post

In conversation with Martin Hairer

It’s a warm summer afternoon in early July and we’re sat in our respective homes, on our laptops, waiting to talk to Martin. We’re both feeling a little nervous as Martin is renowned for being one of the top researchers in the field of stochastic analysis, maybe in the maths community full stop. The notification pops up on screen, Martin has arrived. He greets us with a friendly hello, and instantly puts us at ease with his polite and cheerful demeanour. Martin is speaking to us remotely from a hotel garden in Helsinki; he’s in Finland for the International Congress of Mathematicians (ICM) where this year’s Fields medallists will shortly be announced.

The ICM takes place every 4 years, and is where the Fields, Abacus, Gauss and Chern medals are awarded. This year the congress was due to take place in St Petersburg, but a decision to boycott Russia over the invasion of Ukraine led to a series of satellite conferences being hosted in neighbouring countries. Martin has acted as satellite coordinator and chair of the programme committee for this year’s congress. His role involved inviting plenary speakers, and helping to transition the conference to a hybrid format. But this isn’t Martin’s first involvement with the ICM—he won the Fields medal at the 2014 congress for his work on a regularity theory for stochastic partial differential equations.

Martin Hairer

Photo: King Ming Lam

The long and winding road

The Fields medal is probably the most famous award on Martin’s long list of achievements, but Martin’s first taste of success was developing Amadeus, a popular sound editing software package, which can be used for creating podcasts or original musical scores. He started developing the software as a teenager, and entered it into a competition that he won. At the time it seemed like the perfect outlet to practise his newfound coding skills and indulge his love of music: “I like classical pop rock music basically, you know, like Pink Floyd or Dire Straits or the Beatles”.

Amadeus continues to be something Martin is well known for. It’s hard to understand how Martin has time for all of this: developing software is a full time job in itself; doing it alongside being a prizewinning mathematician is extraordinary. Martin is frank: “I haven’t had much time the last couple of years,” he says, explaining that lately all he has time for is to maintain his software, but “zero time for development”.

Given his early success developing software, was he ever tempted to study computer science at university?

“Yeah, I was actually. I guess the reason why I didn’t study computer science as an undergrad was that I kind of arrogantly believed I already knew it somehow,” Martin says with a sheepish laugh.

Ernst Hairer

Ernst Hairer, professor of mathematics at the University of Geneva. Photo: Renate Schmid, CC BY-SA 2.0 DE

Martin instead decided to study physics as an undergraduate, and went on to do his PhD in the physics department as well. For Martin this was something of a deliberate teenage rebellion—Martin’s father is a distinguished maths professor with his own list of accolades to his name. Perhaps it wasn’t the most potent of rebellions—Martin admits that during his undergraduate he took many classes from the maths department. Further, he says his PhD would’ve been taken under the maths department at most universities.

Even after his PhD, software development was always in the back of his mind. “It was always a plan B. After my PhD, when I decided to continue in academia, I sort of knew it was tough getting permanent jobs in academia. So the plan B was, well, if it doesn’t work out, then I’ll just do software development.”

It’s interesting to hear an accoladed academic talking about job precarity, and reassuring too: this is a concern that many doctoral students contend with, and a reality many postdoctoral researchers have to face. Despite these concerns, Martin did continue in academia, finally moving to the maths department for his postdoc.

Martin really enjoys working in academia. For him it provides one of the best things about being a mathematician: “the freedom to do basically what you want, at least to some extent.” Martin continues: “In some sense you don’t really have a boss as an academic. I mean, officially, inside the university structures you’d always have a head of department or somebody, but they don’t tell you what to do. So you can do what you want to a large extent, and I think that freedom is really quite precious.”

Martin Hairer

Spoilers for Martin’s next paper. Photo: King Ming Lam

Every little thing

For a man whose list of awards is staggeringly long, Martin is generous with his time and easy to talk to, happy to chat about everything we throw at him. The only thing he seems a little shy about talking about is the awards themselves. We asked him what his biggest achievement is, or what prize he is most proud of, expecting him to mention the Fields medal, or perhaps the Breakthrough Prize which he won in 2021. He replies that nothing could top figuring out his regularity theory.

This is something that comes up a few times while we speak to him—the joy of figuring out a problem. Where does that joy come from? Is it the surety in knowing that a theory is finished, that the wrinkles have been smoothed out, or even that the work is ready to be published?

For Martin it’s the moment right at the beginning, when it all starts to click together. “You sort of have a few days, where you’re scribbling around like mad and trying to convince yourself that this actually has a chance of working. I convinced myself relatively quickly that it would work and be a big deal somehow. Obviously it takes a lot of time to work things out, but I knew that it would work out, so it’s not so much putting in the last piece.”

Martin’s work focuses on understanding small scale randomness through large scale behaviour of systems, in particular stochastic partial differential equations. A stochastic partial differential equation (SPDE for those in the know) is a partial differential equation where some of the parameters are random variables, or where the solution can be written in terms of a random variable.

One of the most famous SPDEs Martin works with is the Kardar–Parisi–Zhang (KPZ) equation. Imagine setting a piece of paper on fire, from the bottom edge. The top edge of the burnt part will move up in a way that’s mostly predictable if you’re standing far away, but will look random and jagged if you’re standing very close. The KPZ equation describes situations like this: if we write $h(x,t)$ for the function describing the height of the burnt part at time $t$, it is governed by the equation
\[
\frac{\partial h}{\partial t} = \nu \frac{\partial^2 h}{\partial x^2} + \frac{\lambda}{2} \left(\frac{\partial h}{\partial x}\right)^2 + \eta (x,t),
\]
where $\lambda$ and $\nu$ are constants representing how inflammable the paper is. Importantly here, $\eta(x,t)$ is a random ‘white noise’ term. If we leave it out, we’re left with just a partial differential equation, or PDE. These are tricky, but solvable, and we can work out all sorts of nice things about the solutions. For example, this PDE would have a solution whose graph is smooth. However, including the white noise throws a real spanner in the works, and for decades mathematicians were mostly stumped about how to analyse the behaviour of $h$. Martin found a way to make the equations make sense, which comes down to ‘subtracting infinity from infinity’.

Martin’s regularity theory for SPDEs astounded the maths community—his ideas seemed to come out of the blue. One of the reasons this theory surprised the community is that his breakthrough idea came from physics—using finite series of wavelets (more typically used to encode information in digital files) to understand the behaviour of SPDEs .

Martin’s work continues to focus on stochastic processes. One of his current research projects is exploring discrete systems which are updated according to simple rules. “You try to understand the global large-scale behaviour of these things, and the limits you can get in that way. In some restricted context, people have a pretty good idea of what is happening, and then there are cases which are completely open. There are quite a lot of things there still—it’s one of the big areas of probability theory. The first result of this type would have been the central limit theorem… which goes way back to the 18th century.” Describing how trends in maths evolve, Martin continues: “People have always been interested in these sorts of areas. You have periods of stagnation, and periods where people come up with a technique and a flurry of activity. That’s kind of the way mathematics works.”

Martin Hairer

Distinguish yourself by wearing colourful socks. Photo: King Ming Lam

In my life

For the last five years Martin has been a professor at Imperial College London where he is a researcher, PhD supervisor, and occasional lecturer of a masters course on his breakthrough work. Martin regularly delivers summer schools, which are lecture courses typically attended by PhD students and early career researchers. For Martin, teaching summer schools is a really important part of being a researcher, as sharing knowledge is fundamental to advancing mathematical ideas. “If you’re just doing something alone in a corner and don’t explain it to anyone, there’s not much point, right?”

Two years of the pandemic have changed how mathematics is taught, developed and discussed. After all that time Martin is excited to be back attending events in person. “In the beginning there was a lot of enthusiasm for online seminars and workshops. It works pretty well. You can even give a blackboard talk using your iPad.” But, he adds, “it gets old pretty quickly.”

The lawn outside the Queen's Tower at Imperial College London

The pretty part of Imperial College London

A new hybrid method of working does have its advantages: it makes it easier to collaborate with fellow researchers overseas, to attend conferences in faraway places and it can be more accessible for people who might struggle to come into the office, whether due to childcare responsibilities or health reasons. However for many of us the joy of doing maths is doing it with others, and it’s difficult to mimic the fluid exchange of ideas that occurs in person. “Just catching up with people, to figure out what’s actually going on in the community: you can’t quite figure it out by only listening to talks.”

How does Martin think the pandemic years will affect the future of how mathematics is done? “We seem to be converging on a model,” Martin says, “where it’s an option for the speaker to deliver it remotely or come in in person.” This way of working certainly seems to offer a good compromise of social interaction and accessibility.

Martin with fellow 2014 Fields medallists Artur Avila, Maryam Mirzakhani and Manjul Bhargava (and Maryam's daughter Anahita, who is yet to win a Fields medal).

Martin with fellow 2014 Fields medallists Artur Avila, Maryam Mirzakhani and Manjul Bhargava (and Maryam’s daughter Anahita, who is yet to win a Fields medal). Photo: Wikimedia Commons user Monsoon0, CC BY-SA 4.0

Come together

When Hairer won the Fields medal in 2014, he won alongside Artur Avila (interviewed in Chalkdust issue 02), Manjul Bhargava and Maryam Mirzakhani. Avila was the first South American to win the award, Bhargava the first person of Indian origin and Mirzakhani the first woman, all in the 78th year since the medal’s inception. Since we spoke, this year’s winners have been announced as Hugo Duminil-Copin, June Huh, James Maynard and Maryna Viazovska—the second woman to win the award.

Stem subjects, and in particular maths, have a reputation for struggling with diversity. Does Martin think the ICM has a responsibility to do more to increase gender and minority representation in the sciences? Martin is thoughtful, “it’s tricky,” he says, especially as it goes beyond gender, with under-representation based on ethnicity, socioeconomic and cultural backgrounds too. “There’s a leaky pipeline, the proportion of women gets smaller over time. It’s difficult to find a solution; the one thing that you definitely don’t want to do for something like a Fields medal is to put some kind of quota, because that just devalues it.”

“So the right way to do it is as things percolate up; naturally there will be more diversity.” Martin is hopeful that gender representation is improving in maths. “You see that already in the ICM—if you look at the proportion of female speakers at this ICM it’s about 25%. At the last ICM it was like 15%.”

Waiting for things to percolate up only works if we have the structures in place to support under-represented groups once they reach each stage of the pipeline. This is a difficult task—one important thing is to encourage everyone from a young age that they can do maths, and that it is a viable vocation regardless of background.

Looking back we wonder what advice Martin would have for his 18-year-old self, just embarking on his mathematical journey?

“Maybe I should have paid a bit more attention in the algebra classes.” More seriously, Martin continues with advice to anyone just starting their undergraduate degree: “follow your nose, and do the things which you find interesting.”

Martin Hairer

Photo: King Ming Lam

post

I don’t like science fiction

I have a confession. I’m quite nervous to say this really, but it’s an inescapable fact and I feel the readers of Chalkdust need to know that people like me exist. Here goes…

I don’t like science fiction.

Phew. That’s a huge weight off my shoulders. It’s true though. Despite being a proud nerd, I just can’t get into sci-fi. It’s not that I don’t like fiction (I do) and it’s certainly not that I don’t like science, but together it does nothing for me. And I know I’m missing out on a lot. I will never know that Darth Vader is Luke’s father. Oops. And I will never know how a flux capacitor works. But one thing I don’t need to miss out on is learning about teleportation. Specifically quantum teleportation. Because quantum teleportation exists at the very least in the mathematical world, even if not yet in the physical world.

Continue reading

post

A canal runs through it

For those of you who haven’t already switched, this article will mark a dividing point between two phases of your life: the before, when you would be able to simply walk under a railway bridge; and the after when you will feel compelled to stop and check.

Find your nearest arched railway bridge built out of bricks, the traditional type that looks like cuboids. Go ahead, I’ll be here when you get back. Found one? Great. The chances are, the bricks were in one of two configurations. Either they were laid in rows with the longest brick edges horizontal, as shown in the first photo below, or the rows were all slanted—set at a nonzero angle to the horizontal, as in the second photo.

An arched bridge with horizonal bricks. (Gors-Opleeuw, Belgium)

An arched bridge with sloping bricks. (Winterborne Monkton, Dorset)

Why would you ever build a bridge with bricks that aren’t horizontal? That’s an excellent question. Someone should write a Chalkdust article about that. While they’re doing it, they should also explain why these diagonal bricks have a suboptimal configuration, and how to do it properly.

Now, walls—walls are simple. Walls are practically always built with bricks in rows parallel to the ground because bricks are very strong in compression—they tend to hold up well if you squish them. By contrast, if you push two bricks past each other—ie if they undergo a shearing force—they are more likely to slip past each other in which case, the wall fails. Mortar can help to some extent, but it’s still significantly better to build walls so that the forces involved act perpendicular to the joins between the rows of bricks. If you build a wall with horizontal bricks, like this below,

the forces on each brick all act vertically—neither the brick’s weight nor the contact forces with its neighbours in the rows above and below have any horizontal component. You could remove the mortar and the wall would be perfectly fine (unless something bumped into it). If, however, you build the wall with the long brick edges at an acute angle to the ground, like this,

suddenly the mortar becomes important because the rows of bricks have a tendency to slide past each other. Think about the brick at the lower end of the top row: without mortar, there’s nothing to stop it just sliding off, with the rest of the row following. Even with mortar, the extra shear forces make it more likely to fail.

So, is it possible to build an arch without shear forces? There are three answers to that question: and  . I’ll spend the rest of the article exploring those.

Without loss of generality, I assert that an arch’s sole reason for existence is to support a bridge carrying a railway over a canal.

You can have an arch without shear forces…
…as long as the railway and canal are perpendicular

It’s perfectly possible to build an arch without shear forces, as long as the railway travels at right angles to the canal.

This gives you the simplest type of arch—the archetype, if you like. The bricks are laid in horizontal rows, each row tilted slightly further towards the centre of the canal than the row below it, so that from the viewpoint of a barge, you see something like a semicircle or an ellipse as you approach the bridge, like this: So what happens with the forces here? The weight of the railway acts in vertical planes parallel to the tracks.

It’s easiest to see if we unwrap the cylinder to get a rectangle; the forces become straight lines and the bricks are, quite naturally, laid in lines at right angles to the forces: 

Wrap it back up, and there’s your arch. But what if the canal and railway aren’t perpendicular?

You can’t have an arch without shear forces…
…if you insist on laying the bricks in rows

For a long time, there was only one reasonable solution to having a track cross a river at anything other than right angles: move the track so that it does cross the river at a right angle. With a path or a road that’s, at worst, a bit annoying. Railways, though, don’t do sharp bends: you have to cross the canal at an arbitrary angle. So what’s the problem with that? Why not just build an arch with horizontal bricks like before? The problem is, your arches fall down. The strength of the original arch comes from the two halves leaning against each other, all the way along. If you simply chop off a diagonal from the ends of your arches, there’s nothing holding up the bits that remain—the red areas below:

Fine, then (I hear you say): why not start with a row of bricks perpendicular to the railway? That’s a good idea. Let’s see what happens by doing another unwrapping experiment.

Start with a semi-circular cylinder—you can cut a toilet roll tube in half lengthways if you want to—and chop the ends diagonally (say at 30\degree) to represent the direction of the railway line. Draw grey lines parallel to these end faces along the tube, representing the weight of the railway: this acts in vertical planes parallel to the tracks.

You should have something like this: Before you unroll it, predict what shape you’ll get when you do. I was surprised.

The answer is…Now let’s draw the rows of bricks on.

Find the middle of the top of the arch—midway between the flat edges, and halfway along a line parallel to them. Draw a line through this point, perpendicular to the ends—then as many lines as you like parallel to these representing your bricks:

Hopefully you’ll notice at least one of two things when you fold it back up, or inspect the drawing above:

  • the lines of bricks reaching the ground do so at an angle, rather than parallel to the canal; and
  • the lines of bricks are not perpendicular to the forces except along the first line you drew.

The first point explains why some bridges have bricks at a diagonal. The second of these explains why the answer to the question was ‘no’—there is shear almost everywhere in this type of bridge, and you need to do some clever engineering to counteract it. (‘Do some clever engineering’ is code for ‘build more bridge’.) More to the point, the steeper the angle between the canal and the railway, the more engineering you have to do, and you reach a point where it becomes impossible.

But surely there’s a way to cross a river at a steep angle? Step aside, engineers: we’re going to do some maths.

You can have a bridge without shear forces…
…if you change the shape of the bricks

At this point, we meet the hero of the story: Edward Sang, who is best known for constructing state-of-the-art logarithmic tables and for showing how a spinning top can prove that the Earth rotates.

Around the end of 1835, he presented a proposal to the magnificently named Society for the Encouragement of the Useful Arts, explaining how he would build a bridge. Naturally, he came up with the most mathematician-esque suggestion possible: he redefined the meaning of ‘brick’.

If we want the joins between the stones to be perpendicular to the weights exerted on them, the bricks can’t be rectangular—they need to have curved surfaces. Moreover, there are very precise constraints on the curves where these bricks meet, which I’ll call the joining curves. At every point, they must:

  1. lie in a plane perpendicular to the weight force exerted there; and
  2. lie on the surface of the arch.

We can accomplish that by cunning use of vector calculus. Any curve that lies in two surfaces is (locally) perpendicular to the surfaces’ normal vectors—which in this case are the weight vector at the point and the normal vector to the surface. To find the direction of the curve, we can take the vector product of these. This is not precisely the way that Sang did it; it’s a way I can understand it.

First, we’re going to set up a system of coordinates. Let the $x$-axis (with a unit vector $\mathbf{i}$) run horizontally, perpendicular to the canal, the $y$-axis (with a unit vector $\mathbf{j}$) run horizontally along the centre of the canal, and the $z$-axis (with a unit vector $\mathbf{k}$) be vertical. Here’s what it looks like:

The weight forces lie in vertical planes parallel to the tracks, as before. If our railway runs at an angle $\theta$ to the $x$-axis, these planes satisfy \begin{equation*} y = x \tan(\theta) + y_0, \end{equation*} where $y_0$ is a different constant for each plane. To save on typesetting, I’ll let $\lambda = \tan(\theta)$.

The next question is, what shape should our bridge’s cross-section be? Until now, we’ve only looked at semicircular cylinders, but these are not practical for canals: typically, we might want a railway to cross about three metres above a canal that’s eight metres wide. An arc of a circle is no good (there’s not enough clearance for someone to walk—or, these days, cycle—along the towpath), so our best bet is a semi-ellipse.

A semi-elliptical cylinder with width $2a$ and height $b$ following our coordinate system satisfies the equation \begin{equation*} \frac{x^2}{a^2} + \frac{z^2}{b^2} = 1, \end{equation*} or, if we set $e=b/a$,

$$ e^2x^2 + z^2 = b^2.  \tag{1}$$

Note that $z$ is positive everywhere in our arch, so it can be rewritten as a function of $x$. We won’t do that explicitly, because it’s unnecessarily ugly. However, we will need to know $\mathrm{d} z/ \mathrm{d} x$ in a few paragraphs, so let’s differentiate implicitly here: \begin{equation*} 2e^2 x + 2z \frac{\mathrm{d}z}{\mathrm{d}x} = 0, \quad \text{so} \quad \frac{\mathrm{d}z}{\mathrm{d}x} = -e^2 \frac{x}{z}. \end{equation*} Any point on the surface can then be described by the position vector $x \mathbf{i} + y \mathbf{j} + z(x) \mathbf{k}$. This is important, because we can then find the curves where the force planes intersect the cylinder: we know that $y = \lambda x + y_0$, so our forces follow the curve \begin{equation*} \mathbf{r}(x) = x \mathbf{i} + (\lambda x + y_0)\mathbf{j} + z(x)\mathbf{k}. \end{equation*} The direction of the weight force is parallel to the derivative of this curve at any point. If we differentiate $\mathbf{r}(x)$ with respect to $x$, we get \begin{equation*} \frac{\mathrm{d}\mathbf{r}(x)}{x} = \mathbf{i} + \lambda \mathbf{j} + \frac{\mathrm{d}z(x)}{\mathrm{d}x} \mathbf{k} = \mathbf{i} + \lambda \mathbf{j} – e^2 \frac{x}{z} \mathbf{k}. \end{equation*} Since we’re only interested in the direction, we can multiply through by $z$ to get \begin{equation*} \mathbf{F} \propto z \mathbf{i} + \lambda z \mathbf{j} + -e^2 x\mathbf{k}. \end{equation*} That’s one of the ingredients for our vector product! The other is the normal vector to the surface. That lies in a plane perpendicular to the axis, so it turns out to be the normal vector to the ellipse in (1); with a little work, we get \begin{equation*} \mathbf{n} = e^2 x \mathbf{i} + z \mathbf{k}. \end{equation*} The vector product is \begin{equation*} \mathbf{n} \times \mathbf{F} = -\lambda z^2 \mathbf{i} + (z^2 + e^4 x^2)\mathbf{j} + \lambda e^2 xz \mathbf{k}, \end{equation*} and this is the direction vector of our joining curve at any given $x$ value along it. We eventually want our joining curve in the form $\mathbf{R}(x)$.

Again, we can multiply by anything convenient here, since we only care about the direction: if we pick $-1/(\lambda z^2)$, we get a direction vector of \begin{equation*} \mathbf{i}- \frac{1}{\lambda}\left(1 + \frac{e^4 x^2}{z^2}\right) \mathbf{j} – \frac{e^2 x}{z} \mathbf{k}. \end{equation*} Why would we make it messier like this? Because by making the $\mathbf{i}$ component 1, we have a natural expression for $\mathrm{d}R/\mathrm{d}x$, which we can then integrate. The $\mathbf{i}$ component is trivial—ignoring constants of integration until later, that integrates to $x$. The $\mathbf{k}$ component isn’t much harder: it’s in the form we got from differentiating equation (1), so we know it integrates to $z$, as we would hope. The $\mathbf{j}$ component is not so straightforward, but it’s still possible. To work out \begin{equation*} -\frac{1}{\lambda} \int \left [ 1 + \frac{e^4 x^2}{z^2} \right] \mathrm{d}x, \end{equation*} first (counterintuitively) substitute for the $x^2$: \begin{align*} -\frac{1}{\lambda} \int & \left [ 1 + \frac{e^2 \left(b^2 – z^2\right)}{z^2} \right ] \mathrm{d}x \\ &= -\frac{1}{\lambda} \int \left[ \left(1-e^2\right) + \frac{e^2 b^2}{z^2} \right ]\mathrm{d}x. \end{align*} Now we can substitute for the $z^2$,\begin{equation*} = -\frac{1}{\lambda} \int \left[ \left(1-e^2\right) + \frac{ e^2 b^2}{b^2 – e^2 x^2} \right] \mathrm{d}x, \end{equation*} and integrate, \begin{equation*}= -\frac{1}{\lambda} \left[ \left( 1 – e^2 \right)x + e b\ \mathrm{arctanh} \left( \frac{ex}{b}\right)\right]. \end{equation*} So, finally we get that our curves have the form \begin{align*} \mathbf{R}(x) = \; & \mathbf{r}_0 + x \mathbf{i} \\ & – \frac{1}{\lambda}\left[ \left( 1 – e^2 \right)x + e b\ \mathrm{arctanh} \left( \frac{ex}{b}\right)\right] \mathbf{j} + z \mathbf{k}, \end{align*} where $\mathbf{r}_0$ is any point on the surface of the arch. Written like that, it looks like a bit of a mess. But look below what happens when you plot it—it’s glorious!

And that brings us the Liverpool–Leeds canal, which the west coast mainline crosses at a precipitous $53^\circ$ off of square, near Chorley in Lancashire (///bottle.cuts.space, if you’re a what3words aficionado). And, in the 1840s, they built bridge 74A according to Sang’s principles. As far as I can make out, the masons building bridge 74A didn’t do the calculus. They rolled up paper on a model and drew ellipses in the right place, which I have to concede is a pretty decent pre-WolframAlpha solution to the problem.

You can see this splendid piece of engineering below. If you visit and stop on the towpath, for example to admire the brickwork and take photos, passers-by will engage you with their understanding of the bridge’s history—it was hand-carved by French stonemasons, there’s another one just like it south of Bolton station, probably the same hands—and share your observation that it’s a good bridge.

Next time you’re walking under a bridge, look up. If it’s a good bridge, stop passers-by and explain the principles to them. They might appreciate it.

Bridge 74A on the Liverpool–Leeds canal, constructed like the figure above.

post

Quarto in higher dimensions

When I lecture on the mathematical theory behind board games, Quarto is one of my favourite examples. It’s a commercially available board game for two players: you can see an example board to the right.

Each game piece has four attributes taking one of two values. Each piece is: tall or short; black or white; round or square; and flat-topped or dimpled.

A Quarto board has sixteen spaces arranged in a $ 4 \times 4 $ grid. Players take turns placing pieces on the board and the aim is to be the player who places the fourth in a row, column or diagonal which all match in any one attribute (eg four square pieces or four dimpled pieces).

But there’s a twist! Players don’t choose which piece to play. Instead, they are handed a piece by their opponent. So in fact the aim is to get your opponent to hand you a piece you can use to win.

This is helped by the fact that there are a limited number of pieces. In fact, each combination of attributes is used once. Since there are two possibilities for each attribute and four attributes, that means there are $2^4=16$ pieces.

Having 16 pieces is handy, because a $ 4 \times 4 $ board has $4^2=16$ spaces, meaning that the pieces all fit on the board. I think it wouldn’t be as satisfying for a game that ended in a draw to be left with empty spaces on the board.

Once I was chatting with a student about the similarities between Quarto and noughts and crosses. As he was working on a project about $ 3 \times 3 $ noughts and crosses, it seemed natural to try to modify Quarto to make a $ 3 \times 3 $ version.

To downsize the 16 game pieces, we can ignore one of the attributes—perhaps ‘dimpled or flat-topped’. The three remaining attributes generate $2^3=8$ pieces.

We have a problem! Our $3 \times 3 $ board has $3^2 = 9$ positions, meaning that there is a gap on the board when the game ends in a draw.

Bigger boards

What size boards are possible, if we say that there must be the same number of spaces on the board as there are pieces, and all our pieces are different?

Rather than think about the number of pieces directly, it’s helpful to think about the number of different attributes.

In general, if we have $m$ attributes then we have $2^m$ pieces.

We want this to be the same as the number of spaces on an $n \times n$ board, for some values of $m$ and $n$. For a game we can actually play, we need $m$ and $n$ to be integers.

That is, we require
\[ 2^m=n^2. \]
We can rearrange this to give
\[ m = 2\log_2(n)\text{.}\]
Since the right hand side is multiplied by 2, this tells us that $m$ must be even. To make $m$ an integer, we need $\log_2(n)$ to be an integer, which will only happen if $n$ is a power of 2.

The regular game uses four attributes on a $ 4 \times 4 $ board. Using six attributes would generate 64 pieces and this means using an $ 8 \times 8 $ board. We know this will work as
\[ 2^6 = 8^2 = 64.\]

Of course, we don’t have to stop there. To give another example, fourteen attributes would use a $ 128 \times 128 $ board since
\[ 2^{14} = 128^2 = \text{16,384}\text{.}\]

Keeping track of the pieces

With six attributes, we might imagine adding stripes to half the pieces, and perhaps putting bumps on the side of half. But you might reasonably be wondering how on earth you’d keep track of the 16,384 pieces with fourteen different attributes (weight? smell? attractiveness to cats?).

Rather than try to dream up ever more elaborate ways to distinguish the pieces, a simple system is simply to label each attribute either 0 or 1.

For the standard Quarto set, I choose arbitrarily to label the attributes as follows:

  • white (0), black (1);
  • short (0), tall (1);
  • round (0), square (1);
  • dimpled (0), flat (1).

I can then label the pieces using numbers in this order:

  • colour,
  • height,
  • shape,
  • dimpledness.

This means each piece in the standard set can be represented by a four-digit binary number.

For example, $0100$ would be white, tall, round and dimpled, while $1001$ would be black, short, round and flat-topped.

Now instead of worrying about whether two pieces are the same height, colour or temperature, I can just say the winning condition is met if all pieces in a line share one digit in common. For example, $0100$ and $1001$ are both round, and so both have a third digit $0$.

The standard Quarto pieces are labelled with their binary digits here:

Quarto pieces labelled with binary numbers.

Non-binary thinking

So far we have changed the size of the board, but there is more we can generalise if we wish. Can you imagine three different heights? Three different colours? A ternary number taking values 0, 1 or 2? Then we can design games where attributes take three values.

Since there are now three options for each of our $m$ attributes, we require that
\[ 3^m = n^2\text{,} \]
that is,
\[ m = 2 \log_3(n)\text{.}\]

We still need $m$ to be even, but now we are looking for $n$ to be a power of three.

Good news! We can now design a $ 3 \times 3 $ game of Quarto! This would use $m=2$ attributes and generate nine pieces since $3^m=n^2=9$.

The next viable board would be $ 9 \times 9 $, using four attributes and $ 3^4 = 9^2 = 81 $ pieces.

In general, if each attribute could take $k$ values (each piece is one of $k$ different heights, plays one of $k$ different jingles, etc), then we would be looking for
\[ k^m = n^2,\]
or
\[ m = 2 \log_k(n).\]
That is, we want even $m$ and for $n$ to be a power of $k$.

Higher dimensions

You may have noticed there’s still a pesky number in our equation: the 2, reflecting the fact we are on a two-dimensional game board.

It’s certainly possible to imagine playing on a three-dimensional board. You might have seen a 3D noughts and crosses puzzle, or be imagining Star Trek’s 3D chess.

It’s probably best to think about a $4 \times 4 \times 4 $ Quarto cube as four Quarto boards stacked on top of each other. Before you get your hammer and nails out, you could of course just draw four boards next to each other on a piece of paper. It’s important to remember which board is on top of which other board, though—because diagonal lines moving up the board in the third dimension can get a bit complicated.

A $ 4 \times 4 \times 4 $ cube could be used with 64 pieces having six binary attributes, since
\[ 2^6 = 4^3 = 64.\]
As a human, you might be surprised to learn we don’t have to stop in three dimensions either. Mathematically, it’s perfectly possible to design games in any of $d$ dimensions. Then we would need
\[ k^m = n^d,\]
or
\[ m = d \log_k(n).\]
So we can generate games provided $m$ is divisible by $d$ and $n$ is a power of $k$.

For example, four binary attributes on a $ 2 \times 2 \times 2 \times 2 $ 4-cube (or hypercube) would use $ 2^4 = 2^4 = 16 $ pieces, which is handy because those are the same pieces as standard Quarto.

How do you play on a four-dimensional cube? Start with a $2 \times 2 $ grid, like below. Now draw a second grid next to it and remember this is above the first in 3D space. At this point, we have two small boards that are really layers of a cube:

Two 2x2 grids, the first labelled 'bottom' and the second labelled 'top'

So far it’s not too scary, right? Now draw a second ‘cube’ next to your first (as below). OK, now—squinting a little—just imagine that your new cube is next to the first one in the fourth dimension!

Four 2x2 grids, with columns labelled 'bottom' and 'top' and rows labelled 'Top' and 'Bottom'. Clockwise from top, the lower-left segment of each board is labelled C, nothing, B and A.

Actually, it doesn’t matter if, like me, you can’t visualise this 4D cube! Just like with the 3D cube above we remembered that one square was on top of the other, so we remember the relationships between the squares here. To make it easy, I’ve labelled the top and bottom of the layers in the third dimension using top and bottom and the top and bottom of the layers in the fourth dimension using Top and Bottom.

So the space labelled A in the last diagram is the bottom layer of a 3D cube, and the space labelled B is above it in the third dimension. But also that cube is the bottom layer of a four-dimensional hypercube, and the space labelled C is above A if we move in the fourth dimension. Keeping track of the relationships between the layers avoids the need to think too hard about the fourth dimension, and if you do this you may notice that there are loads of ways to make two-in-a-row.

Game design

Putting all this together, we can come up with a game where pieces have $m$ different attributes, which can each take $k$ different values, on a board in $d$ dimensions, and each side of the board is $ n$ spaces long.

A combination of $m, k, d, $ and $n$ will work if $n = k^p$ for some positive integer $ p$, and if $ m = dp$. For example, choosing $p=7$ when $k=2$ and $d=2$, we get $m=14$ and $n=128$: our Quarto set with 16,384 pieces we discussed earlier.

So I invite you to choose a dimension and number of attributes and generate yourself a game. It may be more complicated and less fun to play than the original set, though!

post

2022 Fields medal winners

When I tell people that I work in mathematics research, I’m often met with a baffled expression. “You can research mathematics? I thought it was already all figured out!” It’s hard enough to explain my own work; you’d be forgiven for thinking that the winners of the Fields medal, the most famous of mathematical achievements, must work on the most incomprehensible and inapplicable mathematics of all.

Aalto University

The 2022 Fields medals were awarded at Aalto University in Helskinki. Image: Wikimedia commons user KFP, CC BY-SA 3.0.

In fact, nothing could be further from the truth. The work of 2022 Fields medal winners James Maynard, June Huh, Maryna Viazovska and Hugo Duminil-Copin shows how modern mathematical techniques can be applied to break open age-old problems, of the kind that anyone can appreciate. Next time I’m asked what mathematics research is about, I will refer to the recipients of the most prestigious maths award in the world for the answer: ingenious solutions to beautiful questions.

James Maynard

James Maynard

James Maynard. Image: Mathematisches Forschungsinstitut Oberwolfach, CC BY-SA 2.0 DE


The very first Fields medal was given at the International Congress of Mathematicians (ICM) in 1936, although the congress itself dates back even further. In fact, it was at the 1912 ICM that Edmund Landau famously stated his four ‘unattackable’ conjectures, which have inspired generations of number theorists, and remain unsolved to this day:

  • There are infinitely many pairs of primes with a difference of 2 (the twin prime conjecture)
  • Every even integer is the sum of two primes (the Goldbach conjecture)
  • There is a prime number between any two consecutive squares
  • There are infinitely many primes of the form $ n^2 + 1 $.

James Maynard works in the field of analytic number theory, which aims to resolve questions like these by estimating the quantity of numbers satisfying these criteria, rather than working with the numbers themselves. In 2016, he proved that there are infinitely many prime numbers that do not contain the digit 7.

When you think about it, this result is truly mindblowing. Since extremely large numbers should contain an extremely large number of 7s (the largest known prime number contains over 2 million!), the probability of containing none at all is absolutely minute.

Among analytic number theorists, he is most well known for his ‘primes in bounded gaps’ theorem. He was able to show that, for some width $ W $, there are infinitely many pairs of primes with a difference of at most $ W $ and if we want to find sets of 3, or 4, or $ m $ primes instead of pairs, there’s always a $ W $ that will give us infinitely many. Although this doesn’t quite imply the famous twin prime conjecture, it is very much in that spirit and the conjecture is firmly in his sights.

June Huh

June Huh

June Huh. Image: Mathematisches Forschungsinstitut Oberwolfach, CC BY-SA 2.0 DE


Given four points on a plane, how many different ways are there to draw a straight line through at least two of them? A quick sketch on paper may suggest that the answer is six, but in fact it depends on the positions of the points: if three are in a row then the answer is four lines, and if they all line up then there is only one.

The question of how many lines may be drawn in general is one of combinatorics: the maths of counting and the specialist subject of June Huh, the first Korean recipient of the Fields medal. One of his most important results tackles the generalisation of the lines and dots problem to higher dimensions.

When we put our points in $n$-dimensional space, we can ask how many lines, planes, and $d$-dimensional hyperplanes they determine. Although the numbers depend on how the points are arranged, June showed that the sequence of numbers displays a pattern known as unimodality: as $ d$ increases, the number of hyperplanes increases to a maximum and then decreases. No bouncing around!

Coloured nodes on a graph

This graph has been coloured so that no two adjacent nodes are the same colour.

June has also made groundbreaking contributions to the field of graph theory, which examines the properties of networks of nodes connected by edges. A famous question asks for the number of ways a graph can be coloured, so that no two adjacent nodes have the same colour. To understand how these numbers behave, graph theorists use chromatic polynomials. June’s results proved that a 40-year-old conjecture about chromatic polynomials is true, including establishing that their coefficients are unimodal.

June’s most important contributions to the field of combinatorics are arguably not the results themselves, but his method of proof, which uses advanced results from the vastly different world of algebraic geometry. By forging a link between these distant fields, he brings the mathematical community closer together, and unveils hints that mathematics still has many secrets left to offer.

Hugo Duminil-Copin

Hugo Duminil-Copin

Hugo Duminil-Copin. Image: Mathematisches Forschungsinstitut Oberwolfach, CC BY-SA 2.0 DE


A magnet at room temperature will stick to a metal sheet. Heat the magnet up, and there will come a moment where it suddenly falls to the ground. This is a dramatic example of a phase transition in physics, but why does it happen?

In order to understand such problems, mathematicians like Hugo Duminil-Copin create mathematical models of the physical systems. In this case, one suitable model is the Ising model, which consists of a lattice of points representing the atoms, and an assignment of $1$s and $-1$s to the points representing the ‘spin’ of the nuclei. The spin of each nucleus is influenced by the spins of its neighbours; by studying the distribution of spins at a particular temperature, we can understand the properties of the magnet itself.

One key problem with these models is that their assumptions are unrealistic. For example, in real magnets the atoms aren’t arranged in perfect grids! One of Hugo’s main results is in demonstrating that phase transitions do not depend on how the atoms are arranged. This means that the assumptions are justified after all.

Hugo has also applied his techniques of statistical physics to seemingly unrelated mathematical conundrums. For example, one of his recent results helps us to count the number of ways to cross a honeycomb from left to right without ever revisiting a cell. This is an easy problem when the honeycomb is small, but extremely tricky for bigger sections. Solving major open problems considered impossible is said to be one of Hugo’s trademarks.

Maryna Viazovska

Maryna Viazovska

Maryna Viazovska. Image: Mathematisches Forschungsinstitut Oberwolfach, CC BY-SA 2.0 DE


Maryna Viazovska’s most famous result is a perfect example of her unique approach. Her work is on sphere packing, which you can find in practice in any supermarket. The most space-efficient way to stack oranges is to fill the first layer, and then work upwards, with the oranges snugly fitting in the gaps from the layer below. Although greengrocers have been stacking oranges like this for centuries, it took almost 400 years to prove that this is the best way, from the initial conjecture of Johannes Kepler to the eventual proof of Thomas Hales. Even then, the proof was long, messy, and required extensive use of computers.

Well-packed oranges

In two-dimensional space, hexagonal packing is the most efficient. In higher dimensional space, it’s less obvious what the most efficient packing arrangement is. Image: Böhringer Friedrich, CC BY-SA 2.5


Maryna’s work is on the higher-dimensional version of the orange problem. Exceptionally space-efficient packing techniques had been discovered for stacking 8-dimensional oranges in 8-dimensional space, and 24-dimensional oranges in 24-dimensional space, known as the $E_8$ lattice and Leech lattice respectively. However, as with our 3-dimensional oranges, it wasn’t clear whether these techniques were optimal.

Researchers knew that the missing link to prove optimality was a ‘magic function’, describing how tightly it’s possible to pack the spheres. The search for this function seemed so hopeless that Thomas Hales admitted ‘I felt that it would take a Ramanujan to find it’. In 2016, this is exactly what Maryna did. She used tools from number theory, including modular forms, to construct the magic function in a way that would have certainly impressed Ramanujan himself. Her technique is all the more impressive for its simplicity, especially when compared to Hales’ lengthy computer-assisted argument.

As the second female recipient and the second Ukrainian recipient of the Fields medal, Maryna is a fantastic ambassador for the global nature of mathematics in the 21st century.

post

Spam calls and number blocking

Nowadays, it seems that everyone is inundated with spam phone calls, which manifest as a spoofed phone number or name to hide the caller’s true identity. At best, answering such calls results in insistent pitches about things like insurance benefits we are missing out on or roofing inspections we are assured we need.

This isn’t terribly different from spam email, for which you might use software to automatically shunt offenders to a junk mailbox, or to /dev/null if you are a particularly frustrated Unix user. For telephone communication, an option is enrolling in ‘no-call lists’ that flag your number as off limits with respect to unsolicited calls from telemarketers and the like. Both of these approaches run the dual risk of allowing an actual spam call to pass through and of mistakenly rejecting a genuine call as spam. In statistics, these types of errors are commonly referred to as false negatives and false positives, respectively. The goal of any spam blocking approach is always to minimise both.

Another and seemingly very popular option is number blocking: the user vets a new call, and, if they deem the number to be spam, they then set their phone to reject all subsequent calls from that number. As successive spam calls are denied, the phone compiles its own increasingly large database of blocked numbers. Some time ago, my phone started to be inundated with spam calls. In reflexively resorting to number blocking, I found myself wondering about the success rate of this method. Aside from the obvious practical significance of the question, it is also mathematically interesting and can be investigated using the tools of combinatorial probability.

Continue reading

post

Formal sums

Some two thousand years ago in ancient Greece, philosophers Aristotle and Zeno asked some interesting thought-provoking questions, including a case of what are today known as Zeno’s paradoxes. The most famous example is a race, known as Achilles and the tortoise. The setup is as follows:

Achilles and a tortoise are having a race, but (in the spirit of fairness) the tortoise is given a headstart. Then Achilles will definitely lose: he can never overtake the tortoise, since he must first reach the point where the tortoise started, so the tortoise must always hold a lead.

Since there are virtually infinitely many such points to be crossed, Achilles should not be able to reach the tortoise in finite time.

This argument is obviously flawed, and to see that we consider the point of view of the tortoise. From the tortoise’s perspective, the problem is equivalent to just Achilles heading towards it at the speed equal to the difference between their speeds in the first version of the problem.

Achilles and the tortoise a distance D apart

Since $\text{distance} = \text{speed}\times\text{time}$, we can say that after time $t$, Achilles has travelled a distance equal to $v_At$ and the tortoise $v_Tt$. The distance between them is
\begin{equation*}
D-v_At+v_Tt = D – (v_A-v_T)t,
\end{equation*}
and so Achilles catches the tortoise—ie the distance between them is $0$—when the time, $t$, is equal to $D/(v_A-v_T)$.

There is another way to see this problem that satisfies better the purpose of this article and directly tackles the problem Aristotle posed. To get to where the tortoise was at the start of the race, Achilles is going to travel the distance $D$ in time $t_1 = D/v_A$. By that time the tortoise will have travelled a distance equal to
\begin{equation*}
D_1 = v_T t_1,
\end{equation*}
which is the new distance between them.

The tortoise is now D+D_1 in front of the original Achilles

Travelling this distance will take Achilles time
\begin{equation*}
t_2 = \frac{D_1}{v_A} = \left(\frac{v_T}{v_A}\right)t_1.
\end{equation*}
Then the tortoise will have travelled a distance
\begin{equation*}
D_2 = v_Tt_2 = \left(\frac{v_T^2}{v_A}\right)t_1
\end{equation*}
and Achilles will cover this distance after time $t_3 = D_2/v_A = (v_T/v_A)^2t_1$.

The tortoise is now D+D_1+D_2 in front of the original Achilles

Repeating this process $k$ times we notice that the distance between Achilles and the tortoise is
\begin{align*}
D_k &= v_T\left(\frac{v_T}{v_A}\right)^{k-1}t_1 \\ &= \left(\frac{v_T}{v_A}\right)^kD.
\end{align*}

Summing up all these distances we get how far Achilles has to move before catching the tortoise: if we call this $D_A$ it’s
\begin{equation*}
D_A =
\lim_{n\to\infty}\sum_{k=0}^{n}D_k =\sum_{k=0}^{\infty}D_k = D \sum_{k=0}^{\infty}\left(\frac{v_T}{v_A}\right)^k.
\end{equation*}
This is probably the simplest example of an infinite convergent sum. In particular, this is the simplest example of a class of sums called geometric series, which are sums of the form
\begin{equation*}
\sum_{k=0}^{n}a^k.
\end{equation*}
If $|a| < 1$, the sum tends to $(1-a)^{-1}$ as $n$ tends to $\infty$ and diverges otherwise, meaning that it either goes to $\pm\infty$, or a limiting value just doesn't exist. By 'doesn't exist', see for example what happens if $a=-1$: we get \begin{equation*} 1 - 1 + 1 - 1 + 1 - 1 + \cdots + (-1)^n \end{equation*} and the sum oscillates between $0$ (if $n$ is odd) and $1$ (if $n$ is even). In our case, $|a|=|v_T/v_A|<1$ so \begin{equation*} D_A = \frac{D}{1-v_T/v_A} = \frac{D v_A}{v_A-v_T}, \end{equation*} which, when divided by the speed of Achilles, $v_A$, gives exactly the time we found before. So Achilles and the tortoise will meet after Achilles has crossed a distance $D_A$ in time $t_A = D_A/v_A$. Thousands of years later, Leonhard Euler was thinking about evaluating the limit of \begin{equation*} \sum_{k=1}^{n}\frac{1}{k^2} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \cdots + \frac{1}{n^2} \end{equation*} as $n\to\infty$, which is named as the Basel problem after Euler’s hometown. This sum is convergent and it equals $\mathrm{\pi}^2/6$, as Euler ended up proving in 1734. He was one of the first people to study formal sums—which we will try to define shortly—and concretely develop the related theory. In his 1760 work De seriebus divergentibus he says

Whenever an infinite series is obtained as the development of some closed
expression, it may be used in mathematical operations as the equivalent of
that expression, even for values of the variable for which the series diverges.

So let’s think about series which diverge. One way a series can diverge is simply by its terms getting bigger. One such example is the sum
\begin{equation*}
\sum_{k=1}^{n}k = 1 + 2 + 3 + 4 + \cdots + n = \frac{n(n+1)}{2},
\end{equation*}
the limit of which when $n\to\infty$ is, of course, infinite.

But now let’s think about the harmonic series,
\begin{equation*}
\sum_{k=1}^{n}\frac{1}{k} = 1 + \frac12 + \frac13 + \frac14 + \cdots + \frac1n.
\end{equation*}
This time, although the terms themselves get smaller and smaller, the series still diverges as $n\to\infty$. But we can still describe the sum and its behaviour. It turns out that
\begin{equation*}
\sum_{k=1}^{n}\frac1k = \ln(n)+\gamma + O(1/n),
\end{equation*}
where $\gamma$ is the Euler–Mascheroni constant, which approximately equals 0.5772, and $O(1/n)$ means ‘something no greater than a constant times $1/n$’. You can see the sum and its approximation here:

Graph

Historically, the development of the seemingly unconventional theory of divergent sums has been debatable, with Abel, who at some point made contributions to the area, once describing them as shameful, calling them “an invention of the devil”. Later contributions include works of Ramanujan and Hardy in the 20th century, about which more information can be found in the latter’s book, Divergent Series.

More recently, a video on the YouTube channel Numberphile was published, and attempted to deduce the ‘equality’
\begin{equation*}
1+2+3+4+\cdots=-\frac{1}{12}.
\end{equation*}
This video sparked great controversy, and indicates one of the dangers of dealing with divergent sums. One culprit here is the Riemann zeta function, which is defined for $\operatorname{Re}(s)>1$ as
\begin{equation*}
\zeta(s)=\sum_{k=1}^{\infty}\frac{1}{k^s}.
\end{equation*}
When functions are only defined on certain domains, it is sometimes possible to ‘analytically continue’ them outside of these original domains. Specifically at $-1$, doing so here gives $\zeta(-1)=-1/12$. The other culprit here is matrix summation—another method to give some value to divergent sums. By sheer (though neat) coincidence, these methods, such as the Cesáro summation method they use in the video, also give $-1/12$!

The main problem is this: at this point we no longer have an actual sum in the traditional sense.

Instead, we have a divergent sum which is formal, and by that, we mean that it is a symbol that denotes the addition of some quantities, regardless of whether it is convergent or not: it simply has the form of a sum.

These sums are not just naive mathematical inventions, instead, they show up in science and technology quite frequently and they can give us good approximations as they often emerge from standard manipulations, such as (as we’ll see) integration by parts.

Applications in physics can be found in the areas of quantum field theory and quantum electrodynamics. In fact, formal series derived from perturbation theory can give very accurate measurements of physical phenomena like the Stark effect and the Zeeman effect, which characterise changes in the spectral lines of atoms under the influence of an external magnetic and electric field respectively.

In 1952, Freeman Dyson gave an interesting physical explanation of the divergence of formal series in quantum electrodynamics, explaining it via the stability of the physical system versus the spontaneous, explosive birth of particles in a scenario where the corresponding series that describes it is convergent. Essentially he argues that divergence is, in some sense, inherent in these types of systems otherwise we would have systems in pathological states. His paper from that year in Physical Review contains more information.

Euler’s motivation

Sometimes, such assignments of formal sums to finite values (constants or functions) can be useful. The fact that they sometimes diverge does not make much difference in the end, if certain conditions are met.

An example that follows Euler’s line of thought as described earlier emerges when trying to find an explicit formula for the function
\begin{equation*}
\operatorname{Ei}(x):=\int_{-\infty}^{x}\frac{\mathrm{e}^t}{t}\, \mathrm{d} t,
\end{equation*}
for which repeated integration by parts yields
\begin{align*}
\operatorname{Ei}(x) = \int_{-\infty}^x\frac{\mathrm{e}^t}{t}\, \mathrm{d} t =& \left[\frac{\mathrm{e}^t}{t}\right]_{-\infty}^{x}-\int_{-\infty}^{x}\mathrm{e}^{t}\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{1}{t}\right) \mathrm{d} t \\
=& \left[\frac{\mathrm{e}^x}{x} – 0\right] + \int_{-\infty}^{x}\frac{\mathrm{e}^{t}}{t^2} \mathrm{d} t \\
= \cdots =& \frac{\mathrm{e}^{x}}{x}\sum_{k=0}^{n-1}\frac{k!}{x^{k}}+O\left(\frac{\mathrm{e}^x}{x^{n+1}}\right),
\end{align*}
where we’ve been able to say $\mathrm{e}^x/x\to 0$ on the second line as $x\to-\infty$. Dividing through by $\mathrm{e}^x$, this allows us to say
\begin{align*}
\mathrm{e}^{-x}\operatorname{Ei}(x) &= \sum_{k=0}^{n-1}\frac{k!}{x^{k+1}}+O\left(\frac{1}{x^{n+1}}\right)\\
&\sim\sum_{k=0}^{\infty}\frac{k!}{x^{k+1}}\text{ as }x\to\infty.
\end{align*}
Now swap $x$ for $-1/x$ in this equation:
\begin{align*}
\mathrm{e}^{1/x}\operatorname{Ei}(-1/x)&\sim\sum_{k=0}^{\infty}k!(-x)^{k+1}\text{ as }x\to0\\
&=-x+x^2-2x^3+6x^4+\cdots.
\end{align*}
As you can see below, this series now diverges as $x\to\infty$, but we still see convergence of the partial (truncated) sums as $x\to0$, even as we add more terms:

Graph

Euler noticed that $\mathrm{e}^{1/x}\operatorname{Ei}(-1/x)$, in its original integral form, solves the equation
\begin{equation*}
x^2\frac{\mathrm{d}y}{\mathrm{d}x}+y=-x
\end{equation*}
(for $x\neq0$). Now here’s the thing: the formal sum
\begin{equation*}
\sum_{k=0}^{\infty}k!(-x)^{k+1},
\end{equation*}
to which $\mathrm{e}^{1/x}\operatorname{Ei}(-1/x)$ is asymptotic as $x\to0$, also (formally) ‘solves’ the same equation for any $x$.

This solution is not unique, and in fact, adding any constant multiple of $\mathrm{e}^{1/x}$ to $\mathrm{e}^{1/x}\operatorname{Ei}(-1/x)$ would still solve the equation; and the resulting solution would still be asymptotic to the same formal sum.

However, the coefficients of the powers of $x$ are unique. So there may be something in the formal sum that can give away the actual solution of the equation (which is often difficult to find via standard methods—unlike formal solutions that are easier to compute like the one above), at least up to some class of solutions and under certain conditions. In fact, this seems to actually be the case, at least for certain classes of formal sums—the ones that attain ‘at most’ factorial over power rate of divergence.

Solving a differential equation

To elaborate further, let’s consider one more example, the differential equation
\begin{equation}
-\frac{\mathrm{d}y}{\mathrm{d}x}+y = \frac{1}{x}, \quad \text{where } y(x)\to 0 \text{ as }x \to \infty.
\label{fs1}
\tag{*}
\end{equation}
Thinking about the boundary condition there, we could substitute in the (formal) sum of powers of $x$ which decay away as $x\to\infty$,
\begin{equation*}
y(x) = \sum_{k=0}^\infty a_k x^{-k-1} = \frac{a_0}{x} + \frac{a_1}{x^2} + \frac{a_2}{x^3} + \cdots.
\end{equation*}
Doing so, we get
\begin{align*}
-\frac{\mathrm{d}}{\mathrm{d}x}\left[\sum_{k=0}^{\infty}a_kx^{-k-1}\right]+\sum_{k=0}^{\infty}a_kx^{-k-1} &= \frac{1}{x}
\\ \implies \sum_{k=0}^{\infty}(k+1)a_kx^{-k-2}+\sum_{k=0}^{\infty}a_kx^{-k-1} &=
\frac{1}{x} \\
\implies a_0x^{-1}+\sum_{k=0}^{\infty}\big[(k+1)a_k+a_{k+1}\big] x^{-k-2} &=\frac{1}{x}.
\end{align*}
Then for our differential equation to be satisfied the coefficients have to satisfy
\begin{equation*}
a_0=1 \qquad \text{and}
\end{equation*}

\begin{equation*}
(k+1)a_k+a_{k+1}=0\implies a_{k+1} = -(k+1)a_k
\end{equation*}
which recursively means that $a_k=(-1)^kk!$ and our formal sum solution is
\begin{equation*}
y(x) = \sum_{k=0}^{\infty}(-1)^k k!x^{-k-1}.
\end{equation*}
Now that we have a sum that solves the equation formally, we can obtain an actual solution assuming that it is asymptotic to the sum we found as $x\to\infty$ by using the repeated integration by parts result
\begin{equation*}
\int_{0}^{\infty}\mathrm{e}^{-xs} s^k \,\mathrm{d} s = k! x^{-k-1} \text{ for }x>0,
\end{equation*}
which implies
\begin{align*}
y(x) &= \sum_{k=0}^{\infty}(-1)^{k}k!x^{-k-1} \\
&= \sum_{k=0}^{\infty}(-1)^k\int_{0}^{\infty}\mathrm{e}^{-xs}s^k \,\mathrm{d} s \\
&= \int_{0}^{\infty}\mathrm{e}^{-xs}\sum_{k=0}^{\infty}(-1)^ks^k \,\mathrm{d} s.
\end{align*}
How is that helpful? Well, for $s:|s|<1$ we know that \begin{equation*} \sum_{k=0}^{\infty}(-1)^ks^{k} = 1-s+s^2+\cdots = \frac{1}{1+s}, \end{equation*} by the formula for geometric series for $a=-s$ from our discussion of Achilles and the tortoise. This is a nice function on the real line, having all the fine properties that we need in order to define \begin{equation*} y(x)=\int_{0}^{\infty}\frac{\mathrm{e}^{-xs}}{1+s}\,\mathrm{d}s, \end{equation*} which is the solution to our differential equation, \eqref{fs1}, we are looking for, and is also asymptotic to the formal sum $\sum_{k=0}^{\infty}(-1)^k k!x^{-k-1}$ as $x\to\infty$: Graph

Notice that any linear combination of these formal sums will result from the same linear combination of the respective convergent (for $s:|s|<1$) series $1-s+s^2-s^3+\cdots$ inside the integral. In conclusion, it is possible to obtain a solution in closed form to a differential equation just by finding a formal power series to which the solution is asymptotic.

Not just reinventing the wheel

The aforementioned example is, of course, quite simple and trying to find a solution in the way we just described might look like we’re reinventing the wheel using modern-era technology. However, the true potential of the method described above can be seen in nonlinear equations, to which we generally cannot find solutions in standard ways. In my own research I used formal sums to study an equation with applications in fluid mechanics.

In one of the first talks I gave about this topic, I remember noticing several of my peers tilting their heads in distrust when I mentioned that the emerging sums are divergent. This reaction was almost expected and for obvious reasons. It took an hour-long talk and several questions later to convince them that the mathematics involved is genuine.

Controversial as it may sound, at first sight, this concept is even more realistic than imaginary numbers, which are simply symbols with properties that we just accept and use. The idea is that, although imaginary, these numbers can demonstrably give us, when interpreted properly, very real results such as solutions to differential equations like
\begin{equation*}
\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+y=0.
\end{equation*}
The same is true for formal sums too.

Why do we assign actual numbers to formal sums in the first place? Because they are sometimes easier to work with and can lead to interesting results (such as solutions to differential equations) if interpreted properly. The underlying mechanisms should be well-defined mathematical processes and well-understood in order to avoid any serious mistakes when working with such sums. An example of erroneous use of such sums is Henri Poincaré’s attempt to solve the three-body problem in order to win the King Oscar prize in 1889. He managed, however, in the next decade to spark the development of chaos theory. But that’s for another time.

post

Things with silly names

When I talk about maths research, one thing I often take care to point out to budding mathematicians is to bear in mind just how cool it is that you could be the first person to discover a new thing, or prove an interesting theorem that nobody else has managed to prove before. And while the sheer joy of mathematical discovery and solving the puzzle should be motivation in itself for the future generation of mathematicians, I also make it very clear that it’s often the case that whoever discovers a thing also gets to pick a name for it.

You could take the boring option and name it after yourself (assuming your name is unusual enough that people would know it was actually you—who knows which Bernoulli was responsible for what?). But a much better option is to name your mathematical concept, object or theorem after something amusing, strange or otherwise silly.

In that spirit, here’s my roundup of mathematical things with silly names—you may have your own favourites, but these are a few I’ve been being amused by recently.

Continue reading

post

On the cover: Some assembly required

Sam Palmer is a creator of mathematical digital art, which he shares through Twitter, on Reddit and on his website. He uses the open source sketching software Processing to generate both static and animated graphics with beautiful geometric themes. Well, a lot of people think they’re beautiful; some people actually find some of his gifs a little unsettling:

Reddit comments: "This is trippy, I love it" and "It's like a brain massage"

Reddit comments: "Excuse me, please don't do that to my eyes" and "Imma go throw up in my closet"

The Chalkdust issue 16 cover art in particular was inspired by a transitional animation of falling Koch snowflakes, in a piece named Some assembly required:

The Koch snowflake is a particularly beautiful famous shape and is one example of the group of mathematical objects known as fractals. Fractals are self-similar objects, and have really interesting mathematical properties, including non-integer dimension. For the particular example of the Koch snowflake, this shape has infinite perimeter but finite area. Other popular fractals include Sierpinski’s triangle, the Menger sponge, and the Mandelbrot set.

The Koch snowflake shape is built up iteratively by starting from an equilateral triangle, and adding smaller equilateral triangles (specifically a ninth of the area of the previous triangle added) in the middle of each side of the shape:

Koch snowflake generation

Fancy trying out the software? The Processing programming language is based on Java, and when you open it up for the first time there are links to some great tutorials and templates. It’s available for Windows, MacOS and Linux, so you have no excuse.

Need inspiration for a title for your artwork? How about The Koch Ness Monster, Kochodile Rock or even BBC News at Ten O’Koch

post

A day in the life: universities

Mathematicians are a diverse bunch. As a group, they come from different experiences and backgrounds; they have different hobbies and aspirations; different preferences for computing software; and, as we recently found out at Chalkdust HQ, wildly different opinions on how to pronounce ‘scone’.

One thing they do all have in common is a love of mathematics. We strongly believe that a mathematician is anybody who does maths and wants to define themselves as a mathematician. And mathematics itself is full of variety and diversity. People can study a whole universe of different things! These things can be entirely unrelated—or even cooler, they can seem entirely unrelated but actually end up having deep connections.

This is why we can find it difficult to describe exactly what mathematics is: because it encompasses so many different things. Maybe we should stick with saying that mathematics is ‘what mathematicians do’. But then…what do mathematicians do?

Fear not: Chalkdust has the answer! Although our definition of mathematician leaves room for all sorts of people, doing all sorts of things, day to day, we’ve decided to focus on those doing mathematics in a university setting, and give you an insight into the daily life of four people at different stages of their mathematical careers:

  • Piper, an undergraduate student at Durham University,
  • Jessica, a PhD student at RWTH Aachen, in Germany,
  • Smitha, a postdoctoral researcher at the University of Liverpool,
  • Helen, a professor of applied mathematics and the head of department for mathematics at UCL.

Piper Lane

I’m Piper, I like random walks on the theoretical beach and contemplating the viability of gravity being represented by a particle. I am a mathematics undergrad, having just completed my first year at Durham University.

Piper at her desk, the screen says "Important maths things"

Piper hard at work

My busiest and thus most exciting day of each week begins at 8am on a Thursday, with a cold shower and a hearty college breakfast. I am usually joined by fellow mathematician friends with whom I converse, and subsequently walk, to the maths and computer science department above the main science site for the first session of the day; a 9am calculus tutorial. I have found tutorials to be useful in helping consolidate the current content, as I am given the opportunity to collaborate with other students through new problems, with as much or as little guidance by the tutor as needed.

Afterwards, I have little time to make my way to the learning centre for calculus and linear algebra lectures, followed immediately by analysis and then linear algebra tutorials back atop the science site hill.

I find a space around the maths department to work for the next hour, often at a bench outside if the weather is nice, at which I can finally eat lunch while watching the designated pre-recorded probability videos for the week. Although there is a lot of content to cover per module, each lecturer provides detailed pre-recorded videos alongside a coherent set of lecture notes, as well as regular live lectures presented on Zoom and/or in person. This gives a range of options for students depending on their circumstances regarding Covid-19 and learning preferences.

Over time I have found that the best methods for my personal learning have been to annotate and edit the existing lecture notes during in-person lectures, and then watch the pre-recorded content videos in my own time afterwards should I fail to understand any part of the lecture.

A lit up computer setup in the dark.

Analysis I vibes

My last class of the day is a two-hour programming tutorial, during which students work through a practical sheet of ascending difficulty depending on the week. This is usually more relaxed due to existing programming experience accelerating my understanding.

Since my day ends around 5pm with limited time between sessions, I head straight back to college for dinner and a rendezvous with my friends. Evenings are often my favourite part of the day, as I still have some time to go for a walk with my friends around the city, enjoying the scenery and the challenge of navigating Durham in the dark. After returning to college, I like to order a toastie from the buttery—the gem of Trevelyan College—before winding down with a cup of tea and a good book.

Jessica Wang

Hi! I am a first-year maths PhD student at RWTH Aachen, Germany. I work in the field of symplectic topology, specifically on billiards. I did my bachelor’s and master’s degrees at Durham University, before coming to Aachen. On a typical day, I go to the office in the mornings and start working, which could be either studying new materials or doing research. Since I am still at the start of my PhD, I have so far spent most of my time here learning background knowledge.

Jess sitting in front of a computer. Wikipedia is on the screen.

Jessica fixing the mistakes on Wikipedia

On Tuesdays, all the members in our geometry and analysis chair have lunch together with our secretary (who deals with all the admin work in our chair) to discuss maths-related stuff such as organising or travelling to conferences, social events etc. The afternoons are pretty much the same as the mornings, except that I go to ballet classes two to three times a week during late afternoons. My supervisor and I would sometimes meet to discuss my progress.

Jess in front of a fancy looking building

Jessica strikes a pose

Apart from being in the office, I attend some seminars and lectures during term time which are closely related to my research. I find the lectures particularly useful since I came from a very different academic background (my master’s thesis was on representation theory of braid groups, which is pretty algebraic).

Lastly, and probably the most fun part of my job, is attending conferences; you get to listen to lots of interesting talks, know more people from your field, and travel a bit around the city where the conference is in!

Smitha Maretvadakethope

The difficulty of trying to capture a day in the life of a postdoc at any institution, whether it’s at the University of Liverpool in the department of mathematical sciences (where I work) or halfway across the world, is that in academia every day is different from the last. So, if you read this and think “that is not a representative sample, Smitha” I challenge you to tweet @chalkdustmag what a typical day looks like for you!

Smitha standing in front of the gate to DAMTP in Cambridge.

Smitha on loan to Cambridge

Picture this:

It is a rainy day and campus is littered with students and teaching staff navigating their hectic time tables. That’s when I arrive on the scene. I arrive in the department and have a friendly chat with the building manager and everyone’s favourite cleaning lady. A warm greeting makes every day brighter.

Once the computer is up and running, the first port of call is the staff common room, a key fixture for everyone’s caffeine and tea addictions. This includes a water cooler for all obligatory water cooler talk.

Hydrated and ready? Time for the real work.

Stage 1: Emails, emails, emails (Part 1).

Stage 2: Chat to office mates.

Stage 3: Actually do some work. Depending on the day this is likely to be meetings, analysis/debugging, reading papers, or academic writing. The most time-consuming.

Has it hit 12pm yet? If yes, it’s time to drop everything and huddle around for lunch. Postdocs from other groups might drop by, PhD students might drop by. Anyone is welcome.
1pm? Back to work.

Stage 4: Emails, emails, emails (Part 2).

Stage 5: Chat to colleagues about emails that are confusing/strange/newsworthy.

Stage 6: Attend/organise seminars/workshops and become a pro at all things Microsoft Teams and Zoom.

Stage 7: Repeat stages 2–3.

Is it past 4pm yet? If yes, move onto stage 8. If not, rinse and repeat stages 6–7 until it is.

Stage 8: Emails, emails, emails. (Part 3)

Stage 9: Turn off that computer and call it a day. A good work–life balance is important self care.

And… that’s it! If I’ve missed anything, please note that it’s not within the scope of this article.

Helen Wilson

My name is Helen Wilson. I’m a professor of applied mathematics and currently the head of department for UCL mathematics.

I have school-age children, so my working hours are unusual. I share the school runs with my husband: today I’m doing the morning one, so at 8.30 I’m walking across town from the school to the station. This is good ‘refocus’ time for me, when I move my head out of the family space and into planning my work day.

Helen sitting at her desk, laptop out

Helen fighting fires

I reach the office soon after 10 and deal with the most urgent emails (I get about 100 a day so I have to stay on top of them). I’ll be giving a tutorial at 11, so I print the example sheet and look over it, and check my whiteboard markers still work.

The tutorial is really enjoyable. The students are giving presentations, which they have worked on in their groups, so my role is very much in the background. Some students give an excellent display of how their question can be solved, and I need to do hardly anything. Others have struggled to find a way through, so I give small nudges to encourage them to work it out live at the board.

Back in my office, I log the tutorial attendance and eat lunch (I almost always eat at the desk these days) and plough through more emails. There’s a freedom of information request, which needs prompt action by law, so I get on with that one sharpish!

In the afternoon I work on teaching allocation for the next academic year. This is a hugely complex process, starting with gathering the views of the teaching staff, and ending with checking everyone’s happy with the changes.

Glasses and a pen on top of some paper

Helen wears glasses now

Last thing, I have a Zoom meeting with my most senior PhD student. He’s had his viva and is working on the minor corrections, so he doesn’t need much from me—but it’s nice to check things are progressing well and also hear how he’s getting on with his new job.

I leave the office around 5 and get home at 6.30, only about half an hour after my husband gets in with the children. It’s all about them for the next few hours, but when they go to bed at about 9 we get some adult time. We eat and chat and then typically both do a bit more work. I save easy, unstressful jobs for this time of day—writing student references, approving expense claims, anything that takes time but doesn’t require serious decision-making.

Lettuce know!

What does a day in your life look like? If you spend a lot of time doing mathematics in a non-university setting, tell us what you do at @chalkdustmag or contact@chalkdustmagazine.com, and your story could be featured in a future issue!