Wikipedia:Reference desk/Archives/Mathematics/2011 April 13

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April 13

Constant (mathematics) vs Mathematical constant

I came across these two whilst cleaning up a few maths articles. Is there any significant difference between these two articles that is preventing the one being merged/subsumed into the other? Could someone with a stronger mathematical background please have a look at them and decide? There is also a merge notice on the first article suggesting a merge with Variable (mathematics). Input much appreciated. Regards. Zunaid 09:48, 13 April 2011 (UTC)

Update: the mathematical constant article seems (to me) to be filled with a gratuitous list of examples rather than an encyclopedic discourse on mathematical constants as phenomena themselves (e.g. explaining in outline how they arise from geometry, number theory or whatever). The meat of the article is all hung on the examples, each of which have their own articles and do not need to be duplicated in this article to pad it out. I would venture that the article could be snipped to only its lead section (even the lead section is more about the examples than the phenomenon itself!), appended by a suitable template table of mathematical constants, without losing anything. And yes, I know this should be a discussion for the talk page, but asking for more eyes from a different source can't do any harm :) Zunaid 09:59, 13 April 2011 (UTC)

The two articles have quite different subjects. Constant (mathematics) is about the concept of constants (as opposed to variables) in mathematical expressions; mathematical constant is an navigation article that gives a summary of the values and properties of specific mathematical constants. There may be some merit in your suggestion of removing the summary sections from mathematical constant, leaving a lead and a table, so that it becomes more of a "List of ..." style article (like list of mathematical symbols) - but, as you say, this proposal should be discussed at Talk:Mathematical constant. Gandalf61 (talk) 10:18, 13 April 2011 (UTC)

There was some discussion about this at Talk:Constant (mathematics) in the past. Sławomir Biały (talk) 11:07, 13 April 2011 (UTC)

They do seem different articles but there's quite a few things that grate I must admit. They have no disambiguation to each other. The mathematical constant article doesn't refer to the main articles on its constants and the very first one I looked at the name 'Archimedes constant' is fairly uncommon compared to pi. There certainly is a problem about why those particular constants were chosen, for instance by citing some book list of useful or interesting constants. We have a Category:Mathematical constants so this need only be a very select group or otherwise it should be a 'list of' article. In the constant (mathematics) article it quite unnecessarily uses differentiation when the level could be usefully kept quite a bit lower, for instance by just extending the quadratic example to give the solution of a quadratic equation. It has no references either. If anyone would like to try a hand at improving articles in Wikipedia these seem like good ones to start on. Dmcq (talk) 11:21, 13 April 2011 (UTC)

Yupana

How were calculation performed on the Yupana?Smallman12q (talk) 12:00, 13 April 2011 (UTC)

It appears from the article that no one knows for sure today. I imagine though it worked something like an abacus but with pebbles or kernels of maize (as described in the article) placed in the depressions to signify various quantities rather than beads being moved. You could easily construct a "pebble abacus" using this idea to see how this might work with the decimal system.--RDBury (talk) 13:46, 13 April 2011 (UTC)

Representation theory - problem understanding lecture notes, request for clarification!

Hello everyone,

I'm a second year student trying to teach myself representation theory in preparation for the coming year, using the following notes (http://tartarus.org/gareth/maths/notes/ii/Representation_Theory.pdf or http://www.dpmms.cam.ac.uk/study/II/RepresentationTheory/2010-2011/Representation_Theory.pdf, hopefully one of those links will work!). However, I'm having trouble understanding something in the explanation of the calculation of the character tables of the symmetric groups S_4 and S_5 on page 21. (You do not need to read the first 20 pages to know what it's going on about, assuming you know some representation theory, just page 20 I think!)

Assuming the use of the notation discussed on page 20 for the permutation representation/characters, it seems to be saying the 4th line of the S4 character table is derived from the product of $\pi _{X_{3}}\times \pi _{X_{2}}$, but it is obviously (just by looking at the values) derived from the product of the characters $\chi _{2}$ and $\chi _{3}$ in the two preceding rows. However, using the notation from the previous page, surely $\chi _{2}$ and $\chi _{3}$ are not the characters $\pi _{X_{2}}$ and $\pi _{X_{3}}$? If I've understood correctly, those characters would be given by looking at the number of elements fixed by something in each conjugacy class in the symmetric group S_4, which would in most cases be zero: certainly fewer than in the case of $\pi _{X_{4}}$, and of course $\pi _{X_{4}}-1$ is the character $\chi _{3}$. How do we rectify this then with the fact that the arrow appears (I can only surmise, since this material is not lectured for my year I wasn't able to attend the lecture itself!) to be saying we can deduce that 4th character from the product $\pi _{X_{3}}\times \pi _{X_{2}}$?

I was hoping that perhaps it was simply a mistake in the writing of the notes (there have certainly been others I've found!), but the same thing appears to have been going on in the character table for S_5: the product of the characters $\chi _{i}$ becomes that of the characters $\pi _{X_{i}}$, for i=2, 3 and 5. Am i missing something? What's going on? Quick response to aid my understanding would be greatly appreciated: thank you in advance! :) Sdnahzzaj (talk) 05:41, 6 May 2011 (UTC)

I think in this context the $\pi _{X_{n}}$ is just supposed to mean the representation corresponding to the character $X_{n}$; the meaning you described may have been used elsewhere in the book but only as a temporary convenience. I agree that the notation is at best confusing. The gist of it, whatever the notation is, is that the new character is the product of two previously found characters, which you can get from the tensor product of the corresponding representations.--RDBury (talk) 14:21, 13 April 2011 (UTC)

It was a typo (probably a cut&paste error, knowing me), which I have now (hopefully) corrected. Please let me know of any other mistakes - you say you have found some. My email address is at the end of the notes. 86.26.0.138 (talk) 21:26, 18 April 2011 (UTC)

Even graph with even edges

I am trying to show that every Eulerian simple graph with an even number of vertices has an even number of edges. I have devised a long argument to show this (of whose validity I am not sure about) and would highly appreciate it if someone can give me a neat clean proof. Can someone also tell me whether my argument is sound or not. Thanks-Shahab (talk) 13:15, 13 April 2011 (UTC)

My proof

Consider the incidence matrix of such a graph and cross out its columns in the following fashion: Consider the first column.
If the first vertex having entry 1 from the top is v_i (and the next is v_j) then cross out all columns
containing 1 in the ith row. The parity of the number of uncrossed columns does not change as we have crossed out even number
of columns.
Next cross out all columns containing 1 in the jth row. Note that there will be at least 1 such column for else there are
multiple edges. This time the parity changes for we cross out odd number of columns corresponding to v_j
Move on to the next uncrossed column after column 1 and keep repeating the procedure. If even number of vertices are yet to
be considered, the parity changes and if odd number of vertices are yet to be considered then it doesn't change.
A stage will come when there is exactly 1 uncrossed column. Since even number of vertices are yet to be considered,
the parity has changed. Crossing out this last column considers two vertices in one go, and thus doesnt change the parity.
Since 0 is even so we had initially an even number of edges.

The diagram File:Spirodecane-ifa.png shows a graph with all vertices having even degree, and an even number of vertices, but 11 edges.--RDBury (talk) 14:44, 13 April 2011 (UTC)

Ok. Thanks-Shahab (talk) 14:45, 13 April 2011 (UTC)

Parameterisation

If you have a closed curve C in the complex plane with a known parameterisation $\gamma (t)=x(t)+iy(t)$ and you also have a known complex valued polynomial function $h(z)$, is a parameterisation for $h(C)$ given by $h(\gamma )$? Cheers for any help. meromorphic [talk to me] 13:19, 13 April 2011 (UTC)

C is a closed curve, parametrised by γ : S¹ → C, where S¹ is the circle. You have a polynomial h : C → C, and you want to parametrised h(C)? Then the image h(C) is parametrised by the composite (h∘γ) : S¹ → C. This is because, by definition (h∘γ)(θ) = h(γ(θ)) for all θ ∈ S¹. The composite might not be a regular parametrisation, but it is a parametrisation. — Fly by Night (talk) 13:56, 13 April 2011 (UTC)

So to put it in simple terms, for a simpleton such as myself, if $h(\gamma )=u(t)+iv(t)$ then u and v are my parametric representation? Also, what is a regular parameterisation? meromorphic [talk to me] 13:59, 13 April 2011 (UTC)

If γ(θ) = x(θ) + iy(θ), then (h∘γ)(θ) = h(x(θ) + iy(θ)). The expression (h∘γ) is a parametrisation for h(C). In your last post, u and v parametrize the real and imaginary parts of h(C) respectively. γ : S¹ → C is a regular parametrisation if dγ/dθ ≠ 0 for all θ. Assuming that γ : S¹ → C is regular then (h∘γ) : S¹ → C. will be regular (d(h∘γ)/dθ ≠ 0) provided C doesn't pass through the points where dh/dz = 0. — Fly by Night (talk) 14:07, 13 April 2011 (UTC)

I'm considering this in relation to a contour integral. If I have $\int _{h(C)}f\,ds$ where s is arc length and C is the circle of, fixed, radius r, centre the origin in the z plane. Clearly, if the circle is parameterised by $\phi$ then $ds=rd\phi$ but does this hold for the parameterisation of $h(C)$ in the w plane? I ask because the specific function I am considering is, trivially, written as a function of $\phi$ and I don't know how to convert this to arc length (and I feel I am not meant to change variable to arc length). On rereading my question, I'm unsure if it's clear or not but can't quite pin point what's niggling me so let me know if you're unsure what I mean. Thank you. meromorphic [talk to me] 21:14, 13 April 2011 (UTC)

It seems to me like you have a few things mixed up. If you have a contour integral, then it is usually expressed as

$\oint _{C}f(z)\,\operatorname {d} z\,.$

If C is the circle, centre 0, radius r then z(θ) = { re^iθ : 0 ≤ θ < 2π }. In that case, we substitute z = re^iθ which means dz = ire^iθ. It follows that

$\oint _{C}f(z)\,\operatorname {d} z=ir\!\int _{0}^{2\pi }\!\!\!\!f\!\left(re^{i\theta }\right)\operatorname {d} \theta \,.$

If you wanted the parametrisation in terms of arc-length then you'd use

$z(\theta )=re^{i\theta /r}\ {\text{where}}\ \,0\leq \theta <2\pi r\,.$

The value of the integral does not depend on the parametrisation, provided that it is regular. So I don't see a need to use arc-length. Does that help at all? It would help me a lot if you gave the problem explicitly, so we were both singing from the same hymn sheet :-) — Fly by Night (talk) 21:34, 13 April 2011 (UTC)

I appreciate that my current question is perhaps a bit vague but, due to the question originating from a piece of coursework, it would be unsporting (and most likely collusion) for me to go into specifics about it; thanks for offering to give me more detailed help though. I'll try reformulating my question and see if it makes a difference to your understanding of it.

I am considering a mapping between the z and w planes given by w=h(z) and the particular set of points I want to consider lie on the circle radius r, centre the origin in the z plane. I then need to compute the contour integral $\int _{h(C)}f\,ds$, where f is a function traditionally of arc length but I have a parameterisation of f that uses an arbitrary variable, say $\theta$, which I am told is given by $\theta =\arg(z)$ in the z plane, and I don't know how to change variable to arrive at the natural, arc length representation of f, so I need to stick with $\theta$. Now, as before, we know that $ds=rd\phi$ in the z plane. My question is, can I use this change of variable for the contour integral or do I need to consider the fact the contour if $h(C)$ not C? meromorphic [talk to me] 22:03, 13 April 2011 (UTC)

I used {{od}} to out-dent your last post, to give more space. What I think me be happening is this: you have the unit circle in the z-plane, denoted by C. Then you have a contour in the w-plane that is the image of the unit circe under a map h : C_z → C_w. You want to integrate some function g(w) along the contour h(C)? Well, we can transform the integrals by substitution:

$\oint _{h(C)}g(w)\,\operatorname {d} w=\oint _{C}{\frac {\operatorname {d} h}{\operatorname {d} z}}\times g(h(z))\,\operatorname {d} z\,.$

Does this seem like what you're trying to do? All I did, was instead of integrating w along h(C), I replaced w by h(z) and integrated along C. Each point of C gives a point on h(C), and that point is w where w = h(z). Obviously dw transforms to give d(h(z)) = (dh/dz)⋅dz. I hope that helps. — Fly by Night (talk) 16:35, 14 April 2011 (UTC)

Self complementary graphs

Resolved

Here is another graph theory question. Let G be a self complementary graph. Prove that G has a cut vertex iff G has a vertex of degree 1.-Shahab (talk) 17:30, 13 April 2011 (UTC)

Are these homework questions? What have you tried so far? Do you have any ideas about how to start? —Bkell (talk) 19:25, 13 April 2011 (UTC)

No these arent homework. Unfortunately I dont go to school, I am trying to read by myself . If G has a cut vertex then G-v is disconnected. Its complement G'-v must be connected (G' is the complement of G). Not sure what can I conclude from here.-Shahab (talk) 03:22, 14 April 2011 (UTC)

Here's what I did: I drew a picture with the cut-vertex v in the middle, connected (somehow) to a blob on the left I called H and a blob on the right I called K. I don't know what H and K look like, but I do know there are no edges between them. So, in the complement, every vertex in H must be joined to every vertex in K. So, if H has h vertices and K has k vertices, the complement has K_h,k (a complete bipartite graph) as a subgraph. That must mean that the original graph also has K_h,k as a subgraph. Where is it? It can't be completely contained in either H or K, because individually they don't have enough vertices. And it can't be partly in H and partly in K, because there are no edges between H and K. So it must include v, and in particular one of h or k must be 1. So one of H or K must consist of only a single vertex, which of course must be of degree 1. —Bkell (talk) 03:51, 14 April 2011 (UTC)

By the way, it is not necessarily true (a priori) that G′ − v must be connected. It is true that G − v must be disconnected, but the complement of a disconnected graph is not necessarily connected. —Bkell (talk) 03:54, 14 April 2011 (UTC)

Hmm, on second thought, maybe it is true that the complement of a disconnected graph is always connected. It is certainly not true that the complement of a connected graph is disconnected, but I guess the implication does work in the opposite direction. —Bkell (talk) 03:57, 14 April 2011 (UTC)

Ah, also, it should be noted that what I said above really proves only one direction of the "if and only if." The other direction (if G has a vertex of degree 1, then it has a cut-vertex) should be pretty easy to establish. —Bkell (talk) 04:02, 14 April 2011 (UTC)

Thanks -Shahab (talk) 14:50, 14 April 2011 (UTC)