Wikipedia:Reference desk/Archives/Mathematics/2006 August 14

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Heisenberg-Weyl group

What's a nice introductory reference for mathematics around Heisenberg-Weyl groups and their matrix representations? --HappyCamper 10:15, 14 August 2006 (UTC)[reply]

You could start with our article on the Heisenberg group. Also try Knapp's paper (PDF) from the May 1996 AMS Notices. A recommended text is Fulton and Harris, Representation Theory: A First Course (ISBN 978-0-387-97495-8), but that's probably overkill for what you want to learn. --KSmrq^T 11:04, 14 August 2006 (UTC)[reply]

P.S.: I cite ISBN-13 book numbers; there is an online converter that can produce the old 10-digit form if so desired. --KSmrq^T 11:04, 14 August 2006 (UTC)[reply]

To be honest, I'm not sure what I need, but I've been fascinated with these things ever since I found them. What I have at hand, is essentially a bunch of operators, and I want to know what "space" they are part of. Essentially, I have the creation and annihilation operators up to order n, say

\left\{1,{\hat {a}},{\hat {a}}^{\dagger },{\hat {a}}^{2},{\hat {a}}^{\dagger }{\hat {a}},{\hat {a}}^{\dagger 2},\cdots \right\}

where n denotes the number of operators. I think I can figure out the associated Lie algebra for this case. Now, suppose I define a bunch of projection operators P_ab = |a><b| which commute with the operators above, a and b are finite. Let's construct a new set of operators, where it's exactly as above, but we multiply each element with each possible projection operator. Now, this is where I get stuck...How do I figure out what is the associated representation? --HappyCamper 01:10, 15 August 2006 (UTC)[reply]

Few weeks ago, we looked for topics and/or pointers to create a NEW second year course for mathematics major with the objective to motivate students toward current research. It appears that Knapp's paper (PDF) from the May 1996 AMS Notices mentioned by KSmrq above should be a good choice. Question: Are there any similar topics and/or pointers that may satisfy the requirements of our NEW course? Thank you in advance. Sorry, this may be out of place.Twma 00:27, 17 August 2006 (UTC)[reply]

Direction of max slope

Using 3D geometry/trig, I found that the direction of max slope on a 3D surface having slopes along 2 directions at right angles of y' and x' is arctan(y'/x') relative to the x direction. This must be a standard result, so where would I have found it, rather than do it myself (though the effort was no doubt good for me)?Semiable 13:17, 14 August 2006 (UTC)[reply]

This is a basic result of multivariate calculus, so you may want to do some reading on this topic. Its coverage in Wikipedia doesn't currently look very satisfactory, but you should have a look at gradient. This result is expressed there differently, but it's the same thing. -- Meni Rosenfeld (talk) 14:23, 14 August 2006 (UTC)[reply]

Thanks for reply. I don't actually see the arctan result, so presumably it's lost in the detail, which I'll flog through sometime.

To get the result, I assumed a plane represented the probably curved surface, this being OK for an infinitesimal displacement. So I wonder - is there a word planirity (or something like) corresponding to linearity in 2D, or is linearity used in all dimensions? Semiable 15:05, 14 August 2006 (UTC)[reply]

Linearity is used in all dimensions. The word "planarity" is occasionally used, but I'm not sure that in this context. The arctan doesn't appear because the direction is given not as an angle, but as a vector (the vector being the gradient). For example, if you have z = 3x + 4y, the direction of steepest slope is along the gradient vector (3, 4), which forms an angle of arctan (4 / 3) = 53.13° with the x-axis. -- Meni Rosenfeld (talk) 16:08, 14 August 2006 (UTC)[reply]

Tex 2 Wiki

Hi
I have a TeX file which I like to add to Wikipedia. Is there any utility to convert the tags between them?
Regards, Arash

The software that Wikipedia runs on uses a pared-back LaTeX markup for mathematical (and other) formulae - to start, enclose the formula with <math></math> tags, then check on m:Help:Math to see which LaTeX commands work. Anything that's not in there, you'll probably have to find a workaround for. —Preceding unsigned comment added by ConMan (talk • contribs)

Binary interval-filling sequence

Does the sequence 0, 1, 1/2, 1/4, 3/4, 1/8, 3/8, 5/8, 7/8, 1/16, ... (or some sensible permutation thereof) have a name? Fredrik Johansson 16:30, 14 August 2006 (UTC)[reply]

Although I could find several applications of this sequence, usually starting at 1/2, sometimes arranged in the obvious way as a binary tree, none of these offered a name. Except for a different start, the numerators are found in the "Josephus problem" (sequence A006257 in the OEIS). The denominators are also listed (sequence A062383 in the OEIS), but neither is brought into relation with this sequence of fractions. --Lambiam ^Talk 17:30, 14 August 2006 (UTC)[reply]

I believe this is a kind of Cantor set. - Rainwarrior 02:20, 15 August 2006 (UTC)[reply]

It is not, Cantor sets are uncountable and that sequence is, of course, countable. (Cj67 17:14, 18 August 2006 (UTC))[reply]

Or maybe it is similar to the Smith-Volterra-Cantor set. At least, the construction is very similar to both of these. The main difference between your set and these sets, is that yours will be uniformly distributed. In which case... wouldn't you just call your set, say: multiples of 1/16 between 0 and 1? (Or something to that effect? Multiples of 1/32, or however deep you build it.) - Rainwarrior 02:23, 15 August 2006 (UTC)[reply]

The set formed by these points is dense. Both the ordinary Cantor set and the Smith-Volterra-Cantor set are nowhere dense. --Lambiam ^Talk 03:14, 15 August 2006 (UTC)[reply]

This is true. So this isn't a kind of cantor set at all, really. - Rainwarrior 04:13, 15 August 2006 (UTC)[reply]

"Binary fractions"? —Tamfang 06:28, 15 August 2006 (UTC)[reply]

No idea how that escaped me... but I still wonder what the best name for the enumeration would be. Fredrik Johansson 07:55, 15 August 2006 (UTC)[reply]

Dyadic rationals, actually. Melchoir 23:21, 15 August 2006 (UTC)[reply]

This is known as the (binary) van der Corput sequence, after the Dutch mathematician J. G. van der Corput who described it in 1935. —Ilmari Karonen (talk) 11:09, 15 August 2006 (UTC)[reply]

Note that the binary vdC sequence is a "sensible permutation" of the original sequence. --Lambiam ^Talk 18:29, 15 August 2006 (UTC)[reply]

Hmm -- as far as I can tell, this is the same thing as the binary Halton sequence. Should the articles be merged? Or is there a distinction I'm missing? --Trovatore 20:15, 15 August 2006 (UTC)[reply]

I find that article quite incomprehensible. Where I come from, sequences are not particularly active; they do not "fill space" or "divide space into segments", nor do they "fill in empty spaces" or "use cycles" that in turn "place draws" (whatever these are) in segments. What I think I do understand is that there are several Halton sequences, all of which have finite lengths. There is an "original" Halton sequence, but it is not explained what it is and how it relates to the other ones. A small example might have worked wonders. In any case, the sequence we have here is infinite. --Lambiam ^Talk 23:23, 15 August 2006 (UTC)[reply]

This paper: (P. L'Ecuyer and C. Lemieux "Recent Advances in Randomized Quasi-Monte Carlo Methods", in Modeling Uncertainty: An Examination of Stochastic Theory, Methods, and Applications, M. Dror, P. L'Ecuyer, and F. Szidarovszki, eds., Kluwer Academic Publishers, 2002, 419–474.), accessed online from [1], states: "Halton sequence This sequence was introduced in 1960 by Halton [37] for constructing point sets of arbitrary length, and is a generalization of the one-dimensional van der Corput sequence [86]." In the definition given there, it is clear that there is an (infinite) Halton sequence of points in (0,1)ⁿ for all finite dimensions n > 0, and that the one-dimensional Halton sequence is the van der Corput sequence. I can't access the original Halton paper (J.H. Halton. "On the efficiency of certain quasi-random sequences of points in evaluating multi-dimensional integrals". Numerische Mathematik, 2:84--90, 1960.) to verify that this conforms to the original definition. In any case, there appears to be a distinction, and my advice is to keep the articles separate, although they should of course cross-refer. --Lambiam ^Talk 23:41, 15 August 2006 (UTC)[reply]

Ah, got it: So the vdC sequence is the one-d H sequence. The Halton sequence article needs a serious rewrite, and someone should write up Sobol sequence too (currently a redirect). --Trovatore 23:47, 15 August 2006 (UTC)[reply]

Right. I'm not really an expert on low-discrepancy sequences, but from what I gather the van der Corput sequence is more or less the granddaddy of them all, with the Halton and Hammersley sequences being generalizations of the van der Corput sequence to multiple dimensions. —Ilmari Karonen (talk) 00:40, 16 August 2006 (UTC)[reply]

I've put a {{cleanup-rewrite}} tag on the Halton sequences article.

Multiplying with zero always equallying zero

100 x 0 = 0

However, if I have 100 apples and multiply them by zero, then I still have 100 left, and so in this sense: 100 x 0 = 100 (as I still have in my possession the one hundred apples I had in the first place).

Not correct?

Question - How do you "multiply" 100 apples by 0? In school-level mathematics, multiplying a by b means taking b groups of a objects, meaning zero groups of 100 apples, meaning 0 apples. Confusing Manifestation 00:08, 15 August 2006 (UTC)[reply]

If you have 100 apples and multiply them by zero, the Bad Apple Fairy comes and takes all your apples. --Lambiam ^Talk 01:49, 15 August 2006 (UTC)[reply]

If I give you a hundred dollar bill once, you get $100.

If I give you a hundred dollar bill twice, you get $200.

If I give you a hundred dollar bill thrice, you get $300.

If I give you a hundred dollar bill zero times, you get $0.

So $100 x 0 = $0 QED. Ohanian 13:39, 15 August 2006 (UTC)[reply]

"QED" signifies the end of a mathematical proof. "quod erat demonstrandum" : "which was to be demonstrated [was demonstrated]". — [Mac Davis] (talk)

Technically you could give the same bill over and over again. I personally like to think in x groups of apples. o groups equals no apples. - Mgm|^(talk) 09:11, 17 August 2006 (UTC)[reply]

"Not correct?"

Yes, not correct.

Think of it this way: the apples came from an orchard. You state, "I have 100 apples" which means that you've already taken them from the orchard once.

100 x 1 = 100

In order to have taken them zero times (the point of your question) you'd have to put them back. Now, how many apples would you have then?

100 x 0 = 0

I believe that your question comes from a misunderstanding of what "multiplying" means. You have, perhaps, confused it, momentarily, with "adding." When you multiply by some number (call it X), what you are saying is, "For each one I had before, I now have X," not "For each one I had before, I now have X more." B00P 21:49, 17 August 2006 (UTC)[reply]