## Someone Else’s Homework: A Probability Question

My friend’s finished the last of the exams and been happy with the results. And I’m stuck thinking harder about a little thing that came across my Twitter feed last night. So let me share a different problem that we had discussed over the term.

It’s a probability question. Probability’s a great subject. So much of what people actually do involves estimating probabilities and making judgements based on them. In real life, yes, but also for fun. Like a lot of probability questions, this one is abstracted into a puzzle that’s nothing like anything anybody does for fun. But that makes it practical, anyway.

So. You have a bowl with fifteen balls inside. Five of the balls are labelled ‘1’. Five of the balls are labelled ‘2’. Five of the balls are labelled ‘3’. The balls are well-mixed, which is how mathematicians say that all of the balls are equally likely to be drawn out. Three balls are picked out, without being put back in. What’s the probability that the three balls have values which, together, add up to 6?

My friend’s instincts about this were right, knowing what things to calculate. There was part of actually doing one of these calculations that went wrong. And was complicated by my making a dumb mistake in my arithmetic. Fortunately my friend wasn’t shaken by my authority, and we got to what we’re pretty sure is the right answer.

## 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.

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.

## A Bunch Of Tweets I’d Thought To Save

I’m slow about sharing them is all. It’s a simple dynamic: I want to write enough about each tweet that it’s interesting to share, and then once a little time has passed, I need to do something more impressive to be worth the wait. Eventually, nothing is ever shared. Let me try to fix that.

Just as it says: a link to Leonhard Euler’s Elements of Algebra, as rendered by Google Books. Euler you’ll remember from every field of mathematics ever. This 1770 textbook is one of the earliest that presents algebra that looks like, you know, algebra, the way we study it today. Much of that is because this book presented algebra so well that everyone wanted to imitate it.

An entry in the amusing and novel proofs. This one is John Conway’s candidate for most succinct published mathematics paper. It’s fun, at least as I understand fun to be.

This Theorem of the Day from back in November already is one about elliptic functions. Those came up several times in the Summer 2017 Mathematics A To Z. This day about the Goins-Maddox-Rusin Theorem on Heron Triangles, is dense reading even by the standards of the Theorem of the Day tweet (which fits each day’s theorem into a single slide). Still, it’s worth lounging about in the mathematics.

Elke Stangl, writing about one of those endlessly-to-me interesting subjects: phase space. This is a particular way of representing complicated physical systems. Set it up right and all sorts of physics problems become, if not easy, at least things there’s a standard set of tools for. Thermodynamics really encourages learning about such phase spaces, and about entropy, and here she writes about some of this.

So ‘e’ is an interesting number. At least, it’s a number that’s got a lot of interesting things built around it. Here, John Golden points out a neat, fun, and inefficient way to find the value of ‘e’. It’s kin to that scheme for calculating π inefficiently that I was being all curmudgeonly about a couple of Pi Days ago.

Jo Morgan comes to the rescue of everyone who tries to read old-time mathematics. There were a lot of great and surprisingly readable great minds publishing in the 19th century, but then you get partway through a paragraph and it might as well be Old High Martian with talk about diminishings and consequents and so on. So here’s some help.

As it says on the tin: a textbook on partial differential equations. If you find yourself adrift in the subject, maybe seeing how another author addresses the same subject will help, if nothing else for finding something familiar written in a different fashion.

And this is just fun: creating an ellipse as the locus of points that are never on the fold line when a circle’s folded by a particular rule.

Finally, something whose tweet origin I lost. It was from one of the surprisingly many economists I follow considering I don’t do financial mathematics. But it links to a bit of economic history: Origins of the Sicilian Mafia: The Market for Lemons. It’s 31 pages plus references. And more charts about wheat production in 19th century Sicily than I would have previously expected to see.

By the way, if you’re interested in me on Twitter, that would be @Nebusj. Thanks for stopping in, should you choose to.

## Reading the Comics, January 3, 2018: Explaining Things Edition

There were a good number of mathematically-themed comic strips in the syndicated comics last week. Those from the first part of the week gave me topics I could really sink my rhetorical teeth into, too. So I’m going to lop those off into the first essay for last week and circle around to the other comics later on.

Jef Mallett’s Frazz started a week of calendar talk on the 31st of December. I’ve usually counted that as mathematical enough to mention here. The 1st of January as we know it derives, as best I can figure, from the 1st of January as Julius Caesar established for 45 BCE. This was the first Roman calendar to run basically automatically. Its length was quite close to the solar year’s length. It had leap days added according to a rule that should have been easy enough to understand (one day every fourth year). Before then the Roman calendar year was far enough off the solar year that they had to be kept in synch by interventions. Mostly, by that time, adding a short extra month to put things more nearly right. This had gotten all confusingly messed up and Caesar took the chance to set things right, running 46 BCE to 445 days long.

But why 445 and not, say, 443 or 457? And I find on research that my recollection might not be right. That is, I recall that the plan was to set the 1st of January, Reformed, to the first new moon after the winter solstice. A choice that makes sense only for that one year, but, where to set the 1st is literally arbitrary. While that apparently passes astronomical muster (the new moon as seen from Rome then would be just after midnight the 2nd of January, but hitting the night of 1/2 January is good enough), there’s apparently dispute about whether that was the objective. It might have been to set the winter solstice to the 25th of December. Or it might have been that the extra days matched neatly the length of two intercalated months that by rights should have gone into earlier years. It’s a good reminder of the difficulty of reading motivation.

Brian Fies’s The Last Mechanical Monster for the 1st of January, 2018, continues his story about the mad scientist from the Fleischer studios’ first Superman cartoon, back in 1941. In this panel he’s describing how he realized, over the course of his long prison sentence, that his intelligence was fading with age. He uses the ability to do arithmetic in his head as proof of that. These types never try naming, like, rulers of the Byzantine Empire. Anyway, to calculate the cube root of 50,653 in his head? As he used to be able to do? … guh. It’s not the sort of mental arithmetic that I find fun.

But I could think of a couple ways to do it. The one I’d use is based on a technique called Newton-Raphson iteration that can often be used to find where a function’s value is zero. Raphson here is Joseph Raphson, a late 17th century English mathematician known for the Newton-Raphson method. Newton is that falling-apples fellow. It’s an iterative scheme because you start with a guess about what the answer would be, and do calculations to make the answer better. I don’t say this is the best method, but it’s the one that demands me remember the least stuff to re-generate the algorithm. And it’ll work for any positive number ‘A’ and any root, to the ‘n’-th power.

So you want the n-th root of ‘A’. Start with your current guess about what this root is. (If you have no idea, try ‘1’ or ‘A’.) Call that guess ‘x’. Then work out this number:

$\frac{1}{n}\left( (n - 1) \cdot x + \frac{A}{x^{n - 1}} \right)$

Ta-da! You have, probably, now a better guess of the n-th root of ‘A’. If you want a better guess yet, take the result you just got and call that ‘x’, and go back calculating that again. Stop when you feel like your answer is good enough. This is going to be tedious but, hey, if you’re serving a prison term of the length of US copyright you’ve got time. (It’s possible with this sort of iterator to get a worse approximation, although I don’t think that happens with n-th root process. Most of the time, a couple more iterations will get you back on track.)

But that’s work. Can we think instead? Now, most n-th roots of whole numbers aren’t going to be whole numbers. Most integers aren’t perfect powers of some other integer. If you think 50,653 is a perfect cube of something, though, you can say some things about it. For one, it’s going to have to be a two-digit number. 103 is 1,000; 1003 is 1,000,000. The second digit has to be a 7. 73 is 343. The cube of any number ending in 7 has to end in 3. There’s not another number from 1 to 9 with a cube that ends in 3. That’s one of those things you learn from playing with arithmetic. (A number ending in 1 cubes to something ending in 1. A number ending in 2 cubes to something ending in 8. And so on.)

So the cube root has to be one of 17, 27, 37, 47, 57, 67, 77, 87, or 97. Again, if 50,653 is a perfect cube. And we can do better than saying it’s merely one of those nine possibilities. 40 times 40 times 40 is 64,000. This means, first, that 47 and up are definitely too large. But it also means that 40 is just a little more than the cube root of 50,653. So, if 50,653 is a perfect cube, then it’s most likely going to be the cube of 37.

Bill Watterson’s Calvin and Hobbes rerun for the 2nd is a great sequence of Hobbes explaining arithmetic to Calvin. There is nothing which could be added to Hobbes’s explanation of 3 + 8 which would make it better. I will modify Hobbes’s explanation of what the numerator. It’s ridiculous to think it’s Latin for “number eighter”. The reality is possibly more ridiculous, as it means “a numberer”. Apparently it derives from “numeratus”, meaning, “to number”. The “denominator” comes from “de nomen”, as in “name”. So, you know, “the thing that’s named”. Which does show the terms mean something. A poet could turn “numerator over denominator” into “the number of parts of the thing we name”, or something near enough that.

Hobbes continues the next day, introducing Calvin to imaginary numbers. The term “imaginary numbers” tells us their history: they looked, when first noticed in formulas for finding roots of third- and fourth-degree polynomials, like obvious nonsense. But if you carry on, following the rules as best you can, that nonsense would often shake out and you’d get back to normal numbers again. And as generations of mathematicians grew up realizing these acted like numbers we started to ask: well, how is an imaginary number any less real than, oh, the square root of six?

Hobbes’s particular examples of imaginary numbers — “eleventenn” and “thirty-twelve” — are great-sounding compositions. They put me in mind, as many of Watterson’s best words do, of a 1960s Peanuts in which Charlie Brown is trying to help Sally practice arithmetic. (I can’t find it online, as that meme with edited text about Sally Brown and the sixty grapefruits confounds my web searches.) She offers suggestions like “eleventy-Q” and asks if she’s close, which Charlie Brown admits is hard to say.

And finally, James Allen’s Mark Trail for the 3rd just mentions mathematics as the subject that Rusty Trail is going to have to do some work on instead of allowing the experience of a family trip to Mexico to count. This is of extremely marginal relevance, but it lets me include a picture of a comic strip, and I always like getting to do that.

## Reading the Comics, December 9, 2017: Zach Weinersmith Wants My Attention Edition

If anything dominated the week in mathematically-themed comic strips it was Zach Weinersmith’s Saturday Morning Breakfast Cereal. I don’t know how GoComics selects the strips to (re?)print on their site. But there were at least four that seemed on-point enough for me to mention. So, okay. He’s got my attention. What’s he do with it?

On the 3rd of December is a strip I can say is about conditional probability. The mathematician might be right that the chance someone will be murdered by a serial killer are less than one in ten million. But that is the chance of someone drawn from the whole universe of human experiences. There are people who will never be near a serial killer, for example, or who never come to his attention or who evade his interest. But if we know someone is near a serial killer, or does attract his interest? The information changes the probability. And this is where you get all those counter-intuitive and somewhat annoying logic puzzles about, like, the chance someone’s other child is a girl if the one who just walked in was, and how that changes if you’re told whether the girl who just entered was the elder.

On the 5th is a strip about sequences. And built on the famous example of exponential growth from doubling a reward enough times. Well, you know these things never work out for the wise guy. The “Fibonacci Spiral” spoken of in the next-to-last panel is a spiral, like you figure. The dimensions of the spiral are based on those of golden-ratio rectangles. It looks a great deal like a logarithmic spiral to the untrained eye. Also to the trained eye, but you knew that. I think it’s supposed to be humiliating that someone would call such a spiral “random”. But I admit I don’t get that part.

The strip for the 6th has a more implicit mathematical content. It hypothesizes that mathematicians, given the chance, will be more interested in doing recreational puzzles than even in eating and drinking. It’s amusing, but I’ll admit I’ve found very few puzzles all that compelling. This isn’t to say there aren’t problems I keep coming back to because I’m curious about them, just that they don’t overwhelm my common sense. Don’t ask me when I last received actual pay for doing something mathematical.

And then on the 9th is one more strip, about logicians. And logic puzzles, such as you might get in a Martin Gardner collection. The problem is written out on the chalkboard with some shorthand logical symbols. And they’re symbols both philosophers and mathematicians use. The letter that looks like a V with a crossbar means “for all”. (The mnemonic I got was “it’s an A-for-all, upside-down”. This paired with the other common symbol, which looks like a backwards E and means there exists: “E-for-exists, backwards”. Later I noticed upside-down A and backwards E could both be just 180-degree-rotated A and E. But try saying “180-degree-rotated” in a quick way.) The curvy E between the letters ‘x’ and ‘S’ means “belongs to the set”. So that first line says “for all x that belong to the set S this follows”. Writing out “isLiar(x)” instead of, say, “L(x)”, is more a philosopher’s thing than a mathematician’s. But it wouldn’t throw anyway. And the T just means emphasizing that this is true.

And that is as much about Saturday Morning Breakfast Cereal as I have to say this week.

Sam Hurt’s Eyebeam for the 4th tells a cute story about twins trying to explain infinity to one another. I’m not sure I can agree with the older twin’s assertion that infinity means there’s no biggest number. But that’s just because I worry there’s something imprecise going on there. I’m looking forward to the kids learning about negative numbers, though, and getting to wonder what’s the biggest negative real number.

Percy Crosby’s Skippy for the 4th starts with Skippy explaining a story problem. One about buying potatoes, in this case. I’m tickled by how cranky Skippy is about boring old story problems. Motivation is always a challenge. The strip originally ran the 7th of October, 1930.

Dave Whamond’s Reality Check for the 6th uses a panel of (gibberish) mathematics as an example of an algorithm. Algorithms are mathematical, in origin at least. The word comes to us from the 9th century Persian mathematician Al-Khwarizmi’s text about how to calculate. The modern sense of the word comes from trying to describe the methods by which a problem can be solved. So, legitimate use of mathematics to show off the idea. The symbols still don’t mean anything.

Rick Detorie’s One Big Happy for the 7th has Joe trying to get his mathematics homework done at the last minute. … And it’s caused me to reflect on how twenty multiplication problems seems like a reasonable number to do. But there’s only fifty multiplications to even do, at least if you’re doing the times tables up to the 10s. No wonder students get so bored seeing the same problems over and over. It’s a little less dire if you’re learning times tables up to the 12s, but not that much better. Yow.

Olivia Walch’s Imogen Quest for the 8th looks pretty legitimate to me. It’s going to read as gibberish to people who haven’t done parametric functions, though. Start with the plane and the familiar old idea of ‘x’ and ‘y’ representing how far one is along a horizontal and a vertical direction. Here, we’re given a dummy variable ‘t’, and functions to describe a value for ‘x’ and ‘y’ matching each value of ‘t’. The plot then shows all the points that ever match a pair of ‘x’ and ‘y’ coordinates for some ‘t’. The top drawing is a shape known as the cardioid, because it kind of looks like a Valentine-heart. The lower figure is a much more complicated parametric equation. It looks more anatomically accurate,

Still no sign of Mark Anderson’s Andertoons and the drought is worrying me, yes.

But they’re still going on the cartoonist’s web site, so there’s that.

## Reading the Comics, November 8, 2017: Uses Of Mathematics Edition

Was there an uptick in mathematics-themed comic strips in the syndicated comics this past week? It depends how tight a definition of “theme” you use. I have enough to write about that I’m splitting the week’s load. And I’ve got a follow-up to that Wronski post the other day, so I’m feeling nice and full of content right now. So here goes.

Zach Weinersmith’s Saturday Morning Breakfast Cereal posted the 5th gets my week off to an annoying start. Science and mathematics and engineering people have a tendency to be smug about their subjects. And to see aptitude or interest in their subjects as virtue, or at least intelligence. (If they see a distinction between virtue and intelligence.) To presume that an interest in the field I like is a demonstration of intelligence is a pretty nasty and arrogant move.

And yes, I also dislike the attitude that school should be about training people. Teaching should be about letting people be literate with the great thoughts people have had. Mathematics has a privileged spot here. The field, as we’ve developed it, seems to build on human aptitudes for number and space. It’s easy to find useful sides to it. Doesn’t mean it’s vocational training.

Lincoln Peirce’s Big Nate on the 6th discovered mathematics puzzles. And this gave him the desire to create a new mathematical puzzle that he would use to get rich. Good luck with that. Coming up with interesting enough recreational mathematics puzzles is hard. Presenting it in a way that people will buy is another, possibly greater, challenge. It takes luck and timing and presentation, just as getting a hit song does. Sudoku, for example, spent five years in the Dell Magazine puzzle books before getting a foothold in Japanese newspapers. And then twenty years there before being noticed in the English-speaking puzzle world. Big Nate’s teacher tries to encourage him, although that doesn’t go as Mr Staples might have hoped. (The storyline continues to the 11th. Spoiler: Nate does not invent the next great recreational mathematics puzzle.)

Jef Mallett’s Frazz for the 7th start out in a mathematics class, at least. I suppose the mathematical content doesn’t matter, though. Mallett’s making a point about questions that, I confess, I’m not sure I get. I’ll leave it for wiser heads to understand.

Mike Thompson’s Grand Avenue for the 8th is a subverted word-problem joke. And I suppose a reminder about the need for word problems to parse as things people would do, or might be interested in. I can’t go along with characterizing buying twelve candy bars “gluttonous” though. Not if they’re in a pack of twelve or something like that. I may be unfair to Grand Avenue. Mind, until a few years ago I was large enough my main method of getting around was “being rolled by Oompa-Loompas”, so I could be a poor judge.

Keith Tutt and Daniel Saunders’s Lard’s World Peace Tips for the 8th does a rounding joke. It’s not much, but I’ve included appearances of this joke before and it seems unfair to skip it this time.

## Calculus For Breakfast

Robert Austin, of the RobertLovesPi blog, got to thinking about one of those interesting mathematics problems. It starts with the equations that describe the volume and the surface area of a sphere.

If the sphere has radius r, then the surface area of the sphere is 4πr2. And the volume is (4/3)πr3. What’s interesting about this is that there’s a relationship between these two expressions. The first is the derivative of the second. The derivative is one of the earliest things one learns in calculus. It describes how much a quantity changes with a tiny change in something it depends on.

And this got him to thinking about the surface area of a cube. Call the length of a cube’s side s. Its surface is six squares, each of them with a side of length s. So the surface area of each of the six squares is s2, which is obvious when you remember we call raising something to the second power “squaring”. Its total surface area then is 6s2. But its volume is is s3. This is why we even call raising something to the third power “cubing”. And the derivative of s3 is 3s2. (If you don’t know calculus, but you suspect you see a pattern here, you’re learning calculus. If you’re not sure about the pattern, let me tell you that the derivative of s4 would be 4s3, and the derivative of (1/3)s2 would be (2/3)s.)

There’s an obvious flaw there, and Austin’s aware of it. But it got him pondering different ways to characterize how big a cube is. He can find one that makes the relationship between volume and surface area work out like he expects. But the question remains, why that? And what about other shapes?

I think that’s an interesting discussion to have, and mean to think about it some more myself. And I wanted to point people who’d be interested over there to join in.

## Proportional Dice

So, here’s a nice probability problem that recently made it to my Twitter friends page:

(By the way, I’m @Nebusj on Twitter. I’m happy to pick up new conversational partners even if I never quite feel right starting to chat with someone.)

Schmidt does assume normal, ordinary, six-sided dice for this. You can work out the problem for four- or eight- or twenty- or whatever-sided dice, with most likely a different answer.

But given that, the problem hasn’t quite got an answer right away. Reasonable people could disagree about what it means to say “if you roll a die four times, what is the probability you create a correct proportion?” For example, do you have to put the die result in a particular order? Or can you take the four numbers you get and arrange them any way at all? This is important. If you have the numbers 1, 4, 2, and 2, then obviously 1/4 = 2/2 is false. But rearrange them to 1/2 = 2/4 and you have something true.

We can reason this out. We can work out how many ways there are to throw a die four times, and so how many different outcomes there are. Then we count the number of outcomes that give us a valid proportion. That count divided by the number of possible outcomes is the probability of a successful outcome. It’s getting a correct count of the desired outcomes that’s tricky.

## Any Requests?

I’m thinking to do a second Mathematics A-To-Z Glossary. For those who missed it, last summer I had a fun string of several weeks in which I picked a mathematical term and explained it to within an inch of its life, or 950 words, whichever came first. I’m curious if there’s anything readers out there would like to see me attempt to explain. So, please, let me know of any requests. All requests must begin with a letter, although numbers might be considered.

Meanwhile since there’s been some golden ratio talk around these parts the last few days, I thought people might like to see this neat Algebra Fact of the Day:

People following up on the tweet pointed out that it’s technically speaking wrong. The idea can be saved, though. You can produce the golden ratio using exactly four 4’s this way:

$\phi = \frac{\cdot\left(\sqrt{4} + \sqrt{4! + 4}\right)}{4}$

If you’d like to do it with eight 4’s, here’s one approach:

And this brings things back around to how Paul Dirac worked out a way to produce any whole number using exactly four 2’s and the normal arithmetic operations anybody knows.

## Silver-Leafed Numbers

In a comment on my “Gilded Ratios” essay fluffy wondered about a variation on the Golden and Golden-like ratios. What’s interesting about the Golden Ratio and similar numbers is that their reciprocal — one divided by them — is a whole number less than the original number. That is, 1 divided by 1.618(etc) is 0.618(etc), which is 1 less than the original number. 1 divided by 2.414(etc) is 0.414(etc), exactly 2 less than the original 2.414(etc). 1 divided by 3.302(etc) is 0.302(etc), exactly 3 less than the original 3.302(etc).

fluffy wondered about a variation. Is there some number x that’s exactly 2 less than 2 divided by x? Or a (presumably) differently number that’s exactly 3 less than 3 divided by it? Yes, there is.

Let me call the whole number difference — the 1 or 2 or 3 or so on, referred to above — by the name b. And let me call the other number — the one that’s b less than b divided by it — by the name x. Then a number x, for which b divided by x is exactly b less than itself, makes true the equation $\frac{b}{x} = x - b$. This is slightly different from the equation used last time, but not very different. Multiply both sides by x, which we know not to be zero, and we get a polynomial.

Yes, quadratic formula, I see you waving your hand in the back there. And you’re right. There are two x’s that will make that equation true. The positive one is $x = \frac12\left( b + \sqrt{b^2 + 4b} \right)$. The negative one you get by changing the + sign, just before the square root, to a – sign, but who cares about that root? Here’s the first several of the (positive) silver-leaf ratios:

Some More Numbers With Cute Reciprocals
Number Silver-Leaf
1 1.618033989
2 2.732050808
3 3.791287847
4 4.828427125
5 5.854101966
6 6.872983346
7 7.887482194
8 8.898979486
9 9.908326913
10 10.916079783
11 11.922616289
12 12.928203230
13 13.933034374
14 14.937253933
15 15.940971508
16 16.944271910
17 17.947221814
18 18.949874371
19 19.952272480
20 20.954451150

Looking over those hypnotic rows of digits past the decimal inspires thoughts. The part beyond the decimal keeps rising, closer and closer to 1. Does it ever get past 1? That is, might (say) the silver-leaf number that’s 2,038 more than its reciprocal be 2,039.11111 (or something)?

No, it never does. There are a couple of ways to prove that, if you feel like. We can take the approach that’s easiest (to my eyes) to imagine. It takes a little algebraic grinding to complete. That is to look for the smallest number b for which the silver-leaf number, $\frac12\left(b + \sqrt{b^2 + 4b}\right)$, is larger than $b + 1$. Follow that out and you realize that it’s any value of b for which 0 is greater than 4. Logically, therefore, we need to take b into a private room and have a serious talk about its job performance, what with it not existing.

A harder proof to imagine working out, but that takes no symbol manipulation, comes from thinking about these reciprocals. Let’s imagine we had some b for which its corresponding silver-leaf number x is more than b + 1. Then, x – b has to be greater than 1. But if x is greater than 1, then its reciprocal has to be less than 1. We again have to talk with b about how its nonexistence is keeping it from doing its job.

Are there other proofs? Most likely. I was satisfied by this point, and resolved not to work on it more until the shower. Updates after breakfast, I suppose.

## Gilded Ratios

I may have mentioned that I regard the Golden Ratio as a lot of bunk. If I haven’t, allow me to mention: the Golden Ratio is a lot of bunk. I concede it’s a cute number. I found it compelling when I first had a calculator that let me use the last answer for a new operation. You can pretty quickly find that 1.618033 (etc, and the next digit is a 9 by the way) has a reciprocal that’s 0.618033 (etc).

There’s no denying that. And there’s no denying that’s a neat pattern. But it is not some aesthetic ideal. When people evaluate rectangles that “look best” they go to stuff that’s a fair but not too much wider in one direction than the other. But people aren’t drawn to 1.618 (etc) any more reliably than they like 1.6, or 1.8, or 1.5, or other possible ratios. And it is not any kind of law of nature that the Golden Ratio will turn up. It’s often found within the error bars of a measurement, but so are a lot of numbers.

The Golden Ratio is an irrational number, but basically all real numbers are irrational except for a few peculiar ones. Those peculiar ones happen to be the whole numbers and the rational numbers, which we find interesting, but which are the rare exception. It’s not a “transcendental number”, which is a kind of real number I don’t want to describe here. That’s a bit unusual, since basically all real numbers are transcendental numbers except for a few peculiar ones. Those peculiar ones include whole and rational numbers, and square roots and such, which we use so much we think they’re common. But not being transcendental isn’t that outstanding a feature. The Golden Ratio is one of those strange celebrities who’s famous for being a celebrity, and not for any actual accomplishment worth celebrating.

I started wondering: are there other Golden-Ratio-like numbers, though? The title of this essay gives what I suppose is the best name for this set. The Golden Ratio is interesting because its reciprocal — 1 divided by it — is equal to it minus 1. Is there another number whose reciprocal is equal to it minus 2? Another number yet whose reciprocal is equal to it minus 3?

So I looked. Is there a number between 2 and 3 whose reciprocal is it minus 2? Certainly there is. How do I know this?

Let me call this number, if it exists, x. The reciprocal of x is the number 1/x. The number x minus 2 is the number x – 2. We’ll pick up the pace in a little bit. Now imagine trying out every single number from 2 to 3, in order. The reciprocals 1/x start out at 1/2 and drop to 1/3. The subtracted numbers start out at 0 and grow to 1. There’s no gaps or sudden jumps or anything in either the reciprocals or the subtracted numbers. So there must be some x for which 1/x and x – 2 are the same number.

In the trade we call that an existence proof. It shows there’s got to be some answer. It doesn’t tell us much about what the answer is. Often it’s worth looking for an existence proof first. In this case, it’s probably overkill. But you can go from this to reasoning that there have to be Golden-Like-Ratio numbers between any two counting numbers. So, yes, there’s some number between 2,038 and 2,039 whose reciprocal is that number minus 2,038. That’s nice to know.

So what is the number that’s two more than its reciprocal? That’s whatever number or numbers make true the equation $\frac{1}{x} = x - 2$. That’s straightforward to solve. Multiply both sides by x, which won’t change whether the equation is true unless x is zero. (And x can’t be zero, or else we wouldn’t talk of 1/x except in hushed, embarrassed whispers.) This gets an equivalent equation $1 = x^2 - 2x$. Subtract 1 from both sides, and we get $0 = x^2 - 2x - 1$ and we’re set up to use the quadratic formula. The answer will be $x = \left(\frac{1}{2}\right)\cdot\left(2 + \sqrt{2^2 + 4}\right)$. The answer is about 2.414213562373095 (and on). (No, $\left(\frac{1}{2}\right)\cdot\left(2 - \sqrt{2^2 + 4}\right)$ is not an answer; it’s not between 2 and 3.)

The number that’s three more than its reciprocal? We’ll call that x again, trusting that we remember this is a different number with the same name. For that we need to solve $\frac{1}{x} = x - 3$ and that turns into the equation $0 = x^2 - 3x - 1$. And so $x = \left(\frac{1}{2}\right)\cdot\left(3 + \sqrt{3^2 + 4}\right)$ and so it’s about 3.30277563773200. Yes, there’s another possible answer we rule out because it isn’t between 3 and 4.

We can do the same thing to find another number, named x, that’s four more than its reciprocal. That starts with $\frac{1}{x} = x - 4$ and gets eventually to $x = \left(\frac{1}{2}\right)\cdot\left(4 + \sqrt{4^2 + 4}\right)$ or about 4.23606797749979. We could go on like this. The number x that’s 2,038 more than its reciprocal is $x = \left(\frac{1}{2}\right)\cdot\left(2038 + \sqrt{2038^2 + 4}\right)$ about 2038.00049082160.

If your eyes haven’t just slid gently past the equations you noticed the pattern. Suppose instead of saying 2 or 3 or 4 or 2038 we say the number b. b is some whole number, any that we like. The number whose reciprocal is exactly b less than it is the number x that makes true the equation $\frac{1}{x} = x - b$. And that leads to the finding the number that makes the equation $x = \left(\frac{1}{2}\right)\cdot\left(b + \sqrt{b^2 + 4}\right)$ true.

And, what the heck. Here’s the first twenty or so gilded numbers. You can read this either as a list of the numbers I’ve been calling x — 1.618034, 2.414214, 3.302776 — or as an ordered list of the reciprocals of x — 0.618034, 0.414214, 0.302276 — as you like. I’ll call that the gilt; you add it to the whole number to its left to get that a number that, cutely, has a reciprocal that’s the same after the decimal.

I did think about including a graph of these numbers, but the appeal of them is that you can take the reciprocal and see digits not changing. A graph doesn’t give you that.

Some Numbers With Cute Reciprocals
Number Gilt
1 .618033989
2 .414213562
3 .302775638
4 .236067977
5 .192582404
6 .162277660
7 .140054945
8 .123105626
9 .109772229
10 .099019514
11 .090169944
12 .082762530
13 .076473219
14 .071067812
15 .066372975
16 .062257748
17 .058621384
18 .055385138
19 .052486587
20 .049875621

None of these are important numbers. But they are pretty, and that can be enough on a quiet day.

## Spaghetti Mathematics

Let’s ease into Monday. Win Smith with the Well Tempered Spreadsheet blog encountered one of those idle little puzzles that captures the imagination and doesn’t let go. It starts as many will with spaghetti.

I’m sure you’re intrigued too. It’s not the case that any old splitting of a strand of spaghetti will give you three pieces you can make into a triangle. You need the lengths of the three pieces to satisfy what’s imaginatively called the Triangle Inequality. That inequality demands the lengths of any two sides have to be greater than the length of the third side. So, suppose we start with spaghetti that’s 12 inches long, and we have it cut into pieces 5, 4, and 3 inches long: that’s fine. If we have it cut into pieces 9, 2, and 1 inches long, we’re stuck.

The Triangle Inequality is often known as the Cauchy Inequality, or the Cauchy-Schwarz Inequality, or the Cauchy-Bunyakovsky-Schwarz Inequality, or if that’s getting too long the CBS Inequality. And some pranksters reorder it to the Cauchy-Schwarz-Bunyakovsky Inequality. The Cauchy (etc) Inequality isn’t quite the same thing as the Triangle Inequality. But it’s an important and useful idea, and the Cauchy (etc) Inequality has the Triangle Inequality as one of its consequences. The name of it so overflows with names because mathematics history is complicated. Augustin-Louis Cauchy published the first proof of it, but for the special case of sums of series. Viktor Bunyakovsky proved a similar version of it for integrals, and has a name that’s so very much fun to say. Hermann Amandus Schwarz first put the proof into its modern form. So who deserves credit for it? Good question. If it influences your decision, know that Cauchy was incredibly prolific and has plenty of things named for him already. He’s got, without exaggeration, about eight hundred papers to his credit. Collecting all his work into a definitive volume took from 1882 to 1974.

Back to the spaghetti. The problem’s a fun one and if you follow the Twitter link above you’ll see the gritty work of mathematicians reasoning out the problem. As with every probability problem ever, the challenge is defining exactly what you’re looking for the probability of. This we call finding the “sample space”, the set of all the possible outcomes and how likely each of them are. Subtle changes in how you think of the problem will change whether you are right.

Smith cleans things up a bit, but preserves the essence of how the answer worked out. The answer that looks most likely correct was developed partly by reasoning and partly by numerical simulation. Numerical simulation is a great blessing for probability problems. Often the easiest way to figure out how likely something is will be trying it. But this does require working out the sample space, and what parts of the sample space represent what you’re interested in. With the information the numerical simulation provided, Smith was able to go back and find an analytic, purely reason-based, answer that looks plausible.

## Another Reasoning Puzzle From ChefMongoose

My friend ChefMongoose had another reasoning problem come to him, and I’m happy to share it further. It’s rather like that famous Singapore Birthday Problem that drove people crazy a couple of months ago. Here’s the problem:

I have a combination lock at work. There are three digits, all in the range 1 – 40; they’re all prime numbers. They’re X+Y, X+2Y, X+3Y — where X and Y are positive integers.

If I told you what X was but not Y, you wouldn’t be able to tell me the combination. If I told you what Y was but not X, you wouldn’t be able to tell me the combination. Now, what’s the combination?

I did work out the puzzle. It did make me notice a couple of strings of uniformly-spaced prime numbers I hadn’t done before, too, such as 3-13-23. (However, 3-13-23 isn’t one of the possible answers, because of the constraints of the problem. There aren’t positive X and Y for which X + Y = 3, X + 2Y = 13, and X + 3Y = 23.)

As with the Singapore Birthday Problem, this is a puzzle based on reasoning about the information we have. Mercifully there aren’t actually many prime numbers below 40, so if you want you can take the brute force approach and find all the strings of uniformly-spaced prime numbers. Then you can find what one matches the rules in ChefMongoose’s second paragraph.

I confess I wasn’t that systematic. I had a strong suspicion what the starting number of the sequence had to be, and then did some tests to be sure. I credit that to just having stared at lot at the smaller prime numbers in my life, so I’d had some intuitive feel for it. That’s a dangerous way to work. My intuitive feel, for example, hadn’t warned me about 3-13-23. But then there aren’t other trios of prime numbers spaced by ten, so that set would be ruled out by the “If I told you what Y was but not X” constraint. But now I know how to get stuff out of ChefMongoose’s work locker, you know, just in case.

## Elevator Mathematics

My friend ChefMongoose sent a neat little puzzle that came in a dream. I wanted to share it.

So! It’s not often that my dreams give me math puzzles. Here’s one: You are on floor 20 of a hotel. The stairs are blocked.

There are four elevators in front of you with display panels saying ‘3’, ‘4’, ‘7’, and ’35’. They will take you up that many floors, then the number will double. Going down an elevator will take you down that many floors, then the number will halve.

(The dream didn’t tell me what will happen if you can’t halve the number. For good puzzle logic, let’s assume the elevator goes down that much, then breaks.)

There is no basement, the hotel has an infinite amount of floors. Your challenge: get to floor 101. Can it be done?

(And I have no idea if it can be done. But apparently I, Riker, and Worf were trying to do it.)

The puzzle caught my imagination. It so happens the dream set things up so that this is possible: I worked out one path, and ChefMongoose found another.

ChefMongoose was right, of course, that something has to be done about halving the floor steps. I’d thought to make it half of either one less than or one more than the count. That is, going down 7 would turn the elevator into one that goes down either 3 or 4 floors. (My solution turned out not to need either.)

It looks lucky that ChefMongoose, Riker, and Worf picked a set of elevator moves, and rules, and starting, and ending floors that had a solution. Is it, though? Suppose we wanted to get to, say, floor 35? … Well, that’s possible. (Up 7, up 14, down 4, down 2.) 34 obviously follows. (Down 1 more.) 36? (Up 7, up 3, up 6.) 38? (Up 35, down 7, down 3, down 4, down 2, down 1.) The universe of reachable floors is bigger than it might seem at first.

The elevator problem had nagged at me with the thought it was related to some famous mathematical problem. At least that it was a type of one. ChefMongoose worked out what I was thinking of, the Collatz Conjecture. But on further reflection that’s the wrong parallel. This elevator problem is more akin to the McNuggets Problem. (When McDonald’s first sold Chicken McNuggets they were in packs of six, nine, and twenty. So what is the largest number of McNuggets that could not be bought by some combination of packages?) The doubling and halving of floor range makes the problem different, though. I am curious if there are finitely many unreachable floors. I’m also curious whether allowing negative numbers — basement floors — would change what floors are accessible.

The Collatz Conjecture is a fun one. It’s almost a game. Start with a positive whole number. If it’s even, divide it in half. If it’s odd, multiply it by three and add one. Then repeat.

If we start with 1, that’s odd, so we triple it and add one, giving us 4. Even, so cut in half: 2. Even again, so cut in half: 1. That’s going to bring us back to 4.

If we start with 2, we know where this is going. Cut in half: 1. Triple and add one: 4. Cut in half: 2. And repeat.

Starting with 3 suggests something new might happen. Triple 3 and add one: 10. Halve that: 5. Triple and add one: 16. Halve: 8. Halve: 4. Halve: 2. Halve: 1.

4 we’re already a bit sick of at this point. 5 — well, we just worked 5 out. That’ll go 5, 16, 8, 4, 2, 1, etc. Start from 6: we halve it to 3 and then we just worked out 3.

7 jumps right up to 22, then 11, then 34 — what a interesting number there — and then 52, 26, 13, 40, 20, 10 and we’ve seen that routine already. 10, 5, 16, 8, 4, 2, 1.

The Collatz Conjecture is that whatever positive whole number you start from will lead, eventually, to the 4, 2, 1 cycle. It may take a while to get there. I was working the numbers in my head while falling asleep the other night and got to wondering what exactly was 27’s problem anyway. (It takes over a hundred steps to settle down, and gets to numbers as high as 9,232 before finishing.)

Nobody knows whether it’s true. It seems plausible that it might be false. We can imagine a number that doesn’t. At least I can imagine there’s some number, let me call it N, and suppose it’s odd. Then triple that and add one, so we get an even number; halve that maybe a couple times until we get an odd number, and triple that and add one and get back the original N. You might have fun trying out numbers and seeing if you can find a loop like that.

Just do that for fun, though. Mathematicians have tested out every number less than 1,152,921,504,606,846,976. (It’s a round number in binary.) They all end in that 4, 2, 1 cycle. So it seems hard to believe that 1,152,921,504,606,846,977 and onward wouldn’t. We just don’t know that’s so.

If you allow zero, then that’s a valid but very short cycle: 0 halves to 0 and never budges. If you allow negative numbers, then there are at least three more cycles. They start from -1, from -5, and from -17. It’s not known whether there are any more in the negative integers.

The conjecture’s named for Lothar Collatz, 1910 – 1990, a German mathematician who specialized in numerical analysis. That’s the study of how to do calculations that meaningfully reflect the mathematics we would like to know. The Collatz Conjecture is, to the best of my knowledge, a novelty act. I don’t know of any interesting or useful results that depend on it being true (or false). It’s just a question easy to ask and understand, and that we don’t know how to solve. But those are fun to have around too.

## What can you see in the number 585?

Iva Sallay’s Find The Factors page I’ve mentioned before, since it provides a daily factorization puzzle. That’s a fun recreational mathematics puzzle even if the level 6’s and sometimes level 5’s will sometimes feel impossible. I wanted to point out, though, there’s also talk about the factoring of numbers, and ways to represent that factoring, that’s also interesting and attractive to look at. It can also include neat bits of trivia about numbers and their representation. In this example 585 presents some interesting facets, including several ways that it’s a palindromic number. If you don’t care for, or aren’t interested in, the factoring puzzles you might find it worth visiting for the trivia alone.

This week I watched an excellent video titled 5 x 9 is more than 45. Indeed 45 is so much more than simply 5 x 9. Every multiplication fact is much more than that mere fact, but Steve Wyborney used 5 x 9 = 45 in his video… Guess what! 585 is a multiple of 45.

As I thought about the number 585, I marveled at some of the hidden mysteries this number holds.

Since 585 is divisible by two different centered square numbers, 5 and 13, I saw that 585 could be represented by this lovely array that has 45 larger squares made up of 13 smaller colorful squares. When you look at the array, do you just see 585 squares or can you see even more multiplication and division facts? If you rotate the array 90 degrees, do the facts change?

What do you see in this…

View original post 268 more words

## A Summer 2015 Mathematics A to Z Roundup

Since I’ve run out of letters there’s little dignified to do except end the Summer 2015 Mathematics A to Z. I’m still organizing my thoughts about the experience. I’m quite glad to have done it, though.

For the sake of good organization, here’s the set of pages that this project’s seen created:

## Calculating Pi Terribly

I’m not really a fan of Pi Day. I’m not fond of the 3/14 format for writing dates to start with — it feels intolerably ambiguous to me for the first third of the month — and it requires reading the / as a . to make sense, when that just is not how the slash works. To use the / in any of its normal forms then Pi Day should be the 22nd of July, but that’s incompatible with the normal American date-writing conventions and leaves a day that’s nominally a promotion of the idea that “mathematics is cool” in the middle of summer vacation. This particular objection evaporates if you use . as the separator between month and day, but I don’t like that either, since it uses something indistinguishable from a decimal point as something which is not any kind of decimal point.

Also it encourages people to post a lot of pictures of pies, and make jokes about pies, and that’s really not a good pun. It plays on the coincidence of sounds without having any of the kind of ambiguity or contrast between or insight into concepts that normally make for the strongest puns, and it hasn’t even got the spontaneity of being something that just came up in conversation. We could use better jokes is my point.

But I don’t want to be relentlessly down about what’s essentially a bit of whimsy. (Although, also, dropping the ’20’ from 2015 so as to make this the Pi Day Of The Century? Tom Servo has a little song about that sort of thing.) So, here’s a neat and spectacularly inefficient way to generate the value of pi, that doesn’t superficially rely on anything to do with circles or diameters, and that’s probability-based. The wonderful randomness of the universe can give us a very specific and definite bit of information.

## When 2 plus 2 Equals 5, plus Another Unsettling Equation

I just wanted to note for folks who don’t read The Straight Dope — the first two books of which were unimaginably important to the teenage me, hundreds of pages of neat stuff to know delivered in a powerful style, that overwhelmed even The People’s Almanac 2 if you can imagine — that the Straight Dope Science Advisory board tried to take on the question of Does 2 + 2 equal 5 for very large values of 2?

Straight Dope Staffer Dex takes the question a bit more literally than I have ever interpreted the joke to be. I’ve basically read it as just justifying a nonsense result with a nonsense explanation, fitting in the spectrum of comic answers somewhere between King Lear’s understanding of why there are seven stars in the Pleiades and classic 1940s style double-talk. But Dex uses the equation to point out how rounding and estimation, essential steps in translating between the real world and the mathematical representation of the world, can produce results which are correct at every step but wrong in the whole, which is worth considering.

Also, in a bit of reading I’m doing and which I might rip off^W^W use as inspiration for some posts around here the (British) author dropped in an equation meant to be unsettling and, yeah, this unsettles me. Let me know what you think:

$3 \mbox{ feet } + 2 \mbox{ tons } = 36 \mbox{ inches } + 2440 \mbox{ pounds }$

I should say it’s not like I’m going to have nightmares about that, but it feels off anyway.

## Denominated Mischief

I’ve finally got around to reading one of my Christmas presents, Alfred S Posamentier and Ingmar Lehman’s Magnificent Mistakes in Mathematics, which is about ways that mathematical reasoning can be led astray. A lot, at least in the early pages, is about the ways a calculation can be fowled by a bit of carelessness, especially things like dividing by zero, which seems like such an obvious mistake that who could make it once they’ve passed Algebra II?

They got to a most neat little erroneous calculation, though, and I wanted to share it since the flaw is not immediately obvious although the absurdity of the conclusion drives you to look for it. We begin with a straightforward problem that I think of as Algebra I-grade, though I admit my memories of taking Algebra I are pretty vague these days, so maybe I missed the target grade level by a year or two.

$\frac{3x - 30}{11 - x} = \frac{x + 2}{x - 7} - 4$

Multiply that 4 on the right-hand side by 1 — in this case, by $\frac{x - 7}{x - 7}$ — and combine that into the numerator:

$\frac{3x - 30}{11 - x} = \frac{x + 2 - 4(x - 7)}{x - 7}$

Expand that parentheses and simplify the numerator on the right-hand side:

$\frac{3x - 30}{11 - x} = \frac{3x - 30}{7 - x}$

Since the fractions are equal, and the numerators are equal, therefore their denominators must be equal. Thus, $11 - x = 7 - x$ and therefore, 11 = 7.

Did you spot where the card got palmed there?

## How To Hear Drums

The @mathematicsprof tweet above links to a paper, by Carolyn Gordon and David Webb and published in American Scientist in 1996, that’s about one of those questions that’s both mathematically interesting and of obvious everyday interest. The question was originally put, in nice compact and real-world-relevant form, in 1966 by Mark Kac: can one hear the shape of a drum?

At first glance the answer may seem, “of course” — you can hear the difference between musical instruments by listening to them. You might need experience, but, after all, you’re not going to confuse a bass drum from a bongo even if you haven’t been in the music store much. At second glance, why would Kac bother asking the question if the answer were obvious? He didn’t need the attention. He had, among other things, his work in ferromagnetism to be proud of (and I should write about that some.) And could you tell one bass drum from another?

The question ties into what’s known as “spectral theory”: given a complicated bundle of information what can you say about the source? One metaphorical inspiration here is studying the spectrum of a burning compound: the wavelengths of light emitted by it give you information about what elements go into the compound, and what their relative abundances are.

The sound of a drum is going to be a potentially complicated set of sound waves produced by the drum’s membrane itself oscillating. That membrane oscillation is going to depend, among other things, on the shape of the membrane, and that’s why we might suppose that we could tell what the shape of the drum is by the sound it makes when struck. But then it might also be that multiple different shapes could produce the exact same sound.

It took to about 1990 to get a definite answer; Gordon and Webb, along with Scott Wolpert, showed that you can get different-shaped drums that sound the same, and very nicely showed an example. In the linked article, Gordon and Webb describe some of the history of the problem and how they worked out a solution. It does require some technical terms that maybe even re-reading several times won’t help you parse, but if you’re willing to just move on past a paragraph that looks like jargon to the rest I believe you’ll find some interesting stuff out, for example, whether you could at least hear the area of a drum, even if you can’t tell what the shape is.