Mathematics desk
< December 1	<< Nov \| December \| Jan >>	December 3 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

December 2

Graph theory illustrations

Hi. I'm writing a paper on a topic in graph theory, and I'm using LaTeX for the first time. I have a number of illustrations I'd like to include, but I'm at a loss as far as what program to use to make such figures. MSPaint seems hopelessly clunky, and I just don't have any fancy graphics software (nor would be likely to recognize it if it bit me). Apparently LaTeX wants the images to be in a ".eps" format? Has anyone here ever dealt with this issue, and can you help me? -GTBacchus^(talk) 02:16, 2 December 2006 (UTC)[reply]

You want graphviz; it's clunky, but when it works, it works.

For what it's worth, .eps is Encapsulated Postscript. Such things as The GIMP can write that format, and convert other graphics to it. Typically for something like graph theory you want vector graphics; Inkscape is generally considered a good editor for those. It also might be useful to try Dia, which is a bit more specialized but simpler. (These are all of course free software.) Does that help? --Tardis 03:46, 2 December 2006 (UTC)[reply]

Inkspace seems easy enough to use, and it even knows how to save files in the ".eps" format. Thank you very much! -GTBacchus^(talk) 04:54, 2 December 2006 (UTC)[reply]

Actually, the graphic shows up when I TeX the document as a .dvi file, but when it tell TeXMaker to make a .pdf, the image doesn't show up.... any idea why that might be? -GTBacchus^(talk) 04:59, 2 December 2006 (UTC)[reply]

Alternately, you may want to try using pdflatex. It lets you dispose of the .dvi and .eps files; you include .pdf figures and it outputs a .pdf document, first try. I don't even remember how to do figures in latex any more, but I remember it was frustrating. Melchoir 06:11, 2 December 2006 (UTC)[reply]

Ah, I see. So if I'm using pdflatex, then I want my images to be .pdf's, and I only use .eps images if I'm TeXing it up as a .dvi? I'll try that... -GTBacchus^(talk) 06:38, 2 December 2006 (UTC)[reply]

If you're doing \usepackage{graphicx} ... \includegraphics{imagefile} you should leave off the extension. That way your same TeX source can find the eps file with that name when you run it as TeX (as you need to do to submit to some sites e.g. arxiv) and find the pdf file when you run it as pdfTeX, without having to change the name in the source to switch between the two. —David Eppstein 06:51, 2 December 2006 (UTC)[reply]

That sounds intriguing, like it could make my life easier at some point, but instead of figuring out what "\usepackage{graphicx}" is, I found a way to convert the .dvi to a .ps and then the .ps to a .pdf, which will do, in this case. Thanks for your advice, everyone. -GTBacchus^(talk) 03:38, 3 December 2006 (UTC)[reply]

How to use graphicx. If you're going to be using eps or pdf illustrations in latex, you probably want to know how to do this anyway, since it provides a very convenient way of resizing images, trimming them when they have bad bounding boxes, etc. —David Eppstein 04:30, 3 December 2006 (UTC)[reply]

As for producing the figures in the first place, I concur with the suggestion of Graphviz as a helpful tool. For example, its directed graph tool, "dot", easily produces the following image:

digraph g { a -> b; a -> c; a -> d; a -> e; b -> d; c -> d; c -> e; d -> e; }

from the input shown. The tool called "neato" will just as easily produce a nicely colored illustration of the pair of Kuratowski subgraphs K₅ and K_3,3 from the following input:

 graph NonPlanar {
   graph [ratio=0.8, bgcolor="#ffffff"]
   node[fontname="Arial", shape="circle"]
   node[group="K5",color="#009246"]
   1[pos="2,2"]; 2; 3; 4; 5
   node[group="K33"]
   node[color="#0081cd"]
   a[pos="1,1"]; b[pos="2,1"]; c[pos="3,1"]
   node[color="#bc1e47"]
   x[pos="1,0"]; y[pos="2,0"]; z[pos="3,0"]
 
   subgraph K5 {
     1 -- {2; 3; 4; 5}
     2 -- {3; 4; 5}
     3 -- {4; 5}
     4 -- {5}
   }
   subgraph K33 {
     a -- {x; y; z}[color="#fec200", style="dashed, setlinewidth(2)"]
     b -- {x; y; z}
     c -- {x; y; z}
   }
 }

Since the output can be requested in SVG form, you can then edit the results in Inkscape to add features or modifications. --KSmrq^T 10:51, 3 December 2006 (UTC)[reply]

Probability Question

A high school gym clas consists of 5 seniors,7 juniors,8 sophomores,6 freshmen. Two students will be chosen at random to assist the teacher in setting up equipment. What is the probability that a freshman will be chosen first and a senior chosen second? Can someone provide the equation for this problem?

Thank you

The probability that a freshman will be chosen first is the number of freshmen divided by the total number of students.

The probability that a senior will be chosen second, given that a freshman was chosen first, is the number of seniors divided by the total number of students left (which is now one less, because one student was already chosen in the first round).

Multiply both probabilities to find the probability of both events happening.

StuRat 13:48, 2 December 2006 (UTC)[reply]

Ambiguity in Mathematics

I am trying to find some examples of ambiguity in Mathematics. One of them would be graph (either vertices and edges or a plot of a function). Any others? Mr.K. 20:01, 2 December 2006 (UTC)[reply]

Lattice (mathematics) (either a regular grid of points or a partial order with meets and joins). —David Eppstein 20:21, 2 December 2006 (UTC)[reply]

Blossom (mathematics) (a construction for spline surfaces or a subgraph in graph matching theory). —David Eppstein 22:57, 2 December 2006 (UTC)[reply]

Kernel (several related meanings as a "measure" of non-injectivity, but also the kernel of an integral transform). Closed form (a finitary expression giving a solution "in closed form", or a differential form whose differential is 0). Primitive element (a generator of a cyclic group, a single element generating a field extension, or a third meaning related to Hopf algebras). Linear space (either a vector space, or a specific kind of geometric incidence structure). --Lambiam ^Talk 02:07, 3 December 2006 (UTC)[reply]

Algebra can either be a mass noun representing a branch of math that studies algebraic structures, or it can be a count noun representing a particular type of algebraic strucure (such as a Boolean algebra or a σ-algebra). Thus: "All algebraists study Algebra, but only some algebraists study algebras." -GTBacchus^(talk) 02:47, 3 December 2006 (UTC)[reply]

I wish to point out that this is not ambiguity in Mathematics. It is just ambiguity in words or terms used in Mathematics. 202.168.50.40 00:35, 4 December 2006 (UTC)[reply]

In other words, ironically, the title of this thread is ambiguous? :) Dugwiki 16:43, 4 December 2006 (UTC)[reply]

It depends if you consider the terminology of the Mathematics a part of Mathematics itself or not.Mr.K. 21:35, 5 December 2006 (UTC)[reply]

See Category:Mathematical disambiguation. PrimeHunter 15:37, 6 December 2006 (UTC)[reply]

Projective plane : why is translation group a group , or why two elations form a new elation

Hello,

this is not homework! It's just something I was wondering about :

Consider a projective plane P. Let L be a line. An elation is by definition a collineation( an automorphism of the projective plane) fixing each points on the line L, but not fixing any other points.

Now let a and b be two elations. Prove that the composition a.b is again an elation, or trivial.

This is important, as it is the only difficult axiom that needs to be proven in order to see that they form a group, together with the trivial collineation. Of course, the composition will still be a collineation fixing all points on the line L, but how can one prove it can't fix any other points as well?

Thank you very much, Evilbu 20:26, 2 December 2006 (UTC)[reply]

Wouldn't your group need an identity element, and shouldn't that be the identity automorphism, which fixes not only the line in question, but every point on the projective plane? Am I missing something basic? -GTBacchus^(talk) 20:53, 2 December 2006 (UTC)[reply]

Yes, you are completely correct, but I was taking that into account, that is what I meant by "trivial collineation", fixing every point.Evilbu 20:55, 2 December 2006 (UTC)[reply]

Ah, I see it now. Sorry. -GTBacchus^(talk) 02:48, 3 December 2006 (UTC)[reply]

Coxeter, in Introduction to Geometry, 2/e (ISBN 978-0-471-50458-0), takes a slightly different approach. He defines

A collineation is a transformation of the plane which transforms collinear points into collinear points.
A perspective collineation with center O and axis o is a collineation which leaves invariant all the lines through O and all the points on o.
A perspective collineation is an elation or a homology according as the center and axis are or are not incident.

Later he gives as an exercise a lemma, not a definition.

A collineation which leaves just the points of one line invariant is an elation.

Also, terminology varies, and foundations vary. The term central collineation is often used to mean the same thing as Coxeter's perspective collineation; the term perspectivity is also seen. A projective plane need not be real or complex; for example, it can be based on a vector space over an arbitrary field. A linear transformation of the form

{\begin{bmatrix}1&0&0\\0&1&0\\0&a&1\end{bmatrix}}

may be called a transvection or a shear, depending on the context, and corresponds to an elation in the associated projective geometry.

Most important, I believe the theorem you want to prove should demand a fixed axis and center. Then given any point pair P and P′ not incident with the axis, there is a unique elation with that axis and center mapping P to P′ (which we can construct geometrically). Furthermore, we can compose these.

Try using these hints to take you further; I think you'll be able to succeed. If not, ask again. --KSmrq^T 10:22, 3 December 2006 (UTC)[reply]

Thanks. But I know that they don't have to real or complex, I'm quite familiar with galois fields. But my question is in the most general case : I'm only assuming the axioms of a projective plane. But I already found the answer and I won't keep it from those who helped me of course :

first one needs to know that given an elation with fixed axis L, the elation is completely determined by its action on just one other point

now let $a,b$ be elations with axis L and assume that a b=u is nontrivial and not an elation

It will still be a perspectivity with axis L, thus a homology. Let its center be c.

now $c^{a}=c^{ub^{-1}}=c^{b^{-1}}$

This means that $a$ and $b^{-1}$ map c onto the same point, so they are equal. Thus a b is tirival, contradicting our assumption Thanks!Evilbu 17:08, 3 December 2006 (UTC)[reply]

question

what is 2+i?

It's just 2+i. The standard representation for complex numbers is as the sum of a real part and an imaginary part. See linked article for more information. -- Meni Rosenfeld (talk) 23:02, 2 December 2006 (UTC)[reply]

i is equal to the square root of a negative 1. i should be treated like a variable when adding or multiplying, unless it is being multiplied by i. If you really want to know more, take a look at complex number. --Ian 00:00, 3 December 2006 (UTC)[reply]

i is equal to the principle square root of -1 and the other square root of -1 is -i (minus i) of course. 220.239.110.225 11:27, 4 December 2006 (UTC)[reply]

Here it is, graphed on the complex plane:

     ^
   3i|
   2i|  
    i|   X
<----+---------->
  -1 | 1 2 3 4 5
     |
  -3i|
  -4i|
     v

StuRat 10:36, 3 December 2006 (UTC)[reply]

2+i is also a length of 5^1/2 and an angle of 30°. -- DLL ^{.. T} 16:46, 3 December 2006 (UTC)[reply]

No, it certainly does not have an angle of 30° (more like 26.565...°). -- Meni Rosenfeld (talk) 18:46, 3 December 2006 (UTC)[reply]

What is 2+i? It's one of the square roots of 3+4i. -GTBacchus^(talk) 08:58, 4 December 2006 (UTC)[reply]