## 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.

• #### RobertLovesPi 8:05 pm on Saturday, 12 March, 2016 Permalink | Reply

Much thanks for the “signal boost” — I appreciate it!

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• #### Joseph Nebus 3:34 am on Monday, 14 March, 2016 Permalink | Reply

Quite happy to be of service. Thank you.

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## 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.

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• #### Joseph Nebus 3:45 am on Tuesday, 16 February, 2016 Permalink | Reply

Oh, dear, they’re staring at me!

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• #### Thumbup 2:34 pm on Tuesday, 16 February, 2016 Permalink | Reply

UH OH! It does look like it!

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• #### Joseph Nebus 5:58 am on Saturday, 20 February, 2016 Permalink | Reply

Maybe the dice need some sunglasses or something, to look cooler.

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• #### Thumbup 6:03 am on Saturday, 20 February, 2016 Permalink | Reply

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• #### howardat58 3:14 pm on Wednesday, 10 February, 2016 Permalink | Reply

Vegetarians clearly have different definitions.

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• #### Joseph Nebus 3:47 am on Tuesday, 16 February, 2016 Permalink | Reply

Well, we have to be a bit more careful than average, is all. There’s troublesome stuff where you’d never expect it.

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• #### Chiaroscuro 4:00 pm on Wednesday, 10 February, 2016 Permalink | Reply

So, let’s make these A/B=C/D for the dice, assuming in-order rolls. 1296 possibilities.

If A=C and B=D, it’ll always work. So that’s 36.

Additionally: If A=B and C=D, it’ll always work (1=1). So that’s 36,. minus the 6 where A=B=C=D.

Then 1/2=2/4 (and converse and inverse and both), 1/2=3/6 (same), 2/4=3/6 (same), 1/3=2/6 (same). 4, 4, 4, 4.so 16 total.

36+30+16=82, unless I’ve missed some. 82/1296, which reduces to 41/648.

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• #### Chiaroscuro 4:02 pm on Wednesday, 10 February, 2016 Permalink | Reply

Ooh! I missed 2/3=4/6. (and converse, and inverse, and both). So another 4, meaning 86/1296.

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• #### Joseph Nebus 3:51 am on Tuesday, 16 February, 2016 Permalink | Reply

There you go. For what it’s worth, I make out the same count of in-order rolls. And a little program I wrote to count them comes up with 86 as well.

Rolls where order doesn’t matter, and you can rearrange the dice to find a pattern that fits … well, obviously there’s more of them.

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## 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.

How about A is for axiom?

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• #### Joseph Nebus 4:18 pm on Saturday, 30 January, 2016 Permalink | Reply

Happy to. I’ll set it on the list.

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Yes!

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• #### KnotTheorist 8:48 pm on Saturday, 30 January, 2016 Permalink | Reply

I enjoyed last year’s Mathematical A-To-Z Glossary, so I’m glad to see you’ll be doing another one!

I’d like to see C for continued fractions.

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• #### Joseph Nebus 12:39 am on Tuesday, 2 February, 2016 Permalink | Reply

Continued fractions … mm. Well, I’ll have to learn more about them, but that’s part of the fun of this. Thank you.

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• #### davekingsbury 6:29 pm on Sunday, 31 January, 2016 Permalink | Reply

Energy = Mass times Twice the Speed of Light … or is that more like Physics?

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• #### Joseph Nebus 12:40 am on Tuesday, 2 February, 2016 Permalink | Reply

$E = mc^2$ is physics, although it’s something that we learned from mathematical considerations. And a big swath of mathematics is the study of physics. There’s a lot to talk about in energy for mathematicians.

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• #### elkement (Elke Stangl) 7:38 am on Monday, 8 February, 2016 Permalink | Reply

Of course I second that :-) What about explaining a Lagrangian in layman’s terms? ;-)

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• #### Joseph Nebus 5:27 am on Wednesday, 10 February, 2016 Permalink | Reply

You know, I think I’ve got a hook on how to explain that. It might even get to include a bit from my high school physics class.

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• #### davekingsbury 9:14 am on Tuesday, 2 February, 2016 Permalink | Reply

Is the equation based on theory or is there a practical mathematics behind it?

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• #### Joseph Nebus 12:03 am on Wednesday, 3 February, 2016 Permalink | Reply

I’m not sure what you mean by theory versus practical mathematics. The energy-mass equivalence does follow, mathematically, from some remarkably simple principles. Those amount to uncontroversial things like the speed of light being a constant, independent of the observer, and that momentum and energy are conserved.

It is experimentally verified, though. We can, for example, measure the mass of atoms before and after they fuse, or fission, and measure the amount of energy released or absorbed as light in the process. The amounts match up as expected. (That’s not the only test to run, of course, but it’s an easy one to understand.) So the reasoning isn’t just good, but matches what we see in the real world.

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• #### davekingsbury 10:36 am on Wednesday, 3 February, 2016 Permalink | Reply

Thanks for your clear explanation. I’m not a scientist. Theory wasn’t the right word, then – I was thinking of empirically verifiable which your 2nd paragraph shows. Are the ‘uncontroversial things in your first paragraph also measurable in the real world?

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• #### Joseph Nebus 8:34 pm on Friday, 5 February, 2016 Permalink | Reply

OK. Well, these are measurable things, in that experiments give results that are what we would expect from the assumptions, and that are inconsistent with what we’d expect from alternate assumptions. For example, we now assume the speed of light (in a vacuum) to be constant. That followed a century of experimentation that finds it does appear to always be constant, and it’s consistent with tests that look to see if there might be something surprising now that we have a new effect to measure or a new tool to measure with. Assumptions about, for example, the way that velocities have to add together in order for this constant-speed-of-light to work have implications for how, say, moving electric charges will produce magnetic fields, and we see magnetic fields induced by moving electric charges consistently with that.

We can imagine our current understanding to be incomplete, and that the real world has subtleties we haven’t yet detected. But I’m not aware of any outstanding mysteries that suggest strongly that we’re near that point.

So, given assumptions that seem straightforward enough, and that match experiment as well as we’re able to measure, physicists and mathematicians are generally inclined to say that these assumptions are correct. Or at least correct enough for the context in which they’re used. This is starting to get into the philosophy of science and the concept of experimental proof and gets, I admit, beyond what I’m competent to discuss with authority.

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• #### davekingsbury 8:43 pm on Friday, 5 February, 2016 Permalink | Reply

Thanks for taking the time (and space) to explain this so clearly and enjoyably to a rookie. No more questions, promise … for now!

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• #### Gillian B 5:39 am on Wednesday, 3 February, 2016 Permalink | Reply

Isomorphism.

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• #### Joseph Nebus 8:12 pm on Friday, 5 February, 2016 Permalink | Reply

Ooh, a challenging one. I’ll give it a try, though.

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• #### gaurish 7:00 am on Monday, 8 February, 2016 Permalink | Reply

Normal subgroup (easy one) or Number (difficult one, Bertrand Russell tried it once).

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• #### Joseph Nebus 5:24 am on Wednesday, 10 February, 2016 Permalink | Reply

Oh, number is easy. Three, for example, is the thing that’s in common among Marx Brothers, blind mice, tricycle wheels, penny operas, and balls in the Midnight Multiball of the pinball game FunHouse. Normal subgroup, now that’s hard.

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• #### gaurish 7:10 am on Monday, 8 February, 2016 Permalink | Reply

Transcendental numbers; Dedikind Domain; matrix; polynomial; quartenions; subjective map; vector.

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• #### Joseph Nebus 5:26 am on Wednesday, 10 February, 2016 Permalink | Reply

There’s some good challenges here! My first reaction was to say I didn’t even know what a Dedekind domain was, although in looking it up I realize that I must have learned of them. I just haven’t thought of one in obviously too long, and I like the chance to learn something just in time to explain it.

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• #### elkement (Elke Stangl) 7:40 am on Monday, 8 February, 2016 Permalink | Reply

C as Conjecture. More of a history of science question: When is an ‘unproven idea’ honored by being called a conjecture?

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• #### Joseph Nebus 5:35 am on Wednesday, 10 February, 2016 Permalink | Reply

Conjecture may work, yes, and fit neatly against axiom trusting that I use that.

I’m not sure there is a clear guide to when an unproven idea gets elevated to the status of conjecture. I suspect it would defy any rationally describable process. I mean about getting regarded as a name-worthy conjecture. There’s conjectures in much mathematical literature and those tend to mean the person writing the paper got a hunch that something might be so, but didn’t have the time or ability to prove it and is happy to let someone else try.

But to be, let’s say, the Stangl Conjecture takes more. I suspect part is that it has to be something that feels likely to be true, and which has some obviously interesting consequence if true (or false). That can’t be all, though. The Collatz Conjecture, as I’ve mentioned, seems to be nothing but an amusing trifle. But then that’s also a conjecture that’s very easy for anyone to understand, and it has some beauty to it. The low importance of it might be balanced by how much fun it seems to be and how everyone can be in on the fun.

I’ll have to do some more poking around famous conjectures, though, and see if I can better characterize what they have in common.

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• #### elkement (Elke Stangl) 7:30 am on Tuesday, 1 March, 2016 Permalink | Reply

Equation or Differential Equation, depending on which letter is still open. I am thinking of the way THE FORMULA is depicted in movies, and I believe that it might imply that anything with an equal sign in it is more like Ohm’s Law – a ‘formula’ you just have to plug numbers into. I am sure you can explain the difference between a simple formula and a differential equation nicely :-) Or use Formula instead if F has not been taken.

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• #### Joseph Nebus 8:57 pm on Tuesday, 1 March, 2016 Permalink | Reply

Hm. I may take you up on differential equation, since the first nominee — Dedekind domains — is taxing my imagination. And I’d slid continued fractions over to F … but I will think about whether I can find a way to put Formula in under another letter.

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• #### Jacob Kanev 11:54 pm on Friday, 4 March, 2016 Permalink | Reply

First things that tumble into my mind: Itô integral, Stratonovitch integral, Kulbach-Leibler divergence, Fisher information, Turing machine, Church’s lemma (is this the correct term in English? And you have ‘C’ already, haven’t you?), grammars (both context sensitive and not), Girsanov transformation (sorry for using ‘G’ twice), filtration (I’d really like a good explanation of this one) (and ‘F’), Banach spaces. Orthogonal. Projection. Distance. Metric. Measure. NP-completeness? Gödel’s theorem? Laws of form (that calculus by George Spencer Brown)?

Might be too nichey, though. You decide.

Lots of regards, Jacob.

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• #### Joseph Nebus 7:36 am on Wednesday, 9 March, 2016 Permalink | Reply

Well, wow. I do have a couple of these letters taken already — I’ve got through ‘F’ penciled in, plus a couple such as ‘I’ taken after that. But I’ll try to get as many of these as I can done in a coherent form. It’s going to be an exciting month ahead.

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## 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.

• #### John Friedrich 1:08 am on Monday, 25 January, 2016 Permalink | Reply

The Golden Ratio is neat for determining the nth Fibonacci sequence number at least.

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• #### Joseph Nebus 10:22 pm on Tuesday, 26 January, 2016 Permalink | Reply

This is true, although the Fibonacci sequence has a similar problem in being pretty but not all that useful. It’s a bit more useful than the Golden Ratio, I’ll grant, and it would be out of character for me to complain about corners of mathematics that are just fun. But fun and important are different things.

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• #### fluffy 7:06 am on Monday, 25 January, 2016 Permalink | Reply

So what about finding an arbitrary $x_N$ such that $N/x = x-N$? Is that solvable?

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• #### fluffy 7:06 am on Monday, 25 January, 2016 Permalink | Reply

And what of the problem of inline LaTeX in a comment? Is that solvable?

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• #### Joseph Nebus 10:27 pm on Tuesday, 26 January, 2016 Permalink | Reply

I’m not sure, although I wonder if you didn’t forget to include the word ‘latex’ after the first \$ in the comment. I think inline LaTeX is supposed to be enabled in these parts.

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• #### fluffy 7:01 pm on Thursday, 28 January, 2016 Permalink | Reply

I didn’t know that was necessary. So I do like $a^3 + b^3 = c^3$?

A preview function would be nice.

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• #### Joseph Nebus 10:36 pm on Thursday, 28 January, 2016 Permalink | Reply

That’s the way, yes. WordPress’s commenting system certainly needs a preview function and an edit button.

(There might be other themes that have preview functions. The one I’m using here is a bit old-fashioned and it might predate a preview. I know it isn’t really mobile-friendly.)

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• #### Joseph Nebus 10:32 pm on Tuesday, 26 January, 2016 Permalink | Reply

It’s certainly possible. Starting from $\frac{N}{x} = x - N$ we get the equation $0 = x^2 - Nx - N$ and that has solutions $x = \frac{1}{2} \left( N \pm \sqrt{N^2 + 4N}\right)$.

I don’t seem able to include the table that would list the first couple of these without breaking the commenting system. But it’s easy to generate from that start. The x_N for a given N gets to be quite close to N+1.

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• #### Eric Mann 11:55 am on Wednesday, 27 January, 2016 Permalink | Reply

These x_N’s are nice little multiples of the (dare I say) golden ratio, yes? They will arise from a rectangle with dimensions of x and N with an embedded N by N square.

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• #### Joseph Nebus 10:35 pm on Thursday, 28 January, 2016 Permalink | Reply

I don’t see any link between these x_N’s or the golden ratio. Could you tell me what you see, please?

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• #### Eric Mann 11:51 am on Wednesday, 27 January, 2016 Permalink | Reply

Well done. I love the inquiry. What do the rectangles look like? Or, is there still a geometric interpretation?

I admit I still find the sequence of rectangles with Fibonacci dimensions and an embedded spiral attractive. I appreciate it in the limit. I am not attached to the ratio being golden with a capitol g, but is it just a curiosity? An attraction? I expect more out of the ratio, rectangle, and spiral.

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• #### Joseph Nebus 10:32 pm on Thursday, 28 January, 2016 Permalink | Reply

Well, rectangles in these gilt-ratio proportions would be longer and skinnier things. For example, you might see a rectangle that’s one inch wide and 20.049(etc) inches long. I doubt anyone could tell the difference between that and a rectangle that’s one inch wide, 20 inches long, though.

I do think it’s just a curiosity, an attractive-looking number. Or family of numbers, if you open up to these sorts of variations. There’s nothing wrong with looking at something that’s just attractive, though. It’s fun, for one thing. And the thinking done about one problem surely helps one practice for other problems. I was writing recently about the Collatz Conjecture. As far as I know nothing interesting depends on the conjecture being true or false, but it’s still enjoyable.

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## 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.

• #### elkement (Elke Stangl) 8:27 am on Tuesday, 19 January, 2016 Permalink | Reply

I’d be very interested in this – but the link to the blog does not work: It’s an a tag without a href attribute.

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• #### Joseph Nebus 11:07 pm on Tuesday, 19 January, 2016 Permalink | Reply

Oh, that’s embarrassing. It looks like I wrote it up originally and missed a quote mark in the attribute tag, and of course WordPress deleted that rather than give me a hint something was wrong. (Well, I should’ve checked the link before an after posting too, admittedly.) Thank you. It ought to be working now.

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## 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.

• #### Chiaroscuro 11:48 pm on Tuesday, 5 January, 2016 Permalink | Reply

You certainly could!

If I told you the locker number.

Which I’ll have to think about mathwise a little further for.

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• #### Joseph Nebus 10:34 pm on Thursday, 7 January, 2016 Permalink | Reply

I’d just assumed it would be the one with your name on it.

Actually, I’d assumed it would be the one just low enough you have to bend uncomfortably far, but not low enough you can sit on a bench or stool, because that’s where everybody’s locker is.

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• #### Geoffrey B 12:09 am on Wednesday, 6 January, 2016 Permalink | Reply

Hm, let’s see how far I can avoid trial-and-error methods here.

If X+Y=2, then X+3Y is even and larger than 2, which means it can’t be prime, ergo contradiction.

It follows that none of X+Y, X+2Y, X+3Y can be equal to 2, hence they can’t be even, which implies that X is odd and Y is even.

Applying the “knowing Y is not enough to find X” rule:

If Y is NOT a multiple of 3, then one of X+Y, X+2Y, X+3Y must be a multiple of 3. The only such prime is 3, and since this is the smallest possible prime after eliminating 2, that implies that X+Y=3, i.e. X=1, Y=2.

But this means that if Y is not a multiple of 3, then knowing Y would be enough to figure out the combination, which contradicts info given.

Hence Y must be a multiple of 2 and 3, i.e. a multiple of 6. It also follows that X cannot be a multiple of 3, else X+Y would be nonprime.

X+3Y is no greater than 40, hence X <= 40-3Y. If Y is 18 or higher, this would make X negative, which isn’t allowed.

If Y = 12, then X <= 4. We already know X is odd and not a multiple of 3, which means X must equal 1. Hence, knowing Y=12 would be enough to find X and solve the problem, not allowed.

This means that Y has to equal 6. Our sequence becomes X+6, X+12, X+18. X is not a multiple of 2 or 3, and must be less than 22. Hence the only possible values for X are 1, 5, 7, 11, 13, 17.

Of these values 7, 13, and 17 give non-prime numbers in the sequence and can be eliminated, so we’re down to three possibilities:

X=1, Y=6 (combination 7, 13, 19)
X=5, Y=6 (11, 17, 23)
X=11, Y=6 (17, 23, 29)

Now, forgetting the “knowing Y is not enough to find X” rule and applying the “knowing X is not enough to find Y” rule: if we knew the value of X out of those three, what could we deduce about Y?

Knowing X>1 would tell us that X+Y > 3, hence none of the numbers in the combination can be a multiple of 3. This then implies that Y must be a multiple of 3. We already established that Y is even (without needing to invoke the “knowing Y doesn’t give us X” value). But if X >= 5, Y can’t be greater than (40-5)/3 < 12, so Y could only be 6.

Hence, knowing X=5 or X=11 would be enough to deduce Y. Contradiction. So the only candidate solution left is X=1, Y=6, combination = 7, 13, 19.

We’ve shown that anything else contradicts the information given, but we still need to confirm that this solution IS legit. Perhaps there’s some other way we haven’t explored in which knowing X=1 would tell us Y=6, or vice versa?

If we’re told X, but not Y, we can’t rule out the solution X=1, Y=2 i.e. combination = 3, 5, 7.

If we’re told Y, but not X, we can’t rule out the solution X=5, Y=6, i.e. combination = 11, 17, 23. (Or the solution where X=11, Y=6.)

So we’re done. The combination is 7, 13, 19. Little bit of case-checking but not too much trial and error.

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• #### Joseph Nebus 10:36 pm on Thursday, 7 January, 2016 Permalink | Reply

Excellently done, yes. Full marks.

As I hinted in the original post, my instinct was that “7 has got to be one of the numbers” and then it was a matter of could I rule out 3-5-7, 3-7-11, or any of the other sequences starting with 7. And then double back to see if there was reason to doubt 7-13-19. Nowhere near as safe, of course.

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• #### davekingsbury 11:15 am on Wednesday, 6 January, 2016 Permalink | Reply

Your posts are a wonderful mystery to me, like reading shamanic texts! I had a maths block at school when I returned from illness and the teacher pilloried me for not understanding hundreds, tens and units. Quite proud of being able to do Sudoku, though …

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• #### Joseph Nebus 10:37 pm on Thursday, 7 January, 2016 Permalink | Reply

Aw, dear, I’m sorry to be mysterious. I hope you do enjoy the experience though. There’s a lot of beauty in the subject, and a lot of it can be understood without needing to do calculations. A numbers-riddle puzzle obviously isn’t one of them, though.

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• #### davekingsbury 10:59 pm on Thursday, 7 January, 2016 Permalink | Reply

Yeah, don’t understand everything but like reading for the gist. Just bought Simon Singh’s Fermat’s Last Theorem and looking forward to it.

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• #### Garfield Hug 11:18 am on Saturday, 9 January, 2016 Permalink | Reply

Ha ha you knew about our math problem 😉

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• #### Joseph Nebus 3:47 am on Saturday, 16 January, 2016 Permalink | Reply

Oh, yes, the Singapore Birthday Problem got to be quite popular for a week or so back around late summer. It was driving several of my friends crazy and they weren’t too happy even after the explanation was given. The New York Times did a good writeup on it that seems to have settled most people’s questions, at least.

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• #### Garfield Hug 3:49 am on Saturday, 16 January, 2016 Permalink | Reply

Lol! It seems USA is buying up our math, science, physics test papers and books for use in their schools. Apparently we are “famous” for setting mind boggling questions😉

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• #### John Quintanilla 1:02 pm on Friday, 15 January, 2016 Permalink | Reply

Reblogged this on Mean Green Math and commented:

I enjoyed this challenge.

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• #### Joseph Nebus 3:54 am on Saturday, 16 January, 2016 Permalink | Reply

Very glad you liked it and I’ll pass the word on to the puzzle’s author.

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## 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.

• #### sheldonk2014 8:41 pm on Wednesday, 20 January, 2016 Permalink | Reply

I don’t know why Joseph but I am not getting your posts
So I came to visit
Just to let you know

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• #### Joseph Nebus 5:01 am on Sunday, 24 January, 2016 Permalink | Reply

That is curious and I’m not sure what’s happening. I do know of one RSS aggregator that’s not been getting the feed. That might be connected, but I don’t know if it isn’t just coincidence since that aggregator isn’t very reliable.

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## 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.

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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

• #### ivasallay 3:10 pm on Friday, 21 August, 2015 Permalink | Reply

Thanks for the reblog! Also thank you for being one of my most consistent supporters.

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• #### Joseph Nebus 9:25 pm on Saturday, 22 August, 2015 Permalink | Reply

Very welcome, and thank you for being reliably interesting.

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## 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.

• #### abyssbrain 10:28 am on Saturday, 14 March, 2015 Permalink | Reply

When I first read about this method for calculating pi before, I have entertained the idea of trying it myself but I quickly discarded that idea, since who knows how long it would take before I would reach pi :)

Btw, I’m also very confused with the American way of writing dates since I’m used to either ddmmyyyy format or yyyymmdd format. So, March 14, 2015 for me is 14/3/2015. I’ve also just posted some of my reasons why I don’t celebrate pi day…

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• #### Joseph Nebus 12:25 am on Monday, 16 March, 2015 Permalink | Reply

Oh, it would just take forever to find pi using needles and ruled lines. You’d do considerably better if you drew a quarter-circle on a square dartboard, and tossed darts at it, counting the ratio of darts that hit inside the quarter-circle to darts outside. At least you’d have a better night of it.

I don’t know why the United States uses the month-day-year format, particularly since it hasn’t got much (any?) use elsewhere in the world. My suspicion is that there probably was a time when both month-day and day-month were common enough in English-speaking nations and the United States settled on one format while the United Kingdom another back in the 19th Century back when stuff standardized.

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• #### abyssbrain 1:08 am on Monday, 16 March, 2015 Permalink | Reply

Well, it seems widespread because of US websites like Google use mmddyyyy format by default and most of the top sites are from the US…

Though I have noticed that they are now slowly changing the date format of many wikipedia articles to ddmmyyyy format.

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• #### Joseph Nebus 9:05 pm on Tuesday, 17 March, 2015 Permalink | Reply

I have noticed what looks like a slow shift in american use to day-month-year format, at least when the month is given its proper name rather than a number. The year-month-day order seems irresistible if you’re determined to stick to writing things as digits, for reasons I have to agree are pretty solid.

Anyway, there does seem to be something logical about sticking to one logical path about whether the thing written first should be the thing most likely to change and the thing written last the least likely, or the other way around.

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• #### LFFL 6:09 pm on Saturday, 14 March, 2015 Permalink | Reply

See. I opened this blog after avoiding it for such a long time and my headache started instantly! The agony.

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• #### Joseph Nebus 12:26 am on Monday, 16 March, 2015 Permalink | Reply

Aw, dear. If it helps any I should have a fresh comic strips review in the next day or so. That’s nice and friendly.

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:)

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## 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.

• #### abyssbrain 1:42 am on Sunday, 8 March, 2015 Permalink | Reply

Then there’s also the classic of Abbott and Castello “proving” that 13 x 7 = 28 :)

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• #### Joseph Nebus 11:37 pm on Monday, 9 March, 2015 Permalink | Reply

You know, I’m not familiar with that sketch offhand.

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• #### abyssbrain 12:24 am on Tuesday, 10 March, 2015 Permalink | Reply

Actually, it’s a part of a film.

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• #### Joseph Nebus 7:50 pm on Thursday, 12 March, 2015 Permalink | Reply

Oh, that’s a great routine, and I hadn’t seen it before. Thank you. (It’s surely from their TV show, though?)

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• #### abyssbrain 1:07 am on Friday, 13 March, 2015 Permalink | Reply

If I remembered correctly, that particular video was from one of their films, but they had also done this sketch once on their tv show.

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• #### Joseph Nebus 3:25 am on Saturday, 14 March, 2015 Permalink | Reply

That makes sense. It’d be uncharacteristic for them to use a good bit only the once, especially since it could be years between anyone in the audience seeing a movie, a radio program where they did it, and a TV show using the same bit.

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• #### Matthew Wright 1:55 am on Sunday, 8 March, 2015 Permalink | Reply

To me British Imperial measures like the long ton (which is only half a smoot longer than a short ton) pretty much sum up the problem the British also have making reliable cars. And landing things on Mars, when one half of the team is using Imperial and the other half metric.

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• #### Joseph Nebus 11:48 pm on Monday, 9 March, 2015 Permalink | Reply

As best I can tell the short ton is an invention of the Americans, so the British aren’t directly at fault for the long ton/short ton divide. Granted that 2240 pounds is a superficially weird number of pounds to put into any unit, but that is at least a nice convenient twenty hundredweights, which admittedly moves the problem back to why a hundredweight is a hundred pounds. In that case it’s because a hundredweight was a nice convenient eight stone, which had been twelve and a half pounds avoirdupois, until King Edward III yielded to the convenience of the wool trade and increase the stone to fourteen pounds (making a sack of cloth, 28 stone, more conveniently measured without cheating on available scales and also a nice (nearly) round 500 Florentian libbrae, and the rest followed from there.) Which is to admit that it’s daft, but every step made sense at the time, which is the best we can ever hope for.

Now, the Imperial/Metric problem with the space probe is interesting because while the difference in units is the proximate cause of the vehicle’s loss, it’s not the real cause. There were hints, from earlier maneuvers, that something was wrong in the way thrusts were being calculated or executed, but those weren’t followed up on. Had they been, a correction would’ve been straightforward. It’s a lesson in the importance of having good project management, and that project management has to include people signaling clearly when they suspect there’s problems and exploring adequately whether these suspicions are well-founded.

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• #### elkement 2:55 pm on Sunday, 8 March, 2015 Permalink | Reply

It was not until recently that I learned how ‘ton’ is used in engineering (related to air conditioning). I learned a lot of – maybe a ton of – new units when trying to respond to questions in the comments section on my blog :-)

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• #### Joseph Nebus 11:54 pm on Monday, 9 March, 2015 Permalink | Reply

I did not know there were custom uses of the ‘ton’ for engineering purposes until just now, and I’m fascinated to see how many different “big mass of the thing we’re measuring” get called tons, now. (Panama Canal Net Ton? Who ordered that?)

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• #### elkement 10:57 am on Tuesday, 10 March, 2015 Permalink | Reply

I hope I understood it correctly finally – but I was baffled about “ton” being used as a unit for (heating or cooling) *power*, rather than weight.

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• #### Joseph Nebus 7:55 pm on Thursday, 12 March, 2015 Permalink | Reply

Yeah, thinking of ton as a unit of power is weird, although I suppose it’s not inherently stranger than describing a distance by the amount of time it’d take to get there. It’s just less familiar.

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## 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?

• #### Little Monster Girl 11:13 pm on Monday, 2 February, 2015 Permalink | Reply

You didn’t do anything to the left side of the equation?

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• #### Joseph Nebus 10:20 am on Tuesday, 3 February, 2015 Permalink | Reply

It’s true nothing’s done on the left-hand side, but that isn’t by itself an error. If we start from the assumption that the original equation is true we can manipulate one side, or the other, or both, into a form that’s more convenient without changing whether or not the whole equation is true. The catch is that somewhere in this is a manipulation that doesn’t preserve the truth of the whole thing.

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• #### howardat58 11:30 pm on Monday, 2 February, 2015 Permalink | Reply

Formally, cross multiply is in order. Of course, they don’t call it that these days.

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• #### Joseph Nebus 10:21 am on Tuesday, 3 February, 2015 Permalink | Reply

Cross-multiplying ought to give a fair shot at avoiding the error, yeah. But I couldn’t blame someone for seeing an equation of the form a/b = a/d and going right to b = d directly.

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• #### ivasallay 7:41 am on Tuesday, 3 February, 2015 Permalink | Reply

Where did (3x – 30)/ (11 – x) = (x + 2)/(x – 7) – 4 come from? It certainly isn’t true for all x.

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• #### Joseph Nebus 10:23 am on Tuesday, 3 February, 2015 Permalink | Reply

Well, that’s just a problem to be solved, to find values of x which make it true. It’s just that along the way to finding those x’s, we end up with a conclusion that 11 equals 7.

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• #### elkement 7:18 pm on Tuesday, 3 February, 2015 Permalink | Reply

I think the trick is to keep in mind that when the numerator is zero then it does not matter if both denominators are different (as long as they are not equal to zero as well).

So if x is equal to 10 the equation is true as both sides are equal to zero although the denominators are 1 and -3, respectively.

The short version is: You must not divide both sides of an equation of zero.

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• #### Joseph Nebus 11:31 pm on Wednesday, 4 February, 2015 Permalink | Reply

That’s it exactly, and I’m delighted by the problem since it is one in which the ever-forbidden division by zero is made nicely non-obvious.

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## 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.

• #### ivasallay 8:29 pm on Thursday, 30 October, 2014 Permalink | Reply

When I attended college in the early 70’s, somebody (probably Mark Kac) did a presentation at the University on “Can You Hear the Shape of a Drum?” I had one professor who was especially intrigued and shared tidbits of the presentation with the class afterwards. It is so nice to get this update. Thank you!

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• #### Joseph Nebus 11:47 pm on Friday, 31 October, 2014 Permalink | Reply

Oh, I’m interested to hear about the history of the problem. Also I’m glad you were able to learn something about how the problem’s turned out.

It’s curious what presentations and seminars will last. One that I remember from about 1995 or 1996 was a guy who was modeling viscous fluid flows, and to give this study a nice punchy real-world application, brought up the Sherwin-Williams “Cover The Earth” logo. If you suppose the paint being dropped on the sphere to be about average, then, does the way it’s dripping tell you the scale of the logo — how big the imitation Earth there would be? And, yes: if I’m not remembering this incorrectly, the logo implies a model Earth that’s the size of a grapefruit.

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• #### howardat58 8:33 pm on Thursday, 30 October, 2014 Permalink | Reply

I read most of it. These guys are very good at explaining stuff. I learned quite a bit. When I was a student I avoided differential equations like the plague !

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• #### Joseph Nebus 11:45 pm on Friday, 31 October, 2014 Permalink | Reply

They did explain it quite well, yes. I think it helps they were writing for an audience that could be supposed to know a fair bit of mathematics; the less background stuff you have to explain the easier it is to get to the interesting point.

I was a poor student the first time I took a differential equations course (that is, the kind for mathematics majors) and that hobbled me through a lot of my work, sad to say.

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## A Forest of 240 Factor Trees

I admit this is a little self-indulgent, but, what the heck. You might also like the factor trees for the extremely factorable number 240 that Ivasallay’s put up at Find The Factors.

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• 240 is a composite number.
• Prime factorization: 240 = 2 x 2 x 2 x 2 x 3 x 5, which can be written (2^4) x 3 x 5
• The exponents in the prime factorization are 4, 1 and 1. Adding one to each and multiplying we get (4 + 1)(1 + 1)(1 + 1) = 5 x 2 x 2 = 20. Therefore 240 has 20 factors.
• Factors of 240: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120, 240
• Factor pairs: 240 = 1 x 240, 2 x 120, 3 x 80, 4 x 60, 5 x 48, 6 x 40, 8 x 30, 10 x 24, 12 x 20, or 15 x 16
• Taking the factor pair with the largest square number factor, we get √240 = (√16)(√15) = 4√15 ≈ 15.492.

Because 240 has so many factors, it is possible…

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## Splitting a Cake with a Missing Piece

I wanted to offer something a little light today, as I’m in the midst of figuring out the next articles in a couple of my ongoing threads and getting ready for a guest posting. So here, from Mathematics Lounge, please enjoy this nice little puzzle about how to cut, into two even pieces, a cake that’s already had a piece cut out of it. It’s got a lovely answer and it’s worth pondering it and why that answer’s true before reading the solution. And there’s another, grin-worthy, solution offered in the comments.

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Problem:
Jeremy and Jane would like to divide a rectangular cake in half, but their friend Bob (who can be a jerk sometimes) has already cut out a piece for himself. Bob’s slice is a rectangle of some arbitrary size and rotation. How can Jeremy and Jane divide the remaining cake into two equal portions, using a single cut with a sufficiently long knife?

Description:
This is an interesting problem with a fairly elegant solution. It is the type of problem that can be posed as a math puzzle/riddle, and figured out on the spot with some ingenuity.

For this problem, we define a single cut as a separation of the area made by a straight line, viewed from above. For example, a cut that crosses a gap (like below) may intersect the cake in two separate places, but still counts as one cut. (This example, of course, clearly does…

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## In the Overlap between Logic, Fun, and Information

Since I do need to make up for my former ignorance of John Venn’s diagrams and how to use them, let me join in what looks early on like a massive Internet swarm of mentions of Venn. The Daily Nous, a philosophy-news blog, was my first hint that anything interesting was going on (as my love is a philosopher and is much more in tune with the profession than I am with mathematics), and I appreciate the way they describe Venn’s interesting properties. (Also, for me at least, that page recommends I read Dungeons and Dragons and Derrida, itself pointing to an installment of philosophy-based web comic Existentialist Comics, so you get a sense of how things go over there.)

And then a friend retweeted the above cartoon (available as T-shirt or hoodie), which does indeed parse as a Venn diagram if you take the left circle as representing “things with flat tails playing guitar-like instruments” and the right circle as representing “things with duck bills playing keyboard-like instruments”. Remember — my love is “very picky” about Venn diagram jokes — the intersection in a Venn diagram is not a blend of the things in the two contributing circles, but is rather, properly, something which belongs to both the groups of things.

The 4th of is also William Rowan Hamilton’s birthday. He’s known for the discovery of quaternions, which are kind of to complex-valued numbers what complex-valued numbers are to the reals, but they’re harder to make a fun Google Doodle about. Quaternions are a pretty good way of representing rotations in a three-dimensional space, but that just looks like rotating stuff on the computer screen.

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John Venn, an English philosopher who spent much of his career at Cambridge, died in 1923, but if he were alive today he would totally be dead, as it is his 180th birthday. Venn was named after the Venn diagram, owing to the fact that as a child he was terrible at math but good at drawing circles, and so was not held back in 5th grade. In celebration of this philosopher’s birthday Google has put up a fun, interactive doodle — just for today. Check it out.

Note: all comments on this post must be in Venn Diagram form.

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## Next In A Continuing Series

For today’s entry in the popular “I suppose everybody heard about this already like five years ago but I just found out about it now”, there’s the Online Encyclopedia of Integer Sequences, which is a half-century-old database (!) of various commonly appearing sequences of integers. It started, apparently, when Neil J A Sloane (a graduate student at Cornell University) needed to know the next terms in a sequence describing a particular property of trees, and he couldn’t find a way to look it up and so we got what I imagine to be that wonderful blend of frustration (“it should be easy to find this”) and procrastination (“surely having this settled once and for all will speed my dissertation”) that produces great things.

It’s even got a search engine, so that if you have the start of a sequence — say, “1, 4, 5, 16, 17, 20, 21” — it can find whether there’s any noteworthy sequences which begin that way and even give you a formula for finding successive terms, programming code for the terms, places in the literature where it might have appeared, and other neat little bits.

This isn’t foolproof, of course. Deductive logic will tell you that just because you know the first (say) ten terms in a sequence you don’t actually know what the eleventh will be. There are literally infinitely many possible successors. However, we’re not looking for deductive inevitability with this sort of search engine. We’re supposing that our sequence starts off describing some pattern that can be described by some rule that looks simple and attractive to human eyes. (So maybe my example doesn’t quite qualify, though their name for it makes it sound pretty nice.) There’s bits of whimsy (see the first link I posted), and chances to discover stuff I never heard of before (eg, the Wilson Primes: the encyclopedia says it’s believed there are infinitely many of them, but only three are known — 5, 13, and 563, with the next term unknown but certainly larger than 20,000,000,000,000), and plenty of stuff about poker and calendars.

Anyway, it’s got that appeal of a good reference tome in that you can just wander around it all afternoon and keep finding stuff that makes you say “huh”. (There’s a thing called Canada Perfect Numbers, but there are only four of them.)

On the title: some may protest, correctly, that a sequence and a series are very different things. They are correct: mathematically, a sequence is just a string of numbers, while a series is the sum of the terms in a sequence, and so is a single number. It doesn’t matter. Titles obey a logic of their own.

## 16,000 and a Square

I reached my 16,000th page view, sometime on Thursday. That’s a tiny bit slower than I projected based on May’s readership statistics, but May was a busy month and I’ve had a little less time to write stuff this month, so I’m not feeling bad about that.

Meanwhile, while looking for something else, I ran across a bit about mathematical notation in Florian Cajori’s A History of Mathematical Notation which has left me with a grin since. The book is very good about telling the stories of just what the title suggests. It’s a book well worth dipping into because everything you see written down is the result of a long process of experimentation and fiddling about to find the right balance of “expressing an idea clearly” and “expressing an idea concisely” and “expressing an idea so it’s not too hard to work with”.

The idea here is the square of a variable, which these days we’d normally write as $a^2$. According to Cajori (section 304), René Descartes “preferred the notation $aa$ to $a^2$.” Cajori notes that Carl Gauss had this same preference and defended it on the grounds that doubling the symbol didn’t take any more (or less) space than the superscript 2 did. Cajori lists other great mathematicians who preferred doubling the letter for squaring, including Christiaan Huygens, Edmond Halley, Leonhard Euler, and Isaac Newton. Among mathematicians who preferred $a^2$ were Blaise Pascal, David Gregory (who was big in infinite series), and Wilhelm Leibniz.

Well of course Newton and Leibniz would be on opposite sides of the $aa$ versus $a^2$ debate. How could the universe be sensible otherwise?

## A Picture Showing Why The Square Root of 2 Is Irrational

Seyma Erbas had a post recently that I quite liked. It’s a nearly visual proof of the irrationality of the square root of two. Proving that the square root of two is irrational isn’t by itself a great trick: either that or the proof there are infinitely many prime numbers is probably the simplest interesting proof-by-contradiction someone could do. The Pythagoreans certainly knew of it, and being the Pythagoreans, inspired confusing legends about just what they did about this irrationality.

Anyway, in the reblogged post here, a proof (by contradiction) that the square root of two can’t be rational is done nearly entirely in pictures. The paper which Seyma Erbas cites, Steven J Miller and David Montague’s “Irrationality From The Book”, also includes similar visual proofs of the irrationality of the square roots of three, five, and six, and if the pictures don’t inspire you to higher mathematics they might at least give you ideas for retiling the kitchen. Miller and Montague talk about the generalization problem — making similar diagrams for larger and larger numbers, such as ten — and where their generalization stops working.

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Yesterday I came a across a new (new to me, that is) proof of the irrationality of $sqrt{2}$. I found it in the paper “Irrationality From The Book,” by Steven J. Miller, David Montague, which was recently posted to arXiv.org.

Apparently the proof was discovered by Stanley Tennenbaum in the 1950′s but was made widely known by John Conway around 1990. The proof appeared in Conway’s chapter “The Power of Mathematics” of the book Power, which was edited by Alan F. Blackwell, David MacKay (2005).

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