31 December 2007

Some football probabilities

Back in September I answered a question about probabilities in Major League Baseball's wild-card race..

John Armstrong (of The Unapologetic Mathematician fame) asked me about NFL football: assuming all teams are evenly matched, what is the probability that a division sends three teams to the playoffs? For those of you who aren't familiar with the NFL's playoff structure, there are two conferences of sixteen teams each (the NFC and the AFC); each conference is divided into four divisions of four. The winner of each division gets into the playoffs, and the two best teams among non-division winners in each conference get in as "wild cards". (For an example, see this year's final standings at nfl.com.)

This is equivalent to the following question: what is the probability that both wild cards in a conference come from the same division? I suspect that this question is inspired by the fact that this actually happened in both conferences this year, in the AFC South and NFC East respectively. (This must be especially frustrating for fans of the fourth-place team in each of those divisions. The fourth-place team in the NFC East was the Philadelphia Eagles. I think I'll avoid Eagles fans today. But that might be hard.)

Anyway, the probability that this event happens in any given conference in any given year is 2/11. Let the teams play out their season. Consider the twelve non-division winners in a given conference, and rank them in order from 1 to 12. (You might think: in the NFL each team only plays sixteen games, so won't there be ties? Perhaps, but the NFL's wild card tiebreakers don't care what division a team is in.) The top two teams get in. The divisions of the twelve teams (from top to bottom) form a word with three N's, three E's, three S's, and three W's, picked uniformly at random. Say without loss of generality that the first letter that the word starts with an N; then the remainder of the word is picked uniformly at random from permutations of NNEEESSSWWW. The chances that starts with an N are 2/11.

Alternatively, there are ${12 \choose 3,3,3,3} = 369600$ words which contain three each of N, E, S, and W. The number of these that begin with the same letter twice is $4{10 \choose 1,3,3,3} = 67200$ -- pick the letter that appears twice at the beginning in one of four ways, then arrange the ten remaining letters however you like. So the probability of the two division winners coming from the same division is 67200/369600, or 2/11. I don't like this approach quite as much, though, because it involves big numbers that cancel out; it seems cleaner to me to not invoke the big numbers in the first place.

By the way, the NFL has had its current structure since 2002, and the years and conferences in which this event has actually occurred are the 2007 AFC South, 2007 NFC East, and 2006 NFC East, making three out of twelve.

The question I asked in September was not the analogue of this one (MLB only has one wild card, so there is no analogue); there I found the probability that the four MLB teams making the playoffs from any league were actually the four teams with the best records. But I don't have a general method to find the solution for some league with arbitrary structure, and the NFL would require slogging through more cases than MLB (two wild cards instead of one) so I won't bother for now. (Or probably ever -- I don't particularly care about that number.)

30 December 2007

Why are knots so common?



29 December 2007





Gravity pods



28 December 2007

On the Mian-Chowla sequence

An interesting problem from Additive Combinatorics by Terence Tao and Van Vu. (This is almost a direct quote; I've modified it slightly in order to be able to render it entirely in HTML.)

Exercise 6.2.5. Let S = {1, 2, 4, 8, 13, 21, 31, ...} be the Sidon set of positive integers constructed by the greedy algorithm (this set is sometimes known as the Mian-Chowla sequence). Show that the kth element of S does not exceed (k-1)3 + 1, and hence the number of elements of S which are less than n is Ω(n1/3) as n goes to infinity.

A Sidon set, for those of you who don't know, is a set for which all the pairwise sums are distinct. Tao and Vu credit this result to Stöhr; see below. A rough translation of the title of that paper is "Solved and unsolved questions on bases of sequences of natural numbers", although my German isn't so good. (I've been looking for mathematics written in German, though, because my department requires me to be able to read two of mathematical French, German, and Russian -- I've already passed the French and I know a little German and no Russian -- so I might try to read this, or at least parts of it.)

First, an example of where these numbers are coming from. Let s1 = 1, s2 = 2, s3 = 4, and so on form the sequence S. Say we know the first four terms, s1 through s4, are 1, 2, 4, 8. Then 9 can't be in the sequence, because 9 + 1 = 8 + 2. 10 can't be in it, because 10 + 2 = 8 + 4. 11 can't be in it, because 11 + 1 = 8 + 4. And 12 can't be in it, because 12 + 4 = 8 + 8. But none of 13 + 1, 13 + 2, 13 + 4, and 13 + 8 are pairwise sums of {1, 2, 4, 8}, so the fifth term is 13. It turns out to be easier, I think, to think in terms of differences instead of sums; we say that 9 can't be in the sequence because 9 - 2 = 8 - 1, and so on. So if s1, ..., sk are known, we choose sk+1 to be the smallest integer N such that the intersection
\{ N-s_i \}_{1 \le i \le k} \cap \{ s_j - s_i \}_{1 \le i < j \le k}

is empty.

The solution is straightforward. Now, by the "greedy algorithm" it's meant that if we know s1 through sk, then sk+1 is the smallest integer greater than sk which doesn't violate the definition of our sequence. Then it's enough to minimize the difference tk = sk+1 - sk; this is the smallest positive integer which does not occur among pairwise differences of elements of {s1, ..., sk}. There are k(k-1)/2 such differences, and they're not necessarily distinct, so the smallest positive integer which doesn't occur is at most (k(k-1)/2) + 1, or ${k \choose 2} + 1$. Then the sk are just the partial sums of the tk, so we get
s_k \le 1 + \sum_{j=1}^k \left( {j \choose 2} + 1 \right) = {k+1 \choose 3} + k + 1

which is asymptotically equal to k3/6.

The same method gives a lower bound -- all the differences sk - si for i < k are distinct, so tk is at least k+1. This gives a lower bound for sk of order k2/2.

The continuation of the sequence can be found in the Encyclopedia of Integer Sequences; it shares its first eight terms with this sequence, with which I confused it at first. The condition for computing this "impostor sequence" is that
\{ N - s_k \} \cap \{ s_j - s_i \}_{1 \le i < j \le k}

be empty. The impostor sequence begins {1, 2, 4, 8, 13, 21, 31, 45, 60, ...} and it's at the term "60" that it first differs from the Mian-Chowla sequence. We have 60-45 = 15, and 15 never occurs as a difference between any of the first eight terms. But 60 - 31 = 31 - 2 = 29, so 60 isn't the ninth term of the Mian-Chowla sequence; that turns out to be 66. The divergence only gets wider after that. But the upper and lower bounds that I gave above still apply!

It appears (from computations of Frank Chu) that the nth term of the Mian-Chowla sequence grows like n2.8. I don't know why this should be, though. However, the sequence I confused it with is the partial sums of the segmented numbers; the nth segmented number is about n lg lg n (this is just a conjecture, but a reasonable one) and so my impostor sequence should grow like (n2 lg lg n)/2, which fits the known values reasonably well. Since my impostor sequence has fairly weak conditions allowing us to extend it, I would conjecture that its growth rate is close to the lower bound, of order n2, and so n2 times a subpolynomial function seems nice. I'd actually conjecture that the Mian-Chowla sequence grows like n3 divided by some subpolynomial function -- but which one? Finally, something about the fact that we're looking at collisions between the difference sets makes me think that there might be a probabilistic model for this problem.

Frank Chu. Notes on the Mian-Chowla Sequence. http://www.cumc.math.ca/2004/mc.pdf
A. Stöhr, Gelöste und ungelöste Fragen über Basen der natürlichen Zahlenreihe, I, II. J. Reine Angew. Math. 194 (1955), 40-65; 111-140.
Terence Tao and Van Vu, Additive Combinatorics. Cambridge: Cambridge University Press, 2006.
Encyclopedia of Integer Sequences: A004978, A005282.

27 December 2007

A couple articles from the Princeton Companion to Mathematics

Differential forms and integration (10 pp.), from the forthcoming Princeton Companion to Mathematics. (From Terence Tao, who has also written other articles for the PCM. I think it would have been rather nice to receive such an article at the outset of the class in which I learned about this stuff, as it would have helped to give some context; although one can often get context from a textbook, it's difficult to do so because all the little explanatory bits are often mingled with the details. The PCM articles I've seen do not suffer from this flaw; indeed, one might describe a lot of them as what the preface of a textbook covering the same material should be.

In a less theoretical, and less general, vein, here's Frank Kelly's The Mathematics of Traffic in Networks, also from the PCM, which I learned about from a post from Tao last week. One thing this mentions is the paradox that under certain conditions, adding roads to a road network can make trips slower, even though each driver tries to minimize their own time spent on the road. (Fortunately, my life is arranged so that I walk almost everywhere; thus at the moment I only appreciate the weirdness of traffic on a theoretical level. But I know it won't always be like that.) I'd be interested to see if this paradox occurs in actual road networks or only in the contrived example that Kelly gives.

The PCM will sell for about $100, so I probably won't buy it when it comes out. But I suspect I will spend some time browsing it when our library gets it.

A postal nomogram




25 December 2007

Merry Christmas

Merry Christmas!



24 December 2007

Why W?

Here's an interesting article I found while poking around the Web for something else: Why W?, by Brian Hayes, from the March-April 2005 issue of American Scientist. W is not the president here, but the inverse function of x → xex. Apparently the name didn't become popular until the Maple software was written, because Maple needed a name for the function.

Empirically speaking, when I type random things I'm curious about into Maple, the W function is one of the functions that most frequently appears in its analytic solutions; Bessel functions are also quite common, which shouldn't be surprising, as are various instances of the hypergeometric series. Since so many familiar functions can be written in terms of the hypergeometric series, this is hardly surprising.

The coefficient of xn in the Taylor series for -W(-x) is nn-1/n!; the number of labelled trees on n vertices is nn-2 (Cayley), so it's not that surprising that W is involved in the enumeration of trees. In fact, W(x) is the generating function for labeled rooted trees. (n rooted trees correspond to each unrooted one, since we can choose the root from among all the verticecs.

Unfortunately most of these trees are very oddly shaped and would not look nice as a Christmas tree.

23 December 2007

Reciprocal fuel economy

Eric de Place notes:
This sounds counterintuitive at first. But the "natural" units for fuel consumption, at least in this case, are not distance per unit of fuel but units of fuel per distance. In some parts of the word fuel usage is given in liters per 100 km; let's say we were to give fuel usage in gallons per 100 miles. (The constant "100" is just there to make the numbers reasonably sized.) Then switching from a 15 mpg car to an 18 mpg car is switching from a car that gets 6.67 gal/100 mi to 5.56 (lower is better); switching from a 50 mpg car to a 100 mpg car is switching from a car that gets 2.00 gal/100 mi to 1.00. (Another interesting consequence -- switching from 50 mpg to 100 mpg has the same effect as switching from 100 mpg to ∞ mpg, i. e. a car that uses no fuel at all.)

The idea is that chopping off the low-fuel-economy tail of the distribution (by legal means) would be a much easier way to reduce oil consumption than trying to make very-high-fuel-economy cars.

But not all incremental achievements are created equal. It was probably a lot harder to get from 1 mpg to 2 mpg than it will be to get from 100 mpg to 101 mpg.

Also, note that a pair of cars that get, say, 20 mpg and 50 mpg will average "35 mpg" in the way that the new regulations for average mileage of a automaker's fleet are written; but for each car to go 100 miles, it'll take a total of seven gallons of fuel, for a fuel economy of 200/7 = 28.6 mpg. (This is the harmonic mean of 20 and 50.) The regulations aren't necessarily flawed -- they probably should be stated in terms of the measures of fuel economy that are most commonly used -- but there's room for possible misunderstanding.

Another place I can think of where the "natural" units are the reciprocal of the ones that are habitually used is in statistical mechanics; there are tons of formulas there that have temperature in the denominator, and for the purposes of statistical mechanics it makes more sense to use inverse temperature. (I've written about this before, I think; it basically comes out of the fact that the partition function involves inverse temperature.) Are there others?

(I found this from Marginal Revolution.)

Tracking an intruder

From Yuliy Baryshnikov: an animation described as follows: "In the 40-by-40 array of sensors, a running intruder is represented as a black dot. Can you spot her? The catch: the sensors are prone to spontaneous signals ("false positives"). These signals are rare (just 2.5% probability), but what a mess they make!"

But of course, if you stare long enough you can see the intruder. The trick, is that the intruder moves slowly, say one grid square per second. If you overlay the grids at time n and n+1 on top of each other, there will be about eight pairs of (kingwise) adjacent squares such that the first square is a false positive at time n and the second square is a false positive at time n+1. (There are roughly 402 × 8 ordered pairs of kingwise adjacent squares.) If we consider it at three consecutive times our expected number of false positives is the 402 × 82 triples of kingwise adjacent squares, times the 40-3 probability that the three squares of such a triple are occupied at times n, n+1, n+2, for an expected 1.6 false positives. And so on -- there are one-fifth as many false positives for each extra consecutive step we look at.

If false positives occurred one-eighth of the time, though, then we'd have no hope; the number of "false positives" when looking at some window of length k would be constant, instead of decreasing exponentially with k. In general, you'd expect that if the intruder has r choices for her position of each step, then as the probability of false positives approaches 1/r the intruder becomes impossible to track.

A quick skim of titles of Baryshnikov's papers doesn't tell me if this is an illustration of something from his research. I'm curious, so I've sent him an e-mail to ask.

22 December 2007

What is power?

From Walter Lewin's 8.01 Lecture 14, about 14 minutes in:

"What is power? Power is work that is done in a certain amount of time. dw/dt, if w is the work, is the instantaneous power at time t. We also know power in terms of political power. That's very different. Political power, you can do no work at all in a lot of time and you have a lot of power. Here in physics, life is not that easy."

This reminds me of the classic claim that, say, bricklayers do more work than lawyers, because work is force times distance, and bricklayers are exerting much larger forces when they lift things, since bricks are heavier than pieces of paper. This is a sort of translingual pun, the two languages being English and scientific jargon.

21 December 2007

When asteroids attack

Asteroid May Hit Mars in Next Month (AP).

Scientists tracking the asteroid, currently halfway between Earth and Mars, initially put the odds of impact at 1 in 350 but increased the chances this week. Scientists expect the odds to diminish again early next month after getting new observations of the asteroid's orbit, Chesley said.

If they expect the odds to diminish, why haven't they set them lower in the first place? I think this may be another issue of mean-median confusion -- say, there's a 1 in 3 chance that the odds will go up to 1 in 25 next month and a 2 in 3 chance it'll go away completely. But the statement seems kind of silly. Apparently efficient market theory doesn't apply to asteroids.

Links for 2007-12-21

Mathematicians solve the mystery of traffic jams, which has been making the rounds lately. Basically, sometimes traffic jams come out of nowhere because someone taps their brakes and the traffic propagates backwards. I remember reading this ten or fifteen years ago. It must be a slow news day.

Thanks to Wing Mui, The trouble with five by Craig Kaplan. Why is it that we can tile the plane with triangles, quadrilaterals, or hexagons -- but not pentagons? And what sort of structure do tilings with some sort of fivefold symmetry have?

Also, a few days ago I wrote about Galileo's argument that the thickness of bones should scale like the 3/2 power of an animal's height, which I learned about from Walter Lewin's lectures, but that this turns out not to be true. Having gotten a bit further into Lewin's lectures, I found out that the real thing we're protecting against is buckling, not crushing.

Greg Mankiw a couple months ago provided a link to a paper on optimal mortgage refinancing by Sumit Agarwaland, John Driscoll, and David Laibson. The Lambert W function (the inverse of f(x) = xex) turns out to be good for something! If you use this, I want a cut of your savings for pointing you to it.

20 December 2007

A sighting of mathematics on Jeopardy!





Weather forecasting

One way I procrastinate and/or pass time is by reading the archives of long-running blogs. Today it's glenn mcdonald's furialog, because I was thinking of buying the My So-Called Life box set, but decided against it because it's $70; mcdonald used to write a music review column called The War Against Silence which I liked, the first installment of which was written five days after the last episode of that tragically short-lived television show aired. mcdonald was a fan of the show, and I was bored this morning, so I thought of his blog. (For the record, he's a programmer by trade but not by training, although this may be irrelevant.)

Anyway, he wrote on 13 January 2005, commenting on weather forecasting:
But what I would hold someone responsible for, if I thought it weren't a pervasive cultural flaw, is the destructive precision with which uncertain predictions are communicated. Weather is merely the most obvious daily public manifestation of a fundamental reluctance, or perhaps an inability, to say what we really know, rather than what we wish we knew.

Similarly, I've watched a bunch of Walter Lewin's introductory physics lectures in the last few days (I've mentioned this before) and he has drummed into my head that a measurement without uncertainty is useless. What is true for measurements is doubly true for predictions.

It would be interesting to see error bars on weather predictions, but we never will. One can get something like this, though, for the tracks of tropical storms at Weather Underground; they show the outputs of various computer models. If these models all show a storm going in the same direction, one can be fairly sure it will actually go in that direction; if the projected paths are spread out, one knows not to be so sure. I wonder if it is possible to determine the accuracy of more mundane weather predictions by, say, tuning into a large number of different TV channels, web sites, and so on which make independent forecasts. If it were possible to get archived weather forecasts I'd take a look; as it is I don't think I can find out what people on July 14, 2006 (say) thought the weather would be like on July 18, 2006, so the only way I could collect data would be to wait, or perhaps to call up some weather-forecasting entity and ask, and I don't care quite that much.

The fundamental reason we don't see error bars on weather forecasts, though, is because they are brought to us for the most part by profit-seeking entities, and such entities really don't want to advertise "we don't know what's going to happen!" -- even though everybody knows this. This is the "pervasive cultural flaw" that mcdonald refers to above.

Also, although I cannot produce data for this, I seem to perceive that at least certain weather forecasting outlets use only even numbers of (Fahrenheit) degrees in their forecasts, not odd numbers. If this is true, I suspect it means that they believe they can't be accurate to within one degree.

19 December 2007

What if gravity reversed itself?

A friend of mine, not long ago, noticed that it was snowing up outside her office window. (When I've seen this happening, it's usually due to something like an internal courtyard in a building, so the air currents get sufficiently weird to push the snow upwards. Don't worry, physics isn't broken.)

This raises the (somewhat silly, I must admit) question: what if gravity had actually reversed itself outside my friends' office? Three scenarios are possible:

  • gravity instantaneously reverses itself;

  • the downward-pointing vector of gravitational acceleration decreases in magnitude, goes through zero, and then becomes an upward-pointing vector;

  • the gravitational acceleration vector always has the same magnitude, but swings around through different angles to point up instead of down.

I claim that the third of these is not reasonable. Basically, this is for reasons of symmetry. Imagine gravity that neither points straight down nor straight up. How would gravity know in which of the many possible "sideways" directions to point? There are only two points that could possibly have any effect on the direction of gravitational acceleration, namely the point that you are at and the center of the Earth. The direction of gravity must be invariant under any isometry which fixes those two points -- thus it must point straight towards the center of the Earth or straight away from it.

Apparently I am more attached to gravity being a central force than to its particular strength or even to the fact that it is attractive, not repulsive.

(I'm not sure whether the first or second of the models suggested above is reasonable. This post is silly enough as it is.)

18 December 2007

Scaling arguments

There's a classic argument, supposedly due to Galileo, for why mice and elephants are shaped differently, and there are no organisms that have the general mammalian body plan but are a hundred feet tall.

The argument goes as follows: assume that all animals have roughly similar skeletons. Consider the femur (thigh bone), which supports most of the weight of the body. The length of the femur, l, is proportional to the animal's linear size (height or length), S. The animal's mass, m, is proportional to the cube of the size, S3. Now, consider a cross-section of the bone; say the bone has cross-sectional area d. The pressure in the bone is proportional to the mass divided by the cross-sectional area, or l3/d2. Bones are probably not much thicker than they have to be (that's how evolution works), and have roughly the same composition across organisms. So l3/d2 is a constant, and so d is proportional to l3/2 or to s3/2. SSimilar arguments apply to other bones. So the total volume of the skeleton, if bones are asusmed to be cylindrical, scales like S7/2 -- and so the proportion of the animal taken up by the skeleton scales like S1/2. What matters here is that if S is large enough then the entire animal must be made up of bone! And that just can't happen.

Unfortunately, the 3/2 scaling exponent isn't true, as I learned from Walter Lewin's 8.01 (Physics I) lecture available at MIT's OpenCourseWare. It's one of the canonical examples of dimensional analysis... but it turns out that it just doesn't hold up. I suspect, although I can't confirm, that this is because elephant bones are actually substantially different from mouse bones. It looks like for actual animals, d scales with l (or perhaps like lα for some α slightly greater than 1), not with l3/2. Lewin uses this as an example of dimensional analysis; he also predicts that the time that it takes an apple to fall from a height is proportional to the square root of that height, which is true, but that's such a familiar result that it seems boring.

(Watching the 8.01 lecture is interesting. The videotaped lectures available on OCW are from the fall of 1999, which is two years before I entered MIT; occasionally the camera operator pans to the crowd of students, and a few of them look vaguely familiar.)

P. S. By the way, this blog is six months old. Thanks for reading!

Edited, Wednesday, 9:14 AM: Mark Dominus points out that the argument is in fact due to Galileo, and can be found in his Discourses on Two new Sciences. This is available online in English translation.

Particularly numerically egregious typos

Language Log's Mark Liberman wrote this morning about The biggest typo in history, from this New York Times article, which referred to 10500 instead of 10500. (They've fixed the mistake now; I wonder if they found it independently or if someone pointed out the Language Log post to them?) The number here refers to the size of the string theory landscape, which I don't pretend to understand.

The Times made a similar mistake in The Death of Checkers, December 9, writing:
The brute-force method is slow, which is its big limit. Schaeffer says he suspects you couldn’t use it to solve chess, because that game — with between 1040 and 1050 possible arrangements of pieces — is far more complicated than checkers, which has 5 × 1020 positions. “Chess won’t be solved in my lifetime,” he predicts. “We need some new breakthrough in technology to do that.

(James Kushner, a reader of this blog, pointed this out to me; I meant to mention it at the time but didn't.)

This particular mistake is actually so common that I barely notice it any more, especially because I usually have a decent idea of what the number should be. But to someone who doesn't know, it could make a big difference in how they perceive the article. For one thing, if you interpret the checkers quote literally, you get the idea that checkers is more complicated than chess: after all, 5 × 1020 = 5010 5100.

Another paper that's been known to make these mistakes is The Tech, MIT's student newspaper. The difference here is that The Tech ought to know better; their entire staff is MIT students, who ought to be familiar with exponents. The Times at least has an excuse. (Disclaimer: I went to MIT as an undergraduate.)

Now if only I could get the Times to write about tilings of the plane, where they could say that an 80 by 80 rectangle has "10800" domino tilings. (That's "ten thousand eight hundred", as opposed to the correct number, which is around "ten to the eight hundredth".) Or just in general to write about statistical mechanics, which is where numbers like this come up.

(The Times article which provoked all this -- on the question of whether the laws of nature are inherently mathematical or whether we've just gotten lucky -- is interesting as well. The way I see it, even if a nonmathematical physics is possible, the mathematical physics is so far ahead that we might as well just act as if physics is ultimately governed by mathematics. But I don't think a nonmathematical physics that has the same explanatory power as the physics we have is possible. That might just be a matter of faith.)

Monkeys can do math. Sort of.

Monkeys and college students equal at mental math? (Reuters)





17 December 2007

What is infinity, anyway?

At this post from A Dialogue on Infinity, Alexandre Borovik writes about an experiment that one of his teachers tried once, in which the calculus teacher at his boarding school tried to build all of calculus in terms of finite elements. The logic here is basically the following: the main use of calculus is to solve differential equations. The differential equations basically come from approximating some discrete calculation as a continuous one. (The example that comes to mind for me is the way in which the equation of the catenary, the shape that a chain hangs in if it's only under the influence of gravity, is usually derived; see, for example, this page, which asks us to "Consider the equilibrium of the short length of chain subtending a distance dx on the x-axis." And then if you're reduced to solving a differential equation numerically (a common state of affairs), this is basically done by this same sort of finite element analysis -- Euler's method and the like.

I'd be interested to see how that worked.

But on the other hand, I'm not sure how valuable it is. Sometimes finite numbers are large enough that one doesn't want to deal with them as finite numbers, and one wants to make infinitesimal approximations. Basically, sums are hard, and integrals are easy. This is the insight behind the Euler-Maclaurin formula, which approximates sums by integrals and then lets you know how good the approximation is. It's hard to come up with the formula for the sum of the first n 8th powers; but do you really need the exact number, or do you just need to know it's about n9/9, which is what integration would tell you?

A commenter at an earlier post from the same blog wrote:
Your example from Computer Science reminds me of something often forgotten or overlooked in present-day discussions of infinite structures: that some of the motivation for the study of the infinite in mathematics in the 19th and early 20th centuries came from physics where (strange as it may seem to a modern mathematician or a computer scientist) the infinite was used as an approximation to the very-large-finite.

I didn't realize that historical point. But somehow that doesn't seem strange to me. This may mean I spent too much time hanging around physicists in my formative years.

16 December 2007

The elevator paradox

Saw Friday's episode of Numb3rs, which I usually skip because the mathematical content is usually laughable from my point of view. But I learned about the elevator paradox, which says that when you're at the bottom of a building, usually the first elevator which comes will be going down, but when you're at the top of a building, it'll usually be going up. The reason, according to the Wikipedia article, is:
Essentially, the explanation seems to be this: a single elevator spends most of its time in the larger section of the building, and thus is more likely to approach from that direction when the prospective elevator user arrives. An observer who remains by the elevator doors for hours or days, observing every elevator arrival, rather than only observing the first elevator to arrive, would note an equal number of elevators traveling in each direction.

But this raises an interesting question: how do we know that you'd have an equal number of elevators traveling in each direction? This is, it appears, basically by conservation of elevators. If in general more elevators go upwards than downwards past a certain point, then the number of elevators below that point will steadily decrease. Surprisingly, none of the online expositions seem to mention this. They seem to think it's obvious that the first elevator passing a point has an equal probability of going up and down. But I've spent enough time with probability to know it's ridiculously counterintuitive.

Also, should Wikipedia have mathematical proofs?, from Slashdot. I haven't thought about this; if I think of anything worth saying I'll post it here, but the link alone didn't seem worth its own post.

15 December 2007

Playing around with sums of squares

An interesting random fact, learned from Arcadian Functor:

12 + 22 + 32 + ... + 242 = 702

This doesn't seem quite so weird if you recall that the sum of the first n squares is n(n+1)(2n+1)/6; then the sum of the first 24 squares is (24)(25)(49)/6.

But are there any other n for which the sum of the first n squares is also a square? Not under a million. I tested. It seems kind of tricky to prove this directly, because knowing the factorization of any of n, n+1, or 2n+1 doesn't tell you much about the factorizations of the others.

Let f(n) = n(n+1)(2n+1)/6; for what proportion of values of n is f(n) squarefree? For random integers it's π2/6. Let g(n) be the number of integers [1,n] such that f(n) is squarefree; then g(1000) = 504, g(10000) = 5029, g(100000) = 50187.

Hmm, it looks an awful lot like g(n) ~ n/2, doesn't it? I have no idea if this is true. It seems quite obvious that g(n) ~ cn for some constant c, but I'm not sure if c is exactly 1/2. (And I'd be surprised if it did come out to be 1/2; these sorts of probabilities are rarely rational.)

Tilings of the plane

For various reasons, I've been thinking about tilings of the plane a lot lately. This reminded me of a paper I haven't thought about in a while, by Cris Moore, Some Polyomino Tilings of the Plane, which "calculates generating functions for the number of tilings of rectangles of various widths by" various polyominoes. Let Tm,n be the number of tilings of an m-by-n board by a given set of tiles; then the bivariate generating function
\sum_{m \ge 0} \sum_{n \ge 0} T_{m,n} x^m y^n

is in general a quite complicated object. (I don't know of any explicit examples of this function, which I think just goes to show that it's complicated!) Even the "diagonal" generating function,
\sum_{n \ge 0} T_{n,n} x^n

which has as the coefficient of xn the number of tilings of an n-by-n square, seems to be on the ugly side. On the contrary,
\sum_{n \ge 0} T_{m,n} x^n

is usually a rational function. In a sense tilings of strips of fixed width are a "one-dimensional" problem -- for example, one could encode tilings of fixed-width strips with a certain set of tiles as words over some finite alphabet, I suspect. The fact that these generating functions are rational seems to be a "folk theorem" -- anybody who's thought about tilings knows it, but I've never seen a proof.

For example, the number of tilings of the 2-by-n rectangle with dominoes are just the Fibonacci numbers! (See Mathworld's pictures.) This is easy to see. Let Tn be the number of ways to tile a 2-by-n rectangle with domnioes. The left edge of a 2-by-n rectangle (that's 2 rows, n columns) tiled with dominoes is either a single vertical domino, in which case there are Tn-1 ways to finish the tiling, or two horizontal ones, in which case there are Tn-2 ways to finish the tiling. So Tn = Tn-1 + Tn-2; noting that T1 = 1, T2 = 2 finishes the proof.

But the number of tilings of an m-by-n rectangle by dominoes is given by the quite difficult formula
T_{m,n} = 2^{mn/2} \prod_{j=1}^m \prod_{k=1}^n \left( \cos^2 {j\pi \over m+1} + \cos^2 {k\pi \over n+1} \right)^{1/4}

which is one of those curious formulas that occur in combinatorics in which some expression that has no business even being an integer is not only an integer, but manages to count something! (Notice that if both m and n are odd, then the number of domino tilings of Tm,n should be zero; the formula takes care of this nicely. Namely, if m and n are both odd then the j = (m+1)/2, k = (n+1)/2 term of the product is zero. (I'm quoting Graham, Knuth, and Patashnik for this formula; it's originally due to Kasteleyn. Anybody who mentions domino tilings is required by law to cite Kasteleyn's paper.) There's a rather strange interpretation of this formula I read somewhere once: given an m-by-n board, you can write the number

in the (j, k) square, and the product of those numbers is the number of domino tilings of the board.

Furthermore, if you consider log Tm,n the product becomes a sum:
\log T_{m,n} = {mn \over 2} \log 2 + {1 \over 4} \sum_{j=1}^m \sum_{k=1}^n \log \left( \cos^2 {j\pi \over m+1} + \cos^2 {k\pi \over n+1} \right)

and we can approximate that sum by an integral, since it's a Riemann sum; thus
\log T_{m,n} \approx {mn \over 2} \log 2 + {mn \over 4\pi^2} \int_0^\pi \int_0^\pi \log(\cos^2 x + \cos^2 y) \thinspace dx \thinspace dy .

The integral isn't elementary, but it can be evaluated numerically; thus we get log Tm,n ≈ .29156 mn.

Perhaps a bit surprisingly, this only depends on the size of the rectangle being tiled, not its shape; that's not quite true, though, because the approximation by an integral is a bit crude. Still, for example, an 80 by 80 rectangle has 1.18 × 10800 domino tilings, while a 40 by 160 rectangle has 4.26 × 10797; in terms of the logarithm (which is how these things should be viewed) those are quite close. There are deep connections to statistical physics here that I don't quite understand, although I could if I put my mind to it; the limit of log Tm,n/mn as m and n get large is called the entropy of a set of tiles (Moore), and I believe something like this is true for any set of tiles, although entropies aren't exactly easy to calculate.

1. Ronald L. Graham, Donald E. Knuth, Oren Patashnik. Concrete Mathematics, 2nd edition. Addison-Wesley, 1994.
2. P. W. Kasteleyn, "The statistics of dimers on a lattice, I: The number of dimer arrangements on a quadratic lattice." Physica 27 (1961) 1209-1225.
3. Cris Moore, "Some Polyomino Tilings of the Plane". Availalable from author's web site, http://www.santafe.edu/~moore/.

14 December 2007

Credit card "points"

I got a credit card today.

Like many credit cards, it comes with rewards "points". I get one point for every $1 I spend. The information that came with the card includes a table that begins as follows, which purports to show "how fast your points... can add up":

Everyday PurchasesAmountPoints


What, are people so stupid that they can't multiply by one?

(Of course, there are much more serious quantitative-illiteracy issues involved with credit cards, namely that people don't realize how much they get gouged on the interest rates, or just how long it'll take them to pay off the thing -- and how much interest they'll pay -- if they make the "minimum payment". But I won't go there.)

Why mathematics doesn't need a Mitchell Report

Yesterday, the Mitchell Report, on steroid usage in baseball, was released. A large number of players were named as users or potential users.

The media coverage of this has routinely mentioned that amphetamines seem to be more common in baseball than steroids; one source I ran into said it's believed that half of baseball players use amphetamines regularly.

Now, when I think of amphetamines, I think of Paul Erdos, and the following story: Ron Graham bet Erdos that he couldn't quit amphetamines for a month, cold turkey. Erdos did. Graham paid up. Erdos said "you've set mathematics back a month".

As far as I know, we mathematicians don't have a drug problem. (Unless you count coffee. I'll freely admit I have a coffee problem.) But I don't think that anybody would say that Erdos should be stripped of any award he won because he was using drugs. The difference is that in sports, the players are competing against each other; thus a level playing field is essential. But in mathematics, we like to believe that we are not competing against each other but cooperating to discover truth; thus we may gladly accept all the chemical help we can get.

13 December 2007

Problem A3 from the Putnam

Here is problem A3 from the 2007 Putnam exam:
Let k be a positive integer. Suppose that the integers 1, 2, 3, ..., 3k+1 are written down in random order. What is the probability that at no time during this process, the sum of the integers that have been written up to that time is a positive integer divisible by 3?

I'm going to give a solution that probably wouldn't get full credit on the Putnam (I'm leaving out some details) -- a lot gets lost in the details.

First, instead of considering these numbers, just consider everything modulo 3. In particular, we want to know what the probability is that if we arrange k+1 +1s, k -1s, and k 0s in random order, no partial sum will be divisible by 3. (The number of permutations of {1, ..., 3k+1} which map to a given sequence modulo 3 is the same for any such sequence, so this is legitimate.) The number of ways to arrange k+1 +1s, k -1s, and k 0s is, of course, ${3k+1 \choose k, k, k+1}$.

Now we consider the simpler problem of arranging k+1 +1s and k -1s so that no partial sum is divisible by 3. Clearly the last partial sum (i. e. the full sum) is 1, and the sequence of partial sums moves upwards or downwards by 1 at each step. Such an arrangement must begin with a +1; if it began with a -1 then the first partial sum would be -1, the last partial sum would be 1, so some intermediate partial sum would have to be zero. This +1 must be followed with another +1 to avoid having the second partial sum be 0; then a -1 so that the third partial sum is not 3; and so on. Thus there is exactly one such arrangement, namely

+1, +1, -1, +1, -1, +1, -1, ..., +1, -1

where the partial sums are

1, 2, 1, 2, 1, 2, 1, ..., 2, 1.

The probability of obtaining an arrangement with no partial sums divisible by three if we pick at random from all arrangements of k+1 +1s and k -1s is thus $1/{2k+1 \choose k}$.

Now, to obtain an arrangement of k+1 +1s, k -1s, and k 0s uniformly at random, we can first place the 0s at random, then the +1s and -1s. There are  ${3k+1 \choose k}$ ways in which the 0s can be placed. Interspersing 0s among some arrangement of +1s and -1s doesn't change the set of partial sums which occur -- unless a 0 precedes all the +1s and -1s. So the probability of placing the 0s such that a "good" arrangement is still possible is ${3k \choose k}/{3k+1 \choose k}$. There is exactly one way to place the +1s and -1s once we've done this that actually gives a "good" arrangement. So the final probability is
f(k) = {{3k \choose k} \over {3k+1 \choose k}} {1 \over {2k+1 \choose k}} = {k!(k+1)! \over (2k)!(3k+1)}.

For example, f(1) = 1/4 = 6/24; this corresponds to the six "good" arrangements of {1, 2, 3, 4}, namely 1342, 1432, 1423, 4312, 4132, 4123. The function f(k) is asymptotic to $\sqrt{\pi k}/(3 \cdot 4^k)$ from Stirling's approximation. I like to interpret the 4 there as meaning that if we are building up such an arrangement of +1s, -1s, and 0s, we have a probability of roughly 1/41/3 of making an acceptable choice at each step; the "1/3" comes from the fact that we actually make 3k choices in building up such an arrangement. This constant is about 0.63, and seems about right; it should be around two-thirds, since at each step except the first, exactly two of {0, +1, -1} are valid ways to extend the sequence.

The simplicity of the problem seems to hinge on the fact that the partial sums have to fluctuate within a very narrow band, and only move by ± 1. If we instead required that no partial sums were divisible by, say, 4, so the partial sums could move around in {1, 2, 3}, this would be trickier. But I think the best way to work that problem would still be to work modulo 3, so each summand is either +1, -1, or 0; the crucial observation I made early on was that if the partial sum was -1 at some point and +1 later it had to go through zero, but you don't get this sort of "continuity" (the observation is a discrete analogue of the Intermediate Value Theorem) in any modulus other than 3.

(This is basically the same as the solution given at the UNL Putnam archive, but I came up with it independently.)

12 December 2007

Teaching courses for non-technical students

I just learned that I have been assigned to TA a course called "Ideas in Mathematics" (which, if I understand correctly, meets some sort of distribution requirement for the college as a whole, so will mostly be non-technically-oriented people) in the spring. I'm looking forward to this -- it'll be a nice change of pace from my usual teaching assignments (calculus) and I have all sorts of half-formed ideas about how mathematics should be presented in a non-technical setting that I'd love to be able to test out.

But I've never taught a course which wasn't calculus. So I ask you, my readers -- what's different about teaching these two kinds of courses?

11 December 2007

Why don't sports teams use randomization?

Why don't sports teams use randomization?, a guest post by Ian Ayres at Freakonomics. It's well-known that the best strategy in a lot of game-theoretic situations is to choose at random what you'll do at each encounter, where the appropriate probabilities can be calculated. (A sports example would be a pitcher choosing what sort of pitch to make, and a batter deciding whether to swing or not.) So why doesn't anybody (so far as we know) use a random number generator to call pitches?

Although this doesn't seem to be brought up in the comments there (at least not yet), I suspect that the reason for this is that sports people are for the most part resistant to change. But more importantly, they have to answer to the media. And when your closer throws a pitch down the middle of the plate and it gets hit out of the park, and you lose, do you really want to explain to the news people that a random number generator told you to do that? There is no other line of work I can think of where it would be quite so obvious which decision led to the bad outcome; more importantly, there is no other line of work with call-in radio shows devoted to dissecting everything that happens.

10 December 2007

6 ways of looking at an arithmetic sequence

Lately I've been reading Ramsey Theory on the Integers by Bruce M. Landman and Aaron Robertson.

The second chapter of the book is devoted to Van der Waerden's theorem, which states that given positive integers k and r, there is some positive integer N such that if integers {1, ..., N} are colored in r colors, there will exist a k-term arithmetic progression in which all the elements are the same color. We let w(k,r) be the smallest such N. For example, w(2,3) = 9. There is a way to color {1, ..., 8} so that there is no three-term monochromatic arithmetic progression, namely

1 2 3 4 5 6 7 8

That's 1, 4, 5, and 8 red, and 2, 3, 6, and 7 blue.) But there's no way to color {1, ..., 9} so that there's no three-term arithmetic progression; the proof of this is basically by trial and error.

A couple months ago, Mark Dominus wrote about some programs that he wrote a while back to try to search for "good" colorings; it's interesting from a programming point of view. As far as I know, there is no fast way to do this; as a result, very few values of w(k,r) are known. (w(2,r) = r+1 trivially; according to Landman and Robertson, the only other known values are w(3,2) = 9, w(3,3) = 27, w(4,2) = 35, w(3,4) = 76, w(5,2) = 178.)

It turns out that we can generalize this question by replacing "arithmetic progressions" with various other families of sets. I was surprised to see how many generalizations of arithmetic progressed there are:

  • Quasi-progressions: these are sequences {x1, ..., xk} where xi - xi-1 is between d and d+n, for some small n. (Obviously any increasing sequence is a quasi-progression for large enough n.

  • Generalized quasi-progressions are like quasi-progressions, but n above is replaced by some function δ(i).

  • Descending waves, which satisfy xi - xi-1 ≤ xi-1 - xi-2.

  • Semi-progressions, which are subsets of arithmetic progressions.

  • Iterated polynomial sequences: sequences of the form {x1, ..., xk} where xi+1 = p(xi) for some polynomial p and the xi form an increasing sequence. The case of arithmetic progressions is the case f(x) = x+d.

  • Solutions to a linear recurrence relation: sequences where xk = c1xk-1 + ... + cnxk-n, where we specify x1 ... xn to get the recurrence started. Arithmetic progressions are just the case xk = 2xk-1 - xk-2.

One conspicuously absent generalization of arithmetic progressions are sequences in which the gaps xi - xi-1 are chosen from some set which isn't an interval -- for example, sequences in which xi - xi-1 is always 2 or 5. (I chose those numbers at random.) I'm not an expert in this subject, so I don't know if this is because such results don't exist, because they turn out to be uninteresting, or because the authors wanted to keep the book to a reasonable length.

Still, who would have thought there were this many ways of thinking of such a simple object as an arithmetic progression?

09 December 2007

N ways to solve a differential equation

Mark Dominus writes about how to solve the differential equation (f(x))2 + (f'(x))2 = 1, which arises in the analysis of LC-circuits.
Go ahead, try it. There are at least four distinct solutions given there (bearing in mind the usual caveats about distinctness of solutions); I won't give any of them. If you want to know how I solved it, read Dominus' post, which ends with:
I would like to add a pithy and insightful conclusion to this article, but I've been working on it for more than a week now, and also it's almost lunch time, so I think I'll have to settle for observing that sometimes there are a lot of ways to solve math problems.
That's something definitely worth remembering.

As to why I ended up using a power series approach, which you'll see explained in the link: I'd been teaching the solution of differential equations by power series at the time, so it seemed natural to do that rather than to look for some kind of trick.

I'm 4!

Today I turn 24.

Now, at some point a couple weeks ago I realized that 24 is 4 factorial. I will probably never be factorial-aged again; depending on who you believe, either one or two people in history have lived to be 120. (However, both of these people are fairly recent; bear in mind that the records that would allow one to verify this sort of thing didn't really start existing until sometime in the 19th century. Furthermore, lifespans are getting longer. So I'd be willing to bet that in my lifetime, more than two people will live to 120. However, I wouldn't bet on that person being me.)

Furthermore, 23 is a prime, and 25 is a square. Are there any numbers n such that n!-1 is prime and n!+1 is square?

Squares are rarer than primes (the density of squares around n is 1/(2√n), the density of primes around n is 1/(log n)), so it makes sense to start with squares. I found that 4!+1, 5!+1, 7!+1 were squares and then I stopped being able to do it in my head. That's enough to look it up in the Encyclopedia of Integer Sequences; where the numbers 25, 121, 5041 form an entire sequence; there are no more solutions to n!+1 = k2 with n < 4 × 109. Wikipedia tells me that this is called Brocard's problem, and that the presence of a finite number of solutions would follow from the abc conjecture.

Primes of the form n!-1 are a lot more common. 4!-1 = 23 is prime; 5! - 1 = 119 isn't; 7! - 1 = 5039 is. The sequence of integers n such that n!-1 is prime is sparse, but not sparse enough that I'd be willing to conjecture it's finite. And in fact, heuristically, the density of primes around n! is 1/(log n!), but log n! is around n log n, so the density of primes around n! is 1/(n log n); the sum
\sum_{n=a}^\infty {1 \over n \log n}

diverges by comparison with the integral
\int_a^\infty {1 \over x \log x} = \lim_{b \to \infty} (\log \log x) - (\log \log b)

and so I conjecture that the sequence is infinite. (Furthermore, this heuristic probably underestimates the number of primes in the sequence, because n! - 1 is automatically not divisible by any prime less than or equal to n.)

To answer the original question: there are almost certainly two solutions, (23, 24, 25) and (5039, 5040, 5041).

08 December 2007

SAGE: open-source mathematics software

"Open Source 'Sage' Takes Aim at High End Math Software", from Slashdot. (The actual site, sagemath.org, is Slashdotted right now.) Here's the article that Slashdot links to (by the way, the links in that article go to malformed URLs, but they're not hard to fix). The leader of the Sage project is William Stein, who with David Joyner wrote an opinion piece in November's Notices stating that mathematical software should be open source. From what I can tell, SAGE includes both new code written specifically for the project and a lot of old code that's been previously used (it appears to incorporate Maxima, GAP, etc.)

I agree with this; although it may very rarely be practical to check all the work that goes into establishing a mathematical result, it should always be at least theoretically possible. It seems acceptable for the internal workings of mathematical software to be closed when it's just students that are using it, or when the serious researchers are just using it to do experiments. But as we start seeing more and more proofs which reduce to "so I reduced the problem to some finite but very long computation, and then I ran this program, and it said my theorem is true!", it becomes more important that someone can actually verify that the program does what the author claims it does. For comparison, see Wolfram on why you don't need to know what's going on inside Mathematica, which is mildly amusing.

I haven't actually used SAGE, but it certainly seems more in line with the open-information ethos of the mathematical community than the proprietary software that's out there. It's a shame I know Maple pretty well, because that creates a disincentive for me to switch if I wanted to.

07 December 2007

Two grading-related thoughts

In the class I TA, I drop the lowest homework grade (of eleven) for each student. But what does this mean? The meaning of "dropping the lowest homework" is clear enough if all homeworks count the same. But there was one homework assignment that was particularly long, and one that was particularly short, so I counted those for 30 and 10 points, respectively, while all the other homeworks were out of 20.

The first solution is to drop the homework for each student with the lowest numerical grade -- but that's just stupid, because for most students that actually did all the homeworks, that would be the single homework assignment that was graded out of 10. I rejected that without even bothering to do the code.

The next solution seemed to be to drop the homework with the lowest percentage grade -- so, for example, if a student had 23/30 on the single 30-point homework and all their other grades were 16/20 or 8/10 or above, then I dropped the 23/30. But isn't it possible that that algorithm might not maximize the score? Indeed, paradoxes like that are possible, as this Mathtrek article and this paper by Daniel Kane show. They remind me of Simpson's paradox, although I suspect there's no deep connection here.

The eventual solution I settled on was to drop the homework for each student that gave them the highest possible percentage; this wasn't hard to code, since I was dropping a single homework, but I wouldn't want to go to that trouble if I were dropping, say, the lowest two homeworks for each student.

Of course, I should have just made all the homeworks count the same amount. And I was even aware of this particular paradox! (The Mathtrek article, as you can see, is dated June 2006; I don't know if I read it then but I certainly had read it before the start of the current semester.

I also played around with the idea of dropping the lowest 20 points worth of homework, which would mean:

  • if the homework with the lowest percentage was a 20-point one, dropping that one;

  • if the homework with the lowest percentage was the 30-point one, rescaling it for that student to be out of 10 points;

  • if the homework with the lowest percentage was the 10-point one, dropping that one, and reducing the weight of the homework with the second lowest percentage grade by ten points.

(This was inspired by a policy that at least one of my undergrad classes had, where the final counted twice as much as each exam, and the lowest exam was dropped, except if the lowest exam was the final, in which case the final was reduced in weight to have the same weight as a midterm.) But I wasn't sure what properties that has, and technically it is not what I said I would do. Of course the algorithm I choose isn't going to change any grades by much, and dropping the lowest homework is really something to take a bit of pressure off the students, but I wanted to stick to my word.

Furthermore, the method I chose has the nice property that no student can come to me and say "my grade would have been better if you'd chosen a different homework to drop".

Also, it's occurred to me that what we tell our students about grades is a lie. In the particular class I'm TAing, we tell the students that the homework counts 20%, the two midterm exams count 25% each, and the final counts 30%. But this ends up not quite being true, because homework grades are less narrowly spread than exam grades! (The mean overall homework grade was 84%, the standard deviation 12%; the mean grade on each exam was in the low sixties with a standard deviation of 16%.) This has actually generated complaints for me from students in the past, who assumed that we were curving each section of the course separately and were disappointed to learn that we curved the total raw score. (These were students who had relatively better homework grades than exam grades, and thus ended up being hurt by this policy.) I'm a probabilist, though, not a statistician, so I won't elaborate on this.

06 December 2007

A cute number-theoretic result

Here's a cute result that I saw a few days ago: let α and β be irrational positive numbers, such that 1/α + 1/β = 1. Then for any positive integer n, exactly one of $n = \lfloor k\alpha \rfloor$ and $n = \lfloor k\beta \rfloor$ has a positive integer solution k, which is unique. That is, the two sequences


contain, between them, each positive integer exactly once. In the case where $\alpha = \sqrt{2}$ and $\beta = 2 + \sqrt{2}$, for example, these sequences are
1, 2, 4, 5, 7, 8, 9, 11, 12, 14, 15, ...
3, 6, 10, 13, 17, 20, 23, 27, 30, 34, 37, ...
which partition the positive integers.

The proof is quite nice. It's enough to show that the number of integers of the form $\lfloor k\alpha \rfloor$ plus the number of integers of the form $\lfloor k\beta \rfloor$ which are less than n is equal to n-1. The number of integers of the form $\lfloor k\alpha \rfloor$ which are less than n is $\lfloor n/\alpha \rfloor$; the number of integers of the form $\lfloor k\beta \rfloor$ which are less than n is $\lfloor n/\beta \rfloor$. Now, we have the inequalities

n/\alpha - 1 < \lfloor n/\alpha \rfloor < n/\alpha

n/\beta - 1 < \lfloor n/\beta \rfloor < n/\beta

(the inequalities are strict since n/α and n/β are irrational.) Adding these together,
n/\alpha + n/\beta - 2 < \lfloor n/\alpha \rfloor + \lfloor n/\beta \rfloor < n/\alpha + n/\beta

and recalling that 1/α + 1/β = 1, we have

and the middle member of the equation must be n-1; that's what we wanted. So the two sequences $\{ \lfloor k\alpha \rfloor \}_{k=1}^{\infty}, \{ \lfloor k\beta \rfloor \}_{k=1}^{\infty}$ contain between them n-1 members which are less than n, and n members which are less
than n+1; thus exactly one of them contains n, for any positive integer n.

The analogous theorem for three sequences of integers, where 1/α + 1/β + 1/γ = 1, doesn't hold; a weaker theorem holds that on average an integer n can be written in exactly one of the forms
\lfloor k\alpha \rfloor, \lfloor k\beta \rfloor, \lfloor k\gamma \rfloor
, though, and actually no integer can be written in all three of these forms; I leave this as an exercise for the reader. What I don't know is whether there is some set α, β, γ for which the original result is salvaged.

05 December 2007

Links for 2007-12-05

04 December 2007

How much land is in the tropics?







If you live in the U. S. and pay any attention to the news, you know that there's a mess going on in the mortgage markets, as companies which have lent to "subprime" borrowers are having trouble. And because of the way that the markets work, this has had effects spread far and wide.

But doesn't "subprime" sound like it should be a mathematical term? It seems like it would either mean:

  • numbers which are just less than a prime (let's say numbers of the form p-1, where p is a prime)

  • numbers which are "almost prime" (say, products of two primes)

But what would a "superprime" be? In the first case, the analogous definition is clearly numbers of the form p+1; in the second case, the only extension I can think of is numbers which are products of 0 primes, i. e. the number 1.

03 December 2007

Best name ever for a talk

"Pretentiousness in analytic number theory", from next week's Mini-conference in arithmetic combinatorics at the Institute for Advanced Study.

(from Secret Blogging Seminar. Incidentally, I was surprised to learn this was at IAS, because I associate Secret Blogging Seminar with Berkeley, and thus with MSRI.)

If I didn't have a final to proctor and grade that day, I'd consider making the trip up to Princeton just because you've got to figure that anybody who can give a talk a title like that has a sense of humor. Plus, I've been reading Tao and Vu's Additive Combinatorics in my spare time and I've been finding it interesting. I'll probably have more substantial things to say about Tao and Vu -- and indeed about a lot of things -- once classes are over here. Hopefully Ben Webster, who made the SBS post and is currently at IAS, will let us know.

02 December 2007

Redistricting maps

I've previously mentioned the shortest splitline algorithm for determining congressional districts in the United States. (This could work in other districting situations, as well.) The algorithm breaks states into districts by breaking them up along the shortest possible lines; for a more thorough description see this.

Well, today I got an e-mail from the good people at rangevoting.org saying that they now had computer-generated maps of their algorithm's redistrictings for all fifty states. These, I assume, supercede their approximate sketches

I'm not entirely sure how good an idea this particular redistricting algorithm is. Basically, assuming that straight lines are the right way to break things up seems to imply that all directions should be treated equally, when actual settlement patterns aren't isotropic. But the beauty of any algorithm which doesn't include any "tunable" parameters -- of which this is an example -- is that there is zero possibility of gerrymandering. If we imagine an algorithm that takes into account "travel time" between points, for example, instead of as-the-crow-flies distance, then how do you define travel time? And next thing you know, you'll see new roads getting built because of how you'd expect them to change congressional districting. As-the-crow-flies distance along the surface of the earth doesn't have these issues.

Not surprisingly, the people at rangevoting.org also support something called range voting, which would basically allow people to give scores to candidates in an election, and the candidate with the highest scores would win. I haven't much thought about it, but it seems like a good idea. And here's their page for mathematicians!

Recitation instructors, TV pundits, and Poincare

I've been reading through the archives of Eliezer Yudkoswsky's Overcoming Bias, and I found an interesting posts:

Focus Your Uncertainty talks about the plight of a novice TV pundit who has to get what the market will do in advance, in order to decide how ey will allocate time to preparing remarks, so that ey can explain what happened after the fact.

I face much the same problem in teaching recitations for calculus classes. My recitations are very much driven by the students, in that I spend most of the time answering their questions. Which questions do I prepare for ahead of time, and which do I just decide I'll figure out on the fly? Here there is another wrinkle -- there are some questions that students aren't that likely to ask about, but which will take a long time to prepare.

In the case of the TV pundit, I'd almost say that having a long explanation for a certain action in the market is in itself evidence for that action being unlikely, basically by Occam's razor. This doesn't carry over to the calculus classes -- the things having long explanations (that is, the hard problems) are exactly the ones the students are most likely to ask about.

Yudkowsky writes, in that post:
Alas, no one can possibly foresee the future. What are you to do? You certainly can't use "probabilities". We all know from school that "probabilities" are little numbers that appear next to a word problem, and there aren't any little numbers here. Worse, you feel uncertain. You don't remember feeling uncertain while you were manipulating the little numbers in word problems. College classes teaching math are nice clean places, therefore math itself can't apply to life situations that aren't nice and clean. You wouldn't want to inappropriately transfer thinking skills from one context to another. Clearly, this is not a matter for "probabilities".

That's something to keep in mind -- in working with probability one doesn't feel uncertain. Sometimes I feel the same way, and this may be because we've thoroughly axiomatized away the uncertainty. I was recently reading Poincare's book Science and Hypothesis(*), which includes a chapter on "the calculus of probabilities" -- a lot more uncertainty seems to permeate this chapter than would a similar chapter written now, because Poincare lived before the Kolmogorov axioms. But this is an interesting philosophical fact about probability -- we are saying things about uncertainty, but we know that they're true. And sometimes, as with the "probabilistic method" in combinatorics, this allows us to prove things about structures that don't involve uncertainty. I leave this to the philosophers.

(*) I was proud of myself for picking this up at a used bookstore for $6.95 -- but Amazon's selling it for $6.69! (Of course, I ought to not be too proud, since Penn's library almost certainly has it.)

01 December 2007

Boiling eggs and dimensional analysis




  • mass (m) has units of kg

  • density (ρ) has units of kg m-3






