Wikipedia:Reference desk/Archives/Mathematics/2012 November 20
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November 20
editHilbert space
editOperator A is generator of a strong continuous semigroup S(t) on Hilbert space H; then the domain A, D(A) is dense in H.
I want to know whether the domain of A*(the dual operator of A), D(A*) is dense in H. — Preceding unsigned comment added by 159.226.25.244 (talk) 02:32, 20 November 2012 (UTC)
- I added a title, please update it to make it better. StuRat (talk) 02:52, 20 November 2012 (UTC)
Dispute over election percentages
editThe reference desk is not the place for an dispute, take it to talk page, WT:WPM or use dispute resolution |
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The following discussion has been closed. Please do not modify it. |
This may be inappropriate, but I see no harm in asking. Is there someone with a strong mathematical background who can look at this discussion and shed some light on it? I thought that I was seeing the situation clearly, but those in the opposing camp, as it were, are so strong in their belief that I fear I am missing something. Thank you. → Michael J Ⓣ Ⓒ Ⓜ 03:23, 20 November 2012 (UTC) The object of rounding is to obtain the closest multiple of some unit. Is 1.45 closer to 1 or 2? Obviously it is closer to 1. But what if we round it to the tenths first? Then we have 1.5. If we round it again, by standard rules of rounding (half-up), that would be rounded to 2. Rounding twice does not give the closest multiple, and is therefore a worse approximation. Or a more obvious example, 4445 rounded to the thousand would be 4000. But round it at the tens (4450), then hundreds (4500), then thousands (5000), and you get 5000. Always round once. Also, note that there's only one person arguing for the double rounding, and his defense rests on "this is the way we've always done it" and claims of "worldwide standards for rounding" (which are correct, but irrelevant, because we're talking about double rounding). KyuubiSeal (talk) 04:47, 20 November 2012 (UTC)
@76: Suppose we do accept double rounding as a valid operation. Mitt Romney won 47.647% of the vote, but say we knew this to less precision - say we know he won about 48% of the vote. We can round once and say he won about 50% of the vote. Now round yet again, following the fives large convention, and you have that he won 100% of the vote - he won the election with the largest landslide ever! (which is absurd) Therefore, by contradiction, the premise is false (QED :) 24.92.74.238 (talk) 02:05, 21 November 2012 (UTC)
O.k. time to call your bluff, '76.189.101.221'. Are you going to produce the necessary evidence that this so-called 'precedent' exists, or are we going to have to get you blocked for trolling? Your choice... AndyTheGrump (talk) 02:53, 21 November 2012 (UTC)
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Circle packing in a square
editIn the problem of circle packing in a square (where n circles of unit radius are packed into the smallest possible square), what is the asymptotic amount of wasted space? Double sharp (talk) 06:47, 20 November 2012 (UTC)
- I assume you mean as n becomes large. You should be able to work that out with a touch of intuition in terms of packing density on an infinite plane. A more interesting problem is when this gets extended to a larger number of dimensions, e.g. Close-packing of equal spheres. — Quondum 09:21, 20 November 2012 (UTC)
Guide to pronouncing (saying) mathematics in German
editHi,
So my German is mediocre, but a particular gap I've noticed is that I have no real idea how to pronounce most things that don't get written down. In particular I'd like to be able to better read academic texts in German (I study mathematics and logic, and a lot of the important texts were in German to begin with) but not knowing how expressions are pronounced makes that feel incomplete (not to mention I might actually want to talk about maths out loud at some point).
Expressions I'm referring to are things like - not so much mathematical terminology (for which there are glossaries everywhere) but the more basic things. My attempt at reading that starts with "a plus b Quadrat über..." and trails off into failure around there. The bottom of [1] has what I'm getting at, but in reverse. Does anyone know of such a resource?
Thanks,
Daniel (‽) 12:22, 20 November 2012 (UTC)
- This link [2] gives a clue. Bo Jacoby (talk) 12:54, 20 November 2012 (UTC).
Here's a suggestion that may or may not pan out. In my experience, it's rare to see mathematics, especially advanced mathematics, sounded out in a reference. That's why I was pleased to see French mathematics sounded out in a French-Russian mathematics dictionary published in Soviet times that I came across in a library. (For example, I learned that the little upside down hat accent is pronounced "tchèche", probably the only occurrence of that word in print anywhere. I have a theory that it originated as a pronunciation of the name of the mathematician Eduard Čech.) If you can find a German-Russian mathematics dictionary and it happens to have been produced by the same people (which is a reasonable bet, given the way Soviet academic publishing was organized), you may be in luck. Search for Немецко-русский математический словарь. 96.46.194.85 (talk) 08:41, 23 November 2012 (UTC)
- As to your "theory" — I strongly doubt it. The tchèche is almost certainly a phonetic rendering of the English word check, as in check mark (called "tick mark" in British English, I think). --Trovatore (talk) 07:16, 26 November 2012 (UTC)
- To the OP: you are being misled by a false friend. In mathematics, German "über" is not the same as English "over" (although the mean the same thing in most other contexts). The way to say in German is "a durch b" or "a geteilt durch b" (a divided by b); "a über b" means , the binomial coefficient, instead. —Kusma (t·c) 09:45, 23 November 2012 (UTC)
Thanks for all your answers. Daniel (‽) 17:29, 24 November 2012 (UTC)
Just to follow up - certainly a bit late - the German-Russian mathematics dictionary by Kaluzhnina has seven pages' worth of formulas sounded out in German. 64.140.122.50 (talk) 05:08, 15 February 2013 (UTC)
Inclusion
editShow that .
such that .
What we want to show is that such that
Is there some obvious next step?--AnalysisAlgebra (talk) 17:44, 20 November 2012 (UTC)
- I probably don't know what I'm talking about but if R is a ring with unity, I think implies that either I or J has a unity, from which the rest trivially follows. -- Meni Rosenfeld (talk) 17:50, 20 November 2012 (UTC)
- I don't think so. ;
- and are ideals;
- has a unity but neither nor do.--AnalysisAlgebra (talk) 18:13, 20 November 2012 (UTC)
- --AnalysisAlgebra (talk) 18:16, 20 November 2012 (UTC)
- When dealing with ideals, if you know that something is the entire ring, usually the only important information is that it contains 1. So for some . Now you have some , and you want to use that equation to show that .--80.109.106.49 (talk) 18:21, 20 November 2012 (UTC)
- . Also, and . So is an expression of the correct form .--AnalysisAlgebra (talk) 18:43, 20 November 2012 (UTC)