## 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 $[-\frac{pi}{2},\frac{\pi}{2}]$.…

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

Asking what the inverse of a function is, is the same as asking what is the function that will undo this function?

What is the inverse of the function $f(x)=x+3$? That is asking the question, if I put a number into this function (call that number a), it will give me another number (call it b). What is the function which, whatever number I put in, when applied to the number that comes out from the first function will be the original number. That is to say:

If f(a)=b. Then what is the function g for which g(b)=a. That would give you g(f(a))=a. g Is then the inverse of f and we can write $g(x)=f^{-1}(x)$. g(x) is the thing that undoes f. Put simply, composing the inverse of a function, with the function will, on the appropriate domain, return the identity (ie. not do anything to the number you put in).…

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

(c) $(f\circ f)(x)$.

(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.…

## 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.…

## Mathematical induction with an inequality

In the tutorial sessions it was clear that one question in particular was causing problems. This is an induction proof with an inequality. The one which we will look at is the inequality:

$2^n>n^3$ for $n\ge 10$

I am going to talk you through it in more detail than would be needed for the formal proof but I want to give some intuition along the way.

So, we start off, as always with the base case. The base case is always the first integer for which the statement is claimed to be true. In this case it is for n=10. Let’s check for n=10. Is it true that?

$2^{10}>10^3$ ?

Well this is:

$1024>1000$

and so we should be happy with that. We’ve proved the base case. Note that you do not then need to check for n=11, or n=12. I have seen many students check a base case for n=1, and then also check for n=2 and n=3.…

## Mathemafrica and the NEF #2

On Monday we worked hard to bring our Maths exhibition exhibition to the Next Einstein Forum venue, and to setup everything.  Find below a few pictures! It was hard, time was short, logistics was tricky, but: the venue is amazing! We got the prime spot of the venue, just right at the main entrance – and the whole NEF organisation is very professional!

Find below some pictures of our “setup day”:

Packing all items (inside and on top of the car).

Mounting the exhibition (roll-ups, images on the wall, screens, touch-screens). A big part of the exhibition is about “Mathematics of Planet Earth”, we will also launch a new competition about open exhibits – so everybody can contribute with own exhibits.

The local NEF tech team help us to fix the images on the wall.

This is the final setup (the computers already turned off). On Tuesday, the exhibition start around 2 pm, in the morning the presidents of Senegal and Rwanda are expected to be at the NEF (and might also visit the exhibition)

This could be a good spot to take pictures… today many people already took selfies in front of our images.…

## Domain of a composite function – part 1

This post was 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.

Consider the two functions below, from which we want to find the domain of $( f \circ g )(x)$

$f(x) = \frac{1}{x+2}$, $g(x) = \frac{x}{x-3}$

We know that f(x) and g(x) cannot be defined at the values x = -2 and x = 3 respectively. This can be written as follows: f(-2) and g(3) are not defined. The domain of a composite function will not allow any values restricted by the domain of the starting function, which is g(x).…

Gallery

## Mathemafrica and the Next Einstein Forum (NEF) #1

I am sitting at the AIMS-Senegal institute in Mbour (about 1.5 hours drive South of Dakar) and together with my team member Sebastian and the AIMS-Senegal staff and students, we are preparing an interactive mathematics exhibition (as part of the IMAGINARY – open mathematics project).. It will be shown as of March 8 at the Next Einstein Forum (NEF), to be held at a huge conference venue just outside Dakar.

There will be many ministers, scientists, politicians, even presidents from (apparently all fifty-four) African countries – and also many international guests – joining the NEF, with the goal to discuss about scientific innovations, collaborations and solutions in Africa!

We have to plan to blog live from the NEF, with insights and views from participants. And of course, we will let you know about our exhibition, about a new competition, we will launch and everything happening around!

Prepare yourself for the NEF at:

gg2016.nef.org/

iameinstein.org

Please find a picture from our first technical setup yesterday at the AIMS Institute.…

## Mathematical induction

One of the concepts that most students seem to struggle with the most in the first year maths course is that of mathematical induction. It feels abstract, yet when you have to prove a concrete statement, it feels like all the assumptions, and cases you look at shouldn’t have any real impact on the thing that you’re trying to prove. I will now try and prove that this is not true (though not by mathematical induction!).

I’m going to start with a ladder brought from a magical ladder supplier. Steps on the ladder are labeled S(1), S(2), S(3) etc. The question is, are there infinitely many steps on the ladder? Well, with the information that I’ve given you so far, you just don’t know, but the manufacturer has given a little leaflet with the it. In the leaflet it says:

“As long as your ladder has a step S(n), we hereby guarantee that it will have a step S(n+1), for any integers n”.…

## Polynomial division

Note: In the following I use the words power, and order somewhat interchangeably, in relation to the exponent of x.

Please also forgive the rather bad formatting for some of the expressions in this text. WordPress and LaTeX are somewhat unforgiving.

One of the topics which seems to have caused the most problems in the assignments for MAM1000W so far this year is that of polynomial division. Thus I want to go through an example here to show you exactly what we’re doing when we think about performing such a calculation.

If you think you already know what to do, but want to have some more practice, plug in some random polynomials into this page and make sure that you get the same steps as them.

In fact, the first question might be: what’s the point? Often the expression looks more complicated in the end than it does at the beginning.…