Chalkdust, Issue 04 by Chalkdust

autumn 2016

chalkdust

Editorial Director Rafael Prieto Curiel The team Rob Becke Atheeta Ching Alex Doak Eleanor Doman Aryan Ghobadi Nikoleta Kalaydzkieva Rudolf Kohulák Anna Lambert Ma hew Scroggs Pietro Servini Belgin Seymenoğlu Ehsan Shahriari Adam Townsend Ma hew Wright Cartoonist Tom Hockenhull Cover adapted from Central adratic by John Crabtree, used with permission of UCL Art Museum, University College London chalkdustmagazine.com contact@chalkdustmagazine.com @chalkdustmag chalkdustmag chalkdustmag

Big (literally!) things have happened at Chalkdust since our last issue. We spent quite a few hours building a giant sphere out of recycled paper cups and it was a fantastic experience since it allowed us to visualise an interesting maths problem and to construct something with our own hands. Its construction involved most of our team: modelling the problem, collecting the many paper cups required, designing the whole sphere, stapling, taking pictures and writing up the results is not something that a single person can do. This summer we also had the opportunity to meet Professor Andrea Bertozzi and she, too, puts her success down to working in a team, collaborating as much as possible with people not only from her own field but also from others. The strengths and eﬀorts of everyone combine to produce great results. We have put together this issue hoping to make you smile with the flowchart that helps you pick the branch of mathematics that most suits you (and sorry if you like stats or physics as I do!). We wish to intrigue you with the mathematics of crime science and help you when you are faced with a maths problem. This issue will reach many of our readers in its printed version but it will reach even more people through our website, where every page of this magazine is also published and available for free. If you like it, don’t forget to also read our weekly blog. Chalkdust is a space open to contributions from you, so if you have any idea that you want to share with us, please don’t hesitate to get in touch.

Chalkdust Magazine, Department of Mathematics, UCL, Gower Street, London WC1E 6BT, UK.

Rafael Prieto Curiel Editorial Director

Acknowledgements We would like to thank our sponsors for their support, which helps us keep this project free for our readers. Thanks to Prof. Robb McDonald, Dr Luciano Rila and all the staﬀ at the UCL Department of Mathematics, as well as to all of this issue’s authors and contributors. Also, special thanks to Natasha Reith-Banks for helping us promote our project. We would like to thank Sam Brown for all of his work on our LATEX templates. Most of all, we would like to thank you, our readers, for your continued enthusiasm for our magazine. 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 2016. 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

chalkdustmagazine.com

You might think that maths and psychedelic hallucinations tend not to mix very well. But you would be mistaken! There are a series of visual hallucinations known as form constants that are highly geometric, and a mathematical model of them has provided us with some fascinating insight into how our visual cortex (the part of the brain that processes the information we receive from our eyes) works. These hallucinations were first observed in patients who had taken mescaline, a psychedelic drug produced from a cactus found in South An artist’s impression of reportings of patient’s hallucinations. America. Form constants have subsequently been reported in a number of other altered states such as sensory deprivation, waking/falling asleep states, near death experiences and by individuals with synaesthesia. Some people even report seeing these patterns after closing their eyes and applying firm pressure to both eyelids for a few seconds! The mathematical model we referred to was described in a paper by Bressloff et al., and is based on anatomical features of our brain. It seems that the visual cortex has certain symmetry properties, such as reflective, translational and even a novel shift-twist symmetry. Its electrical activity can be represented mathematically and—a bit of group theory, some eigenvectors and a couple of transformations later—has steady state solutions to the resulting equations that are remarkably similar to the observed hallucinogenic experiences. Groovy! Disclaimer: Chalkdust does not advocate pressing hard on your eyelids.

A visualisation of the steady state solutions to the mathematical model.

Written in collaboration with Samuel Mills

Pikachu adapted from picture by Matt Levya, CC BY 2.0; Hallucination pictures taken with kind permission from PC Bressloﬀ, JD Cowan, M Golubitsky, PJ Thomas and MC Wiener. What geometric visual hallucinations tell us about the visual cortex. Neural Computation 14(3) (2002), 473–91.

autumn 2016

chalkdust

In conversation with

ANDREA BERTOZZI Rafael Prieto Curiel & Anna Lambert

prediction, robotics, big data, image processing, fluid dynamics. Andrea Bertozzi, professor of applied mathematics at the University of California, Los Angeles (UCLA), has worked in all these areas and more. Her curiosity and many collaborations have made a real impact in our modern world. We meet Bertozzi at the lobby of her London hotel for breakfast. It’s a typical week for her: she has already given a lecture at the University of Oxford on image processing and is preparing to give yet another talk about big data at the British Library. However, she has still found the time to welcome us warmly and share some fascinating anecdotes from her varied research career.

As an applied mathematician, I am being paid to learn new things.

Bertozzi has always been surrounded by science: her father is a nuclear physicist at the Massachuse s Institute of Technology (MIT), and her sister a chemist at Stanford. “As a young kid, my parents would take us to the science museums so that we could pick one area to study seriously as an adult. But the problem was that I liked everything! So that’s why I moved into applied maths. With that you can do everything, from science to engineering and even social sciences.” Indeed, her career as an applied mathematician has seen her accomplish that goal. While she began her research with largely theoretical work on nonlinear partial diﬀerential equations, she has more recently studied collaborative robots, image processing, big data and crime sciences. chalkdustmagazine.com

chalkdust She was always an outstanding student: when challenged to work on a project during her undergraduate degree at Princeton, she remembers that she had handwri en about 10 pages as the solution for a fluids problem. Her undergraduate advisor, Andrew Majda, convinced her to write it up as a paper; this became her first publication, a single author paper at the age of 22, when she was in her first year of graduate school. It was the first of many to come: Bertozzi has wri en nearly 200 scientific papers and been cited thousands of times. Amazingly, during 2015, she published a scientific paper every three weeks, and gave a talk every 10 days.

Andrea Bertozzi

Experimental apparatus used in Bertozzi’s lab to study particle laden flow. Opening a gate to an oil reservoir causes oil and other particles to rush out and a laser sheet illuminates the profile of the mixture.

Not many researchers work in so many branches of mathematics and science, but it is precisely this that gives Bertozzi such a vast array of methods to tackle whatever problem she is currently working on. When presented with a new and difficult challenge, she describes the feeling when “suddenly a light bulb comes on: if we write that equation from image processing in a different way, we are solving an equation found in incompressible fluid dynamics. We are actually solving the Euler equations with the Navier–Stokes boundary conditions. So if we add diffusion we will have the right equations with the right boundary conditions!” That realisation led to a whole new method in image inpainting (the process of reconstructing parts of an image that might be deteriorated or corrupted). The ability to make such surprising connections is a crucial part of interdisciplinary research. She describes her research process as “going out there and talking to people. When you work in other fields, you need to work with a person who knows the right questions to be asking, believes in the scientific process and has some data.” For Bertozzi, science is like “walking around with a jigsaw puzzle with a missing piece, and sometimes you meet the person who holds that piece”. Perhaps inspired by those childhood science museum visits, Bertozzi now runs an experimental laboratory in addition to doing her mathematical work. This is unusual for a mathematician, but in keeping with Bertozzi’s broad outlook on the world. In her lab, her group studies collaborating robots (where robots have to perform a joint task, forcing them to coordinate) and complex fluid flow—allowing Bertozzi to pair her mathematical models with experimental observations. She even has a couple of patents, one for inpainting and one for data fusion mapping estimation (a method that gives more geographically accurate probability density estimates).

Research is going out there and talking to people.

She lived through the computer revolution in the early years of her career. As a college student she had a summer job with a management consultancy firm where she worked with first generation commercial desktop computers writing billing so ware, installing hardware and 5

Without programming, you cut yourself oﬀ from twothirds of the problems you could be working on. autumn 2016

chalkdust working on clients’ datasets. Although Bertozzi le the business world to pursue an academic career, her interest in computing remained. She enthuses about the opportunities that scientific programming creates: “you are handed a tool that allows you to visualise the mathematics you have been studying—so all of a sudden we knew what questions we should be asking. Computers had a big impact on my research.” Now she insists that all her students learn programming, even those working on proving theorems, “because otherwise you cut yourself off from two-thirds of the problems you could be working on”. Her incursion into crime research is perhaps the most surprising of the many fields she has worked on. While she was initially hesitant about working on social issues, due in part to the availability of reliable data, a meeting with the anthropologist Jeffrey Brantingham was enough to change her mind. “We started working with more than 48,000 automobile the s registered by the Los Angeles Police Department, with the location, time and details recorded. At the time I had no idea what sort of maths we could use and how interesting it could be, but I gave it a try.” It certainly paid off. Her group developed several models to predict where and when a crime is likely to occur based on past events. One particular model by George Mohler and Bertozzi used a mathematical structure called a self-exciting point process (similar to a Poisson process, but with history playing an additional role). This looked very promising, and within a year of its publication, the Santa Cruz Police Department used it to deploy officers to the areas with the highest probability of a crime occurring. The results were astonishing: in the first month they reported a 27% decrease in the crime rate. The so ware developed using this model, PredPol, is currently used in more than 30 cities worldwide to make be er policing decisions, prevent crime and keep cities safer. As a result of the success it has had in reducing crime, this research has been covered many times on TV and in the media, but for her it is a superficial thing. “Ge ing the stuff on the news was kind of fun, sure. You’re a rock star for a day, but as a mathematician, what is really exciting is when you figure something out. I remember discovering under-compressible shocks in the 90s and that was one of the most exciting days. The impact you have is fulfilling in a different way.”

On TV, you’re a rock star for a day, but as a mathematician, what is really exciting is when you figure something out.

chalkdustmagazine.com

Bertozzi describes her multidisciplinary approach as “having a foot in mathematics but having another foot in many other fields”. Her perception is that “sometimes mathematicians live in a bubble, doing things only in the world of mathematics, without seeing how science works in other fields. But to be collaborative you need to see how others do science.” 6

chalkdust Bertozzi has more than 120 co-authors on research papers, including top researchers in their own fields: Stanley Osher and Guillermo Sapiro (experts in image processing), Paul Weiss (expert in atomic scale measurements and control), and Bob Behringer and Anne-Marie Cazabat (experts in so condensed ma er). One day she is working with robotics and the next she is meeting with the chief of the Los Angeles Police Department. Bertozzi a ributes her academic success to being lucky: “I had to be given the right problem to start with and I had to meet the right people.” She extends this selfless a itude towards her group: she says proudly that “this work would not have been done without these really talented young people”. To date, she has supervised nearly 30 postgraduate students and over 40 postdocs—and she sees much of her role today as one of a human resources developer: “I feel very strongly that it is not just about me. I can only do maths for 24 hours a day, but when you work with others, you really multiply that time. That is when you can have an impact on society. I care deeply about developing the next generation and helping them to become great scientists.” And so, our interview ended, allowing Bertozzi to head over to the British Library to give her talk on big data and do exactly that: share her knowledge and experience with tomorrow’s scientists. Rafael Prieto Curiel & Anna Lambert Rafael and Anna are PhD students at UCL. Rafael studies mathematics and crime, and Anna is working on mathematical models of bioreactors. @rafaelprietoc

@anna_lambert

rafaelprietoc.wordpress.com

My favourite set In mathematics, a set is a collection of items. We have spread some of our favourites throughout this issue. We’d really love to hear about yours! Send them to us at contact@chalkdustmagazine.com, @chalkdustmag or chalkdustmag, and you might just see them on our blog!

The Cantor set Belgin Seymenoğlu

If you want to construct my favourite set, start with the interval [0,1]. Next, remove the open middle third interval. This gives you two line segments: [0,1/3] and [2/3,1]. Again, delete the middle third for each remaining interval (which leaves you with four new intervals). Now repeat the final step ad infinitum.

Once you’re done, you’re le with the Cantor set (also called the Cantor comb). But what does the Cantor set look like? Infinitely many discrete points? An infinite collection of line segments? The answer is a bit of both, and that’s because the dimension of my favourite set is neither zero nor one: it’s ln 2/ ln 3 ≈ 0.63093. The other wonderful feature of this set is that it’s a fractal, so if you zoom in on a tiny portion, you get the Cantor set again! 7

autumn 2016

& WHAT’S

WHAT’S

HOT NOT

Maths is a fickle world. Stay à la mode with our guide to the latest trends.

Agree? Disagree? @chalkdustmag chalkdustmag

NOT Paper Möbius strips

HOT

As Dirichlet says, Möbius strips should be band.

Mathematical crochet Hyperbolic surfaces! Klein bottles! Impress fellow passengers with your DIY manifolds!

Gravitational waves HOT 1916: predicted; 2016: discovered. Einstein: always right.

HOT

Dirichlet’s counting problem Still unsolved since Issue 03.

Fermat’s Last NOT Theorem More like Fermat’s Worst Theorem. We get it: Wiles solved it… in 1995! Next!

HOT Comic books about maths Who wouldn’t want to look at drawings of Hilbert, Feynman, Lovelace & Babbage with speech bubbles?

Films about maths Oh look, the socially awkward man with wild hair has locked himself away to draw on windows. SEEN IT!

NOT Katrina and the NOT Waves How do you even walk on sunshine? Light has no mass… or does it??

Autograph

NOT

Autograph repair, Autograph REPLACED.

GeoGebra Is there a new teacher who doesn’t ❤ this? Plot away with this free software for all platforms.

HOT

Pictures CC-BY-SA Gravitational waves: Alejandro Mos Riera; Wiles: Klaus Barner. Background: Flickr user Harmon. CC-BY-NC Crochet: Flickr user Pandaeskimo. Permission GeoGebra. Public domain Fermat. Chalkdust Comic.

chalkdustmagazine.com

More free fashion advice online at chalkdustmagazine.com 8

chalkdust

Problem solving 101 Wikimedia user Scouten, CC-BY-SA 3.0

Stephen Muirhead

the outside looking in, maths problem solving can seem like a kind of magic. Here is a typical image: a lone genius, peering at a vexing problem, rubs their chin, paces up and down; then a bolt of inspiration hits and the solution falls neatly into place.

And while it’s true that inspiration can strike the lucky few, for the rest of us this is no more than an illusion (and o en a carefully cultivated illusion at that). In reality, problem solving is usually much more prosaic, nothing more than a careful application of well-known, and o en quite elementary, techniques. So what are these elementary techniques? In this article, I’ll look at some of the simplest and easiest to understand. Happily, they are also some of the most powerful and widely applicable. These techniques will be explained by way of example problems; I strongly encourage you to a empt the problems yourself before reading the solutions.

Get some intuition Faced with an unfamiliar problem, it can o en be diﬀicult to know where to begin. An important first step is to develop some intuition. This will o en allow you to guess the correct answer—an important step in itself—or, if you’re lucky, point you in the direction of a full solution. A good way to develop intuition is to play around with the parameters of the problem, looking for particular easy cases in which the solution is clear. Let’s look at an example. 9

autumn 2016

chalkdust Problem 1. Two people take turns placing identical plates on a square table in such a way that they do not touch; the first who cannot place another plate loses. Which player has a winning strategy? At first glance this problem appears underspecified. How large is the table? What shape and size are the plates? This seems an additional diďŹ&#x20AC;iculty to surmount. In another light, it also suggests an opportunity to gain intuition, by varying these parameters. For example, suppose the table were very, very small. Then clearly the first player has a winning strategy, because there would only be space for a single plate. Although far from a proof, this gives a strong hint that the first player always has a winning strategy, since otherwise there must be a critical threshold of table size at which the solution to the problem fundamentally changes, and this is rather implausible.

To develop this idea further, you might next consider a table that is only slightly larger than the size of a single plate, and determine a winning strategy in that case. We will see the solution to this problem later. For now, letâ&#x20AC;&#x2122;s see a second example.

Problem 2. Would it be quicker to run along a moving walkway (a travelator) to the end and back again, first with the flow and then against the flow, or to run the same path along the ground?

Once again the problem appears underspecified, and once again this suggests a strategy for guessing the answer, by playing around with the parameters. Suppose that the travelator were moving very fast, faster than you could run. Then you could never run successfully against the flow, and so it is certainly quicker to run on the ground. On the other hand, if the travelator had zero speed, then the two routes would take the same amount of time. This strongly hints at the correct answer: that taking the travelator is always slower, unless it is not moving.

Flight to Melbourne

FINAL CALL â&#x2C6;&#x17E;

chalkdustmagazine.com

chalkdust

Find and exploit the structure of the problem Maths is essentially the study of pa erns, and a problem will generally be easier if you can identify and exploit some structure. A common example is symmetry—geometric, algebraic, etc—and exploiting symmetry in an appropriate way o en leads to very simple and elegant proofs. Symmetry comes in many varieties, but in general you should look for transformations of the problem (rotation in space, permutation of variables, etc) that leave the important features unchanged. Problem 3. If the outer square on the le has area 1, what is the area of the inner square? Given this, calculate the proportion of the figure on the right that is coloured blue. There is strong symmetry in both of the above pictures. Aside from the obvious reflective and rotational symmetries, there are also subtler versions more relevant to the quantities we wish to calculate. In the first image, the area of the inner square does not change if we rotate it, leaving the outer square fixed. Why is this useful? Well, consider rotating the inner square 45 degrees. Then by decomposing the outer square into quarters, shown here on the le , it is clear that the inner square has half the area of the outer square. In the problem’s second picture, notice that the same image is repeated at a smaller scale inside the inner square: this can be thought of as nested symmetry. This implies that the proportion of the whole square that is coloured blue is the same as the proportion of the inner square that is coloured blue. Combining this with the first observation, we come to the equation

Total blue area

Blue area in outer ring

1 2

× Total blue area.

In other words, Total blue area = 2 × Blue area in outer ring. Since the blue area in the outer ring is 1 − π/4 (subtract the area of the circle from the area of the square), and the area of the whole square is 1, the proportion of blue in the figure is 2 − π/2. 11

autumn 2016

chalkdust Problem 4.

Calculate the values of ! " √ 1 1+ and 2 + 2 + 2 + 2 + ···. 1 + 1+ 1 1 1+···

These expressions look completely diﬀerent to the squares and circles we saw in the previous problem, but mathematically the problems have a similar structure, just in an algebraic rather than geometric se ing. In particular, they have nested symmetry that we can exploit in the same way as before. Let the values we wish to calculate be X and Y. Then, exploiting the nested symmetry, X and Y must satisfy the equations √ 1 X=1+ and Y = 2 + Y . X (Some care is needed to justify working with the limits, but we will ignore this point.) These equations are now easy to solve, giving the values √ 1+ 5 X= and Y = 4. 2 Problem 5.

How soon a er midnight do a clock’s hour and minute hands first meet?

This problem can be readily solved using simultaneous equations, but we can also exploit the symmetry of the clock to give a simpler solution. Imagine watching the clock for a full 12 hours from midnight until midday, and notice that the hands start together and end together. Between these times, the symmetry of the clock implies that the times at which the clock hands meet are evenly spaced. To see this, consider the moment when the hands first meet and imagine rotating the clock face until the hands are upright. Then, the movements of the clock hands from this moment on are identical to their movements from midnight, and so the situation is symmetrical. How many times do the hands meet between midnight and midday? Well, the hour hand goes once round the clock in this period, and the minute hand goes round 12 times. Hence they meet 11 further times between midnight and midday inclusive, and the first time they meet is therefore 1 Midnight + × 12 hours ≈ 1.05 am. 11

12 9

3 6

Finally, let’s return to the problem involving placing plates on a square table (Problem 1), and use symmetry to give the first player a winning strategy. Such a strategy needs to ensure that, no ma er the current state of the table, if the second player can place a plate, the first player can also do so. This suggests that a strategy of “copying” the moves of the second player might work. Indeed, if the first player places the first plate in the dead centre of the table, they can play any further moves by rotating the second player’s most recent move by 180 degrees. It’s easy to check that this guarantees a win for the first player. chalkdustmagazine.com

chalkdust

Look for quantities that do not change A question that o en perplexes newcomers to mathematics is how to go about proving that something does not exist, or that a procedure is impossible. Of course, proving that something exists (for example, an integer x such that x 2 = 144) is simple: just construct it explicitly. But to prove the opposite would seem to require checking an infinite number of possibilities. The technique of finding quantities that do not change—invariants—or more generally quantities that change in a controlled manner—monovariants—gives rise to elegant and simple solutions to problems of this sort. Let’s see a typical example.

Problem 6. Prove that a 5 × 13 rectangle cannot be cut up into pieces and rearranged into an 8 × 8 square. You could a empt all the diﬀerent ways of cu ing up the rectangle, and show that none of them can be arranged into the square. It wouldn’t take long to realise that the number of ways is infinite. Instead, we construct a quantity that doesn’t change, no ma er how the rectangle is cut up. If this quantity is diﬀerent for the rectangle and the square, we will have shown that no cu ing procedure can change one into the other. The total area of a shape gives a natural invariant: when a piece is cut in two, no ma er how, the total area stays the same. Since the rectangle’s area is 65 and the square’s is 64, we have proved that the task cannot be done. Apart from proving that things are impossible, invariants can also be used to show that a given procedure always produces the same result, no ma er how it is undertaken. Let’s see an example.

Problem 7. Suppose that the numbers 1 to 20 are wri en on a blackboard. At each turn, you may erase two numbers a and b and write the sum a + b in their place. A er 19 such turns, there will be one number le on the board. What will this number be?

If you try this a few times, you’ll soon see that the final number is 210 no ma er which way you do the erasing. To prove this, notice that there is a simple quantity that does not change no ma er how you erase: the total sum of all the numbers currently wri en on the board. Hence, the total sum at the end of the process—which is just the final number—must be equal to the total sum of the numbers at the start of the process, which is 1 + 2 + · · · + 20 = 210. A monovariant, which changes in a predictable manner, might be a quantity that always increases, or one that always alternates between 0 and 1. Consider the following example. 13

autumn 2016

chalkdust Problem 8. Three coins lie well spaced out on a table. Each turn, you flick one of the coins so that it passes between the other two without touching them. If you do this 99 times, is it possible for you to return each of the coins to their original positions? Suppose we try to return the coins to their original positions and fail each time. Could we prove it is impossible? It’s not obvious how to use an invariant here: we are asking precisely that the coins end up in an identical position, and so it seems that no invariant could prove this impossible. But remark that we want to return the coins a er an odd number of flicks. This suggests looking for a quantity that always alternates between two values, say from 0 to 1 and then back to 0, etc. If we do this, we will have proved the impossibility of returning the coins a er 99 flicks.

Such a quantity is the cyclic ordering of the coins: in other words, if the coins are labelled A, B and C, it is either the ordering ABC or ACB, depending on which way the labels go when taken in a clockwise order. When we make a flick, this cyclic order always switches. Hence a er 99 flicks the coins will lie in a diﬀerent cyclic order, and so cannot be in their original positions.

In all these examples the invariant/monovariant was very intuitive, but o en this will not be the case. A big part of the art is finding an appropriate invariant. Consider the following problem, which is one of my favourites. Problem 9. Three amoebae are placed at positions (0, 0), (1, 0) and (0, 1) on a two-dimensional la ice. Each turn, one of the amoebae reproduces by spli ing in two, with one oﬀspring moving to the site directly right and the other oﬀspring moving to the site directly above, but only if both of these sites are currently vacant (if they are not, the amoeba stays put). Is it possible that eventually the three initial sites will all be free of amoebae?

1 8

1 16

1 32

1 64

1 4

1 8

1 16

1 32

1 2

1 4

1 8

1 16

1 2

1 4

1 8

chalkdustmagazine.com

A quick play around may convince you that this is impossible—the grid gets clogged up pre y quickly—but it is very hard to be sure. To prove this conclusively, we construct a quantity that does not change, and show that vacating the three initial sites would imply a change in this quantity. The idea is to assign a score to each square that contains an amoeba. Since the amoebae split in two, and the oﬀspring occupy the squares to the right and directly above, it is natural to score the board with a geometric series, as shown on the le . This ensures that the total score of all the amoebae never changes. 14

chalkdust Now consider that the initial score is 1 + 1/2 + 1/2 = 2. What would be the final score, if we were able to vacate the three initial sites? Well, clearly the final score must be strictly less than the total sum of scores on the board outside the three initial sites. Summing up over the whole board as on the right, this number is 3 4 5 6 x= + + + + ··· . 4 8 16 32

1 8

1 16

1 32

1 64

1 4

1 8

1 16

1 32

1 2

1 4

1 8

1 16

1 0

1 2

1 4

1 8

There are several ways to compute the value of x, for instance by decomposing it as 3 4 5 6 2x = + + + + ··· 2 # 4 8 16 $ # $ 3 3 4 5 1 1 1 = + + + + ··· + + + + ··· 2 4 8 16 4 8 16 # $ 3 1 1 1 1 = +x+ + + + ··· , 2 2 2 4 8 and using the formula 1=

1 1 1 + + + ··· , 2 4 8

to get that the total score, x, is exactly 2, precisely the value of the initial score. So we have a contradiction (but only just!).

Consider the extremes The final technique we will look at involves extremes: that is, mathematical objects that are the biggest, smallest, quickest, etc. There are two essentially distinct ways in which extremes are useful in problem solving. The first is to prove that something cannot exist, arguing by contradiction as follows. Begin by assuming the thing exists, and deduce that there must be a most extreme version of that object (note that this may not always be the case: there is no largest prime, for example). Then, use this to construct a version that is even more extreme, resulting in a contradiction. Let’s see a classic example. Problem 10. p/q.

Prove that

√

2 is irrational, ie no integers p and q exist such that

√

√ We argue by contradiction. Suppose that 2 were rational and hence could be wri en as a fraction. Then consider the fraction in its reduced form: that is, the fraction p/q such that p and q are integers with no common factor (this is the most extreme version of the fraction, in this context). Now we deduce that √ p 2= =⇒ p2 = 2q 2 , q 15

autumn 2016

chalkdust which implies that p is a multiple of 2. Substituting p = 2r, we get (2r)2 = 2q 2 =⇒ q 2 = 2r 2 , which implies that q is also a multiple of 2. Hence the fraction p/q was not in reduced form (ie we can find a fraction that is even more reduced), which establishes the contradiction. The second way in which extremal reasoning is useful is to prove that something does exist, by constructing it explicitly. In order to come up with a construction, it is usually helpful to start with extremal objects, and base the construction around them. Problem 11. In a ‘round robin’ tennis tournament of N ! 3 players, each player plays each other exactly once. If no one won all their games, prove that we can find three players such that Player A beat Player B, Player B beat Player C, and Player C beat Player A. The tricky thing about this problem is that there are many ways the tournament could have played out, and it is diﬀicult to know where to look for players with the desired property. One idea is to base the construction around the player that won the most games, the extremal player in this context—call this Player A (if there are several such players, just pick one of them arbitrarily).

C B?

What do we know about Player A? Well, certainly they beat some of the players (in fact, at least half). Also, because no player won all their games, at least one player beat Player A: pick one of these and call them Player C. Now consider all the players that Player A beat. All that is le is to deduce that at least one of them also beat Player C, because then this player can be Player B. But this must be true, because if it were not, Player C would have beaten more players than Player A (all the players that Player A beat, plus Player A), contradicting the assumed extremality of Player A.

Problem solving 102 With some practice, the above basic techniques will get you a fair way in being able to tackle a variety of maths problems. But they are by no means the end of the story. Beyond these, there are many other general techniques that are widely applicable and extremely useful; a good reference is Paul Zeitz’s The Art and Cra of Problem Solving, from which several of the problems in this article have been taken. Of course, each subfield of mathematics also has its own techniques that are more or less specific to it. One of the great pleasures of maths is that the well of techniques is very deep, and hopefully is a long way from running dry. Stephen Muirhead Stephen is a postdoc at the University of Oxford, working in probability theory. He has taught problem solving at UCL and helped train the Australian maths Olympiad team. stephen.muirhead@maths.ox.ac.uk chalkdustmagazine.com

chalkdust

Have you been taking your weekly dose of chalkdust?

How long should a deuce last?

Mathematical crochet?!

Is Donald Duck good at maths?

What are gravitational waves?

Are you making your tea correctly?

How many cups make a sphere?

How is the Highway Code wrong?

Is our voting system broken?

All these, and more, have been answered on our weekly blog, covering topics from cosmology to sport, ďŹ&#x201A;uid dynamics to sociology, history to politics, as well as great puzzles. Read it every week and sign up to our monthly newsletter at

chalkdustmagazine.com

Motorway: Klaus with K, CC-BY-SA 3.0. Highway Code: Paul Downey, CC BY-SA 2.0. Flags: glasnevinz, CC BY-SA 2.0. Gravitational waves: Charly W. Karl, CC BY-ND 2.0. Crochet: Cheryl, CC BY-SA 2.0.

autumn 2016

chalkdust

The doodle theorem, and beyond… Colin Wright

of the things I like about recreational maths is how we can start with a simple game, play around a bit, poke in the corners, and suddenly fall down a deep hole into some serious mathematics. In this article we start with some well-trodden ground, which some readers will find familiar. However, we quickly find that all is not as it seems, and we soon stumble over a veritable pot of gold. To see how, read on…

A simple game There’s a game many children are introduced to, one way or another. It takes several forms—one is this: There’s a drawing on the right: see if you can reproduce it without going over any lines you’ve already drawn, and without li ing your pencil from the paper. As it happens, this was exactly the challenge that faced the residents of Königsberg centuries ago: could they take their Sunday a ernoon stroll, crossing each of the seven bridges in the city exactly once? Some have it that the puzzle required that they return to their starting point, while others didn’t add that extra condition, but over time all the residents came to agree that it was impossible.

Can you reproduce this drawing?

And that’s where normal people leave this sort of puzzle. They try for ages, decide they can’t do it, and put it down. On the other hand, mathematicians then take up the challenge: is it really impossible? chalkdustmagazine.com

chalkdust Can we prove it to be impossible?

The seven bridges of Königsberg

That was done, as many readers will know, by Euler in 1736, and is marked by some as the birthplace of graph theory. Euler reasoned that if it were possible to walk about and cross every bridge exactly once, then every piece of land you visit has to have an even number of bridges—one to enter and one to exit for every visit. The plan of Königsberg, however, clearly shows that every piece of land has an odd number of bridges. So it is clearly impossible.

And so it is when trying to reproduce a drawing. To be able to draw the first diagram (or any other diagram) without retracing lines, and in a single continuous stroke, every vertex, every meeting place, must have an even number of lines entering it. If not, we will enter and depart some number of times, and finally enter and have no way out. Of course, it’s OK for the starting point and ending point to have an odd number of lines, but those are the only two. Returning to our original puzzle, we can see that we must start at one of the bo om two points and end at the other, but since every other vertex (meeting place) has an even number of lines, it must be doable.

We must start and end at one of the bo om points.

Have you seen the trap? I’ll give you a moment. It’s subtle until you see it, and some people still don’t really understand, even when it’s pointed out. Have you seen it? Take a moment. So what is the trap? I have followed in Euler’s steps and shown that if a diagram can be drawn then all but at most two of the vertices must have an even number of lines. I then asserted—without proof—that the converse was also true. I said that if every vertex has an even number of lines meeting it, then it will be drawable. “Well,” say many people. “It’s obvious, isn’t it? With an even number at every vertex, every time you enter you can leave again, so nothing can go wrong. Only stands to reason.” Ah, but it’s not true. Even though most people get le with the impression that this is true (and most of the time when this is shown to children the implication is obvious) it actually isn’t. At this point some of you will be splu ering, but others will be nodding along. On the right is an example of a diagram where all the vertices have even degree (an even number of lines) and yet it cannot be drawn in a single stroke.

Every vertex has an even degree, yet the picture cannot be drawn with a single stroke.

Now some people complain that I’ve tricked them, and of course they were only thinking of con19

autumn 2016

chalkdust nected examples. Well, fair enough, if you were only thinking of connected examples then maybe it is true. But is it? Really? How do you know? And that’s where we need to have a proof. So here’s a conjecture: Every connected network in which every vertex has even degree can be completely traversed in a single journey that returns to its starting point and does not retrace any paths. Now you’ll notice here that I’ve added the condition of returning to where we started, because that now means that we can’t have the alternate condition of two vertices having odd degree. It just makes life simpler, and it’s not hard to prove the more general case from this one. So how do we prove this? There are several proofs, and people diﬀer as to which they think is the easiest and cleanest. The one I like is the following. Take a walk, and return to where you started. You never get stuck when you do this, because at each vertex you use lines in pairs, so every time you come in, there’s always a way out. If you covered every path, you’re done. If not, do it again, but when you come across one of the paths you didn’t take last time, take it now as a diversion, and wander about on paths you didn’t cover the first time until you return to that branch point, and then complete your original journey. You’ve now covered more of the network. Keep doing this, keep augmenting your journey, until finally you’ve covered everything. That’s not a formal proof, but it gives the right idea. You need to check that the things I’ve told you to do are always possible, but that’s not too hard. So in this case the converse is true: you can draw the diagram if and only if every vertex has even degree.

Here’s another example of a doodle that can be drawn in a single stroke.

It’s interesting to note that neither the statement nor the proof actually require that the network be drawn on a plane as a diagram. It’s true for networks in general, which is nice. But now, let’s consider the case where it is drawn on a plane. Here’s an example on the le . Every vertex has even degree, so we can draw the network, returning to our starting point, without retracing steps, and without retracing a line we’ve already drawn.

Adding a twist So far so good. Claim: we can draw this in a single pencil line without crossing over a line we’ve already drawn. Try it. To the right there’s an example where it’s gone wrong. In this a empt we’ve made a good start, but got ourselves into a position where as we approach the next vertex we can see that we will have to cross over a line we’ve already drawn. So that’s not allowed. Is it possible? Try for yourself… chalkdustmagazine.com

A failed a empt to draw the doodle without crossing a line.

chalkdust Again, if any vertex has an odd degree then we won’t be able to do this, but now the claim is that the converse holds, and it’s no longer quite so obvious. A few examples and you might convince yourself that it works, but just because it works on a few examples, that doesn’t mean it’s always true. So is it always true? Can we make every vertex look something like the picture to the le ? Can we draw the lines through a vertex like this?

Really?

Well, yes, we can make every vertex look something like that, and the proof is quite similar to the one we gave above for the original problem. We won’t go into it here, we’ll leave that as an exploration for the interested reader.

A diﬀerent challenge Instead, let’s make another observation about a network drawn on the plane, where all the vertices have even degree. Because of our first theorem we know that it can be drawn in a single line. So take any such ‘doodle’, and try to colour it in, checkerboard style. There’s an example on the right. You will find, I claim, that every doodle can be coloured like this using just two colours. Try it. Again, try a few examples. Get someone else to draw a doodle, and try to colour that. Try to draw one that can’t be coloured. You’ll find that you can’t, no ma er how hard you try. Claim: every doodle can be two-coloured.

An example of a checkerboardstyle coloured doodle

Again, at this point normal people will say “huh, that’s curious” and then move on. Mathematicians, on the other hand, will wonder why this is true. Indeed, they will start by wondering whether it is true. Sure, it worked for all the small examples they tried, but what if there’s a doodle with a million billion regions, will it still work for that? Good question. Yes. Here’s a proof. We’ve already seen (although not proved here) that any doodle can be drawn in a single, continuous pencil stroke such that • no line is retraced; • we finish where we started; and • we never cross over a line we’ve already drawn.

A doodle with its vertices slightly “exploded”: the doodle is simply a distorted circle.

That means the vertices can be sort of ‘exploded’ into small regions of their own, and the path enters and exits the region without ever touching the other parts of the path that visit that vertex. Now the doodle is simply a distorted circle, and that means it has an inside, and an outside. We can shade the inside, and now we have a two-colouring of the original doodle. 21

autumn 2016

chalkdust

Into the unknown Of course, colouring regions like this restricts us to living in the plane, but there’s a way to release ourselves from this limitation. What we do is think of the regions on the doodle as countries, and put a capital city in each country. Then if two countries share a border, we join their respective capitals with a road. In this way we end up with a new network, a network of roads. Colouring the regions corresponds to colouring the capitals, and the rule is now that if two capitals are joined by a road then they must get diﬀerent colours. Below we can see a general example, not just one that’s made from a doodle. From doodles to networks So now instead of a map we have a network, and instead of colouring regions, we are colouring vertices. The reason this is useful is that networks are an abstract concept, not limited to living in the plane. With a map in the plane, for example, it’s impossible to have five regions that all border each other, but with a network we can simply declare that we have five vertices and edges between them all. We have gained flexibility and generality, but we have now lost our proof that certain networks/doodles are two-colourable.

But we can rescue that. In our new network, imagine moving from vertex to vertex, travelling around the network and returning to where we started. If the network vertices can be two-coloured (and remember, they correspond to the regions in the original doodle) then we must alternate colours: black, white, black, white, and so on. So any wandering around the new network must consist of an even number of steps. So now we have a simple way of deciding whether or not a network can have its vertices two-coloured: check to see if every circuit is of even length. That sounds like a huge job, but actually there is a really easy way of doing it. Colour one vertex red, then all its neighbours green, then all their neighbours red, and so on. If you succeed then you’ve proved that all the circuits are of even length, because when you traverse a circuit you keep alternating colour. Here’s a coloured map with its capitals connected.

But if that process goes wrong it’s easy enough to show that there must be an odd length circuit. We’ll leave that as an exercise for the dedicated individual.

So where is this pot of gold? So we have shown that there is an easy way to see if a network can be drawn without li ing the pencil oﬀ the paper. The same test, to check whether every vertex is of even degree, even works for networks not necessarily limited to the plane. We’ve also seen that there is an easy way to discover whether a network can have its vertices twocoloured: it’s possible if and only if every circuit is of even length, and we can test that just by chalkdustmagazine.com

chalkdust trying it! If it fails, we find an odd length circuit. If it succeeds, then, well, it’s succeeded. So there is a simple test to see if we can visit every edge of a network, and there is a simple test to see if we can two-colour the vertices of a network. What next? Well, we can ask if there is a simple test to see whether we can create a cycle that visits every vertex of a network exactly once. Or we can ask if there is a simple test to see whether it’s possible to three-colour the vertices of a network. And here is the fun part: no one knows. The first question—finding a cycle that visits every vertex—is called finding a Hamilton cycle, and we know of no easy way either to find one, or to prove that there isn’t one. The second question is simply called three-colouring, and there is currently no known efficient way of knowing whether an arbitrary network can be three-coloured or not. We just don’t know. People have investigated these questions for a long time and they have discovered something really cool. If you can solve one of these problems, you can use it to solve the other. So in some sense these two problems, even though they look totally different, are kind of the same. And there’s more. These two problems are not an isolated pair. There are hundreds of problems that are all basically equivalent. And because they are equivalent, solving any one of them will thereby solve them all, and similarly, showing that any one of them has no solution will as effectively show that none of them have solutions. This question—whether these problems have efficient solutions or not—is known as the P vs NP problem, one of the Millennium Prize Problems published by the Clay Mathematics Institute, and for which they have offered a one million dollar prize. Well, that escalated quickly. Colin Wright Colin has a PhD in mathematics and is now probably best known for the presentation Juggling: Theory and Practice. He likes using the Twi er Android app to complain about the Twi er Android app. Ben Sparks

@ColinTheMathmo

solipsys.co.uk/new/ColinWright.html

Odd sums

1+3 1+3+5 1+3+5+7 ? What is ? What is ? 7 + 9 + 11 5+7 9 + 11 + 13 + 15 sum of first n odd numbers What is ? sum of next n odd numbers What is

Answers at chalkdustmagazine.com

autumn 2016

Where do you belong? (mathematically speaking)

(1 + δ)n ≈ 1 + nδ + O(δ2) for |δ|<< 1 How do you feel about the real BAD world?

Do you have issues looking past the third dimension?

APPLIED NO

YES

Are you patient with computers?

Has Disney’s key to success been love?

Is your mug a doughnut?

YES Always wondered how planes fly?

YES YES

NUMBER THEORY

GAME THEORY

chalkdustmagazine.com

REALLY??!

Tetris or chess?

FINANCE

Are you a sucker for primes? YES

YES

NO, IT’S ALL ABOUT MONEY

(1 + = 1 + nδ

Which is correct?

The physics department is right by the stats building!

YES

δ)n

PURE

NOT TOO BAD It would be wise to reconsider!

YES

Do you like stats?

START

GEOMETRY

ANALYSIS

chalkdust

Proof by storytelling Jane Hissey, janehissey.co.uk

Jack Hodkinson

%n&

are many equivalent ways of% defining the binomial coefficients r (pronounced ‘n & n choose r ’). In this article, though, r is defined simply as the number of ways of choosing a subset % & of r things from a set of n things. Note that this definition does not give us a way to calculate nr ; and, if you already know how to evaluate the binomial coefficients, you should put the formula out of your mind and let yourself be surprised by the beautiful way in which identities can be proved without resorting to bashing out messy fractions of factorials. The following proofs are known as proofs by double counting, or, perhaps more endearingly, story proofs. The idea is that you can prove that two expressions are equal by telling two different stories that describe the size of the same set in two different ways. First, a simple example: n # $ ' n k=0

= 2n .

You can prove that two expressions are equal by telling two diﬀerent stories.

(1)

This identity is commonly proved by% induction, but there is an elegant proof that anybody, & equipped with the above definition of nr , can understand. 25

autumn 2016

chalkdust Let’s tell a story. There are n light switches on a wall. In how many ways can they be configured? The le -hand side of (1) counts this: it starts by counting the number of ways in which zero light switches can be chosen to be on, then does the same for one, then two, and so on, until it counts the number of ways in which all n light switches can be turned on. But the right-hand side counts the same sum: each of the n light switches could be either on or oﬀ, so for each light switch there are two choices and so the total number of configurations is 2n . Thus the le -hand side and the right-hand side give the size of the same set, and (1) is true. Another identity that is susceptible to a story proof is Vandermonde’s convolution: $ # $ r # $# ' m n m+n = . k r−k r

(2)

k=0

Proving it algebraically requires you to play around with the product of two sums, which, as you can imagine, is both messy and also not particularly insightful.

Having said that, (2) does admit a very nice proof through storytelling. The clue as to what story% to &tell is the right-hand side. The binomial coeﬀicient m+n suggests r that we ought to consider choosing r things from m + n things. Consider a set of m + n people—m males and n females—from which we would like to select a commi ee of r people. The right-hand side counts this in an obvious way: it neglects gender and just chooses A choice of r people from m men and r people from m + n. The le -hand side is less obn women. vious. It first counts how many% ways % &% & &% n &we can choose zero males and r females ( m0 nr ), then one male and r − 1 females ( m1 r−1 ), all the way %m&%n& up to r males and no females ( r 0 ). In this way, it counts every possible commi ee. So the le -hand side and the right-hand side count the same set and so are equal. Not only is the story proof more elegant than the algebraic proof, it is actually shorter, and gives some intuition as to why the identity is true! Both examples so far are well known, but this Not only is the story proof more method extends to situations that are less common. elegant than the algebraic proof, Let’s think about Vandermonde’s convolution in the it is actually shorter. case where m = n = r and, instead of commi ees, think% about For this it’s necessary to note & % teams. & n that nr = n−r . You can convince yourself that this is so by considering that the number of ways of choosing r things from n things is the same as the number of ways of choosing r things to exclude from n things. In this case, Vandermonde’s convolution becomes # $ n # $2 ' n 2n = . (3) k n k=0

How can we interpret this? chalkdustmagazine.com

chalkdust Well, the right-hand side is the number of ways of choosing a team of n people from 2n people, n of whom are male and n of whom are female. The le -hand side is the number of ways of choosing a team of n people comprised of zero males, then one male, then two males, and so on, until we enumerate each possibility. So (3) is true, but Vandermonde had already told us that. What more interesting stories can we tell? Well, every team should have a captain: so in how many ways can we do this? Considering the right-hand side first, we can extend it as follows. %2n& Choose n people (we can do this in n ways), then choose a captain from them (we can do this in n ways), so the number of ways of choosing a team %of &n people from 2n, with a captain, is n 2n n . To count all of the possible teams in a diﬀerent way, let’s consider teams which have k men in them, A choice of n team members from n men and n women, and then sum over all the possible values including a captain. of k: 0, 1, 2, %. .&. , n. Say that my team % has & n k males in it, and that my captain is male. I can choose the men in nk ways, the women in n−k %& % n & % &2 ways, then choose my captain in k ways. Noting that nk = n−k , we find that there are k nk % &2 teams with k males and a male captain. Similarly, there are (n − k) nk teams with k males and a % n &2 female captain, and we can rewrite this as (n−k) n−k . So the total number of teams with k males %n&2 % n &2 and a captain of any gender is k k + (n − k) n−k . Thus we sum over all values of k, recovering the identity: ( # $ # $2 ) # $ n ' n 2 n 2n k + (n − k) =n , k n−k n k=0

or, by noting that the second part of the sum is in fact the same as the first part, just in reverse order, we can write this more succinctly as: # $2 # $ n ' n 2n 2k =n . (4) k n k=0

Now that we have seen various applications of telling stories, we are ready for the big league. Consider the following identity: # $2 # $ n ' 2 n 2 2(n − 1) k =n . (5) k n−1 k=0

It goes without saying that (5) would be an absolute mess to prove algebraically, but is there a nice story proof? It turns out that yes, there is, and that it is even cleaner than the proof of (4). If you would like to have a go at proving the identity yourself, stop reading now! 27

autumn 2016

chalkdust Let’s count the number of ways of forming a team of n people from a set of 2n people, n of whom are male and n female. The twist is that our team must include a female captain, and that we must also designate a male timekeeper, who must not be on the team. Thus the right-hand side of (5) can easily be interpreted: we first choose one of n females to be captain and put her on the team, then one of n males to be timekeeper. Finally we choose the remaining n − 1 people we need on the team from the 2(n − 1) people le . How does the le -hand side count the same thing? Consider a team % & that contains k females. We first choose the % nk &females %& in nk ways, then the remaining n − k males in n−k = nk ways. Then we choose a captain from the k females on the team in k ways, and a scorer from the k males not chosen% in& k ways. Thus the right-hand side is equal to the sum of k 2 nk over all k. So (5) is true. The identity in (5) was proved using the intuition gained from the right-hand side, but what happens if we try to garner information from the le -hand side? It turns out that a completely diﬀerent story proof is possible. What feeling do we get from the le -hand side? Well, each binomial coeﬀicient is squared, so it seems natural to consider all the possible teams with the same number (n) of males and females. That is, we reinterpret the le -hand side as counting the total number of teams with the same number of male members as female members, which also have a designated male and female captain.

A team made by choosing a female and male captain, then choosing n − 1 of the remaining 2(n − 1) people. Of the n − 1 people chosen, add the picked females and the unpicked males to the team. chalkdustmagazine.com

A team made of n members, including a female captain.A male timekeeper, not on the team, is also chosen.

A team made up of k females and n− k males, with one of the k chosen females as captain and one of the k males not chosen as timekeeper.

How can we reinterpret the right-hand side to align with this? We first choose a male captain and a female captain, and put both of these people on our team. We can do this in n2 ways. Then% we choose half of the remaining people. We & can do this in 2(n−1) ways. This choice doesn’t seem to have n−1 anything to do with choosing an equal number of males and females, but here’s the trick: from our choice, we generate a new selection by swapping all the males that we’ve chosen for all the males we haven’t chosen, and use that selection instead. A li le thought shows that for each selection of half of the remaining people, we get exactly one team with equal numbers of males and females, and vice versa, as shown in the diagram. Thus the number of teams with the same number of% males& and females as well as a male and female captain is n2 2(n−1) n−1 . So we have proved (5) in two diﬀerent ways, 28

chalkdust both by telling stories, and both more succinctly than we could have done algebraically! Now that we have established that story proofs are viable methods for proving identities, perhaps you would like to try your hand at a few identities yourself? n ' k=2

$ r # $# ' n n−k k=0

r−k

= 2r

# $ n k(k − 1) = n (n − 1) × 2n−2 k

# $ n r

n ' k=0

1 k 2 = n (n + 1)(2n + 1) 6

The author would like to thank Dominic Rowland for introducing him to the idea of story proofs, and would like to recommend the book Introduction to Combinatorics, published by the United Kingdom Mathematics Trust (UKMT) and authored by Gerry Leversha and Dominic Rowland, which expands on these techniques. Jack Hodkinson Jack is an undergraduate studying mathematics at Corpus Christi College, University of Cambridge. When he’s not telling stories, Jack can be found volunteering for the UKMT, composing choral music, and playing the organ. Heppy Longworth

jrh206@cam.ac.uk

My favourite set

The Mandelbrot set Andrea Bertozzi

My research group (see the interview on pages 4–7) developed an eﬀicient algorithm for tracking the boundaries of sets on diﬀerent scales. I thought we could apply it to fractals so we demonstrated it on the Mandelbrot set. The figure on the le is the Mandelbrot set computed on a 100×100 grid and the figure on the right is the same set computed on a refined grid with 1000×1000 pixels. 29

autumn 2016

chalkdust

Hedy Lamarr

Hollywood star and secret inventor James Grime

golden age of Hollywood was a time of classic movies and classic movie stars. A time of ‘frankly my dear’, ‘play it again’ and ‘whoops Mr Parson’ (I made the last one up). Yet only one star was billed as ‘The most beautiful woman in the world’. This was Hedy Lamarr: an Austrian-born actress, former wife of an arms dealer, international movie star, and occasional inventor. Her most celebrated invention was something without which today’s mobile phone and Wi-Fi technology would not be possible: frequency-hopping. Frequency hopping is the idea of transmi ing radio signals while rapidly changing frequencies. This makes any transmission impossible to intercept, jam, or accidentally cross with your neighbour’s order to the Chinese takeaway. It’s also a technology currently estimated to be worth over $100bn a year. Yet Hedy never earned a penny from her idea, giving the patent to the US Navy, who allowed it to expire while Hedy lived a modest Hollywood retirement.

Frequency-hopping spread spectrum Modern frequency-hopping involves multiplying a data signal by a code sequence. The code sequence occurs at a much faster rate than the data signal, which causes the la er to be spread among a far greater range of frequencies. Although spreading increases the range of frequencies a message will use, not all of that range will chalkdustmagazine.com

chalkdust be in use at any one time, allowing many people to use the same range of frequencies simultaneously. It is this property of spread-spectrum that makes mobile phone technology possible. The combined signal is then used to modulate a carrier signal and transmi ed via radio. The original data signal is recovered by re-multiplying the non-modulated signal by a locally generated copy of the code sequence, causing the code sequence to cancel itself out. Without knowledge of the code sequence, the transmission should be indistinguishable from white noise. Ideally, code sequences must appear as noise-like as possible, yet be easily reproducible at the receiving end. Such sequences are called pseudorandom sequences. For example, an m-sequence (or maximum length sequence) starts with a binary string of length n and iterFrequency hopping atively generates a sequence of strings. The next string is the sum of the last two digits (using modulo-2 addition of the previous string), followed by the first n − 1, which remain unchanged. For example, when n = 4,

Continuing in this way, the procedure eventually generates all binary strings of length n (except for the zero string), before returning to the start. However, Hedy’s original idea was pre-digital, when streaming was just something rivers did, and Spotify was a natural part of adolescence. Instead, Hedy had a much more militaristic application in mind: as a method to control submarine torpedoes remotely. So how did a Hollywood star know so much about munitions?

The arms dealer Hedy’s love of inventing came from her father, a bank director in Vienna, yet Hedy still dropped out of school at sixteen in order to become an actress or, in her own words, “I wanted to be famous”. This is a sensible ambition and nothing bad will come of it. Hedy landed her first starring role at the age of eighteen in the film Ecstasy. This was a sensitive look at a young woman trapped in a loveless marriage to an older man, but who eventually finds new love, new freedom and finally gains her independence. It is 33

This image is being used with permission from ﬀfmovieposters.com

Poster of the film Ecstasy autumn 2016

chalkdust mostly remembered because Hedy gets her nips out and pretends to have some sex. As the first non-pornographic movie to include a sex scene, Ecstasy was ahead of its time, so its time decided to condemn it. Instead of fame, Hedy found infamy—which is like fame but with creepier fans. One such fan was Fritz Mandl, arms dealer to the Nazis and massive jerk. I don’t know what first a racted Hedy to third-richest-man-in-Austria Fritz Mandl, but once they married, Hedy found herself living a life of luxury. These luxuries included three hunting lodges, cars, planes, a yacht, servants and eating from gold plates. Something even Donald Trump might think was a li le gauche. Yet it was a gilded cage, with a possessive and jealous husband. Hedy was no longer allowed to act, while Fritz supposedly spent a fortune trying to find and destroy every copy of Ecstasy. The stories of Hedy’s escape—drugging a maid and escaping in her uniform—may have been an invention of MGM Studios, but by the time she arrived in America, Hedy Mandl was now Hedy Lamarr—a fully minted movie star. Within a couple of decades, Hedy had starred in over twenty movies and divorced six times. Men wanted Hedy Lamarr, women wanted to be Hedy Lamarr. Dogs wanted to be pe ed by Hedy Lamarr, cats were not fussed. Yet when Hedy was not busy making movies or divorcing people, she found time to invent.

Name that tune It is said that Hedy came up with frequency-hopping while playing a game at the piano with her composer friend George Antheil. The game involved one of them playing a tune on the piano for the other to guess and then jump in on duet. To make this game more diﬀicult, George would keep changing the key, forcing Hedy to follow. This game of two piano players remaining synchronised while constantly changing keys gave Hedy her idea. If a transmitter and receiver could remain synchronised while hopping from frequency to frequency, then without knowledge of the pa ern of hops, the signal could not be blocked (also known as jammed). So Hedy decided to invent a radio-controlled torpedo— something she knew a bit about thanks to the many dinner parties she endured entertaining the Nazi generals who were guests of Fritz Mandl. Radio control and frequency-hopping would allow a ship to correct the direction of a torpedo for a more accurate hit. Hedy’s idea chalkdustmagazine.com

George Antheil

chalkdust was entirely original but, according to patent law, Hedy would need to demonstrate some form in which the idea could be put into practice. This would become George Antheil’s biggest contribution, thanks to his experience with self-playing pianos.

Patents and the player piano Self-playing pianos were extremely popular in the days before radio. A tune was represented by holes on a roll of paper, which was then loaded into the player piano. The holes in the paper passed over a vacuum that would pull down the keys and cause the piano to play. Player pianos are very cool—if slightly ghostly. As a young composer, Antheil had wri en Ballet Mécanique, a musical experience involving several synchronised self-playing pianos, as well as saws, hammers, electric bells and three aeroplane propellers. This was early experimental electronic music, a Skrillex of its time and about as well received. The 1926 debut in Paris ended in boos and a riot.

A player piano

So when Hedy needed a practical demonstration of her idea, George thought of the self-playing piano. Hedy and George’s patent described a ship and torpedo that would rapidly, and synchronously, change frequency in a pa ern controlled by ribbons—perforated paper like that of a player piano—which would determine both frequency and duration. With hundreds of hops per minute, switching unpredictably between frequencies, the signal would be impossible for the enemy to jam. Whimsically, George chose to use 88 diﬀerent frequencies, one for every key on a piano.

Hedy and George’s patent

In August 1942, Hedy and George’s patent for a ‘Secret Communication System’ was approved. However, the US Navy ultimately rejected the idea, citing that the mechanism would be too heavy to fit in the torpedoes. This is where Hedy and George had made their mistake. In an a empt to 35

autumn 2016

chalkdust be er explain their idea, they had mentioned the player piano. Although their system could have been reduced to the size of a dollar, it appears that the navy imagined torpedoes containing parts from player pianos themselves. Mysteriously, the navy acquired the patent for frequency-hopping, which was then filed away and le to expire. Instead, Hedy was told that she could serve the nation be er by using her celebrity status to sell war bonds. Meanwhile, far from the glamour of Hollywood, the mathematical foundation of spread-spectrum technology was being established by a man called Claude Shannon, who proved that communication could be made more reliable, and with less chance of error, by increasing the range of frequencies.

The father of information theory Shannon was a mathematician and electrical engineer working for Bell Labs in New York. His work was finally made public in his 1948 paper A Mathematical Theory of Communication. This was a landmark paper that became the foundation of information theory, and even coined the term ‘bit’ as a unit of information. The capacity, C, of an analogue channel, subject to additive white Gaussian noise, can be calculated from the physical properties of that channel using the Shannon– Hartley theorem:

C = B log2 (1 + S/N), where B is the range of frequencies, or bandwidth, of the Tekniska Museet, CC BY-SA 2.0 channel in hertz; S is the average received signal power Claude Shannon over the bandwidth, measured in wa s; N is the average power of the interfering Gaussian noise over the bandwidth, also measured in wa s; and S/N is known as the signal-to-noise ratio. The Shannon–Hartley theorem shows that a channel with li le noise can transmit high amounts of data nearly error-free, while to do the same on a noisy channel you need only increase the bandwidth. This is the principle on which spread spectrum works today. A channel’s capacity is a theoretical limit only and represents the best that can be achieved with any code and modulation method. Modern techniques can now achieve data rates close to that limit. chalkdustmagazine.com

chalkdust

Final recognition Tragically, as Hedy’s beauty faded so did her movie career. Similarly, her contribution to the growing technology of frequency-hopping and spread-spectrum went unacknowledged for decades. Finally, a campaign in the 1990s resulted in Hedy Lamarr and George Antheil receiving the 1997 Pioneer Award from the Electronic Frontier Foundation for their contributions to the technology. On hearing she was to receive the award, Hedy remarked to her son “it’s about time”. Hedy died in 2000, aged 85. Some of her ashes were placed in the Vienna Central Cemetery. Recently, a memorial was added there that, when viewed correctly, creates an image of Hedy. The memorial is made from 88 stainless steel rods—to represent the 88 frequencies in Hedy’s original patent—and features a last few words from Hedy: “Films have a certain place and a certain time period. Technology is forever.”

Papergirl, CC BY-SA 4.0

James Grime James is a mathematician and public speaker, and can be mostly found travelling the world and talking about Enigma machines. James is also a presenter on the popular YouTube channel numberphile, as well as his own channel singingbanana. @jamesgrime

My favourite set

Numbers

My favourite set is:

Matthew Wright

{{}, {{}}, {{}, {{}}}, {{}, {{}}, {{}, {{}}}}, {{}, {{}}, {{}, {{}}}, {{}, {{}}, {{}, {{}}}}} or in other words, the number 5. In set theory, numbers are constructed as follows: Zero is defined to be the empty set: 0 = {} = ∅; one is the set containing the empty set: 1 = {{}} = {∅}; and the number n is defined as the set containing all the previous numbers: n = {0, 1, ....., n − 1}. As you can see, this definition causes things to get complicated quite quickly!

Did you know... …that our T-shirts are now green. Show oﬀ that you’re mathematically keen! Get yours at chalkdustmagazine.com. 37

autumn 2016

chalkdust

Moonlighting agony uncle Professor Dirichlet answers your personal problems. Want the Prof’s help? Contact deardirichlet@chalkdustmagazine.com

Dear Dirichlet,

mer holiday us ﬁnancially, and the only sum for gh tou bit a n bee has r This yea nny East Midlands. week camping in the not-so-su a s wa rd aﬀo to e abl re we we blissful trips to the won’t stop talking about their My work colleagues, however, y’d stop nt to seem jealous but I wish the wa ’t don I . ope Eur rn the sou beaches of on? w can I move the conversation chatting about their suntans. Ho the pale, Selly Oak

— Beyond

! DIRICHLET SAYS:

I, too, sometimes crave the idyllic sands of southern Europe, but alas, circumstances require me to spend most of my days answering personal problems from my country house. I have, however, found a technique which gives the appearance of spending a few weeks in Marbs. On a sunny weekend, ﬁnd a police constable. Ask to borrow her nightstick. Then simply divide the sunsinh with her cosh, and what do you have?...tanh, or as I like to call it, “fake tan”.

Dear Dirichlet,

I am going to a conference in Tou louse in a few months’ time, and I have just discovered that my French crush, who grew up nearby, is also atte nding. I think it might be a prime opportunity for us to get to know each other bet ter, if you catch my drift. But I don’t want to me ss it up and waste my chance . You seem like a well-conferenced chap: how do I make the most of this opportuni ty?

— Je t ’aime, J’habite dans le nord

de l ’Angleterre

DIRICHLET SAYS: As I have often experienced on my visits to Paris, no one can resist the charm of the French. Let me share a technique that I have often, after some analysis, found to be useful. WLOG place yourself at xn and your crush at xm . Then for any ε>0, ensure that there exists an N such that n,m>N, |xn -xm |<ε. This will ensure that your elements will become arbitrarily close to each other as the sequence progresses. When your ε is sufﬁciently small, seal the deal by whispering “voulez-vous Cauchy avec moi?”. !

chalkdustmagazine.com

chalkdust

Dear Dirichlet, I have been trying to be self-su fficient this year with my vegeta ble patch down the allotments by the corn exchan ge, on Windy Lane (turn left at the bus stop), just by the freight railway, not far from the nuclear power plant. I say power plant: it’s more of a test site. I am hoping to make a nice soup from the pod s that are growing down at the bottom of the garden , among the birds and the bees, at the end facing the sewer works, downwind from the (old) gas station, from where you can see the butterfly farm. I say butterfly farm ; it’s more of a butterfly farn. For this recipe, I need the same weight (mass) of (single) cream and vegetables. However, to know how much (single) cream to buy , I need to know how many pea s will grow. Help!

— Keen gardener, Allotments, Downwi nd from (old) gas station, nr. But terfly Farn !

DIRICHLET SAYS: Sounds like you’ve got an N-Pea problem! This will

be hard to solve. Perhaps you can seek the advice of your local mathematician. He’s sitting in his favourite chair in his house on Orchard Close, backing onto the brook through the manor house that used to belong to Sir Alfred, overlooking the abandoned barn, not too far from the (second largest) ford. (I say ford: it’s more of a collapsed bridge.)

Dear Dirichlet,

r, so imagine how ions for my year abroad next yea I was really worried about my opt sorted and ng to New York! I’ve got my visa goi am I t tha r hea to s wa I ited super-exc side of the road, in America: driving on the wrong life on up d rea to ng tryi n bee e hav there any special , how to take a taxicab... But are never using the metric system Big Apple? w about living downtown in the behavioural customs I should kno

South — Worried about distance, Bicester

DIRICHLET SAYS: The Manhattan norm is given, for a vector x, by n

||x||l = Σ |xr |. r=l

More Dear Dirichlet, including two seasonal specials, online at chalkdustmagazine.com 39

autumn 2016

chalkdust

Prime jewellery

Jennifer Balakrishnan

Vicky Neale

was recently given a copy of Cra ing Conundrums: Puzzles and Pa erns for the Bead Crochet Artist by Ellie Baker and Susan Goldstine. This was pre y exciting for me, as although I knew nothing about bead crochet (I’d never heard of it), I’m a mathematician who enjoys exploring mathematical ideas through cra . So naturally I rushed out and bought lots of beads and thread, and a very tiny crochet hook (1.5mm, if you’re really interested). And then I realised that it would probably have helped if I knew how to crochet—which I didn’t. Still, it’s important not to be deterred by minor obstacles, so I pressed on. I had several evenings where the process was: get started, crochet for a bit, realise it wasn’t right, undo it, and repeat until bedtime. (With hindsight, at least some of the problem was that I started with a 1mm crochet hook and this led to tension problems, in every sense of that phrase.) But eventually I figured it out, and made a few bracelets for friends and family using designs from the book—Escher-inspired tessellations, mostly. Even as I crocheted lizards, I found myself wondering: what would a bead crochet bracelet illustrate about the distribution of the primes? And since you’re now wondering about this too (if you weren’t already), let me show you a picture.

Jennifer Balakrishnan

Primes modulo 6 chalkdustmagazine.com

chalkdust I should probably explain what you’re looking at. To make a bracelet like this, you start by threading on all the beads you need for your project, in the right order. So I sat with a list of primes up to 1,000, and a pile of gold beads (for primes) and a pile of white beads (for non-primes), and got threading. Then you start crocheting in a straight line, incorporating one bead in each stitch (apologies to people who actually know about crochet, I don’t know the technical terminology). When you’ve reached the ‘circumference’ of the bracelet, you join the most recent stitch to the first, to form a small circular loop, and then you spiral round on that to build up a long tube. When it’s the right length (when you’ve used up all your beads), there’s a clever gra ing technique to join one end to the other pre y seamlessly to form a beautiful torus/bracelet. The bracelet above has a circumference of 6. I’ve marked the start with black beads. If you look very carefully to the le of those, you can see two adjacent gold beads, corresponding to the primes 2 and 3 (these are the only adjacent gold beads because…). Then there’s a white bead for 4, then a gold for 5, then it goes round the back, and the next prime we can see in this photo is 11, down and to the le of 5. Since the circumference of the bracelet is 6, each loop round takes 6 beads. Do you see there’s a li le flurry of primes with five gold beads in a straight line, corresponding to the arithmetic progression of primes 5, 11, 17, 23, 29? Neat, huh? If you look carefully, you might notice that the primes on a bracelet of circumference 6 seem to fall into two spirals (apart from 2 and 3). I’ll let you ponder on that one… Naturally, I wanted to look at the primes on a bracelet of circumference 7, to compare and contrast, so I rustled one up.

Jennifer Balakrishnan

Primes modulo 7

The start is marked by the dark blue beads, and just to the le you can see two adjacent gold beads, for the primes 2 and 3. The sequence continues upwards on this photo, so 4 is above 3, then we have the prime 5, then it loops round the back, and the next prime we can see is 11, immediately le of 3 in the photo. Two things strike me about the 7-bracelet compared with the 6-bracelet. One is that the gold beads no longer land in two nice neat spirals. The other is that there seem to be a lot of pairs of primes that diﬀer by 8. Intriguing. 41

autumn 2016

chalkdust Now, I could tell you things about the distributions of the primes (both known results and unsolved problems) based on these bracelets, but really the point of this article is to encourage you to make your own bracelets, because you’ll find out a whole lot more that way than by my writing things, and you’ll end up with a piece of jewellery at the end of it! I’ve thought moderately carefully about prime numbers, and I still feel that I got a richer insight into their behaviour by threading on beads and crocheting them into a bracelet than I have done by staring at lists of primes. I recommend it! Just one tip. Fired up by my success with the primes, I made a 7-bracelet of squares. This was not a good idea. By the time you get to squares in the 900s, they’re pre y thin on the ground, so it was quite boring to crochet: lots and lots of unchanging purple beads… Also, I don’t think it looks very pre y. I’m a huge fan of the square numbers, and they give rise to a lot of interesting mathematics, but disappointingly they did not give rise to an interesting bracelet.

Jennifer Balakrishnan

Squares modulo 7

So, now it’s your turn. What will you bead crochet first?

Jennifer Balakrishnan

Vicky Neale Vicky is the Whitehead Lecturer at the Mathematical Institute and Balliol College, University of Oxford. Her job involves enthusing about mathematics to anyone who will stand still for long enough. In her spare time, she enjoys various forms of cra , and her cats are always keen to help. @VickyMaths1729

Odd squares How many square numbers are there whose digits are all odd? Source: Maths Jam Answers at chalkdustmagazine.com

Chalkdust wants YOU! Got an exciting idea? Want to share it with us? Send in your article and it could be published on our weekly blog, or even in the next Chalkdust magazine! Get in touch with us at

contact@chalkdustmagazine.com

chalkdustmagazine.com

chalkdust

mathematical crochet You will need:

Wool, crochet hooks, beads, scissors

Instructions

1 2

Learn how to crochet.

Make lots of cool mathematical objects, such as:

Spheres

Hyperbolic surfaces

Prime number jewellery (see pages 40â&#x20AC;&#x201C;42)

Klein bottle hats

Sphere: Ms Premise-Conclusion, CC BY-NC-ND 2.0; hyperbolic surface: pandaeskimo, CC BY-NC 2.0; bracelet: Jennifer Balakrishnan; Klein bottle: Anna Lambert

Tube map platonic solids and FrĂśbel stars: more How to Make at chalkdustmagazine.com 43

autumn 2016

chalkdust

Emma Bell

The Wizard of Oz, the Scarecrow shows us how intelligent he has become by (mis)quoting Pythagoras’ theorem:

The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side. Homer Simpson does a similar thing when he puts on a pair of glasses and tries to convince himself that he is smart. It would seem that the lasting legacy of Pythagoras of Samos is the formula linking the sides of a right-angled triangle. It could, however, be argued that the actual legacy of Pythagoras is much greater—it’s more than the formula used in contrived situations of ladders being rested against walls or finding the answer to that most fundamental of questions: would the pencil stick out of the top of the pot? His legacy is around us every day… Our knowledge of Pythagoras’ life is a blend of legend and hearsay. His teachings were of a deliberately oral nature because the Pythagoreans argued that questions could not be posed to books in the same way that they could be asked of a person. By bringing together secondary texts and triangulating the facts, we can be certain that Pythagoras (and the Pythagoreans who followed him) were determined to find order in disorder. Any discoveries a ributed to Pythagoras were made as a collective—a hive mentality that the Pythagoreans prided themselves on. The Pythagoreans believed that numbers were at the heart of everything. They held the ‘quaternary’ in extremely high regard, breaking down their understanding of their surroundings into fours—the Tetractys. chalkdustmagazine.com

chalkdust The Tetractys: Numbers: 1, 2, 3, 4; Seasons: spring, summer, autumn, winter; Ages: childhood, adolescence, maturity, old age; Society: man, family, village, city; Dimensions: point, line, surface, solid; Elements: fire, air, water, earth, Their laws led to a beautiful and harmonious universe—Cosmos—by adding order to chaos. This is perfectly illustrated using music. Folklore tells of Pythagoras walking past a blacksmith’s workshop and hearing diﬀerent tones produced by the smithy hammering a rod of metal. While there may be truth in this tale, we do know for definite that the Pythagoreans experimented with an instrument called a monochord.

A monochord

This simple instrument comprises of a single string, held taut, with a moveable bridge. When the string is plucked, it produces a note. Versions of this instrument are still used today. One notable advocate of the ‘diddley bow’ is Seasick Steve, who has a song dedicated to it. Investigations by the Pythagoreans took the form of creating order from disorder— they interpreted the multitude of possible positions of the bridge as disorder, but by applying the limiting factor of number to the placement of the bridge, harmony and order could be found. They knew that the shorter the string, the higher the pitch of the note produced. By methodically comparing Bengt Nyman, CC BY 2.0 Galilea, CC BY-SA 3.0 the sounds produced by various string Seasick Steve and Pythagoras: a striking similarity? lengths, they found that the notes created by lengths that could be expressed as ratios using small numbers were the most pleasant and harmonious. The Pythagoreans took these findings as proof of the interrelation of number, harmony and beauty— that mathematics and music were two sides of the same sheet of paper. They defined their musical notes using the ratios between them, rather than giving them individual ‘names’. 45

autumn 2016

chalkdust • An octave would have the ratio 1:2. • A fi h would have the ratio 2:3, which could be calculated by the harmonic mean of the endpoints of the octave. • A fourth would have the ratio 3:4, which could be calculated by the arithmetic mean of the endpoints of the octave. All of these intervals were expressed using the first four natural numbers so intrinsic to the Pythagorean school of thought. By transposing these ratios onto the musical scale that we are familiar with today (known as the diatonic scale), a beautiful symmetry can be seen. Taking an arbitrary string length of 24cm, we can see how dividing the string into intervals produces pleasing harmonies.

It is interesting to note that the geometric mean was ignored—it would produce a number in √ the ratio that would be most discordant! In our 24cm example, the geometric mean would be 12 2. The discomfort that Pythagoreans felt with regards to irrational numbers is infamous. Examples of these intervals can be heard in well-known tunes, such as in the first two notes of each of these melodies: Fourth: Super Mario theme; fi h: Star Wars theme; octave: Somewhere Over the Rainbow. It’s a shame that the Scarecrow’s intelligence could not be portrayed eﬀectively using Dorothy’s opening song: its link to Pythagoras is certainly more universal than a 2 + b 2 = c 2 . Emma Bell Emma is a maths teacher at Franklin College in Grimsby, UK. @El_Timbre

emma.bell@franklin.ac.uk

Did you know... …that in four dimensions, it is impossible to knot a rope? chalkdustmagazine.com

chalkdust

Warning Chalkdust T-shirts will not teach you circus tricks. But they will make you look as cool as Luciano. Now available in blackboard green (only). Get yours at

chalkdustmagazine.com

Two lines Let A and B be two straight lines. The y-intercept of A is the gradient of B; and the y-intercept of B is the gradient of A. What are the coordinates of the point where A and B meet? Source: mscroggs.co.uk Answers at chalkdustmagazine.com

My favourite set

The empty set Rob Beckett

My favourite set is the empty set (â&#x2C6;&#x2026;). The empty set is the only set of its kind and contains no elements. The empty set is a subset of any set but the only subset of the empty set is itself. The empty set is not nothing, but is a set that contains nothing. If I had a bag with multicoloured counters in, these are the elements of a non-empty set. On the other hand, if I had a bag with no counters in, there are no elements in the set and this is an example of the empty set. 47

autumn 2016

#4 Set by Humbug

3 N

Rk R8

Re Rd

kR kk kj

k8 ke kd

jR jj

j9 j8

Rules Although many of the clues have multiple answers, there is only one solution to the completed crossnumber. As usual, no numbers begin with 0. Use of Python, OEIS, Wikipedia, etc is advised for some of the clues. To enter, send us the sum of the across clues via the form on our website ( chalkdustmagazine.com) by 7 January 2017. Only one entry per person will be accepted. Winners will be notified by email and announced on our blog by 21 January 2017. One randomly-selected correct answer will win a ÂŁ100 Maths Gear goody bag, including non-transitive dice, a Festival of the Spoken Nerd DVD, solids of constant width 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, with free UK shipping. Find out more at mathsgear.co.uk chalkdustmagazine.com

chalkdust

Across 1 This number is a multiple of one of the two-digit answers in the crossnumber and shares no factors with the other two-digit answers. 3 The sum of 15D, 9A and 6D. 5 142 less than 5D. 8 The product of the three largest twodigit answers in this crossnumber. 9 This number is equal to the number of digits in its factorial. 11 A Sierpiński number.* 12 The sum of the reciprocals of this number’s non-zero digits is one. 13 Twice 7D. 16 A counterexample to the conjecture that every odd number can be wri en in the form p + 2a 2 , where p is 1 or a prime and a is an integer. 17 A second counterexample to the conjecture in 16A. 19 A multiple of 717. 23 Equal to another answer in this crossnumber. 25 A palindrome. 26 Less than 15D. 27 12A backwards. 30 The sum of this number’s digits is 5. 32 The first le er of each digit of this number is the same as the last le er of the previous digit. 33 A number that is three times the sum of its digits. 34 Integer part of the square root of a Fibonacci number. 35 The sum of 9A, 27A, 27D and 28D. 36 1 12 times 29D. 37 One less than a multiple of 30A.

Down 1 A multiple of 1A. 2 A factor of 1A. 3 A positive integer whose square is the sum of 50 consecutive squares. 4 All the digits of this number are the same. 5 Half the diﬀerence between 5A and 13A. 6 The middle two digits of 3A. 7 808 more than 5A. 10 The largest number that is not the sum of two abundant numbers.† 14 The remainder when this number to the power of 91 is divided by 18,793,739 is 12A. 15 1D more than 26A. 18 An anagram of 31D. 19 The sum of this number’s digits is one less than 6D. 20 The lowest common multiple of 7D and 5A. 21 Equal to another answer in this crossnumber. 22 The smallest number that appears eight times in Pascal’s triangle. 24 The maximum number of regions that can be formed by joining 27 points on a circle with straight lines. 27 The number of straight lines that go through at least two points of a 10×10 grid of points. 28 The sum of four consecutive positive cubes. 29 Five less than a palindrome. 30 The sum of 18D and 31D. 31 An anagram of 18D. 34 A number whose square root is equal to the sum of its digits.

(4)

(4) (4) (5) (2) (6) (4) (4) (4)

(4) (9) (9) (4) (4) (4) (4) (6)

(2) (5) (4) (4) (4)

(4) (4) (4) (9) (4) (2) (4) (5) (6)

(4) (4) (4) (6) (9) (4) (5)

(4)

(4) (4) (4) (4) (2)

*A Sierpiński number is an odd number, k, such that for all integers n, k × 2n + 1 is not prime. †

Abundant numbers are numbers which are smaller than the sum of their proper divisors.

autumn 2016

chalkdust

On the cover:

Mathematics—queen of the arts? UCL Art Museum, University College London

John Crabtree

the brief tradition of Chalkdust cover articles there is a developing discussion of how mathematics and art are related.

Art is simply the making of representations. Art happens when a person has an idea or a vision that exists in their imagination (the mind’s eye) and is impelled to communicate said idea by making a visible manifestation (representation) of it in the material world. The idea or vision on its own is not art. Art occurs amid the struggle to make a representation of the idea that the artist can show to other people. Art may be relatively ‘fine’ or popular, conceptual or objective, highbrow or applied, yet still fall within this definition. Judgements about the quality of art are made largely by consensus among the cognoscenti in a given art milieu. These judgements are subject to change over time as the perception of works of art are always modified by the current ‘cultural environment’ and fashion. So artists and mathematicians share the ‘having of ideas’. But what then? Mathematicians communicate their ideas—yes. But ideas in maths take the form of theorems or conjectures about numbers, space or other abstract entities. The quality of these ideas is first assessed by proof. Can the idea be shown to be true? And second, if the idea is true, is it interesting? That is, does it usefully contribute to the mass of existing mathematics? Communication of mathematical ideas may require the invention of new symbols or diagrammatic forms, etc, but these are in the nature of being a new language, not art. In my view then, art and mathematics share the magical process of ‘idea ge ing’ but essentially UCL Art Museum, University College London The image Central adratic explains itself, I diﬀer in where they go with those ideas. If maths hope, as a celebration of analytic geometry. is to be considered an art, it would have to be a sort of ‘super-art’ or art ‘to a higher power’. Easier, I think, to class mathematics as the science of number, space, shape and structure, etc—the abstract entities that exist in our minds. Imagine an intelligent alien’s perception of our arts and our mathematics. Our art would be more chalkdustmagazine.com

chalkdust or less incomprehensible, depending on how alien the being was; but our maths would be as true for the alien as it is for us. Furthermore, good mathematics will not diminish with time or go out of fashion. There is an aﬀinity between some mathematicians and some artists. Certainly, it is a most pernicious error that scientific and artistic talent exclude each other—an idea unfortunately common among school counsellors. The common ground between art and science/maths that leads us to the ‘ge ing of ideas’ is the activity we call play. The thing of it—the thrilling thing, the magical thing—is the moment when one discovers a new idea, or pa ern, or conceptual framework, or whatever: the eureka moment! And are these moments not usually approached through playing in the mind with new combinations and orderings of existing mental constructs?

In ray tracing, each ray is used to decide the colour of a pixel on the image plane.

I had one of my most memorable eureka moments sometime in 1971 while si ing on a dead tree in Epping forest. At that time, I had been collaborating on an automatic projective line drawing program with ‘hidden line removal’, going where Autocad later arrived. I was considering algorithmic approaches to colouring surfaces in projective drawings. I realised that if I thought of objects in the scene as being represented mathematically as arrays of vertices and planes in some coordinate space, then I could solve for the equation of the line going from an eyepoint through a particular pixel in the image plane and into the scene (as in the diagram). From the equation of the line, I could find the closest surface along the path and then compute the colour and illumination value John Crabtree for that pixel based on the defined colour on the Spheres was created using ray tracing. surface, along with its relationship to any light source or other light-emi ing surfaces. And so I had invented ray tracing—the foundation of all computer generated synthetic imaging for special eﬀects in cinema, television and gaming. Of course, I neither invented it first nor alone—and I certainly had neither the persistence nor vision 51

autumn 2016

chalkdust to pursue ray tracing to practical or rewarding development. But its discovery was a thrill, as were the few simple pictures I made using the technique in a primitive manner on the pen plo er available. As spheres have become my most persistent motif, I will end with two more related works that play on the division and articulation of spherical surfaces: Sphere Architecture and Star Sphere.

John Crabtree

Sphere Architecture (le ) and Star Sphere (right) both play on the division and articulation of spheres.

John Crabtree John, born in O awa, Canada in 1948, studied Fine Art at Walthamstow School of Art and then at The Slade, UCL from 1967 to 1975. He was a teaching assistant in etching at The Slade from 1975 to 1979, then head of print studio and later digital arts coordinator at Byam Shaw School of Art from 1979 to 2002. johncrcrabtree@hotmail.co.uk

My favourite set

ℵ1

Matthew Scroggs

If you could count forever, you would reach infinity. You might be surprised to learn that there are bigger things than this infinity. My favourite set, ℵ1 (aleph one), is the smallest set that is bigger than the counting infinity. Cantor’s diagonal argument (if you’ve not heard of it, Google it!) can be used to show that there are more real numbers than natural numbers, and therefore that one infinite thing is bigger than another. But ℵ1 is even weirder than this. It is not known whether or not ℵ1 and the real numbers are the same size. In fact, this is not just unknown, but it cannot be proven either way using the standard axioms of set theory! (The suggestion that they are the same size is called the continuum hypothesis.) So my favourite set is bigger than the smallest infinity—but we can’t work out by how much. chalkdustmagazine.com

We love it when our readers write to us. Here are some of the best emails and tweets we’ve been sent. Send your comments by email to contact@chalkdustmagazine.com, on Twitter @chalkdustmag, or by post to Chalkdust Magazine, Department of Mathematics, UCL, Gower Street, London WC1E 6BT, UK.

$ ¯ # ¯ W Life interrupts a crossnumber. When you just want to find the appropriate mean, Or how many ways you can move a queen, Or a five-digit cube that ends in nineteen, Instead, you load the washing machine. Life interrupts a crossnumber. When you just want to add sixteen-A to four-D, Or factorise a number ending in three, Or find the right section of Fibonacci, Instead, you give the children their tea. At last you finish the crossnumber. You find that sequence of curious primes, The twelve-digit number works all down the line, You add up your answers and check them three times, And then... you get back to the washing line. Sam Hartburn, Canterbury

Chalkdust was absolutely superb once again. Hannah Fry, London

@FryRsquared

Nice article in Spring ’16 Chalkdust. Maths degree in ’08—ﬁrst time I’ve heard of fractional calculus! James Owens, Edinburgh

@SamHartburn

$ ¯ # ¯ W

Keep up the good work! I really like Chalkdust and have the pdfs handy on my phone (though I do often silently disagree with some of your “what’s nots”).

@JamesOwens

Alistair, Guildford

Thanks to Chalkdust, I knew how to fractionally integrate at last night’s conference dinner. And it certainly wasn’t improper! Lisa Chalmers, London

@MsChalmersMaths

autumn 2016

chalkdust

The Buckingham π theorem and the atomic bomb Photo courtesy of the National Nuclear Security Administration / Nevada Site Oﬃce

Diego Antonio Carranza Ortiz

16 July 1945, the first nuclear test, ‘Trinity’, was carried out and with it the nuclear age began. The explosion was huge, but the actual calculation of the amount of energy released was rather diﬀicult due to the large number of physical and chemical processes involved in the detonating reaction; even the rough estimates were far from accurate. It was not until the publication of the photographs of the explosion that scientists became aware of its magnitude. With just these photographs and some clever mathematical arguments, British physicist GI Taylor, Soviet physicist LI Sedov and Hungarian–American mathematician John von Nuemann estimated independently an energy of about 17 kilotons of TNT. Taylor published this result in 1950, with the US Army not at all thrilled that this sensitive piece of information was now in the public domain. Although the estimates of Taylor, Sedov and von Neumann required the use of some complex mathematics, dimensional analysis and the Buckingham π theorem allow us to come to the same conclusion with a minimum amount of knowledge of physics. In order to understand the importance of the Buckingham π theorem, we have to realise that since the pioneering work of Isaac Newton, physics and maths have been intimately connected. From that point onwards, understanding nature was not only a ma er of knowing that a projectile would fall to earth some metres away, but also of being able to predict the exact point it would reach and the time it would spend in the air. Mathematics has proved to be a very powerful ally when describing the laws ruling the universe, but the diﬀiculty involved can escalate very quickly as one moves away from idealised situations, which are sometimes not realistic at all. If we want chalkdustmagazine.com

chalkdust to tackle some serious or cu ing edge problems it is very likely that we will have to draw upon computers. However, if one simply wants to gain an insight into what is happening physically speaking, some basic linear algebra can give us valuable information. This aim lies at the heart of the Buckingham π theorem.

Units and dimensions When we speak about the mass or speed of an object, it is not enough to give a number but we must also specify the units in which the quantity is measured. For example, we can say that the mass is 5kg and speed is 30 miles per hour; of course, equivalent units such as pounds or km/h are equally valid. As you can observe, velocity consists of two units: one expressing length (km) and the other time (h). These are known as dimensions. In modern physics, all quantities can be expressed in terms of seven fundamental and independent dimensions. For our purposes it will be suﬀicient to consider only three: time (T ), length (L) and mass (M ). For instance, density (ρ) has dimension mass over cubic length and energy (E ) has dimension mass times squared length divided by squared time. From now on, we will denote dimensions using square brackets, [ ], so the previous examples can be represented as: [ρ] =

M = ML−3 , L3

[E ] =

ML2 = ML2 T −2 . T2

It is important to note that although diﬀerent systems of units can be employed, the physical dimensions remain the same. It is not diﬀicult to realise that there are certain combinations of quantities such that when their dimensions are considered they cancel each other out. Taking a time t, a length ℓ and an acceleration a—which has dimensions of length divided by squared time—the combination at 2 /ℓ is dimensionless. These kind of quantities will be of great importance, since the Buckingham π theorem is expressed in terms of them.

The simple pendulum Now that we have a clearer notion of what physical dimensions are, we are ready to understand the Buckingham π theorem. There are several ways in which it can be stated, some of them actually quite sophisticated, but for this article we will use a simple version. To help illustrate it, we will consider the classical example of the simple pendulum—a mass a ached to an inextensible string moving from side to side, as shown overleaf. Let’s assume we want to study a physical system (the pendulum) involving n variables denoted as a1 , a 2 , . . . , an . In principle, all of these variables are related via some function in such a way that any one of them, say a1 , can be obtained from the others. 55

Edgar Buckingham (1867– 1940), a er whom the Buckingham π theorem is named. autumn 2016

chalkdust Mathematically, this can be wri en as a1 = f (a 2 , a 3 , . . . , an ). The function f can be quite simple or really complicated depending on how many variables are included in the problem: indeed, the trickiest step is to identify the relevant ones. In the case of the pendulum, we want to understand how the mass m moves as time t passes, so it is natural to include both. Likewise, the length of the string ℓ must play an important role, as well as θ, the angle it makes with the vertical. And we must also remember that the pendulum’s movement is possible thanks to the action of gravity, which is taken into account through the acceleration g produced by the Earth. So we have five variables to consider. The dimensions of our variables are as follows: [m] = M, [ℓ ] = L, [t ] = T, [θ ] = 1 and [g] = LT −2 . Firstly, we note that since θ is a pure number, its dimensions are denoted simply as 1. Secondly, we see that although the dimensions associated with each variable are diﬀerent, those corresponding to g can be expressed in terms of the others. On the other hand, the dimensions of m, ℓ and t are fundamental, so there is no way that any of them can be expressed in terms of the others (you can think of this as being like a basis of a vector space). We say that they are dimensionally independent.

Model of a simple pendulum

So, returning to the general case, assume that we list the n variables in such a way that a1 , a 2 , . . . , aj are dimensionally independent, while the remaining ones, aj+1 , . . . , an , can be wri en as combinations of the others. Having done this, the π quantities are constructed as follows. Take aj+i and divide it by a product of powers of a1 , . . . , aj : aj+i , for i = 1, . . . , n − j. . . . azj i

πi =

q a1i ar2i

The powers qi , ri , . . . , zi must then be chosen such that πi has no dimensions. For the case of the pendulum, we have five variables, of which three are independent: hence n = 5 and j = 3. So there are two π quantities: π1 =

θ mq1 ℓ r1 t s1

π2 =

g mq2 ℓ r2 t s2

→ →

[π1 ] = [π2 ] =

1 Mq 1 Lr1 T s1

L1−r2 . Mq 2 T s2 +2

We want both π1 and π2 to be dimensionless, so for π1 it is evident that q1 = r1 = s1 = 0 do the job: this means that θ is by itself a π quantity. On the other hand, the powers appearing in π2 must satisfy the following system of linear equations: chalkdustmagazine.com

chalkdust ⎧ ⎪ ⎨s 2 + 2 = 0 r2 − 1 = 0 ⎪ ⎩ q2 = 0.

Its solution is easy to obtain: q2 = 0, r2 = 1 and s 2 = −2. Thus the dimensionless quantities governing the motion of the pendulum are: π1 = θ and π2 =

gt 2

ℓ

The Buckingham π theorem then establishes that our physical system can be completely described in terms of this set of dimensionless quantities. In general, this means that there exists a function F such that F (π 1 , π 2 , . . . , π n−j ) = 0. In our case we have, F

θ,

gt 2

ℓ

= 0.

The Buckingham π theorem proves to be a priceless—and simple—tool

The first interesting thing to observe is that the mass no longer appears. This seems counterintuitive, but it is explained by the fact that under the action of gravity all bodies fall with the same acceleration regardless of their mass. The other point concerns the function F, expressing the relationship between π 1 and π 2 . The Buckingham π theorem only establishes that such a function exists, but it does not provide any further information about the form of F. This must be obtained in some other way: through an experiment, for example. The theorem’s aim is diﬀerent: the advantage it gives is that it tells us which parameters are important (gt 2 /ℓ ) and which ones can be ignored (the mass). Actually, since the motion of our pendulum is oscillatory, we can guess a trigonometric function like sine or cosine must be involved. For those who have previously studied the pendulum, the form of F might not have come as a great surprise since the exact solution for its oscillation about a small angle is #! 2 $ gt θ = θ max sin ,

ℓ

where θ max is the angle the string makes with the vertical just before the mass is released. Taking into account that only a small amount of physical intuition was necessary, that is not bad for a first a empt.

Blast wave We return now to perhaps the most famous example of how the Buckingham π theorem can be used to understand a complex physical system: the energy released by a nuclear explosion like 57

autumn 2016

chalkdust the Trinity test. Of course, a similar process can also be applied to estimate the energy released by any huge explosion, like a star ejecting its outer layers into space during a supernova event. A phenomenon of this type passes through diﬀerent stages, each one of them characterised by particular physical processes, but for the sake of simplicity, we can think of it simply as a hemispherical fireball expanding into the atmosphere. First, we have to identify the relevant variables in our problem. As the explosion is a dynamical process, time t must play a role, as must the radius of the fireball R. One might also think that the mass m of the bomb should be included, but it is in fact equivalent to the energy E (recall the famous Einstein equation E = mc 2 ), so it is unnecessary to include both. An additional variable, which is perhaps not as evident at first sight, is the density ρ of the surrounding environment. The reason is A photo of the Trinity test, taken 0.025s a er the exsimple: the shock wave will expand more plosion. easily in a rarefied atmosphere than in a dense one. Of course, there can be lots of additional variables, but as long as they do not have a great relevance to the problem, we can neglect them. Once we are confident in our choice of variables, we can start to construct the π quantities. We have four quantities with dimensions: [R ] = L, [t ] = T, [ρ] = ML−3 and [E ] = ML2 T −2 . The dimensionally independent variables this time correspond to R, t and ρ. (Of course, just as for the basis of a vector space, the choice of independent variables is not unique and we could have chosen R, t and E to play this role: this would lead to the same final result.) Since we have four variables and three of them are dimensionally independent, we will have only one π quantity: π1 =

E q R t r ρs

→

[π1 ] =

L3s−q+2 . T r+2 M s−1

To apply the Buckingham π theorem this must be dimensionless and so we have the following system of equations: ⎧ ⎪ ⎨3s − q + 2 = 0 r+2=0 ⎪ ⎩ s − 1 = 0. This system is only satisfied when q = 5, r = −2 and s = 1, so the theorem guarantees that the energy released by the bomb can be expressed as # 2$ Et F = 0, R 5ρ chalkdustmagazine.com

chalkdust where F is some function. Again, we do not know what F is, but on this occasion we do not need any further information about it: the quantity E t 2 /R 5 ρ must satisfy the above equation at all times. The only way in which this is possible is if it is constant. Therefore, we set it equal to C and solve for the energy to obtain E=C

ρR 5 . t2

In reality, the constant C must be obtained by carrying out an experiment or using a physical model. Surprisingly, its typical values have been found to range between 0.9 and 1.1, so se ing C = 1 is a reasonable assumption. So what is our estimate for the emi ed energy of the Trinity test? From the photo on the previous page, we can observe that 0.025s a er the explosion the radius of the fireball was approximately R = 130m. We also know that the density of air at 20C is ρ = 1.20kg/m3 . Inserting these values into our formula, we compute an energy of 7.13 × 1013 J, which is similar to 17.1 kilotons of TNT, the same as Taylor’s result. To sum up, we have found a formula to estimate the energy released in a nuclear explosion without a single bit of nuclear physics, but only some guesswork, elementary maths and the invaluable help of the Buckingham π theorem. The range of applicability of the Buckingham π theorem is not restricted to these few examples, but using more powerful versions it proves to be a priceless—and simple—tool in the study of thermodynamic and gravitational systems. Although it is usually underestimated, it makes explicit some of the deepest connections between physics and mathematics, as well as highlights some fundamental properties inherent to natural phenomena. Diego Antonio Carranza Ortiz Diego is a PhD student at een Mary University of London, working on the mathematical aspects of the general theory of relativity using conformal methods. @diego_carranza_

d.a.carranzaortiz@qmul.ac.uk

Did you know... …that it is not known whether or not there is a cuboid whose sides and diagonals are all of integral length?

My favourite set

Mahut–Isner set Mattheas Recht

My favourite set is the final set in the match between Nicolas Mahut and John Isner at Wimbledon in 2010. The set lasted over 11 hours with a final score of 70–68 to Isner. 59

autumn 2016

This issue features the top ten colours of chalk. To vote on the top ten parts of a circle go to chalkdustmagazine.com. At 10, it's perfect for colouring in sideburns: brown chalk.

A re-entry at nu mber 9, it's yellow chalk by UCL's second most successful band, Coldplay.

Down ten places to 7: it's magenta chalk.

Getting its first Big Break at number 8: snooker chalk.

At 6, our favourite character from Scrubs: it's Sarah Chalke from Scrubs.

At 5 this week, it's very handy when you can't find a rubber: black chalk.

At 4, a very useful colour when you're drawing the new Chalkdust T-shirt or the District line: dark green chalk.

Thousands of people buying Prince brand chalk have brought purple chalk back into this week's top ten.

colour At 2, a very useful g the in aw when you're dr the or Poddington Peas ded map: District line on a fa light green chalk. Topping the Pops this week: the ever popular white chalk. Can you imagine a lecture without white chalk? Neither can we. Because we've never seen a whiteboard. Sarah Chalke photo: Photo Larry, CC BY 3.0

chalkdustmagazine.com