Wikipedia:Reference desk/Archives/Mathematics/2007 November 19

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November 19

How to accurately represent a banked racetrack turn (basic question about geometric angles)

DONE! -- this task was figured out. See the results at:

• Image:Track banking.svg
• Image:Track banking 2xx.svg
• Image:AVUS_track_banking.svg

Thanks, Guroadrunner (talk) 06:23, 19 November 2007 (UTC) _

For the article on the motorsports track called the AVUS, I would like to make a graphical representation of its 43-degree banking ("The Wall of Death") in comparison to other major race tracks, such as Indy's 11-degree banked turns.

Basically it would look like the image at right.

Mathematically, I want accurate representations of lines that would graphically represent a 43 degree angle, a 31 degree angle and an 11 degree angle.

Any help or software recommendations to generate this image? Guroadrunner (talk) 00:02, 19 November 2007 (UTC)

Well, first off, I'm going to assume you want all three lines to go through (0,0). And that you want the slope to represent the angle of however many degrees you want (11, 31, and 43). To graph them you will need to employ a trigonometric function tangent (tan) of the angle in degrees (make sure your calculator is in degree mode not radians!. so from point-slope form of a line we get $y-y_{1}=m(x-x_{1})$ where (x_1,y_1) is the point we want the line to pass through (which is (0,0) in this case). So the above simplifies to $y=mx$. Now to find m we need y/x from our angles, which is where "tan" comes in; tan is defined as $\operatorname {tan} \theta ={\frac {y}{x}}$ where theta "$\theta$" is the angles (in degrees in this case)/ So $m=\operatorname {tan} \theta$ and $y=(\operatorname {tan} \theta )x$. A math-wiki (talk) 00:23, 19 November 2007 (UTC)

Thank you for the response. Is there any computer program that outputs a graphical representation of this using the trigonomy formula given, or preferably a simpler method using geometry or perhaps at an algebra I level? I do want all three straight lines to start at (0,0). Guroadrunner (talk) 00:51, 19 November 2007 (UTC)

I can make it for you, but what do you want it to look like (colour, labels, size, etc..) You can also do it yourself using gnuplot or inkscape or any other graphing program really. You also don't need to worry about trig since the tan($\theta$) is a constant. i.e. tan 11 appox.= 0.194380309 meaning one formula would be y=(0.19438)x. —Cronholm¹⁴⁴ 01:53, 19 November 2007 (UTC)

Thank you for the offer. I found this site http://nces.ed.gov/nceskids/createagraph/

and used it to make this: http://nces.ed.gov/nceskids/createagraph/default.aspx?ID=d5ef31c91fc049aa97ed65dd8ceac2d3

and it worked pretty well. However, inkscape looks really great and I look forward to learning it. Guroadrunner (talk) 02:17, 19 November 2007 (UTC)

Cool, looks good! One caveat on Inkscape, it isn't a formal graphing program so much as a graphic editor(a very useful one though), but it is sufficient for the creation of lines at a specific angle, like you were requesting. —Cronholm¹⁴⁴ 02:37, 19 November 2007 (UTC)

Separability in Topology

I know that the set of all continuous functions from the reals to the reals are separable because the set of all polynomials are dense in this set by the Weierstrass Approximation Theorem. My question is what about the set of all functions from the reals to reals (meaning $\mathbb {R^{R}}$)? Is this set separable? If yes, what is the dense countable subset? Any proofs perhaps?

A Real Kaiser (talk) 03:12, 19 November 2007 (UTC)

I'm going to say yes, it's separable. But that's just because I put the trivial topology on it. Did you have a different one in mind? 134.173.93.150 (talk) 05:08, 19 November 2007 (UTC)

To me, the obvious topology is the product topology, which is separable according to separable space. No proof is given there. Algebraist 11:48, 19 November 2007 (UTC)

Dugundji proves (in VIII.7) that a product of at most 2^{$\aleph _{0}$} separable spaces is separable (his actual claim is an if and only if statement). The proof selects a countable dense subset in each factor and requires the existence of an injection of the index set of the product into the unit interval. If the compact-open topology on functions R→R (the first discrete, the second standard) is used instead, it's not obviously separable. Since one natural topology on a function space is compact-open, I thought this is what the OP meant. Michael Slone (talk) 14:31, 19 November 2007 (UTC)

Note that compact-open on R→R is just pointwise convergence if you put the discrete topology on the first R. It would also be natural to consider the compact-open topology on functions R→R with the standard topology on both factors; this would be the topology of uniform convergence on compact sets. Tesseran (talk) 00:07, 20 November 2007 (UTC)

Yes, I found that in Munkres after I posted. Thanks for pointing this out. I'm a little out of practice in point-set. Michael Slone (talk) 02:40, 20 November 2007 (UTC)

Jacques Tits

There has been a request for information on the pronunciation of the surname of mathematician Jacques Tits to be added to his article. No-one has yet come up with anything definitive at Talk:Jacques Tits. Nurg (talk) 04:50, 19 November 2007 (UTC)

I've generally heard it pronounced Teats, but I'm not sure if that's correct or simply a way for the English to avoid smirking. I couldn't find any references anywhere. Tom 100%_{(talk,contribs)} 16:20, 19 November 2007 (UTC)

"Teats" to prevent smirking? Sounds counterintuitive to me. =)–King Bee ^{(τ • γ)} 17:05, 19 November 2007 (UTC)

With a name like Jacques Tits, you'll never stop the English smirking ;p DuncanHill (talk) 17:22, 19 November 2007 (UTC)

At a mathcamp I worked at, there were mathematicians with the names Butts, Cox, and Pusey. Good times. Black Carrot (talk) 21:20, 21 November 2007 (UTC)

Finite dimensional algebra

Howdy,

I am trying to understand a particular class of algebras, but the original definition is fairly imprecise and unclear. I worked a few completely trivial examples, but an example that was merely just plain trivial has me stumped. The question is a bit long, so a quick summary is:

• Here is a "nearly natural" finite dimensional algebra of finite representation type. Can it be written down naturally? Can its indecomposable modules be explicitly checked?

Take k to be a (finite) field, x an indeterminate, R=k[x]/(x^2), A = R^4 with R-basis { e11, e12, e21, e22 } and R-linear multiplication the same as for matrix units (eij*ekl = delta(j,k)*eil), except e12*e21=x*e11 and e21*e12=x*e22.

• Is there a short, clean, and clear (re)presentation of this algebra as a quotient of a free algebra, a subalgebra of a matrix k-algebra, or a path algebra?

This algebra is supposed to be Morita equivalent to the principal 3-block of the symmetric group on four points (when k=Z/3Z), and so theoretically I know a lot about its module theory. In particular, A should have two irreducible modules and six indecomposable modules, all of which are submodules of A.

• Is there a simple, concrete way to verify the indecomposable modules?

Ideally these concrete calculations should assist in larger examples where I have fewer known results to apply. In particular, pulling the calculations back to k[S4] would be counterproductive to my purposes, and might lose some of the intuition that the field is arbitrary.

A fuller version of the question is at [1] and the currently empty follow-up discussion is at [2] or at any sci.math usenet location. JackSchmidt (talk) 21:06, 19 November 2007 (UTC)

Formula for quadratics based on very little info

Hi. I was doing a homework problem today and I need help understanding how to do it. Basically, the teacher gave us a picture, which is shaped like a parabola (what are the odds *rolls eyes*), and we have to come up with a formula from the info in there. The only info (yes, I am sure) is:

• The x-intercepts (0 and 24)
• The vertex (12,40)

Now, I've tried doing it in vertex form (can never get the intercepts right), and in intercept form (go figure - I can't get the vertex right). Is there a way to figure this out or did she give us too little information? The formula would be appreciated. Thanks! 96.225.64.203 (talk) 21:43, 19 November 2007 (UTC)

Well remember that you can write the formula of a parabola as $y=a(x-b)(x-c)$. where b and c are the x-intercepts. That just means that you need to find the vertex, so plug in the x and y values and solve for a. Then take your values for a, b and c in the formula I gave and simplify and you're done. I'll give you one hint to help you out: a should be negative. Donald Hosek (talk) 21:50, 19 November 2007 (UTC)

In fact, you even have slightly more information than you need (you can readily deduce that the abscissa of the vertex is 12, you don't need to have it explicitly specified). I guess that by "intercept form" you mean $a(x-x_{1})(x-x_{2})$, as suggested by Donald, and it leads to one way to solve it. By "vertex form" I guess you mean $a(x-x_{0})^{2}+y_{0}$, which you can also solve since you know $(x_{0},y_{0})$, and substituting x=0 allows you to find a. Finally, you can simply write the formula as $ax^{2}+bx+c$; you know that the parabola passes through (0, 0), (12, 40) and (24, 0), and this gives you $c=0$, $144a+12b+c=40$ and $576a+24b+c=0$, a system of linear equations which you can solve (this is a special case of the Lagrange polynomial). -- Meni Rosenfeld (talk) 22:21, 19 November 2007 (UTC)