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May 27

Heat equation on the ice rink

[moved here from the science desk]

Is it possible to solve the heat equation on this shape analytically (not just numerically)? On an ellipse or rectangle, the solution is relatively simple (in the former case we use polar coordinates). But an ice rink is most conveniently described in a piecewise fashion, as the union of two semi-circles with a rectangle. In particular, does knowledge of the solutions on the ellipse and rectangle help?

Homogeneous boundary conditions are okay, since to me, they seem to be a reasonable physical assumption. Also I'm not sure about how the cooling of the ice physically works but if coolant coils pass under the whole ice sheet, I would think that those heat sources would be relatively easy to incorporate.

I would also be interested if there were a better equation than the heat equation for this particular physical problem.--Jasper Deng (talk) 02:14, 27 May 2015 (UTC)[reply]

Properly, though, an ice rink is not two ellipses and a rectangle. It's a rectangle with rounded corners. Unlike in an ellipse, the short and long ends of an ice rink are both perfectly straight. You'd have to describe it as a rectangle less the four corners, which are pretty close to the space left over from a circle inscribed inside a square. --Jayron 32 02:31, 27 May 2015 (UTC)[reply]

An expansion is always possible in eigenfunctions of the Laplace operator, but the eigenfunctions for such a domain are not analytically determinable in terms of standard special functions. You can infer something about the eigenvalues by the shape of the domain; for instance, the first eigenvalue is related to the area of the domain by an isoperimetric type inequality. But even knowing that the domain decomposes into somewhat nice pieces doesn't give much information about the corresponding eigenfunctions. The standard solutions all rely on separation of variables, and so in particular are dependent on having a region which is "nice" in some orthogonal coordinate system: these are the coordinate systems in which the Laplacian is separable. In two dimensions, there are five such coordinate systems: Cartesian coordinate system, Polar coordinate system, Parabolic coordinate system, Bipolar coordinates, Elliptic coordinates. (There are also periodic solutions on elliptic curves that appear in number theory, and have solutions given by Jacobian theta functions. But that's not going to help here either.) Sławomir Biały (talk) 02:56, 27 May 2015 (UTC)[reply]

I was hoping that someone had already investigated eigenfunctions of the Laplacian on this domain, because I think ice rinks are a very important practical application of the heat equation, but it looks like any such orthogonal expansion is going to rely on nontrivial functions anyways.--Jasper Deng (talk) 03:11, 27 May 2015 (UTC)[reply]

In his blog, Terrence Tao has talked about scarring in the Bunimovich stadium, which has implications for the Laplacian eigenvalue distribution. An ice hockey rink is a generalization of this shape, but I'd guess that the same sort of chaos, and thus scarring, would occur in it, too.--Mark viking (talk) 03:25, 27 May 2015 (UTC)[reply]

Simple statistical proof

A very sexist population prefers boys to girls. Every family tries various birthing strategies to have more sons than daughters, such as repeatedly giving birth until they get one son. Prove that no strategy can have any effect on the sex ratio.

I know this is easy to prove using the optional stopping theorem, but I'm trying to explain this to a friend without a math background. Is there a simpler, more intuitive proof that a smart layman who knows high school math could understand? It doesn't have to be absolutely rigorous, but has to be rigorous enough to be convincing. Thanks! --Bowlhover (talk) 03:18, 27 May 2015 (UTC)[reply]

You are assuming that this preference nonetheless cannot change the actual probability of having girls or boys, and that you're ignoring abortions and other ways one gender could be physically removed, right?--Jasper Deng (talk) 04:24, 27 May 2015 (UTC)[reply]

(edit conflict) Do some of the birthing strategies include Sex-selective abortion? If so, that would alter the ratio. For example, in China, the gender difference is about 2.5% in favor of males, (see Demographics of China which doesn't seem like much, but is significantly different than the Human sex ratio of the world, which is about 0.5% in favor of males. Indeed, China itself is such a huge population, they are actually driving the world average a big portion of the world average. --Jayron 32 04:25, 27 May 2015 (UTC)[reply]

This needn't be true even if you disallow abortion. For example, if different mothers have different chances of having male children, which I think is actually the case, then stopping at the first girl will result in more boys in total than having children indiscriminately. So your friend's suspicion of the claim may well be justified.

To prove this you need to assume that the sex of each baby is an independent random coin flip, and at that point I'd just say that you've assumed your conclusion. If that isn't obvious enough, you could replace the many couples with a single (immortal) couple trying all of the strategies in sequence. E.g., instead of a bunch of couples each having kids until one is a boy, the one couple has kids until one is a boy, then has more kids until one is a boy, etc. It seems intuitively obvious that this will give the same sex ratio as when there are many couples, and that the ratio will be the chance of flipping heads vs tails. -- BenRG (talk) 04:52, 27 May 2015 (UTC)[reply]

I'm not sure the last part is that obvious. An intuitive proof of that would answer the OP. -- Meni Rosenfeld (talk) 15:51, 27 May 2015 (UTC)[reply]

If we assume that the probability of conceiving a boy is the same for all couples, is independent of previous conceptions and does not change over time then we can combine a set of parallel sequences of conceptions into one single sequence (placing them end to end, or interleaving them in chronological order, or whatever scheme suits us) without affecting probabilities. If we drop any of these assumptions (e.g. if some couples are more likely to conceive boys than others; if a boy is more likely to be conceived after a boy; if first children are more likely to be boys) then we cannot combine sequences without affecting probabilities, and conversely there are possible strategies for conceiving more boys. Gandalf61 (talk) 16:34, 27 May 2015 (UTC)[reply]

The part where you can combine multiple couples into a single one is clear. The part where a single couple can have no strategy (assuming each birth is a fair coin toss), not so much. It seems to me that a proof of that would not be much different from the proof of the optional stopping theorem, which in turn does not seem completely trivial. -- Meni Rosenfeld (talk) 18:20, 27 May 2015 (UTC)[reply]

Here is another intuitive "proof": Consider an population of a million people with one fair coin who play the following game: A person tosses the coin till it lands on head (analogous to having a son) and then passes the coin to the next person, who does the same, till each person has had their turn. Now which-person-gets-their-hand-on-the-coin-when obviously depend upon the exact sequence of heads and tails, but overall all that has happened is that a fair coin has been tossed (about) 2 million times. So we should expect a million heads and a million tails.

And as a corollary it should be obvious that no individual can expect to beat the odds either by handing over the coin as soon as they get a head, since everyone is using that strategy and if it worked, the difference would show up in the total number of heads. Also nothing in the above proof depends upon the coin being "fair", although they do have to use the same coin (as BenRG and Gandalf61 have pointed out). Abecedare (talk) 01:43, 28 May 2015 (UTC)[reply]

I think people above are over-complicating things. The proposition hinges on the assumption that the births are independent events with respect to sex of the child. That is, the probability of getting a boy/girl in any individual birth is unaffected by the outcome of any other birth. (This is the definition of an independent event.) We also assume that the probability of a boy/girl birth is the same for all births, and doesn't depend on other factors like which mother/time of year/age of father/etc. Once you make those assumptions, it doesn't matter how the births are arranged, because each is unaffected by any other and has a constant probability. At that point, to get the ratio of boy to girl births, you can just use the (frequentist) definition of probability: a probability of an outcome is equivalent to the frequency of that outcome in a large population of (independent) trials. Strategy doesn't matter, because it doesn't figure into the calculation. - Of course, that's assuming sex at birth is an independent event and every birth has the same probability of a boy/girl birth. If that's not the case, then you possibly can change the sex ratio by strategy. But you would need to clarify details on how the assumptions are broken to figure out how. -- 162.238.240.55 (talk) 13:16, 28 May 2015 (UTC)[reply]

I think you are simultaneously over-complicating and over-simplifying things. It seems we all agree the problem can be reduced to fair coin tossing (can be generalized to unfair coins, but to examine the core issue it's enough to consider a fair coin). So we can simply go with that rather than repeating the domain-specific problem.

With that behind us, I maintain that it's not obvious that no strategy can work, and that the proof is not trivial. You should keep in mind that if a strategy is used, then given the total number of tosses, the different tosses are no longer independent of each other, and the probability of each particular toss to result in heads is no longer 1/2. So I don't think you can just naively apply "everything is independent".

The frequentist definition of probability assumes that we repeat the experiment a number of times independent of the results each time. It does not apply directly to a case that the number of experiments depends on their outcomes. You'd need to actually prove that. -- Meni Rosenfeld (talk) 16:48, 28 May 2015 (UTC)[reply]

Exactly... this is the difference between Gaussian statistics and Poisson statistics. Nothing about the universe changes when we switch models; but the fact that we may optionally change our behaviors because of prior outcomes can affect the distribution. As I recall, the central limit theorem (or more specifically, de Moivre's other theorem) says that the Poisson distribution approaches the Gaussian distribution as the number of events increase. Therefore, I would go so far as to say that any strategy that seeks to keep n small actually has a better chance of escaping from the expected value. Hence, a strategy exists that may actually have an effect on the outcome distribution - although we have very little control over which direction this effect manifests! In the extreme example: if we deduce a strategy that permits only one baby to be born, ever, then we are guaranteeing a ratio of male-to-female of (either) 1:0 or 0:1. This is not equal to 50% - it is quite far from the expected value! - And what is more, we have guaranteed this strong deviation from the expected value by selecting our strategy! However, this specific strategy can't enable us to control the outcome (0:1 or 1:0)!

Altogether, though, I think it provides enough justification for me to question the premise. I don't think you can prove that there does not exist any strategy that might affect the ratio: I have provided a counterexample. Perhaps the original premise needs to be restated in more formal language before we try to prove anything. Nimur (talk) 22:43, 28 May 2015 (UTC)[reply]

Thanks everyone! I think BenRG's simple proof is convincing enough, even if it's not absolutely rigorous. I also think it's fascinating that the strategy of producing children until getting a son decreases the proportion of boys in the world, assuming different mothers have different chances of producing sons.

@Nimur: You're right that I didn't phrase the question too well. I meant that the expected number of boys is equal to the expected number of boys, given any strategy. Your strategy where only one child is ever born has this property as well, so it's not a counterexample. --Bowlhover (talk) 08:24, 29 May 2015 (UTC)[reply]

If every pregnancy has an equal chance of producing a boy or a girl, it should be clear that no substitution of one pregnancy for another can change the ratio. Therefore, from n births there will be m boys on average, no matter what.

However, the human sex ratio is potentially affected by various factors; for example our article mentions hepatitis B, which I didn't know. Clearly if someone carries a genetic disease that kills one sex but not the other (sex-linked trait) that person's ratio will be different. In most of these situations, the way to "beat" the game is to have those who have had a son go on and produce more kids in the hope that they were unusually prone to have sons. Of course, such is monstrously intrusive and minimally effective! Wnt (talk) 00:29, 30 May 2015 (UTC)[reply]

Indeterminate forms with tetration

Go to tetration. Are these (using ^^ for tetration) indeterminate forms??

1^^(-1)
(-1)^^0

Note that (-1)^^1 is not indeterminate; it's unambiguously -1. Georgia guy (talk) 23:36, 27 May 2015 (UTC)[reply]

They are not indeterminate forms, in the usual sense of that phrase, which is a closed-end list of exactly seven expressions. You could try to make an abstract definition of indeterminate-form-in-general if you wanted to (and no doubt someone has), but no such definition has been generally adopted, probably because it has not been seen as useful.

However, none of the definitions in our tetration article assign a value to either of these expressions. --Trovatore (talk) 23:47, 27 May 2015 (UTC)[reply]

The question of them being indeterminate forms is not particularly meaningful even in the broadest sense, since that only sensibly generalizes to when both arguments are continuous. Georgia guy, do you mean are they well-defined? Tetration could be extended to zero and negative "exponents" by saying that we add a base for every increment of the "exponent", so we ask "what number do we have to treat as ⁰x so that ¹x = x^(⁰x)? The natural answer is ⁰x = 1 (ignoring quibbles about x = 0). Similarly, ⁻¹x = 0 for all x. But ⁻²x and further cannot be defined sensibly. So I get (under this extension), that 1^^(−1) = ⁻¹1 = 0, and (−1)^^0 = ⁰(−1) = 1. —Quondum 01:06, 28 May 2015 (UTC)[reply]

To me it looks like both cases are covered in the existing article. It explicitly says 1^^(-1) is not unique in the section about extension to negative heights, and a unique value of zero for odd n seems to follow from the section about stacks with height zero. Both these are definitions, however, and so should be addressed by whoever is defining tetration as you use it. (This is the kind of thing where you'd really have to RTFM for the software library) Wnt (talk) 17:34, 28 May 2015 (UTC)[reply]