## Derivatives of the function y=x|x|

This is an example of a very simple real function that is only differentiable a finite number of times.

Let $f:\mathbb{R} \to \mathbb{R}$ be a function defined by $f(x)=x|x|$. The aim is to show that this function is differentiable but that it is not twice differentiable. Notice that $f(x)$ can also be written as
$$f(x) = \left\{ \begin{array}{l l} x^2 & \quad \text{if x\geq0}\\ -x^2 & \quad \text{if x<0} \end{array} \right.$$ The graph of this function looks as follows

It is clear that $f(x)$ is differentiable for $x\!>\!0$ and $x\!<\!0$ with derivative $f^{\prime}(x)=2x$ and $f^{\prime}(x)=-2x$ respectively. We only need to check that $f(x)$ is differentiable at $x=0$.

By definition a function $g:I \to \mathbb{R}$ where $I \subset \mathbb{R}$ is differentiable at a point $x \in I$ if the limit
$$\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}$$
exists. $g$ is called differentiable if it is differentiable at every point $x \in I$.

Let’s check that this limit exists for $f$ when $x=0$
$$\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0}\frac{(0+h)|0+h|}{h}$$
$$=\lim_{h\to 0}\frac{h|h|}{h}=\lim_{h\to 0}|h|=0$$

So $f(x)$ is indeed differentiable at $x=0$ and we can write the derivative of $f(x)$ as $f^{\prime}(x)=2|x|$. The graph of the derivative of $f$ looks as follows

The graph of 2|x| and the derivative of x|x|

We see that $f^{\prime}(x)$ is differentiable when $x \neq 0$ but when we try to find the derivative of $f^{\prime}$ at $x=0$ we have
$$\lim_{h\to 0^{+}}\frac{f^{\prime}(0+h)-f^{\prime}(0)}{h}=\lim_{h\to 0^{+}}\frac{2|h|}{h}=2$$
$$\lim_{h\to 0^{-}}\frac{f^{\prime}(0+h)-f^{\prime}(0)}{h}=\lim_{h\to 0^{-}}\frac{2|h|}{h}=-2$$
The left and right limits are not the same and therefore $f^{\prime}$ is not differentiable at $x=0$. The conclusion is that $f$ is differentiable but not twice differentiable.

It is not much more difficult to show (by induction is possibly the easiest way) that the function given by $h(x)=x^{n}|x|$ is $n$-times differentiable but not $n+1$ times differentiable.

## Solutions of Polynomials

For a general quadratic equation $ax^{2}+bx+c=0$ the solutions are given by

$$x=\dfrac{-b+\sqrt{b^{2}-4ac}}{2a}$$ and

$$x=\dfrac{-b-\sqrt{b^{2}-4ac}}{2a}$$

– it may not necessarily be the most efficient method of calculating the solutions to a quadratic since in some cases factorisation is much faster, but it gets the job done and if you’re feeling a little bit lazy and can’t be bothered to factorise a quadratic then this equation practically does the job for you.

Naturally at some point in my mathematics studies (probably during my A Levels) I asked myself the question – does a formula exist that I can use to solve cubic equations? Naively (I later realised), I tried to find a formula using similar techniques to how the quadratic formula is derived. Of course I failed (but it’s amazing how much you can learn even when you fail). However, formulae do exist that give the solutions to a general cubic equation (equation of general cubic).

It is worth noting that any cubic $ax^{3}+bx^{2}+cx+d=0$ can be transformed into a cubic of the form $x^{3}+px+q=0$ by using the transformation $x=y-\dfrac{b}{3a}$. The solutions to this new cubic (given here without derivation) are

$$x=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}$$

$$x=j\left(\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}\right)+j^{2}\left(\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}\right)$$

and

$$x=j^{2}\left(\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}\right)+j\left(\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}\right)$$

where $j=\mathrm{e}^{\frac{2\pi i}{3}}$

The big problems here are that these formulae are much bigger than the formulae for the solutions to a quadratic, it certainly doesn’t roll off the tongue quite as easily, and also it is very cumbersome to work with. Nonetheless we have our equations, whether or not they are easy to work with may not be relevant – the fact that they exist is what matters here.

So what about quartic equations $ax^{4}+bx^{3}+cx^{2}+dx+e=0$ Are there any formulae for the solutions of a quartic? Yes there are … but the expressions are extremely long; you can see them at the wikipedia page on quartic equations. Again, we have the formulae and we at least know that they exist so we can be satisfied with that for now. Since general quadratic, cubic and quartic polynomials have closed formulae to describe their solutions they are said to be “solvable by radicals”

But things stop there. There are no such formulae for quintic (degree 5) polynomials or indeed any polynomial of degree greater than or equal to 5. It’s not that the formulae have yet to be found – they just don’t exist. The reason why they don’t exist takes us into Galois Theory where it can be proved that general polynomials of degree 5 and greater are not solvable by radicals – in other words, no formulae exist that give the solutions to general polynomials of degree 5 and above.

This is not to say that these polynomials cannot be solved at all, nor that their solutions don’t exist – it is quite easy to construct a quintic polynomial where all of the solutions can easily be found by (educated or uneducated) guesswork – but not using a systematic method or any series of ‘standard formulae’. To solve a general quintic (or higher degree) polynomial we will usually have to resort to numerical methods unless we have some additional knowledge about the polynomial or unless we are lucky enough to be able to guess some solutions!

## Topology – An Informal Introduction

A square and a circle, as I imagine you probably know, look completely different – and you would certainly be right to say that they are different. Mathematically, however, a circle and a square share many structural properties and it may come as a surprise to some that in some cases mathematicians may go so far as to not even bother to distinguish between the two – in other words, the circle and the square can be considered one and the same thing. This is where topology comes in…

Topology is a very abstract area of geometry that simplifies many problems that would be very difficult if not impossible to solve. A formal introduction to topology requires knowledge of some basic set-theory and a knowledge of some analysis and metric spaces is usually helpful. I want to give a simple and informal introduction to this fascinating area of mathematics.

Given a set, $X$, a topological space is a couple formed from the set $X$ and a collection $\tau$ of subsets (called a topology) of $X$ that satisfy

• $\emptyset \in \tau$ and $X \in \tau$
• $\tau$ is closed under finite intersections
• $\tau$ is closed under arbitrary unions

Anything (yes anything) that satisfies the above definition is a topological space. The set $X$ may be a region of the $x$-$y$ plane, a $3$-dimensional region of space, or even an abstract collection of objects – the level of abstractness of topological spaces can be demonstrated in the following example of a topological space

Let $X=\{Amy, Boris, Charlie\}$.

One possible topology, $\tau$, for $X$ could be $\{\emptyset, \{Amy\}, \{Amy, Boris\}, X\}$.

What has all this got to do with the square and the circle? Well two topological spaces can be considered the same (topologically indistinguishable) if a homeomorphism exists between the two. Homeomorphism is mathematical jargon but in simple terms it is a way of “moulding one space into the shape of another space.” Homeomorphisms can be informally visualised by imagining a circular piece of plasticine. It wouldn’t be too difficult to re-shape the plasticine into the shape of a square and vice-versa – this way of deforming the plasticine obeys the rules of a homeomorphism and therefore the circle and the square are topologically equivalent.

A graph of a circle and a square showing topological equivalence

One possible function that takes the square onto the circle is $f(x,y)=\dfrac{(x,y)}{\sqrt{x^{2}+y^{2}}}$

The word topology doesn’t usually manage to make its way out of university maths departments, yet topology does have practical uses. If you have seen a London underground map before then you have seen a practical application of topology – if you have not seen an underground map before then you can see one here. In reality the tube tracks do not run in perfect straight lines but by representing the tube tracks as straight lines the whole thing is much easier to understand – the map of the London underground is topologically identical to the real thing. In other words, even though they are two different things, they have identical structures.

There are many other examples of topologically equivalent pairs of objects such as

• the interval $(0,1)$ and the real-line $\mathbb{R}$
• a sphere and a cube
• $\mathbb{R}^{2}$ and the punctured sphere ($S^{2}$ with a point removed)

There are also many examples of pairs of objects that have significant structural differences and are not topologically equivalent such as

• A sphere and a torus – these are not homeomorphic because the torus has a hole but the sphere does not; both are shown below for comparison.
• $\mathbb{R}^{2}$ and $S^{2}$ are not homeomorphic
• the letter $A$ is not homeomorphic to the letter $T$
 Sphere Torus

## Leeds City Toastmasters August – 1st August 2013

The evening of Thursday 1st August 2013 saw a brilliant meeting at Leeds City Toastmasters – and I was presenting! I was doing a project from one of the Toastmasters manuals – Speech #4 from the Speaking to Inform advanced manual; my speech was titled “A Beautiful Equation”.

The meeting was full of energy and I have to say was one of the best meetings that I have ever attended at Toastmasters during my more than two year membership to date. I took the opportunity to record my speech and here are a few of the highlights from the speech. To say the speech was less than 10 minutes long didn’t make things any easier – with such a tight time-limit it is crucial to only say what you need to say to keep things flowing and to cut out any superfluous material.

I loved giving the speech, though I was a bit worried when I turned up and couldn’t find a lead to connect me up to the projector until about a minute before the meeting was due to start and then about 10 seconds into my speech I realised I hadn’t connected my slide-clicker up properly! But no speech is perfect and it’s all part of the fun. I’m looking forward to the next meeting and of course, my next speech!