The Universe of Discourse

Recommended reading: Matt Levine’s Money Stuff

Lately my favorite read has been Matt Levine’s Money Stuff articles from Bloomberg News. Bloomberg's web site requires a subscription but you can also get the Money Stuff articles as an occasional email. It arrives at most once per day.

Almost every issue teaches me something interesting I didn't know, and almost every issue makes me laugh.

Example of something interesting: a while back it was all over the news that oil prices were negative. Levine was there to explain what was really going on and why. Some people manage index funds. They are not trying to beat the market, they are trying to match the index. So they buy derivatives that give them the right to buy oil futures contracts at whatever the day's closing price is. But say they already own a bunch of oil contracts. If they can get the close-of-day price to dip, then their buy-at-the-end-of-the-day contracts will all be worth more because the counterparties have contracted to buy at the dip price. How can you get the price to dip by the end of the day? Easy, unload 20% of your contracts at a bizarre low price, to make the value of the other 80% spike… it makes my head swim.

But there are weird second- and third-order effects too. Normally if you invest fifty million dollars in oil futures speculation, there is a worst-case: the price of oil goes to zero and you lose your fifty million dollars. But for these derivative futures, the price could in theory become negative, and for short time in April, it did:

If the ETF’s oil futures go to -$37.63 a barrel, as some futures did recently, the ETF investors lose $20—their entire investment—and, uh, oops? The ETF runs out of money when the futures hit zero; someone else has to come up with the other $37.63 per barrel.

One article I particularly remember discussed the kerfuffle a while back concerning whether Kelly Loeffler improperly traded stocks on classified coronavirus-related intelligence that she received in her capacity as a U.S. senator. I found Levine's argument persuasive:

“I didn’t dump stocks, I am a well-advised rich person, someone else manages my stocks, and they dumped stocks without any input from me” … is a good defense! It’s not insider trading if you don’t trade; if your investment manager sold your stocks without input from you then you’re fine. Of course they could be lying, but in context the defense seems pretty plausible. (Kelly Loeffler, for instance, controversially dumped about 0.6% of her portfolio at around the same time, which sure seems like the sort of thing an investment adviser would do without any input from her? You could call your adviser and say “a disaster is coming, sell everything!,” but calling them to say “a disaster is coming, sell a tiny bit!” seems pointless.)

He contrasted this case with that of Richard Burr, who, unlike Loeffler, remains under investigation. The discussion was factual and informative, unlike what you would get from, say, Twitter, or even Metafilter, where the response was mostly limited to variations on “string them up” and “eat the rich”.

Money Stuff is also very funny. Today’s letter discusses a disclosure filed recently by Nikola Corporation:

More impressive is that Nikola’s revenue for the second quarter was very small, just $36,000. Most impressive, though, is how they earned that revenue:

During the three months ended June 30, 2020 and 2019 the Company recorded solar revenues of $0.03 million and $0.04 million, respectively, for the provision of solar installation services to the Executive Chairman, which are billed on time and materials basis. …

“Solar installation projects are not related to our primary operations and are expected to be discontinued,” says Nikola, but I guess they are doing one last job, specifically installing solar panels at founder and executive chairman Trevor Milton’s house? It is a $13 billion company whose only business so far is doing odd jobs around its founder’s house.

A couple of recent articles that cracked me up discussed clueless day-traders pushing up the price of Hertz stock after Hertz had declared bankruptcy, and how Hertz diffidently attempted to get the SEC to approve a new stock issue to cater to these idiots. (The SEC said no.)

One recurring theme in the newsletter is “Everything is Securities Fraud”. This week, Levine asks:

Is it securities fraud for a public company to pay bribes to public officials in exchange for lucrative public benefits?

Of course you'd expect that the executives would be criminally charged, as they have been. But is there a cause for the company’s shareholders to sue? If you follow the newsletter, you know what the answer will be:

Oh absolutely…

because Everything is Securities Fraud.

Still it is a little weird. Paying bribes to get public benefits is, you might think, the sort of activity that benefits shareholders. Sure they were deceived, and sure the stock price was too high because investors thought the company’s good performance was more legitimate and sustainable than it was, etc., but the shareholders are strange victims. In effect, executives broke the law in order to steal money for the shareholders, and when the shareholders found out they sued? It seems a little ungrateful?

I recommend it.

• Matt Levine on Bloomberg.com
• Free email newsletter signup

Levine also has a Twitter account but it is mostly just links to his newsletter articles.

[Other articles in category /ref] permanent link

A maybe-interesting number trick?

I'm not sure if this is interesting, trivial, or both. You decide.

Let's divide the numbers from 1 to 30 into the following six groups:

Choose any two rows. Chose a number from each row, and multiply them mod 31. (That is, multiply them, and if the product is 31 or larger, divide it by 31 and keep the remainder.)

Regardless of which two numbers you chose, the result will always be in the same row. For example, any two numbers chosen from rows B and D will multiply to yield a number in row E. If both numbers are chosen from row F, their product will always appear in row A.

[Other articles in category /math] permanent link

Flappers

Gulliver's Travels (1726), Part III, chapter 2:

I observed, here and there, many in the habit of servants, with a blown bladder, fastened like a flail to the end of a stick, which they carried in their hands. In each bladder was a small quantity of dried peas, or little pebbles, as I was afterwards informed. With these bladders, they now and then flapped the mouths and ears of those who stood near them, of which practice I could not then conceive the meaning. It seems the minds of these people are so taken up with intense speculations, that they neither can speak, nor attend to the discourses of others, without being roused by some external action upon the organs of speech and hearing… . This flapper is likewise employed diligently to attend his master in his walks, and upon occasion to give him a soft flap on his eyes; because he is always so wrapped up in cogitation, that he is in manifest danger of falling down every precipice, and bouncing his head against every post; and in the streets, of justling others, or being justled himself into the kennel.

When I first told Katara about this, several years ago, instead of “the minds of these people are so taken up with intense speculations” I said they were obsessed with their phones.

Now the phones themselves have become the flappers:

Y. Tung and K. G. Shin, "Use of Phone Sensors to Enhance Distracted Pedestrians’ Safety," in IEEE Transactions on Mobile Computing, vol. 17, no. 6, pp. 1469–1482, 1 June 2018, doi: 10.1109/TMC.2017.2764909.

Our minds are not even taken up with intense speculations, but with Instagram. Dean Swift would no doubt be disgusted.

[Other articles in category /book] permanent link

How are finite fields constructed?

Here's another recent Math Stack Exchange answer I'm pleased with.

OP asked:

I know this question has been asked many times and there is good information out there which has clarified a lot for me but I still do not understand how the addition and multiplication tables for !!GF(4)!! is constructed?

I've seen [links] but none explicity explain the construction and I'm too new to be told "its an extension of !!GF(2)!!"

The only “reasonable” answer here is “get an undergraduate abstract algebra text and read the chapter on finite fields”. Because come on, you can't expect some random stranger to appear and write up a detailed but short explanation at your exact level of knowledge.

But sometimes Internet Magic Lightning strikes  and that's what you do get! And OP set themselves up to be struck by magic lightning, because you can't get a detailed but short explanation at your exact level of knowledge if you don't provide a detailed but short explanation of your exact level of knowledge — and this person did just that. They understand finite fields of prime order, but not how to construct the extension fields. No problem, I can explain that!

I had special fun writing this answer because I just love constructing extensions of finite fields. (Previously: [1] [2])

For any given !!n!!, there is at most one field with !!n!! elements: only one, if !!n!! is a power of a prime number (!!2, 3, 2^2, 5, 7, 2^3, 3^2, 11, 13, \ldots!!) and none otherwise (!!6, 10, 12, 14\ldots!!). This field with !!n!! elements is written as !!\Bbb F_n!! or as !!GF(n)!!.

Suppose we want to construct !!\Bbb F_n!! where !!n=p^k!!. When !!k=1!!, this is easy-peasy: take the !!n!! elements to be the integers !!0, 1, 2\ldots p-1!!, and the addition and multiplication are done modulo !!n!!.

When !!k>1!! it is more interesting. One possible construction goes like this:

1. The elements of !!\Bbb F_{p^k}!! are the polynomials $$a_{k-1}x^{k-1} + a_{k-2}x^{k-2} + \ldots + a_1x+a_0$$ where the coefficients !!a_i!! are elements of !!\Bbb F_p!!. That is, the coefficients are just integers in !!{0, 1, \ldots p-1}!!, but with the understanding that the addition and multiplication will be done modulo !!p!!. Note that there are !!p^k!! of these polynomials in total.

2. Addition of polynomials is done exactly as usual: combine like terms, but remember that the the coefficients are added modulo !!p!! because they are elements of !!\Bbb F_p!!.

3. Multiplication is more interesting:

   a. Pick an irreducible polynomial !!P!! of degree !!k!!. “Irreducible” means that it does not factor into a product of smaller polynomials. How to actually locate an irreducible polynomial is an interesting question; here we will mostly ignore it.

   b. To multiply two elements, multiply them normally, remembering that the coefficients are in !!\Bbb F_p!!. Divide the product by !!P!! and keep the remainder. Since !!P!! has degree !!k!!, the remainder must have degree at most !!k-1!!, and this is your answer.

Now we will see an example: we will construct !!\Bbb F_{2^2}!!. Here !!k=2!! and !!p=2!!. The elements will be polynomials of degree at most 1, with coefficients in !!\Bbb F_2!!. There are four elements: !!0x+0, 0x+1, 1x+0, !! and !!1x+1!!. As usual we will write these as !!0, 1, x, x+1!!. This will not be misleading.

Addition is straightforward: combine like terms, remembering that !!1+1=0!! because the coefficients are in !!\Bbb F_2!!:

$$\begin{array}{c|cccc} + & 0 & 1 & x & x+1 \\ \hline 0 & 0 & 1 & x & x+1 \\ 1 & 1 & 0 & x+1 & x \\ x & x & x+1 & 0 & 1 \\ x+1 & x+1 & x & 1 & 0 \end{array} $$

The multiplication as always is more interesting. We need to find an irreducible polynomial !!P!!. It so happens that !!P=x^2+x+1!! is the only one that works. (If you didn't know this, you could find out easily: a reducible polynomial of degree 2 factors into two linear factors. So the reducible polynomials are !!x^2, x·(x+1) = x^2+x!!, and !!(x+1)^2 = x^2+2x+1 = x^2+1!!. That leaves only !!x^2+x+1!!.)

To multiply two polynomials, we multiply them normally, then divide by !!x^2+x+1!! and keep the remainder. For example, what is !!(x+1)(x+1)!!? It's !!x^2+2x+1 = x^2 + 1!!. There is a theorem from elementary algebra (the “division theorem”) that we can find a unique quotient !!Q!! and remainder !!R!!, with the degree of !!R!! less than 2, such that !!PQ+R = x^2+1!!. In this case, !!Q=1, R=x!! works. (You should check this.) Since !!R=x!! this is our answer: !!(x+1)(x+1) = x!!.

Let's try !!x·x = x^2!!. We want !!PQ+R = x^2!!, and it happens that !!Q=1, R=x+1!! works. So !!x·x = x+1!!.

I strongly recommend that you calculate the multiplication table yourself. But here it is if you want to check:

$$\begin{array}{c|cccc} · & 0 & 1 & x & x+1 \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & x & x+1 \\ x & 0 & x & x+1 & 1 \\ x+1 & 0 & x+1 & 1 & x \end{array} $$

To calculate the unique field !!\Bbb F_{2^3}!! of order 8, you let the elements be the 8 second-degree polynomials !!0, 1, x, \ldots, x^2+x, x^2+x+1!! and instead of reducing by !!x^2+x+1!!, you reduce by !!x^3+x+1!!. (Not by !!x^3+x^2+x+1!!, because that factors as !!(x^2+1)(x+1)!!.) To calculate the unique field !!\Bbb F_{3^2}!! of order 27, you start with the 27 third-degree polynomials with coefficients in !!{0,1,2}!!, and you reduce by !!x^3+2x+1!! (I think).

The special notation !!\Bbb F_p[x]!! means the ring of all polynomials with coefficients from !!\Bbb F_p!!. !!\langle P \rangle!! means the ring of all multiples of polynomial !!P!!. (A ring is a set with an addition, subtraction, and multiplication defined.)

When we write !!\Bbb F_p[x] / \langle P\rangle!! we are constructing a thing called a “quotient” structure. This is a generalization of the process that turns the ordinary integers !!\Bbb Z!! into the modular-arithmetic integers we have been calling !!\Bbb F_p!!. To construct !!\Bbb F_p!!, we start with !!\Bbb Z!! and then agree that two elements of !!\Bbb Z!! will be considered equivalent if they differ by a multiple of !!p!!.

To get !!\Bbb F_p[x] / \langle P \rangle!! we start with !!\Bbb F_p[x]!!, and then agree that elements of !!\Bbb F_p[x]!! will be considered equivalent if they differ by a multiple of !!P!!. The division theorem guarantees that of all the equivalent polynomials in a class, exactly one of them will have degree less than that of !!P!!, and that is the one we choose as a representative of its class and write into the multiplication table. This is what we are doing when we “divide by !!P!! and keep the remainder”.

A particularly important example of this construction is !!\Bbb R[x] / \langle x^2 + 1\rangle!!. That is, we take the set of polynomials with real coefficients, but we consider two polynomials equivalent if they differ by a multiple of !!x^2 + 1!!. By the division theorem, each polynomial is then equivalent to some first-degree polynomial !!ax+b!!.

Let's multiply $$(ax+b)(cx+d).$$ As usual we obtain $$acx^2 + (ad+bc)x + bd.$$ From this we can subtract !!ac(x^2 + 1)!! to obtain the equivalent first-degree polynomial $$(ad+bc) x + (bd-ac).$$

Now recall that in the complex numbers, !!(b+ai)(d + ci) = (bd-ac) + (ad+bc)i!!. We have just constructed the complex numbers,with the polynomial !!x!! playing the role of !!i!!.

[ Note to self: maybe write a separate article about what makes this a good answer, and how it is structured. ]

[Other articles in category /math/se] permanent link

What does it mean to expand a function “in powers of x-1”?

A recent Math Stack Excahnge post was asked to expand the function !!e^{2x}!! in powers of !!(x-1)!! and was confused about what that meant, and what the point of it was. I wrote an answer I liked, which I am reproducing here.

You asked:

I don't understand what are we doing in this whole process

which is a fair question. I didn't understand this either when I first learned it. But it's important for practical engineering reasons as well as for theoretical mathematical ones.

Before we go on, let's see that your proposal is the wrong answer to this question, because it is the correct answer, but to a different question. You suggested: $$e^{2x}\approx1+2\left(x-1\right)+2\left(x-1\right)^2+\frac{4}{3}\left(x-1\right)^3$$

Taking !!x=1!! we get !!e^2 \approx 1!!, which is just wrong, since actually !!e^2\approx 7.39!!. As a comment pointed out, the series you have above is for !!e^{2(x-1)}!!. But we wanted a series that adds up to !!e^{2x}!!.

As you know, the Maclaurin series works here:

$$e^{2x} \approx 1+2x+2x^2+\frac{4}{3}x^3$$

so why don't we just use it? Let's try !!x=1!!. We get $$e^2\approx 1 + 2 + 2 + \frac43$$

This adds to !!6+\frac13!!, but the correct answer is actually around !!7.39!! as we saw before. That is not a very accurate approximation. Maybe we need more terms? Let's try ten:

$$e^{2x} \approx 1+2x+2x^2+\frac{4}{3}x^3 + \ldots + \frac{8}{2835}x^9$$

If we do this we get !!7.3887!!, which isn't too far off. But it was a lot of work! And we find that as !!x!! gets farther away from zero, the series above gets less and less accurate. For example, take !!x=3.1!!, the formula with four terms gives us !!66.14!!, which is dead wrong. Even if we use ten terms, we get !!444.3!!, which is still way off. The right answer is actually !!492.7!!.

What do we do about this? Just add more terms? That could be a lot of work and it might not get us where we need to go. (Some Maclaurin series just stop working at all too far from zero, and no amount of terms will make them work.) Instead we use a different technique.

Expanding the Taylor series “around !!x=a!!” gets us a different series, one that works best when !!x!! is close to !!a!! instead of when !!x!! is close to zero. Your homework is to expand it around !!x=1!!, and I don't want to give away the answer, so I'll do a different example. We'll expand !!e^{2x}!! around !!x=3!!. The general formula is $$e^{2x} \approx \sum \frac{f^{(i)}(3)}{i!} (x-3)^i\tag{$\star$}\ \qquad \text{(when $x$ is close to $3$)}$$

The !!f^{(i)}(x)!! is the !!i!!'th derivative of !! e^{2x}!! , which is !!2^ie^{2x}!!, so the first few terms of the series above are:

$$\begin{eqnarray} e^{2x} & \approx& e^6 + \frac{2e^6}1 (x-3) + \frac{4e^6}{2}(x-3)^2 + \frac{8e^6}{6}(x-3)^3\\ & = & e^6\left(1+ 2(x-3) + 2(x-3)^2 + \frac34(x-3)^3\right)\\ & & \qquad \text{(when $x$ is close to $3$)} \end{eqnarray} $$

The first thing to notice here is that when !!x!! is exactly !!3!!, this series is perfectly correct; we get !!e^6 = e^6!! exactly, even when we add up only the first term, and ignore the rest. That's a kind of useless answer because we already knew that !!e^6 = e^6!!. But that's not what this series is for. The whole point of this series is to tell us how different !!e^{2x}!! is from !!e^6!! when !!x!! is close to, but not equal to !!3!!.

Let's see what it does at !!x=3.1!!. With only four terms we get $$\begin{eqnarray} e^{6.2} & \approx& e^6(1 + 2(0.1) + 2(0.1)^2 + \frac34(0.1)^3)\\ & = & e^6 \cdot 1.22075 \\ & \approx & 492.486 \end{eqnarray}$$

which is very close to the correct answer, which is !!492.7!!. And that's with only four terms. Even if we didn't know an exact value for !!e^6!!, we could find out that !!e^{6.2}!! is about !!22.075\%!! larger, with hardly any calculation.

Why did this work so well? If you look at the expression !!(\star)!! you can see: The terms of the series all have factors of the form !!(x-3)^i!!. When !!x=3.1!!, these are !!(0.1)^i!!, which becomes very small very quickly as !!i!! increases. Because the later terms of the series are very small, they don't affect the final sum, and if we leave them out, we won't mess up the answer too much. So the series works well, producing accurate results from only a few terms, when !!x!! is close to !!3!!.

But in the Maclaurin series, which is around !!x=0!!, those !!(x-3)^i!! terms are !!x^i!! terms intead, and when !!x=3.1!!, they are not small, they're very large! They get bigger as !!i!! increases, and very quickly. (The !! i! !! in the denominator wins, eventually, but that doesn't happen for many terms.) If we leave out these many large terms, we get the wrong results.

The short answer to your question is:

Maclaurin series are only good for calculating functions when !!x!! is close to !!0!!, and become inaccurate as !!x!! moves away from zero. But a Taylor series around !!a!! has its “center” near !!a!! and is most accurate when !!x!! is close to !!a!!.

[Other articles in category /math/se] permanent link

I screw up buying a marker

Toph left the cap off one of her fancy art markers and it dried out, so I went to get her a replacement. The marker costs $5.85, plus tax, and the web site wanted a $5.95 shipping fee. Disgusted, I resolved to take my business elsewhere.

On Wednesday I drove over to a local art-supply store to get the marker. After taxes the marker was somehow around $8.50, but I also had to pay $1.90 for parking. So if there was a win there, it was a very small one.

But also, I messed up the parking payment app, which has maybe the worst UI of any phone app I've ever used. The result was a $36 parking ticket.

Lesson learned. I hope.

[Other articles in category /oops] permanent link

Zuul crurivastator

Today I learned:

• There is a genus of ankylosaurs named Zuul after “demon and demi-god Zuul, the Gatekeeper of Gozer, featured in the 1984 film Ghostbusters”.

• The type species of Zuul is Zuul crurivastator, which means “Zuul, destroyer of shins”. Wikipedia says:

  The epithet … refers to a presumed defensive tactic of ankylosaurids, smashing the lower legs of attacking predatory theropods with their tail clubs.

  My eight-year-old self is gratified that the ankylosaurids are believed to attack their enemies’ ankles.

• The original specimen of Z. crurivastator, unusually-well preserved, was nicknamed “Sherman”.

Here is a video of Dan Aykroyd discussing the name, with Sherman.

[Other articles in category /bio] permanent link

More trivia about megafauna and poisonous plants

A couple of people expressed disappointment with yesterday's article, which asked were giant ground sloths immune to poison ivy?, but then failed to deliver on the implied promise. I hope today's article will make up for that.

Elephants

I said:

Mangoes are tropical fruit and I haven't been able to find any examples of Pleistocene megafauna that lived in the tropics…

David Formosa points out what should have been obvious: elephants are megafauna, elephants live where mangoes grow (both in Africa and in India), elephants love eating mangoes [1] [2] [3], and, not obvious at all…

Elephants are immune to poison ivy!

Captive elephants have been known to eat poison ivy, not just a little bite, but devouring entire vines, leaves and even digging up the roots. To most people this would have cause a horrific rash … To the elephants, there was no rash and no ill effect at all…

It's sad that we no longer have megatherium. But we do have elephants, which is pretty awesome.

Idiot fruit

The idiot fruit is just another one of those legendarily awful creatures that seem to infest every corner of Australia (see also: box jellyfish, stonefish, gympie gympie, etc.); Wikipedia says:

The seeds are so toxic that most animals cannot eat them without being severely poisoned.

At present the seeds are mostly dispersed by gravity. The plant is believed to be an evolutionary anachronism. What Pleistocene megafauna formerly dispersed the poisonous seeds of the idiot fruit?

A wombat. A six-foot-tall wombat.

I am speechless with delight.

[Other articles in category /bio] permanent link

Were giant ground sloths immune to poison ivy?

The skin of the mango fruit contains urushiol, the same irritating chemical that is found in poison ivy. But why? From the mango's point of view, the whole point of the mango fruit is to get someone to come along and eat it, so that they will leave the seed somewhere else. Posioning the skin seems counterproductive.

An analogous case is the chili pepper, which contains an irritating chemical, capsaicin. I think the answer here is believed to be that while capsaicin irritates mammals, birds are unaffected. The chili's intended target is birds; you can tell from the small seeds, which are the right size to be pooped out by birds. So chilis have a chemical that encourages mammals to leave the fruit in place for birds.

What's the intended target for the mango fruit? Who's going to poop out a seed the size of a mango pit? You'd need a very large animal, large enough to swallow a whole mango. There aren't many of these now, but that's because they became extinct at the end of the Pleistocene epoch: woolly mammoths and rhinoceroses, huge crocodiles, giant ground sloths, and so on. We may have eaten the animals themselves, but we seem to have quite a lot of fruits around that evolved to have their seeds dispersed by Pleistocene megafauna that are now extinct. So my first thought was, maybe the mango is expecting to be gobbled up by a giant gound sloth, and have its giant seed pooped out elsewhere. And perhaps its urushiol-laden skin makes it unpalatable to smaller animals that might not disperse the seeds as widely, but the giant ground sloth is immune. (Similarly, I'm told that goats are immune to urushiol, and devour poison ivy as they do everything else.)

Well, maybe this theory is partly correct, but even if so, the animal definitely wasn't a giant ground sloth, because those lived only in South America, whereas the mango is native to South Asia. Ground slots and avocados, yes; mangos no.

Still the theory seems reasonable, except that mangoes are tropical fruit and I haven't been able to find any examples of Pleistocene megafauna that lived in the tropics. Still I didn't look very hard.

Wikipedia has an article on evolutionary anachronisms that lists a great many plants, but not the mango.

[ Addendum: I've eaten many mangoes but never noticed any irritation from the peel. I speculate that cultivated mangoes are varieties that have been bred to contain little or no urushiol, or that there is a post-harvest process that removes or inactivates the urushiol, or both. ]

[ Addendum 20200715: I know this article was a little disappointing and that it does not resolve the question in the title. Sorry. But I wrote a followup that you might enjoy anyway. ]

[Other articles in category /bio] permanent link

Ron Graham has died

Ron Graham has died. He had a good run. When I check out I will probably not be as accomplished or as missed as Graham, even if I make it to 84.

I met Graham once and he was very nice to me, as he apparently was to everyone. I was planning to write up a reminiscence of the time, but I find I've already done it so you can read that if you care.

Graham's little book Rudiments of Ramsey Theory made a big impression on me when I was an undergraduate. Chapter 1, if I remember correctly, is a large collection of examples, which suited me fine. Chapter 2 begins by introducing a certain notation of Erdős and Rado: !!\left[{\Bbb N\atop k}\right]!! is the family of subsets of !!\Bbb N!! of size !!k!!, and

$$\left[{\Bbb N\atop k}\right] \to \left[{\Bbb N\atop k}\right]_r$$

is an abbreviation of the statement that for any !!r!!-coloring of members of !!\left[{\Bbb N\atop k}\right]!! there is always an infinite subset !!S\subset \Bbb N!! for which every member of !!\left[{S\atop k}\right]!! is the same color. I still do not find this notation perspicuous, and at the time, with much less experience, I was boggled. In the midst of my bogglement I was hit with the next sentence, which completely derailed me:

After this I could no longer think about the mathematics, but only about the sentence.

Outside the mathematical community Graham is probably best-known for juggling, or for Graham's number, which Wikipedia describes:

At the time of its introduction, it was the largest specific positive integer ever to have been used in a published mathematical proof.

One of my better Math Stack Exchange posts was in answer to the question Graham's Number : Why so big?. I love the phrasing of this question! And that, even with the strange phrasing, there is an answer! This type of huge number is quite typical in proofs of Ramsey theory, and I answered in detail.

The sense of humor that led Graham to write “danger of no confusion” is very much on display in the paper that gave us Graham's number. If you are wondering about Graham's number, check out my post.

[Other articles in category /math] permanent link

Addendum to “Weirdos during the Depression”

[ Previously ]

Ran Prieur had a take on this that I thought was insightful:

I would frame it like this: If you break rules that other people are following, you have to pretend to be unhappy, or they'll get really mad, because they don't want to face the grief that they could have been breaking the rules themselves all this time.

[Other articles in category /addenda] permanent link

Weird constants in math problems

Michael Lugo recently considered a problem involving the allocation of swimmers to swim lanes at random, ending with:

If we compute this for large !!n!! we get !!f(n) \sim 0.4323n!!, which agrees with the Monte Carlo simulations… The constant !!0.4323!! is $$\frac{(1-e^{-2})}2.$$

I love when stuff like this happens. The computer is great at doing a quick random simulation and getting you some weird number, and you have no idea what it really means. But mathematical technique can unmask the weird number and learn its true identity. (“It was Old Man Haskins all along!”)

A couple of years back Math Stack Exchange had Expected Number and Size of Contiguously Filled Bins, and although it wasn't exactly what was asked, I ended up looking into this question: We take !!n!! balls and throw them at random into !!n!! bins that are lined up in a row. A maximal contiguous sequence of all-empty or all-nonempty bins is called a “cluster”. For example, here we have 13 balls that I placed randomly into 13 bins:

In this example, there are 8 clusters, of sizes 1, 1, 1, 1, 4, 1, 3, 1. Is this typical? What's the expected cluster size?

It's easy to use Monte Carlo methods and find that when !!n!! is large, the average cluster size is approximately !!2.15013!!. Do you recognize this number? I didn't.

But it's not hard to do the calculation analytically and discover that that the reason it's approximately !!2.15013!! is that the actual answer is $$\frac1{2(e^{-1} - e^{-2})}$$ which is approximately !!2.15013!!.

Math is awesome and wonderful.

(Incidentally, I tried the Inverse Symbolic Calculator just now, but it was no help. It's also not in Plouffe's Miscellaneous Mathematical Constants)

[ Addendum 20200707: WolframAlpha does correctly identify the !!2.15013!! constant. ]

[Other articles in category /math] permanent link