Here is the first:

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As an Actuarial Science student I was aiming for 70% last year. I clearly remember that at orientation I asked some of the older ActSci students at orientation what they had done when they scored below what they needed to. I was so shocked, and a little scared when the group I asked said they never had. I wasn’t worries at this stage though because I thought I’d done well at maths at school, and I’d do well at maths here.…

This year I will be not only teaching, but entirely in charge of the UCT first year mathematics for scientists courses, known as MAM1000W. I have a number of changes I plan on making, not so much to the syllabus, but to the extra activities associated with the course, in an attempt to make it as rich and deep a learning experience as I can.

The first step of this has been altering the structure of the resource book. The resource book is a PDF which will be sent to all first years taking the course. Historically, it contains a little about the course content, a bit about how your marks will be calculated, a bit about good practice in terms of how to work, and then the second half is filled with tutorial questions.…

So far we have seen that we have a way to translate a system of linear equations into a matrix. We can manipulate the matrix in ways which correspond to operations on the equations which keep the important information in the system of equations the same (ie. the solution of the equations before and after the operations is the same). We have seen a couple of examples of when we can read off the solution from the matrix having performed the operations. So far the order with which we perform the operations feels a bit arbitrary, although we know that we would like to get the matrix into reduced row echelon form. There is however a very systematic way of going about this, and the term for the process is called Gauss Reduction.

Here is a detailed view of what Gauss Reduction will give us:

Gauss Reduction:

To solve a system of linear equations:

1) First find the augmented coefficient matrix of the system of equations.…

Solving a system of linear equations is not technically difficult: just eliminate the variables in a systematic fashion. When there are only two or three variables, this is easy to manage. But for a bigger system, things can quickly get confusing. We need to develop a systematic method.

The first thing to notice is that the names of the variables don’t matter. Consider, for example, the two systems

and

It’s clear that if we ignore the names of the variables — and versus and — these two systems are the same. The reason we can tell that they’re the same is because the {\em coefficients} of the variables are the same and the numbers on the right hand side are the same. These are really the only things about a system of linear equations that matter, and so what we can do is strip the system down to its bare bones and rewrite it like this:

This is an augmented coefficient matrix (in general, a rectangular array of numbers, like the above, is called a matrix; a matrix with an additional vertical line, which plays the same role as the equals signs in the original equations, is augmented).…

• Part I – an ActSci student whose grades kept going down, until they figured out the secret.
• Part II and III – Tutorial partners and useful videos.
• Part IV – Webassign and coming back from tough times
• Part V – “the methodology of thinking and applying helps me to this day”

First semester

Some introductory notes about the course and random links

• Intro to first year mathematics
• Why study mathematics?
• Tutorials
• Revision
• Ways to represent a function
• The first week of lectures
• Preamble and introduction
• MAM1000 Bootcamp

Introductory topics

• All you’ve ever wanted to know about absolute values (and weren’t afraid to ask)
• Why radians?
• Sets
• Absolute values and inequalities
• Polynomial division

MAM1019H notes

• Introduction to sets
• Subsets
• Power sets
• Cartesian products

Proof methods

• Mathematical induction
• Mathematical induction with an inequalityProof by induction winning explanation
• Proof by contradiction – part 1
• Proof by contradiction – part 2

Functions, Continuity and limits

• Domain of a composite function – part 1
• Domain of a composite function – part 2
• Plotting rational functions.

In this section we are going to develop a new set of methods to solve a type of problem we are relatively familiar with. We will find a way to translate between methods we know well, but which turn out not to be very efficient, methods which are graphically very intuitive, but not very calculationally useful, and methods which are computationally extremely powerful, but appear rather abstract compared with the other two ways of looking at these problems. These three methods which we will utilise in detail in the coming sections are shown in the following diagram:

As we go through I will try and show how we can go between these apparently different formalisms.…

Remember that in three dimensions, a line can be defined by the intersection of two planes as in the intersection of the blue and the green planes defining the red line:

Each plane is specified by a single equation, and thus a line is specified by two equations (one for each plane). Here we will see that sometimes you just need one equation to specify a line, if you are using vectors, and sometimes it will seem that you need three equations, if you are using a parametric equation.

Let’s take a line, and specify some point on it.…

So we saw in the last post that we can write the cross product of two vectors, which itself gives a vector, in terms of the determinant of a 3 by 3 array. We can use this to both find a vector perpendicular to two given vectors (unless they are parallel to one another) and also to find the area of a parallelogram formed by two vectors (the area of which is zero if the vectors are parallel to one another).

The second of these is easy enough to do in two dimensions, but in three dimensions that’s not an easy prospect. Using the cross  (otherwise called the vector) product makes this easy.…