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## Create a MathLapse animation and win prizes from the IMAGINARY conference 2016

The following is taken from the IMAGINARY website.

A “MathLapse” (ML) is a new educational and artistic format, which highlights the link between mathematics and real-world phenomena. The name MathLapse is inspired by the timelapse-technique in physics: By re-scaling time, phenomena are visualized which we cannot directly observe.

A ML is short, simple, self-contained, creative and illustrates a single mathematical idea through true or virtual animated images. The content of ML is diverse. For example it can be a geometrical animation or a time-lapse, which go along with mathematical equations and concise explanations.

Everybody is invited to submit a MathLapse on the IMAGINARY platform. The jury will review all submissions and give prizes to the best MathLapses.

A first MathLapse-Festival will be organized at IC16, where the winners will be announced and their MathLapse movies will be screened. It is not necessary to participate at the IC16 conference.

Here is the trailer for the competition

Katzengold:

Primelapse:

See what you can come up with!…

## 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.…

## Circular base, semi-circular top, triangular cross-section

For those who were particularly confused with the shape we discussed today in class, I’ve created a little animation which may help…or not…

Remember a single cross-sectional slice looked like:

 How clear is this post?

## Sticky Post – Read this first. Categories and Links in Mathemafrica

The navigability of Mathemafrica isn’t ideal, so I have created this post which might guide you to what you are looking for. Here are a number of different categories of post which you might like to take a look at:

Always write in a comment if there is anything you would like to see us write about, or you would like to write about.

 How clear is this post?

## Picturing volumes of revolution

One of the homework questions this week was the following:

Let $R=\{(x,y)\in \mathbb{R}^2:y\ge 0, \cos x\le y\le \sin x\,\,and\,\,0\le x\le\pi\}.$

a) Sketch the region R and find its area.
b) Find the volume of the solid obtained by rotating the region R around the y-axis.

The first thing to do is to sketch the graphs of $y=\cos x$ and $y=\sin x$. Once you’ve done that, the other parts of the inequalities should be clear. It should look like the red region in the following plot:

Now we have to imagine bringing out a third axis, perpendicular to the picture above, ie. coming out towards us. We then want to rotate the red form here about the vertical axis. This we can imagine doing in the following animation:

Given this form we can then think about either taking horizontal cross-sections through it, which will give us thin annuli, or we can take vertical, circular slices to give us thin shells. Adding these together and integrating should give us the same answer whichever way we choose to slice it, but one way will be considerably easier.…

## Some more volume visualisations

Here is an animation which may help you imaging a shape which has a circular base, with parallel slices perpendicular to the base being equilateral triangles:

The same thing, where the slices are squares.

And here is the region in the (x,y) plane between $y=\sqrt{x}$, the x-axis and the line x=1. rotated about the y-axis. Here a thin shell is drawn in the volume, then pulled out. Then it is replaced, then the volume is filled with shells, and each of them is pulled out of the volume vertically. This is to give you an idea about how to visualise the method of cylindrical shells.

 How clear is this post?

## Pascal’s triangle, fractals and the Sierpinski triangle

I mentioned today in class that something rather special happens if you colour the even and odd numbers in Pascal’s triangle differently.

A quick reminder that Pascal’s triangle looks like:

For the first 9 layers.

In the animation below I have simply put an “o” for an odd number and put it in red, and “e” for an even number and put it in brown. Here I am starting with the Pascal triangle with a single entry, and building up in the animation two lines at a time, up to 100 lines. You see that a rather interesting pattern emerges.

The Mathematica code for this is:

tt = Table[With[{n = nn, p = nn},
Graphics[Table[Text[Style[With[{x = Binomial[n – j, n – i]}, If[OddQ[x], Style[o, Red], e]],Round[200/Sqrt[p]]], {Sqrt[3] (i – j/2), 3 j/2}], {i, n}, {j, i}], ImageSize -> 1000]], {nn, 1, 100, 2}]

and then exported as a .gif. (see here for a simplified version and some more Pascal’s triangle code).…

Gallery

## How Pringles are made (or alternatively hyperbolic paraboloids)

We are currently looking at functions of 2 variables and their graphs. Today we looked at cross-sections through a couple of different surfaces to try and figure out what they looked like in three dimensions. We did this by looking at slices in different directions and then worked out how they all fitted together. In the following animation I have taken horizontal slices of the function:

$f(x,y)=x^2-y^2$

Remember the graph of this is the set of points $\{(x,y,z)\in{\mathbb R}^3|z=f(x,y) and (x,y)\in D\}$

We can take horizontal slices of the surface by fixing the z-value and seeing how x and y are constrained. For instance, let’s fix the z-value to 0. Then we have:

$0=x^2-y^2$

This actually gives us two functions in the $xy$-plane: $y=\pm x$. This of course is just given by two lines of gradient +1 and -1 which pass through the origin in the $xy$-plane. In three dimensions then, if we slice through our surface at $z=0$ we should find these two lines which look like:

How about at $z=1$?…

## The Fundamental Theorem of Calculus part 1 (part ii)

OK, so up to now we can’t actually use the FTC (Fundamental Theorem of Calculus) to calculate any areas. That will come from the FTC part 2.

For now, let’s take some examples and see what the FTC is saying. I’ll restate it here:

The Fundamental Theorem of Calculus, part 1

If $f$ is continuous on $[a,b]$ then the function $g$ defined by:

$g(x)=\int_a^x f(t) dt$,     for $a\le x\le b$

is continuous on $[a,b]$ and differentiable on $(a,b)$, and $g'(x)=f(x)$.

——

Let’s look at some examples. We’re going to take an example that we can calculate using a Riemann sum. Let’s choose $f(x)=x^2$.

If we integrate this from $0$ to some point $x$ – ie. calculate the area under the curve, we get:

$\int_0^x t^2 dt=\frac{x^3}{3}$.

Make sure that you can indeed get this by calculating the Riemann sum.

So, what does the FTC part 1 tell us? It says that if we take the derivative of this area, with respect to the upper limit, $x$, then we get back $f(x)$.…