contents
3
Page 3 model
9
What’s hot and what’s not
17
Who’s your font soulmate?
25
A canal runs through it Colin Beveridge is the troll under the bridge
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Dear Dirichlet
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Quarto in higher dimensions Peter Rowlett trials the new three-column format
32
Puzzles
34
On the cover: computer-generated fractal art Read more about Sam Palmer’s artwork
36
The crossnumber Win a £100 Maths Gear goody bag
38
A day in the life... ...of four mathematicians at different stages of their careers
44
Things with silly names eg Katie Steckles
49
Dog leads, mirrors, and Hermann Minkowski Donovan Young crashes his cone into another cone
56
Cryptic crossword
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Spam calls and number blocking Michael Wendl has recently been involved in an accident that wasn’t his fault
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How to make sci-fi fans angry
68
The big argument Is Matlab better than Python?
70
Formal sums Nik + Alexandrakis = ?
76
Top ten: mathematical clothes
4 In conversation with Martin Hairer Bea Taylor and Jakob Stein talk to the 2014 Fields medallist
10 I don’t like science fiction Peach Semolina takes a quantum leap
64 Fields medal winners 2022 Albert Wood did not win a Fields medal autumn 2022 ⋅
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Dear reader, We don’t know about you but we are all very excited for hot mathematician autumn. The team Thùy Dương “TD” Ðặng Madeleine Hall Ellen Jolley Vivienne Leech Sophie Maclean Jonathan Pain Matthew Scroggs Belgin Seymenoğlu Jakob Stein Bea Taylor Adam Townsend Clare Wallace Cover artwork Sam Palmer
d c a b l n e
chalkdustmagazine.com contact@chalkdustmagazine.com @chalkdustmag chalkdustmag chalkdustmag @chalkdustmag@mathstodon.xyz Chalkdust Magazine, Department of Mathematics, UCL, Gower Street, London WC1E 6BT, UK.
For strolling through the park in our berets, for sipping chai lattes, for kicking up clouds of fallen leaves. For gambolling through fields, and daydreaming under bridges. For singing along to Taylor Swift in the car. For the days to get gradually shorter, and shorter. For gazing at the stars, and daydreaming about quantum teleportation. For finding some hygge, snuggled up beside the fire, with our fave magazine for the mathematically curious. For tossing Chalkdust onto the fire, to keep warm a little longer. For putting our ear muffs on, when the news comes on TV. For blocking everyone from calling us, to save on the phone bill. For musing why the ham has gone hairy in the fridge. Don’t worry, it’s only fifty sleeps until Christmas. Kind regards, The Chalkdust team
Acknowledgments We would like to thank: all our authors for writing wonderful content; our sponsors for allowing us to continue making the magazine; Helen Wilson, Helen Higgins, Luciano Rila and everyone else at UCL’s Department of Mathematics; King Ming Lam for taking pictures of Martin Hairer; everyone at Achieve Fulfilment for their help with distribution. The emoji used throughout this issue are Twemoji, licensed under CC-BY 4.0. ISSN 2059-3805 (Print). ISSN 2059-3813 (Online). Published by Chalkdust Magazine, Dept of Mathematics, UCL, Gower Street, London WC1E 6BT, UK. © Copyright for articles is retained by the original authors, and all other content is copyright Chalkdust Magazine 2022. All rights reserved. If you wish to reproduce any content, please contact us at Chalkdust Magazine, Dept of Mathematics, UCL, Gower Street, London WC1E 6BT, UK or email contact@chalkdustmagazine.com
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Why don’t stars collapse? Since ancient times, humans have gazed up at the night sky and seen much the same picture—a constant amid the turmoil down on Earth. We now know that chaos lies behind that calm persona: stars are in fact massive, hot balls of turbulent gas, continuously forging new elements through nuclear fusion. Given everything they have going on behind the scenes, have you ever wondered why they seem so. . . stable? Of course, as with all the beautiful things in the universe, we ultimately have maths to thank for this. Stars are held together by just two simple differential equations: 𝐺𝑀𝑟 d𝑃 = −𝜌 𝟤 d𝑟 𝑟
and
d𝑀𝑟 d𝑟
= 𝟦π𝑟 𝟤 𝜌
where 𝑟 represents the distance of a given point to the centre of the star.
The first equation is a statement of hydrostatic equilibrium: the pressure force of the gas pushing the star apart must exactly match the gravitational forwards pulling it in on itself. 𝑃 is the pressure, 𝑀𝑟 is the mass of the sphere contained by the radius 𝑟 , and 𝜌 is the (non-constant) density of the star. The left-hand side, therefore, represents pressure force, while the right-hand side is the gravitational force (familiar perhaps from Newton’s law of gravitation). The second equation effectively just defines 𝑀𝑟 (the right-hand side is simply the surface area of a sphere, multiplied by the density). In fact, with only two more differential equations (and an equation of state, such as the ideal gas law), we can describe a star’s full structure, including temperature and luminosity: d𝑇 𝟥 𝜅𝜌 𝐿𝑟 , =− d𝑟 𝟦𝑎𝑐 𝑇 𝟥 𝟦π𝑟 𝟤
d𝐿𝑟 d𝑟
= 𝟦π𝑟 𝟤 𝜌𝜀.
These have a bit more physics involved but they express the principles of thermal equilibrium and energy equilibrium. Isn’t it amazing that we can break such a massive, complex, intimidating structure down to just a few lines with a bit of mathematical modelling?
Adam Block, CC BY-SA 3.0
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IN CONVERSATION WITH
Martin Hairer Bea Taylor and Jakob Stein talk to the Fields medallist about his life, his work and his advice to his younger self.
King Ming Lam
I
t’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 (see pages 64–67). 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. 𝟦
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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 ▲ Ringo Starr and three lesser musicians 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. 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. Renate Schmid, CC BY-SA 2.0 DE
▲ Ernst Hairer, professor of mathematics at the University of Geneva
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.” autumn 2022 ⋅
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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 King Ming Lam of systems, in particular stochastic partial differ▲ Spoilers for Martin’s next paper ential 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 𝟤
𝜕 𝟤 ℎ 𝜆 𝜕ℎ 𝜕ℎ = 𝜈 𝟤 + ( ) + 𝜂(𝑥, 𝑡), 𝜕𝑡 𝟤 𝜕𝑥 𝜕𝑥
where 𝜆 and 𝜈 are constants representing how inflammable the paper is. Importantly here, 𝜂(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 𝟨
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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.”
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 Wikimedia Commons user Dmnk.saman, CC BY-SA 4.0 schools, which are lecture courses ▲ The pretty part of Imperial College London 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.” 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. autumn 2022 ⋅
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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 DuminilCopin, 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.”
Wikimedia Commons user Monsoon0, CC BY-SA 4.0
▲ 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)
“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 underrepresented 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.”
Bea Taylor Bea is a PhD student in the computer science department at UCL. Since joining the Chalkdust team she’s found that the other editors find the phrase ‘it needs more white space’ very triggering.
Jakob Stein Jakob is a postdoc and mathematician from London and works mainly in differential geometry. In his spare time, he likes to draw, and think about mathematics in art.
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what’s
Maths is a fickle world. Stay à la mode with our guide to the latest trends.
hot and what’s not
▲ HOT
▼ NOT
Artwork by artificial intelligence
Shooting for the moon ”Nasa ground control to Artemis… we’re going to have to delay this launch by a few days….”
Not to brag, but we got priority access to Dall.E ~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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Artwork that makes any sense
Landing among the stars See our page 3 model
“I need you to draw a badly-assembled ham sandwich being cut in half so that all layers are cut equally… got it?”
▲ HOT Dropping a new paper in front of 30 people
▼ NOT
▲ HOT Chalkdust slowly becoming more and more like Cosmo Issue 17 is full of sexy numbers, interviews with theorems and ‘LaTeX: expectation vs reality’ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Chalkdust slowly becoming more and more like Cosmo At least they haven’t given Prof Dirichlet’s slot to Scott Mills yet #lovetheshow
▼ NOT
CHALKD UST
A great turnout to witness the outcome of your many years of academic work
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
▲ HOT
Dropping paper in front of 30 million people Ah well, it’s hardly a cardinal sin… (he’s C of E)
▼ NOT
scorpio the ne n: w you
▼ NOT
Agree? Disagree? Tell us on Twitter @chalkdustmag + more advice online
Pipe flow The hosepipe ban is still in force down south, after a punishingly hot summer ~~~~~~~~~~~~~~~~~~~~~~~~~~~~
▲ HOT
Miss Flo
Supporting maths students with their mental health
At least Holly & Phil will be pleased we moved on
▲ HOT autumn 2022 ⋅
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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.
Quantum plumbing Even the least discerning among you will have worked out by now that this article is about quantum teleportation. And I promise that I will get onto that soon (I’m familiar with Chekhov’s gun). But first, there is a bit of groundwork that needs doing. I’m going to assume that the average reader has no knowledge of quantum mechanics, so we must start at the beginning. What on earth is it?! At its core, quantum mechanics is the study of very small things, usually subatomic particles and other things on a subatomic scale. I’m sure that’s not as exciting as you were expecting. Unfortunately, marketing and PR experts have got their hands on the word ‘quantum’ and warped it beyond recognition. But part of the incredible nature of quantum mechanics is how real and everyday it is, and yet it has so many ethereal and otherworldly consequences. A quick side note on nomenclature (though I’m sure an expert would disagree with me): ‘quantum physics’ and ‘quantum mechanics’ are terms which are largely used interchangeably, at least at the level on which we’ll be discussing the subject. Quantum decorators, quantum plumbers, and quantum interior designers alas don’t exist. Yet. The first idea introduced in most quantum mechanics courses is the concept of quantisation of energy. Put simply, it means that particles can only absorb energy (as radiation) in specific packets, named quanta. Hence the name. ▲ A quantum plumber Another key tenet of quantum mechanics is the idea of wave–particle duality. This asserts that every quantum entity can be treated as either a wave or as a particle, and these two types of object are equivalent. But these two concepts are not relevant for teleportation. The key principle we need is the uncertainty principle.
autumn 2022 ⋅
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Schrödinger’s ET The uncertainty principle has broken into the realms of popular science via the famous thought experiment: Schrödinger’s cat. In this, you are asked to imagine a box in which there is a cat, some radioactive material, and acid. The box is set up with equipment so that if the radioactive material decays, the acid is released and the cat dies. The uncertainty comes in because radioactive decay is a random event and so, without opening the box, the cat could be either dead or alive and only by opening the box do you learn which. I can’t go any further without asking you all to please make sure this remains a thought experiment as I strongly oppose animal cruelty. Do it with an alien or something instead. (Does it count as animal cruelty? Am I going to get sued for this? Do aliens actually exist?) Schrödinger’s thought experiment suggests that quantum mechanics works in much a similar way to the cat in the box and, up to a point, it does. Quantum entities can be in one of a discrete set of states just as the cat can be dead or alive. It’s also true that until you measure the quantum entity, you don’t know which state it’s in, just as you don’t know whether the cat is dead or alive until you open the box. But this is where the model starts breaking: the cool thing about quantum mechanics is that the reason you don’t know what state an entity is in until you measure it is that, unlike the cat, it isn’t in any state until you measure it.
▲ Schrödinger’s cat isn’t
happy about being forced This takes a bit of time to wrap your head around so I’ll explain it in a into a box with some acid bit more detail. Up until you measure a quantum entity, it exists in a and radioactive material superposition (ie a combination) of all possible states. When measured, the entity collapses into a given state and remains in that state from then on. There are set probabilities of it collapsing into each state on measurement and this is dependent on the composition of the superposition of the states.
This is objectively awesome.
Quantum of solace What does this look like in practice? A single bit of quantum information (a qubit) is denoted by the quantum state | 𝜓 ⟩. I’m going to mainly deal with two dimensions here because we don’t need any more for the scope of this article, but it is possible to consider more than two. Let’s consider an twodimensional orthogonal basis: {|𝟢⟩ , |𝟣⟩}. Don’t worry if this seems scary, hopefully it’ll become clear in good time. Those of you familiar with linear algebra will have heard the term ‘orthogonal basis’ before. In linear algebra, vectors are orthogonal if they are perpendicular to each other. A basis is a set of vectors which can be added together in different combinations to form any other vector. An orthogonal basis is therefore a set of perpendicular vectors from which any other vector can be formed, and as a side effect of the orthogonality, this can be done in precisely one way. The definition of an orthogonal basis in quantum mechanics is just the same. |𝟢⟩ and |𝟣⟩ are two quantum states which can be added to form any other quantum state in a unique way. 𝟣𝟤
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This isn’t where the parallels between quantum mechanics and linear algebra end though. In fact, qubits can be described as vectors. So that is what we shall do.
| 𝜓⟩
We can represent our basis as 𝟣 |𝟢⟩ = ( ) , 𝟢
𝟢 |𝟣⟩ = ( ) . 𝟣
▲ A qubit
This means that a general qubit can be written
▼ Qbert
𝑎 | 𝜓 ⟩ = 𝑎 |𝟢⟩ + 𝑏 |𝟣⟩ = ( ) , 𝑏
where 𝑎 and 𝑏 are real numbers. Quantum states are normalised, so in two dimensions, this means that 𝑎𝟤 + 𝑏 𝟤 = 𝟣. One final bit of notation here (it’s a surprise tool that will help us later). We can define ⟨𝜙 | as: †
⟨𝜙 | = |𝜙⟩ = 𝑎∗ ⟨𝟢| + 𝑏 ∗ ⟨𝟣| = (𝑎∗
𝑏∗ ) ,
Steven Miller, CC BY 2.0
where 𝑎∗ is the complex conjugate of 𝑎. Linear algebra fans will recognise that † is the adjoint of a vector. Linear algebra haters need not worry—the word ‘adjoint’ will never be spoken of again (in this article). The above is a general linear algebra result, but for quantum states specifically, 𝑎 and 𝑏 are real so we can simplify this to ⟨𝜙 | = (𝑎 𝑏) .
Lights, camera, action! Now we have our qubits, we can do things with them! We generally refer to ‘things we do to qubits’ as ‘actions’ (apparently ‘things we do to qubits’ isn’t snappy enough). Don’t ask me how these actions are carried out. Just trust me that they are. Some of these actions can be represented by matrices, and to find the effect of the action on a qubit, we simply multiply the vector representing the qubit by this matrix. Do you want some examples? Of course you want some examples. Here are a few of my favourites: • First up, we have an action which I shall refer to as 𝑋 . 𝑋 swaps the |𝟢⟩ and |𝟣⟩ components and is represented by the matrix 𝟢 ( 𝟣
𝟣 ). 𝟢
• The second action keeps the |𝟢⟩ component as is, but flips the sign of the |𝟣⟩ component. I shall call this action 𝑍 , because this is how it is generally known. 𝑍 is represented by the matrix 𝟣 𝟢 ). ( 𝟢 −𝟣
• The final action I want to share with you is so special that it is named after a person: the Hadamard gate. It’s not a coincidence, by the way, that the use of the word ‘gate’ gives logic and computing vibes. I’ll let you think about why that might be for yourself. The Hadamard gate, 𝐻 , is defined by the (normalised) matrix 𝟣 𝟣 𝟣 ). ( √𝟤 𝟣 −𝟣 Graphically, this is actually a reflection in the mirror line at an angle of π/𝟪 to the 𝑥 axis. autumn 2022 ⋅
𝟣𝟥
A state of collapse As I mentioned before, we can work out the probability that the qubit (say 𝜓 ) is in a certain state (say 𝜙 ) when measured. To do this, we take the inner product: ⟨𝜙 | 𝜓 ⟩ (I told you that notation would be useful!) and that equals the probability. A quick demonstration, for those who are new to this: |𝜓 ⟩ = 𝑎 |𝟢⟩ + 𝑏 |𝟣⟩ |𝜙⟩ = 𝑐 |𝟢⟩ + 𝑑 |𝟣⟩
⟹ ⟨𝜙 |𝜓 ⟩ = (𝑐
𝑎 𝑑) ( ) = 𝑐𝑎 + 𝑑𝑏. 𝑏
Because of our normalisation, this is a real number less than or equal to 1.
Once we’ve conducted this measurement (just like with the actions, please don’t ask me how that happens in practice), the qubit collapses into whichever state the result of the measurement is. If it makes more sense to you, you can think of this as it becoming a qubit that, when measured, gives the same result as the initial measurement with probability 1. If that makes less sense to you, then forget I said it.
Spooky action at a distance There’s one last weird quantum-y thing we need to learn about before we can (finally) talk about quantum teleportation and this is, yet another, really exciting thing. I’m really trying not to swear in this article, but honestly this stuff is !#? ! cool. This final thing is quantum entanglement. Up until now, we’ve been dealing with single qubits (remember that |𝜓 ⟩ = 𝑎 |𝟢⟩ + 𝑏 |𝟣⟩ is a single qubit, in much the same way that the coordinate (𝑥, 𝑦) describes a single point). But what if you have two qubits? We first have to expand our basis to allow for this. Our new basis needs to consider the possibility that both the first and the second qubits could be |𝟢⟩ or |𝟣⟩. Our new basis is therefore { |𝟢⟩ |𝟢⟩ , |𝟢⟩ |𝟣⟩ , |𝟣⟩ |𝟢⟩ , |𝟣⟩ |𝟣⟩ }
We’re going to write | 𝜓 ⟩ | 𝜙⟩ as |𝜓 𝜙⟩, just to tidy things up a bit. Our basis now becomes { |𝟢𝟢⟩ , |𝟢𝟣⟩ , |𝟣𝟢⟩ , |𝟣𝟣⟩ }
What about a general vector |𝜓 𝜙⟩ = 𝑎 |𝟢𝟢⟩ + 𝑏 |𝟢𝟣⟩ + 𝑐 |𝟣𝟢⟩ + 𝑑 |𝟣𝟣⟩ ? Here, there are two cases. Firstly, |𝜓 𝜙⟩ could be a product vector. This means there are vectors | 𝜓 ′ ⟩ = 𝛼 |𝟢⟩ + 𝛽 |𝟣⟩ and | 𝜙 ′ ⟩ = 𝛾 |𝟢⟩ + 𝛿 |𝟣⟩ such that | 𝜓 ′ ⟩ | 𝜙 ′ ⟩ = |𝜓 𝜙⟩, ie we want ( 𝛼 |𝟢⟩ + 𝛽 | 𝟣⟩ )( 𝛾 |𝟢⟩ + 𝛿 | 𝟣⟩ ) = 𝑎 |𝟢𝟢⟩ + 𝑏 |𝟢𝟣⟩ + 𝑐 |𝟣𝟢⟩ + 𝑑 |𝟣𝟣⟩ .
Multiplying out gives
𝛼𝛾 |𝟢𝟢⟩ + 𝛼𝛿 |𝟢𝟣⟩ + 𝛽𝛾 |𝟣𝟢⟩ + 𝛽𝛿 |𝟣𝟣⟩ = 𝑎 |𝟢𝟢⟩ + 𝑏 |𝟢𝟣⟩ + 𝑐 |𝟣𝟢⟩ + 𝑑 |𝟣𝟣⟩ .
So we need 𝛼𝛾 = 𝑎, 𝛼𝛿 = 𝑏 , 𝛽𝛾 = 𝑐 and 𝛽𝛿 = 𝑑 . Some quick simultaneous equations will convince you that there are only certain values of 𝑎, 𝑏, 𝑐 and 𝑑 for which such 𝛼, 𝛽, 𝛾 and 𝛿 exist. If a vector is not a product vector, then it is referred to as ‘entangled’. 𝟣𝟦
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We can now apply actions on multiple qubits at once. For example, the action 𝐶𝑋 acts on a pair of qubits by applying 𝑋 to the second qubit if and only if the first qubit is |𝟣⟩.
What if we apply an action to entangled vectors? This is where the magic happens. If two qubits are entangled, then an action applied to one qubit will have an effect on the second qubit. I’m going to repeat that, for dramatic effect. If two qubits are entangled, then an action applied to one qubit will have an effect on the second qubit. Mind = blown.
A quick side note: Some smarty-pantses among you may now be about to protest. You might say that if that’s true then information could travel faster than the speed of light. You could separate the two qubits, and changing one would cause an instantaneous change to the other, no matter far away they are. This is true, but your inference that this means information is transmitted that fast is sadly wrong. There’s a thing called the no-signalling principle, which means that it’s impossible for this change to be detected without additional information. I’ll let you discover the joys of that for yourself though, because now we have all the tools we need to quantum teleport.
Beam me up, Scotty
??
Suppose there are two people, let’s call them Adam and Bea. Adam wants to teleport a qubit |𝛼⟩ = 𝑎 |𝟢⟩+𝑏 |𝟣⟩ to Bea. Adam hasn’t measured his state so he doesn’t know what the state is. How does he do the teleportation? Firstly, Adam and Bea will need to have met beforehand and each taken one qubit from an entangled state. Not just any old state though. Specifically, this one (known at the Bell state): | 𝜙 + ⟩ = ( |𝟢𝟢⟩ + |𝟣𝟣⟩ )/√𝟤. This state is entangled, but if you don’t believe me you can check for yourself, as a fun little exercise! Adam takes the first of these qubits (we’ll call it | 𝜙𝟣+ ⟩) and Bea takes the second (we’ll call it | 𝜙𝟤+ ⟩). Adam adds his mystery qubit before the qubit he got from the entangled state. Between them, Adam and Bea have the state |𝛼⟩| 𝜙 + ⟩ = 𝟣 ( 𝑎 |𝟢⟩ + 𝑏 |𝟣⟩ )( |𝟢𝟢⟩ + |𝟣𝟣⟩ ) √𝟤
= 𝟣 ( 𝑎 |𝟢𝟢𝟢⟩ + 𝑎 |𝟢𝟣𝟣⟩ + 𝑏 |𝟣𝟢𝟢⟩ + 𝑏 |𝟣𝟣𝟣⟩ ). √𝟤
Remember, Adam has the first two qubits and Bea has the third. Adam now applies 𝐶𝑋 to his two qubits. The new state is 𝐶𝑋 (|𝛼⟩| 𝜙𝟣+ ⟩)| 𝜙𝟤+ ⟩
=
𝟣 (𝑎 |𝟢𝟢𝟢⟩ + 𝑎 |𝟢𝟣𝟣⟩ + 𝑏 |𝟣𝟣𝟢⟩ + 𝑏 |𝟣𝟢𝟣⟩ ). √𝟤
Next, Adam applies 𝐻 to the first of his qubits (the one that was originally | 𝛼⟩). A bit of fiddly multiplying out, tells us that the state is now
𝟣 (𝑎 |𝟢𝟢𝟢⟩ + 𝑎 |𝟣𝟢𝟢⟩ + 𝑎 |𝟢𝟣𝟣⟩ + 𝑎 |𝟣𝟣𝟣⟩ + 𝑏 |𝟢𝟣𝟢⟩ − 𝑏 |𝟣𝟣𝟢⟩ + 𝑏 |𝟢𝟢𝟣⟩ − 𝑏 |𝟣𝟣𝟣⟩ ). 𝟤
?
?
▲ Adam and Bea each take one qubit from the Bell state
? Q Q?
?
▲ Adam adds Q = |𝛼⟩ to his qubit
Q? 𝐶𝑋
? ?
?
▲ Adam applies 𝐶𝑋 to his qubits
𝐻
? ?
?
▲ Adam applies 𝐻 to the first of his qubits
autumn 2022 ⋅
𝟣𝟧
Recognise this? No? How about if I write it like this: 𝟣 ( |𝟢𝟢⟩ ( 𝑎 |𝟢⟩ + 𝑏 | 𝟣⟩ ) + | 𝟢𝟣⟩ ( 𝑏 |𝟢⟩ + 𝑎 | 𝟣⟩ ) + | 𝟣𝟢⟩ ( 𝑎 |𝟢⟩ − 𝑏 | 𝟣⟩ ) + | 𝟣𝟣⟩ ( − 𝑏 |𝟢⟩ + 𝑎 | 𝟣⟩ ))? 𝟤
My qubits are 11
𝟣 ( |𝟢𝟢⟩ |𝛼⟩ + |𝟢𝟣⟩ (𝑋 𝟤
?
▲ Adam measures his qubits ? QQ
Q
What about now? Still no? Okay, I’ll put you out of your misery. The new state is actually
𝑍𝑋
▲ Bea applies 𝑋 , 𝑍 , 𝑍 𝑋 , or nothing to her qubit to get Adam’s original qubit
|𝛼⟩) + |𝟣𝟢⟩ (𝑍 |𝛼⟩) + |𝟣𝟣⟩ (𝑋 𝑍 |𝛼⟩) ).
Now, finally, Adam measures his qubits. Remember, this collapses the state. So if Adam measures 𝟢𝟢, then Bea’s qubit must be |𝛼⟩. Similarly, if Adam measures 𝟢𝟣, then Bea’s qubit must be 𝑋 |𝛼⟩. If Adam measures 𝟣𝟢, Bea’s qubit must be 𝑍 |𝛼⟩, and if Adam measures 𝟣𝟣, Bea’s qubit must be 𝑋 𝑍 |𝛼⟩. Adam tells Bea what his measurement is. She then knows what her qubit is in terms of |𝛼⟩, and simply applies the inverse of the action acting on it. The inverse of 𝑋 is conveniently 𝑋 , and similarly the inverse of 𝑍 is 𝑍 . The inverse of 𝑋 𝑍 is just 𝑍 𝑋 . After Bea has done that, the teleportation is complete! She now has the qubit Adam started with, despite being arbitrarily far away!
Around the world in eighty milliseconds? There we have it! Teleportation! Allons-y! Geronimo! OK, so I’ll admit that knowing this is all very well but I’ve not told you how to actually do this in reality, which is largely because I don’t actually know. Plus even if I had explained how to do this in practice, at best, you could teleport quantum entities one by one. I’m sure some of you were hoping to learn how to teleport, say, animals but that brings in all sorts of ethical and philosophical quandaries on top of the practicalities. I started by saying how I wasn’t missing out on science fiction because I had mathematics, and I won’t ask whether or not you agree. I know the audience I’m talking to. Sci-fi nerds among you may in fact feel victorious. Perhaps you would point out that at least in science fiction it is possible to teleport people, and that if I want to truly experience teleportation, I have no choice but to engage with sci-fi. But to that, I have one thing to say. That sounds excruciatingly painful. And teleporting sounds pretty uncomfortable too.
▲ (In James Doohan’s voice:) “Captain, I cannae get ethical approval for beaming you up.”
Peach Semolina To protect the author from the science fiction community, we have given them the pseudonym Peach Semolina and are not publishing Sophie Maclean’s real name. Ah. Well at least we didn’t give away her email address or Twitter handle.
c sophiethemathmo@gmail.com a @sophiethemathmo 𝟣𝟨
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quiz
Who’s your font soulmate?
𝟣
First things first: how do you meet?
R You run into each other when exploring an abandoned castle ♠ Your eyes meet across the room at a ball hosted by your neighbour ▲ You both reach for the last souvlaki
𝟤
What would most impress you on a first date?
R An outstanding collection of suits of armour ♠ A perfect display of manners ▲ An in-depth knowledge of ancient myths
𝟥
What would your Mastermind specialist subject be?
R Castles of Europe ♠ The British peerage ▲ Triangles
𝟦
Where should your date take you to dinner?
R Somewhere dark and atmospheric ♠ Somewhere busy and fashionable – with space for your chaperones, of course ▲ A picnic – pass the olives Mostly R: 𝔐𝔞𝔱𝔥𝔣𝔯𝔞𝔨 Gothic, traditional and a little bit dramatic, your ideal date is Mathfrak. It’s got it all: perfect for Lie algebras and for writing a ‘keep out’ sign on your castle door. First date? See you at the cemetery!
𝟧
What’s your party trick?
R Your uncanny affinity with bats ♠ Your flawlessly elegant dancing skills ▲ Your ability to cut spanakopita into perfect triangles
Mostly ♠: MAT HCAL Your genteel sensibilities make you a perfect match for Mathcal. Just like you, it’s elegant, sophisticated, and obsessed with the marital prospects of everyone it knows. If you want to impress, brush up on your quadrille steps: one, two, three, four…
𝟨
How would you react to someone flirting with your date?
R Invite them to a ‘friendly’ tour of your family crypt ♠ Challenge them to a duel ▲ Triangles
Mostly ▲: GREEK Maybe you’re a big fan of feta cheese, or maybe you just really love triangles. Either way, the Greek alphabet is the one for you. Just like a straight line, we hope your relationship can be extended indefinitely, and that you both live hap𝜋 ly ever after! autumn 2022 ⋅
𝟣𝟩
◀ A canal runs through it Colin Beveridge barges in and admires some curious railway bridges
F
or 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
An arched bridge with horizontal bricks (Gors-Opleeuw, Belgium) Johan Neven, CC BY 2.0
𝟣𝟪
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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 photo below, or the rows were all slanted—set at a nonzero angle to the horizontal, as in the photo on the next page.
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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. autumn 2022 ⋅
𝟣𝟫
If you build a wall with horizontal bricks, like this below,
YES! You can have an arch
without shear forces. . . . . . as long as the railway and canal are perpendicular
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.
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:
So, is it possible to build an arch without shear forces? There are three answers to that question: YES! NO !
and YES! . 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. 𝟤𝟢
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Wrap it back up, and there’s your arch. But what if the canal and railway aren’t perpendicular?
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NO! You can’t have an arch
You should have something like this:
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?
Before you unroll it, predict what shape you’ll get when you do. I was surprised. The answer is. . .
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:
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:
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°) 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. autumn 2022 ⋅
𝟤𝟣
Hopefully you’ll notice at least one of two things when you fold it back up, or inspect the drawing above:
constraints on the curves where these bricks meet, which I’ll call the joining curves. At every point, they must:
• the lines of bricks reaching the ground do so at an angle, rather than parallel to the canal; and
1. lie in a plane perpendicular to the weight force exerted there; 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.
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 𝒊) run horizontally, perpendicular to the canal, the 𝑦 -axis (with a unit vector 𝒋) run horizontally along the centre of the canal, and the 𝑧 -axis (with a unit vector 𝒌 ) be vertical. Here’s what it looks like:
YES! 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 𝟤𝟤
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The weight forces lie in vertical planes parallel to the tracks, as before. If our railway runs at an angle 𝜃 to the 𝑥 -axis, these planes satisfy 𝑦 = 𝑥 tan(𝜃) + 𝑦𝟢 ,
where 𝑦𝟢 is a different constant for each plane. To save on typesetting, I’ll let 𝜆 = tan(𝜃). 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
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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 𝑥 𝟤 𝑧𝟤 + = 𝟣, 𝑎𝟤 𝑏 𝟤 or, if we set 𝑒 = 𝑏/𝑎, 𝑒𝟤 𝑥 𝟤 + 𝑧 𝟤 = 𝑏𝟤 .
(f)
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 d𝑧/d𝑥 in a few paragraphs, so let’s differentiate implicitly here: 𝟤𝑒 𝟤 𝑥 + 𝟤𝑧
d𝑧 = 𝟢, d𝑥
so
𝑥 d𝑧 = −𝑒 𝟤 . d𝑥 𝑧
That lies in a plane perpendicular to the axis, so it turns out to be the normal vector to the ellipse in equation (f); with a little work, we get 𝒏 = 𝑒 𝟤 𝑥𝒊 + 𝑧𝒌.
The vector product is 𝒏 × 𝑭 = −𝜆𝑧 𝟤 𝒊 + (𝑧 𝟤 + 𝑒 𝟦 𝑥 𝟤 )𝒋 + 𝜆𝑒 𝟤 𝑥𝑧𝒌,
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 𝑹(𝑥).
Again, we can multiply by anything convenient here, since we only care about the direction: if we pick −𝟣/(𝜆𝑧 𝟤 ), we get a direction vector of 𝒊−
𝑒𝟤𝑥 𝑒𝟦𝑥 𝟤 𝟣 𝒌. (𝟣 + 𝟤 ) 𝒋 − 𝜆 𝑧 𝑧
Why would we make it messier like this? Because by making the 𝒊 component 1, we have a natural expression for d𝑅/d𝑥 , which we can then integrate.
Any point on the surface can then be described by the position vector 𝑥𝒊 + 𝑦𝒋 + 𝑧(𝑥)𝒌 . This is important, because we can then find the curves where the force planes intersect the cylinder: we know that 𝑦 = 𝜆𝑥 + 𝑦𝟢 , so our forces follow the curve
The 𝒊 component is trivial—ignoring constants of integration until later, that integrates to 𝑥 . The 𝒌 component isn’t much harder: it’s in the form we got from differentiating equation equation (f), so we know it integrates to 𝑧 , as we would hope.
The direction of the weight force is parallel to the derivative of this curve at any point. If we differentiate 𝑟(𝑥) with respect to 𝑥 , we get
𝑒𝟦𝑥 𝟤 𝟣 − ∫ [𝟣 + 𝟤 ] d𝑥, 𝜆 𝑧
𝒓(𝑥) = 𝑥𝒊 + (𝜆𝑥 + 𝑦𝟢 )𝒋 + 𝑧(𝑥)𝒌.
d𝑧(𝑥) d𝒓(𝑥) 𝑥 = 𝒊 + 𝜆𝒋 + 𝒌 = 𝒊 + 𝜆𝒋 − 𝑒 𝟤 𝒌. d𝑥 d𝑥 𝑧
Since we’re only interested in the direction, we can multiply through by 𝑧 to get 𝑭 ∝ 𝑧𝒊 + 𝜆𝑧𝒋 + −𝑒 𝟤 𝑥𝒌.
That’s one of the ingredients for our vector product! The other is the normal vector to the surface.
The 𝒋 component is not so straightforward, but it’s still possible. To work out
first (counterintuitively) substitute for the 𝑥 𝟤 : 𝑒 𝟤 (𝑏 𝟤 − 𝑧 𝟤 ) 𝟣 − ∫ [𝟣 + ] d𝑥 𝜆 𝑧𝟤
𝟣 𝑒𝟤 𝑏𝟤 = − ∫ [(𝟣 − 𝑒 𝟤 ) + 𝟤 ] d𝑥. 𝜆 𝑧
Now we can substitute for the 𝑧 𝟤 ,
𝑒𝟤 𝑏𝟤 𝟣 = − ∫ [(𝟣 − 𝑒 𝟤 ) + 𝟤 ] d𝑥, 𝜆 𝑏 − 𝑒𝟤𝑥 𝟤 autumn 2022 ⋅
𝟤𝟥
and integrate, 𝟣 𝑒𝑥 = − [(𝟣 − 𝑒 𝟤 ) 𝑥 + 𝑒𝑏 artanh ( )] . 𝜆 𝑏
So, finally we get that our curves have the form 𝑹(𝑥) = 𝒓𝟢 + 𝑥𝒊 𝑒𝑥 𝟣 − [(𝟣 − 𝑒 𝟤 ) 𝑥 + 𝑒𝑏 artanh ( )] 𝒋 + 𝑧𝒌, 𝜆 𝑏
where 𝒓𝟢 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° 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 on the right. 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.
Colin Beveridge Colin is a mathematician with nothing to prove. He wrote Cracking Mathematics and The Maths Behind, and blogs regularly at flyingcoloursmaths.co.uk
d colinbeveridge.co.uk a @icecolbeveridge 𝟤𝟦
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feature
Bridge 74A on the Liverpool–Leeds canal, constructed like the figure opposite.
autumn 2022 ⋅
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Moonlighting agony uncle Professor Dirichlet answers your personal problems. Want the prof’s help? Contact c deardirichlet@chalkdustmagazine.com
Dear Dirichlet,
When I was growing up, I really enjoyed being part of the Girl Gui des, so I’ve recently signed my eight-year-old daughter up to the local unit for under10s. I think she’s having lots of fun but now I can’t stop her wan dering off randomly every time we go out . She’s not very quick but I’m just worried she’s going to get herself seriou sly lost... aren’t they supposed to be lea rning orienteering or something?
— Sasha Badges, Tamworth
Whatever you do, do not buy her flying lessons! If my suspicions are correct, in 2D I guarantee she will always come home (Polya’s recurrence theorem) but in 3D there is no hope! Why do I say this? Small steps, random walks... I’m afraid you’ve got yourself a Brownieinmotion.
■
dirichlet says:
Dear Dirichlet,
g classes over In preparation for going away to uni, I’ve been taking cookin nd but my weeke next pans the summer. I’m going to buy some pots and enough big a halls room has limited storage and I’m worried about getting s for small pan which won’t take up all the space. Know any good recipe
— Blanche Talmonds, Worcester
pots?
Funnily enough, I have the perfect recipe to solve both of your problems. Take all your existing recipes and place them in order of size. Now buy the smallest pan that’s bigger than everything you want to cook. That’s it! I call it ‘lim soup’. (Salt quanto basta.) ■
𝟤𝟨
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dirichlet says:
d ea r d ir ic hlet
Dear Dirichlet,
ort vehicles these They don’t half have some strange designs for public transp given a sharp been has days. I see the 253 is no longer a New Routemaster, but namics? My only pointy nose at the front and back, a bit like a ship. Is this aerody nose took out a trip on it ended in disaster when, turning a tight corner, the front traffic light. Needless to say I walked home. — Oyster lover, London
Sounds like you got on the rhombus! Diamond time again I have written to bus designers offering my expertise just for my letters be obtusely ignored. Kite frankly – and we shouldn’t go (trap)ezionum – at least without the nose they have what they should’ve started with: a wrecked angle. Although I don’t ask square this happened, I think in Manchester they do it well – I always hype up (n)arrowyellowtrams.
■
dirichlet says:
Dear Dirichlet,
My niece got her GCSE results back last week and I am thrilled to say they are good... probably? Honestly I can ’t tell any more, they’re just a bunch of numbers. Can you tell me what the y mean in terms of the letters we’ re used to? (See attached.)
— Ed Agsell , Wrotham
I can confirm that the numbers attached do not form a valid credit card number or a sequence from the TV show Lost. Stand in an open space and rotate anti clockwise. Now ask your niece to hold her results sheet and to walk clockwise around you. If you now look at the marksheet as it passes, aha! ABCs as promised! A simple consequence of a ‘retro grade’ orbit.
■
dirichlet says:
Dear Dirichlet,
g UCNCF with A fancy brunch bar has opened up next door and they’re offerin GESG. Any ideas? Otherwise I’ll just order the toast.
— Colin Door, Framwellgate Moor
■
dirichlet says:
That’ll be Caesar salad with scrambled eggs. Bit weird.
autumn 2022 ⋅
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Quarto in higher dimensions Peter Rowlett is gonna need a bigger board
W
hen 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 × 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 combina𝟤𝟪
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tion of attributes is used once. Since there are two possibilities for each attribute and four attributes, that means there are 𝟤𝟦 = 𝟣𝟨 pieces. Having 16 pieces is handy, because a 𝟦 × 𝟦 board has 𝟦𝟤 = 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 × 𝟥 noughts and crosses, it seemed natural to try to modify Quarto to make a 𝟥×𝟥 version. To downsize the 16 game pieces, we can ignore one of the attributes—perhaps ‘dimpled or flat-topped’. The three remaining attributes generate 𝟤𝟥 = 𝟪 pieces. We have a problem! Our 𝟥 × 𝟥 board has 𝟥𝟤 = 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 𝟤𝑚 pieces.
We want this to be the same as the number of spaces on an 𝑛 × 𝑛 board, for some values of 𝑚 and 𝑛. For a game we can actually play, we need 𝑚 and 𝑛 to be integers.
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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. That is, we require 𝟤𝑚
=
𝑛𝟤 .
We can rearrange this to give 𝑚 = 𝟤 log𝟤 (𝑛).
Since the right hand side is multiplied by 2, this tells us that 𝑚 must be even. To make 𝑚 an integer, we need log𝟤 (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 × 𝟦 board. Using six attributes would generate 64 pieces and this means using an 𝟪 × 𝟪 board. We know this will work as 𝟤𝟨 = 𝟪𝟤 = 𝟨𝟦.
Of course, we don’t have to stop there. To give another example, fourteen attributes would use a 𝟣𝟤𝟪 × 𝟣𝟤𝟪 board since 𝟤𝟣𝟦 = 𝟣𝟤𝟪𝟤 = 16,384.
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?).
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 overleaf. autumn 2022 ⋅
𝟤𝟫
0001 0011 0101
0000 0010 0100
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
𝑚 = 𝟤 log𝟥 (𝑛).
0111
0110
that is,
𝟥𝑚 = 𝑛𝟤 ,
1001 1011
1010
1000
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 × 𝟥 game of Quarto! This would use 𝑚 = 𝟤 attributes and generate nine pieces since 𝟥𝑚 = 𝑛 𝟤 = 𝟫.
1101
1100
The next viable board would be 𝟫 × 𝟫, using four attributes and 𝟥𝟦 = 𝟫𝟤 = 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
𝟥𝟢
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1111
1110
or
𝑘 𝑚 = 𝑛𝟤 , 𝑚 = 𝟤 log𝑘 (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 twodimensional 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 × 𝟦 × 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 × 𝟦 × 𝟦 cube could be used with 64 pieces having six binary attributes, since 𝟤𝟨 = 𝟦𝟥 = 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 or
𝑘 𝑚 = 𝑛𝑑 ,
𝑚 = 𝑑 log𝑘 (𝑛).
So we can generate games provided 𝑚 is divisible by 𝑑 and 𝑛 is a power of 𝑘 .
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For example, four binary attributes on a 𝟤 × 𝟤 × 𝟤 × 𝟤 4-cube (or hypercube) would use 𝟤𝟦 = 𝟤𝟦 = 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 × 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:
top
bottom
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! top
bottom
Top
c
a
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.
4D ↕ Bottom
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.
b
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!
Peter Rowlett Peter is a reader in the mathematics group at Sheffield Hallam University, where among other things he teaches a module on game theory and recreational mathematics. His research investigates teaching and learning of mathematics at university level. He blogs at the Aperiodical.
d peterrowlett.net a @peterrowlett autumn 2022 ⋅
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PUZZLES
Looking for a fun puzzle but not got time to tackle the crossnumber? You’re on the right page.
Pirme nerbums Each of the entries in the grid is a prime number that is an anagram of the number given in the clue. For example, 1 across could be 179, 197, 719, or 971 as these are the anagrams of 179 that are prime. 1
2
4
3
5
6
9
8
7 10
12 14
11
13
15
16
17
18 19
20 23
21 24
22
25 27
26
Across
Down
1 3 5 6 8 10 12 14 16 18 19 21 23 25 26 27
1 2 3 4 7 9 11 12 13 14 15 17 20 21 22 24 25 27
179 467 199 37 778 13 113 157 149 139 13 133 79 124 379 337
127 779 16 37 17 778 113 113 19 356 11 124 17 137 337 79 29 337
Arrange the digits Put the numbers 1 to 9 (using each number exactly once) in the boxes so that the sums are correct. The sums should be read left to right and top to bottom ignoring the usual order of operations. For example, 𝟦 + 𝟥 × 𝟤 is 14, not 10.
+ − = 0 𝟥𝟤
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× × ×
+ − = 3
÷ ÷ ÷
+ − = 0
=𝟦
= 𝟥𝟧 = 𝟣𝟪
+ × = 96
÷ ÷
−
+ − = 0
− × ×
÷ + = 7
=𝟧
=𝟣 =𝟧
puzzles
Crossing words Arrange the letters above the grid in the empty squares so that each of the long rows and columns contains a word. AITV K
O
F
E
S
T E
L
EIRR T
T
Z
O
T
O M
ACCEEEEIINRRSTTT
AIILMNORU S
T
S
T
T
S
P
E
L
T D
S
C
A
R
A
V
X
A
N
B
T
F
S S
N
D
S
G
I hate that number The integers from 1 to 9 (inclusive) have been written on slips of paper and put into a hat. Nine people— Annie, Betty, Ceri, Daisy, Eorann, Ffion, Grace, Harmony, and me—are going to take it in turns to pick one of the numbers. (All the exclamation marks in this puzzle are expressions of excitement, not factorials.) First, Annie picks her number and says: “Hooray! A prime number!” Betty then picks her number and says: “A square number, not bad.” Ceri picks next: “Interesting, the sum of digits of the square of this number is 7. Daisy says: “I hope I pick my favourite number.” Then, after picking a number, she exclaims: “Oh no! This means Ceri must have already picked my favourite number.” Eorann picks a number: “It’s a factor of 14! I love 14!” Ffion picks next: “Look Eorann, it’s another factor of 14!” Grace’s turn: “It’s a factor of 12! I love 12 nearly as much as Eorann loves 14!” Harmony takes a number: “I’m very happy, I got nine!” I then refuse to take a number, because I know it’s going to be my least favourite number. What is my least favourite number? autumn 2022 ⋅
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𝟥𝟦
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o n the cover
◀ On the cover...
Sam Palmer is a creator of mathematical digital art, which he shares through Twitter (a @sjpalmer1994), on Reddit (reddit.com/u/sjpalmer94), and on his website ( d sjpalmer.art). 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:
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. You can see some of the frames opposite. 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:
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 f BBC News at Ten O’Koch? autumn 2022 ⋅
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The crossnumber sponsored by
#15, set by Humbug
1
2
4
3
5 9
8 10
11
12 14
17
18 23
15 19
20
21
24
28 30 34
35
38 41
39
31 37
36 40
42
43
45 48
22
25
29 33
46 49
13
16
27
26
32
7
6
44
47 50
Four of the across clues and four of the down clues are false. All the other clues are true. There is only one solution to the completed crossnumber. Solvers may wish to use the OEIS, Python, a slide rule, etc to (for example) obtain a list of cube numbers, but no programming should be necessary to solve the puzzle. As usual, no numbers begin with 0. To enter, send us the sum of all the digits in the row marked by arrows by 14 April 2023 via the form on our website ( d chalkdustmagazine.com). Only one entry per person will be accepted. Winners will be notified by email and announced on our blog by 1 May 2023. One randomly-selected correct answer will win a £100 Maths Gear goody bag, including non-transitive dice, a Festival of the Spoken Nerd DVD, a dodecaplex puzzle and much, much more. Three randomly-selected runners up will win a Chalkdust T-shirt. Maths Gear is a website that sells nerdy things worldwide. Find out more at d mathsgear.co.uk 𝟥𝟨
⋅ chalkdust
puzzles
Across
Down
Four of these clues are false
Four of these clues are false
1 One less than 1D.
(3)
1 One less than 1D
(3)
3 A multiple of 11111.
(5)
2 Two times 10A.
(3)
6 Each digit of this number (excluding the first) is one less than the previous digit.
(3)
3 Two times 38D.
(4)
8 A multiple of 111.
(3)
4 A factor of 1A.
(2)
9 111 more than 8A.
(3)
5 Two times 1D.
(4)
6 Two times 7D.
(3)
7 A prime number.
(3)
10 The highest common factor of 48A and this (3) number is 6. 12 An even number.
(3)
14 Two times 4D.
(2)
16 A factor of 19A.
(2)
17 The sum of this number’s digits is 11.
(4)
19 Greater than 1A.
(3)
21 The sum of this number’s digits is 13.
(4)
23 A square number.
(2)
25 One quarter of 23A.
(2)
26 Two times 28A.
(2)
27 The product of 24D and 25D.
(5)
28 Not a prime number.
(2)
29 Each digit of this number is a factor of 38D. (2) 30 One quarter of 29A.
(2)
32 An anagram of 17A.
(4)
34 15 less than 10A.
(3)
36 An anagram of 21A.
(4)
38 A multiple of 3 whose final digit is the same (2) as the final digit of 12A. 39 A square number.
(2)
41 Two times 41D.
(3)
43 The sum of this number’s digits is 10.
(3)
45 Two times 47A.
(3)
47 A multiple of 9.
(3)
48 Two times 42D.
(3)
49 A multiple of 11111.
(5)
50 A multiple of 55.
(3)
11 Two times 13D.
(3)
13 A multiple of 111.
(3)
15 Two times 14A.
(2)
16 A number with four distinct prime factors.
(2)
17 A multiple of 11111.
(5)
18 A square number.
(2)
20 The product of 19A and 34A.
(5)
21 Five more than 29D.
(2)
22 A multiple of 11111.
(5)
24 30 more than than 6D.
(3)
25 A prime number.
(3)
29 Five more than 21D.
(2)
31 A factor of 25D.
(2)
33 Two times 11D.
(3)
34 A factor of an anagram of 6D.
(2)
35 Greater than 15D.
(2)
37 Two times 33D.
(3)
38 Two times 3D.
(4)
40 A multiple of 1111.
(4)
41 Two times 48A.
(3)
42 Two times 41A.
(3)
43 A multiple of 55.
(3)
44 A number less than 200.
(3)
46 The sum of the digits in the row marked by arrows.
(2)
autumn 2022 ⋅
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Piper undergraduate student
Jessica PhD student
Smitha postdoc
Helen professor
𝟥𝟪
⋅ chalkdust
a day in the life
a d ay in the life
M
athematicians 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 autumn 2022 ⋅
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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.
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.
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. 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 𝟦𝟢
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▲ Piper hard at work
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.
a d ay in the life
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.
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. 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!
▲ 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
▲ Jessica strikes a pose
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Smitha Maretvadakethope
Hydrated and ready? Time for the real work.
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!
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.
▲ Smitha on loan to Cambridge
And. . . that’s it! If I’ve missed anything, please note that it’s not within the scope of this article.
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. 𝟦𝟤
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▲ Everything flows (especially coffee)
a d ay in the life
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. 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.
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.
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. Last thing, I have a Zoom meeting with my most senior PhD student. He’s had his viva
▲ Helen wears glasses now
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! f autumn 2022 ⋅
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Things with silly names Katie Steckles doesn’t understand why any mathematical phenomenon would ever have a not-silly name
W
hen 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.
▲ A hedgehog 𝟦𝟦
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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.
▲ An astroid
▲ A deltoid
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Hedgehog (geometry) The joy of a Wikipedia page with ‘(geometry)’ in the title is that it usually means there’s something that shares a name with a non-geometrical thing, making it necessary to disambiguate. In which case, hedgehog is the perfect name for this object and nobody can persuade me otherwise.
Most of the joy of mathematical hedgehogs is in the excellent compound phrases you can make. We can also define the middle hedgehog of a convex shape, and a hypocycloid, made by rolling a general smaller circle around inside a larger one, can form a self-crossing hedgehog (making me think of the road safety adverts from the late 90s).
The ham sandwich theorem A hedgehog (in geometry) is a curve in the plane with a particular property. You can consider a curve as being defined by an envelope of lines. A nice example is the curve created by joining points on two axes in pairs, creating a series of lines that gradually change slope, like in the bottom-left of the opposite page. Here the axes range from 0 to 10, and a point 𝑝 on one axis is connected to 𝟣𝟣 − 𝑝 on the other. The resulting curve is a parabolic curve, and the actual curve itself is referred to as the envelope of the set of straight lines: a curve which is tangent to some member of the family of curves at every point. In order for such a curve to be a hedgehog—in particular, a ‘sufficiently well-behaved hedgehog’— the curve has to have one tangent line in each oriented direction. That is, for every possible direction you could draw a line pointing in on a 2D plane, there’s some part of the curve that’s pointing in that direction. Annoyingly, this leads you to the conclusion that a boring circle shape is actually a hedgehog (as in fact is any convex curve—one that doesn’t turn inward at any point). But it’s not the only hedgehog! And the more interesting ones are good fun. For example, the star shape made by joining four of the parabola we saw earlier is called an astroid, and is an example of a non-convex hedgehog. A similar shape called a deltoid can be made by tracing a point on a circle that’s rolling around the inside of a larger circle with three times the radius— and this is an example of a projective hedgehog.
It’s even better when the object a theorem is named after is something as mundane as a ham sandwich. This theorem is also from geometry, and concerns the possibility of cutting objects in half. Imagine you have a cake sitting on a table, and you’re looking at it from above (so you’re effectively considering a two-dimensional cake, assuming the cake is of uniform thickness). If you wanted to share the cake with someone else, you could cut it in half using a straight line and have the same amount each. If you had two cakes sitting on the same table, you could also make a single straight cut that cuts both cakes in half exactly. But once you introduce a third cake, you can’t guarantee this will be possible:
? It turns out the limit to this is the number of dimensions—in our two-dimensional space, we can take any two objects (of any size, shape or relative position) and find the unique line that cuts them both in half. In three dimensions, we’d use a two-dimensional plane to cut things in half, but we could have three objects—in the classical case, two bits of bread and a piece of ham—such that we can find a 2D plane that cuts all three objects exactly in half. The obautumn 2022 ⋅
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jects can be arranged in any way you like, be different relative sizes and even intersect each other, and the ham sandwich theorem says this is still possible.
This kind of shape is called a convex hull—the smallest shape that encloses all the shapes and is mathematically convex. We can add in a third circle on the end of the line and draw a shape that encloses all three:
but if we’re looking to minimise this surface area, we could maybe arrange the circles differently. If we put the three circles at the corners of an equilateral triangle so they’re all touching, maybe the perimeter of this shape is shorter? It turns out this is generally true in 𝑛-dimensional space: using an (𝑛 − 𝟣)-dimensional slicing line or plane, we can cut 𝑛 objects exactly in half all at the same time. The calculations to determine exactly where that line or plane is might be a little more difficult, but the theorem guarantees that such a slice exists—so you can share your sandwich knowing that you’ve each got the same amount of bread and ham.
The sausage catastrophe Continuing with the food theme, I’ve always enjoyed the name of this particular geometrical occurrence, which concerns arrangements of spheres in different numbers of dimensions, and in particular the question of the surface area bounding a set of spheres. Imagine a single sphere in 2D—a circle drawn on a page. The smallest region containing that circle is just the circle itself. But if I add a second circle touching it, in order to bound them I need to draw a shape which has rounded ends and straight edges—strictly called a stadium—whose perimeter will be the circumference of two halfcircles, plus four times the radius of the circle:
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Well, no. It turns out that for up to four circles (and up to four spheres, in higher dimensions) arranging the spheres into a long sausage gives the most efficient overall surface area. . . up to a point. For 2 dimensional circles, 3 dimensional spheres, 4D hyperspheres, and 5, 6, 7, 8,. . . dimensional whatever you call them above that, the sausage continues to be the most efficient shape all the way up to 55 dimensions. And then suddenly—it isn’t. For four 56-dimensional spheres, arranging them in a straight line doesn’t give the smallest surface area—there’s actually a different arrangement that gives a more efficient packing. For 57 and 58, it’s back to a sausage again. Then for 59: not a sausage. The most efficient arrangement continues to fluctuate (up to 65, beyond which sausage is always not the best) in a way that mathematicians found so distressing they named it the sausage catastrophe.
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The hairy ball theorem No roundup of mathematical things with daft names would be complete without a mention of the most hilarious topological result, the hairy ball theorem. Despite being from the world of algebraic topology, which often comes with hefty dense terminology, it’s a surprisingly easy-tounderstand concept.
▲ A failed attempt to comb the hair on a ball flat
▼ A successful attempt to comb the hair on a torus flat
A vector field, which consists of a vector pointing in a direction attached at every point of a space, can be thought of as a collection of tiny hairs attached to the surface (particularly if the space you’re considering is a two-dimensional surface like a manifold, or in this case, a sphere). A ball with a hair attached at each point—a hairy ball, if you will—can be considered as a model for a sphere with a vector field. The conditions we require for the vector field to be well-behaved (ie continuous) are exactly those you’d get if the vectors were made out of real actual hairs with a thickness—they don’t lie exactly on top of each other, and move in the right kind of way—and you can manipulate the vector field by combing the hairs. The theorem states that there is no continuous tangent vector field on even-dimensional 𝑛spheres—that is to say, if we try to ‘comb’ the vector field so that all the vectors lie flat, we won’t be able to do this without at least one point where the hairs point directly upwards. You can see a failed attempt above right. It’s famously the reason why the hair on a person’s head often forms a ‘cowlick’ (a bit at the crown of the head where it sticks up); it also has a nice consequence which is that at any given time, the vector field describing the movement of air on the Earth’s surface must have at least one point
where it has no horizontal component to the motion—that is, there’s always somewhere on Earth where it isn’t currently windy. The theorem only applies in certain cases—the sphere needs to be even-dimensional (by dimension here we mean the number of dimensions its surface is made of, so a normal sphere sitting in 3D space is considered to be two-dimensional). It is possible to have a continuous tangent vector field on some shapes which aren’t spheres—for example, a 2-torus (doughnut) shape can be combed flat, by pointing all the hairs in the same direction: that of a circle running around the torus, like you can see above. I’m sure there are many other mathematical theorems, constructions and ideas which have even sillier names than this—and I encourage you to find out about them and share them with as many people as you can. And, if you want a really good motivation for new mathematical research, consider the possibility that if you discover or prove a new piece of maths, you might even get to choose what it’s called.
Katie Steckles Katie Steckles is a mathematician based in Manchester, who talks about maths in schools, at festivals and shows, on BBC radio and TV and on the internet. She blogs at d aperiodical.com and has a YouTube channel at m katiesteckles.
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On this page, you can find out what we think of recent books, films, games, and anything else vaguely mathematical. Full reviews of many of the items featured here can be found at d chalkdustmagazine.com
Cell Tower
d andrewt.net/puzzles/cell-tower A really fun daily word game that involves dividing a grid of letters into words between four and eight letters long. More fun, more challenging, and more addictive than Wordle.
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Electromagnetic Field 2022 An almost perfect festival. See you at the maths village in 2024!
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Big Feastival 2022 Some excellent geometric mirror balls. Very welcoming to mathematicians.
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King’s Lynn MathsJam King’s Lynn was recently added to the list of places that host this monthly event on the second-last Tuesday of each month: the first Jam was a lot of fun.
ggggh San Diego Nice location for a conference. More walkable than LA.
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Choco Leibniz orange Elevates the rich tea biscuit with a Christmassy chocolate topping. Orange zest is tasty but more than four is a bit much.
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Clopen Mic Night #3: Clopen Sept Another hour of very enjoyable mathematical entertainment.
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Placeholder content Often very amusing, but we must be careful not to leave it in the magazine.
gghii Moving house Leaves very little time for writing content for Chalkdust.
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l goD Dog leads, orrim mirrors, and mreH Hermann Minkowski Donovan Young looks at the shapes made when two cones collide.
... it seemed to us a garden full of flowers. In it, we enjoyed looking for hidden pathways and discovered many a new perspective that appealed to our sense of beauty, and when one of us showed it to the other and we marvelled over it together, our joy was complete. David Hilbert’s obituary for Hermann Minkowski (1910)
Y
ou know those extendable dog leads—the ones where the dog can go really far? Imagine you’re standing still and the dog on the other end is running around. How fast is the lead unspooling? At this point I’ll issue a spoiler alert and encourage you to go away and think about this problem for a while, and when you’re satisfied, turn the page and read on. autumn 2022 ⋅
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Vectors Vectors are objects that have a size and a direction, and are widely used in geometry, physics, and elsewhere. In this article, we use two-dimensional vectors, for example 𝑥 𝒓 = ( ) = (𝑥, 𝑦). 𝑦
This is the vector that represents a movement of 𝑥 units horizontally and 𝑦 units vertically. 𝑦
𝒓 𝑥
Using Pythagoras’s theorem, we can write the magnitude of a vector as |𝒓| = √𝑥 𝟤 + 𝑦 𝟤 .
We write 𝒓̂ for the unit vector pointing in the same direction as 𝒓 , ie 𝒓 𝒓̂ = . |𝒓|
We can compute the dot product of two vectors by multiplying the pairs of components together then adding up the results, ie 𝑥 𝑥 ( 𝟣 ) ⋅ ( 𝟤 ) = 𝑥 𝟣 𝑥 𝟤 + 𝑦 𝟣 𝑦𝟤 . 𝑦𝟤 𝑦𝟣
Now that you’ve had time to think about the question yourself, we can take a look together. If the dog is running around you in a circle, the lead isn’t unspooling at all. If the dog is running directly away from you, in any direction, we would expect the rate of unspooling to be the greatest (for a fixed speed). In fact, we only care about the component of the dog’s velocity in the direction radially away from the origin. That component is the unspooling rate. The story I’m about to tell in this article has nothing to do with dogs, but hinges on the answer to this problem, which, expressed mathematically, says that in a small time d𝑡 , in which the 𝑥 -coordinate of the dog moves from 𝑥 to 𝑥 + d𝑥 (and similarly 𝑦 to 𝑦 + d𝑦 ), exactly this much lead, d√𝑥 𝟤 + 𝑦 𝟤 =
𝑥 d𝑥 + 𝑦 d𝑦 𝟤 𝟤 √𝑥 + 𝑦
,
is unspooled. In vector language this may be expressed as d|𝒓| = 𝒓̂ ⋅ d𝒓,
where I have used the dot product to select the component of the vector d𝒓 which is in the direction of 𝒓 .
Hocus focus! No smoke, but a curved mirror If this article isn’t about dog walking, what is it about? It’s about some famous (and some perhaps not-so-famous) shapes for mirrors and lenses. To start with, we want to find a curved mirror with the following property: a laser pointer shone from the origin would be reflected through the point (𝑐, 𝟢), no matter the direction that the laser pointer is pointedf . Already we can see that this mirror must wrap around the person at the origin, so it will be a closed surface. It will also have rotational symmetry around the 𝑥 -axis, which means we can reduce the problem to a closed curve in the 𝑥𝑦 plane which is then rotated about the 𝑥 -axis to produce the surface of the mirror. f If you already know the answer, bear with me. I hope to shed a slightly different perspective on a well-known curve.
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Let (𝑥, 𝑦) be a point on the curve. The vector (d𝑥, d𝑦) is tangent to the curve at that point. The law of reflection states that the angles the incoming and outgoing rays make with the tangent are the same.
𝑦 (𝑥, 𝑦)
(d𝑥, d𝑦)
The vector which joins the origin to (𝑥, 𝑦) is (𝑥, 𝑦), while that which joins (𝑥, 𝑦) to (𝑐, 𝟢) is (𝑐 −𝑥, −𝑦). The equal angles can be expressed this way, using the dot product:
or
̂ ̂ (𝑥, 𝑦) ⋅ (d𝑥, d𝑦) = (𝑐 − 𝑥, −𝑦) ⋅ (d𝑥, d𝑦), 𝑥 d𝑥 + 𝑦 d𝑦 𝑥𝟤
√
+ 𝑦𝟤
=−
(𝑥 − 𝑐) d𝑥 + 𝑦 d𝑦 𝟤 𝟤 √(𝑥 − 𝑐) + 𝑦
(𝑐, 𝟢)
𝑥
▲ All lasers shone from the origin should be reflected through the point (𝑐, 𝟢)
.
But wait! We know this expression from thinking about dog leads! This is d√𝑥 𝟤 + 𝑦 𝟤 = −d√(𝑥 − 𝑐)𝟤 + 𝑦 𝟤 ,
or
d (√𝑥 𝟤 + 𝑦 𝟤 + √(𝑥 − 𝑐)𝟤 + 𝑦 𝟤 ) = 𝟢.
This is one of the easiest differential equations you will ever see: it says that 𝟤 𝟤 𝟤 𝟤 √𝑥 + 𝑦 + √(𝑥 − 𝑐) + 𝑦 = 𝑘,
where 𝑘 is a constant. We asked for the curve with the reflective property explained earlier, and we’ve found a curve with another property: the sum of the distances from the points (𝟢, 𝟢) and (𝑐, 𝟢) to any point (𝑥, 𝑦) on the curve is a constant 𝑘 .
If you are familiar with the conic sections, I hope you don’t feel too cheated right now, because the curve we’ve just found is called an ellipse, and it is usually thought of as coming from the intersection between a cone and a tilted plane. It is famous for having these properties.
Conic sections The conic sections are a family of curves that can be obtained by taking slices through a cone. A circle can be obtained by taking a horizontal slice through the cone. circle An ellipse can be obellipse tained by taking a nonhorizontal slice that is parabola less steep than the side hyperbola of the cone. A parabola can be obtained by taking a slice with the same steepness as the cone. A hyperbola can be obtained by taking a slice steeper than the steepness of the cone.
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Spacetime diagrams and intersecting light cones I want to reconsider our solution 𝟤 𝟤 𝟤 𝟤 √𝑥 + 𝑦 = 𝑘 − √(𝑥 − 𝑐) + 𝑦 ,
𝑧 = √𝑥 𝟤 + 𝑦 𝟤
and think of it as the intersection of the two surfaces
and
𝑧 = √𝑥 𝟤 + 𝑦 𝟤 , 𝑧 = 𝑘 − √(𝑥 − 𝑐)𝟤 + 𝑦 𝟤 .
What are these surfaces? They are cones, both with an opening angle of 45 degrees. The first opens upwards from the origin. The second opens downwards from (𝑐, 𝟢, 𝑘). This is interesting: we have found a famous conic section, the ellipse, from the intersection of two cones, neither of which is the cone used in the usual conic section construction! What do these cones represent?
(𝑐, 𝟢, 𝑘)
(𝟢, 𝟢, 𝟢)
𝑧 = 𝑘 − √(𝑥 − 𝑐)𝟤 + 𝑦 𝟤
▲ When viewed from above, the intersection of these two cones is an ellipse
Another way to think about light, rather than as rays, is as wavefronts. These are circles which begin as points at their source, and then expand out homogeneously at a fixed speed (the speed of light)—like ripples on a pond when a rain drop falls in. In this language, our mirror problem is asking that wavefronts emanating from the origin are reflected and ‘zero back in’ on the point (𝑐, 𝟢). Hermann Minkowski invented a lovely way to visualise expanding wavefronts, a device now called a spacetime diagram. He needed this to think about Einstein’s theory of relativity, but our use will be slightly more mundane (but no less beautiful!). In a spacetime diagram, we place the 𝑥𝑦 plane in its usual position (ie horizontal). The vertical axis is for time. What would an expanding circular wavefront look like in such a diagram? In the 𝑥𝑦 plane, ie when the time 𝑡 = 𝟢, we would have a point at the origin—the nascent wavefront waiting to expand outwards. At some time 𝑇 later, ie on the plane 𝑡 = 𝑇 , we would have a circle of radius the speed of light times 𝑇 . So an expanding wavefront traces out a cone. We’ll assume natural units—this means we take the speed of light to be 1 and the opening angle of the light cone is 45 degrees. I hope you now see that our intersecting cones are just the light cones of the outgoing and reflected wavefronts. Presto chango, our ‘𝑧 ’ coordinate is time. We’ve ‘lifted’ the ellipse from a plane curve to a three-dimensional one. This new curve represents the history of when the various parts of the outgoing wavefront encountered the elliptical mirror and were reflected. The time it takes for any part of the wavefront to go from the original point (the origin) to the final point at (𝑐, 𝟢) is constant (and equal to 𝑘 ). The ancient Greeks, and most famously Apollonius of Perga, knew a lot about the conic sections, but now I think they had the wrong cones. Let’s see how much more we can do!
Two more mirrors... In the case of the ellipse, the reflected ray travelled from the point of reflection (𝑥, 𝑦) to the point (𝑐, 𝟢). A close cousin of the ellipse is obtained by asking that the reflected ray, continued backwards from the point of reflection, passes through (𝑐, 𝟢), where now we must impose 𝑐 < 𝟢. 𝟧𝟤
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▼ The backwards continuation of the reflected ray (dashed red line) goes through the point (𝑐, 𝟢)
The law of reflection now says
𝑦
or (d𝑥, d𝑦)
(𝑥 − 𝑐) d𝑥 + 𝑦 d𝑦
or (𝑐, 𝟢)
𝑥 𝑡 = √𝑥 𝟤 + 𝑦 𝟤
̂ (𝑥̂ − 𝑐, 𝑦) ⋅ (d𝑥, d𝑦) = (𝑥, 𝑦) ⋅ (d𝑥, d𝑦),
𝑡 = 𝑘 + √(𝑥 − 𝑐)𝟤 + 𝑦 𝟤
√(𝑥
− 𝑐)𝟤
+ 𝑦𝟤
=
𝑥 d𝑥 + 𝑦 d𝑦 𝟤 𝟤 √𝑥 + 𝑦
,
d (√(𝑥 − 𝑐)𝟤 + 𝑦 𝟤 − √𝑥 𝟤 + 𝑦 𝟤 ) = 𝟢.
The solution of this is
𝟤 𝟤 𝟤 𝟤 √(𝑥 − 𝑐) + 𝑦 − √𝑥 + 𝑦 = 𝑘.
This curve is called an hyperbola and is also a conic section, traditionally gotten by cutting the cone with a steeply tilted plane. But now, with our new eyes, we can see it as the intersection of the two light cones, (𝟢, 𝟢, 𝟢)
(𝑐, 𝟢, 𝑘)
▲ When viewed from above, the intersection of these two cones is an hyperbola
𝑡 = √𝑥 𝟤 + 𝑦 𝟤
and
𝑡 = −𝑘 + √(𝑥 − 𝑐)𝟤 + 𝑦 𝟤 .
There is another curve which we can think of as a sort of midway point between the ellipse and the hyperbola. For this mirror we ask that the reflected ray is always horizontal. The law of reflection says
or
or
̂ (𝑥, 𝑦) ⋅ (d𝑥, d𝑦) = (𝟣, 𝟢) ⋅ (d𝑥, d𝑦), 𝑥 d𝑥 + 𝑦 d𝑦 𝟤 𝟤 √𝑥 + 𝑦
𝑦 (d𝑥, d𝑦)
= d𝑥,
d (√𝑥 𝟤 + 𝑦 𝟤 − 𝑥) = 𝟢.
The solution of this is
𝑥
▲ All of the reflected rays are horizontal
𝟤 𝟤 √𝑥 + 𝑦 − 𝑥 = 𝑘.
This curve is a parabola (which is also a conic section): this time, the cone is cut with a plane tilted at the opening angle of the cone. Here, the traditional plane-cuts-cone construction, and our intersecting light cone nearly coincide. We can consider our solution above as the intersection of the light cone 𝑡 = √𝑥 𝟤 + 𝑦 𝟤 and the plane 𝑡 = 𝑘 + 𝑥 . autumn 2022 ⋅
𝟧𝟥
How is this plane a light cone? Well, when we asked for the rays to be reflected horizontally, we were asking for the wavefronts to be vertical lines rather than concentric circles. This can be thought of as taking the limit of the ellipse, where the second focus (𝑐, 𝟢) is moved infinitely far down the 𝑥 -axis, so that the reflected rays reach that point at some infinite time in the future. This means that the light cone representing the light leaving the origin intersects only a small portion of the reflected light cone, small enough to be approximated by a plane. The difference with the usual construction is that here the parabola we’re interested in isn’t the one in the plane cutting the cone (that’s a parabola too), but rather the one projected down onto the 𝑥𝑦 plane.
𝑡 = √𝑥 𝟤 + 𝑦 𝟤
𝑡 =𝑘+𝑥
(𝟢, 𝟢, 𝟢)
▲ When viewed from above, the intersection of these two light cones is a parabola
... and two lenses
As I mentioned earlier, the Greeks knew about these optical properties of conic section mirrors. In the late 10th century, in Persia, Ibn Sahl furthered that knowledge using refraction, rather than reflection, to change the course of the light rays. If you don’t know what refraction is, it is the bending of light rays when they travel between two different materials. In his book On the burning instruments, Sahl gives the first account of the law of refraction. 𝒓𝟣̂
𝜃𝟣 𝜃𝟤
𝒇 𝒓𝟤̂
▲ Sahl’s law
In modern language, the law considers the unit vectors 𝒓𝟣̂ and 𝒓𝟤̂ along the incoming and bent rays. It says that the ratio of their components parallel to the interface (between the two materials) is a constant. That constant depends on the two materials in question, and we now know that it is equal to the ratio of the speeds of light, 𝑐𝟣 and 𝑐𝟤 , in each one. It reads as follows, 𝒇 ⋅ 𝒓𝟣̂ sin 𝜃𝟣 𝑐𝟣 = = , 𝒇 ⋅ 𝒓𝟤̂ sin 𝜃𝟤 𝑐𝟤 where 𝒇 is a vector parallel to the interface.
If you know it, you probably know it as Snell’s Law—but that’s the historic forces of Eurocentrism at work: it was known centuries before its rediscovery in the 1600s. We should start calling it Sahl’s Law. In his book, Sahl finds what we might like to call a ‘refractive parabola’. It’s a lens that takes horizontal rays, and bends them so that they all go through a given point (or, if we run time backwards, takes rays emanating from a point and bends them to be horizontal). If you wanted to make such a lens, what shape should the glass be ground to? Let’s find out! Now we’ll be a bit more cavalier, to emphasise the power of the method, and skip over the detail we used when finding the mirrors. We’ll just take the two light cones representing our parallel and focused rays, and bosh them together. Where they intersect (projected down onto the 𝑥𝑦 plane) is our desired curve. The parallel rays have a light ‘plane’ (as we saw for the parabola), given by 𝑡 = 𝑛𝑥. 𝟧𝟦
⋅ chalkdust
𝑦
𝑥
▲ Parallel lasers are all refracted to the origin
feature 𝑡 = 𝑛𝑥
The 𝑛 is 𝟣 over the speed of light in glass, in units where the speed of light in air is 𝟣. It is usually called the index of refraction. Since light moves slower in the glass, 𝑛 > 𝟣, and so this light plane is not tilted at 45 degrees as was the case for the parabolic mirror, but more steeply, according to the fact that these wavefronts move more slowly. The light cone for the rays focusing on the origin is 𝑡 = 𝑘 − √𝑥 𝟤 + 𝑦 𝟤 . Voilà—the shape of our lens is
(𝟢, 𝟢, 𝑘)
𝑡 = 𝑘 − √𝑥 𝟤 + 𝑦 𝟤
▲ When viewed from above, the intersection of these two light cones is an hyperbola
and happens to be an hyperbola (for 𝑘 < 𝟢).
Wait, we forgot to use Sahl’s Law! I told you we were going to be cavalier. You can go ahead and use it to derive a differential equation like we did earlier for the mirrors, but you’ll find the same answer. The law of refraction is all taken care of by the light cones! 𝑦
(𝑐, 𝟢)
𝑥
▲ Rays from (𝑐, 𝟢) appear to be coming from the origin
𝑛𝑥 = 𝑘 − √𝑥 𝟤 + 𝑦 𝟤 ,
What would a ‘refractive hyperbola’ look like? This would be a sort of lens which would take rays emanating from one point within the glass, and bend them as they exited so that they appeared as if they were emanating from a different point.
𝑡 = 𝑘 + √𝑥 𝟤 + 𝑦 𝟤
𝑡 = 𝑛√(𝑥 − 𝑐)𝟤 + 𝑦 𝟤
▲ When viewed from above, the intersection of these two light cones is...
▼ ... the refractive hyperbola (for 𝑘 = 𝟦, 𝑛 = 𝟣.𝟣𝟧 and 𝑐 = −𝟨.𝟧)
We’ll take the first point to be (𝑐, 𝟢), and the apparent point the origin. Our light cones are 𝑡 = 𝑛√(𝑥 − 𝑐)𝟤 + 𝑦 𝟤 and 𝑡 = 𝑘 + √𝑥 𝟤 + 𝑦 𝟤 . Where they intersect is our magical curve. This one wasn’t something known to the Greeks, or to Sahl, as far as I know. When I first began thinking about the conic sections this way, I just wanted to see whether I could construct them, starting with their focusing properties. When I realised I could re-express them as conic intersections I was surprised, and delighted. When I then realised these were the light cones of the original and reflected wavefronts, to paraphrase Hilbert, “my joy was complete”. Donovan Young Donovan is a maths teacher who is transfixed by the science and mathematics of light and vision.
autumn 2022 ⋅
𝟧𝟧
1
Cryptic crossword
7 10
11
#4, set by Humbug
24 25 26 28 29
4
5
9
8 12
13 14
15
16
17
18
19
20
21 22
23
24
25 27
26
29
Across
13 14 16 17 18 19 20 22
3 6
28
1 3 7 10 12
𝟧𝟨
2
Down
XL age limit for 4 & 21. (5) 1 The point in Banach’s theorem for first nine (5) Balls. Advanced zero with FF first embodied #FF0000. (7) 2 Odd rehash should be equal to 14. (1,1,1) Confused cats to sing wordless phrases. (4) 4&21 9, 15, 5 & 23, and 28 were awarded this (6,5) 2.71828 and 9.81, for example. (2) prize for scarecrows? He and I gathered around broken kilns. This (8) 5&23 Probability theorist is getting by (7-5) year’s 4 & 21s were awarded here. without gravity after mild uni mess. Information hidden in old Atari. (4) 6 They confused last email CC group. (5) Every other alpha’s usually equal to 2. (1,1,1) 8 Mixed CD at Hulk’s magazine. (9) Even split is irrational. (2) 9 She packed spheres by initially zinging ovals (9) Smog endlessly enters sky, making it this. (5) against empty Kia. Representation of most common 50. (5) 11 Magma blown up by a ray of energy. (5) Thanks for the article. (2) 15 Regularly though, he linked geometry and (3) Opening of Oswestry Hospital’s maternity unit. (3) combinatorics. SIAM journal on scientific computing is a bit of (4) 16 “Perhaps edit parses?” Sophie implied (5) quasi-science. initially. Is this okay? Confused Paris and East Germany? This is how (8) 17 Start moving coefficient of friction up a level. (7) fast a plane will take you. 20 Interval without endpoints and centre of (6) I circle moon. (2) interval’s support. Berry hidden in a cairn. (4) 21 see 4. This analytic number theorist is darn confused (7) 23 see 5. after former prime minister. 27 1101’s inverse was something in 12 in 2022. (1,1,1) Latin manor in a mess. (5)
⋅ chalkdust
feature
These illustrations were made using DALL⋅E 2, a machine learning software for generating art. See
d
openai.com/dall-e-2
Spam calls and number blocking Michael Wendl really wants those spammers to stop calling him.
N
owadays, 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 obviautumn 2022 ⋅
𝟧𝟩
ous practical significance of the question, it is also mathematically interesting and can be investigated using the tools of combinatorial probability.
Off his phone bat ▼ An example of a simple 2-set collision problem in terms of matching heads versus tails (𝑘 = 𝟤) for two sets of two coins each. set 2
H
1/
T
1/2 2
set 1
2
1/
The reason spammers are able to place so many calls is that they use sophisticated software to automatically make ‘robocalls’, often spoofing numbers and even names that geographically target the recipient. This tactic, commonly referred to as ‘neighbour spoofing’, is a powerful way of increasing the odds of a recipient actually answering their phone, since the call appears to be local. In order to cast this problem in mathematical terms, consider a catalogue of 𝑘 possible telephone numbers and suppose that the recipient’s phone already has its own database of 𝑚 of these numbers blocked which we denote 𝐵 = {𝐵𝟣 , 𝐵𝟤 , … , 𝐵𝑚 }. Further, suppose that spam software makes 𝑛 calls to the recipient’s phone, choosing each spoofed calling number, 𝐶 = {𝐶𝟣 , 𝐶𝟤 , … , 𝐶𝑛 }, randomly from this same catalogue of 𝑘 telephone numbers. A successful block occurs when one random variable from 𝐶 matches, or ‘collides’, with one from 𝐵.
T
1/2
H
▲ The elementary approach of treating each match trial as being independent, where dashed lines indicate lack of a match. Multiplying the probabilities together gives 1/16.
▼ An approach where all possible pairs are enumerated, where boxes indicate instances having no matches between the two sets. 2/16 (or 1/8) of the options have no matches.
Such problems are thus often referred to as matching or collision problems. A famous example is the birthday problem: what is the the probability that at least TTTT THTT HTTT HHTT two people within a group share a common birthday? (You can read more about this in Peter Rowlett’s piece TTTH THTH HTTH HHTH ‘Counting caterpillars’ in issue 09 of Chalkdust). The TTHT THHT HTHT HHHT first investigation into the 2-set problem was a 1987 artiTTHH THHH HTHH HHHH cle by Tony Crilly and Shekhar Nandy in the Mathematical Gazette entitled ‘The birthday problem for boys and girls’, which counted birthday matches between a boy and girl from their respective sets. The collision problem appears in an enormous array of real-world scenarios and there is now quite a literature on 2-set and multi-set problems. However, our proposition is yet slightly different from all of these... Let’s try and calculate the probability of no matches between 𝐵 and 𝐶 , denoted as 𝑃𝟢 (this is equal to one minus the complimentary scenario: that we have one or more matches). Suppose we’re interested in matching heads and tails between two sets of two coins each, so 𝑚 = 𝑛 = 𝑘 = 𝟤.
Specifically, if we treat each coin-to-coin comparison between sets as an independent trial, the overall state of no matches requires each coin in the left set to not match each coin in the right set, namely 𝑃𝟢 = 𝑃(set 1 coin 1 ≠ set 2 coin 1) ⋅ 𝑃(set 1 coin 1 ≠ set 2 coin 2) ⋅
𝑃(set 1 coin 2 ≠ set 2 coin 1) ⋅ 𝑃(set 1 coin 2 ≠ set 2 coin 2).
𝟧𝟪
⋅ chalkdust
feature
We might be tempted to calculate this as follows: assume each of these four trials has a Bernoulli probability of 𝟣/𝟤, meaning that the overall no-match probability is (𝟣/𝟤)𝟦 = 𝟢.𝟢𝟨𝟤𝟧. Yet, if we actually write down all 4 possible head–tail combinations for the left set and the corresponding ones for the right set, then count how many of the 𝟦 × 𝟦 = 𝟣𝟨 possible outcomes result in no overall match, we plainly see two hits, meaning 𝑃𝟢 = 𝟤/𝟣𝟨 = 𝟢.𝟣𝟤𝟧. These two calculations, which seemingly address the same question, differ by a factor of two! The latter calculation is, of course, correct, since we made exhaustive counts. This exercise indicates that match trials between individual variables are not strictly independent of one another, a complication not uncommon in probability problems. Nevertheless, we can still generalise the elementary approach in order to investigate conditions under which its error might become acceptably small compared to the exact solution. Let’s think of this elementary approach as an approximation. Reasoning along the same lines, a single trial between one specific variable in each set has a mismatch probability of 𝟣 − 𝟣/𝑘 and, since there are 𝑚𝑛 such trials, the overall mismatch probability is 𝟣 𝑚𝑛 𝑃𝟢𝐴 = (𝟣 − ) ≈ e−𝑚𝑛/𝑘 , 𝑘
where superscript 𝐴 denotes approximation based upon the assumption of independent trials. The final expression is an additional asymptotic approximation, which is reasonable when 𝑚, 𝑛 and 𝑘 are sufficiently large (as they tend to be in a number blocking problem). This can be seen by comparing expansion of the binomial to the series expansion of the exponential.
Put your phone away Using structured counting, we can calculate the probability, 𝑃𝟢 , of no matches arising between set 𝐵 and set 𝐶 . First, let’s consider the sampling methods by which sets 𝐵 and 𝐶 are actually obtained. Set 𝐶 is sampled with replacement, meaning the software plucks each new spoofed telephone number randomly with equal probability from the catalogue of 𝑘 numbers. Hence the spam-caller-side sample space is of size 𝑘×𝑘×⋯×𝑘 = 𝑘 𝑛 when 𝑛 spam calls have been placed. In many 2-set modelling scenarios, both sets are selected in this way. Examples include Crilly and Nandy’s boy–girl birthday probability discussed above and numerous real-world calculations, like the probability of no collision-enabling common altitudes for aircraft flying between two opposing airports. However, at this point, our scenario departs somewhat from these classic 2-set modelling scenarios. Namely, values in 𝐵, the list of numbers blocked on the phone, cannot be repeated, which is to say they are selected without replacement. The recipient blocks the first randomised spam number that appears, after which that number is, by definition, never seen again, then the second, which is likewise never seen again, and so forth. As such the blocking-side sample space is enumerated by the permutation 𝑘(𝑘 − 𝟣)(𝑘 − 𝟤) ⋯ (𝑘 − 𝑚 + 𝟣), which is known as the ‘falling factorial’ and denoted by (𝑘)𝑚 . Consequently, the total sample space is then of size 𝑘 𝑛 (𝑘)𝑚 . With that, we have the denominator of probability 𝑃𝟢 . . . and we are halfway there! The numerator is a little more complicated, but we can use some visual assistance from graphs. We can represent each random variable by a vertex. Connections between vertices are either solid or dashed lines, solid lines represent a ‘match’ and dashed lines represent ‘no match’. Edges for random variables in 𝐵 (the blue vertices) are necessarily dashed, while edges connecting the random variables in 𝐶 (the green vertices) can be of both kinds. The connection between 𝐵 and 𝐶 is likewise necessarily dashed. autumn 2022 ⋅
𝟧𝟫
This diagram again suggests the falling factorial, with this particular one representing (𝑘)𝟣𝟢 possible ways of realising no collisions, since there are a total of 10 non-repeating values connected by nine dashed edges. Note that the last two vertices in 𝐶 , ie 𝐶𝟥 and 𝐶𝟦 , do not add to the count because they have identical values to their adjacent neighbour 𝐶𝟤 . This enumeration is readily generalised to an arbitrary graph having 𝑚 and 𝑗 uniquely valued vertices in 𝐵 and 𝐶 , respectively, as (𝑘)𝑚+𝑗 .
But, how should we account for the various different possible arrangements of solid and dashed edges on the caller side? Consider the arrangements below gathered under the banner of 𝑆𝟤 (𝟦, 𝟤), which depict the seven ways that four elements can be partitioned into two nonempty subsets. The arrangement marked by the asterisk appears in the 𝑛 = 𝟦 example above. This diagram raises the question, would we have to account for all 15 possible ways of partitioning the four caller vertices shown in the previous diagram to obtain a complete and correct count?
𝐵𝟣
𝐶𝟣
𝐵𝟥
𝐶𝟥
𝐵𝟤 𝐵𝟦 𝐵𝟧 𝐵𝟨
𝐶𝟤 𝐶𝟦
caller side 𝑛 = 𝟦
𝐵𝟩 𝐵𝟪
recipient side 𝑚 = 𝟪
▲ An example with eight blocked numbers, and four spoofed numbers made by the caller, three of which are identical.
It turns out that what we need here is a general way to tally the numbers of ways of partitioning a set of 𝑛 things into 𝑗 non-empty subsets, which is precisely what is quantified by the Stirling numbers of the second kind, 𝑆𝟤 (𝑛, 𝑗). Multiplying this quantity by the falling factorial gives 𝑆𝟤 (𝑛, 𝑗)(𝑘)𝑚+𝑗 , which represents the grand total number of graphs that do not collide when there are 𝑗 unique values on the caller side. Summing the resulting product over the 𝑛 possible multiplicities for the caller side yields the required numerator. Bringing the numerator and denominator together, we find the probability of no matches to be 𝑛 𝟣 ∑ 𝑆𝟤 (𝑛, 𝑗)(𝑘)𝑚+𝑗 . 𝑃𝟢 = 𝑛 𝑘 (𝑘)𝑚 𝑗=𝟣
𝑆𝟤 (𝟦, 𝟤) = 𝟩
𝑆𝟤 (𝟦, 𝟣) = 𝟣
Requisite 𝑆𝟤 values are generated by the recurrence 𝑆𝟤 (𝑛 + 𝟣, 𝑗) = 𝑗 ⋅ 𝑆𝟤 (𝑛, 𝑗) + 𝑆𝟤 (𝑛, 𝑗 − 𝟣) using the initial/edge conditions 𝑆𝟤 (𝟢, 𝟢) = 𝟣 and 𝑆𝟤 (𝑛, 𝟢) = 𝑆𝟤 (𝟢, 𝑛) = 𝟢.
*
𝑆𝟤 (𝟦, 𝟥) = 𝟨
𝑆𝟤 (𝟦, 𝟦) = 𝟣
▲ All the possible ways of partitioning the callerside set of four spoofed numbers into one, two, three and four non-empty subsets.
𝑃𝟢 =
Before getting to actual analysis, let me quickly illustrate the extraordinarily large numbers that combinatorial situations can yield, which are belied by the visually unremarkable decimal values in the resultant probability. For a small, but otherwise arbitrary toy example of 𝑚 = 𝟫, 𝑛 = 𝟣𝟣, and 𝑘 = 𝟦𝟪, we find
𝟣𝟫𝟥𝟤𝟣𝟤𝟣𝟥𝟢𝟣𝟥𝟨𝟫𝟨𝟫𝟪𝟩𝟣𝟣𝟨𝟥𝟣𝟧𝟦𝟤𝟢𝟥𝟢𝟪𝟦𝟪𝟢𝟢 ≈ 𝟢.𝟣𝟢𝟣𝟪𝟩, 𝟣𝟪𝟫𝟨𝟨𝟢𝟨𝟪𝟪𝟥𝟥𝟧𝟧𝟥𝟫𝟦𝟢𝟢𝟩𝟫𝟪𝟨𝟣𝟩𝟪𝟩𝟩𝟪𝟩𝟤𝟨𝟦𝟢𝟢
where the numerator and denominator each exceed 30 digits. 𝟨𝟢
⋅ chalkdust
feature
Blow your phone trumpet Blocking can be examined for any situation where 𝑚, 𝑛, and 𝑘 can be characterised, so let’s look at my own phone as an example. I’ve experienced a few different neighbour spoofing schemes. The most specific schemes I’ve come across display actual names of local establishments, eg banks, medical offices, car dealerships, and restaurants. I must admit to having been tricked into answering some of these kinds of calls. For my locale, I will estimate a corresponding value of 𝑘 = 𝟧𝟢𝟢 possible phone numbers in the catalogue. A less sophisticated scheme I have also experienced displays a telephone number, where my 3-digit area code and 3-digit exchange prefix are duplicated, but the trailing 4 digit line number varies. Here, 𝑘 ≈ 10,000. The least specific neighbour-spoofing for north American telephones only uses the same area code, so we have 𝑘 ≈ 7,920,000 due to various reserved and disqualified numbers within the larger set of 𝟣𝟢𝟩 possibilities. This is the weakest scheme, as presumably recipients aren’t likely to answer these calls. To help us consider number blocking over a one week period, I vetted my phone calls for a week and received a vast 𝑛 = 𝟩𝟩 spam calls. The solution for 𝑃𝟢 that we calculated gives the probability of no matches. Let’s call the probability of one or more matches, ie successful blocks, the blocking probability, 𝑃𝐵 = 𝟣 − 𝑃𝟢 . The graph below shows blocking probabilities for the various neighbour-spoofing schemes described above as a function of the recipient’s blocking aggressiveness, ie size 𝑚 of their blocked number list.
blocking probability (𝟣 − 𝑃𝟢 )
𝟣 𝑘 = 𝟦𝟪
𝟢.𝟪
(neighbour-spoofing
𝑘 = 𝟧𝟢𝟢
(neighbour-spoofing by dupli-
using a local name)
cating area code and prefix)
(toy example)
𝟢.𝟨
𝑘 = 10,000
𝟢.𝟦
𝑘 = 7,920,000
𝟢.𝟤 𝟢
(neighbour-spoofing by duplicating only area code) author’s phone
1
10
100
1000
10000
number of already-blocked spoof numbers (𝑚)
▲ Blocking probabilities calculated by the exact and approximate solutions for various levels of neighbourspoofed spam calls. The blue curves depict the toy enumeration example above (solid line is exact, dashed line is approximate), for which 𝑛 = 𝟣𝟣 and 𝑘 = 𝟦𝟪. All other curves apply to the number blocking problem, for which 𝑛 = 𝟩𝟩. For these, the exact and approximate results are visually indistinguishable.
Two somewhat surprising aspects stand out. First, what was clearly a huge error in the approximate solution for 𝑚 = 𝑛 = 𝑘 = 𝟤 in the coin matching example is dramatically decreased for the toy enumeration example of 𝑘 = 𝟦𝟪 and subsequently all but disappears for our actual number blocking analysis, autumn 2022 ⋅
𝟨𝟣
where 𝑘 ⩾ 𝟧𝟢𝟢 (green, orange, and pink curves). Neglecting event dependence and applying asymptotic approximation quickly become acceptable as the parameters of this problem increase. This is because the former results from a rapidly growing sample space in which the outcomes of joint events are increasingly unaffected by one another, while the latter is explained by comparing associated expansions, as mentioned above. As a result, our telephone problem appears to be well described by the approximate result, 𝑃𝟢𝐴 ≈ e−𝑚𝑛/𝑘 . Of course, we could not have known this without first developing the exact combinatorial solution to the problem. Note that brute force numerical evaluation would also have been limited, as suggested by the fact that even our toy example of 𝑘 = 𝟦𝟪 weighs in at ∼ 𝟣𝟢𝟥𝟦 possible combinations to check. The second observation echoes the nonintuitive result of the birthday problem. Namely, matching, which is to say ‘number blocking’, commences with high probability even for relatively small collections of pre-blocked numbers in the recipient’s phone. For example, it reaches 90% at 𝑚 = 𝟣𝟧 for specific neighbour spoofing (𝑘 = 𝟧𝟢𝟢) and at 𝑚 = 𝟤𝟫𝟧 for area code and prefix spoofing (𝑘 = 𝟣𝟢𝟢𝟢𝟢). In other words, in this particular case, there is very high probability of blocking one or more new calls when only 𝑚/𝑘 ≈ 𝟥% of the available numbers have been blocked! Agreement between these two scenarios is not a coincidence, as some quick algebra on the expo▲ Now that you’ve stopped getting so many spam nential solution shows: 𝑚/𝑘 = − ln(𝟣 − 𝑃𝐵 )/𝑛. calls, you can stop leaving your phone in the garden My own phone, having 455 blocked numbers, would seem to be well-positioned against the better neighbour-spoofing attacks (green and orange curves), but not the weakly targeted calls (pink curve). Luckily I don’t tend to answer the weaklytargeted calls anyway so this isn’t too much of a concern.
Phonomenal Of course, our analysis here is still limited: it fails to account for many spammers simultaneously using multiple different spoofing schemes, nor does it quantify the distribution of the number of calls blocked. Nevertheless, this is another example in a seemingly endless succession showing the remarkable power of mathematics to provide meaningful insights into nontrivial, real world problems. In his essay The unreasonable effectiveness of mathematics in the natural sciences, Eugene Wigner quipped that this insight mathematics offers borders on the mysterious—I have long been inclined to agree.
Michael C Wendl Mike is at Washington University in St Louis. Between his work in cancer bioinformatics and theoretical mechanics, he spends time playing the accordion.
c mwendl@wustl.edu 𝟨𝟤
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Sci-fi fans angry Method
You will need
• Roll the die six times. Use the numbers you roll to fill in each blank. • Read out the statements you have formed.
• A science fiction fan • A 6-sided die
I don’t think the
1 2 3 4 5 6
TIE fighter dragon lack of racial diversity food they eat fictional language time travel
1 2 3 4 5 6
1 2 3 4 5 6
Revenge of the Sith The New Hope An Empire Strikes Back Return of the Jedi The Last Jedi The Wrath of Khan
The best bit was when
I thought the new
1 2 3 4 5 6
1 2 3 4 5 6
was very realistic.
Sting Salvador Dali The Worm William Shatner Sigourney Weaver Timothee Chalamet’s eyebrow
was the best actor in Dune.
was the best Star Wars film.
Dwayne The Rock Johnson Captain Scarlet Spock Wicket W Warrick Darth Vader Keanu Reeves
Ghostbusters Total Recall The Matrix Planet of the Apes Battlestar Galactica Chalkdust
said
1 2 3 4 5 6
“I’ll be back.” “GOODNIGHT ORSON.” “PAIN!” “Yub nub.” “I am your father, Luke.” “My name is Neil.”
was better than the old one.
autumn 2022 ⋅
𝟨𝟥
Fields medal winners 2022 Albert Wood explains why this year’s medallists were awarded their prizes.
W
hen 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.
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 Wikimedia common user KFP, CC BY-SA 3.0 that anyone can appreciate. ▲ The 2022 Fields medals were awarded at Aalto University in Helsinki 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. 𝟨𝟦
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feature
James Maynard 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) Mathematisches Forschungsinstitut Oberwolfach, CC BY-SA 2.0 DE
• Every even integer is the sum of two primes (the Goldbach conjecture)
▲ James Maynard
• There is a prime number between any two consecutive squares • There are infinitely many primes of the form 𝑛𝟤 + 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 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.
Mathematisches Forschungsinstitut Oberwolfach, CC BY-SA 2.0 DE
▲ June Huh
autumn 2022 ⋅
𝟨𝟧
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!
▲ 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 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 repMathematisches Forschungsinstitut Oberwolfach, CC BY-SA 2.0 DE resenting the atoms, and an assignment of 𝟣s and −𝟣s to the points representing the ‘spin’ of the nuclei. The spin ▲ Hugo Duminil-Copin 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, 𝟨𝟨
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feature
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’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.
Mathematisches Forschungsinstitut Oberwolfach, CC BY-SA 2.0 DE
▲ Maryna Viazovska
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 𝐸𝟪 lattice and Leech lattice respectively. However, as with our 3-dimensional oranges, it wasn’t clear whether these techniques were optimal.
Böhringer Friedrich, CC BY-SA 2.5
▲ 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.
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.
Albert Wood Albert Wood is a postdoctoral researcher working with geometric flows: physics-inspired motion of curves, surfaces, and their higher-dimensional generalisations. His quest for mathematical truth has most recently taken him to Taiwan, where he takes breaks from research by cycling up nearby volcanoes.
autumn 2022 ⋅
𝟨𝟩
Is Matlab better than Python? YES
argues ELLEN JOLLE Y
argues MADELEINE HALL
while people still think Python is better than Matlab
for people who think Matlab is better than Python:
Whoever said that the best things in life are free had clearly never tried scientific computing.
There are so many reasons why Python is better than Matlab: [it’s open-source, it has named arguments, it doesn’t think everything is a matrix, . . . ]. As well as all of these obvious ones. . .
I have tried coding with Python, and I really wanted it to work out, but everything I wanted to do was always harder in Python than Matlab. Why are vectors called lists? What exactly was unexpected about my indent? Good for you if you are over these conceptual humps but personally I have neither the time nor the interest. I like my coding languages to meet me where I’m at and Matlab does just that. Learning vector and matrix manipulation in Matlab is an effortless task for a mathematician—it is effectively written in code just as it is on paper. Not in Python. Import this, import that, import who, import what?! I just want an easy life where machines understand what I tell them without me having to do anything, is that so much to ask? If you like Python, then truly I am happy for you, and I hope you enjoy all your fancy jobs and feeling clever by pretending to find counting from zero easier than counting from one. Oh, and if you’re about to tell me I’ll never get a job if I don’t learn Python. . . well, I did, with MathWorks, so how do you like that? end 𝟨𝟪
NO
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reasons.append(Python is not just simply a scripting language for mathematics. It’s also an imperative and functional language which people can use for anything and everything—from crawling web servers, to controlling external devices, to making UIs. Fun fact: the original Google algorithm was written in Python.) reasons.append(It indexes from zero which is obviously correct since zero is CLEARLY the first integer.) reasons.append(I don’t need to LITTER my functions or loops with silly little end statements.) reasons.append(It’s named after Monty Python which is way more fun than Matlab which is named after ‘Matrix Laboratory’ (BORING).) And finally, reasons.append(Python doesn’t give me traumatic flashbacks to my PhD research.)
YOUR CORRESPONDENCE
16
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Look what has arrived just in time for Edinburgh @MathsJam tonight. Hamish loves reading @chalkdustmag — @soupie666 This one’s our favourite—we love
Ou r f av e
you too, Hamish! I would really like a paper copy of issue 14 for my daughter’s birthday. There are several articles she would love! Would that be possible? Yours hopefully, Juliet Happy birthday, Juliet’s daughter!
Thanks for your great work! My mum and I both really enjoy reading it and doing the puzzles. — Asha Elizabeth Roberts
@chalkdustmag I do love the smell of a brand new Chalkdust magazine. — @dtayler
Thanks, Asha! (Thasha.) We really enjoyed reading this too.
Get in touch! Did you love one of our articles? Come across
another maths thing with a silly name? Have you made a sci-fi fan angry? We’d love to hear about it! a @chalkdustmag l @chalkdustmag c contact@chalkdustmagazine.com
Dear Sir Graham, I write with regret to request a vote of no confidence in our leader, the prime minister. – She’s lost control of the party – She has lost market confidence – She is actually divisible by 13 Dear Boris — I’m afraid we can’t help you with this. Try 0800 00 1922.
autumn 2022 ⋅
𝟨𝟫
F ormal sums Nik Alexandrakis explains what they are and what they can tell us
S
𝟩𝟢
ome two thousand years ago in ancient Greece, philosophers Aristotle and Zeno asked some interesting thoughtprovoking 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:
Since there are virtually infinitely many such points to be crossed, Achilles should not be able to reach the tortoise in finite time.
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
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
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must first reach the point where the tortoise started, so the tortoise must always hold a lead.
feature
in the first version of the problem. To be more precise, we shall formulate this problem in mathematical language, and we do so by considering the speed of Achilles, 𝑣𝐴 , the speed of the tortoise 𝑣𝑇 (and let’s say 𝑣𝐴 > 𝑣𝑇 ), and the initial distance 𝐷 which we expect to get shorter and shorter.
and Achilles will cover this distance after time 𝑡𝟥 = 𝐷𝟤 /𝑣𝐴 = (𝑣𝑇 /𝑣𝐴 )𝟤 𝑡𝟣 .
𝐷𝟤
𝐷𝟣
𝐷
Repeating this process 𝑘 times we notice that the distance between Achilles and the tortoise is
𝐷
𝑣 𝑘−𝟣 𝐷𝑘 = 𝑣𝑇 ( 𝑇 ) 𝑡𝟣 𝑣𝐴
𝐷 − 𝑣𝐴 𝑡 + 𝑣𝑇 𝑡 = 𝐷 − (𝑣𝐴 − 𝑣𝑇 )𝑡,
Summing up all these distances we get how far Achilles has to move before catching the tortoise: if we call this 𝐷𝐴 it’s
Since distance = speed × time, we can say that after time 𝑡 , Achilles has travelled a distance equal to 𝑣𝐴 𝑡 and the tortoise 𝑣𝑇 𝑡 . The distance between them is
and so Achilles catches the tortoise—ie the distance between them is 𝟢—when the time, 𝑡 , is equal to 𝐷/(𝑣𝐴 − 𝑣𝑇 ).
𝑣 𝑘 = ( 𝑇 ) 𝐷. 𝑣𝐴
𝑛 ∞ ∞ 𝑣 𝑘 𝐷𝐴 = lim ∑ 𝐷𝑘 = ∑ 𝐷𝑘 = 𝐷 ∑ ( 𝑇 ) . 𝑛→∞ 𝑣 𝑘=𝟢
𝑘=𝟢
𝑘=𝟢
𝐴
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 𝑡𝟣 = 𝐷/𝑣𝐴 . By that time the tortoise will have travelled a distance equal to
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
𝐷𝟣 = 𝑣𝑇 𝑡𝟣 ,
If |𝑎| < 𝟣, the sum tends to (𝟣 − 𝑎)−𝟣 as 𝑛 tends to ∞ and diverges otherwise, meaning that it either goes to ±∞, or a limiting value just doesn’t exist. By ‘doesn’t exist’, see for example what happens if 𝑎 = −𝟣: we get
which is the new distance between them.
𝐷
𝐷𝟣
Travelling this distance will take Achilles time 𝑡𝟤 =
𝐷𝟣 𝑣 = ( 𝑇 ) 𝑡𝟣 . 𝑣𝐴 𝑣𝐴
Then the tortoise will have travelled a distance 𝑣𝟤 𝐷𝟤 = 𝑣𝑇 𝑡𝟤 = ( 𝑇 ) 𝑡𝟣 𝑣𝐴
𝑛
∑ 𝑎𝑘 .
𝑘=𝟢
𝟣 − 𝟣 + 𝟣 − 𝟣 + 𝟣 − 𝟣 + ⋯ + (−𝟣)𝑛
and the sum oscillates between 𝟢 (if 𝑛 is odd) and 𝟣 (if 𝑛 is even). In our case, |𝑎| = |𝑣𝑇 /𝑣𝐴 | < 𝟣 so 𝐷𝐴 =
𝐷𝑣𝐴 𝐷 = , 𝟣 − 𝑣𝑇 /𝑣𝐴 𝑣𝐴 − 𝑣𝑇
which, when divided by the speed of Achilles, autumn 2022 ⋅
𝟩𝟣
𝑣𝐴 , gives exactly the time we found before. So Achilles and the tortoise will meet after Achilles has crossed a distance 𝐷𝐴 in time 𝑡𝐴 = 𝐷𝐴 /𝑣𝐴 .
Thousands of years later, Leonhard Euler was thinking about evaluating the limit of 𝑛
𝟣 𝟣 𝟣 𝟣 𝟣 ∑ 𝟤 =𝟣+ + + +⋯+ 𝟤 𝟦 𝟫 𝟣𝟨 𝑘 𝑛 𝑘=𝟣
as 𝑛 → ∞, which is named as the Basel problem after Euler’s hometown. This sum is convergent and it equals π𝟤 /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 𝑛
∑𝑘 = 𝟣+𝟤+𝟥+𝟦+⋯+𝑛 =
𝑘=𝟣
𝑛(𝑛 + 𝟣) , 𝟤
the limit of which when 𝑛 → ∞ is, of course, infinite. But now let’s think about the harmonic series, 𝑛
𝟣 𝟣 𝟣 𝟣 𝟣 =𝟣+ + + +⋯+ . 𝑘 𝟤 𝟥 𝟦 𝑛 𝑘=𝟣 ∑
This time, although the terms themselves get smaller and smaller, the series still diverges as 𝑛 → ∞. But we can still describe the sum and its behaviour. It turns out that 𝑛
𝟣 = ln(𝑛) + 𝛾 + 𝑂(𝟣/𝑛), 𝑘 𝑘=𝟣 ∑
where 𝛾 is the Euler–Mascheroni constant, which 𝟩𝟤
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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: 𝟥 𝟤 𝟣 𝟢
𝟢
𝟤
𝟦
𝑛
𝟨
ln(𝑛) + 𝛾 𝑛 ∑𝑘=𝟣 𝟣/𝑘
𝟪
𝟣𝟢
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’ 𝟣+𝟤+𝟥+𝟦+⋯=−
𝟣 . 𝟣𝟤
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 Re(𝑠) > 𝟣 as ∞
𝟣 𝜁 (𝑠) = ∑ 𝑠 . 𝑘 𝑘=𝟣
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 𝜁 (−𝟣) = −𝟣/𝟣𝟤. The other culprit here is matrix summation—another method to give some value to divergent sums. By sheer (though neat) coinci-
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dence, these methods, such as the Cesàro 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.
described earlier emerges when trying to find an explicit formula for the function
for which repeated integration by parts yields 𝑥
𝑥
An example that follows Euler’s line of thought as
𝑥
e𝑡 e𝑡 d 𝟣 d𝑡 = [ ] − ∫ e𝑡 ( ) d𝑡 𝑡 𝑡 d𝑡 𝑡 −∞ −∞ −∞ 𝑥 𝑡 𝑥 e e = [ − 𝟢] + ∫ 𝟤 d𝑡 𝑥 −∞ 𝑡
Ei(𝑥) = ∫
=⋯=
𝑛−𝟣
𝑘! e𝑥 e𝑥 ∑ + 𝑂 ( 𝑛+𝟣 ) , 𝑥 𝑘=𝟢 𝑥 𝑘 𝑥
where we’ve been able to say e𝑥 /𝑥 → 𝟢 on the second line as 𝑥 → −∞. Dividing through by e𝑥 , this allows us to say 𝑛−𝟣
e−𝑥 Ei(𝑥) = ∑
𝑘=𝟢 ∞
∼∑
𝑘=𝟢
𝑘!
𝑥 𝑘+𝟣 𝑘!
𝑥 𝑘+𝟣
𝟣 + 𝑂 ( 𝑛+𝟣 ) 𝑥
as 𝑥 → ∞.
Now swap 𝑥 for −𝟣/𝑥 in this equation: ∞
e𝟣/𝑥 Ei(−𝟣/𝑥) ∼ ∑ 𝑘!(−𝑥)𝑘+𝟣 as 𝑥 → 𝟢 𝑘=𝟢
= −𝑥 + 𝑥 𝟤 − 𝟤𝑥 𝟥 + 𝟨𝑥 𝟦 + ⋯ .
As you can see below, this series now diverges as 𝑥 → ∞, but we still see convergence of the partial (truncated) sums as 𝑥 → 𝟢, even as we add more terms: 𝟣 𝟢.𝟧
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.
𝑥
e𝑡 d𝑡, −∞ 𝑡
Ei(𝑥) ≔ ∫
−𝟢.𝟥
−𝟢.𝟤
−𝟢.𝟣
−𝟢.𝟧 −𝟣
𝟢.𝟣
𝟢.𝟤
𝑥
𝟢.𝟥
e𝟣/𝑥 Ei(−𝟣/𝑥) 𝟣
∑𝑘=𝟢 𝑘!(−𝑥)𝑘+𝟣 𝟫
∑𝑘=𝟢 𝑘!(−𝑥)𝑘+𝟣
Euler noticed that e𝟣/𝑥 Ei(−𝟣/𝑥), in its original inautumn 2022 ⋅
𝟩𝟥
tegral form, solves the equation 𝑥𝟤
Doing so, we get
d𝑦 + 𝑦 = −𝑥 d𝑥
(for 𝑥 ≠ 𝟢). Now here’s the thing: the formal sum ∞
∑ 𝑘!(−𝑥)𝑘+𝟣 ,
𝑘=𝟢
to which e𝟣/𝑥 Ei(−𝟣/𝑥) is asymptotic as 𝑥 → 𝟢, also (formally) ‘solves’ the same equation for any 𝑥.
This solution is not unique, and in fact, adding any constant multiple of e𝟣/𝑥 to e𝟣/𝑥 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.
−
∞
∞
d 𝟣 [∑ 𝑎 𝑥 −𝑘−𝟣 ] + ∑ 𝑎𝑘 𝑥 −𝑘−𝟣 = d𝑥 𝑘=𝟢 𝑘 𝑥 𝑘=𝟢 ∞
∞
⟹ ∑(𝑘 + 𝟣)𝑎𝑘 𝑥 −𝑘−𝟤 + ∑ 𝑎𝑘 𝑥 −𝑘−𝟣 = 𝑘=𝟢
𝑘=𝟢
∞
𝟣 𝑥
𝟣 ⟹ 𝑎𝟢 𝑥 −𝟣 + ∑ [(𝑘 + 𝟣)𝑎𝑘 + 𝑎𝑘+𝟣 ]𝑥 −𝑘−𝟤 = . 𝑥 𝑘=𝟢
Then for our differential equation to be satisfied the coefficients have to satisfy 𝑎𝟢 = 𝟣
and
(𝑘 + 𝟣)𝑎𝑘 + 𝑎𝑘+𝟣 = 𝟢 ⟹ 𝑎𝑘+𝟣 = −(𝑘 + 𝟣)𝑎𝑘
which recursively means that 𝑎𝑘 = (−𝟣)𝑘 𝑘! and our formal sum solution is ∞
𝑦(𝑥) = ∑(−𝟣)𝑘 𝑘!𝑥 −𝑘−𝟣 . 𝑘=𝟢
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 → ∞ by using the repeated integration by parts result ∞
−𝑥𝑠 𝑘 −𝑘−𝟣 for 𝑥 > 𝟢, ∫ e 𝑠 d𝑠 = 𝑘!𝑥 𝟢
which implies
Solving a differential equation
∞
𝑦(𝑥) = ∑(−𝟣)𝑘 𝑘!𝑥 −𝑘−𝟣 𝑘=𝟢 ∞
To elaborate further, let’s consider one more example, the differential equation −
d𝑦 𝟣 +𝑦 = , d𝑥 𝑥
∞
𝑎 𝑎 𝑎 𝑦(𝑥) = ∑ 𝑎𝑘 𝑥 −𝑘−𝟣 = 𝟢 + 𝟣𝟤 + 𝟤𝟥 + ⋯ . 𝑥 𝑥 𝑥 𝑘=𝟢
𝟩𝟦
⋅ chalkdust
𝑘=𝟢 ∞
𝟢
∞
= ∫ e−𝑥𝑠 ∑(−𝟣)𝑘 𝑠 𝑘 d𝑠.
where 𝑦(𝑥) → 𝟢 as 𝑥 → ∞. (f)
Thinking about the boundary condition there, we could substitute in the (formal) sum of powers of 𝑥 which decay away as 𝑥 → ∞,
∞
= ∑(−𝟣)𝑘 ∫ e−𝑥𝑠 𝑠 𝑘 d𝑠 𝟢
𝑘=𝟢
How is that helpful? Well, for 𝑠 ∶ |𝑠| < 𝟣 we know that ∞
∑(−𝟣)𝑘 𝑠 𝑘 = 𝟣 − 𝑠 + 𝑠 𝟤 + ⋯ =
𝑘=𝟢
𝟣 , 𝟣+𝑠
feature
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 𝑦(𝑥) = ∫ 𝟢
∞
e−𝑥𝑠 d𝑠, 𝟣+𝑠
which is the solution to our differential equation, (f), we are looking for, and is also asymptotic to ∞ the formal sum ∑𝑘=𝟢 (−𝟣)𝑘 𝑘!𝑥 −𝑘−𝟣 as 𝑥 → ∞: 𝟢.𝟦
𝑦(𝑥)
𝟣 ∑𝑘=𝟢 (−𝟣)𝑘 𝑘!𝑥 −𝑘−𝟣 𝟧 ∑𝑘=𝟢 (−𝟣)𝑘 𝑘!𝑥 −𝑘−𝟣
𝟢.𝟥 𝟢.𝟤 𝟢.𝟣 𝟢
𝟢
𝟤
𝟦
𝑥
𝟨
𝟪
𝟣𝟢
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 + ⋯ 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 d𝟤 𝑦 + 𝑦 = 𝟢. d𝑥 𝟤 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.
Nik Alexandrakis Nik Alexandrakis is a fourth-year PhD student at Lancaster University. Apart from differential equations, he is sometimes interested in environmental and animal rights activism. He hates writing but usually likes the outcome. autumn 2022 ⋅
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MATHEMATICAL CLOTHES ! At 9, it’s A Pair of Geometry Set Earrings by The Pogues.
𝟗𝟗 d (
𝟕𝟕
𝟓𝟓
etsy.com/shop/Bojanglies)
At 7, it’s a wearable version of everyone’s favourite one-sided shape: a Klein bottle hat
At 5, it’s A Pair of Pi Earrings by The Pogues.
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At 8, it’s a wearable version of everyone’s favourite one-sided shape: a Möbius scarf.
At 6, it’s great for doing quick calculations on your paper skirt: a pencil skirt.
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At 3, it’s no one’s favourite LATEX package: \usepackage{realhats}.
𝟏
Topping the pops this issue, it’s the only item you need in your wardrobe: a Chalkdust T-shirt.
At 4, it’s A Pair of Mathematical Socks by The Pogues. Read our full reviews of socks at d chalkdustmagazine.com
At 2, it’s A Pair of Curly Braces by The Pogues.
𝟏𝟎 𝟏𝟎 𝟖𝟖
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𝟒𝟒
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