself. Let’s see how this works. The proof that you were given is in black, the annotations are in blue, and after I’ve been through the proof, I will expand on it in a simplified form.

Theorem: The function f is differentiable at x=a if and only if there is a constant m and a function E of x, defined for all $x \not = a$ , such that

$f(x)=f(a)+m(x-a)+E(x)(x-a)$ for all $x \not = a$ – (eq 1)

and,

$\lim\limits_{x\to a}E(x)=0$ .

(If both these conditions are satisfied, then $f'(a)=m$ .)

What we are doing here is giving another definition of differentiability (at a particular point, a).…

By Jonathan Shock| April 22nd, 2016|Courses, First year, MAM1000, Uncategorized, Undergraduate|1 Comment

Proof by induction for a non-mathematician – a competition: Vote for the winner!

I set a voluntary assignment for my course a few weeks back. Students had just learned about proof by induction, and I tend to find that this is a subject which many get confused by. I think that one of the best ways to really understand a topic is to try and teach it to someone else, so I set up an exercise which was to write an explanation of Proof by Induction for a young high school student. We had around 100 entries, which took a while to read through! Of these 100 entries there were four which stood out (and many which were also very good). We (myself, and two senior tutors) have been unable to come up with an outright winner. That’s where you come in!

Please take a look at the following entries, and vote for the one that you think is best at this link: https://www.surveymonkey.com/r/9K3P8BJ.…

By Jonathan Shock| April 12th, 2016|Competition, Courses, First year, MAM1000, Undergraduate|2 Comments

The squeeze (or sandwich) theorem

Let’s say I ask you how tall Craig is going to be when he’s 15 years old, and let’s say, given his genetic information, you just can’t tell what height he will be at that age. However, you do know that he’s going to be taller than Lisa, and shorter than Khangelani up until the age of 15 (they are all the same age). With this information alone, you haven’t learnt much about the height of Craig when he’s 15. However, what if you can figure out, using your clever genetic detective work, that at the age of 15 Lisa and Khangelani will be the same height? The only way for this to be true, is if Craig (whose height is sandwiched in between that of Lisa and Khangelani) is also the same height at that age.

That’s really all the sandwich theorem is. Let’s look at a mathematical example.…

By Jonathan Shock| April 9th, 2016|Courses, First year, MAM1000, Undergraduate|4 Comments

Continuity – (Part Two).

Definition:

(i) A function $f$ is said to be continuous from the right at a if

$\lim\limits_{x \to a^{+}} f(x) = f(a)$

We can see that, as the function approaches a certain x-value from the right, $f$ is defined and

$\lim\limits_{x \to a^{+}} f(x) \equiv f(a)$

And as the function approaches a certain x-value from the left, $f$ is not defined, i.e;

$\lim\limits_{x \to a^{-}} f(x) \neq f(a)$ .

Therefore, we say that the function is continuous from the right at this point, but is discontinuous from the left.

(ii) A function $f$ is said to be continuous from the left at a if

$\lim\limits_{x \to a^{-}} f(x) = f(a)$

Here, it is clear from the graph that the function is continuous from the left as x approaches 3. This is because the function is defined at x = 3 and,

$\lim\limits_{x \to 3^{-}} f(x) \equiv f(3)$

However, from the right,

$\lim\limits_{x \to 3^{+}} f(x) \neq f(3)$

So, we can say that the function is continuous from the left, but discontinuous from the right, at the point x = 3.

We may also have a function where it is neither continuous at a point from the left or from the right but is defined elsewhere.…

By Azhar Rohiman| March 31st, 2016|Courses, First year, MAM1000, Undergraduate|4 Comments

Continuity – (Part One).

Definition:

A function $f(x)$ is continuous at a given point x = a if those three conditions below are met “simultaneously”:

(i) $f(a)$ is defined. (i.e; a is in the domain of $f$ )

(ii) $\lim\limits_{x \to a} f(x)$ exists.

(iii) $\lim\limits_{x \to a} f(x) = f(a).$

NOTE:

If any one of the three conditions is false, then $f$ is discontinuous at a, or it has a discontinuity at a.

Let’s now look at the different cases where $f(x)$ may not be continuous at x = a.

(i) $f(a)$ is defined but $\lim\limits_{x \to a} f(x)$ does not exist.

At a = 0, the function is not continuous despite $f(a)$ is defined (Here, $f(a)$ is equal to -1). This is because the two one-sided limits are not equal and as a consequence, the limit does not exist. This is called a jump discontinuity.

(ii) $\lim\limits_{x \to a} f(x)$ exists, but $f(a)$ is not defined.

Assume a is the x-value where there is a hole in the graph. We can see that the limit from the right of a and the limit from the left of a are equal.…

By Azhar Rohiman| March 31st, 2016|Courses, First year, MAM1000, Uncategorized, Undergraduate|5 Comments

Proof by contradiction – part 2

So, in the last post we proved that $\sqrt{2}$ is irrational, by trying to see what the consequences would be if it were rational. We first said that if it were rational, then we should be able to write it in a simplest form $\frac{p}{q}$ where p and q had no common factors, and then showed that in fact this was impossible, so our original proposition was indeed true (as trying to prove otherwise gave us a contradiction).
Now we are going to look at another example, which looks very similar, but here our contradiction will be a little different to last time. The fact is that, unlike much of what you would have done at highschool, there isn’t such a roadmap to how to do things here – you have to figure out for yourself where the contradiction comes in. In these posts I will point them out for you, but in general, you need to build your intuition about where something looks a bit dodgy and a contradiction is raising its head.…

By Jonathan Shock| March 22nd, 2016|Courses, First year, MAM1000, Undergraduate|5 Comments

Proof by contradiction – part 1

Proof by contradiction may at first seem completely weird! I give you something to prove, and you seem to ignore me and try and prove that what I want you to prove is wrong!

Actually, this isn’t nearly as strange as it first seems, and it can work in contexts other than mathematics. The idea stems from the fact that a statement is either true, or false (well, if you listen to Gödel, then you have to be a bit careful, but it’s reasonable enough for now). The process is the following:

You want to prove that a statement is true.
You say “what would happen if the statement were actually false?”
You explore the consequences of it being false.
If you find that it gives you a contradiction (something which you claimed to be true, but which you now see isn’t true), then you know that in fact the original statement can’t be false…so it must be true, and you’ve proved it.

…

By Jonathan Shock| March 20th, 2016|Courses, First year, MAM1000, Uncategorized, Undergraduate|7 Comments

Can we find the inverse of a function which is not one-to-one? (part two)

So, in the last post we had seen that while the sin function is not one-to-one and thus doesn’t have an inverse, so long as we restrict it to a given domain, you will find that it is invertible. The domain that we found (indeed chose), was between $[-\frac{\pi}{2},\frac{\pi}{2}]$ . It’s inverse was a function with domain $[-1,1]$ . The name of the inverse is arcsin(x). How can we use this to help us to solve problems?

Well, what if I asked you to solve:

$sin(x)=\frac{1}{2}$

You might think that because we have found the inverse of sin, that we can simply say that the solution to this is:

$x=arcsin\frac{1}{2}$

Well, because arcsin is itself a one-to-one function, restricted to the domain $[-1,1]$ this will clearly give us a single number (the answer is about 0.52):

Is that it then? Well, let’s look at the graph of sin(x) and see if this is the only solution to $sin(x)=\frac{1}{2}$ :

In fact, clearly there are an infinite number of solutions to the equation $sin(x)=\frac{1}{2}$ and we have just caught the one within the region $[-\frac{pi}{2},\frac{\pi}{2}]$ .…

By Jonathan Shock| March 16th, 2016|Courses, First year, MAM1000, Uncategorized, Undergraduate|1 Comment

Domain of a composite function – part 2

This is the second part of a post, written by Muhammad Azhar Rohiman, a first year student on MAM1000W at UCT. This post came about when he asked me a question related to domains of composite functions, and it was clear that on first learning about such topics, there are some simple misunderstandings. I suggested that he write a couple of paragraphs explaining what he had learnt, and the following is, I think, a very clear explanation of some of the ideas and pitfalls of this topic. The first part of the post is here.

Consider the two functions below, from which we want to find the domain of

(a) $(f\circ g)(x)$ .

(b) $(g\circ f)(x)$ .

(d) $( g \circ g )(x)$ .

$f(x) = x + \frac{1}{x}$

$g(x) = \frac{x+8}{x+2}$

(a) The functions f(x) and g(x) cannot be defined at the values x = 0 and x = -2 respectively. Therefore, we can write this as follows: f(0) and g(-2) are not defined.…

By Jonathan Shock| March 12th, 2016|Courses, First year, MAM1000, Undergraduate|1 Comment

Arbitrary functions as the sum of odd and even functions

Let’s take a function h(x), whose domain is the real numbers. We are simply going to start by writing h(x) in a slightly strange way. We will write it as:

$h(x)=\dfrac{h(x)+h(-x)+h(x)-h(-x)}{2}$

This might seem an odd thing to do – we have essentially added zero to the original function (in the form h(-x)-h(-x)). However, we can see that we can split this as:

$h(x)=\dfrac{h(x)+h(-x)}{2}+\dfrac{h(x)-h(-x)}{2}$

It’s exactly the same thing we started with, right? But now it’s written in a peculiar way. Now let’s call the two fractions f(x) and g(x) respectively. So:

$f(x)=\dfrac{h(x)+h(-x)}{2}$

and

$g(x)=\dfrac{h(x)-h(-x)}{2}$

So our original function can be written as h(x)=f(x)+g(x). If you plug in f(x) and g(x) above you will see that we have said nothing which is not trivial in any of this. However, the interesting part comes when we look at the properties of f(x) and g(x). What is f(-x)?

$f(-x)=\dfrac{h(-x)+h(x)}{2}=\dfrac{h(x)+h(-x)}{2}=f(x)$

But this is just the defining property of an even function, so f(x) is even.…

By Jonathan Shock| March 10th, 2016|Courses, First year, MAM1000, Undergraduate|1 Comment

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