Wikipedia:Reference desk/Archives/Mathematics/2012 April 10

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April 10

two parts

first, is having a designated driver, even one who doesn't prefer the role, a nash equilibium? (since ifhe starts drinking he'll kill us all?)

secondly, if this is the case I would like a joke, a play on sobriety and 'equilibrium' but one that's funny. I can't think of one. 94.27.165.121 (talk) 01:04, 10 April 2012 (UTC)

my work so far

I know you like to see what I've done on my own, here is what I have: "What keeps anyone from getting drunk first? Nash equilibrium." I know, it doesn't really work. (I intended this one to be about social acceptability). Or: "The students all wanted to get drunk, but nobody wanted to be the one to get beer. They were in a Nash equilibrium".

This doesn't really work as a pun.

I've been trying for some time but I just can't do it on my own. Hope you can help here.

"What keeps professors from getting drunker than their students? Nash equilibrium..." this one 'sounds right' but doesn't make any sense. (it's meaningless). I'd like something along these lines but that works... 94.27.165.121 (talk) 01:12, 10 April 2012 (UTC)

Take a look at our article Nash equilibria. — Fly by Night (talk) 01:49, 10 April 2012 (UTC)

I read that article, which gives both formal and informal definitions, but does not use it in a joke. 134.255.115.229 (talk) 10:17, 10 April 2012 (UTC)

Well, if dealing with driving, you might want to include the old car maker, Nash Motors. Let's see: "Two drunks drove an antique Nash convertible into a guardrail, and it was left hanging over the edge of a cliff. Neither wanted to try to get out, for fear they would send the car over. They were in a Nash equilibrium." StuRat (talk) 05:14, 10 April 2012 (UTC)

That one's good, though a bit long. I wonder if it's possible to simplify by factor out the Nash Motors term (making it a simple instead of double pun) since 'equilibrium' is fine to pun on vs. 'precariousness' (edge of cliff) - you don't have to add nash motors to it to make the joke quadratic. To me, this is as good as maintaining one's equilibrium i.e. sobriety. Could you try simplifying the joke, Stu? 134.255.115.229 (talk) 10:17, 10 April 2012 (UTC)

They got themselves into a Nash equilibrium after they lost their equilibrium. Great idea, Stu, bringing in the old Nash cars! Duoduoduo (talk) 15:43, 10 April 2012 (UTC)

Thanks. I usually try to add as many dimensions to my puns as possible. The last line is also supposed to sound like "They were in a Nash Metropolitan" (this is why I didn't specify the model, but note that that is a model which came in a convertible.) StuRat (talk) 21:22, 10 April 2012 (UTC)

Volumes of parallelepipeds in higher dimensions.

One of the basic things we are taught in Calc III is the area of the parallelogram formed by two vectors ${\vec {x}}_{1},{\vec {x}}_{2}\in \mathbb {R} ^{2}$ is simply given by $\left\|{\vec {x}}_{1}\times {\vec {x}}_{2}\right\|$. Additionally, the area of the parallelepiped formed by the three vectors ${\vec {x}}_{1},{\vec {x}}_{2},{\vec {x}}_{3}\in \mathbb {R} ^{3}$ is simply given by $({\vec {x}}_{1}\times {\vec {x}}_{2})\cdot {\vec {x}}_{1}$. First, I began thinking about how to simply find the area of the parallelogram formed by two vectors ${\vec {x}}_{1},{\vec {x}}_{2}\in \mathbb {R} ^{n}$. If $n=4$, for instance, we cannot use the formula above since we cannot calculate a cross product in four dimensional space (however, we can easily compute dot products). Thus, I formulated the following: ${\text{Volume}}=\left\|{\vec {x}}_{1}\right\|\left\|{\vec {x}}_{2}\right\|\sin \theta =\left\|{\vec {x}}_{1}\right\|\left\|{\vec {x}}_{2}\right\|{\sqrt {1-\cos ^{2}\theta }}$. However, $\cos \theta$ can be expressed in terms of the dot product. Thus, ${\text{Volume}}=\left\|{\vec {x}}_{1}\right\|\left\|{\vec {x}}_{2}\right\|{\sqrt {1-\left({\frac {{\vec {x}}_{1}\cdot {\vec {x}}_{2}}{\left\|{\vec {x}}_{1}\right\|\left\|{\vec {x}}_{2}\right\|}}\right)^{2}}}={\sqrt {\left\|t{\vec {x}}_{1}\right\|^{2}\left\|{\vec {x}}_{2}\right\|^{2}-({\vec {x}}_{1}\cdot {\vec {x}}_{2})^{2}}}$. This formula works for all $n$, and can be shown to equal $\left\|{\vec {x}}_{1}\times {\vec {x}}_{2}\right\|$ when $n=2$. Doing this same process for three-dimensional parallelepipeds proved more difficult since the required angles were not as readily found. Additionally, if we wanted to to extend this further by finding the volume of a four-dimensional parallelepiped defined by four linearly independent vectors, how would one even do that? Even in the base case of vectors in $\mathbb {R} ^{4}$ I wouldn't know how to compute this "four-volume". How would one formulate a generic formula for the $n$-volume of the parallelepiped formed by $n$ vectors in $m$-dimensional space? — Trevor K. — 04:38, 10 April 2012 (UTC) — Preceding unsigned comment added by Yakeyglee (talk • contribs)

I think it's just determinant of the matrix of the vectors determining the hyperparallelepiped. Widener (talk) 06:01, 10 April 2012 (UTC)

So for example, a hyperparallelepiped formed by the four vectors in $\mathbb {R} ^{4}$ $(1,0,0,0),(1,1,0,0),(0,0,2,1),(0,0,1,2)$ would have volume $\det {\begin{bmatrix}1&1&0&0\\0&1&0&0\\0&0&2&1\\0&0&1&2\end{bmatrix}}=3$. Widener (talk) 06:09, 10 April 2012 (UTC)

As for finding the n-volumes of lower dimensional parallelepipeds in higher dimensional m-space, just find an orthonormal basis for the subspace of dimension n containing your parallelepiped (using Gram-Schmidt or some other way) and then it is the determinant of the matrix which transforms this basis into the vectors of your parellelepiped (you could do a coordinate transformation into the standard basis for $\mathbb {R} ^{n}$ to make things easier). In other words, I don't think there is a general formula (or if there is it would be complicated), the way you do this is more algorithmic. Widener (talk) 06:18, 10 April 2012 (UTC)

The general name is parallelotope; follow the link for a more general approach to your problem. You may find it worthwhile to explore geometric algebra, which replaces the cross product (${\vec {a}}\times {\vec {b}}$) with the more useful wedge product (${\vec {a}}\wedge {\vec {b}}$). — Quondum^☏ 07:47, 10 April 2012 (UTC)

The volume of k-dimensional parallelotope in n-dimensional space, formed by $x_{1},\cdots ,x_{k}\in \mathbb {R} ^{n}$ is ${\sqrt {\det {\begin{bmatrix}x_{1}\cdot x_{1}&x_{1}\cdot x_{2}&\cdots &x_{1}\cdot x_{k}\\x_{2}\cdot x_{1}&x_{2}\cdot x_{2}&\cdots &x_{2}\cdot x_{k}\\\cdots &\cdots &\cdots &\cdots \\x_{k}\cdot x_{1}&x_{k}\cdot x_{2}&\cdots &x_{k}\cdot x_{k}\end{bmatrix}}}}$ 77.125.72.170 (talk) 09:34, 10 April 2012 (UTC)

I've never heard of "hyper-volume". You just call it volume. The hyper- prefix is usually attached to manifolds of codimension one, e.g. hyper-plane, hyper-sphere or hyper-surface. For example, we don't talk about "hyper-distance". The oriented volume of parallelotope spanned by $v_{1},\ldots ,v_{n}$ is, as mentioned, just the volume of the matrix whose first column is $v_{1},$ whose second column is $v_{2},$ etc. This definition has the advantage that you don't need an inner product to have been defined, e.g. the dot product. In general, you can define a volume form on any orientable manifold. — Fly by Night (talk) 12:47, 10 April 2012 (UTC)

"hypervolume" has ~59k google hits. While I take your point re:codimension one, hypervolume is a term used in many contexts, perhaps more likely outside of the pure math domain. Commonly, it is used to emphasize that one is dealing with objects in more than three dimensions. The fact that you haven't heard of it doesn't mean it's not commonly used ;) 128.118.84.113 (talk) 15:10, 10 April 2012 (UTC)

I get 7,500 hits on Google Scholar. Many of these article seem to be from computer scientists, and not mathematicians. It's unsurprising that they misapply the the prefix hyper. Don't forget that this is, after all, the mathematics reference desk. I suppose I could always start calling the dot product the potato product. I would, of course, like to hear the opinion of other mathematicians. Maybe I have become blinkered in my years of study. — Fly by Night (talk) 20:47, 10 April 2012 (UTC)

Semantics schmemantics. Widener (talk) 21:22, 10 April 2012 (UTC)

No one talks about hyper-distance because that makes no fucking sense. --128.62.87.144 (talk) 13:06, 11 April 2012 (UTC)

I've read of hypervolume in a math context quite a few times. Also, I googled "potato product" and got lots of hits -- it's possible that they're not all math hits for dot product synonyms, though. Duoduoduo (talk) 17:16, 11 April 2012 (UTC)

Our article Four-dimensional space#Hypersphere uses the term "hypervolume". Duoduoduo (talk) 17:53, 11 April 2012 (UTC)

I'd be interested to see who introduced "hypervolume" into the article. You'll notice that the volume form article doesn't mention hypervolume, and neither do any of these books:

• Nomizu, K.; Sasaki, T. (2008). Affine Differential Geometry. Cambridge University Press. ISBN 0521064392.
• An-Min, L.; Udo, S.; Guosong, Z. (1993). Global Affine Differential Geometry of Hypersurfaces. Walter De Gruyter. ISBN 3110127695.
• Buchin, S. (1983). Affine Differential Geometry. Routledge. ISBN 0677310609.

It should be noted that affine differential geometry is all about volume, and volume forms on submanifolds. Obviously some people will use "hypervolume", just like some people pronounce Euler as You-ler. But those in the know don't. — Fly by Night (talk) 21:53, 11 April 2012 (UTC)

Yes, we probably shouldn't use the term "hypervolume". Although I acknowledge the term does exist, it is not in very wide use professionally. Sławomir Biały (talk) 22:48, 11 April 2012 (UTC)

Notation, conditional probability

A quick question: Do the notations $P(A|B,C)$ and $P(A|B\cap C)$ mean exactly the same thing? Thanks! --91.186.78.4 (talk) 09:46, 10 April 2012 (UTC)

Trying it out with Venn diagrams gives me the feeling I got it exactly wrong: P(A|B,C)= P(A|B U C)? Which version, if any, is the correct one? --91.186.78.4 (talk) 12:22, 10 April 2012 (UTC)

I think the problem is that the latter notations are unambiguous, but the former IS ambiguous. My instinct (based on reading many probability texts), is to interpret the commma as an "and", reading it as "probability of A, given B and C". So I would opt for considering $P(A|B,C)=P(A|B\cap C)$. On the other hand, one could also reasonably use the comma to mean "or", which would give you the second option. We could probably give you better answers if you give us some examples of where you've seen the comma notation. SemanticMantis (talk) 15:02, 10 April 2012 (UTC)

The comma always means "and", as does $\cap$. Duoduoduo (talk) 15:53, 10 April 2012 (UTC)

I agree, in this context a comma is understood to mean "and". So in principle, the answer to the original question is "yes".

However, I think the former notation is a bit more general. I think $\cap$ would be used only if B and C are events. A comma can also be used when B and C are variables, and then you're referring to the conditional distribution (a function of 3 variables) with a more condensed and fancy notation. -- Meni Rosenfeld (talk) 17:11, 10 April 2012 (UTC)

Good point, thanks. SemanticMantis (talk) 22:09, 10 April 2012 (UTC)

Thanks very much, everyone, for your answers to my question! --NorwegianBlue ^talk (Original Poster) 17:44, 11 April 2012 (UTC)

Octahedral and icosahedral symmetry

I'm asking this question for The Doctahedron: is it possible to have a polyhedron with both octahedral and icosahedral symmetry? (I don't think it's possible, but I'm not completely sure.) Double sharp (talk) 11:10, 10 April 2012 (UTC)

If there were, I'm betting on myself it would be the tetrahedron, but I'm looking higher than that. 68.173.113.106 (talk) 13:52, 10 April 2012 (UTC)

But it doesn't have any higher symmetry than T_d. Double sharp (talk) 12:56, 11 April 2012 (UTC)

Looking at List of spherical symmetry groups, I don't see anything that refines both octahedral symmetry and icosahedral symmetry. 98.248.42.252 (talk) 14:46, 11 April 2012 (UTC)

Would you settle for the six-dimensional cartesian product of an octahedron and an icosahedron? —Tamfang (talk) 16:21, 11 April 2012 (UTC)

I don't think that was the intention of the question. (On an unrelated note, is it possible to find the Cartesian product of any two polytopes as long as at least one of them has a hypercubic symmetry?) Double sharp (talk) 13:21, 12 April 2012 (UTC)

You can find the Cartesian product of any two sets. Rckrone (talk) 16:01, 12 April 2012 (UTC)

The order-4 dodecahedral honeycomb has octahedral symmetry about each vertex and icosahedral symmetry about each cell. —Tamfang (talk) 04:39, 13 April 2012 (UTC)

mathematical conjectures formerly assumed true

I would like a list of conjectures that had generally been assumed true (for example, perhaps in number theory it was true for a great many integers, which had been checked, or whatever) but have been disproved. I'm specifically looking for conjectures everyone 'assumed' were true, not just ones that had been unresolved. Ones where the eventual result (disproof) was a genuine surprise to all. 188.6.92.6 (talk) 13:00, 10 April 2012 (UTC)

1. It was assumed by the Pythagoreans that any distance could be expressed by a rational number. The proof by Hippasus of Metapuntum that the diagonal of the unit square is irrational was a genuine surprise to all.
2. It was assumed that Euclid's Fifth Axiom could actually be proved. Non-euclidian geometry were a genuine surprise to all.
3. It was assumed that the 17-gon could not be constructed by ruler and compass. Gauss' solution was a genuine surprise to all.
4. It was assumed that algebraic equations could be solved by radicals. The abel-ruffini theorem was a genuine surprise to all.
5. It was assumed that true statements could be proved. Gödel's incompleteness theorem was a genuine surprise to all.

This list is not complete. Bo Jacoby (talk) 13:44, 10 April 2012 (UTC).

hey, thanks! Anything in the past few decades? (when computers could have checked millions, giving statistical/probabilistic weight to a conjecture, or a clear graph for the first few million items, which leads one to think it would have a certain limit or go on indefinitely, but this changes much later, or has a final term, etc. basically, since numerical analysis). thanks again for the above list. 188.6.92.6 (talk) 14:42, 10 April 2012 (UTC)

It doesn't exactly correspond to your question, but perhaps you will find this discussion interesting. --Iae (talk) 15:28, 10 April 2012 (UTC)

Was it really assumed that the 17-gon could not be constructed, or was it just considered an open question for all n that had not been constructed? I thought the latter. Duoduoduo (talk)

I was wondering that too - wasn't it more of a surprise when Gauss proved that there were non-constructable polygons? Also, Gauss himself did not produce the first explicit construction of the 17-gon, at least according to our article. AndrewWTaylor (talk) 16:06, 10 April 2012 (UTC)

We learn from Euclid how to construct regular n-gons for n=3,4,5,6,8,10,12,15,16,20,24,30 but not for n=7,9,11,13,14,17,18,19,21,22,23,25,26,27,28,29. It is no surprise if the n-gons of the second list are non-constructable. It was a surprise that the 17-gon turned out to be constructible. Making an explicit construction based on the algebraic expression of cos(2π/17) is trivial. Bo Jacoby (talk) 11:15, 11 April 2012 (UTC).

From my post about the Goldbach conjecture question of April 7:

from Pythagorean triple#Elementary properties of primitive Pythagorean triples:

There exist infinitely many Pythagorean triples with square numbers for both the hypotenuse c and the sum of the legs a+b. The smallest such triple[10] has a = 4,565,486,027,761; b = 1,061,652,293,520; and c = 4,687,298,610,289. Here a+b = 2,372,15922 and c = 2,165,01722. This is generated by Euclid's formula with parameter values m = 2,150,905 and n = 246,792.

Presumably this was assumed impossible before the smallest example was found.

Also, from Discover Magazine, May 2002, p. 82:

The equation w⁴ + x⁴ + y⁴ = z⁴ is harder. In 1988, after 200 years of mathematicians' attempts to prove it impossible, Noam Elkies of Harvard found the counterexample, 2,682,440⁴ + 15,365,639⁴ + 18,796,760⁴ = 20,615,673⁴.

(It turns out that this is not the smallest counterexample.) Duoduoduo (talk) 16:26, 10 April 2012 (UTC)

List_of_conjectures#Disproved_.28no_longer_conjectures.29 might be a place to start, or to add to from this discussion.John Z (talk) 01:26, 11 April 2012 (UTC)

Unique factorisation was, I think, thought to extend to quadratic integers, such that it was taken for granted and used in other proofs (which were therefore false). I can't remember where I read this, and the article doesn't mention it, so perhaps someone could confirm. IBE (talk) 20:04, 11 April 2012 (UTC)

Not only was it assumed that all algebraic equations could be solved by radicals; it was assumed that any real solution could be algebraically expressed in radicals of real numbers. Cardano's solution gave real solutions of the cubic in terms of cube roots of what we now call imaginary numbers in the case in which there are three real roots, and we now know that this is the best that you can do. See casus irreducibilis. Duoduoduo (talk) 22:19, 11 April 2012 (UTC)

Let pi(x) be the prime counting function, it was a surprise that pi(x) - Li(x) changes sign infinitely often; it also went against all numerical evidence at the time (see the page skewe's number). Another example is that Merten's Conjecture turned out false, this too had a large amount of computational evidence backing it. I'll see if I can't think of any others Phoenixia1177 (talk) 11:33, 14 April 2012 (UTC)

One might argue that showing the reals and the naturals have different cardnality were an example of this (though I'm not sure how well that fits) In the same vein, the existence of nonmeasurable sets (though, again, this is off the mark since it is more a philosophical thing than what you're talking about) I'm not sure if it was shocking, or not, but I would imagine that Rice's Theorem went against the grain of what most people anticipated; or, at least, what felt natural to them. Going back to the nonmeasureable thing, again, the Banach-Tarski paradox; people believed the opposite, but I don't think anybody really thought about it in a formal way (it was just assumed, the same way the Jordan Curve theorem was). You could probably also add to this list Weirstrass (sp?) everywhere continous nowhere diff. function. Probably others, but these aren't quite what you're looking for I suspect:-) Phoenixia1177 (talk) 11:51, 14 April 2012 (UTC)