Wikipedia:Reference desk/Archives/Mathematics/2011 April 15

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April 15

Hawaiian earring

The Hawaiian earring article says:

The Hawaiian earring looks very similar to (but is not homeomorphic to) the wedge sum, of countably infinitely many circles; that is, the rose with infinitely many petals.

What exactly is the difference? Michael Hardy (talk) 02:12, 15 April 2011 (UTC)

In the Hawaiian earring, any neighborhood of the point joining the circles contains all but finitely many of them; the same cannot be said of the wedge sum. Also, I think the wedge sum is not compact, but I'm not sure on that.--RDBury (talk) 04:31, 15 April 2011 (UTC)

It isn't: an open cover consisting of small open neighbourhoods of each circle has no proper subcover. Algebraist 07:11, 15 April 2011 (UTC)

More generally: The wedge of circles is a CW complex with an infinite number of 1-cells, one for each circle. And a CW complex is compact if and only if it has a finite number of cells. Aenar (talk) 00:58, 16 April 2011 (UTC)

Thank you. Michael Hardy (talk) 16:04, 15 April 2011 (UTC)

I have now added the explanation to the article. Michael Hardy (talk) 16:48, 15 April 2011 (UTC)

Also, as indicated in the article, the fundamental group of the Hawaiian earring is more complicated than the fundamental group of the wedge of circles -- the latter is just the free group on countably infinitely many generators by van Kampen's theorem. Aenar (talk) 00:50, 16 April 2011 (UTC)

lim

What technique should I use to evaluate $\lim _{x\rightarrow \infty }{\frac {x^{\ln {x}}}{(\ln {x})^{x}}}$? Widener (talk) 04:09, 15 April 2011 (UTC)

Use $x=e^{\ln {x}}$ to put the numerator and the denominator as e to some function of x. Compare those functions. Invrnc (talk) 04:24, 15 April 2011 (UTC)

(ec) Oh I see now. Thanks for your help! Widener (talk) 04:32, 15 April 2011 (UTC)

prime number

NEW FORM OF PRIME NUMBER

Of the Form (P-2)n +2 ,P = Prime Number

n=0,1,2,4,6,8,10,12,14,16,18,20,22,24,26,28,30….

(After 0 and 1 all are even number in this series)

When n=0,and n=1,(P-2)n +2=2 and 3(Both 2&3 are prime)

(2-2)X0 +2 =2

(3-2)X1 +2 =3

(5-2)X2 +2=8

(7-2)X4 +2=22

(11-2)X8 +2 =56

(13-2)X10+2=90…

(Any problem in my calculation? )

No gaps between two prime

The only TWO Ghosh prime is 2,3

As of 11 April 2011 these are the only

— Preceding unsigned comment added by Prime007 (talk • contribs) 08:42, 15 April 2011 (UTC)

Yes, there are problems with your calculation. For n = 0 you get (p − 2)⋅n + 2 = 2, which is prime. For n = 1 you get (p − 2)⋅n + 2 = p, which is also prime. For even n ≥ 2, (p − 2)⋅n + 2 will always be composite, independent of the choice of prime p. There is no hope of finding any more prime numbers in the sequence. Although you should already know that since I explained this on your talk page. I would recommend you go back to the drawing board. — Fly by Night (talk) 11:50, 15 April 2011 (UTC)

The case n even is silly. If you relax the condition that n be even, then there are infinitely many primes of this form, by the Dirichlet theorem. But this is already a well-known result (for over 150 years). Sławomir Biały (talk) 20:40, 15 April 2011 (UTC)

Exam Question - graph theory

Hello,

The following is an old exam question I'm trying to complete; the whole question is meant to be doable in half an hour ideally, so the fact I'm unable to do the second part is becoming quite irritating!

Q: Let G be a connected cubic graph drawn in the plane with each edge in the boundary of two distinct faces. Show that the associated map of G is 4-colourable iff G is 3-edge-colourable. (Done)

Is the above statement true if the plane is replaced by the torus, and all faces are required to be simply connected? Give a proof, or a counterexample.

Now I'm not making any real progress with the second part in all honesty: i feel like there should be a counterexample, as we haven't done much work previously on graphs on 3D surfaces, so it seems unlikely we'd be able to concoct a decent proof in 15 minutes or so. However, I'm having trouble constructing counterexamples which have only simply connected faces: it doesn't seem immediately obvious to me when this condition is satisfied, and anything else I manage to construct seems to fail the condition of being 3-regular. The answer can't be that hard! Could anyone help suggest anything? Thanks very much Tasterpapier (talk) 12:15, 15 April 2011 (UTC)

By joining the single arrows together and the double arrows together, one obtains a torus with seven mutually touching regions; therefore seven colors are necessary

What is meant by "the associated map of G"? Does that mean that you are coloring the faces of G? If so, there is a well known way to draw seven mutually touching regions on the torus (see right), which I think provides a counterexample. If instead the phrase means that you are coloring the vertices of G, then this example doesn't work, because G wouldn't be cubic. —Bkell (talk) 23:07, 15 April 2011 (UTC)

(See also Four color theorem#Generalizations.) —Bkell (talk) 23:09, 15 April 2011 (UTC)

Yes, sorry, it does mean colouring the faces - I spotted that example, but I wasn't completely sure as to whether it was 3-edge-colourable: is that obviously the case, or should I just sit down and try to randomly drop some colourings down on the edges until I find a configuration which works? Tasterpapier (talk) 20:50, 16 April 2011 (UTC)

Well, in the picture here, if you classify edges as "vertical," "short horizontal," or "long horizontal" (where "horizontal" actually means "somewhat diagonal"), then at every vertex there is precisely one edge from each of these three classes. So you can assign one color to all the vertical edges, a second color to all the short horizontal edges, and a third color to all the long horizontal edges to get a proper 3-edge-coloring. —Bkell (talk) 21:51, 16 April 2011 (UTC)

Ah, well spotted, thank you very much! How I'm expected to come up with something similar to that construction in just 10 minutes I'm not completely sure... Ah well, at least i'll know for the future! Tasterpapier (talk) 01:00, 17 April 2011 (UTC)

There is probably a simpler counterexample than this one. After all, this one requires seven colors, when all we really needed was something that requires five colors. —Bkell (talk) 01:59, 17 April 2011 (UTC)

mathmatics/trigonomatry

cos 4x+cos 3x+cos 2x /sin 4x+sin 3x+ sin 2x=cot 3x — Preceding unsigned comment added by Khanjaved2297 (talk • contribs) 16:21, 15 April 2011 (UTC)

What about it? Are you trying to prove an identity, or solve an equation, or what?? Can you write words? Michael Hardy (talk) 16:47, 15 April 2011 (UTC)

:-) — Fly by Night (talk) 16:50, 15 April 2011 (UTC)

Here is my suspicion. A textbook says something like PROVE THE FOLLOWING IDENTITIES, and then there is a list of them, and one of the items in the list is

${\frac {\cos 4x+\cos 3x+\cos 2x}{\sin 4x+\sin 3x+\sin 2x}}=\cot 3x.$

But it could have said SOLVE THE FOLLOWING EQUATIONS, and then had a similar list.

Those are different problems. But you omitted the words! That is what some students do when they don't understand what's going on because they're not paying attention because they're only there to pass a course and get a grade, rather than because they want to learn the topic.

It would be easier to sympathize with your request if you had said "How can this identity be proved?", since then it would seem as if you're asking a question, rather than being a stenographer attempting to pass the question along to us, and getting it wrong because you didn't care. Michael Hardy (talk) 16:56, 15 April 2011 (UTC)

(edit conflict) It looks like you're trying to prove the following identity:

${\frac {\cos(4x)+\cos(3x)+\cos(2x)}{\sin(4x)+\sin(3x)+\sin(2x)}}=\cot(3x)\,.$

Try using the identities sin(α ± β) = sin(α)cos(β) ± sin(β)cos(α) and cos(α ± β) = cos(α)cos(β) ∓ sin(α)sin(β). — Fly by Night (talk) 16:57, 15 April 2011 (UTC)

(edit conflict × 2) It's an identity (assuming those parts left and right of the slash are in brackets), so I'm assuming you have to show it. I'd start by looking at the double angle formulae, the triple angle formulae, and quadruple angle formulae, either double double or directly, and going from there. Resolving it all down to functions of x should work, I'm not sure if there's an easier way. Grandiose (me, talk, contribs) 16:58, 15 April 2011 (UTC)

That solution is too complicated. From the cotangent half-angle formula

$\cot \left({\frac {\alpha +\beta }{2}}\right)={\frac {\cos \alpha +\cos \beta }{\sin \alpha +\sin \beta }},$

one can draw a conclusion about

${\frac {\cos(4x)+\cos(2x)}{\sin(4x)+\sin(2x)}}.$

This turns out to be equal to

${\frac {\cos(3x)}{\sin(3x)}},$

and hence also equal to the left side of the proposed identity. Michael Hardy (talk) 17:18, 15 April 2011 (UTC)

It would depend on what you're allowed to assume. In my course, at least, the cotangent half-angle formula wasn't standard, so could be problematic. I can't recall whether the tan one was. Of course, we know little about the OP, and it certainly is a lot quicker. Grandiose (me, talk, contribs) 17:26, 15 April 2011 (UTC)

Lest I be misunderstood: the contangent half-angle formula that I mention here is not the cotangent half-angle formula; it's one of several cotangent half-angle formulas. Michael Hardy (talk) 02:20, 16 April 2011 (UTC)

Lazy students who don't want to waste their time learning all those stupid addition formulae, half angle formulae, etc. etc. are adviced to master complex numbers:

Sin(2x) + Sin(3x) + Sin(4x) = Im[exp(2 i x) + exp(3 i x) + exp(4 i x)] =

Im[exp(2 i x) (1- exp(3 i x))/(1-exp(i x)] =

Im[i/2 exp(2 i x)(1 + exp(-i x))(1 - exp(3 i x))/sin(x)] =

Im[2 exp(3 i x) cos(1/2 x) sin(3/2 x)/sin(x)] =

Before simplifying, we can immediately write down the expression for the numerator, as we only have to change "Im" to "Re":

Cos(2x) + Cos(3x) + Cos(4x) = Re[2 exp(3 i x) cos(1/2 x) sin(3/2 x)/sin(x)]

So, we see that the numerator equals the denominator times cot(3x). Count Iblis (talk) 22:45, 16 April 2011 (UTC)

Topology Question

This question is motivated by the local triviality axiom of vector bundles. Let the base space be the real line R along with the Euclidean topology. Let T be the total space, and let π : T → R be a continuous surjection. The local triviality axiom says that for each x in R; there exists an open neighbourhood of x, say U, and a positive integer, say n, such that π^–1(U) is homeomorphic to U × Rⁿ.

I know there are other proofs; but as a way to prove the fact that n is constant for all x, I tried to argue about filling R with interlocking open intervals. Once again: I know there are other proofs, but this is just some recreational mathematics.

Consider a point x₁ ∈ R. There is an open interval U₁ = { x ∈ R : | x − x₁ | < ε₁ } such that π^–1(U₁) is homeomorphic to U₁ × R^n₁ for some n₁. Then we define x₂ = x₁ + ε₁. There is an open interval U₂ = { x ∈ R : | x − x₂ | < ε₂ } such that π^–1(U₂) is homeomorphic to U₁ × R^n₂ for some n₂. However, since U₁ ∩ U₂ ≠ ∅ it follows that π^–1(U₁ ∪ U₂) is homeomorphic to (U₁ ∪ U₂) × R^n_1,2 for some n_1,2. After that, we repeat the process. We define x₃ = x₂ + ε₂, and so on and so forth.

There are two posibilities. The x_k have no limit point and this procedure fills the entire interval (x₁ − ε₁, _∞). That proves the proposition and is quite dull. On the other hand, the x_k have a limit point. But then I define x_1,1 to be that limit point and define x_1,2, x_1,3,… in exactly the same way as I did the x₂, x₃,… above.

There are two posibilities. The x_1,k have no limit point and this procedure fills the entire interval (x₁ − ε₁, _∞). That proves the proposition. On the other hand, the x_1,k have a limit point. But then I define x_1,1,1 to be that limit point and define x_1,1,2, x_1,1,3,… in exactly the same way as I did the x_1,2, x_1,3,… above.

I want to know if repeating this process ad infinitum will fill up the interval (x₁ − ε₁, _∞). Assuming the worst case scenario: The sequence (x_k) has a limit point, and then (x_1,k) has another limit point, etc. That gives a sequences of limit points; which, in the worse case scenario will have a limit point. But even then, I carry on, and start a sequence of limit points of a sequence of limit points. At each step, the limit point is greater than the limit point of the last step.

Intuitively, because I always start a new sequence at a point greater that the limit point of the last sequence, and I can make as many sequences as I like; it seems that I should be able to fill the interval. Whenever I get stuck at a limit point, I define a new starting point just after the limit point and start a new sequence; when I get stuck, etc… The sequences themselves are indexed by N and the set of sequences in indexed by N, so the whole space of sequences is indexed by N^N, which is the space of maps from N to N. If I can fill the interval then I can do the same in the negative direction and show that the fibers of vector bundles over R with the Euclidea topology all have the same dimension.

Maybe an argument about the separation axioms might help me prove this? — Fly by Night (talk) 21:42, 15 April 2011 (UTC)

Polyhedra and topology

Hi. Please take a look at these pictures. Both are views looking down on a square pyramid which has tetrahedral indentations made in two of its faces. I hope you can visualise that OK. In terms of node (vertex) and edge topology, the two are exactly equivalent. In other words, you could number the vertices appropriately and then check off identical vertex-to-vertex edges for each. However, as polyhedra they are not topologically equivalent (you can't deform one into the other). I suppose you could say that these polyhedra are degenerate/disallowed because their faces are not all simple polygons. Nevertheless, I would like to understand the exact criteria which must hold in order that topological node-edge equivalence guarantees topological polyhedra equivalence. Is it enough that all the faces are simply-connected, for example? Apologies if I have misused any technical terminology in my question. I hope it is still clear what I am trying to ask. 86.176.210.249 (talk) 22:12, 15 April 2011 (UTC)

Aren't all polyhedra just homeomorphic to the sphere? Or is there some finer notion of topological equivalence of polyhedra that applies here? Sławomir Biały (talk) 22:40, 15 April 2011 (UTC)

The fact that all polyhedra are in a sense equivalent to spheres is not relevant to my question. I hope you can see that my two polyhedra are inequivalent in a different way: one cannot be distorted into the other while respecting vertices and edges. In contrast, a cube and a parallelepiped, for example, are equivalent in the sense that I mean. Perhaps "topologial" was the wrong word to use. "Common sense" suggests that all polyhedra are equivalent (in my sense) if their vertices and edges are equivalent (in my sense), but I have given a counterexample, albeit of an exotic type. 86.176.210.249 (talk) 23:56, 15 April 2011 (UTC)

You need to be careful with your language. Mathematics is a very precise business. Both Sławomir and Bkell answered your question as stated. It seems that the equivalence you want is a linear transformation; which is incredibly restrictive. To see this, the space of linear transformations of the plane is four dimensional. The space of homeomorphisms of the plane is infinite dimensional. — Fly by Night (talk) 00:14, 16 April 2011 (UTC)

No, if I understand correctly, the equivalence the questioner is talking about is not required to be a linear transformation; a cube is also equivalent to a square frustum, for example. —Bkell (talk) 05:52, 16 April 2011 (UTC)

I'm not completely sure what I'm about to say is correct, because I'm pretty fuzzy on my polytopes knowledge, but I think the concept you're looking for is called combinatorial isomorphism. I don't know much more than the name, but perhaps with a name you can find more information. —Bkell (talk) 22:57, 15 April 2011 (UTC)

(edit conflict) When you speak of the "node (vertex) and edge topology", you're talking about graph isomorphisms, and they are isomorphic graphs. Sławomir is correct when he says that they are both homeomorphic to the sphere. When you say that "you can't deform one into the other", that usually means that they are not homotopic, which is clearly false since they're both homeomorphic. It seems that Bkell has the equivalence relation that you seek. — Fly by Night (talk) 00:07, 16 April 2011 (UTC)

• It seems that my concept of polyhedra being "equivalent" may have caused confusion. Let me try to explain it again. Take a parallelepiped made out of rubber sheet. The edges are drawn as black lines. Simply by distorting the rubber sheet you can deform it into a cube, so that all the black lines become edges of the cube, without cutting the sheet, splitting it, passing it through itself, etc. Therefore the two are, in my sense "equivalent". However, if you make the left-hand polyhedron in my graphic out of rubber sheet, and mark the edges with black lines, then you can't distort it into the right-hand polyhedron, so that all the edges and vertices correspond. Therefore the two are not equivalent. Some of the material referenced looks too technical for me. If my general question is too complicated to answer in this thread, I would most of all like to know whether, if two polyhedra have simply-connected faces and are graph-isomorphic in terms of vertices and edges, then are they also equivalent in the sense I have described? 86.179.4.74 (talk) 00:53, 16 April 2011 (UTC)

It may not help to answer the question, but we can formulate your notion of topological equivalence as follows: given two embeddings of a planar graph into the sphere, we ask whether the embedding are isotopic. Sławomir Biały (talk) 01:32, 16 April 2011 (UTC)

The link that Sławomir was looking for was this. An isotopy is a special kind of homotopy. But you need to realise that straight lines don't usually go to straight lines under isotopies. To use your rubber sheet example: you could twist it and stretch it so that the "edges" become wiggly curves. — Fly by Night (talk) 01:49, 16 April 2011 (UTC)

For my purposes it does not matter that straight lines could become wiggly. I'm concerned with the possibility of a rubber-sheet-style deformation of one polyhedron to another, but the fact that the rubber sheet could also potentially be deformed to all manner of wiggly-edged pseudo-polyhedra doesn't matter. 86.179.4.74 (talk) 02:34, 16 April 2011 (UTC)

Hmmm. Take a look at these. Again, these are views looking down on a square pyramid with tetrahedral indentations on the faces. The edges and vertices are graph-isomorphic, all the faces are simply connected, and yet the polyhedra are not "equivalent" in the sense relevant here. This is quite contrary to my original intuition, which I'm beginning to realise is pretty hopeless as far as this topic is concerned. So, what about convex polyhedra? Surely all graph-isomorphic convex polyhedra must be "equivalent", right?? 86.181.173.166 (talk) 03:57, 16 April 2011 (UTC).

Yes, I believe that is true. By Steinitz's theorem, three-dimensional convex polyhedra correspond precisely to 3-connected planar graphs; Whitney proved in 1932 that any two planar embeddings of a 3-connected graph are equivalent. The meaning of "equivalent" here is that any graph isomorphism φ between the two planar embeddings preserves not only vertex-edge incidences, which is required of any graph isomorphism, but also vertex-face incidences, i.e., you can identify faces in the first embedding with faces in the second embedding such that the vertices adjacent to any face in the first embedding are mapped by φ to precisely those vertices adjacent to the corresponding face in the second embedding. I think this is what you mean by "equivalent," too. —Bkell (talk) 05:49, 16 April 2011 (UTC)

Thank you! Yes, correspondence of faces (as well as edges and vertices) is what I mean. 86.181.201.80 (talk) 11:27, 16 April 2011 (UTC)