Wikipedia:Reference desk/Archives/Mathematics/2009 May 29

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May 29

How many different types of six sided dice are there?

A six sided dice(die) is defined as a cube where every element of the numeric set {1,2,3,4,5,6} appears as a numeric symbol on a face of the dice(die) and each face have no more than one numeric symbol.

Two six sided dice is considered different from each other if one of the die cannot be made to appear identical to the other dice through actions of rotation, translation and shrinking/expanding.

I read the wikipedia article about dice but they do not mention how many type of six sided dice there are. 122.107.207.98 (talk) 12:33, 29 May 2009 (UTC)

Rotate the cube so that you face number 1. There are 5 choices for the number on the opposite side. The remaining 4 numbers are arranged around the cube in any permutation (of which there are 4! = 24) but each permutation is equivalent to 3 others by rotation. The total number is thus 5 × 24 / 4 = 30. — Emil J. 12:41, 29 May 2009 (UTC)

How does this change if you assume that the numeric symbols have 4 different and distinct orientations (as e.g. the Arabic numeral 1-6 in many typefaces) and you distinguish dice by that criterion? At worst you get 4⁶ permutations, but I suspect many of these are again rotations. --Stephan Schulz (talk) 12:59, 29 May 2009 (UTC)

(5*6)*(4^5)=30*1024=30720.Julzes (talk) 13:08, 29 May 2009 (UTC)

(e/c) Using the same method, you can rotate the cube so that you face number 1, oriented upwards. This constraint precludes any further rotation of the cube, hence you get 5! permutations to distribute the other numbers to the remaining faces, and 4⁵ choices of their orientation. The total is 122880. — Emil J. 13:10, 29 May 2009 (UTC)

(e/c, again) And while we are at it, let us consider the standard dice with dots, such as here: the labels for 1, 4, and 5 are rotation invariant, whereas 2, 3, and 6 have two distinct orientations. Using again the same method, there are 30 choices how to assign the numbers as in the first case above; this fixes the rotation of the cube, hence for each of them there are 2³ choices of orientation, hence there are 240 different dice in total. — Emil J. 13:22, 29 May 2009 (UTC)

Is this some sort of style question? Repeat mine before me? Confuse the heck out of people I think.

Good source, though.Julzes (talk) 15:30, 29 May 2009 (UTC)

If you are unfamiliar with the abbreviation, e/c means WP:edit conflict. That is, I was not replying to you but expanding my previous answer, I have only seen your post after I wrote mine. It is a pure coincidence that you used similar words as me. — Emil J. 17:26, 29 May 2009 (UTC)

Correct in putting that factor of four back in. More intricate is the question if dots, rather than digits, are used (like dominoes, with 1,4,5 rotation invariant, etc.).Julzes (talk) 13:16, 29 May 2009 (UTC)

Closer, perhaps, to the literal question, as this is how six-sided dice are usually.Julzes (talk) 13:20, 29 May 2009 (UTC) But not OP question, I see.Julzes (talk) 13:22, 29 May 2009 (UTC)

With normal six-sided dice, the pip counts on opposite faces add up to 7. I believe this results in only two possible configurations, called left-handed and right-handed. Only one of these is actually used but I'm not sure which. 207.241.239.70 (talk) 01:01, 30 May 2009 (UTC)

This page explains which is which. With a right handed dice, if 1 is on top and 2 is facing you then 3 is on the right hand side of the dice (so the faces numbered 1-2-3 run anticlockwise around their common vertex); with a left handed dice 3 is on the left (so the numbers 1-2-3 run clockwise around their common vertex). That page also says that standard Western dice are right handed (other sources agree), whereas Chinese dice are left handed. In the image on the right (from our dice article) the two dice on the bottom row and the centre die in the middle row all appear to be right handed, but the left hand die in the top row appears to be left handed. Gandalf61 (talk) 09:18, 30 May 2009 (UTC)

For each of the two "handed" configurations there are three faces (the two-, three-, and six-spot) that can have two orientations, so each configuration has 8 possible orientations making a total of 16 possible variants. SpinningSpark 10:44, 30 May 2009 (UTC)

Analytical Continuation

Why can't the sum over n>=0 of z^(n!) be analytically continued beyond the unit circle, if that is correct?Julzes (talk) 13:27, 29 May 2009 (UTC)

For the same reasons as the example in lacunary function#A simple example. (The sketch of a proof there is also a rather lacunary, but you should get the basic idea. Abel's theorem is the main missing link.) — Emil J. 13:34, 29 May 2009 (UTC)

I like the joke. I see the broad category of functions for which it's true but not the precise reason. So Abel's theorem would resolve the issue?Julzes (talk) 13:49, 29 May 2009 (UTC)

The full argument goes like this. Assume that $\sum _{n=0}^{\infty }z^{n!}$ has an analytic continuation f which extends outside the unit disk. Then the domain of f includes an open neighbourhood of a point $w=e^{it}$. Since such points with rational t are dense on the unit circle, we can assume that t is rational.

Assume first that t = 0. Then $\sum _{n=0}^{\infty }w^{n!}=\sum _{n=0}^{\infty }1=\infty$, and since $w^{n!}\geq 0$, we have $\lim _{r\to 1-}\sum _{n=0}^{\infty }r^{n!}=\infty$ by Abel's theorem. This implies $f(1)=\lim _{r\to 1-}f(r)=\infty$, a contradiction.

For general rational t we cannot use Abel's theorem directly as $w^{n!}$ are no longer real and nonnegative. However, we can find N such that Nt is an integer. Then $w^{n!}=1$ for all n ≥ N, and likewise $(rw)^{n!}=r^{n!}$, thus the same argument as above shows that $\lim _{r\to 1-}\sum _{n=N}^{\infty }(rw)^{n!}=\infty$. The remainder $\sum _{n=0}^{N-1}z^{n!}$ is a polynomial, i.e., an entire function, thus $f(w)=\lim _{r\to 1-}f(rw)=\infty$, again a contradiction. — Emil J. 17:20, 29 May 2009 (UTC)

Ok. Thx. I had promised to resolve this higher up the page.Julzes (talk) 17:38, 29 May 2009 (UTC)

While we're on the subject of that article, can someone explain the "elementary result" to me? It says:

"$\lim _{k\to \infty }{\frac {\lambda _{k}}{\lambda _{k-1}}}>1+\delta \,$

where δ > 0 is an arbitrary positive constant"

What is the point of that delta? From what I can tell, the limit just has to be greater than 1+δ for some value of δ. Isn't that just equivalent to it being greater than 1? If we call the limit L, just set δ=(L-1)/2 and you're done. I must be missing something here... --Tango (talk) 14:40, 29 May 2009 (UTC)

I actually doubt that that is the correct statement of Hadamard's theorem (the cited one you refer to, not necessarily one that has been named after him). Should be liminf, not lim.Julzes (talk) 15:11, 29 May 2009 (UTC)

And think about that a little.Julzes (talk) 15:13, 29 May 2009 (UTC)

I think that the delta serves no purpose there, it seems like a misguided attempt to make it evident that the limit is strictly greater than 1. — Emil J. 17:20, 29 May 2009 (UTC)

Seems to be like a misguided attempt to make the next bit about it following the progression (1+δ)ⁿ fit in nicely. It should, however say =1+δ, not >1+δ, if it wants to avoid having to define δ separately later. --Tango (talk) 18:04, 29 May 2009 (UTC)

I've reconsidered and think you're right, but its having a limit? Is that the theorem? Seems overly specified that there be a limit.Julzes (talk) 17:34, 29 May 2009 (UTC)

Yeah, having thought about it liminf would make more sense. --Tango (talk) 18:04, 29 May 2009 (UTC)

Hmmm... our article has a link at the bottom to here which does state the theorem in that manner. It says it will explain it in a later chapter, though, which isn't in the pdf (I'm not sure if it is available online elsewhere, I haven't looked). Very odd... --Tango (talk) 18:36, 29 May 2009 (UTC)

Slightly (!) related question is what is the cardinality of the choices of argument of values z on the unit circle such that the partial sums do not eventually fill the plane? 0, aleph-0, c?Julzes (talk) 14:10, 29 May 2009 (UTC) I think I may have overstated the truth earlier.Julzes (talk) 17:40, 29 May 2009 (UTC) And does it fill (have a point in each neighborhood of) the plane at all? Is this a "good" math question?Julzes (talk) 17:43, 29 May 2009 (UTC) I note that the question wrt random walks does not get an airing at that article, but I have to look at its referred articles. Also, is it relevant?Julzes (talk) 17:48, 29 May 2009 (UTC) Well, I guess a section on Weiner processes and the implication that Hausdorff dimension 4/3 is closer to the formation of a good question.Julzes (talk) 17:55, 29 May 2009 (UTC)

Oh, is the countable cloud of points formed by the partial sums (not the path from consecutive ones to another) primarily but not wholly a certain Hausdorff dimension between 0 and 1, and what is the dimension, etc.?Julzes (talk) 18:10, 29 May 2009 (UTC)

To answer the first question: there is the maximum possible cardinality, 2^ω, of choices of z on the unit circle such that the partial sums are not dense in the plane. More generally, if {s_n | n ∈ ω} is any sequence of complex numbers such that |s_n+1 − s_n| = 1 for each n, then there exists a point z with |z| = 1 such that $\sum _{k<n}z^{k!}=s_{n}+O(\log n)\,$. It can be shown as follows. Let s_n+1 − s_n = e^2πiα_n+1 for some α_n ∈ [0,1], put $a_{n}=\lfloor n\alpha _{n}\rfloor$, $x=\sum _{n=0}^{\infty }{\frac {a_{n}}{n!}}$, and z = e^2πix. We have

$n!x=\sum _{k\leq n}{\frac {n!}{k!}}a_{k}+{\frac {a_{n+1}}{n+1}}+\sum _{k\geq n+2}{\frac {n!}{k!}}a_{k}.$

The first term is an integer, and the last term is a nonnegative number bounded by

$\sum _{k\geq n+2}{\frac {n!}{(k-1)!}}<{\frac {1}{n}},$

hence

$z^{n!}=e^{2\pi in!x}=e^{2\pi i(\alpha _{n+1}+O(1/(n+1)))}=s_{n+1}-s_{n}+O(1/(n+1)),$

which gives

$\sum _{k<n}z^{k!}=\sum _{k<n}(s_{k+1}-s_{k})+O{\Bigl (}\sum _{k<n}{\frac {1}{k+1}}{\Bigr )}=s_{n}+O(\log n).$

A more interesting question is whether there exist a point z such that the partial sums are dense in the plane. As for the last question: the Hausdorff dimension of a countable set is always 0, due to σ-additivity of Hausdorff measure. — Emil J. 12:58, 1 June 2009 (UTC)