Wikipedia:Reference desk/Archives/Mathematics/2008 August 10

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August 10

Which world map projection is most accurate in terms of scale?

Is it something like this one? Also, how accurate is Image:BlankMap-World3.svg in terms of scale? Thank you! Louis Waweru  Talk  01:00, 10 August 2008 (UTC)

Erm. What do you mean by 'scale' - did you mean length? Map_projection#Projections_by_preservation_of_a_metric_property might be useful if you haven't already read it.

The 'net of a sphere' is very good for length (as the number of segments increases) I've seen this as a projection but don't know its name.. 87.102.45.156 (talk) 01:46, 10 August 2008 (UTC)

Out of interest in trying to understand this map, I have read that article a few times over the last couple of years, and I am really still confused by the concepts it discusses. I don't really have an interest in cartography, and I'm not very good at math...thank you for the link, but...whoa. I left another comment below, could I borrow your brain to help find a solution to the issue mentioned there? Louis Waweru  Talk  02:40, 10 August 2008 (UTC)

You can - but I think others have answered your question - in general any equal area projection will be applicable with no further work needed. You can just cut and paste one country on top of another and the comparision will be valid.

As to map projections - there's a lot of different choices.. it confuses me too - it seems more of an art than a science..

As an aside if you are doing a country size comparison - the very best way to do it would be to use an 3D image of a sphere (ie the earth) and slide one country on top of another - I'd guess you would need a computer wizard to help you do this (I would too).87.102.45.156 (talk) 21:29, 10 August 2008 (UTC)

Well, it certainly isn't the so-called "Peters projection", which was a serious crock. Not a terrible map in itself if you understand what it's good for and what it isn't, but the stuff Peters wrote about it and its comparison to the Mercator projection was just bad bad nonsense.

It is not possible for a flat map of a spherical object to be accurate everywhere in terms of linear scale. To say which one is the most accurate, we have to know in what way you're adding up the distortions — there isn't any single answer. --Trovatore (talk) 01:59, 10 August 2008 (UTC)

I forgot to include the link.. http://gwydir.demon.co.uk/jo/solid/other.htm - the 'net of the sphere' of which I spoke. Also see Dymaxion map which get better as the number of sides increase - though as that happens the projection becomes increasingly disjoint.87.102.45.156 (talk) 02:07, 10 August 2008 (UTC)

I am asking for the sake of creating something like this Image:Comparison Size with Russia map.GIF. I was suggesting to the user who made it that the projection used there isn't approriate for that type of comparison. Then I realized I didn't know which projection would be "best" or most suitable for what he's trying to do. Sorry for looking for the a single "best" answer, but if you could just think in terms of his goal, what would you suggest? Louis Waweru  Talk  02:40, 10 August 2008 (UTC)

Oh, so what you want to compare is area. For that the Peters map is fine. But so are lots of others. See equal-area projection for a list of them. --Trovatore (talk) 03:22, 10 August 2008 (UTC)

Right, "preserving area" is what he needs to do. Great, thank you! Anyone interested here are some editable ones: Mollweide & Gall-Peters Louis Waweru  Talk  05:45, 10 August 2008 (UTC)

You don't need to use an area-preserving projection for a size comparison like that; you can just take any image of the country and scale it such that its area in pixels is proportional to its actual land area. As long as the proportionality constant is the same for all the images, it doesn't matter where they came from. (They don't even need to resemble the country—you could use circles or rectangles or something.) If you have software that will produce national outlines in arbitrary projections on demand, I'd probably go for a Lambert azimuthal equal-area projection centered on the middle of each country in turn. -- BenRG (talk) 13:38, 10 August 2008 (UTC)

That's true, but then you're going to have to count pixels. You can't just use the number of pixels in the map—a country like Vietnam is going to take up much less of the area of a square map than a country like Suriname. —Bkell (talk) 16:52, 11 August 2008 (UTC)

What's a nonstandard integer?

Thanks, Wanderer57 (talk) 02:47, 10 August 2008 (UTC)

The integers are extended to ratonals and irrationals through the use of decimal expansion. I.e., 1/7 = 0.1428571428... and π = 3.14159.... Non-standard integers are integers, not rationals or irrationals; however, they are also expressed by decimal expansions. I.e., 312...123. These numbers have the same properties as standard integers, such as being a unique product of primes. The usual operations can be applied to them, too. On the other hand, they are larger than any standard integer. I believe it is a matter of perspective weather they are finite or non-finite.

More technically, they arise out of the surreal numbers. For more information, see this introduction. — gogobera (talk) 03:16, 10 August 2008 (UTC)

No, this is not quite right. Nonstandard integers are elements of a model of Peano arithmetic that is not isomorphic to the standard model (the familiar natural numbers). You don't get them from decimal expansions (they have decimal expansions, but these have digits in nonstandard positions, so you're not really getting anywhere from this approach). The surreals are a different system altogether; I don't know of any canonical way to recover a nonstandard model of PA as a subset of the surreals. --Trovatore (talk) 03:29, 10 August 2008 (UTC)

We have a pretty good article on them. I'm afraid I don't know much model theory, so I can't really give a formal statement of what they are, but the idea is to extend arithmetic to infinite numbers in a way that preserves as much of arithmetic as possible. So that, for instance, every third nonstandard integer is divisible by three, and the others aren't. This doesn't mean that you can actually do arithmetic on them, since for the most part they won't be representable as a finite sequence of digits. The ellipsis in the earlier post hides a lot. It would be a bit more accurate to say that you can do algebra on them. Also, it isn't exactly true that each nonstandard integer is a product of primes, since most of them are divisible by infinitely many primes, and there's no obvious way to extend products to include infinitely many terms. However, it is true that every nonstandard integer greater than one is either a prime or divisible by at least two primes, that if a=bc and a prime p divides a then p divides b or c or both, and so on. It is also true that, if a nonstandard integer is divisible by only finitely many primes, then it's the product of some powers (possibly nonstandard) of those primes. One of the trickiest things about this is keeping track of what you're allowed to transfer over and what you aren't. Black Carrot (talk) 22:29, 10 August 2008 (UTC)

Hmm, actually the article you pointed to is on the hyperreals, which are the analogous concept for the real numbers, not for the integers. Certainly many of the same ideas apply to both concepts, but they aren't really the same. It's not clear to me, for example, that every nonstandard model of PA can be embedded into a model of the theory of the reals with plus and times (I'm also not sure that it can't; could be an interesting question (or, could be trivial; I haven't thought about it that much)). --Trovatore (talk) 08:33, 11 August 2008 (UTC)

Non-standard arithmetic is our article on nonstandard models of PA. There's not much there though. Algebraist 11:42, 11 August 2008 (UTC)

I pointed to the article on the hyperreals because it's the only one that says anything comprehensible. I don't know if every model of PA can be embedded in some nonstandard model of the reals, but the countable model described in that article is plenty to get a hands-on feel for how it works, to the extent that that's possible with something that can't be explicitly constructed. The nonstandard integers there are Z*, the ultrapower construction applied to the set of integers. Black Carrot (talk) 01:05, 12 August 2008 (UTC)

I think you can do the embedding trivially, in essentially the same way as for the standard natural numbers. First adjoin negatives to get an integral domain, then take its field of fractions. Then adjoin square roots repeatedly (you might need Zorn here, I'm not sure) until you've got a real closed field, i.e. a model of the theory of the reals. Algebraist 10:53, 12 August 2008 (UTC)

Unrepresented Nations and Peoples Organization, as an example of Russell's paradox

Is the Unrepresented Nations and Peoples Organization—an "international organization" for peoples who "lack representation internationally"—an example of Russell's paradox? I think they should find a new name! − Twas Now ( talk • contribs • e-mail ) 05:56, 10 August 2008 (UTC)

It's cute, but it's not Russell, because you're not asking whether the organization itself is an unrepresented nation or people.

It does remind me of a .sig I saw on Usenet somewhere, that went something like this

Consider the set of all sets that haven't been considered. Hey wait! They're all gone!

Don't remember who wrote it -- you could probably find it on Google Groups. But again, it's not Russell. --Trovatore (talk) 06:30, 10 August 2008 (UTC)

Right. The name would have to be "Unrepresented Organizations Organization" I guess. I will ask for the name of whatever this is called at another Ref Desk. − Twas Now ( talk • contribs • e-mail ) 07:04, 10 August 2008 (UTC)

See Groucho Club. Bo Jacoby (talk) 14:09, 10 August 2008 (UTC).

Try the Unreferenced Reference Desk. siℓℓy rabbit (talk) 14:23, 10 August 2008 (UTC)

Horizontal force

I have this problem I'm supposed to do for homework- the homework really has nothing to do with physics but for some reason they put in this problem- "A 41,000 pound container is toppled against one wall of a facility. If the container makes a 52 degree angle with the floor, what is the magnitude of the horizontal force pressing against the wall?" I'm not asking for an answer- just an explanation about how to solve it since it really has nothing to do with what I was learning in the class. Nadando (talk) 07:24, 10 August 2008 (UTC)

First find the center of gravity of the container. Then draw an extended free body diagram for the container with 4 forces: 1) the upward normal force exerted by the floor originating from the point of contact between the container and the floor, 2) the horizontal friction force exerted by the floor originating from the same point, 3) the horizontal normal force (opposite in direction to force 2) exerted by the wall originating from the point of contact between the container and the wall, and 4) gravity, originating from the center of mass. In this situation, the sum of forces and the sum of torques (using either point of contact on the container for the pivot) equals 0. Then solve the system. Of course, if the container is not rigid, then more complex methods will be necessary. 99.141.162.75 (talk) 08:31, 10 August 2008 (UTC)

There is a fifth force on the container - the vertical friction force exerted by the wall. In the absence of other information about the dimensions or shape of the container, I think you must assume that its centre of gravity is vertically above the mid-point between the wall and the point of contact on the floor. I think you also have to assume that the coefficient of friction between the container and the wall is the same as that between the container and the floor, otherwise you have too many unknowns. Gandalf61 (talk) 09:23, 10 August 2008 (UTC)

Is this not trigonometry? 82.13.18.200 (talk) 13:04, 10 August 2008 (UTC)

Some knowledge of trigonometry is required in the solution, when taking moments, but the problem as a whole is a problem in statics. Gandalf61 (talk) 13:27, 10 August 2008 (UTC)

There is no fifth force if the container is rigid. 99.141.162.75 (talk) —Preceding unsigned comment added by 76.238.5.141 (talk) 04:36, 11 August 2008 (UTC)

I am assuming a rigid container. You can only say there is no vertical friction force exerted by the wall if you assume a smooth wall i.e. with a coefficient of friction of 0. My assumption, on the other hand, is that the wall has the same coefficient of friction as the floor. I see nothing in the question to suggest one assumption is more likely to be correct than the other. I guess we can conclude that the question is ambiguous. Gandalf61 (talk) 09:01, 11 August 2008 (UTC)

There aren't enough equations to uniquely determine the fifth force because the container is not necessarily "just about to slip" according to the OP (though I agree that the problem is a bit ambiguous to begin with). Thus, one can only assume that friction ≤ μ_sn. I should have said that from the start, nonetheless. 76.224.123.69 (talk) 19:31, 11 August 2008 (UTC)

Indian numbering system

I'm not sure whether this question belongs here or to Miscellaneous desk. Feel free to move it if you think it shouldn't be here. I was reading the article about the Indian numbering system and I was wondering which of the large numbers listed (1,00,000 and above) are understood by most people? I mean that almost all Americans can count to a trillion or a quadrillion, but you rarely hear "quintillion." I've seen lakh a lot in common usage and crore somewhat less, but I've never seen arawb or anything higher either on or off the web. Admiral Norton ^(talk) 16:58, 10 August 2008 (UTC)

Everything that you see on that page (and other related pages) is correct as far as I know. First of all, this numbering system is only understood by people in India, Pakistan, and other listed countries. So if you are asking about the people in THOSE countries, then almost the entire population knows the highest number to be khawrab. Everyone knows hazar, lakh, crore, arawb, and kharawb. But I have never heard or seen a person even claiming to know about higher denominations. I am surprised that you didn't find arawb or kharawb on the net. One thing to remember (while you search) is that some amount of such content on the internet, will be in actual Urdu or Hindi and you can't really search using English strings. So I am sure that these names have at least some existence on the web, maybe not as widespread as a thousand but still, it should be there.

The page also describes how larger denominations are denoted (using 1 lakh crore, for example) and this is so common that this IS actually the accepted standard today. Even on TV or newspapers, one would see 1 crore crore rather than a neelam which means that it is accepted as proper Urdu, Hindi, Punjabi, etc. And just to let you know, you can't really compare this to the American system because the American system (in English) is based on some logic. Even if I didn't explicitly know quadrillion to be the next denomination after trillion, I should be able to deduce it (using my knowledge of the English language and how it is based on Latin) as a reasonably educated person. On the other hand, the Indian system is not based on ANY such system as far as I know. Maybe my knowledge of the roots of these languages is limited but knowing about lakh, arawb, kharawb would not have allowed me to guess the next denomination. I must admit that before reading this article here, I myself didn't know that higher denominations even existed much less what are they called. By the way, I was born and raised in Pakistan and am fluent in reading and writing several regional languages.--A Real Kaiser...NOT! (talk) 05:39, 11 August 2008 (UTC)

Efficiently determining whether any vector in a span satisfies a system of inequalities

I have a set of $m$ vectors in $\mathbb {R} ^{n}$, where $m<n$, and I have a system of linear inequalities (for instance, perhaps I want a vector whose components are all non-negative, so the inequalities are $\mathbf {e} _{i}\cdot \mathbf {v} \geq 0$, where $\mathbf {e} _{i}$ are the standard basis vectors for $\mathbb {R} ^{n}$). How can I efficiently determine whether any vector in the subspace spanned by the $m$ given vectors satisfies all the inequalities? —Keenan Pepper 20:07, 10 August 2008 (UTC)

You probably want a nonzero vector or strict inequalities, else it's trivial. This feels like it ought to be a standard problem in computational geometry, but a small amount of web searching failed to turn up anything. So here are my own thoughts. You can easily convert this to the problem of finding whether there exists any (nonzero) vector in $\mathbb {R} ^{m}$ satisfying linear constraints—just substitute $\mathbf {v} =\sum a_{i}\mathbf {u} _{i}$ in your constraints, where $\mathbf {u} _{1},\ldots ,\mathbf {u} _{m}$ are your vectors, then interpret $(a_{1},\ldots ,a_{m})$ as a vector in $\mathbb {R} ^{m}$. A positive scalar multiple of a solution is also a solution, so you could convert this in various ways to a question about affine constraints in $\mathbb {R} ^{m-1}$. An intersection of affine inequalities (half-spaces) is either empty or a (possibly infinite) convex polytope. The vertices of the polytope are defined by the intersection of $m-1$ half-space boundary planes. So you could take all groups of $m-1$ inequalities, try to solve them as a system of equalities, and if there's a unique solution test it against the full set of inequalities. If you find a point satisfying all the inequalities you're done. If you don't it might mean there are no solutions, but it might also mean that the solution polytope has no vertices (e.g. it's defined by $x\geq 0$ and $x\leq 1$). How about this: instead of reducing to $\mathbb {R} ^{m-1}$ work with the linear constraints in $\mathbb {R} ^{m}$, but add some extra affine constraints to keep everything finite (for example you could add the $2m$ constraints $-1\leq \mathbf {v} \cdot \mathbf {e} _{i}\leq 1$, bounding everything to a hypercube, or the $m+1$ constraints $\mathbf {v} \cdot \mathbf {e} _{1}\geq -1,\mathbf {v} \cdot (\mathbf {e} _{i}-\mathbf {e} _{i+1})\leq 1,\mathbf {v} \cdot \mathbf {e} _{m}\leq 1$, bounding everything to an irregular simplex). Then take the constraints in groups of $m$ and test nonzero unique solutions. This is pretty inefficient, though, something like $O(k^{m-1}m^{3})$ for $k$ constraints. And it doesn't work if the inequalities are strict. And I don't know if it would work in the presence of roundoff errors. -- BenRG (talk) 23:01, 10 August 2008 (UTC)

I'd probably just get an orthogonal basis for the vectors and plug them in as equal to zero along with all the inequalities into a linear programming package. Not marvelously efficient as the orthogonal basis would probably be huge but it would be easy on the brain and computers are cheap. Dmcq (talk) 13:55, 11 August 2008 (UTC)

I don't understand how orthogonalizing the basis would help. Why is an orthogonal basis more useful than a non-orthogonal basis for this application? —Keenan Pepper 01:01, 12 August 2008 (UTC)

Probably because standard linear programming software assumes you're working with one. Algebraist 10:46, 12 August 2008 (UTC)

Sorry Orthogonal complement would have been a better way of putting it. Setting a set of basis vectors for the orthogonal complement equal to zero means the result is confined to the subspace formed by the original vectors. Dmcq (talk) 19:29, 12 August 2008 (UTC)