Pi and the Mandelbrot set

For those studying calculus, can you see what Pi has to do with the Mandelbrot set? I’ll give you a hint, it has to do with certain integrals which you have almost certainly studied in class. We will have a post explaining exactly why this number pops up in this particular place, soon!

If you don’t know about the Mandelbrot set, take a look here, then have a look at the video above.

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By | October 5th, 2015|English, Level: intermediate|1 Comment

UCT MAM1000 lecture notes part 50 – linear algebra part iii

Gauss reduction

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.…

UCT MAM1000 lecture notes part 49 – linear algebra part ii

Matrices

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

 

\begin{array}{cc}  x + y &=3\\  2x-y &= 4  \end{array}

 

and

 

\begin{array}{cc}  u + v &=3\\  2u-v &= 4  \end{array}

 

It’s clear that if we ignore the names of the variables — x and y versus u and v — 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:

 

\left(  \begin{array}{cc|c}  1&1&3\\  2&-1&4  \end{array}  \right)

 

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).…

First year maths lecture notes subject links

Advice from former MAM1000W students

First semester

Some introductory notes about the course and random links

Introductory topics

MAM1019H notes

Proof methods

Functions, Continuity and limits

UCT MAM1000 lecture notes part 48 – linear algebra part i

These notes are taken from the resource book and were originally written by Dr Erwin. I will be editing and adding to them throughout. Most mistakes within them can thus be presumed to be mine rather than Dr Erwin’s.

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:

matrices.001

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

Khan academy in Xhosa

For those who don’t know, a huge amount of Khan academy has been translated into Xhosa. The videos can be found here.

For instance, here’s the binomial theorem:

as translated by Zwelithini Mxhego. It looks like these videos haven’t been viewed all that many times, though a huge amount of work has clearly gone into them. If you think you know people who might benefit from these, please do spread the word!

 

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By | September 28th, 2015|English, Level: intermediate|3 Comments

UCT MAM1000 lecture notes part 45 – 3D geometry and vectors part viii

We will discuss mostly three dimensions here, but what we have will be applicable to any number of dimensions (greater than or equal to 1). We want to be able to describe a straight line – a one dimensional object, infinitely long in both directions. We will see that vectors give us a perfect language with which to do this.

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:

int

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.…

UCT MAM1000 lecture notes part 44 – 3D geometry and vectors part vii

In the following, I’m going to miss out quite a few details which I think are very nicely laid out in Stewart. I will try and add a slightly more pedagogical tone to some of it, and some nice diagrams along the way.

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.…

UCT MAM1000 lecture notes: More complex numbers practice

I’ve been asked a few times for more practice questions on complex numbers. This is where Wolfram Alpha can be your friend (like it’s not already!).

I’ll just give a few examples of questions from the tut on complex numbers which you could have solved using Wolfram Alpha, and from this you will be able to set up your own questions.

For instance, question 48 c) Find the roots of z^5=1+\sqrt(3) I can be solved in Wolfram Alpha with the command:

 

Solve[z^5==1+sqrt[3]I,z]

 

Moreover it will solve this for you, give you the five roots and plot them in the complex plane. So now you can come up with any root question you can possibly think of. There’s an infinite number of questions to start you off. You can thank me later!

If you want to convert between the trigonometric form and the exponential form, you can use the two commands:

 

TrigToExp[Sin[x]+2 I Cos[x]]

ExpToTrig[Exp[I z+3]]

 

Though remember the definition of the hypergeometric trig functions from a previous tut.…

UCT MAM1000 lecture notes part 43 – 3D geometry and vectors part vi

Determinants

The idea of determinants have been about since around the 3rd century when it first appeared in an ancient Chinese book of Mathematics called The Nine Chapters on the Mathematical Art. It was used originally to define certain properties of systems of linear equations, as we will see later in the section on linear algebra, however for now we will simply use it as a particular way to easily calculate the cross product. Let’s take a two by two array of numbers and define the determinant for this.

 

\left|\begin{array}{cc}a & b \\ c & d \\\end{array}\right|=ad-bc

 

The vertical lines on the left and right are the sign that the we are taking a determinant. For now this is just a definition and we will work with it in what follows. Don’t worry too much about where it comes from, but we will see later where it comes from and we will see now why it is useful.…