Wikipedia:Reference desk/Archives/Mathematics/2012 November 20

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November 20

Hilbert space

Operator 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

The reference desk is not the place for an dispute, take it to talk page, WT:WPM or use dispute resolution
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) There is a small section on double rounding and the efforts to avoid it in Java in the article Rounding#Double rounding. There seems to be some literature about avoiding it in places other than computing which could be put in there. I just did a quick google and I saw a class action against some insurers who claimed that double rounding premiums was within the rules and required for what they were doing - they lost their case and had to change the rules to avoid double rounding. Dmcq (talk) 09:21, 20 November 2012 (UTC) I've added in a bit about that case. There's lots of books about avoiding it for other things too. Dmcq (talk) 09:49, 20 November 2012 (UTC) I can sympathize with anon's position. Saying "you can never round from already-rounded numbers" is wrong - of course you can, you just end up with something different than the single-rounded value. Rounding is an approximation, and double-rounding gives a worse approximation than single-rounding; but this doesn't disqualify its use, as with any approximation you need to balance the accuracy with the convenience. As I understand, the root problem is that the percentage is reported in the media and in the article with 2 decimal places, and the infobox is to display 1 decimal place. Yes, if you go back to the raw numbers and round from that, you get a more accurate 1-dp figure. But that is harder to derive and verify (you'd have to go back to the raw numbers every time rather than using widely-circulated figures). A policy saying the double-rounded value should be used, which is much more scalable and results in only a trivial inaccuracy in edge cases, makes sense. The thing is, I didn't see anon linking to such a policy; if it exists, it should be debated but followed while in place. If not, a new policy should be established, and whatever it is, should be applied to all articles retrospectively. -- Meni Rosenfeld (talk) 17:42, 20 November 2012 (UTC) If there is such a policy, I have not seen any indication of where it is to be found (or rather, I have seen a vague one, but did not find it when I tried to follow that indication). In any case, I suppose you're right that it doesn't make much difference and there's some minimal rational basis for such a thing. However, I think would be strongly counterintuitive to have a policy (or even guideline) that mandates giving a less accurate answer just because it's easier to check. If someone is willing to go back to the raw numbers and figure out the right answer, surely that's an improvement, even if admittedly a minimal one. I would even say IAR in that circumstance. --Trovatore (talk) 00:50, 21 November 2012 (UTC) And now the disputed discussion on the 2012 US presidential election page has been locked! with a comment saying that "there is consensus" with the used method! It seems there is rather consensus AGAINST that method! As the user above mentions, "double-rounding gives a worse approximation than single-rounding; but this doesn't disqualify its use, as with any approximation you need to balance the accuracy with the convenience": the result is worse but it's just a matter of "convenience". THe thing is, once the results have been made final and are no longer evolving, there is no justification to follow this 2-step rounding: a result made final at 47.646, can be easily displayed with 2 decimals as 47.65 and with 1 decimal as 47.6. The two values can be entered simultaneously in the article: there is no need to first enter the 47.65 value and then base the 1 decimal number on that estimate. This is not rocket science and seems utterly extraordinary if this is how things are done in all election pages! Patphilly (talk) 23:19, 20 November 2012 (UTC) I see no need for new policy - it is a basic Wikipedia principle that we base articles on published sources, not on our own article content - and the double rounding is doing the latter. Incidentally, I can see no logical reason to be giving the results to two different precisions anyway - if we are calculating to two decimal places, why not be consistent and show that everywhere? AndyTheGrump (talk) 23:57, 20 November 2012 (UTC) Patphilly, please educate yourself on what consensus is. It is not at all about vote-counting. As Wikipedia's consensus guidelines tell us, "consensus is not about seeing who has the most people on their side." And what Meni Rosenfeld said was, "A policy saying the double-rounded value should be used, which is much more scalable and results in only a trivial inaccuracy in edge cases, makes sense." --76.189.101.221 (talk) 00:08, 21 November 2012 (UTC) Yet more nonsense. The overwhelming consensus is that double-rounding is wrong - and there is nothing whatsoever that makes an opinion posted on the Maths reference desk any more significant that one posted anywhere else. And there is no policy that says that we double-round election results - in fact there haven't even been links posted to previous discussions, despite the repeated claims by the IP that there is a 'precedent' for this. AndyTheGrump (talk) 00:17, 21 November 2012 (UTC) As I stated previously, please educate yourself on consensus, and if you'd like to change the long-standing, established process, which involves every election article, you are welcome to initiate that effort via the election articles project, where it was created. But I'd suggest not calling what they tell you "nonsense" and "garbage" as you have done so frequently in other talk page discussions. Your uncivil behavior has resulted in numerous editing bans and I wouldn't want to see that happen again. Good luck to you. --76.189.101.221 (talk) 00:31, 21 November 2012 (UTC) I can't find this WikiProject discussion you speak of. I went to Wikipedia talk:WikiProject Elections and Referendums and entered "rounding" in the archive search box. There was one hit, which was for the word "round" rather than "rounding", and had to do with a round of elections. --Trovatore (talk) 00:46, 21 November 2012 (UTC) I know perfectly well what 'consensus' means, you patronising little jerk - and it doesn't mean 'whatever you agree with'. Now, are you going to produce evidence that there is a 'policy' supporting your position - because if you don't I think it is safe to assume that it doesn't exist, and you have been trolling all along. AndyTheGrump (talk) 00:42, 21 November 2012 (UTC) Extra stages of rounding is something to be avoided. The best figures available in a reliable source should be used directly rather than rounded to two figures and then to one. The best figures will already have been rounded to some number of decimal places so there will always be double rounding if their figure is to a higher precision but sticking in extra stages unnecessarily is simply a silly thing to do and does not comply with the WP:CALC policy of obviousness. And by the way a local consensus cannot override policy even if someone could produce some evidence of such a consensus which does not seem to be the case. Dmcq (talk) 00:47, 21 November 2012 (UTC) So much for "Be bold". How is such a policy (still no evidence of its existence) overridden anyways? Also, you can tell when double rounding affects the answer (only when it ends in a 5), so even if your original source is rounded, you can still get the correct answer. And I agree with AndyTheGrump's request for evidence, although not necessarily his tone. (Seriously though, everyone wants that evidence, so if you want to be taken seriously, show it quick.) And he's right, you are being patronizing. Stop that. Also, at one point, you cite Meni Rosenfeld as the ultimate arbiter of this discussion. He's not the only mathematician in this thread. See also argument from authority. Lastly, don't remove warnings from your talk page, and say "oh I wasn't being aggressive looks like I'll take this off". He referred me to WP:OWNTALK, where I read I was wrong. KyuubiSeal (talk) 01:15, 21 November 2012 (UTC) Andy, please be more careful with your language. There is no clear evidence to support your presumption that the ip is little. --Anthonyhcole (talk) 02:02, 21 November 2012 (UTC) Anthonyhcole, you have a long history of being banned from editing, like AndyTheGrump. Being uncivil by attempting to bully or humiliate other editors will only lead to you getting banned yet again. --76.189.101.221 (talk) 02:49, 21 November 2012 (UTC) Look more carefully at his history. Most of the blocks have been misunderstandings, and were promptly unblocked afterward. KyuubiSeal (talk) 02:55, 21 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) That's more of a "rounding at the wrong place" issue though, not specifically double rounding. 53% rounds to 100% as well. KyuubiSeal (talk) 02:24, 21 November 2012 (UTC) 100% of the result was Obama ;-) Dmcq (talk) 13:09, 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) AndyTheGrump, your most recent editing ban for personal attacks/harassment was for 2 weeks. It looks like you're shooting for 3 this time. Let me know when you initiate that effort to get me blocked for not adhering to yet another one of your non-sensical threats. I'll look forward to seeing how that turns out for you. But keep in mind the wise words of an admin I read recently: Beware the boomerang. Now let's see who's bluffing. --76.189.101.221 (talk) 17:59, 21 November 2012 (UTC) So, still no link to this putative policy right? Dbrodbeck (talk) 18:17, 21 November 2012 (UTC)

Circle packing in a square

In 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

Hi,

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 ${\frac {a+b^{2}}{(-4)^{n}}}=|{\sqrt {f(x)}}|$ - 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 ${\frac {a}{b}}$ in German is "a durch b" or "a geteilt durch b" (a divided by b); "a über b" means $a \choose b$, 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

Show that $R=I+J\implies I\cap J\subset IJ$.

$\forall r\in R\exists i\in Ij\in J$ such that $r=i+j$.

What we want to show is that $\forall a\in I\cap J\exists i_{1},..,i_{n}\in I,j_{1},...,j_{n}\in J$ such that $a=i_{1}j_{1}+...+i_{n}j_{n}$

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 $R=I+J$ 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. $\mathbb {Z} =5\mathbb {Z} +7\mathbb {Z}$;

$5\mathbb {Z}$ and $7\mathbb {Z}$ are ideals;

$\mathbb {Z}$ has a unity but neither $5\mathbb {Z}$ nor $7\mathbb {Z}$ do.--AnalysisAlgebra (talk) 18:13, 20 November 2012 (UTC)

$1=3\times 7-4\times 5$--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 $i+j=1$ for some $i,j$. Now you have some $a\in I\cap J$, and you want to use that equation to show that $a\in IJ$.--80.109.106.49 (talk) 18:21, 20 November 2012 (UTC)

$a=a\cdot 1=a(i+j)=ai+aj$. Also, $a\in I$ and $a\in J$. So $ai+aj$ is an expression of the correct form $(i_{1}j_{1}+...+i_{n}j_{n})$.--AnalysisAlgebra (talk) 18:43, 20 November 2012 (UTC)