Mini-monomath

July 19, 2014

The title of this post is a nod to Terry Tao’s four mini-polymath discussions, in which IMO questions were solved collaboratively online. As the beginning of what I hope will be a long exercise in gathering data about how humans solve these kinds of problems, I decided to have a go at one of this year’s IMO problems, with the idea of writing down my thoughts as I went along. Because I was doing that (and doing it directly into a LaTeX file rather than using paper and pen), I took quite a long time to solve the problem: it was the first question, and therefore intended to be one of the easier ones, so in a competition one would hope to solve it quickly and move on to the more challenging questions 2 and 3 (particularly 3). You get an average of an hour and a half per question, and I think I took at least that, though I didn’t actually time myself.

What I wrote gives some kind of illustration of the twists and turns, many of them fruitless, that people typically take when solving a problem. If I were to draw a moral from it, it would be this: when trying to solve a problem, it is a mistake to expect to take a direct route to the solution. Instead, one formulates subquestions and gradually builds up a useful bank of observations until the direct route becomes clear. Given that we’ve just had the football world cup, I’ll draw an analogy that I find not too bad (though not perfect either): a team plays better if it patiently builds up to an attack on goal than if it hoofs the ball up the pitch or takes shots from a distance. Germany gave an extraordinary illustration of this in their 7-1 defeat of Brazil.

I imagine that the rest of this post will be much more interesting if you yourself solve the problem before reading what I did. I in turn would be interested in hearing about other people’s experiences with the problem: were they similar to mine, or quite different? I would very much like to get a feel for how varied people’s experiences are. If you’re a competitor who solved the problem, feel free to join the discussion!

If I find myself with some spare time, I might have a go at doing the same with some of the other questions.

What follows is exactly what I wrote (or rather typed), with no editing at all, apart from changing the LaTeX so that it compiles in WordPress and adding two comments that are clearly marked in red.

Problem Let a_0<a_1<a_2<\dots be an infinite sequence of positive integers. Prove that there exists a unique integer n\geq 1 such that

a_n <\frac{a_0+a_1+\dots+a_n}n\leq a_{n+1}\ .
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ECM2016 — your chance to influence the programme

June 29, 2014

UPDATE: I HAVE NOW GONE BACK TO MODERATING COMMENTS ONLY IF THEY ARE FROM PEOPLE WHO HAVE NOT HAD A COMMENT ACCEPTED IN THE PAST. SO IF YOU HAVE A SUGGESTION TO MAKE FOR AN ECM2016 SPEAKER, PLEASE EMAIL A MEMBER OF THE COMMITTEE DIRECTLY RATHER THAN COMMENTING HERE.

Just before I start this post, let me say that I do still intend to write a couple of follow-up posts to my previous one about journal prices. But I’ve been busy with a number of other things, so it may still take a little while.

This post is about the next European Congress of Mathematics, which takes place in Berlin in just over two years’ time. I have agreed to chair the scientific committee, which is responsible for choosing approximately 10 plenary speakers and approximately 30 invited lecturers, the latter to speak in four or five parallel sessions.

The ECM is less secretive than the ICM when it comes to drawing up its scientific programme. In particular, the names of the committee members were made public some time ago, and you can read them here.

I am all in favour of as much openness as possible, so I am very pleased that this is the way that the European Mathematical Society operates. But what is the maximum reasonable level of openness in this case? Clearly, public discussion of the merits of different candidates is completely out of order, but I think anything else goes. In particular, and this is the main point of the post, I would very much welcome suggestions for potential speakers. If you know of a mathematician who is European (and for these purposes Europe includes certain not obviously European countries such as Russia and Israel), has done exciting work (ideally recently), and will not already be speaking about that work at the International Congress of Mathematicians in Seoul, then we would like to hear about it. Our main aim is that the congress should be rewarding for its participants, so we will take some account of people’s ability to give a good talk. This applies in particular to plenary speakers.
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Elsevier journals — some facts

April 24, 2014

Update: figures now in from Imperial. See below.

Further update: figures in from Nottingham too.

Further update: figures now in from Oxford.

Final update: figures in from LSE.

A little over two years ago, the Cost of Knowledge boycott of Elsevier journals began. Initially, it seemed to be highly successful, with the number of signatories rapidly reaching 10,000 and including some very high-profile researchers, and Elsevier making a number of concessions, such as dropping support for the Research Works Act and making papers over four years old from several mathematics journals freely available online. It has also contributed to an increased awareness of the issues related to high journal prices and the locking up of articles behind paywalls.

However, it is possible to take a more pessimistic view. There were rumblings from the editorial boards of some Elsevier journals, but in the end, while a few individual members of those boards resigned, no board took the more radical step of resigning en masse and setting up with a different publisher under a new name (as some journals have done in the past), which would have forced Elsevier to sit up and take more serious notice. Instead, they waited for things to settle down, and now, two years later, the main problems, bundling and exorbitant prices, continue unabated: in 2013, Elsevier’s profit margin was up to 39%. (The profit is a little over £800 million on a little over £2 billion.) As for the boycott, the number of signatories appears to have reached a plateau of about 14,500.
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A few analysis resources

March 12, 2014

This will be my final post associated with the Analysis I course, for which the last lecture was yesterday. It’s possible that I’ll write further relevant posts in the nearish future, but it’s also possible that I won’t. This one is a short one to draw attention to other material that can be found on the web that may help you to learn the course material. It will be an incomplete list: further suggestions would be welcome in the comments below.

A good way to test your basic knowledge of (some of) the course would be to do a short multiple-choice quiz devised by Vicky Neale. If you don’t get the right answer first time for every question, then it will give you an idea of the areas of the course that need attention.

Terence Tao has also created a number of multiple-choice quizzes, some of which are relevant to the course. They can be found on this page. The quiz on continuity expects you to know the definitions of adherent points and limit points, which I did not discuss in lectures.
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How do the power-series definitions of sin and cos relate to their geometrical interpretations?

March 2, 2014

I hope that most of you have either asked yourselves this question explicitly, or at least felt a vague sense of unease about how the definitions I gave in lectures, namely

\displaystyle \cos x = 1 - \frac{x^2}{2!}+\frac{x^4}{4!}-\dots

and

\displaystyle \sin x = x - \frac{x^3}{3!}+\frac{x^5}{5!}-\dots,

relate to things like the opposite, adjacent and hypotenuse. Using the power-series definitions, we proved several facts about trigonometric functions, such as the addition formulae, their derivatives, and the fact that they are periodic. But we didn’t quite get to the stage of proving that if x^2+y^2=1 and \theta is the angle that the line from (0,0) to (x,y) makes with the line from (0,0) to (1,0), then x=\cos\theta and y=\sin\theta. So how does one establish that? How does one even define the angle? In this post, I will give one possible answer to these questions.
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Differentiating power series

February 22, 2014

I’m writing this post as a way of preparing for a lecture. I want to discuss the result that a power series \sum_{n=0}^\infty a_nz^n is differentiable inside its circle of convergence, and the derivative is given by the obvious formula \sum_{n=1}^\infty na_nz^{n-1}. In other words, inside the circle of convergence we can think of a power series as like a polynomial of degree \infty for the purposes of differentiation.

A preliminary question about this is why it is not more or less obvious. After all, writing f(z)=\sum_{n=0}^\infty a_nz^n, we have the following facts.

  1. Writing S_N(z)=\sum_{n=0}^Na_nz^n, we have that S_N(z)\to f(z).
  2. For each N, S_N'(z)=\sum_{n=1}^Nna_nz^{n-1}.

If we knew that S_N'(z)\to f'(z), then we would be done.

Ah, you might be thinking, how do we know that the sequence (S_N'(z)) converges? But it turns out that that is not the problem: it is reasonably straightforward to show that it converges. (Roughly speaking, inside the circle of convergence the series \sum_na_nz^{n-1} converges at least as fast as a GP, and multiplying the nth term by n doesn’t stop a GP converging (as can easily be seen with the help of the ratio test). So, writing g(z) for \sum_{n=1}^\infty na_nz^{n-1}, we have the following facts at our disposal.

  1. S_N(z)\to f(z)
  2. S_N'(z)\to g(z)

Doesn’t it follow from that that f'(z)=g(z)?
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Recent news concerning the Erdos discrepancy problem

February 11, 2014

I’ve just learnt from a reshare by Kevin O’Bryant of a post by Andrew Sutherland on Google Plus that a paper appeared on the arXiv today with an interesting result about the Erdős discrepancy problem, which was the subject of a Polymath project hosted on this blog four years ago.

The problem is to show that if (\epsilon_n) is an infinite sequence of \pm 1s, then for every C there exist d and m such that \sum_{i=1}^m\epsilon_{id} has modulus at least C. This result is straightforward to prove by an exhaustive search when C=2. One thing that the Polymath project did was to discover several sequences of length 1124 such that no sum has modulus greater than 2, and despite some effort nobody managed to find a longer one. That was enough to convince me that 1124 was the correct bound.

However, the new result shows the danger of this kind of empirical evidence. The authors used state of the art SAT solvers to find a sequence of length 1160 with no sum having modulus greater than 2, and also showed that this bound is best possible. Of this second statement, they write the following: “The negative witness, that is, the DRUP unsatisfiability certificate, is probably one of longest proofs of a non-trivial mathematical result ever produced. Its gigantic size is comparable, for example, with the size of the whole Wikipedia, so one may have doubts about to which degree this can be accepted as a proof of a mathematical statement.”

I personally am relaxed about huge computer proofs like this. It is conceivable that the authors made a mistake somewhere, but that is true of conventional proofs as well. The paper is by Boris Konev and Alexei Lisitsa and appears here.

Taylor’s theorem with the Lagrange form of the remainder

February 11, 2014

There are countless situations in mathematics where it helps to expand a function as a power series. Therefore, Taylor’s theorem, which gives us circumstances under which this can be done, is an important result of the course. It is also the one result that I was dreading lecturing, at least with the Lagrange form of the remainder, because in the past I have always found that the proof is one that I have not been able to understand properly. I don’t mean by that that I couldn’t follow the arguments I read. What I mean is that I couldn’t reproduce the proof without committing a couple of things to memory, which I would then forget again once I had presented them. Briefly, an argument that appears in a lot of textbooks uses a result called the Cauchy mean value theorem, and applies it to a cleverly chosen function. Whereas I understand what the mean value theorem is for, I somehow don’t have the same feeling about the Cauchy mean value theorem: it just works in this situation and happens to give the answer one wants. And I don’t see an easy way of predicting in advance what function to plug in.

I have always found this situation annoying, because a part of me said that the result ought to be a straightforward generalization of the mean value theorem, in the following sense. The mean value theorem applied to the interval [x,x+h] tells us that there exists y\in (x,x+h) such that f'(y)=\frac{f(x+h)-f(x)}h, and therefore that f(x+h)=f(x)+hf'(y). Writing y=x+\theta h for some \theta\in(0,1) we obtain the statement f(x+h)=f(x)+hf'(x+\theta h). This is the case n=1 of Taylor’s theorem. So can’t we find some kind of “polynomial mean value theorem” that will do the same job for approximating f by polynomials of higher degree?

Now that I’ve been forced to lecture this result again (for the second time actually — the first was in Princeton about twelve years ago, when I just suffered and memorized the Cauchy mean value theorem approach), I have made a proper effort to explore this question, and have realized that the answer is yes. I’m sure there must be textbooks that do it this way, but the ones I’ve looked at all use the Cauchy mean value theorem. I don’t understand why, since it seems to me that the way of proving the result that I’m about to present makes the whole argument completely transparent. I’m actually looking forward to lecturing it (as I add this sentence to the post, the lecture is about half an hour in the future), since the demands on my memory are going to be close to zero.
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How to work out proofs in Analysis I

February 3, 2014

Now that we’ve had several results about sequences and series, it seems like a good time to step back a little and discuss how you should go about memorizing their proofs. And the very first thing to say about that is that you should attempt to do this while making as little use of your memory as you possibly can.

Suppose I were to ask you to memorize the sequence 5432187654321. Would you have to learn a string of 13 symbols? No, because after studying the sequence you would see that it is just counting down from 5 and then counting down from 8. What you want is for your memory of a proof to be like that too: you just keep doing the obvious thing except that from time to time the next step isn’t obvious, so you need to remember it. Even then, the better you can understand why the non-obvious step was in fact sensible, the easier it will be to memorize it, and as you get more experienced you may find that steps that previously seemed clever and nonobvious start to seem like the natural thing to do.

For some reason, Analysis I contains a number of proofs that experienced mathematicians find easy but many beginners find very hard. I want to try in this post to explain why the experienced mathematicians are right: in a rather precise sense many of these proofs really are easy, in the sense that if you just repeatedly do the obvious thing you will solve them. Others are mostly like that, with perhaps one smallish idea needed when the obvious steps run out. And even the hardest ones have easy parts to them.
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Introduction to Cambridge IA Analysis I 2014

January 11, 2014

This term I shall be giving Cambridge’s course Analysis I, a standard first course in analysis, covering convergence, infinite sums, continuity, differentiation and integration. This post is aimed at people attending that course. I plan to write a few posts as I go along, in which I will attempt to provide further explanations of the new concepts that will be covered, as well as giving advice about how to solve routine problems in the area. (This advice will be heavily influenced by my experience in attempting to teach a computer, about which I have reported elsewhere on this blog.)

I cannot promise to follow the amazing example of Vicky Neale, my predecessor on this course, who posted after every single lecture. However, her posts are still available online, so in some ways you are better off than the people who took Analysis I last year, since you will have her posts as well as mine. (I am making the assumption here that my posts will not contribute negatively to your understanding — I hope that proves to be correct.) Having said that, I probably won’t cover exactly the same material in each lecture as she did, so the correspondence between my lectures and her posts won’t be as good as the correspondence between her lectures and her posts. Nevertheless, I strongly recommend you look at her posts and see whether you find them helpful.

You will find this course much easier to understand if you are comfortable with basic logic. In particular, you should be clear about what “implies” means and should not be afraid of the quantifiers \exists and \forall. You may find a series of posts I wrote a couple of years ago helpful, and in particular the ones where I wrote about logic (NB, as with Vicky Neale’s posts above, they appear in reverse order). I also have a few old posts that are directly relevant to the Analysis I course (since they are old posts you may have to click on “older entries” a couple of times to reach them), but they are detailed discussions of Tripos questions rather than accompaniments to lectures. You may find them useful in the summer, and you may even be curious to have a quick look at them straight away, but for now your job is to learn mathematics rather than trying to get good at one particular style of exam, so I would not recommend devoting much time to them yet.
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