Wikipedia:Reference desk/Archives/Mathematics/2010 November 28

= November 28 =

anti-Delaunay
From Delaunay triangulation:


 * The problem of finding the Delaunay triangulation of a set of points in n-dimensional Euclidean space can be converted to the problem of finding the convex hull of a set of points in (n + 1)-dimensional space, by giving each point p an extra coordinate equal to |p|2, taking the bottom side of the convex hull, and mapping back to n-dimensional space by deleting the last coordinate. ...

(Bottom here means the set of facets whose normal has a negative component in the extra dimension.) I've always wondered: do the top facets make anything interesting? —Tamfang (talk) 02:59, 28 November 2010 (UTC)


 * I doubt it.... One thing I did figure out, if P is your set of points, and C is the subset that lies on the n-dimensional convex hull of P, then these top facets should be a triangulation of C.  It feels something like an "opposite"-Delaunay triangulation, in that (in a few examples I tried) it seems to produce triangulations as far as possible from being Delaunay.  Maybe if you did some kind of transform like v maps to $$\frac v{|v|^2}$$ this triangulation of C could be translated into something recognizable.  Eric.  82.139.80.250 (talk) 00:08, 29 November 2010 (UTC)

Group theory example
Hello. Can someone give me an example of an infinite non abelian group all of whose conjugacy classes are finite? Thanks-Shahab (talk) 05:19, 28 November 2010 (UTC)


 * I think the infinite dihedral group is an example. Eric.  82.139.80.250 (talk) 08:16, 28 November 2010 (UTC)
 * So is the direct product of any finite nonabelian group with any infinite abelian group. Algebraist 11:52, 28 November 2010 (UTC)
 * OK. Thanks-Shahab (talk) 03:36, 29 November 2010 (UTC)

Delta-epsilon limits revisited
What are some recommended strategies or readings (preferably very detailed almost to the point of repetitiveness, and ending up at a rigorous understanding but not necessarily starting rigorously) for understand how to make and use the delta-epsilon proofs of limits. I tried Spivak's Calculus but he is too rigorous in this respect and doesn't give me the detail I need (he skips a lot of steps or assumes the reader can see immediately how he got from two steps). I already understand the other aspects of calculus pretty well but this is something that has me puzzled a bit still. Thanks. 24.92.78.167 (talk) 18:47, 28 November 2010 (UTC)


 * Start by mastering this: to say that
 * $$ \lim_{x\to a} f(x) = L \, $$
 * means that &fnof;(x) can be made as close as desired to L by making x close enough, but not equal, to a. The number &epsilon; is how close one wishes to make &fnof;(x) to L; the number &delta; is how close one will make x to a in order to achieve that. Michael Hardy (talk) 20:33, 28 November 2010 (UTC)
 * means that &fnof;(x) can be made as close as desired to L by making x close enough, but not equal, to a. The number &epsilon; is how close one wishes to make &fnof;(x) to L; the number &delta; is how close one will make x to a in order to achieve that. Michael Hardy (talk) 20:33, 28 November 2010 (UTC)
 * means that &fnof;(x) can be made as close as desired to L by making x close enough, but not equal, to a. The number &epsilon; is how close one wishes to make &fnof;(x) to L; the number &delta; is how close one will make x to a in order to achieve that. Michael Hardy (talk) 20:33, 28 November 2010 (UTC)

Maybe some specific examples of what you didn't understand in Spivak would help here. Michael Hardy (talk) 20:35, 28 November 2010 (UTC)


 * OK, for example it says for what δ will |f(x) - L| < ε for 0 < |x-a| < δ, where f(x)=x4 and L=a4. I get $$\frac{\epsilon}{4a^3}$$ but in the back of the book it has a crazy solution with absolute values and lots of terms in the denominator and the min(x) function. 24.92.78.167 (talk) 02:03, 29 November 2010 (UTC)


 * Not really a good exercise if you're still struggling with the logic. Keep in mind that it took about 150 years from when Newton and Leibniz invented calculus for people like Cauchy and Weierstrass to put on a rigorous footing, so it shouldn't be surprising that a typical college student will take a while to get it. On the other hand some ways of teaching it are better than others and the approach not to take is to just start with the epsilons and deltas and assume students will catch up eventually. If you are planing on being a math major then this is something you should master, if only because you'll find more difficult concepts down the road and it's good practice. If you're not going to be a math major then you should know that many people manage to lead perfectly happy, productive lives without understanding this at all. But if you're the kind of person who skips to the 'difficult' section of a sudoku book then you may derive a certain amount of satisfaction in figuring this out that will be worth the effort spent.--RDBury (talk) 08:18, 29 November 2010 (UTC)


 * Let's see ... if |x-a| < δ then a-δ < x < a+δ. Then we have
 * $$f(a-\delta) - f(a) = (a-\delta)^4 - a^4=-4a^3\delta+6a^2\delta^2-4a\delta^3+\delta^4$$
 * $$f(a+\delta) - f(a) = (a+\delta)^4 - a^4=4a^3\delta+6a^2\delta^2+4a\delta^3+\delta^4$$
 * The sign and relative magnitude of these differences depends on the sign of a, but we can combine both differences into a "worst case" scenario to give
 * $$|f(a\pm\delta) - f(a)| \le 4|a|^3\delta+6a^2\delta^2+4|a|\delta^3+\delta^4$$
 * and if we want this absolute difference to be less than ε then we must find δ such that
 * $$4|a|^3\delta+6a^2\delta^2+4|a|\delta^3+\delta^4 < \epsilon$$
 * Your solution assumes that the terms involving second and higher powers of δ can be omitted. This is fine for a non-rigorous approach, but to be rigorous we either need to solve this quartic inequality exactly (tricky) or we need to make some further assumption about the relative sizes of a and δ. For example, if we assume δ < |a| then we have
 * $$4|a|^3\delta+6a^2\delta^2+4|a|\delta^3+\delta^4 < 15|a|^3\delta$$
 * and so
 * $$\delta < \frac{\epsilon}{15|a|^3} \Rightarrow \epsilon > 15|a|^3\delta > 4|a|^3\delta+6a^2\delta^2+4|a|\delta^3+\delta^4 \ge |f(a\pm\delta) - f(a)|$$
 * Combining the two conditions together, we have
 * $$\delta < \min(|a|,\frac{\epsilon}{15|a|^3})$$
 * (modulo schoolboy errors in algebra !) Gandalf61 (talk) 11:14, 29 November 2010 (UTC)

Derivative
I've noticed that circumference is the derivative of area of a circle. WHat is the significance of this (besides the fundamental theorem of calculus)? Does this mean that in general, the area of a graph is the integral of the function/relation defining its boundary? I can see this is probably true for an ellipse, but what about an open boundary like a parabola or hyperbola? What about a really weird, closed, shape like a blob? 24.92.78.167 (talk) 23:29, 28 November 2010 (UTC)
 * For a circle, the derivative of the area A=πr2 as function of the radius r is the circumference L = dA/dr = 2πr because the ringshaped area dA is that of a rectangle of length L = circumference and width = dr = tiny increment of radius. dA = L dr. A similar result is not true for an ellipse because the elliptical ring does not have constant width. Bo Jacoby (talk) 00:29, 29 November 2010 (UTC).

That is to be expected. There are a number of different ways to see this. Here's one. It also works with the volume and surface area of a sphere, for the same reason. Michael Hardy (talk) 00:47, 29 November 2010 (UTC)
 * This principle works if the boundary moves at the same rate at all points. In some cases, different parts of the boundary move at different rates, and then you've got a more complicated situation. Michael Hardy (talk) 00:49, 29 November 2010 (UTC)


 * By the way, Archimedes first computed the volume of a sphere by means of mechanical arguments (more or less equivalent to integration by parts in modern formalism). Then he computed in a more conventional and rigorous way (using the method of exhaustion) the area of the spherical surface. He later explained the first method in the famous letter to Eratosthenes (the treatise The Method of Mechanical Theorems). It is interesting that there he also observes that the formulas for the volume and for the surface area are easily seen to be equivalent, in that the volume is one third of radius times surface area -this, comparing the sphere with a cone, not by means of a derivative ;-).--pm a  12:06, 5 December 2010 (UTC)