MAM1000W – Page 7 – Mathemafrica

UCT MAM1000 lecture notes part 29 – complex numbers part vii

So, we know how to take the exponential of any complex number now. We do it by converting the exponential into the exponential of the real and imaginary parts separately, and then use the relationship between $e^{ia}$ and the $\cos$ and $\sin$ functions to write everything in terms of functions of real numbers, which we know how to deal with. How about the trigonometric functions applied to complex numbers? Well, we have a pretty good hint already from how we got from the exponential of complex numbers to trigonometric functions of real numbers. In fact we’re just going to give the answer, but you can work it out using Taylor series as well. For a complex number $z$ :

$\cos z=\frac{e^{iz}+e^{-iz}}{2}$

$\sin z=\frac{e^{iz}-e^{-iz}}{2i}$

The first thing to check is that this is true when $z$ is a real number. It looks pretty strange at first site, especially the definition of $\sin$ because there’s an $i$ sticking out in the denominator like a sore thumb!…

By Jonathan Shock| August 26th, 2015|Courses, English, First year, Level: intermediate, MAM1000, Uncategorized, Undergraduate|3 Comments

UCT MAM1000 lecture notes part 28 – complex numbers part vi

So last time we discovered that there was this amazing link between exponentials and trigonometric functions where the bridge between them was precisely complex numbers.

We now know how to take the exponential of any complex number and it is given by the exponential of a real number and a sum of terms which contain trig functions of real numbers:

$e^{a+ib}=e^a(\cos b+i\sin b)$

We can see also that now we have a third way to write a complex number. If you have a number in modulus argument form like:

$r(\cos\theta+i\sin\theta)$

Then we can also write this as $re^{i\theta}$ . This is an alternative way of writing the modulus argument form.

In this form it also becomes more obvious that moduli multiply and arguments add. If we have two complex numbers:

$z=|z|e^{i\theta}$ and $w=|w|e^{i\phi}$ Then:

$zw=|z||w|e^{i(\theta+\phi)}$

The fact that we can now take the exponential of any complex number is very powerful. The point is that in order to calculate this function, all we need to be able to do is to take exponentials and trig functions of real numbers, and that we can do.…

By Jonathan Shock| August 26th, 2015|Courses, English, First year, Level: intermediate, MAM1000, Uncategorized, Undergraduate|3 Comments

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 Jonathan Shock| 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.…

By Jonathan Shock| August 23rd, 2015|Courses, English, First year, Level: intermediate, MAM1000, Uncategorized, Undergraduate|4 Comments

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.…

By Jonathan Shock| August 16th, 2015|Courses, English, First year, Level: intermediate, MAM1000, Uncategorized, Undergraduate|2 Comments

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.…

By Jonathan Shock| August 16th, 2015|Courses, English, First year, Level: intermediate, MAM1000, Undergraduate|3 Comments

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:

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

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 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. This is because while the polynomial is only an approximation, we want that certain properties between the two hold exactly at $x=a$ .…

By Jonathan Shock| August 16th, 2015|Courses, English, First year, Level: intermediate, MAM1000, Uncategorized, Undergraduate|1 Comment

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 $x=2.5$ ?…

By Jonathan Shock| August 16th, 2015|Courses, English, First year, Level: intermediate, MAM1000, Undergraduate|1 Comment

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.…

By Jonathan Shock| August 16th, 2015|Courses, English, First year, Level: intermediate, MAM1000, Uncategorized, Undergraduate|1 Comment

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.…

By Jonathan Shock| August 14th, 2015|Courses, English, First year, Level: intermediate, MAM1000, Uncategorized, Undergraduate|1 Comment

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