My 2018 Mathematics A To Z: Fermat’s Last Theorem

Today’s topic is another request, this one from a Dina. I’m not sure if this is Dina Yagodich, who’d also suggested using the letter ‘e’ for the number ‘e’. Trusting that it is, Dina Yagodich has a YouTube channel of mathematics videos. They cover topics like how to convert degrees and radians to one another, what the chance of a false positive (or false negative) on a medical test is, ways to solve differential equations, and how to use computer tools like MathXL, TI-83/84 calculators, or Matlab. If I’m mistaken, original-commenter Dina, please let me know and let me know if you have any creative projects that should be mentioned here.

Art by Thomas K Dye, creator of the web comics Newshounds, Something Happens, and Infinity Refugees. His current project is Projection Edge. And you can get Projection Edge six months ahead of public publication by subscribing to his Patreon. And he’s on Twitter as @Newshoundscomic.

Fermat’s Last Theorem.

It comes to us from number theory. Like many great problems in number theory, it’s easy to understand. If you’ve heard of the Pythagorean Theorem you know, at least, there are triplets of whole numbers so that the first number squared plus the second number squared equals the third number squared. It’s easy to wonder about generalizing. Are there quartets of numbers, so the squares of the first three add up to the square of the fourth? Quintuplets? Sextuplets? … Oh, yes. That’s easy. What about triplets of whole numbers, including negative numbers? Yeah, and that turns out to be boring. Triplets of rational numbers? Turns out to be the same as triplets of whole numbers. Triplets of real-valued numbers? Turns out to be very boring. Triplets of complex-valued numbers? Also none too interesting.

Ah, but, what about a triplet of numbers, only raised to some other power? All three numbers raised to the first power is easy; we call that addition. To the third power, though? … The fourth? Any other whole number power? That’s hard. It’s hard finding, for any given power, a trio of numbers that work, although some come close. I’m informed there was an episode of The Simpsons which included, as a joke, the equation . If it were true, this would be enough to show Fermat’s Last Theorem was false. … Which happens. Sometimes, mathematicians believe they have found something which turns out to be wrong. Often this comes from noticing a pattern, and finding a proof for a specific case, and supposing the pattern holds up. This equation isn’t true, but it is correct for the first nine digits. An episode of The Wizard of Evergreen Terrace puts forth , which apparently matches ten digits. This includes the final digit, also known as “the only one anybody could check”. (The last digit of 3987¹² is 1. Last digit of 4365¹² is 5. Last digit of 4472¹² is 6, and there you go.) Really makes you think there’s something weird going on with 12th powers.

For a Fermat-like example, Leonhard Euler conjectured a thing about “Sums of Like Powers”. That for a whole number ‘n’, you need at least n whole numbers-raised-to-an-nth-power to equal something else raised to an n-th power. That is, you need at least three whole numbers raised to the third power to equal some other whole number raised to the third power. At least four whole numbers raised to the fourth power to equal something raised to the fourth power. At least five whole numbers raised to the fifth power to equal some number raised to the fifth power. Euler was wrong, in this case. L J Lander and T R Parkin published, in 1966, the one-paragraph paper Counterexample to Euler’s Conjecture on Sums of Like Powers. and there we go. Thanks, CDC 6600 computer!

But Fermat’s hypothesis. Let me put it in symbols. It’s easier than giving everything long, descriptive names. Suppose that the power ‘n’ is a whole number greater than 2. Then there are no three counting numbers ‘a’, ‘b’, and ‘c’ which make true the equation . It looks doable. It looks like once you’ve mastered high school algebra you could do it. Heck, it looks like if you know the proof about how the square root of two is irrational you could approach it. Pierre de Fermat himself said he had a wonderful little proof of it.

He was wrong. No shame in that. He was right about a lot of mathematics, including a lot of stuff that leads into the basics of calculus. And he was right in his feeling that this stuff was impossible. He was wrong that he had a proof. At least not one that worked for every possible whole number ‘n’ larger than 2.

For specific values of ‘n’, though? Oh yes, that’s doable. Fermat did it himself for an ‘n’ of 4. Euler, a century later, filed in ‘n’ of 3. Peter Dirichlet, a great name in number theory and analysis, and Joseph-Louis Lagrange, who worked on everything, proved the case of ‘n’ of 5. Dirichlet, in 1832, proved the case for ‘n’ of 14. And there were more partial solutions. You could show that if Fermat’s Last Theorem were ever false, it would have to be false for some prime-number value of ‘n’. That’s great work, answering as it does infinitely many possible cases. It just leaves … infinitely many to go.

And that’s how things went for centuries. I don’t know that every mathematician made some attempt on Fermat’s Last Theorem. But it seems hard to imagine a person could love mathematics enough to spend their lives doing it and not at least take an attempt at it. Nobody ever found it, though. In a 1989 episode of Star Trek: The Next Generation, Captain Picard muses on how eight centuries after Fermat nobody’s proven his theorem. This struck me at the time as too pessimistic. Granted humans were stumped for 400 years. But for 800 years? And stumping everyone in a whole Federation of a thousand worlds? And more than a thousand mathematical traditions? And, for some of these species, tens of thousands of years of recorded history? … Still, there wasn’t much sign of the solving the problem. In 1992 Analog Science Fiction Magazine published a funny short-short story by Ian Randal Strock, “Fermat’s Legacy”. In it, Fermat — jealous of figures like René Descartes and Blaise Pascal who upstaged his mathematical accomplishments — jots down the note. He figures an unsupported claim like that will earn true lasting fame.

So that takes us to 1993, when the world heard about elliptic integrals for the first time. Elliptic curves are neat things. They’re polynomials. They have some nice mathematical properties. People first noticed them in studying how long arcs of ellipses are. (This is why they’re called elliptic curves, even though most of them have nothing to do with any ellipse you’d ever tolerate in your presence.) They look ready to use for encryption. And in 1985, Gerhard Frey noticed something. Suppose you did have, for some ‘n’ bigger than 2, a solution . Then you could use that a, b, and n to make a new elliptic curve. That curve is the one that satisfies . And then that elliptic curve would not be “modular”.

I would like to tell you what it means for an elliptic curve to be modular. But getting to that point would take at least four subsidiary essays. MathWorld has a description of what it means to be modular, and even links to explaining terms like “meromorphic”. It’s getting exotic stuff.

Frey didn’t show whether elliptic curves of this time had to be modular or not. This is normal enough, for mathematicians. You want to find things which are true and interesting. This includes conjectures like this, that if elliptic curves are all modular then Fermat’s Last Theorem has to be true. Frey was working on consequences of the Taniyama-Shimura Conjecture, itself three decades old at that point. Yutaka Taniyama and Goro Shimura had found there seemed to be a link between elliptic curves and these “modular forms”, which are a kind of group. That is, a group-theory thing.

So in fall of 1993 I was taking an advanced, though still undergraduate, course in (not-high-school) algebra at Rutgers. It’s where we learn group theory, after Intro to Algebra introduced us to group theory. Some exciting news came out. This fellow named Andrew Wiles at Princeton had shown an impressive bunch of things. Most important, that the Taniyama-Shimura Conjecture was true for semistable elliptic curves. This includes the kind of elliptic curve Frey made out of solutions to Fermat’s Last Theorem. So the curves based on solutions to Fermat’s Last Theorem would have be modular. But Frey had shown any curves based on solutions to Fermat’s Last Theorem couldn’t be modular. The conclusion: there can’t be any solutions to Fermat’s Last Theorem. Our professor did his best to explain the proof to us. Abstract Algebra was the undergraduate course closest to the stuff Wiles was working on. It wasn’t very close. When you’re still trying to work out what it means for something to be an ideal it’s hard to even follow the setup of the problem. The proof itself was inaccessible.

Which is all right. Wiles’s original proof had some flaws. At least this mathematics major shrugged when that news came down and wondered, well, maybe it’ll be fixed someday. Maybe not. I remembered how exciting cold fusion was for about six weeks, too. But this someday didn’t take long. Wiles, with Richard Taylor, revised the proof and published about a year later. So far as I’m aware, nobody has any serious qualms about the proof.

So does knowing Fermat’s Last Theorem get us anything interesting? … And here is a sad anticlimax. It’s neat to know that can’t be true unless ‘n’ is 1 or 2, at least for positive whole numbers. But I’m not aware of any neat results that follow from that, or that would follow if it were untrue. There are results that follow from the Taniyama-Shimura Conjecture that are interesting, according to people who know them and don’t seem to be fibbing me. But Fermat’s Last Theorem turns out to be a cute little aside.

Which is not to say studying it was foolish. This easy-to-understand, hard-to-solve problem certainly attracted talented minds to think about mathematics. Mathematicians found interesting stuff in trying to solve it. Some of it might be slight. I learned that in a Pythagorean triplet — ‘a’, ‘b’, and ‘c’ with — that I was not the infinitely brilliant mathematician at age fifteen I hoped I might be. Also that if ‘a’, ‘b’, and ‘c’ are relatively prime, you can’t have ‘a’ and ‘b’ both odd and ‘c’ even. You had to have ‘c’ and either ‘a’ or ‘b’ odd, with the other number even. Other mathematicians of more nearly infinite ability found stuff of greater import. Ernst Eduard Kummer in the 19th century developed ideals. These are an important piece of group theory. He was busy proving special cases of Fermat’s Last Theorem.

Kind viewers have tried to retcon Picard’s statement about Fermat’s Last Theorem. They say Picard was really searching for the proof Fermat had, or believed he had. Something using the mathematical techniques available to the early 17th century. Or that follow closely enough from that. The Taniyama-Shimura Conjecture definitely isn’t it. I don’t buy the retcon, but I’m willing to play along for the sake of not causing trouble. I suspect there’s not a proof of the general case that uses anything Fermat could have recognized, or thought he had. That’s all right. The search for a thing can be useful even if the thing doesn’t exist.

Did The Greatest Generation Hosts Get As Drunk As I Expected?

I finally finished listening to Benjamin Ahr Harrison and Adam Pranica’s Greatest Generation podcast reviews of the first season of Star Trek: Deep Space Nine. (We’ve had fewer long car trips for this.) So I can return to my projection of how their drinking game would turn out.

Their plan was to make more exciting the discussion of some of Deep Space Nine‘s episodes by recording their reviews while drinking a lot. The plan was, for the fifteen episodes they had in the season, there would be a one-in-fifteen chance of doing any particular episode drunk. So how many drunk episodes would you expect to get, on this basis?

It’s a well-formed expectation value problem. There could be as few as zero or as many as fifteen, but some cases are more likely than others. Each episode could be recorded drunk or not-drunk. There’s an equal chance of each episode being recorded drunk. Whether one episode is drunk or not doesn’t depend on whether the one before was, and doesn’t affect whether the next one is. (I’ll come back to this.)

The most likely case was for there to be one drunk episode. The probability of exactly one drunk episode was a little over 38%. No drunk episodes was also a likely outcome. There was a better than 35% chance it would never have turned up. The chance of exactly two drunk episodes was about 19%. There drunk episodes had a slightly less than 6% chance of happening. Four drunk episodes a slightly more than 1% chance of happening. And after that you get into the deeply unlikely cases.

As the Deep Space Nine season turned out, this one-in-fifteen chance came up twice. It turned out they sort of did three drunk episodes, though. One of the drunk episodes turned out to be the first of two they planned to record that day. I’m not sure why they didn’t just swap what episode they recorded first, but I trust they had logistical reasons. As often happens with probability questions, the independence of events — whether a success for one affects the outcome of another — changes calculations.

There’s not going to be a second-season update to this. They’ve chosen to make a more elaborate recording game of things. They’ve set up a modified Snakes and Ladders type board with a handful of spots marked for stunts. Some sound like fun, such as recording without taking any notes about the episode. Some are, yes, drinking episodes. But this is all a very different and more complicated thing to project. If I were going to tackle that it’d probably be by running a bunch of simulations and taking averages from that.

Real actual episode that was really actually made and really actually aired for real. I’m going to go ahead and guess that it hasn’t aged well.

Also I trust they’ve been warned about the episode where Quark has a sex change so he can meet a top Ferengi soda magnate after accidentally giving his mother a heart attack because gads but that was a thing that happened somehow.

How Drunk Can We Expect The Greatest Generation Podcast Hosts To Get?

Among my entertainments is listening to the Greatest Generation podcast, hosted by Benjamin Ahr Harrison and Adam Pranica. They recently finished reviewing all the Star Trek: The Next Generation episodes, and have started Deep Space Nine. To add some fun and risk to episode podcasts the hosts proposed to record some episodes while drinking heavily. I am not a fun of recreational over-drinking, but I understand their feelings. There’s an episode where Quark has a sex-change operation because he gave his mother a heart attack right before a politically charged meeting with a leading Ferengi soda executive. Nobody should face that mess sober.

At the end of the episode reviewing “Babel”, Harrison proposed: there’s 15 episodes left in the season. Use a random number generator to pick a number from 1 to 15; if it’s one, they do the next episode (“Captive Pursuit”) drunk. And it was; what are the odds? One in fifteen. I just said.

Space-drunk engineer Jim Shimoda throwing control chips around in the moment that made him a Greatest Generation running joke. In the podcast’s context this makes sense. In the original context this made us all in 1987 grit our teeth and say, “No, no, this really is as good a show as we need this to be shut up shut up shut up”.

The question: how many episodes would they be doing drunk? As they discussed in the next episode, this would imply they’d always get smashed for the last episode of the season. This is a straightforward expectation-value problem. The expectation value of a thing is the sum of all the possible outcomes times the chance of each outcome. Here, the possible outcome is adding 1 to the number of drunk episodes. The chance of any particular episode being a drunk episode is 1 divided by ‘N’, if ‘N’ is the number of episodes remaining. So the next-to-the-last episode has 1 chance in 2 of being drunk. The second-from-the-last has 1 chance in 3 of being drunk. And so on.

This expectation value isn’t hard to calculate. If we start counting from the last episode of the season, then it’s easy. Add up , ending when we get up to one divided by the number of episodes in the season. 25 or 26, for most seasons of Deep Space Nine. 15, from when they counted here. This is the start of the harmonic series.

The harmonic series gets taught in sequences and series in calculus because it does some neat stuff if you let it go on forever. For example, every term in this sequence gets smaller and smaller. (The “sequence” is the terms that go into the sum: , and so on. The “series” is the sum of a sequence, a single number. I agree it seems weird to call a “series” that sum, but it’s the word we’re stuck with. If it helps, consider: when we talk about “a TV series” we usually mean the whole body of work, not individual episodes.) You can pick any number, however tiny you like. I can then respond with the last term in the sequence bigger than your number. Infinitely many terms in the sequence will be smaller than your pick. And yet: you can pick any number you like, however big. And I can take a finite number of terms in this sequence to make a sum bigger than whatever number you liked. The sum will eventually be bigger than 10, bigger than 100, bigger than a googolplex. These two facts are easy to prove, but they seem like they ought to be contradictory. You can see why infinite series are fun and produce much screaming on the part of students.

No Star Trek show has a season has infinitely many episodes, though, however long the second season of Enterprise seemed to drag out. So we don’t have to worry about infinitely many drunk episodes.

Since there were 15 episodes up for drunkenness in the first season of Deep Space Nine the calculation’s easy. I still did it on the computer. For the first season we could expect drunk episodes. This is a number a little bigger than 3.318. So, more likely three drunk episodes, four being likely. For the 25-episode seasons (seasons four and seven, if I’m reading this right), we could expect or just over 3.816 drunk episodes. Likely four, maybe three. For the 26-episode seasons (seasons two, five, and six), we could expect drunk episodes. That’s just over 3.854.

The number of drunk episodes to expect keeps growing. The harmonic series grows without bounds. But it keeps growing slower, compared to the number of terms you add together. You need a 31-episode season to be able to expect at four drunk episodes. To expect five drunk episodes you’d need an 83-episode season. If the guys at Worst Episode Ever, reviewing The Simpsons, did all 625-so-far episodes by this rule we could only expect seven drunk episodes.

Still, three, maybe four, drunk episodes of the 15 remaining first season is a fair number. They shouldn’t likely be evenly spaced. The chance of a drunk episode rises the closer they get to the end of the season. Expected length between drunk episodes is interesting but I don’t want to deal with that. I’ll just say that it probably isn’t the five episodes the quickest, easiest suggested by taking 15 divided by 3.

And it’s moot anyway. The hosts discussed it just before starting “Captive Pursuit”. Pranica pointed out, for example, the smashed-last-episode problem. What they decided they meant was there would be a 1-in-15 chance of recording each episode this season drunk. For the 25- or 26-episode seasons, each episode would get its 1-in-25 or 1-in-26 chance.

That changes the calculations. Not in spirit: that’s still the same. Count the number of possible outcomes and the chance of each one being a drunk episode and add that all up. But the work gets simpler. Each episode has a 1-in-15 chance of adding 1 to the total of drunk episodes. So the expected number of drunk episodes is the number of episodes (15) times the chance each is a drunk episode (1 divided by 15). We should expect 1 drunk episode. The same reasoning holds for all the other seasons; we should expect 1 drunk episode per season.

Still, since each episode gets an independent draw, there might be two drunk episodes. Could be three. There’s no reason that all 15 couldn’t be drunk. (Except that at the end of reviewing “Captive Pursuit” they drew for the next episode and it’s not to be a drunk one.) What are the chances there’s no drunk episodes? What are the chances there’s two, or three, or eight drunk episodes?

There’s a rule for this. This kind of problem is a mathematically-famous one. We get our results from the “binomial distribution”. This applies whenever there’s a bunch of attempts at something. And each attempt can either clearly succeed or clearly fail. And the chance of success (or failure) each attempt is always the same. That’s what applies here. If there’s ‘N’ episodes, and the chance is ‘p’ that any one will be drunk, then we get the chance ‘y’ of turning up exactly ‘k’ drunk episodes by the formula:

That looks a bit ugly, yeah. (I don’t like using ‘y’ as the name for a probability. I ran out of good letters and didn’t want to do subscripts.) It’s just tedious to calculate is all. Factorials and everything. Better to let the computer work it out. There is a formula that’s easy enough to work with, though. That’s because the chance of a drunk episode is the same each episode. I don’t know a formula to get the chance of exactly zero or one or four drunk episodes with the first, one-in-N chance. Probably the only thing to do is run a lot of simulations and trust that’s approximately right.

But for this rule it’s easy enough. There’s this formula, like I said. I figured out the chance of all the possible drunk episode combinations for the seasons. I mean I had the computer work it out. All I figured out was how to make it give me the results in a format I liked. Here’s what I got.

The chance of these many drunk episodes	In a 15-episode season is
0	0.355
1	0.381
2	0.190
3	0.059
4	0.013
5	0.002
6	0.000
7	0.000
8	0.000
9	0.000
10	0.000
11	0.000
12	0.000
13	0.000
14	0.000
15	0.000

Sorry it’s so dull, but the chance of a one-in-fifteen event happening 15 times in a row? You’d expect that to be pretty small. It’s got a probability of something like 0.000 000 000 000 000 002 28 of happening. Not technically impossible, but yeah, impossible.

How about for the 25- and 26-episode seasons? Here’s the chance of all the outcomes:

The chance of these many drunk episodes	In a 25-episode season is
0	0.360
1	0.375
2	0.188
3	0.060
4	0.014
5	0.002
6	0.000
7	0.000
8 or more	0.000

And things are a tiny bit different for a 26-episode season.

The chance of these many drunk episodes	In a 26-episode season is
0	0.361
1	0.375
2	0.188
3	0.060
4	0.014
5	0.002
6	0.000
7	0.000
7	0.000
8 or more	0.000

Yes, there’s a greater chance of no drunk episodes. The difference is really slight. It only looks so big because of rounding. A no-drunk 25 episode season has a chance of about 0.3604, while a no-drunk 26 episodes season has a chance of about 0.3607. The difference comes from the chance of lots of drunk episodes all being even worse somehow.

And there’s some neat implications through this. There’s a slightly better than one in three chance that each of the second through seventh seasons won’t have any drunk episodes. We could expect two dry seasons, hopefully not the one with Quark’s sex-change episode. We can reasonably expect at least one season with two drunk episodes. There’s a slightly more than 40 percent chance that some season will have three drunk episodes. There’s just under a 10 percent chance some season will have four drunk episodes.

Real actual episode that was really actually made and really actually aired for real. I don’t know when I last saw it. I’m going to go ahead and guess that it hasn’t aged well.

There’s no guarantees, though. Probability has a curious blend. There’s no predicting when any drunk episode will come. But we can make meaningful predictions about groups of episodes. These properties seem like they should be contradictions. And they’re not, and that’s wonderful.

The Summer 2017 Mathematics A To Z: Klein Bottle

Gaurish, of the For The Love Of Mathematics blog, takes me back into topology today. And it’s a challenging one, because what can I say about a shape this involved when I’m too lazy to draw pictures or include photographs most of the time?

Art courtesy of Thomas K Dye, creator of the web comic Newshounds. He has a Patreon for those able to support his work. He’s also open for commissions, starting from US$10.

In 1958 Clifton Fadiman, an open public intellectual and panelist on many fine old-time radio and early TV quiz shows, edited the book Fantasia Mathematica. It’s a pleasant read and you likely can find a copy in a library or university library nearby. It’s a collection of mathematically-themed stuff. Mostly short stories, a few poems, some essays, even that bit where Socrates works through a proof. And some of it is science fiction, this from an era when science fiction was really disreputable.

If there’s a theme to the science fiction stories included it is: Möbius Strips, huh? There are so many stories in the book that amount to, “what is this crazy bizarre freaky weird ribbon-like structure that only has the one side? Huh?” As I remember even one of the non-science-fiction stories is a Möbius Strip story.

I don’t want to sound hard on the writers, nor on Fadiman for collecting what he has. A story has to be about people doing something, even if it’s merely exploring some weird phenomenon. You can imagine people dealing with weird shapes. It’s hard to imagine what story you could tell about an odd perfect number. (Well, that isn’t “here’s how we discovered the odd perfect number”, which amounts to a lot of thinking and false starts. Or that doesn’t make the odd perfect number a MacGuffin, the role equally well served by letters of transit or a heap of gold or whatever.) Many of the stories that aren’t about the Möbius Strip are about four- and higher-dimensional shapes that people get caught in or pass through. One of the hyperdimensional stories, A J Deutsch’s “A Subway Named Möbius”, even pulls in the Möbius Strip. The name doesn’t fit, but it is catchy, and is one of the two best tall tales about the Boston subway system.

Besides, it’s easy to see why the Möbius Strip is interesting. It’s a ribbon where both sides are the same side. What’s not neat about that? It forces us to realize that while we know what “sides” are, there’s stuff about them that isn’t obvious. That defies intuition. It’s so easy to make that it holds another mystery. How is this not a figure known to the ancients and used as a symbol of paradox for millennia? I have no idea; it’s hard to guess why something was not noticed when it could easily have been It dates to 1858, when August Ferdinand Möbius and Johann Bendict Listing independently published on it.

The Klein Bottle is newer by a generation. Felix Klein, who used group theory to enlighten geometry and vice-versa, described the surface in 1882. It has much in common with the Möbius Strip. It’s a thing that looks like a solid. But it’s impossible to declare one side to be outside and the other in, at least not in any logically coherent way. Take one and dab a spot with a magic marker. You could trace, with the marker, a continuous curve that gets around to the same spot on the “other” “side” of the thing. You see why I have to put quotes around “other” and “side”. I believe you know what I mean when I say this. But taken literally, it’s nonsense.

The Klein Bottle’s a two-dimensional surface. By that I mean that could cover it with what look like lines of longitude and latitude. Those coordinates would tell you, without confusion, where a point on the surface is. But it’s embedded in a four-dimensional space. (Or a higher-dimensional space, but everything past the fourth dimension is extravagance.) We have never seen a Klein Bottle in its whole. I suppose there are skilled people who can imagine it faithfully, but how would anyone else ever know?

Big deal. We’ve never seen a tesseract either, but we know the shadow it casts in three-dimensional space. So it is with the Klein Bottle. Visit any university mathematics department. If they haven’t got a glass replica of one in the dusty cabinets welcoming guests to the department, never fear. At least one of the professors has one on an office shelf, probably beside some exams from eight years ago. They make nice-looking jars. Klein Bottles don’t have to. There are different shapes their projection into three dimensions can take. But the only really different one is this sort of figure-eight helical shape that looks like a roller coaster gone vicious. (There’s also a mirror image of this, the helix winding the opposite way.) These representations have the surface cross through itself. In four dimensions, it does no such thing, any more than the edges of a cube cross one another. It’s just the lines in a picture on a piece of paper that cross.

The Möbius Strip is good practice for learning about the Klein Bottle. We can imagine creating a Bottle by the correct stitching-together of two strips. Or, if you feel destructive, we can start with a Bottle and slice it, producing a pair of Möbius Strips. Both are non-orientable. We can’t make a division between one side and another that reflects any particular feature of the shape. One of the helix-like representations of the Klein Bottle also looks like a pool toy-ring version of the Möbius Strip.

And strange things happen on these surfaces. You might remember the four-color map theorem. Four colors are enough to color any two-dimensional map without adjacent territories having to share a color. (This isn’t actually so, as the territories have to be contiguous, with no enclaves of one territory inside another. Never mind.) This is so for territories on the sphere. It’s hard to prove (although the five-color theorem is easy.) Not so for the Möbius Strip: territories on it might need as many as six colors. And likewise for the Klein Bottle. That’s a particularly neat result, as the Heawood Conjecture tells us the Klein Bottle might need seven. The Heawood Conjecture is otherwise dead-on in telling us how many colors different kinds of surfaces need for their map-colorings. The Klein Bottle is a strange surface. And yes, it was easier to prove the six-color theorem on the Klein Bottle than it was to prove the four-color theorem on the plane or sphere.

(Though it’s got the tentative-sounding name of conjecture, the Heawood Conjecture is proven. Heawood put it out as a conjecture in 1890. It took to 1968 for the whole thing to be finally proved. I imagine all those decades of being thought but not proven true gave it a reputation. It’s not wrong for Klein Bottles. If six colors are enough for these maps, then so are seven colors. It’s just that Klein Bottles are the lone case where the bound is tighter than Heawood suggests.)

All that said, do we care? Do Klein Bottles represent something of particular mathematical interest? Or are they imagination-capturing things we don’t really use? I confess I’m not enough of a topologist to say how useful they are. They are easily-understood examples of algebraic or geometric constructs. These are things with names like “quotient spaces” and “deck transformations” and “fiber bundles”. The thought of the essay I would need to write to say what a fiber bundle is makes me appreciate having good examples of the thing around. So if nothing else they are educationally useful.

And perhaps they turn up more than I realize. The geometry of Möbius Strips turns up in many surprising places: music theory and organic chemistry, superconductivity and roller coasters. It would seem out of place if the kinds of connections which make a Klein Bottle don’t turn up in our twisty world.

Reading the Comics, April 24, 2017: Reruns Edition

I went a little wild explaining the first of last week’s mathematically-themed comic strips. So let me split the week between the strips that I know to have been reruns and the ones I’m not so sure were.

Bill Amend’s FoxTrot for the 23rd — not a rerun; the strip is still new on Sundays — is a probability question. And a joke about story problems with relevance. Anyway, the question uses the binomial distribution. I know that because the question is about doing a bunch of things, homework questions, each of which can turn out one of two ways, right or wrong. It’s supposed to be equally likely to get the question right or wrong. It’s a little tedious but not hard to work out the chance of getting exactly six problems right, or exactly seven, or exactly eight, or so on. To work out the chance of getting six or more questions right — the problem given — there’s two ways to go about it.

One is the conceptually easy but tedious way. Work out the chance of getting exactly six questions right. Work out the chance of getting exactly seven questions right. Exactly eight questions. Exactly nine. All ten. Add these chances up. You’ll get to a number slightly below 0.377. That is, Mary Lou would have just under a 37.7 percent chance of passing. The answer’s right and it’s easy to understand how it’s right. The only drawback is it’s a lot of calculating to get there.

So here’s the conceptually harder but faster way. It works because the problem says Mary Lou is as likely to get a problem wrong as right. So she’s as likely to get exactly ten questions right as exactly ten wrong. And as likely to get at least nine questions right as at least nine wrong. To get at least eight questions right as at least eight wrong. You see where this is going: she’s as likely to get at least six right as to get at least six wrong.

There’s exactly three possibilities for a ten-question assignment like this. She can get four or fewer questions right (six or more wrong). She can get exactly five questions right. She can get six or more questions right. The chance of the first case and the chance of the last have to be the same.

So, take 1 — the chance that one of the three possibilities will happen — and subtract the chance she gets exactly five problems right, which is a touch over 24.6 percent. So there’s just under a 75.4 percent chance she does not get exactly five questions right. It’s equally likely to be four or fewer, or six or more. Just-under-75.4 divided by two is just under 37.7 percent, which is the chance she’ll pass as the problem’s given. It’s trickier to see why that’s right, but it’s a lot less calculating to do. That’s a common trade-off.

Ruben Bolling’s Super-Fun-Pax Comix rerun for the 23rd is an aptly titled installment of A Million Monkeys At A Million Typewriters. It reminds me that I don’t remember if I’d retired the monkeys-at-typewriters motif from Reading the Comics collections. If I haven’t I probably should, at least after making a proper essay explaining what the monkeys-at-typewriters thing is all about.

Ted Shearer’s Quincy from the 28th of February, 1978. So, that FoxTrot problem I did? The conceptually-easy-but-tedious way is not too hard to do if you have a calculator. It’s a buch of typing but nothing more. If you don’t have a calculator, though, the desire not to do a whole bunch of calculating could drive you to the conceptually-harder-but-less-work answer. Is that a good thing? I suppose; insight is a good thing to bring. But the less-work answer only works because of a quirk in the problem, that Mary Lou is supposed to have a 50 percent chance of getting a question right. The low-insight-but-tedious problem will aways work. Why skip on having something to do the tedious part?

Ted Shearer’s Quincy from the 28th of February, 1978 reveals to me that pocket calculators were a thing much earlier than I realized. Well, I was too young to be allowed near stuff like that in 1978. I don’t think my parents got their first credit-card-sized, solar-powered calculator that kind of worked for another couple years after that. Kids, ask about them. They looked like good ideas, but you could use them for maybe five minutes before the things came apart. Your cell phone is so much better.

Bil Watterson’s Calvin and Hobbes rerun for the 24th can be classed as a resisting-the-word-problem joke. It’s so not about that, but who am I to slow you down from reading a Calvin and Hobbes story?

Garry Trudeau’s Doonesbury rerun for the 24th started a story about high school kids and their bad geography skills. I rate it as qualifying for inclusion here because it’s a mathematics teacher deciding to include more geography in his course. I was amused by the week’s jokes anyway. There’s no hint given what mathematics Gil teaches, but given the links between geometry, navigation, and geography there is surely something that could be relevant. It might not help with geographic points like which states are in New England and where they are, though.

Zach Weinersmith’s Saturday Morning Breakfast Cereal for the 24th is built on a plot point from Carl Sagan’s science fiction novel Contact. In it, a particular “message” is found in the digits of π. (By “message” I mean a string of digits that are interesting to us. I’m not sure that you can properly call something a message if it hasn’t got any sender and if there’s not obviously some intended receiver.) In the book this is an astounding thing because the message can’t be; any reasonable explanation for how it should be there is impossible. But short “messages” are going to turn up in π also, as per the comic strips.

I assume the peer review would correct the cartoon mathematicians’ unfortunate spelling of understanding.

The End 2016 Mathematics A To Z: Normal Numbers

Today’s A To Z term is another of gaurish’s requests. It’s also a fun one so I’m glad to have reason to write about it.

Normal Numbers

A normal number is any real number you never heard of.

Yeah, that’s not what we say a normal number is. But that’s what a normal number is. If we could imagine the real numbers to be a stream, and that we could reach into it and pluck out a water-drop that was a single number, we know what we would likely pick. It would be an irrational number. It would be a transcendental number. And it would be a normal number.

We know normal numbers — or we would, anyway — by looking at their representation in digits. For example, π is a number that starts out 3.1415926535897931159979634685441851615905 and so on forever. Look at those digits. Some of them are 1’s. How many? How many are 2’s? How many are 3’s? Are there more than you would expect? Are there fewer? What would you expect?

Expect. That’s the key. What should we expect in the digits of any number? The numbers we work with don’t offer much help. A whole number, like 2? That has a decimal representation of a single ‘2’ and infinitely many zeroes past the decimal point. Two and a half? A single ‘2, a single ‘5’, and then infinitely many zeroes past the decimal point. One-seventh? Well, we get infinitely many 1’s, 4’s, 2’s, 8’s, 5’s, and 7’s. Never any 3’s, nor any 0’s, nor 6’s or 9’s. This doesn’t tell us anything about how often we would expect ‘8’ to appear in the digits of π.

In an normal number we get all the decimal digits. And we get each of them about one-tenth of the time. If all we had was a chart of how often digits turn up we couldn’t tell the summary of one normal number from the summary of any other normal number. Nor could we tell either from the summary of a perfectly uniform randomly drawn number.

It goes beyond single digits, though. Look at pairs of digits. How often does ’14’ turn up in the digits of a normal number? … Well, something like once for every hundred pairs of digits you draw from the random number. Look at triplets of digits. ‘141’ should turn up about once in every thousand sets of three digits. ‘1415’ should turn up about once in every ten thousand sets of four digits. Any finite string of digits should turn up, and exactly as often as any other finite string of digits.

That’s in the full representation. If you look at all the infinitely many digits the normal number has to offer. If all you have is a slice then some digits are going to be more common and some less common. That’s similar to how if you fairly toss a coin (say) forty times, there’s a good chance you’ll get tails something other than exactly twenty times. Look at the first 35 or so digits of π there’s not a zero to be found. But as you survey more digits you get closer and closer to the expected average frequency. It’s the same way coin flips get closer and closer to 50 percent tails. Zero is a rarity in the first 35 digits. It’s about one-tenth of the first 3500 digits.

The digits of a specific number are not random, not if we know what the number is. But we can be presented with a subset of its digits and have no good way of guessing what the next digit might be. That is getting into the same strange territory in which we can speak about the “chance” of a month having a Friday the 13th even though the appearances of Fridays the 13th have absolutely no randomness to them.

This has staggering implications. Some of them inspire an argument in science fiction Usenet newsgroup rec.arts.sf.written every two years or so. Probably it does so in other venues; Usenet is just my first home and love for this. In a minor point in Carl Sagan’s novel Cosmos possibly-imaginary aliens reveal there’s a pattern hidden in the digits of π. (It’s not in the movie version, which is a shame. But to include it would require people watching a computer. So that could not make for a good movie scene, we now know.) Look far enough into π, says the book, and there’s suddenly a string of digits that are nearly all zeroes, interrupted with a few ones. Arrange the zeroes and ones into a rectangle and it draws a pixel-art circle. And the aliens don’t know how something astounding like that could be.

Nonsense, respond the kind of science fiction reader that likes to identify what the nonsense in science fiction stories is. (Spoiler: it’s the science. In this case, the mathematics too.) In a normal number every finite string of digits appears. It would be truly astounding if there weren’t an encoded circle in the digits of π. Indeed, it would be impossible for there not to be infinitely many circles of every possible size encoded in every possible way in the digits of π. If the aliens are amazed by that they would be amazed to find how every triangle has three corners.

I’m a more forgiving reader. And I’ll give Sagan this amazingness. I have two reasons. The first reason is on the grounds of discoverability. Yes, the digits of a normal number will have in them every possible finite “message” encoded every possible way. (I put the quotes around “message” because it feels like an abuse to call something a message if it has no sender. But it’s hard to not see as a “message” something that seems to mean something, since we live in an era that accepts the Death of the Author as a concept at least.) Pick your classic cypher `1 = A, 2 = B, 3 = C’ and so on, and take any normal number. If you look far enough into its digits you will find every message you might ever wish to send, every book you could read. Every normal number holds Jorge Luis Borges’s Library of Babel, and almost every real number is a normal number.

But. The key there is if you look far enough. Look above; the first 35 or so digits of π have no 0’s, when you would expect three or four of them. There’s no 22’s, even though that number has as much right to appear as does 15, which gets in at least twice that I see. And we will only ever know finitely many digits of π. It may be staggeringly many digits, sure. It already is. But it will never be enough to be confident that a circle, or any other long enough “message”, must appear. It is staggering that a detectable “message” that long should be in the tiny slice of digits that we might ever get to see.

And it’s harder than that. Sagan’s book says the circle appears in whatever base π gets represented in. So not only does the aliens’ circle pop up in base ten, but also in base two and base sixteen and all the other, even less important bases. The circle happening to appear in the accessible digits of π might be an imaginable coincidence in some base. There’s infinitely many bases, one of them has to be lucky, right? But to appear in the accessible digits of π in every one of them? That’s staggeringly impossible. I say the aliens are correct to be amazed.

Now to my second reason to side with the book. It’s true that any normal number will have any “message” contained in it. So who says that π is a normal number?

We think it is. It looks like a normal number. We have figured out many, many digits of π and they’re distributed the way we would expect from a normal number. And we know that nearly all real numbers are normal numbers. If I had to put money on it I would bet π is normal. It’s the clearly safe bet. But nobody has ever proved that it is, nor that it isn’t. Whether π is normal or not is a fit subject for conjecture. A writer of science fiction may suppose anything she likes about its normality without current knowledge saying she’s wrong.

It’s easy to imagine numbers that aren’t normal. Rational numbers aren’t, for example. If you followed my instructions and made your own transcendental number then you made a non-normal number. It’s possible that π should be non-normal. The first thirty million digits or so look good, though, if you think normal is good. But what’s thirty million against infinitely many possible counterexamples? For all we know, there comes a time when π runs out of interesting-looking digits and turns into an unpredictable little fluttering between 6 and 8.

It’s hard to prove that any numbers we’d like to know about are normal. We don’t know about π. We don’t know about e, the base of the natural logarithm. We don’t know about the natural logarithm of 2. There is a proof that the square root of two (and other non-square whole numbers, like 3 or 5) is normal in base two. But my understanding is it’s a nonstandard approach that isn’t quite satisfactory to experts in the field. I’m not expert so I can’t say why it isn’t quite satisfactory. If the proof’s authors or grad students wish to quarrel with my characterization I’m happy to give space for their rebuttal.

It’s much the way transcendental numbers were in the 19th century. We understand there to be this class of numbers that comprises nearly every number. We just don’t have many examples. But we’re still short on examples of transcendental numbers. Maybe we’re not that badly off with normal numbers.

We can construct normal numbers. For example, there’s the Champernowne Constant. It’s the number you would make if you wanted to show you could make a normal number. It’s 0.12345678910111213141516171819202122232425 and I bet you can imagine how that develops from that point. (David Gawen Champernowne proved it was normal, which is the hard part.) There’s other ways to build normal numbers too, if you like. But those numbers aren’t of any interest except that we know them to be normal.

Mere normality is tied to a base. A number might be normal in base ten (the way normal people write numbers) but not in base two or base sixteen (which computers and people working on computers use). It might be normal in base twelve, used by nobody except mathematics popularizers of the 1960s explaining bases, but not normal in base ten. There can be numbers normal in every base. They’re called “absolutely normal”. Nearly all real numbers are absolutely normal. Wacław Sierpiński constructed the first known absolutely normal number in 1917. If you got in on the fractals boom of the 80s and 90s you know his name, although without the Polish spelling. He did stuff with gaskets and curves and carpets you wouldn’t believe. I’ve never seen Sierpiński’s construction of an absolutely normal number. From my references I’m not sure if we know how to construct any other absolutely normal numbers.

So that is the strange state of things. Nearly every real number is normal. Nearly every number is absolutely normal. We know a couple normal numbers. We know at least one absolutely normal number. But we haven’t (to my knowledge) proved any number that’s otherwise interesting is also a normal number. This is why I say: a normal number is any real number you never heard of.

Reading the Comics, June 26, 2015: June 23, 2016 Plus Golden Lizards Edition

And now for the huge pile of comic strips that had some mathematics-related content on the 23rd of June. I admit some of them are just using mathematics as a stand-in for “something really smart people do”. But first, another moment with the Magic Realism Bot:

A watchmaker realises that a golden lizard is controlling mathematics.

— Magic Realism Bot (@MagicRealismBot) June 25, 2016

So, you know, watch the lizards and all.

Tom Batiuk’s Funky Winkerbean name-drops E = mc² as the sort of thing people respect. If the strip seems a little baffling then you should know that Mason’s last name is Jarr. He was originally introduced as a minor player in a storyline that wasn’t about him, so the name just had to exist. But since then Tom Batiuk’s decided he likes the fellow and promoted him to major-player status. And maybe Batiuk regrets having a major character with a self-consciously Funny Name, which is an odd thing considering he named his long-running comic strip for original lead character Funky Winkerbean.

Tom Batiuk’s Funky Winkerbean for the 23rd of June, 2016. They’re in the middle of filming one or possibly two movies about the silver-age comic book hero Starbuck Jones. This is all the comic strip is about anymore, so if you go looking for its old standbys — people dying — or its older standbys — band practice being rained on — sorry, you’ll have to look somewhere else. That somewhere else would be the yellowed strips taped to the walls in the teachers lounge.

Charlie Podrebarac’s CowTown depicts the harsh realities of Math Camp. I assume they’re the realities. I never went to one myself. And while I was on the Physics Team in high school I didn’t make it over to the competitive mathematics squad. Yes, I noticed that the not-a-numbers-person Jim Smith can’t come up with anything other than the null symbol, representing nothing, not even zero. I like that touch.

Ryan North’s Dinosaur Comics rerun is about Richard Feynman, the great physicist whose classic memoir What Do You Care What Other People Think? is hundreds of pages of stories about how awesome he was. Anyway, the story goes that Feynman noticed one of the sequences of digits in π and thought of the joke which T-Rex shares here.

π is believed but not proved to be a “normal” number. This means several things. One is that any finite sequence of digits you like should appear in its representation, somewhere. Feynman and T-Rex look for the sequence ‘999999’, which sure enough happens less than eight hundred digits past the decimal point. Lucky stroke there. There’s no reason to suppose the sequence should be anywhere near the decimal point. There’s no reason to suppose the sequence has to be anywhere in the finite number of digits of π that humanity will ever know. (This is why Carl Sagan’s novel Contact, which has as a plot point the discovery of a message apparently encoded in the digits of π, is not building on a stupid idea. That any finite message exists somewhere is kind-of certain. That it’s findable is not.)

e, mentioned in the last panel, is similarly thought to be a normal number. It’s also not proved to be. We are able to say that nearly all numbers are normal. It’s in much the way we can say nearly all numbers are irrational. But it is hard to prove that any numbers are. I believe that the only numbers humans have proved to be normal are a handful of freaks created to show normal numbers exist. I don’t know of any number that’s interesting in its own right that’s also been shown to be normal. We just know that almost all numbers are.

But it is imaginable that π or e aren’t. They look like they’re normal, based on how their digits are arranged. It’s an open question and someone might make a name for herself by answering the question. It’s not an easy question, though.

Missy Meyer’s Holiday Doodles breaks the news to me the 23rd was SAT Math Day. I had no idea and I’m not sure what that even means. The doodle does use the classic “two trains leave Chicago” introduction, the “it was a dark and stormy night” of Boring High School Algebra word problems.

Stephan Pastis’s Pearls Before Swine is about everyone who does science and mathematics popularization, and what we worry someone’s going to reveal about us. Um. Except me, of course. I don’t do this at all.

Ashleigh Brilliant’s Pot-Shots rerun is a nice little averages joke. It does highlight something which looks paradoxical, though. Typically if you look at the distributions of values of something that can be measured you get a bell cure, like Brilliant drew here. The value most likely to turn up — the mode, mathematicians say — is also the arithmetic mean. “The average”, is what everybody except mathematicians say. And even they say that most of the time. But almost nobody is at the average.

Looking at a drawing, Brilliant’s included, explains why. The exact average is a tiny slice of all the data, the “population”. Look at the area in Brilliant’s drawing underneath the curve that’s just the blocks underneath the upside-down fellow. Most of the area underneath the curve is away from that.

There’s a lot of results that are close to but not exactly at the arithmetic mean. Most of the results are going to be close to the arithmetic mean. Look at how many area there is under the curve and within four vertical lines of the upside-down fellow. That’s nearly everything. So we have this apparent contradiction: the most likely result is the average. But almost nothing is average. And yet almost everything is nearly average. This is why statisticians have their own departments, or get to make the mathematics department brand itself the Department of Mathematics and Statistics.

How My Mathematics Blog Was Read, For January 2015

And after reaching 20,000 views on the final day of December, 2014, could I reach 21,000 views by the end of January? Probably I could have, but in point of fact I did not. I am not complaining, though: I finished the month with 20,956 page views all told, after a record 944 pages got viewed by somebody, somewhere, for some reason. This is a record high for me, going well past the 831 that had been the January 2013 and December 2014 (tied) record. And likely I’ll reach 21,000 in the next couple days anyway.

According to WordPress, this was read by 438 distinct visitors, reading 2.16 views per visitor on average. That isn’t quite a record: January 2013 remains my high count for visitors, at 473, but it’s still, all told, some pretty nice numbers especially considering I don’t think I had my best month of blog-writing. I can’t wait to get some interesting new topics in here for February and see that they interest absolutely nobody.

The new WordPress statistics page is still awful, don’t get me wrong, but it has been getting a little bit better, and it does offer some new data I couldn’t gather easily before. Among them: that in January 205 I received 197 likes overall — a high for the past twelvemonth, which is as far as I can figure out how to get it, and up from 128 in December — and 51 comments, up from December 29, and also a high for the twelvemonth.

The three countries sending me viewers were, once again, the big three of the United States (594), Canada (56), and the United Kingdom (52), with Austria sending in 32 viewers, and Germany and Argentina ending 22 each. And India, for a wonder sent me a noticeable-to-me 18 readers, although on a per capita basis that still isn’t very many, I admit.

There was a bumper crop of single-reader countries, though, up from last month’s six: Belgium, Estonia, Finland, Greece, Hungary, Indonesia, Iraq, Japan, Kuwait, Libya, Mexico, Paraguay, Slovakia, and the United Arab Emirates each found only one person viewing anything around here. Greece and Mexico are repeats from December.

This month’s most popular articles were mostly comic strip posts, although they were a pretty popular set; none of these had fewer than 35 views per, which feels high to me. The top posts of the last 30 days, then, were:

1. Reading the Comics, January 6, 2015: First of the Year Edition, in which I included drawing a sloppy `2′ as a snoring `Z’ as somehow connected to mathematics.
2. Reading the Comics, January 24, 2015: Many, But Not Complicated Edition, which includes an explanation for why margins of errors on surveys are always like three or four percent.
3. Reading the Comics, January 11, 2015: Standard Genres And Bloom County Edition, in which I reveal my best guess for Jon Bon Jovi’s shorts size in the late 80s.
4. 20,000: My Math Blog’s Statistics, because my narcissism is apparently quite popular?
5. Reading the Comics, January 17, 2015: Finding Your Place Edition, where, again, I can flog that thing about a watch as a compass.
6. How Many Trapezoids I Can Draw, which also reveals how many trapeziums I think are different in interesting ways.
7. A bit more about Thomas Hobbes, and his attempt to redefine the very nature of mathematics, which didn’t succeed in quite the way he wanted.

Among the interesting search terms that brought people here the past month have been ([sic] on all of them):

• science fiction and trapazoids (Somebody should totally write the definitive SFnal treatment of trapezoids, I agree.)
• food. stotagre nebus (I feel strangely threatened by this.)
• a group of student offer at least one of mathematics,physics, and statistcs , 14 of them offer mathematics, 12 offer physics,and 16 offer statistics.7 offer statistics and maths 6 offer maths and physics, 4 offer physics and statistics only, while 5 offer all the three subject (Help?)
• hiw to draw diffrent trameziums
• soglow otto radio (Pretty sure I used to listen to that back on WRSU in my undergrad days.)
• if a calendar has two consecutive months with friday the 13th which would they be (February and March, in a non-bissextile — that is, non-leap — year)
• how to measure a christmas tree made of triangles and trapeziums (I would use a tape measure, myself)

So if I would summarize January 2015 in my readership here, I would say: tramezium?

In A Really Big Universe

I’d got to thinking idly about Olbers’ Paradox, the classic question of why the night sky is dark. It’s named for Heinrich Wilhelm Olbers, 1758-1840, who of course was not the first person to pose the problem nor to give a convincing answer to it, but, that’s the way naming rights go.

It doesn’t sound like much of a question at first, after all, it’s night. But if we suppose the universe is infinitely large and is infinitely old, then, along the path of any direction you look in the sky, day or night, there’ll be a star. The star may be very far away, so that it’s very faint; but it takes up much less of the sky from being so far away. The result is that the star’s intensity, as a function of how much of the sky it takes up, is no smaller. And there’ll be stars shining just as intensely in directions that are near to that first star. The sky in an infinitely large, infinitely old universe should be a wall of stars.

Oh, some stars will be dimmer than average, and some brighter, but that doesn’t matter much. We can suppose the average star is of average brightness and average size for reasons that are right there in the name of the thing; it makes the reasoning a little simpler and doesn’t change the result.

The reason there is darkness is that our universe is neither infinitely large nor infinitely old. There aren’t enough stars to fill the sky and there’s not enough time for the light from all of them to get to us.

But we can still imagine standing on a planet in such an Olbers Universe (to save myself writing “infinitely large and infinitely old” too many times), with enough vastness and enough time to have a night sky that looks like a shell of starlight, and that’s what I was pondering. What might we see if you looked at the sky, in these conditions?

Well, light, obviously; we can imagine the sky looking as bright as the sun, but in all directions above the horizon. The sun takes up a very tiny piece of the sky — it’s about as wide across as your thumb, held at arm’s length, and try it if you don’t believe me (better, try it with the Moon, which is about the same size as the Sun and easier to look at) — so, multiply that brightness by the difference between your thumb and the sky and imagine the investment in sunglasses this requires.

It’s worse than that, though. Yes, in any direction you look there’ll be a star, but if you imagine going on in that direction there’ll be another star, eventually. And another one past that, and another past that yet. And the light — the energy — of those stars shining doesn’t disappear because there’s a star between it and the viewer. The heat will just go into warming up the stars in its path and get radiated through.

This is why interstellar dust, or planets, or other non-radiating bodies doesn’t answer why the sky could be dark in a vast enough universe. Anything that gets enough heat put into it will start to glow and start to shine from that light. The stars will slow down the waves of heat from the stars behind them, but given enough time, it will get through, and in an infinitely old universe, there is enough time.

The conclusion, then, is that our planet in an Olbers Universe would get an infinite amount of heat pouring onto it, at all times. It’s hard to see how life could possibly exist in the circumstance; water would boil away — rock would boil away — and the planet just would evaporate into dust.

Things get worse, though: it’s not just our planet that would get boiled away like this, but as far as I can tell, the stars too. Each star would be getting an infinite amount of heat pouring into it. It seems to me this requires the matter making up the stars to get so hot it would boil away, just as the atmosphere and water and surface of the imagined planet would, until the star — until all stars — disintegrate. At this point I have to think of the great super-science space-opera writers of the early 20th century, listening to the description of a wave of heat that boils away a star, and sniffing, “Amateurs. Come back when you can boil a galaxy instead”. Well, the galaxy would boil too, for the same reasons.

Even once the stars have managed to destroy themselves, though, the remaining atoms would still have a temperature, and would still radiate faint light. And that faint light, multiplied by the infinitely many atoms and all the time they have, would still accumulate to an infinitely great heat. I don’t know how hot you have to get to boil a proton into nothingness — or a quark — but if there is any temperature that does it, it’d be able to.

So the result, I had to conclude, is that an infinitely large, infinitely old universe could exist only if it didn’t have anything in it, or at least if it had nothing that wasn’t at absolute zero in it. This seems like a pretty dismal result and left me looking pretty moody for a while, even I was sure that EE “Doc” Smith would smile at me for working out the heat-death of quarks.

Of course, there’s no reason that a universe has to, or even should, be pleasing to imagine. And there is a little thread of hope for life, or at least existence, in a Olbers Universe.

All the destruction-of-everything comes about from the infinitely large number of stars, or other radiating bodies, in the universe. If there’s only finitely much matter in the universe, then, their total energy doesn’t have to add up to the point of self-destruction. This means giving up an assumption that was slipped into my Olbers Universe without anyone noticing: the idea that it’s about uniformly distributed. If you compare any two volumes of equal size, from any time, they have about the same number of stars in them. This is known in cosmology as “isotropy”.

Our universe seems to have this isotropy. Oh, there are spots where you can find many stars (like the center of a galaxy) and spots where there are few (like, the space in-between galaxies), but the galaxies themselves seem to be pretty uniformly distributed.

But an imagined universe doesn’t have to have this property. If we suppose an Olbers Universe without then we can have stars and planets and maybe even life. It could even have many times the mass, the number of stars and everything, that our universe has, spread across something much bigger than our universe. But it does mean that this infinitely large, infinitely old universe will have all its matter clumped together into some section, and nearly all the space — in a universe with an incredible amount of space — will be empty.

I suppose that’s better than a universe with nothing at all, but somehow only a little better. Even though it could be a universe with more stars and more space occupied than our universe has, that infinitely vast emptiness still haunts me.

(I’d like to note, by the way, that all this universe-building and reasoning hasn’t required any equations or anything like that. One could argue this has diverted from mathematics and cosmology into philosophy, and I wouldn’t dispute that, but can imagine philosophers might.)

Weightlessness at the Equator (Whiteboard Sketch #1)

The mathematics blog Scientific Finger Food has an interesting entry, “Weightlessness at the Equator (Whiteboard Sketch #1)”, which looks at the sort of question that’s easy to imagine when you’re young: since gravity pulls you to the center of the earth, and the earth’s spinning pushes you away (unless we’re speaking precisely, but you know what that means), so, how fast would the planet have to spin so that a person on the equator wouldn’t feel any weight?

It’s a straightforward problem, one a high school student ought to be able to do. Sebastian Templ works the problem out, though, including the all-important diagram that shows the important part, which is what calculation to do.

In reality, the answer doesn’t much matter since a planet spinning nearly fast enough to allow for weightlessness at the equator would be spinning so fast it couldn’t hold itself together, and a more advanced version of this problem could make use of that: given some measure of how strongly rock will hold itself together, what’s the fastest that the planet can spin before it falls apart? And a yet more advanced course might work out how other phenomena, such as tides or the precession of the poles might work. Eventually, one might go on to compose highly-regarded works of hard science fiction, if you’re willing to start from the questions easy to imagine when you’re young.

scientific finger food

At the present time, our Earth does a full rotation every 24 hours, which results in day and night. Just like on a carrousel, its inhabitants (and, by the way, all the other stuff on and of the planet) are pushed “outwards” due to the centrifugal force. So we permanently feel an “upwards” pulling force thanks to the Earth’s rotation. However, the centrifugal force is much weaker than the centri petal force, which is directed towards the core of the planet and usually called “gravitation”. If this wasn’t the case, we would have serious problems holding our bodies down to the ground. (The ground, too, would have troubles holding itself “to the ground”.)

Especially on the equator, the centrifugal and the gravitational force are antagonistic forces: the one points “downwards” while the other points “upwards”.

How fast would the Earth have to spin in order to cause weightlessness at the…

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Listening To Vermilion Sands

My Beloved is reading J G Ballard’s Vermillion Sands; early in one of the book’s stories is a character wondering if an odd sound comes from one of the musical … let’s call it instruments, one with a 24-octave range. We both thought, wow, that’s a lot of range. Is it a range any instrument could have?

As we weren’t near our computers this turned into a mental arithmetic problem. It’s solvable in principle because, if you know the frequency of one note, then you know the frequency of its counterpart one octave higher (it’s double that), and one octave lower (it’s half that). It’s not solvable, at this point, because we don’t have any information about what the range is supposed to be. So here’s roughly how we worked it out.

The note A above middle C is 440 Hertz, or at least you can use that for tuning ever since the International Standards Organization set that as a tuning standard in 1953. (As with any basically arbitrary standard this particular choice is debatable, although, goodness but this page advocating a 432 Hertz standard for A doesn’t do itself any favors by noting that “440 Hz is the unnatural standard tuning frequency, removed from the symmetry of sacred vibrations and overtones that has declared war on the subconscious mind of Western Man” and, yes, Nikola Tesla and Joseph Goebbels turn up in the article because you might otherwise imagine taking it seriously.) Anyway, it doesn’t matter; 440 is just convenient as it’s a number definitely in hearing range.

So I’m adding the assumption that 440 Hz is probably in the instrument’s range. And I’ll work on the assumption that it’s right in the middle of the range, that is, that we should be able to go down twelve octaves and up twelve octaves, and see if that assumption leads me to any problems. And now I’ve got the problem defined well enough to answer: is 440 divided by two to the twelfth power in human hearing range? Is 440 times two to the twelfth power in range?

I’m not dividing 440 by two a dozen times; I might manage that with pencil and paper but not in my head. But I also don’t need to. Two raised to the tenth power is pretty close to 1,000, as anyone who’s noticed that the common logarithm of two is 0.3 could work out. Remembering a couple approximations like that are key to doing any kind of real mental arithmetic; it’s all about turning the problem you’re interested in into one you can do without writing it down.

Another key to this sort of mental arithmetic is noticing that two to the 12th power is equal to two to the second power (that is, four) times two to the tenth power (approximately 1,000). In algebra class this was fed to you as something like “a^{x + y} = (a^x)(a ^y)”, and it’s the trick that makes logarithms a concept that works.

Getting back to the question, 440 divided by two twelve times over is going to be about 440 divided by 4,000, which is going to be close enough to one-tenth Hertz. There’s no point working it out to any more exact answer, since this is definitely below the range of human hearing; I think the lower bound is usually around ten to thirty Hertz.

Well, no matter; maybe the range of the instrument starts higher up and keeps on going. To see if there’s any room, what’s the frequency of a note twelve octaves above the 440-Hertz A?

That’s going to be 440 Hertz times 4,000, which to make it simpler I’ll say is something more than 400 times 4000. The four times four is easy, and there’s five zeroes in there, so, that suggests an upper range on the high side of 1,600,000 Hertz. Again, I’m not positive the upper limit of human hearing but I’m confident it’s not more than about 30,000 Hertz, and I leave space below for people who know what it is exactly to say. There’s just no fitting 24 octaves into the human hearing range.

So! Was Ballard just putting stuff into his science fiction story without checking whether the numbers make that plausible, if you can imagine a science fiction author doing such a thing?

It’s conceivable. It’s also possible Ballard was trying to establish the character was a pretentious audiophile snob who imagines himself capable of hearing things that no, in fact, can’t be discerned. However, based on the setting … the instruments producing music in this story (and other stories in the book), set in the far future, include singing plants and musical arachnids and other things that indicate not just technology but biology has changed rather considerably. If it’s possible to engineer a lobster that can sing over a 24 octave range, it’s presumably possible to engineer a person who can listen to it.

The Liquefaction of Gases – Part I

I know, or at least I’m fairly confident, there’s a couple readers here who like deeper mathematical subjects. It’s fine to come up with simulated Price is Right games or figure out what grades one needs to pass the course, but those aren’t particularly challenging subjects.

But those are hard to write, so, while I stall, let me point you to CarnotCycle, which has a nice historical article about the problem of liquefaction of gases, a problem that’s not just steeped in thermodynamics but in engineering. If you’re a little familiar with thermodynamics you likely won’t be surprised to see names like William Thomson, James Joule, or Willard Gibbs turn up. I was surprised to see in the additional reading T O’Conor Sloane show up; science fiction fans might vaguely remember that name, as he was the editor of Amazing Stories for most of the 1930s, in between Hugo Gernsback and Raymond Palmer. It’s often a surprising world.

carnotcycle

On Monday 3 December 1877, the French Academy of Sciences received a letter from Louis Cailletet, a 45 year-old physicist from Châtillon-sur-Seine. The letter stated that Cailletet had succeeded in liquefying both carbon monoxide and oxygen.

Liquefaction as such was nothing new to 19th century science, it should be said. The real news value of Cailletet’s announcement was that he had liquefied two gases previously considered ‘non condensable’.

While a number of gases such as chlorine, carbon dioxide, sulfur dioxide, hydrogen sulfide, ethylene and ammonia had been liquefied by the simultaneous application of pressure and cooling, the principal gases comprising air – nitrogen and oxygen – together with carbon monoxide, nitric oxide, hydrogen and helium, had stubbornly refused to liquefy, despite the use of pressures up to 3000 atmospheres. By the mid-1800s, the general opinion was that these gases could not be converted into liquids under any circumstances.

But in…

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Why I Don’t Believe It’s 1/e

The above picture, showing the Leap-the-Dips roller coaster at Lakemont Park before its renovation, kind of answers why despite my neat reasoning and mental calculations I don’t really believe that there’s a chance of something like one in three that any particular board from the roller coaster’s original, 1902, construction is still in place. The picture — from the end of the track, if I’m not mistaken — dates to shortly before the renovation of the roller coaster began in the late 90s. Leap-the-Dips had stood without operating, and almost certainly without maintenance, from 1986 (coincidental to the park’s acquisition by the Boyer Candy company and its temporary renaming as Boyertown USA, in miniature imitation of Hershey Park) to 1998.

The result of this period seems almost to demand replacing every board in the thing. But we don’t know that happened, and after all, surely some boards took it better than others, didn’t they? Not every board was equally exposed to the elements, or to vandalism, or to whatever does smash up wood. And there’s a lot of pieces of wood that go into a wooden roller coaster. Surely some were lucky by virtue of being in the right spot?