HOLO-MATH’s website is short on firm details, but it seems to be something to do with using Microsoft’s HoloLens VR goggle thingies to make interactive VR maths “experiences”. Here’s the blurb:

HOLO-MATH is an international project to produce immersive live experiences in mathematical sciences using the latest mixed reality technology.

It’s the first project to use state of the art technology for scientific knowledge transfer in a museum environment and on a large scale.

The experiences are presented in science museums/centers and at special events. They are targeted at groups of 20 participants led by human guides and virtual avatars. New forms of augmented visualization and interaction are core features. The audio-visual experience is of the highest quality.

In different HOLO-MATH experiences, participants will be able to play, discover, experiment and learn about science history and current research.

There’s more information on holo-math.org, and some pictures of be-goggled guests at the project’s launch on the hashtag #holomath.

]]>Consider the following question: **How many ways are there to connect $2n$ points on a circle so that each point is connected to exactly one other point?**

For example, for $n = 2$ there are four points on the circle – say, $P_1,P_2,P_3,P_4$ arrange clockwise. If $P_1$ is mapped to $P_2$, then $P_3$ can only be mapped to $P_4$. Likewise, if $P_1$ is sent to $P_4$, then $P_3$ can only be sent to $P_2$. If $P_1$ and $P_3$ are connected, then mapping $P_2$ to $P_4$ creates an intersection point.

This shows that for $n = 2$, the answer is $2$.

What is the answer to this question for general $n$? For $n = 3$, the answer turns out to be $5$.

The problem can be solved by setting a recurrence relation: denote by $C_n$ the number of ways to connect $2n$ points $P_1, P_2, \ldots , P_{2n}$ on a circle so that each point is connected to exactly one other point.

For $n = 0$, there is nothing to do, and so $C_0 = 1$. Assume $P_1$ is mapped to $P_k$ for some $k = 2, 3, \ldots, 2n$. Then, none of the points $P_2,\ldots, P_{k−1}$ can be mapped to one of the points $P_{k+1},\ldots , P_{2n}$ because this will create an intersection point. Furthermore, $k$ must be an even number, otherwise one of the points $P_2, \ldots, P_{k−1}$ will be left out.

Therefore, we have \begin{align} C_n = C_0 \cdot C_{n−1} + C_1 \cdot C_{n−2} + \cdots +C_{n−2} \cdot C_1 + C_{n−1} \cdot C_0\\ = \sum_{j=0}^{n-1}C_j \cdot C_{n−1−j} . \end{align} ‘Solving’ this non-linear recursion leads to the formula \[ \frac{1}{n+1}{2n \choose n} \] For $n \geq 1$, the sequence starts $1,2,5,14,42,\ldots$ $C_n$ is called the *$n$ ^{th} Catalan number*.

These numbers were introduced by the French and Belgian mathematician Eugène Charles Catalan. As it turns out, Catalan numbers count various combinatorial structures such as the ‘number of trees’.

A **tree** is a graph that satisfies the following conditions:

- two nodes can only be connected through one edge (no multiple edges between pairwise nodes);
- no loop at a node (no node can be connected to itself via an arc);
- the edges of the graph are not oriented – they are simple edges;
- there is always a path to get from one node to another (the graph is connected);
- there is no cycle (a path that starts and ends at a same node) in the graph.

In other words, a mathematical tree is a graph in which any two nodes are connected by a unique path.

But how many trees are there? Mathematicians and computer scientists had already started answering this question since the work of the German mathematician Carl Wilhelm Borchardt in 1860, before the work of Cayley, Pólya, Prüfer and many others.

Nowadays, the study of trees involves determining a vast number of properties, such as the number of possible trees given a number of nodes.

We highlight three types of trees that are counted via Catalan numbers.

If we choose one node of the tree and call it the **root**, we have what is called a **rooted tree**. Thinking of the root of a tree as a common ancestor to all other nodes in the tree, we define the notion of a **successor** – a node further along the same path in the tree. $C_n$ enumerates:

A binary tree is a rooted tree in which each node has two successors or no successor at all; the number of possible such trees with $2n+1$ nodes is the $n$^{th} Catalan number.

A pruned binary tree is a rooted tree obtained from a binary tree by leaving out all nodes with no successors; the number of possible such trees with $n$ nodes is the $n$^{th} Catalan number.

A plane tree is a rooted tree in which a node can have any number of ordered successors; the number of possible such trees with $n+1$ nodes is the $n$^{th} Catalan number.

Besides trees, amongst other structures counted by $C_n$, we have *the number of Dyck paths of length $2n$.* A Dyck path is made from ↗ and ↘ unit-length steps in the $xy$-plane that start and end on the $x$-axis but never go below it.

Also, (and just for fun), you could try to compute the following integral: \[ \frac{2^{1+2n}}{\pi} \int_{-1}^{1} x^{2n}(1-x^2)^{1/2} \, \mathrm{d}x \] You would find the answer is… the Catalan numbers again!

*If you’d like to write about some maths that you find interesting, get in touch!*

The Further Maths Support Programme is an organisation in the UK that supports students wishing to take an A-level in Further Maths. Since this isn’t offered in all schools and colleges, the Programme helps organise tuition for people who can’t do it through their school, but also encourages students at younger ages to consider taking the A-level through workshops and university visit days. They also run excellent training courses for teachers, and have a number of resources on their website for students and teachers, including problem solving materials, videos, podcasts and maths competitions.

According to a recent blog post by maths teacher Jo Morgan, a government review has made the FMSP’s future precarious. Their funding through the Department for Education will be removed next April, and they’ll be replaced by the “Level 3 Maths Support Programme”. The L3MSP will support Core Maths as well as A level mathematics and further mathematics, but will focus on only certain geographical areas, meaning many will lose access to the resources currently provided.

Two of the programmes previously supported by the same funding have already had their funding stopped – the Core Maths Support Programme, and Underground Maths – but the FMSP hasn’t finished yet, and Jo hopes that by contacting the DfE we could convince the government to continue funding it. As they point out in the blog post, the FMSP has made a huge difference to the numbers of students taking maths and has had a direct impact in classrooms supporting teachers all over the UK.

So what do we do? Start a petition? Tweet the DfE to tell them? Over to you, readers.

Friends of the Aperiodical, nerd-comedy troupe *Festival of the Spoken Nerd*, are currently on tour around the UK. As part of their show, questionably titled *You Can’t Polish a Nerd*, Matt Parker attempts to calculate the value of $\pi$ using only a length of string and some meat encased in pastry. He’s previously done this on YouTube, and the idea was inspired by the Aperiodical’s 2015 Pi Approximation Challenge, and in particular my own attempt to approximate $\pi$ with a (more conventional) pendulum.

For our $\pi$ approximate-off, we wanted to derive values for $\pi$ using a suite of methods that mostly didn’t involve measuring the circumference and diameter of a circle. This included evaluating as many terms of some infinite series as the length of our room booking permitted, and a recreation of the famous Buffon’s Needle experiment. But surely the most satisfying method is to just swing a heavy ball on a bit of string.

As many people learn in high school, the formula governing how long a pendulum takes to do one complete swing is \[ T = 2\pi\sqrt{\frac l g} \] where $l$ is the length of the pendulum and $g$ is the acceleration due to gravity (about $9.8ms^{-2}$). Slightly counterintuitively, the swing time doesn’t depend on the weight of the bob (or if you prefer, the mass of the weight), nor on how far back you swing it from. So just from timing a swing and measuring the string, you can solve the equation to get a numerical value for $\pi$.

But all that calculation sounds pretty boring. We can choose the length of the string, so why not choose it so that it cancels the $l$ and the 2, to give $T=\pi$? Conveniently that requires a length of $g/4 \approx 2.45$ metres (if you’ll excuse my abuse of dimensionality), which is just about doable in a tallish room or a stairwell or a theatre stage. When we tried this timing ten swings to reduce the impact of imperfect human reaction time, we ended up with a value of 3.133. (I speculate that the overwhelming source of inaccuracy is the difficulty in measuring the length of the string from the precise point of pivot at the top to the precise centre of mass of the bob/weight/pie.) On his video, Matt, using a fancy slow-mo camera, got a value of 3.128 seconds per swing. The Aperiodical has yet to take its annual pilgrimage to the *Festival*, so it remains to be seen how successfully the feat is accomplished live.

You might think that using a pendulum in an attempt to find $\pi$ without using circles is cheating a bit: surely a pendulum’s swing time depends on $\pi$ because it moves along the arc of a circle? Well, no. In fact the opposite is true. I lied a bit when I said the time doesn’t depend on how far back you pull the pendulum. The formula above is only an approximation for small initial angles, when a pendulum approximates simple harmonic motion. The full formula is this:

So in fact the more of a circle you make your pendulum trace out, the *worse* your approximation will be. Circles *kill pi*.

If you’d like to see Matt attempt this live on stage, the Spoken Nerd show is still on tour until the end of November, and as well as swinging food, it includes songs, live experiments, comedy and many nerdy references.

Calculating π with a pendulum, by Matt Parker on YouTube

Aperiodical’s π approximation challenge, on YouTube

Festival of the Spoken Nerd tour

]]>

At the start of his HLF lecture on Asymptotic Group Theory on Thursday morning, Fields medalist Efim Zelmanov described the ‘group’ as: “the great unifying concept in mathematics,” remarking “if you go for a trip, and you are allowed to take only two or three mathematical concepts with you, give serious consideration to this one.” Very loosely defined, a group is a set of things (its ‘elements’) that you can ‘multiply’ together, with this multiplication behaving in certain helpful ways. Think of numbers being added, functions composed together or rotations and reflections of a shape being carried out one after the other. I doubt any mathematician would accuse Zelmanov of overstating their importance in mathematics.

In his talk he discussed residually finite groups. These are groups which are infinite in size but still just a little bit finite-y. In technical terms, the group has a set of homomorphisms with finite kernels having trivial intersection. Although the group is too large to see all at once, as Zelmanov put it, we have “photos from all sides of the group”. He contrasted this to “hopelessly infinite groups”, for which no such photo album is possible.

A common way to look at a group is to find a set of ‘generators’: these are elements of the group which you can multiply together to create any element of a group (the elements ‘generate’ the entire group). Some infinite groups can’t be generated from a finite set — consider trying to find a set of rational numbers that you can multiply together to create any rational number. Those that can be generated from a finite set are unexcitingly called ‘finitely generated’. Of course, finite groups are also finitely generated.

Zelmanov considered under what circumstances finitely generated groups can be proved to be finite. One immediate way this won’t happen is if one of the generators is not periodic: if you keep multiplying it by itself you keep getting new elements forever, never ‘looping back’ to the original generator. (Imagine starting with 1 and continually adding 1…) The Burnside problem asks whether there are any other ways to make a finitely-generated, yet infinite, group. In 1991, Zelmanov proved that for residually finite groups, there aren’t. However, this isn’t the case for the ‘hopelessly infinite’ groups.

In his lecture Zelmanov, accompanied by his excellent hand-drawn slides, discussed this before moving on to related topics such as the growth of groups (if you start with a generating set, and create new elements by multiplying them together, how quickly does the set grow?) and ‘approximate groups’ (which, as the name suggests, are things that are like, but not quite, groups).

]]>The text reads:

An orchestra of 120 players takes 40 minutes to play Beethoven’s 9th Symphony. How long would it take for 60 players to play the Symphony? Let P be the number of players and T the time playing.

Well, once you’re done laughing, we’ve done some investigative journalism and found the origin of this question. And it turns out it’s quite nice!

I wrote this!! How did you get this??? I am a maths teacher in Nottingham UK. Wrote this 10 years ago. Here is the original whole worksheet pic.twitter.com/jYX55GSBKz

— Claire Longmoor (@LongmoorClaire) October 11, 2017

The question is from a worksheet developed by maths teacher Claire Longmoor (who is, based on current evidence, brilliant) ten years ago. Claire put together a selection of example questions with relationships in direct and inverse proportion, and deliberately included the orchestra question as an example of something where it doesn’t work that way. It’s a nice activity to help reinforce the difference, and in context the question works nicely.

Other examples on the sheet include a bricklaying example with creditably diverse gender representation, a car with terrifyingly low fuel efficiency, good cow names and a delightful insight into the bygone world of fruit picking.

]]>Marcus du Sautoy has tweeted about a mathematics and music project he’s involved in, called **The Sound of Proof**. Five classical proofs from Euclid’s Elements have been interpreted by composer Jamie Perera into musical pieces, and they’ve put together an app/game to see if you can work out which one corresponds to which.

They’ll be announcing the results at an event as part of Manchester Science Festival in October. The project is a collaboration with PRiSM, the research arm of the Royal Northern College of Music in Manchester.

]]>The Lectures are now available on the LMS’s YouTube channel, along with many of the previous years’ videos.

]]>Aperiodipal and number ninja, Stand-up Mathematician Matt Parker, has set up a petition on the UK parliament petitions website to change the awful, awful tourist board official symbol for a football ground (US readers: imagine I’m saying ‘soccer stadium’). In Matt’s words:

The football shown on UK street signs (for football grounds) is made entirely of hexagons. But it is mathematically impossible to construct a ball using only hexagons. Changing this to the correct pattern of hexagons and pentagons would help raise public awareness and appreciation of geometry.

To end this madness, Matt needs 10,000 signatures for the petition to be responded to by the government (and 100,000 for it to considered for debate in parliament). It’s currently around the 3,000 mark – so it’s plausible that he might do it. It’s also had coverage in The Independent already, and Matt’s YouTube video on the campaign already has over 100,000 views.

To sign, you simply need to be a British citizen or UK resident, and fill in your details on the site (you’ll need a valid postcode). Ban this hexagonal filth!

Update the UK Traffic Signs Regulations to a geometrically correct football, on UK Parliament Petitions

]]>