Plotting functions of complex numbers: Not examinable

Just to get a bit of a picture of what taking a function of a complex number means, we can play a bit of a game (I use this term in the loosest sense). Normally we think of functions as going from a real number to another real number. \sin(x) takes a real number x and gives you another real number. We can plot this on a graph by plotting a two dimensional set of data which tells you about the value that \sin(x) takes for every x along the real line. We are very used to this idea of a function. However, a function of a complex number is more difficult to visualise.


Complex numbers themselves live in 2 dimensions (they have a real part and an imaginary part) and when you apply a function to them, very often the result is another complex number which also lives in a 2 dimensional space.…

By | August 26th, 2015|English, Level: intermediate, Uncategorized|0 Comments

UCT MAM1000 notes part 24 – complex numbers part ii

So, last time we discovered that numbers are maybe not quite as real as we thought that they were, and that we can have numbers which don’t obviously correspond to something in the real world (though we’ll discover later that they are a way to jump between islands of reality).

In the resources on Vula you will find some great notes on complex numbers, so I want this to be an additional resource, and not an alternative. This means that sometimes we will look at things from a slightly different perspective than in the resource book.

Let’s start off discussing a bit more about the complex plane.

When you learnt about integers, one of the first things that you learnt to do was to put them in order. 3 came after 2 and 7 came after 6. You could put them all in a line. When you learnt about the negative numbers, it was quite clear that this line which had previously started with zero simply went backwards in the other direction, and you could count backwards to whatever large negative integer you wanted.…

UCT MAM1000 lecture notes part 23 – Thursday August 20th

Complex numbers

These will be an addition to the notes already on Vula on complex numbers. Please refer to that document as well as I will be taking a slightly alternative approach on occasion.

A philosophical detour

Before we get on to talking about imaginary numbers and complex numbers, let’s try and break down our preconceptions about numbers in general. We look at the world around us and see many things which we categorise. We see a computer, a piece of paper, we see other people, we see our hands. These are labels that we use to categorise the world around us, but these objects seem very physical and very real. We rarely question their existence, though if one wants to take the Cartesian view, we should also question the reality we are in. We are not going to go that far, but let’s try and ask about the existence of numbers.…

UCT MAM1000 lecture notes part 22 – Wednesday 19th of August

So, last time we looked at the Maclaurin expression for e^x. The exponential function was particular easy because its derivative is equal to the function itself every time. Let’s look at a slightly more involved example where this is not true: f(x)=\sin x about x=0. Again, we start with the table of derivatives:


\left(  \begin{array}{ccc}  \text{ i} & f^{(i)}\text{(x)} & f^{(i)}\text{(0)} \\  0 & \sin (x) & 0 \\  1 & \cos (x) & 1 \\  2 & -\sin (x) & 0 \\  3 & -\cos (x) & -1 \\  4 & \sin (x) & 0 \\  5 & \cos (x) & 1 \\  \end{array}  \right)


Now the values of the derivatives are not always the same. They are zero every other term, and they change in sign when they are not zero. This leads to a very elegant expression for the \sin function expanded around x=0:


\sin x\approx\sum_{i=0}^n \frac{(-1)^ix^{2i+1}}{(2i+1)!}


An important point is that here the terms get smaller and smaller as you take more and more of them, so if, for instance, you want to know the value of \sin 2.4 you can plug it into the right hand side, take a finite number of terms and you will get an approximation:


\sum_{i=0}^n \frac{(-1)^i2.4^{2i+1}}{(2i+1)!}


The higher value you choose for n the more accurate will be your answer, but we can see that we can now, in theory calculate the \sin of any number with pen and paper, so long as you have enough patience and will-power.…

UCT MAM1000 lecture notes part 21 – Tuesday 18th August

So we’ve now looked at a couple of different functions and found polynomials which approximate the functions to different levels of accuracy. Let’s try and come up with a general method of formulating this. Let’s say that we have some function f(x) and we want to approximate it close to x=a. We will then assume that we can write the polynomial approximation as:


\sum_{i=0}^n c_i (x-a)^i


Note that previously we wrote a_i but it’s good to get used to slightly changeable notation. The context is what should tell you the meaning.

We want to have that:


f(x)\approx \sum_{i=0}^n c_i (x-a)^i


We will first ask that the value of the polynomial is equal to the value of the function at x=a. We do this by setting x=a in both sides of the above. Note that we are being slightly ambiguous in what we mean by the approximation here because in a moment we will go from a \approx sign to an = sign.…

UCT MAM1000 lecture notes part 20 – part iii – Monday 17th August

Now we’re going to take a more complicated expression and approximate it by a polynomial function. The function we’re going to look at is f(x)=2 \sin x+\frac{\cos 3x}{2}, but we could choose any function which is well behaved close to where we want to approximate it (there is a much more precise way to phrase this, but for the current discussion, this is enough).

This function looks like:


OK, so how are we going to go about approximating this function? Well, let’s ask about approximating it close to the point x=2.5 (this value is arbitrary and we could have asked for any value). What would be the most naive approximation we could make? Well, if we have a function which is a constant, and equal to the original function at x=2.5 then that’s a start. At least it matches the value of the function at that point, if nothing else. What is the value of this function at x=2.5?…

UCT MAM1000 lecture notes part 20 – part ii – Monday 17th August

So, we saw in the last section that we could work out the polynomial expression for (1+x)^5 both using combinatorics as well as using calculus. We had also found previously that for small x we could just take the first couple of terms of the polynomial and it was a good approximation of the function itself, depending on how accurate we needed the answer. For instance, for small x, 1+5x is a reasonable approximation. One thing to note is that the value of these two functions is exactly the same at x=0 and the derivative of both functions is the same at x=0. The second derivatives are not the same, but had we taken the next term in the polynomial, the second derivatives would have matched as well.

OK, let’s take this one step further and repeat everything exactly as before, but now with the power, not as 5 but as 5.2 and see what happens.…

UCT MAM1000 lecture notes part 20 – part i – Monday 17th August

The beginnings of Taylor polynomials and approximations


We’ll start off with some of the discussion we had in class today about treating the binomial expansion as a way to approximate a function.


We know that (x+y)^5=\sum_{r=0}^5{_5C_r} x^r y^{5-r}. We can then see that if we have a function like (1+x)^5 we can simply replace the y in the above expression by 1 and we get (writing out the sum explicitly):


(1+x)^5=1+5x+10 x^2+10x^3+5 x^4+x^5


If you can’t remember these binomial coefficients offhand, just write out the first 6 lines of Pascal’s triangle and read them off.

ok, so what if I asked you to give me an approximate value for (1+0.01)^5. This isn’t so easy because you’ll have to multiply 1.01 by itself five times. How about if we use the polynomial form from the binomial expansion? Well, we can see that this is:


(1+0.01)^5=1+5(0.01)+10 (0.01)^2+10(0.01)^3+5 (0.01)^4+(0.01)^5


But we can also see that each term gets smaller and smaller.…

UCT MAM1000 lecture notes part 19

The Binomial expansion


So, we’re going to start using some of the combinatorics we’ve just learned to answer some questions which without these techniques would seem completely infeasible.

What is (x+y)^{34}? This seems like an almost impossible task; you’d have to write out (x+y)(x+y)(x+y)..., 34 times, then multiply them all out and it would become incredibly messy. However, everything we’ve done in the last section will allow us to see precisely what this, and any other expression of this form, is given by. Let’s start with a simpler example:

How about (x+y)^4? This is a simpler example and one that we could think of doing by hand, but we will show that there is a very general way to get any expression of this form. We start by writing this out in long form:




We are now going to label the x‘s and y‘s, though the labels (which will be indices) will just be dummy labels and we will remove them in the end.…

By | August 10th, 2015|Courses, First year, MAM1000, Uncategorized, Undergraduate|1 Comment

UCT MAM1000 Lecture notes part 9

Areas between curves

We know how to calculate the area between a curve and the horizontal (x) axis of a graph. We have learnt a number of very sophisticated techniques which allow us to get an analytic answer.

It is a fairly simple extension to study not the area between a curve and the horizontal axis, but between two curves. You can see here the link between the two:


We can see in the figure below how we can approximate the area by a sum of rectangles. In this case we have used the left point approximation. Each rectangle here has \Delta x=0.02 and the heights are f(x_i)-g(x_i) where the x_i are the left points of each rectangle, in this case =0.02(i-1). So x_1=0, x_2=0.02, etc.


We started off studying integration by taking the limit of a Riemann sum. We can do exactly the same thing here, taking the simplest possible Riemann sum, with rectangular regions which stretch between the two curves (see the diagrams on page 448 in Stewart) and then taking the limit to find an expression in terms of integrals.…