Wikipedia:Reference desk/Archives/Mathematics/2010 June 8

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June 8

Ratios

I've been given 100bob to buy 100animals; A cow costs 5bob, a goat 2bob and a chicken is 10cents. I should not spend more or less than the 100bob given. Can a genius help me buy the animals-How many in each category? —Preceding unsigned comment added by 196.201.217.229 (talk) 08:17, 8 June 2010 (UTC)

Its a simultaneous equation with more variables than constraints, so there will be multiple solutions. First numbers of animals: cow+goat+chicken=100. Second cost: 5*cow+2*goat+0.1*chicken=100 (assuming 100 cents to the bob). There is a third unspecified but assumed constraint: That each of cow, goat, chicken be positive integers. -- SGBailey (talk) 12:02, 8 June 2010 (UTC)

BTW, in the above equation it is best to assume spherical cows in vacuum. -- SGBailey (talk) 12:03, 8 June 2010 (UTC)

Method 1. Post your question on the reference desk and let some nerd solve it for you. (Oops, that's just what you did!)

Method 2. Program the computer to select the solution(s) by brute force.

Method 3. Write down the two equations 5x+2y+0.1z−100 = x+y+z−100 = 0 with the three unknowns x, y and z. Eliminating z gives the linear inhomogenous diophantine equation 49x+19y = 900. The corresponding homogenous equation 49x+19y = 0 has the solutions x = 19n, y = −49n for integer n. The intermediate equation 49x+19y = 1 is solved by Euclid's algorithm: 49=2·19+11, 19=11+8, 11=8+3, 8=2·3+2, 3=2+1, and so 1 = 3−2 = 3−(8−2·3) = 3·3−8 = 3·(11−8)−8 = 3·11−4·8 = 3·11−4·(19−11) = 7·11−4·19 = 7·(49−2·19)−4·19 = 7·49−18·19. Multiplying by 900 gives 49·6300−19·16200 = 900. So the integers x = 6300+19n, y = −16200−49n, z = 10000+30n satisfy the two original equations. I leave it to you to determine n such that x≥0, y≥0, and z≥0. Bo Jacoby (talk) 13:00, 8 June 2010 (UTC).

Should be 49x + 19y = 900, hence x = 6300 +190n; y = −16200 − 490n; z = 10000 + 300n. Then n ≤ -34 to make y positive, but n ≥ -33 to keep x and z positive. Gandalf61 13:33, 8 June 2010 (UTC)

You are right. Congratulations! I am wrong. Shit! The problem has no solution. Bo Jacoby (talk) 13:52, 8 June 2010 (UTC).

You all should have stuck to method 2. The problem has a solution. The mistake in Gandalf's unsigned correction is that it should be $x=6300+19n$ etc. -- Meni Rosenfeld (talk) 14:57, 8 June 2010 (UTC)

Ah, yes. Third time lucky ! Gandalf61 (talk) 15:14, 8 June 2010 (UTC)

Thanks Meni, I will correct my errors above. I should stick to method 1! Bo Jacoby (talk) 08:36, 9 June 2010 (UTC).

What happens at x=Pi/2

I'm trying to find Sin(Tan(x)) at x=Pi/2 but the graphing function on the calculator goes crazy. Is it zero, or 1 or -1 ? 122.107.192.187 (talk) 10:01, 8 June 2010 (UTC)

Guess what is Tan(x) at x=Pi/2 first... --CiaPan (talk) 10:10, 8 June 2010 (UTC)

tan(x) = sin(x) / cos(x). At x = π/2, cos(x) = 0, and so tan(x) = sin(x)/cos(x) is undefined. Hence sin(tan(x)) is undefined as well. When x approaches π/2, tan(x) = sin(x)/cos(x) approaches plus or minus infinity because the denominator gets closer and closer to zero (the sign depends on the sign of cos(x)). Hence sin [in sin(tan(x))] will "see" an angle that rapidly gets closer and closer to (plus or minus) infinity, and hence will oscillate fureously. --Andreas Rejbrand (talk) 18:59, 8 June 2010 (UTC)

For purposes of trigonometry it makes sense to add to the complex plane a single point at infinity, rather than +∞ and −∞, so that one should say that tan (π/2) = ∞. And sine and cosine oscillate between +1 and −1 as their argument approaches ∞ along the real axis, and can behave in various other ways as one approaches ∞ in other ways within the complex plane. This is an essential singularity. So sine and cosine have no value there unless maybe you want to do some moderately exotic point-set topology or something. Michael Hardy (talk) 21:10, 8 June 2010 (UTC)

The article on the topologist's sine curve might be tangentially (no pun intended) relevant here. —Ilmari Karonen (talk) 14:54, 14 June 2010 (UTC)

Homomorphism

By definition, homomorphism must be a function (between groups which preserve the same morphological structure), rather than a relation only. My question is: how about relations (between groups which preserve the same morphological structure)? One may expect to deal also with relations which aren't necessarily functions, because: literally, the very concept of "homomorphism" - is simply: "homo - morphism", i.e. it's supposed to express the idea of preserving the "same morphological" structure, while this requirement can be met also by a relation which isn't a function, i.e. a relation R such that for every a,b,c,d: if aRb and cRd then (a+c)R(b*d), the signs: +,* being the group operations. So when mathematics distinguishes between isomorphisms and homomorphisms, why doesn't it add a third concept: relations between groups which preserve the same morphological structure? Is this third concept not interesting enough? or not useful enough? or what? ... HOOTmag (talk) 10:48, 8 June 2010 (UTC)

I don't understand what you mean. Can you explain more precisely, or give an example? How, say, would you define a "relation-homomorphism" between two groups? Algebraist 12:03, 8 June 2010 (UTC)

Signifying the group operations by + and *, I'm talking about a "homomorphic" relation R such that for every a,b,c,d: if aRb and cRd then (a+c)R(b*d).

Note that it's a simple generalization of the concept of homomorphic functions, i.e.:

• For every relation R and a function F satisfying that every x,y satisfy xRy iff F(x)=y, R is a "homomorphic" relation (as defined above) iff F is a homomorphic function.

Here is a simple axample for a "homomorphic" relation R from addition to multiplication: every two integers x,y satisfy xRy iff both x is negative and the last digit of y is 1.

HOOTmag (talk) 08:15, 9 June 2010 (UTC)

The most obvious obstacle is the appropriate formulation of the notion of a "relation-homomorphism" between two groups, as Algebraist points out. The reason that this is an obstacle lies simply in the fact that the objects $f(xy)$, f(x) and f(y) need no longer be elements; rather they are sets of cardinality possibly greater than one. How would you propose to make precise the relation $f(xy)=f(x)f(y)$? PST 12:40, 8 June 2010 (UTC)

I think he must be thinking about things like having an odd or even relation between the group of numbers adding modulo 4 and modulo 6 so we have {(0,0) (0,2) (0,4) (2,0) (2,2) (2,4) (1,1) (1,3) (1,5) (3,1) (3,3) (3,5)} as the relation. Anything straightforward like that for groups one would relate them to a quotient group. Perhaps there might be some useful reason in a different context to get them to not form subsets like this where all of one subset is related to all of a corresponding subset of the other and yet some structure is preserved if one goes one way or the other. Dmcq (talk) 12:50, 8 June 2010 (UTC)

Signifying the group operations by + and *, I'm talking about a "homomorphic" relation R such that for every a,b,c,d: if aRb and cRd then (a+c)R(b*d).

Note that it's a simple generalization of the concept of homomorphic functions, i.e.:

• For every relation R and a function F satisfying that every x,y satisfy xRy iff F(x)=y, R is a "homomorphic" relation (as defined above) iff F is a homomorphic function.

Here is a simple axample for a "homomorphic" relation R from addition to multiplication: every two integers x,y satisfy xRy iff both x is negative and the last digit of y is 1.

HOOTmag (talk) 08:15, 9 June 2010 (UTC)

Such relations are occasionally used in textbooks on groups, for example Proposition 2.2.3 in W.R. Scott Group Theory. Once the machinery of quotient groups is built up the need for them seems to disappear. There may be theorems that could be proved more easily using this kind of relation but mathematicians tend to think in terms of functions.--RDBury (talk) 15:47, 8 June 2010 (UTC)

Signifying the group operations by + and *, I'm talking about a "homomorphic" relation R such that for every a,b,c,d: if aRb and cRd then (a+c)R(b*d).

Note that it's a simple generalization of the concept of homomorphic functions, i.e.:

• For every relation R and a function F satisfying that every x,y satisfy xRy iff F(x)=y, R is a "homomorphic" relation (as defined above) iff F is a homomorphic function.

HOOTmag (talk) 08:15, 9 June 2010 (UTC)

Yes it looks like I ignored that because it didn't make much sense to me. Have you an example of this? Dmcq (talk) 09:32, 9 June 2010 (UTC)

Yes, here is an example: let's define a relation R - the following way: every two complex numbers x,y satisfy xRy iff both of them are real numbers or both of them are imaginary numbers. Then R is a "homomorphic" relation from addition to substraction , but not from addition to multiplication. HOOTmag (talk) 10:02, 9 June 2010 (UTC)

So x and y have this R relationship if they are both multiples of the same basis vector for 3d vectors, and if you add two multiples of the same basis vector you get another multiple of the basis vector. That's what equivalence classes are about and are the basis of quotient groups for group theory. Though of course you don't have any relationship for (1,2,0) defined you can extend what you've got to cover all the vectors fairly easily by not considering basis vectors. Have you some example that doesn't fit into that mould easily? Dmcq (talk) 11:51, 9 June 2010 (UTC)

Looks like you edited your example between my looking at it and pressing edit, th only difference is you now have 2d vectors. Dmcq (talk) 11:56, 9 June 2010 (UTC)

Here is an axample not invovling equivalence classes: let's define a relation R - the following way: every two integers x,y satisfy xRy iff both x is negative and the last digit of y is 1. Then R is a "homomorphic" relation from addition to multiplication. HOOTmag (talk) 18:15, 9 June 2010 (UTC)

So y would have to be both negative and end in a 1 for your relation to work. Why do you think the relation is interesting? I'm not seeing how this captures any essence of going from addition to multiplication. For that one would say aRx and bRy implies (a+b)R(x×y). Dmcq (talk) 22:38, 9 June 2010 (UTC)

Why do you think that y would have to be both negative and end in a 1 for my relation to work? In my example, every two integers x,y satisfy xRy iff both x is negative and the last digit of y is 1. So, y doesn't have to be negative, just as x doesn't have to end in a 1. Note that x just has to be negative, and y just has to end in a 1. HOOTmag (talk) 22:50, 9 June 2010 (UTC)

I see you've changed the text in your questions again without saying so clearly in a contribution at the end. I would have to keep on reading everything over and over again to cope and I really don't feel like doing that. Please state your changes clearly when you make them. My comment was about your set of relations xRy yRz implies xyRyz which you have now removed. As to the latest incarnation category theory and functors is what to look at. Dmcq (talk) 23:33, 9 June 2010 (UTC)

You might check if arbitrary subgroups F of H×K work for you, or perhaps those F such that for every h in H there is some k with (h,k) in F. F is a homomorphism if and only if for every h in H there is exactly one k in K with (h,k) in F, so a "GeneralMapping" (in GAP terminology) is just a subgroup F with no restrictions. You would write hFk if (h,k) in F. If h1, h2 in H and k1, k2 in K and (h1,k1) in F and (h2,k2) in F, then (h1,k1)*(h2,k2) = (h1*h2,k1*k2) is in F too. I believe this is the relation you want. JackSchmidt (talk) 13:32, 10 June 2010 (UTC)

Newline formatting in Wikipedia LaTeX

Hi, for each newline of math I wish to do on Wikipedia, I start and end the line in the

<math></math>

tags. Ideally I'd like to be able to wrap the entire block of math in one set of tags, and use a newline character like \\. Is this possible? —Preceding unsigned comment added by 62.31.58.156 (talk) 11:49, 8 June 2010 (UTC)

You may use e.g. \begin{array} .... \end{array}:

${\begin{array}{l}line\,1=something,\\line\,2=something\ else\\line\,3=whatever\end{array}}$

CiaPan (talk) 13:52, 8 June 2010 (UTC)

Measures of complexity (Chemistry)

Please see Wikipedia:Reference_desk/Science#Molecular_compexity - a measure of complexity of connected graphs by the look of things, and applied to chemical stick structures. eg [1]

If anyone can answer my original question - great.. But also is there an article (mathematical) that relates to this .. I'm also looking for a measure of complexity that gives lower absolute figures than the ones linked to on the science desk for the same structure.(see sci page for reason why). /thanks/ 87.102.13.111 (talk) 20:05, 8 June 2010 (UTC)

Meaning of eigenvalue, eigenvector

Is there an alternative meaning of the word combination: eigenvalue, eigenvector? For example which is stronger in meaning?

• self value, self vector
• own value, own vector,

This is important for better Arabic translations (There are no good translation yet, not to rely on generic translation as Google..), thanks. --Email4mobile (talk) 20:13, 8 June 2010 (UTC)

I'd be surprised there's no Arabic translation already. Self sounds better than own, how about innate? The Languages reference desk might be of help. Dmcq (talk) 21:03, 8 June 2010 (UTC)

Eigen is a German word that can have either of those meanings. Paul Halmos famously objected to its use and wrote a book title Finite-Dimensional Vector Spaces in which he used the terms proper value and proper vector.

To me self seems closer to the intended meaning. An eigenvector is a vector that is mapped to a scalar multiple of itself. Michael Hardy (talk) 21:06, 8 June 2010 (UTC)

"characteristic value" and "characteristic vector" are often used, that might be more translatable. 129.67.37.143 (talk) 21:31, 8 June 2010 (UTC)

Indeed, Arabic lacks for many modern scientific equivalent terms and perhaps the main reason was that most universities are teaching in English except some few as in Syria. The article was already created with a "self value" Arabic meaning but later Ar-Wikipedians argued about the term and I found it much reliable to go directly to the reference desk as a temporary assessment. Don't be afraid I won't prosecute you ;) but would rather say awful thanks :). Email4mobile (talk) 21:46, 8 June 2010 (UTC)

Characteristic sounds good to me and would link them to the characteristic polynomial. Dmcq (talk) 21:48, 8 June 2010 (UTC)

I really think all this talk about what "sounds better" is irrelevant. There is undoubtedly a standard term in Arabic, and the right thing to do is to find out what it is. Ask somebody at the math department at University of Cairo. --Trovatore (talk) 21:51, 8 June 2010 (UTC)

Well, Trovatore; this site [2], would agree with the translation (eigen value =self value); however, such site also lacks for other scientific terms as in monomial, or even mistranslates sometimes. In such case, I believe a temporary sensible translation is better than no article at all.Email4mobile (talk) 23:02, 8 June 2010 (UTC)

Whatever you think best, of course. I can't possibly contribute at Arabic WP. I'm just saying that I think there must surely be a standard term, and in your shoes I would try to find out what it is. Given that searching has failed, the next step would seem to be to contact someone who routinely works with the concept in Arabic. --Trovatore (talk) 03:20, 9 June 2010 (UTC)

Ditto for me, I'd copy someone else like that dictionary rather than trying to do my own thing. Monomial is a much less common term and nowhere near as useful as eigenvalue so it is no surprise to me they have one but not the other. People outside of maths refer to eigenvalues. Dmcq (talk) 08:09, 9 June 2010 (UTC)

It means that a linear vector function of a vector space, when restricted to a subspace, reduces to multiplication by a factor. This explanation does not suggest a name for the factor. The English word eigenvalue was borrowed, rather than translated, from the German word Eigenwert, probably because finding a good translation was not easy. A strange name warns the reader that this is a strange concept. The associations from familiar words like characteristic or own or self or proper may be misleading. Why not borrow Eigenwert into Arabic rather than inventing a word? Bo Jacoby (talk) 08:22, 9 June 2010 (UTC).

If Email4mobile is a research mathematician creating new work, he is free to try that and see if it flies. As an encyclopedist, though, he has to go with the prevailing usage, whatever it is. He seemed to be suggesting that there may not be a prevailing usage in Arabic. I just don't find that believable; the concept is too widespread for that, even if classes on the subject are taught more often in English than in Arabic. --Trovatore (talk) 21:24, 9 June 2010 (UTC)

Yes, it seems rather unbelievable to me that there would not be an established translation of "eigenvector" to Arabic, or indeed to any language used to teach undergraduate mathematics somewhere. (What could be possible is that there might not be a unique dominant translation, but rather several competing ones, although I'd find even that unlikely.) My suggestion to Email4mobile would be to pick up an Arabic textbook on linear algebra; such a thing must surely exist somewhere, and will contain a translation of "eigenvalue" (not to mention an explanation of the concept that you can cite in the article). —Ilmari Karonen (talk) 15:08, 14 June 2010 (UTC)