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UCT MAM1000 lecture notes part 27 – complex numbers part v

We’re about to make one of the most profound links that we will obtain through complex numbers. This is going to show how complex numbers are a bridge between different areas that you already know about, but never knew had anything to do with one another.

We know about exponential functions and how they have very special properties related to their derivatives. $e^x$ is a function which is practically defined as the function which is equal to its derivative. We also know that exponential functions tell us about growth, and we will see this in more detail when we come on to differential equations.

We know that trigonometric functions are to do with triangles, and circles, and angles and they tend to be periodic. They tell us how things vary in a way where they come back to where they started after some time.

Exponential functions and trigonometric functions couldn’t really look much more different if they tried.…

By Jonathan Shock| August 23rd, 2015|Courses, English, First year, Level: intermediate, MAM1000, Uncategorized, Undergraduate|1 Comment

UCT MAM1000 lecture notes part 26 – complex numbers part iv

OK, so we saw something pretty interesting last time when we multiplied together complex numbers using the modulus argument form.

Remember that for two complex numbers which we will write as $z_1=r_1(\cos\theta_1+\sin\theta_1 i)$ and $z_2=r_2(\cos\theta_2+\sin\theta_2 i)$ , where $r_i$ are the moduli, and $\theta_i$ are the arguments of $z_i$ . If we multiply them together then we get:

$z_1 z_2=r_1 r_2 (\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2))$

Well, what would happen if the two complex numbers were the same? ie. if we have $z=r(\cos\theta+i\sin\theta)$ and we want $z^2$ ?

Well, then clearly:

$z^2=r^2(\cos 2\theta+i\sin 2\theta)$ .

What if we then multiplied this by $z$ one more time:

$z^3=z^2 z=r^3(\cos (2\theta+\theta)+i\sin(2\theta+\theta))=r^3(\cos 3\theta+i\sin 3\theta)$

hmm, do we already see a pattern emerging? Let’s say that we have a complex number with modulus 1. Complex numbers of the form:

$z=\cos\theta+i\sin\theta$

Are clearly modulus 1. We know that the modulus is the square root of the sum of the squares of the real and imaginary parts of a complex numbers so $|z|=\sqrt{\cos^2\theta+\sin^2\theta}=1$ .

ok, so how about if we have $z^n$ where $n$ is an integer?…

By Jonathan Shock| August 23rd, 2015|Courses, English, First year, Level: intermediate, MAM1000, Uncategorized, Undergraduate|5 Comments

UCT MAM1000 lecture notes part 25 – complex numbers part iii

So we saw last time that we can take a complex number and put it in a 2 dimensional plane called the complex plane, where its horizontal distance from the origin is given by its real part, and the vertical distance from the origin is given by its imaginary part. We can thus think of the real and imaginary parts as the Cartesian coordinates of that point.

It turns out that there is another way to represent a complex number, but rather than using the real and imaginary parts to specify it, we will use two other pieces of information.

If I tell you that a complex number is a distance $|z|$ away from the origin in the complex plane, then this leaves you with a whole circle of possibilities. All the points on the circle of radius $|z|$ about the origin are the same distance from the origin. But if I also give you an angle subtended between the x-axis and the line joining the complex number and the origin, read anti-clockwise from the x-axis, this will completely pin down the point in the complex plane.…

By Jonathan Shock| August 23rd, 2015|Courses, English, First year, Level: intermediate, MAM1000, Uncategorized, Undergraduate|5 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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