Elsevier journals — some facts

April 24, 2014

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


\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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DBD2 — success of a kind

January 9, 2014

Yesterday, as I was walking to my office in the morning, I planned to write a post in which I was going to say that Polymath9 had basically been a failure, though not a failure I minded about, since it hadn’t had any significant negative consequences. Part of the reason I wanted to say that was that for a few weeks I’ve been thinking about other things, and it seems better to “close” a project publicly than to leave it in a strange limbo.

When I got to my office, those other things I’ve been thinking about (the project with Mohan Ganesalingam on theorem proving) commanded my attention and the post didn’t get written. And then in the evening, with impeccable timing, Pavel Pudlak sent me an email with an observation that shows that one of the statements that I was hoping was false is in fact true: every subset of \{0,1\}^n can be Ramsey lifted to a very simple subset of a not much larger set. (If you have forgotten these definitions, or never read them in the first place, I’ll recap them in a moment.)

How much of a disaster is this? Well, it’s never a disaster to learn that a statement you wanted to go one way in fact goes the other way. It may be disappointing, but it’s much better to know the truth than to waste time chasing a fantasy. Also, there can be far more to it than that. The effect of discovering that your hopes are dashed is often that you readjust your hopes. If you had a subgoal that you now realize is unachievable, but you still believe that the main goal might be achievable, then your options have been narrowed down in a potentially useful way.

Is that the case here? I’ll offer a few preliminary thoughts on that question and see whether they lead to an interesting discussion. If they don’t, that’s fine — my general attitude is that I’m happy to think about all this on my own, but that I’d be even happier to discuss it with other people. The subtitle of this post is supposed to reflect the fact that I have gained something from making my ideas public, in that Pavel’s observation, though simple enough to understand, is one that I might have taken a long, or even infinite, time to make if I had worked entirely privately. So he has potentially saved me a lot of time, and that is one of the main points of mathematics done in the open.
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A little paradox

December 9, 2013

This post is intended as a footnote to one that I wrote a couple of years ago about the meaning of “implies” in mathematics, which was part of a series of posts designed as an introduction to certain aspects of university mathematics.

If you are reasonably comfortable with the kind of basic logic needed in an undergraduate course, then you may enjoy trying to find the flaw in the following argument, which must have a flaw, since I’m going to prove a general statement and then give a counterexample to it. If you find the exercise extremely easy, then you may prefer to hold back so that others who find it harder will have a chance to think about it. Or perhaps I should just say that if you don’t find it easy, then I think it would be a good exercise to think about it for a while before looking at other people’s suggested solutions.
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