In the meanwhile, here’s a little discussion about the glass of water problem.

Each time we add or subtract 50%, we are multiplying the quantity of water by 1/2 or 3/2. If we began with 1 glass’ worth, at each stage, we’ll have a quantity of the form 3^m/2ⁿ with m,n>0  Of course that can never equal 1, but we can get very close if m/n is very close to log₃ 2 = 0.63092975357145743710…

Unfortunately, there’s a serious problem: m/n has to hit the mark pretty closely in order for 3^m/2ⁿ to get really close to 1, and to get within “one molecule”s worth, m and n have to be huge indeed. 

How huge? Well, let’s see: an 8 oz. glass of water contains about 10²⁵ molecules; to get within 1/10²⁵ of 1, we need m=31150961018190238869556, n=49373105075258054570781 !!  One immediate problem is that if you make a switch about 100,000 times a second, this takes about  as long as the universe is old!

But there’s a more serious issue.

In a glass of water, there’s a real, specific number of molecules. Each time we add or subtract 50%, we are knocking out a factor of 2 from this number. Once we’re out of factors of 2, we can’t truly play the game any more, because we’d have to be taking fractions of water molecules. (For example, if we begin with, say, 100 molecules, after just two steps we’d be out of 2’s since 100=2*2*some other stuff.

But even though there are a huge number of water molecules in a glass of water, even if we arrange it so that there are as many 2’s as possible in that number, there just can’t be that many: 2⁸³ is about as good as we can do (of course, we won’t have precisely 8 ounces any more, but still.)

If we are only allowed 83 or so steps, the best we can do is only m= 53, n = 84 (Let’s just make the glass twice as big to accommodate that), and, as Byon noted, 3^53/2^84 is about 1.0021– not that close, really!

We HAVEN’T yet fully answered the coffee and cream question: a follow up post will be coming soon!

1) Send us your candidates for an interesting fact about the number 2012; the winner will receive a handsome Math Prize! As mentioned on the podcast, already its larger prime factor, 503, has a neat connection to the primes 2,3,5, and 7.

2) So what is it about the tetrahedral numbers, and choosing things? In particular, why is the Nth tetrahedral number (aka the total number of gifts on the Nth day of Christmas) is exactly the same as the number of ways of choosing 3 objects out of (N+2)? Not hard, really, to prove, but can you find a simple or intuitive explanation?

3) Finally, about those M&M’s. Maybe I exaggerated a little bit when I claimed this problem holds all the secrets of the thermodynamics of the universe, but I don’t see how! Many classic equations, such as Newton’s Law of Cooling or the Heat Equation, the laws of thermodynamics, and fancier things as well, can all be illustrated by shuffling red and blue M&M’s around. What I don’t understand is how anything got done before M&M’s were invented!

(Since we’ve been offline for a week or so, due to a tremendous ice storm that has paralyzed the town, we add a special bonus: the very first Math Factor episode ever aired, from January 25, 2004.)

This week we pose two interesting e related puzzles:

1) Obtain a large bowl with N strands of spaghetti; grab two loose ends and tie them together. Repeat, until all the loose ends are paired. You will now have a bowl full of loops of spaghetti. On average, what is the expected number of loops?

2) N people walk into a room; each of their (unique) names has been written on a nametag and placed into a bowl. If each person picks a nametag at random, what is the probability that no one gets the right name?

In both cases, the interesting thing is what happens as N increases without bound.

When we were done taping, I remarked to Kyle that, well, surely that’s the end of e related stuff for a while. But I just remembered one of the best e related puzzles of all. We’ll add it here as a bonus:

3) Someone has written counting numbers, one on each of N cards. You don’t have any idea what the largest number is. The cards are shuffled and arranged in a line face down.

You turn the cards over, discarding the cards one by one; you may stop at any time. Your goal is to pick the card with the largest number. (You can’t go back and retrieve a discarded card, and you can’t continue once you stop).

Your strategy, then, is to flip over some number M of cards just to see what the field is like, then taking the first card better than any of the cards in your test sample.

You don’t want M to be too small– you need to get a feel for how big the numbers might be; but you don’t want M to be too big— you don’t want to actually waste the biggest number in your test.

Amazingly, the optimal M works 1/e (almost 37%!) of the time. What is this M, and why does this work?

Amazingly, an optimal path for the Princess is to swim in a half circle of radius 1/8 that of the lake, then dash out to the edge.
We’ll give an analytic proof, but we could give a totally synthetic (geometric) proof as well.

An Analytic, Calculussey Proof

First of all, we should ask:

At a given moment, how fast can the Princess swim towards the edge of the lake?

Let’s not worry about our actors changing directions— this doesn’t really affect our thinking (remember we assumed they could change directions instantly). So the Beast will be dashing around the lake and the princess taking some route to get away.

Let’s scale things so that the lake has radius 1, and the Beast’s speed is 1. At time t then, he will have travelled a distance of t, and the Princess will have traveled one-fourth as far, ¼t.

At a given moment, how fast can the Princess swim towards the edge of the lake?

Consider what happens over a very very small interval of time, of size Δt, when the Princess is r away from the center of the lake.

The Beast travels Δt along the shore. The Princess has to keep the beast on the opposite side of the lake, and so has to swim proportionally less,
r Δt around.

How much can she increase r, if she is travelling a total of ¼ Δt, and needs only to go around a distance of r Δt? It is well worth learning how to think like an 18th Century Mathematician: essentially we have a right triangle, and can use the Pythagorean Theorem. (We don’t really have exactly a right triangle, but the difference is negligible, and as Δt decreases, is less and less important, until irrelevant in the limit.)

By the Pythagorean Theorem, an increment of change of radius Δ r is √( (¼ Δ t)² – (r Δ t)²)

(In the limit, this is good enough; all the differences are vanishingly small compared to Δ, as Δ decreases.)

In other words, in the limit,

dr = √ (¼²–r²) dt

Using a little calculus, and remembering that at t = 0, we have r = 0 as well, we can check that

r = ¼ sin 4t

Remember that t is not only measuring time, but also the monster’s distance around the lake; the lake is radius 1, so this is also the radian measure around the lake. In other words, the initial part of the Princess’ path, in polar coordinates is:

r(θ) = ¼ sin 4θ

a circle as promised!