Wikipedia:Reference desk/Archives/Mathematics/2011 March 17

= March 17 =

Derivatives
Ok this is a really elementry question that has bugged me years ago and has come back now that I'm concentrating on calculus again. Let f:U->R, where U is open in R, the real numbers. When we say the derivative at x we take the limit of the usual quotient as the usual increment h tends to 0. But the term f(x+h) is not defined for all h. So by the derivative at x do we take the limit of the function k:A->R, where A is the set of all h such that x+h is U, and A doesn't contain 0, and k is defined by k(h) = (f(x+h)-f(x))/h. What I'm trying to get at is that the usual limiting quotient has an ill defined domain. Thanks. Money is tight (talk) 13:54, 17 March 2011 (UTC)


 * Since U is open, for all x in U, there is an open neighborhood V of U containing x. Now f(x+h) is defined for h in an open subset containing 0. Invrnc (talk) 14:16, 17 March 2011 (UTC)


 * (edit conflict) I don't see what you gain by changing from f to k—the function f is defined only on the set U, and the function k is defined only on the set A (which is simply U − x), so it seems to me that if you're uncomfortable with f you should be equally uncomfortable with k. But you shouldn't be uncomfortable with either of them. Since the set U is open, for every point x in U there is an open interval containing x that is completely contained within U. Since you are considering a limit as h approaches zero, you can simply focus on that open interval and ignore everything outside of it. In other words, it is true that f(x + h) is not defined for all h, but it is defined for all h that are close enough to 0, so the limit as h approaches 0 is well defined. Just ignore the values of h that are too far away from 0—you wouldn't care about them in the limit anyway. —Bkell (talk) 14:23, 17 March 2011 (UTC)
 * That's exactly my problem, the domain of the limiting quotient is ill defined if we just say its domain is some open ball containing x and in U. I know the resulting limit is the same but I like definite definition. Money is tight (talk) 14:37, 17 March 2011 (UTC)
 * Okay, I'll try to be more definite. Since $$x\in U$$, there exists some $$\eta>0$$ such that $$(x-\eta,x+\eta)\subseteq U$$. We are examining the limit $$\lim_{h\to0}{f(x+h)-f(x)\over h}$$, which is defined to be equal to a real number $$L$$ if and only if, for every $$\epsilon>0$$, there exists some $$\delta>0$$ such that $$\left|{f(x+h)-f(x)\over h}-L\right|<\epsilon$$ whenever $$|h|<\delta$$. For a particular value of $$\epsilon$$, if one value for $$\delta$$ satisfies this condition, then any smaller positive value for $$\delta$$ will also satisfy this condition. So, if we are aiming to prove that the limit is $$L$$ using a delta-epsilon proof, we can always choose $$\delta\le\eta$$; then we will never run into a problem where the difference quotient is undefined. —Bkell (talk) 16:22, 17 March 2011 (UTC)

Monty Hall problem implications
I can't get two issues. First of all, it would be naturally to assume that it's not profitable for the host to offer some door (since the TV show would loose the car), as such the player will most likely choose any door, not suggested by host. Secondly, why the host would open door 3 if I chose door 1? Or they are not labelled? Thanks. —Preceding unsigned comment added by 89.76.224.253 (talk) 15:24, 17 March 2011 (UTC)
 * It's the rules of the game. The host has to offer you a choice, he has to open one of the remaining doors (and of those one that does not have the car), and he then has to offer you the choice of keeping "your" door or changing to the remaining one. How the doors are labeled is irrelevant - what is important is that the host knows where the car is, and never opens the car door. --Stephan Schulz (talk) 15:35, 17 March 2011 (UTC)
 * Keep in mind that these are the rules of the simplest game that results in the usual answer to the problem. They are not the rules by which Monty actually played. --Trovatore (talk) 01:48, 18 March 2011 (UTC)
 * Also, gameshows are happy when contestants win. Winners make for better ratings, and the increase in ad revenue more than makes up for the cost of the prizes.--203.97.79.114 (talk) 11:27, 18 March 2011 (UTC)

Invariance of Differential Forms
I've got a quadratic differential form &omega; = A&thinsp;du2 + 2B&thinsp;du&thinsp;dv + C&thinsp;dv2, where A, B and C are functions of u and v. It's a fact that &omega; is invariant under orientation preserving diffeomorphisms of uv−space. I'm interested in the function g(u,v) = AC − B2, which is the determinant of the matrix representation on &omega;. Am I right in saying that g would not be invariant under orientation preserving diffeomorphisms of uv−space? I'm assuming that the functions A, B and C will transform one way, and that du and dv will transform another way; so that they cancel each other's transformations. Or, is it possible that g is invariant under orientation preserving diffeomorphisms of uv−space? (I really, really hope it is). — Fly by Night  ( talk )  16:50, 17 March 2011 (UTC)
 * Assuming I understand the question, the invariant gadget would be the volume form $$\sqrt{g} du\wedge dv$$. Again see the article metric tensor.  Sławomir Biały  (talk) 17:17, 17 March 2011 (UTC)
 * Stated another way, g is an invariant density of weight 2, meaning that it transforms by the square of the Jacobian determinant of the diffeomorphism. In particular, if the diffeomorphism is symplectic, then g is invariant as a scalar.  Sławomir Biały  (talk) 17:34, 17 March 2011 (UTC)


 * Concretely: In plain old R^2, consider the diffeomorphism that multiplies each coordinate by 2. This also doubles du and dv, so in order to preserve &omega; each of the coefficients has to shrink to a quarter of its former value. Then $$g$$ shrinks by $$\tfrac{1}{16}$$, so not invariant. –Henning Makholm (talk) 17:50, 17 March 2011 (UTC)


 * No, &omega; is not the volume form. I just used g to denote its matrix's determinant. It's more of a general question. I have an explicit form but, as I said, it's not the volume form. It's kernel directions identify the asymptotic directions. Actually, if the surface is given as the graph of a function then
 * $$\omega = \frac{f_{uu} \, \operatorname{d}u^2 + 2f_{uv} \, \operatorname{d}u \, \operatorname{d}v + f_{vv} \, \operatorname{d}v^2}{|f_{uu}f_{uv}-f_{uv}^2|^{1/4}} \, . $$
 * — Fly by Night  ( talk )  18:00, 17 March 2011 (UTC)
 * I was referring to g, not &omega;, above. The form $$\sqrt{g}\,du\wedge dv$$ is a volume form in the uv-plane, and it is invariant assuming that the quadratic differential form &omega; is.  In any event, if you need to regard g itself as invariant, then you need to regard it as a density (that is, a section of a certain line bundle), not a scalar function.   Sławomir Biały  (talk) 18:15, 17 March 2011 (UTC)
 * You've lost me, what is g above? What is &omega; above? I don't need g to be invariant; it either is or it isn't. I would like it to be, and it would mean I've done what I want to do. What does it mean to regard g as a density? A section of a line bundle is a vector field... Like I said, you've lost me. — Fly by Night  ( talk )  18:18, 17 March 2011 (UTC)
 * $$g=AC-B^2$$ is the determinant of the matrix representation of the quadratic form &omega; in the uv coordinate system, per the notation you established. This transforms by the square of the Jacobian determinant of the coordinate transformation.  So it is not an invariant scalar, but is an invariant section of the line bundle $$\wedge^2 T^*\mathbb{R}^2\otimes\wedge^2 T^*\mathbb{R}^2$$.  See density bundle.   Sławomir Biały  (talk) 18:23, 17 March 2011 (UTC)
 * Should there really be four copies of the cotangent bundle, or should it be $$\scriptstyle \wedge^2 T\mathbb{R}^2 \, \otimes \, \wedge^2 T^*\mathbb{R}^2$$? — Fly by Night  ( talk )  19:04, 17 March 2011 (UTC)
 * It's four copies if the cotangent bundle, by dimensional analysis of the original coefficients of omega. Sławomir Biały  (talk) 19:09, 17 March 2011 (UTC)

Binary operation
'''the binary operation ₴ is defined on a set A=(4,6,8,) by a ₴ b= ½ (a-b) if a= 4,b=6 and c=8, evaluate the following; a ₴ (b+c) —Preceding unsigned comment added by 41.218.231.127 (talk) 17:14, 17 March 2011 (UTC)
 * What does ₴ do to a and b? It subtracts the second from the first, and then halves the result. If you want to know what a ₴ (b + c) is then subtract the second argument from the first, and then half the result. That will give you the general answer. Then just substitute in the values given for a, b and c. — Fly by Night  ( talk )  18:02, 17 March 2011 (UTC)
 * Having said that, I think you might have given the wrong set for A. A binary operation must be closed, i.e. it takes two elements of A and gives you an element of A as the answer. If A = {4, 6, 8, &hellip;}, then we have problems, for example 4 ₴ 6 = ½(4 – 6) = –1. But –1 is not in A. If we extend A to the set of all even numbers we still have problems because 2m ₴ 2n = ½(2m – 2n) = m – n; which need not be even. Even if we extend A to the set of all whole numbers we still have problems because m ₴ n = ½(m – n); which is a whole number if and only if m and n are both even numbers. The best you can get away with is A = Q, i.e. A is the set of rational numbers. — Fly by Night  ( talk )  03:18, 18 March 2011 (UTC)
 * It is sufficient if the members of the set A are binary fractions. A = {a2&minus;n : a is an integer, n is a positive integer}. Bo Jacoby (talk) 11:24, 18 March 2011 (UTC).

Rigorous theory of "div, grad, curl, and all that"
Standard textbooks on vector calculus sweep under the carpet the fact that the expressions for things like div, grad, curl, Laplacian, etc. are invariant under rotations, and other things of interest to mathematicians but in which engineers are happy to just trust the mathematicians. Which are the most elementary among books giving a fully rigorous account of such matters? Michael Hardy (talk) 18:09, 17 March 2011 (UTC)


 * I've had good experiences with Marsden and Tromba's textbook "Vector calculus", although I probably wouldn't think of it as "fully rigorous". It seems to have about the right amount of rigor for a solid undergraduate treatment.  Unfortunately, I don't have a copy on my shelf, so I can't say if it discusses precisely these issues.   Sławomir Biały  (talk) 18:20, 17 March 2011 (UTC)
 * Thank you. Michael Hardy (talk) 03:49, 18 March 2011 (UTC)
 * Tom Apostol's old two-volume calculus book covered this stuff pretty rigorously. Again though, it's been a while (high school) since I looked at it, so I don't remember what was in it.  75.57.242.120 (talk) 10:38, 18 March 2011 (UTC)
 * I checked both Apostol and my own recommendation, and neither of them seems to discuss the behavior of div, grad, and curl under rotations. The best I could find aimed at an elementary level is (don't laugh) Schaum's outlines of vector analysis.  They prove in detail that the gradient of a radial function is a rotationally-invariant vector field, and leave it as an exercise to show that the curl and divergence of a rotationally invariant vector field are rotationally invariant, and that the expression for the gradient is rotationally invariant.    Sławomir Biały  (talk) 13:20, 18 March 2011 (UTC)

This is the reason why in physics class we don't use math textbooks to teach this topic. It would be way too cumbersome, you won't learn the efficient techniques to derive the expressions in different coordinate systems. E.g. what you'll find in many physics lecture notes is an argument like this. One considers the expression:

$$df = \frac{\partial f}{\partial x^{i}}dx^{i}$$

and then one notes that because df is a scalar and the $$dx^{i}$$ are the components of a contravariant vector, the quotient theorem implies that the $$\frac{\partial f}{\partial x^{i}}$$ are the components of a covariant vector. Tranforming to different coordinate systems is then trivial. E.g. to transform to spherical coordinates, you can take the expresson for df and take dx^1 = dr, dx^2 = dtheta and dx^3 = dphi and then you re-write that as an inner product with the vector (dr, r dtheta, r sin(theta), dphi) from which you read-off the components of the gradient in spherical coordinates. So, this is a one line derivation, to be contrasted to the two page derivation given in many math textbooks. Count Iblis (talk) 15:31, 19 March 2011 (UTC)