mm-320 === Subject: : Re: Stats Question > I have two groups (control and experiment). I want to give them a pre-test, > perform the experiment, then give them a post-test. What stats test would I > use to compare the increase of scores? At first i thought it would be a > two-tailed t-test, but I'm not 100% sure. > You might be right but it is impossible to say from the information > given. If: > 1 the samples are not too small > 2 the observations have an approximately normal distribution > 3 it is possible that either group would have a larger increase > then you are right. Otherwise you need to consider a non-parametric test Non-parametric? What would this be - a chi-square test? === Subject: : Re: Stats Question >I have two groups (control and experiment). I want to give them a > pre-test, >perform the experiment, then give them a post-test. What stats test > would I >use to compare the increase of scores? At first i thought it would be a >two-tailed t-test, but I'm not 100% sure. >>You might be right but it is impossible to say from the information >>given. If: >>1 the samples are not too small >>2 the observations have an approximately normal distribution >>3 it is possible that either group would have a larger increase >>then you are right. Otherwise you need to consider a non-parametric test > Non-parametric? What would this be - a chi-square test? Wilcoxon rank sum === Subject: : algebra homomorphisms When does a k-algebra homomorphism give rise to a ring homomorphism. For example I am trying to classify when a matrix M in Mat_n(k) is semisimple. I get an isomorphism k[M] = k[x]/(g) where g is the minimum polynomial of M. Now, the k[x]-module isomorphism is also a ring isomorphism, then semisimplicity follows iff g has distinct irred. factors. When else does this hold? === Subject: : Re: LatLong inside polygon The OP's problem was: > Here is the problem: > I have a LatLong pair: upper-left corner and lower-right. > The upper-right and lower-left can of course be extracted > from these. Each side in the square can be several hundreds of kilometers apart. > Now, how do I tell if a LatLong is inside this square? I find this > rather > tricky, especially if one of the poles is inside the square. The problem does need clarification, and given the four corners, ABCD on the Terrestrial Sphere, I would define the boundary of your region to be made from Great Circles joining those corners. There is no difficulty in this as long as no two points are at opposite ends of a diameter. By using Great circles rather than lines of latitude or longitude you ensure that there are no special difficulties with the poles; Great circles are entirely symmetrical with respect to any direction whatever. My assumption that we are talking about the terrestrial sphere, rather than a more accurate Ellipsoid of revolution (= Globe with flatenned poles) is not absolutely vital, by the way. What I would do is turn your lat/long pairs + Earth's radius into xyz Cartesian Coordinates. You can then use cross-products to define a vector normal to each great circle, giving you four unit vectors, each facing towards the interior of your region. You will have to take care with the order of the points to achieve this. Next run through your list of candidate points, taking dot products against each unit vector. A positive answer means it is within the correct half-space of that Great circle, and four positive answers mean it is inside the whole region. As a quick check, choose a kind of mid-point of the four corners, as follows: P is the midpoint of diagonal AC Q is the midpoint of diagonal BD M (our midpoint) is the mid-point of PQ Had ABCD been in a plane, we would have P=Q=M, and if ABCD is not __too__ big compared with the radius of the Earth, or if they are __nearly__ on a small circle, they won't be that far apart. - - Ken, __O -<,_ (_)/ (_) Virtuale Saluton. === Subject: : Re: matrices | |> Let A and B be two n times n matrices over a field k. Assume that for |> all x,y in k, det(xA+yB)=0. Does it follow that there is a non-zero |> element v in k^n,such that Av = Bv=0? | |Let's suppose k is infinite (I imagine things will break down |for finite fields). Then det(A + tB) = 0 as a matrix over k(t). |Thus there is a nonzero vector over k(t) killed by A + tB and we |may assume that its entries are in k[t]. Call this vector |v = v_0 + t v_1 + ... t^m v_m. Then (A + tB) v = 0, |so A v_0 = 0, A v_1 = - B v0, ...., A v_m = - B v_{m-1} and 0 = B v_m. |This gives one an idea to find a counterexample. Let u, u' and u'' |be linearly independent vectors in some vector space. Define a |transformation on this space by setting A u = 0, A u' = u'', |B u = - u'' and B u' = 0 (and A u'' = anything etc.). Then |(A + tB)(u + t u') = 0 (with t an indeterminate). As far as I can |see one can cook up A and B so that they only annihlate |multiples of u and u' etc. But A + tB is singular. am i making a silly mistake? what's the matter with a := 1 0 0 0 and b := 0 1 0 0 over any field? (if there's a convention that makes the example work then use it.) -- [e-mail address jdolan@math.ucr.edu] === Subject: : Re: skewed dualities: functions and sets > | the adjoint functor theorem of category theory explains this, in a > | way. it says that left adjoint functors preserve colimits (such as > | unions) and, dually, right adjoint functors preserve limits (such as > | intersections). a functor is a left adjoint functor if it has a right > | adjoint functor, and vice versa. inverse image is both a left > | adjoint and a right adjoint functor. the left adjoint that it has (by > | virtue of being a right adjoint) is called image, while the right > | adjoint that it has (by virtue of being a left adjoint) is probably > | not well enough known even to have a familiar name; i'll call it > | anti-image for now. the anti-image of a subset s of the domain of a > | function is the subset of the co-domain containing just those points > | whose inverse images are subsets of s. anti-image preserves > | intersections but not unions. > | > |zzzzzzzzzzzzz ... > as i said, you probably have to learn a fair amount of category theory > to appreciate this, so you're only demonstrating that you're an idiot. No, I'm only demonstrating that I recognize a preposterous gasbag when I see one. === Subject: : Re: skewed dualities: functions and sets >> | the adjoint functor theorem of category theory explains this, in a >> | way. it says that left adjoint functors preserve colimits (such as >> | unions) and, dually, right adjoint functors preserve limits (such as >> | intersections). a functor is a left adjoint functor if it has a right >> | adjoint functor, and vice versa. inverse image is both a left >> | adjoint and a right adjoint functor. the left adjoint that it has (by >> | virtue of being a right adjoint) is called image, while the right >> | adjoint that it has (by virtue of being a left adjoint) is probably >> | not well enough known even to have a familiar name; i'll call it >> | anti-image for now. the anti-image of a subset s of the domain of a >> | function is the subset of the co-domain containing just those points >> | whose inverse images are subsets of s. anti-image preserves >> | intersections but not unions. >> | >> |zzzzzzzzzzzzz ... > as i said, you probably have to learn a fair amount of category theory >> to appreciate this, so you're only demonstrating that you're an idiot. >No, I'm only demonstrating that I recognize a preposterous gasbag when I >see one. Now is that fair, calling him a preposterous gasbag just because he used theorems from category theory to explain why f^{-1} does not commute with intersections? Hmm, maybe it is. ************************ === Subject: : Re: Copyrights, fair use, and Internet realities >I post, even though I'm a famous mathematician. I checked -- Anonymous is the author of 17 items in MathSciNet. (Most of them biographies or other historical surveys.) dave === Subject: : Re: Copyrights, fair use, and Internet realities Discussion, linux) >>I post, even though I'm a famous mathematician. > I checked -- Anonymous is the author of 17 items in MathSciNet. > (Most of them biographies or other historical surveys.) Yes, but he's most prodigious in peer reviews. -- All intelligent men are cowards. The Chinese are the world's worst fighters because they are an intelligent race[...] An average Chinese child knows what the European gray-haired statesmen do not know, that by fighting one gets killed or maimed. -- Lin Yutang === Subject: : Plotting f(x,y)=0 Do you know a freeware to plot a function given only as (a polynome) f(x,y)=0? The otherwise fine MathGV can't. I helped myself by running a double loop on x and y, print the values where |f| Do you know a freeware to plot a function given only > as (a polynome) f(x,y)=0? The otherwise fine MathGV > can't. I helped myself by running a double loop > on x and y, print the values where |f| and feeding it into a statistic software to print > as scatterplot. Evidently this runs into different > convergence speeds at different points and won't > look very good. > (Shareware will do either :-) Surf worked well when I last used it and it's even open source: === Subject: : Convergence Hello. I am reading a survey of the theory of counting lattice points in polyhedra. The authors introduce a so called generating function in d complex variables in the the following way: If P is a polyhedron in R^d, let f(x_1,...,x_d) = sum_{m in P integral} x_1^m_1*...*x_d^m_d, where m=(m_1,...,m_d), and where integral means that all entries in m are integral. The authors claim that this series converges absolutely for any x. But what does this mean? I know what is meant by convergence of series in one complex variable. How does this extend to more variables? Finally: What if P=[0,oo) in R. Then f(x)=sum_{i=0}^oo x = 1/(x-1) if |x|<1 but it does not converge for |x|>=1. I must have missed a point. Can anybody help me? For the exact formulation one may check the survey online at http://www.math.lsa.umich.edu/~barvinok/functions.ps (The problem arises in Theorem 3.1 in this paper). -- Michael Knudsen === Subject: : Re: Cantor's Diagonal Argument : > In sci.logic, |-|erc > : > Ouch!! Doesn't this mean b = c(j)? and is it correctly derived? > for all n, exists j, b_1 = c(j)_1, b_2 = c(j)_2, ... b_n = c(j)_n This is equivalent to : for all n, b_n = c(j)_n >> You have to prove that there is such a j. > note the exists j > for all n, exists j, b_1 = c(j)_1, b_2 = c(j)_2, ... b_n = c(j)_n > admitedly this is different to : > exists j, for all n, b_n = c(j)_n > I showed that no matter how many digits you examine, the diag > up to that many digits is covered on the list. > Herc > You have to show that > for all n, there exists a j, such that for all i, > b_n_i = c(j)_i Yes, that's what he would need to show. It might be helpful though to express what he has shown in terms of what sort of nunmbers are involved, because it's often easier to see things expressed in simple categories instead of in something like predicate calculus. And to show how the same thing can be proved about simpler lists and rather more obvious non-members. So what does for all n, exists j, b_i = c)j)_i for i in 1....n mean. It means If x is the number represented by b_1.....b_n, then some number whose expression has those n digits as the first n is on the list. If the list includes all numbers whose expansion terminates after a finite number of terms this is trivially true, since x itself is on the list. Now a list of all rational numbers certainly includes all numbers with finite expansions, so it's clear that if the list is the list of all rational numbers and b (represented as b_1....) is any real number then for any n we can find the number b_1....b_n on the list. Think about the well known fact, established by the ancient Greeks, that there is no rational number whose square is 2. Mow let C(j) be the complete set of rationals between 0 and 1 indexed by natural numbers j, and let b be the real number such that (b+1) is the positive square root of 2. It's clear that for all n there exists a j such that c(j)_i = b_i for i in 1...n, but b is not any of the c(j). So if one believes that proving this theorem about some countable list C of numbers and some number b suggests that b is in the list C, one has to think that the square root of two is rational (indeed one has to think that all real numbers are rational, since the abilty to have a (not necessarily terminating) expansion in any base from 2 upwards is a property of all real numbers. M. [my real email address has no no in it] === Subject: : 3-D exact solution of the Laplace equation in an irregular domain Hello All: I am looking for an exact solution to the Laplace equation in three dimensions for an irregular and finite domain. For 2-D situations, the use of conformal mapping techniques allows one to find an exact solution in irregular and finite domains. For instance, the solution to the Laplace equation between two eccentric cylinders is well known and can be found,for instance, in Carlslaw and Jaeger's book on Conduction of Heat in Solids. However, I have not been successful in finding solutions in 3-D cases of the Laplace Equation. For regular domains such as a cubical domain, that is not a problem. A separation of variables technique would do the trick. But, irregular domains are what I am interested in. Any help would be appreciated. Andrei === Subject: : abstract algebra, normality Theorem: if N is normal in G, and H is a subgroup of G then H/N is normal in G/N if and only if H is normal in G Proof: (<=) easy. (=>) i have that (xN)(hN)(xN)^-1 = (xhx^-1) N = h'N for some h' in H. so xhx^-1 = h'n for some n in N. now what? we need to get h'n in H, but how is this necessarily the case? N is normal, but what does that mean? h'n = n'h' for some n' in N, but so what? please give me a hint on how to finish this up... thanks. === Subject: : Re: abstract algebra, normality If H contains N, then you've got the other direction. My comments are below. >Theorem: >if N is normal in G, and H is a subgroup of G then >H/N is normal in G/N if and only if H is normal in G >Proof: >(<=) easy. >(=>) i have that (xN)(hN)(xN)^-1 = (xhx^-1) N = h'N for some h' in H. >so xhx^-1 = h'n for some n in N. >now what? >we need to get h'n in H, but how is this necessarily the case? N is Is it part of the assumption that N is a subset of H? If so, once you've shown that xhx^-1 = h' N, then it is in H as well (since N is contained in H). >normal, but what does that mean? h'n = n'h' for some n' in N, but so >what? >please give me a hint on how to finish this up... >thanks. Hope this helps, Brian === Subject: : Re: abstract algebra, normality Adjunct Assistant Professor at the University of Montana. >Theorem: >if N is normal in G, and H is a subgroup of G then >H/N is normal in G/N if and only if H is normal in G You are assuming N is contained in H, surely... >Proof: >(<=) easy. >(=>) i have that (xN)(hN)(xN)^-1 = (xhx^-1) N = h'N for some h' in H. >so xhx^-1 = h'n for some n in N. >now what? >we need to get h'n in H, but how is this necessarily the case? N is >normal, but what does that mean? h'n = n'h' for some n' in N, but so >what? For H/N to make sense, N must be contained in... -- ============================================================== ======== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ============================================================== ======== Arturo Magidin magidin@math.berkeley.edu === Subject: : Radical division Hello everyone, while working on the limits of a function in my Calculus book, I came across the division of a radical: (2x^2 + 1)^1/2 / x The book indicates that the division yields: (2 + 1/x^2)^1/2 Can anybody give me a hint or briefly explain what operatios took place to obtain such result? Best regards, Ivan === Subject: : Re: Radical division They used (x^2)^(1/2) = x (true for nonnegative values of x). (2x^2 + 1)^1/2 / x = (2x^2 + 1)^(1/2) * (x^-2)^(1/2) = (2x^2 * x^-2 + 1 * x^-2)^(1/2) = ( 2 + 1/(x^2) )^(1/2) Hope this helps, Brian >Hello everyone, while working on the limits of a function in my >Calculus book, I came across the division of a radical: >(2x^2 + 1)^1/2 / x >The book indicates that the division yields: (2 + 1/x^2)^1/2 >Can anybody give me a hint or briefly explain what operatios took >place to obtain such result? >Best regards, >Ivan === Subject: : Re: Radical division > Hello everyone, while working on the limits of a function in my > Calculus book, I came across the division of a radical: > (2x^2 + 1)^1/2 / x > The book indicates that the division yields: (2 + 1/x^2)^1/2 > Can anybody give me a hint In the denominator: x = (x^2)^(1/2) -- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.html r c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: : Re: Radical division > In the denominator: > x = (x^2)^(1/2) This assumes x is positive... I wonder if the original problem mentioned that. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: : Re: Cantor Paradox > Cantor's diagonal construction is a challenge. > He says to all the world, in effect, give me any list of reals, > numbered by the naturals, and I will show that there is a real number > not in the list. Yes. Nathan has has given a list of reals, numbered by the naturals. Cantor's argument shows there is a real not in the list. > In fact, a minor variation on his construction will easily find > countably many numbers not in the list, so that there are at least as > many numbers not in the list as in the list. > Until you produce such a list, you have not challenged him. I am not the one claiming the definable numbers are countable. Others have claimed that Cantor's number is not definable. I think these people are shooting themselves in the foot. If Cantor's missing number is undefinable in this situation, how can it be definable in any situation? Russell - 2 many 2 count === Subject: : Re: Cantor Paradox > Wow, sounds like Nathan has rediscovered Richard's Paradox (1905). Not > bad for an 11-year-old, if he really did think of himself. > He's not an 11-year-old. Whether he's ever read Richard's Paradox or > not, he clearly has the background to choose the issues carefully > here. > (I hadn't seen the paradox before, so I appreciate having a name > attached to it.) I think he just turned 13 at the end of last month... === Subject: : Beall Conjecture On June 11, 2000, I posted a brief solution to the Beall conjecture. There is a good bit of money riding on the proof. Perhaps my post merits a bit more attention === Subject: : Re: Beall Conjecture X-ID: S3AWJ6ZbYescX3dct9niUT8q2ko87Y53yI1taRak5ruZCMEi1-iHEQ jrtyler27@yahoo.com (John Tyler) schrieb: >On June 11, 2000, I posted a brief solution to the Beall conjecture. >There is a good bit of money riding on the proof. Perhaps my post >merits a bit more attention But you have to spell the name correctly - or there won't be a dime! Thomas === Subject: : Re: Beall Conjecture >On June 11, 2000, I posted a brief solution to the Beall conjecture. >There is a good bit of money riding on the proof. Perhaps my post >merits a bit more attention On the other hand, perhaps it doesn't. It's so hard to know about this kind of thing. Lee Rudolph === Subject: : Games/Puzzles With Advanced Mathematics? I am just curious, what are some good examples of puzzles and games which use *advanced* mathematics? What advanced mathematics means exactly here is subjective. I am more interested in examples of games and puzzles for the general audience. I am aware that some advanced math has been used to analyze some famous games, for example. (Although precise examples slip my mind now.) I wonder specifically here about easy-to-understand puzzles/games where the trick to solving the puzzle or successfully playing the game uses some math way beyond most people, and somewhat (at least) beyond most mathematicians. Leroy Quet === Subject: : Re: Games/Puzzles With Advanced Mathematics? > I am just curious, > what are some good examples of puzzles and games which use *advanced* > mathematics? I think there was a tiling game with Penrose tiles. -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: : Re: Games/Puzzles With Advanced Mathematics? Mime-version: 1.0 Content-type: text/plain; charset=US-ASCII Content-transfer-encoding: 7bit > I am just curious, what are some good examples of puzzles and games which use > *advanced* mathematics? > What advanced mathematics means exactly here is subjective. > I am more interested in examples of games and puzzles for the general > audience. > I am aware that some advanced math has been used to analyze some famous games, > for example. (Although precise examples slip my mind now.) > I wonder specifically here about easy-to-understand puzzles/games where the trick to solving the puzzle or successfully playing the game uses some math > way beyond most people, and somewhat (at least) beyond most mathematicians. Rubiks' Cube; group theory. Bob H === Subject: : Re: Help showing system is not Hamiltonian. > I could seriously use some hints on where I am going wrong with the > following. > I have a system of non-linear differential equations: > dx/dt = X(x,y) = (a - by)(x - x^2) > dy/dt = Y(x,y) = - (c - dx)(y - y^2) > where a, b, c, d are positive constants and 0 < x, y < 1. > I need to show that this system is *not* Hamiltonian unless a = c = 1 > and b = d = 2; > To show that a system *is* Hamiltonian, I believe it is necessary and > sufficient to show that > partial_x X + partial_y Y = 0, for all values of 0< x, y < 1 > Now, in this case we have > partial_x X = a - 2ax - by + 2bxy > partial_y Y = -c + 2cy + dx - 2dxy > Letting a = c = 1, b = d = 2 we see that, indeed, partial_x X + > partial_y Y = 0 and the system is thus Hamiltonian. > BUT it seems to me that for all values of a, if a = c, b = d = 2a the > system is also Hamiltonian. Specifically if a = c = 2, b = d = 2a = 4 we > still have partial_x X + partial_y Y = 0 and the system is thus > Hamiltonian. > I believe that the system is *not* Hamiltonian for any other values of > a, b, c & d but this still seems to contradict what I have to show. > A hint on what am I missing would be greatly appreciated! Assuming your definition is correct: (a - by)(1 - 2x) - (c - dx)(1 - 2y) = 0 a(1 - 2x) - by(1 - 2x) - (c - dx) + 2y(c - dx) = 0 a(1 - 2x) - (c - dx) - by(1 - 2x) + 2y(c - dx) = 0 a(1 - 2x) - (c - dx) + y(-b(1 - 2x) + 2(c - dx)) = 0 Because y is arbitrary (within its prescribed interval), we find that: a(1 - 2x) - (c - dx) = 0 -b(1 - 2x) + 2(c - dx) = 0 a(1 - 2x) - (c - dx) = 0 a - 2ax - c + dx = 0 a - c + (-2a + d)x = 0 Since x is also arbitrary in its interval, we have: a - c = 0, or a = c. -2a + d = 0, or d = 2a -b(1 - 2x) + 2(c - dx) = 0 -b(1 - 2x) + 2(a - 2ax) = 0 -b(1 - 2x) + 2a(1 - 2x) = 0 (2a-b)(1 - 2x) = 0 Since the second term is arbitrary, we find that: 2a = b Thus, 2a = b = 2c = d > 0 is the full solution. A quick google search did not seem to say much about a differential equation being Hamiltonian but rather discussed the Hamiltonian operator and related equations. > BraileTrail > -- === Subject: : Re: Cantor Paradox :-) > Otherwise you can take the set {x in R|there's P so that |-E!x Px}, but this wouldn't garantee that the sought list exists because the set > {(P, x)| E!yPy is provable and Px} isn't necessarily a function. This can be used to show Godel's second incompleteness theorem for ZF. Supose we can prove ZF to be consistent. I claim we can prove that the mentioned above set is a function, which by the diagonal arguent leads to contradiction. Suppose on the contrary that there are P and x != y so that Px, Py hold and (E!zPz) holds. Since we proved ZF is consistent this leads to a contradiction. Posted Via Usenet.com Premium Usenet Newsgroup Services ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: : Re: Cantor Paradox :-) > Enumerate the set S = { P | |= E!x Px }. The existence of the set S > and its enumeration does not depend on whether the set S is recursive. The set you're looking for is S = {x in R| there's P so that E!x Px} which is a set if the truth value of E!x P(x) is recursive in P. It isn't. The truth of a proposition doesn't depend on its form in any way expressible by a formula. Otherwise you can take the set {x in R|there's P so that |-E!x Px}, but this wouldn't garantee that the sought list exists because the set {(P, x)| E!xPx is provable} isn't necessarily a function. Posted Via Usenet.com Premium Usenet Newsgroup Services ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: : Re: Cantor Paradox :-) Discussion, linux) >> Enumerate the set S = { P | |= E!x Px }. The existence of the set > S >> and its enumeration does not depend on whether the set S is > recursive. The set you're looking for is S = {x in R| there's P so that E!x Px} > which is a set if the truth value of E!x P(x) is recursive in P. It's also a set if E!x P(x) is not recursive. Honest. Check the axioms of ZFC. See whether or not the separation axiom mentions that the formula must be recursive. We're doing plain old mathematics in this discussion, not computability theory. -- Jesse F. Hughes That's the base tautological space where by tautological space I mean a region of truth. -- James S. Harris does philosophy of mathematics. JSH is a renaissance man. === Subject: : Re: Cantor Paradox :-) <87ishg5xks.fsf@phiwumbda.org> <844950ljl35rem8ko6jg0aqskgd5sqnpg5@4ax.com> Discussion, linux) > Then in the course of a long thread it was determined that > the problem was with my newsreader (Agent) parsing the > after a little investigation one of the guys there has > determined that it's not a problem with Agent, it's a > problem with the news server I'm using! > (How could that be? It seems that Agent doesn't thread > things based on looking at the actual headers, because it > doesn't have them yet at the time it does the threading. > It does the threading on the basis of XOVER information > (an optional extension), and giganews is returning > Not that there's any reason anyone should care, but > since I've announced I'd found a bug in Agent I thought > I should clarify that I really hadn't. Well, I care. Honestly, I care ever so much. Actually, that does make a lot more sense. As you say, Agent is a popular client. It would be shocking if it had such an obvious flaw. It's somewhat less shocking if your newsserver isn't serving the right information. -- You lack the ability to reason, but instead get an idea in your head and hold on to it against all evidence. I don't find you credible, and reject your claims, as coming from a flawed source. === Subject: : Re: Cantor Paradox :-) >>[...] >> Not that there's any reason anyone should care, but >> since I've announced I'd found a bug in Agent I thought >> I should clarify that I really hadn't. >Well, I care. Honestly, I care ever so much. >Actually, that does make a lot more sense. As you say, Agent is a >popular client. It would be shocking if it had such an obvious flaw. Would be surprising but not impossible, because most software I assumed for a long time that the problems with posts from a few outliers like you must be at your end. >It's somewhat less shocking if your newsserver isn't serving the right >information. I got a free upgrade for not finding a bug! (Turns out that that particular news server has been telling people it's Agent's fault, and the Agent people are happy with me for allowing them to prove conclusively that Agent is not the problem. Not that I see what I did that they couldn't have done, but what the heck, from now on I'm going to search hard for non-bugs in all the programs I use.) ************************ === Subject: : Re: Packing squares into a 1 x Zeta[2] rectangle > Since the sum of the reciprocals of the squares of the positive integers is > pi^2/6, the question arises as to whether squares with sides 1, 1/2, 1/3, > etc can be packed into a rectangle of size 1 by pi^2/6. A picture of such a packing appears at > http://www.pisquaredoversix.force9.co.uk/Tiling.htm > Nice! > Why are 8 and 10 where they are? > I'd have thought that 8 would slot ringht next to 7 rather than 6? > Similarly, the 10 should fit next to 9? You are quite right. The picture at the above link is one I made about 5 years ago using an algorithm that I no longer remember. I have not made any pictures yet of tilings produced by the current algorithm. -- Clive Tooth http://www.clivetooth.dk === Subject: : Re: Packing squares into a 1 x Zeta[2] rectangle > Since the sum of the reciprocals of the squares of the positive integers is > pi^2/6, the question arises as to whether squares with sides 1, 1/2, 1/3, > etc can be packed into a rectangle of size 1 by pi^2/6. > A picture of such a packing appears at > http://www.pisquaredoversix.force9.co.uk/Tiling.htm > I know of no proof that such a packing is possible or impossible. > uses exact arithmetic throughout. The algorithm is simple: >... I wonder if such a packing has been shown to exist for a slightly relaxed problem, such as for a rectangle of 1 by (pi^2/6 + epsilon). As for the picture, it reminds me visually (if not mathematically) of the graphical representation of the continued fraction of pi^2/6, where each term is a real itself represented by a continued fraction of real-terms, ad infinitum. Does anyone know about the more general problem of packing n-cubes of side-lengths 1, 1/2, 1/3,... into an n-box of n-volume zeta(n)? Leroy Quet === Subject: : Re: Packing squares into a 1 x Zeta[2] rectangle > Since the sum of the reciprocals of the squares of the positive integers is > pi^2/6, the question arises as to whether squares with sides 1, 1/2, 1/3, > etc can be packed into a rectangle of size 1 by pi^2/6. > A picture of such a packing appears at > http://www.pisquaredoversix.force9.co.uk/Tiling.htm > Nice! > I know of no proof that such a packing is possible or impossible. > uses exact arithmetic throughout. The algorithm is simple: > The program maintains a list of rectangles. Each member of this list is > itself a list containing exactly two members: the short side of a rectangle > followed by the long side of the rectangle. The program does not store the > position or orientation of any rectangle. The list is stored in ascending > short side order. Initially the list contains one entry: {1, pi^2/6}. > The program inserts the squares in decreasing size order, starting with the > 1x1 square. At each stage, to insert the square with side 1/n, the program > finds the rectangle with the shortest short side which will accommodate the > square. [In the case that there are two or more such rectangles with equal > short sides, the program will pick the youngest one.] > Suppose that this rectangle is a x b, where 1/n<=a<=b. The program deletes > this rectangle from the list and inserts two new rectangles (b-1/n) x a and > (a-1/n) x 1/n into the list. > Is there any special handling of the rational vs. transcendental edges? No. Mathematica handles all that. I simply tell it that the sides of the first rectangle are 1 and Zeta[2] and it knows that Zeta[2] is pi^2/6. > I'd have thought that performance-wise, trying to fit things in rational- > edged rectanges would be quicker, as you don't have to hit arbitrary > precision decimals, only integers which have no precision issues. > What kind of time are we talking about for the program -- > minutes, hours or days? About 11 hours for the N=1,000,000 run. > - If any side of one of these new rectangles is zero, the rectangle is not > inserted. > - Each new rectangle is stored in {short side, long side} form. > Thus, the list of rectangles grows by one per inserted square, except in the > rare cases in which a rectangle of exactly the right width is found. > A trivial optimisation, which took me a long time to notice, is that if the > number of squares to be inserted is decided at the beginning, as N, say, > then any rectangles narrower than 1/N can be discarded as soon as they > appear. This has a dramatic effect on the memory requirements and running > time of the program. In the case of N=10^6 the maximum length of the > rectangle list is only 7,493 which occurs just after inserting the square > with side 1/55,205. Thereafter the length of the list slowly decreases. > After inserting the last square, with side 10^-6, there are just 7 > rectangles in the list. Their dimensions are, roughly: > 0.000001000264 x 0.000371508565 [1,000,000] > 0.000001000289 x 0.000006561855 [ 999,736] > 0.000001000332 x 0.000011950546 [ 999,710] > 0.000001000424 x 0.000099911008 [ 999,668] > 0.000001001705 x 0.000013366392 [ 999,575] > 0.000001001771 x 0.000555285230 [ 998,297] > 0.000621399972 x 0.000621402936 [ 998,232] > The number in square brackets is the square insertion step on which the > rectangle was created. > Note that each rectangle, except the last, has a short side just exceeding > 10^-6. The last one is huge in comparison to the others. Also, it is almost > square, having an aspect ratio of about 1.00000477 . > Fascinating. How many exact fits did you manage during the run? I am not sure. But here is a list of all the exact fits [side of square=short side of a rectangle] for n<10000, together with the factorisation of n: 1/1, exact, factor[1]={} 1/6, exact, factor[6]={{2,1},{3,1}} 1/20, exact, factor[20]={{2,2},{5,1}} 1/36, exact, factor[36]={{2,2},{3,2}} 1/180, exact, factor[180]={{2,2},{3,2},{5,1}} 1/182, exact, factor[182]={{2,1},{7,1},{13,1}} 1/220, exact, factor[220]={{2,2},{5,1},{11,1}} 1/420, exact, factor[420]={{2,2},{3,1},{5,1},{7,1}} 1/468, exact, factor[468]={{2,2},{3,2},{13,1}} 1/560, exact, factor[560]={{2,4},{5,1},{7,1}} 1/588, exact, factor[588]={{2,2},{3,1},{7,2}} 1/1056, exact, factor[1056]={{2,5},{3,1},{11,1}} 1/1170, exact, factor[1170]={{2,1},{3,2},{5,1},{13,1}} 1/1190, exact, factor[1190]={{2,1},{5,1},{7,1},{17,1}} 1/1302, exact, factor[1302]={{2,1},{3,1},{7,1},{31,1}} 1/1320, exact, factor[1320]={{2,3},{3,1},{5,1},{11,1}} 1/1380, exact, factor[1380]={{2,2},{3,1},{5,1},{23,1}} 1/1476, exact, factor[1476]={{2,2},{3,2},{41,1}} 1/1640, exact, factor[1640]={{2,3},{5,1},{41,1}} 1/1664, exact, factor[1664]={{2,7},{13,1}} 1/1806, exact, factor[1806]={{2,1},{3,1},{7,1},{43,1}} 1/1989, exact, factor[1989]={{3,2},{13,1},{17,1}} 1/2100, exact, factor[2100]={{2,2},{3,1},{5,2},{7,1}} 1/2106, exact, factor[2106]={{2,1},{3,4},{13,1}} 1/2160, exact, factor[2160]={{2,4},{3,3},{5,1}} 1/2184, exact, factor[2184]={{2,3},{3,1},{7,1},{13,1}} 1/2340, exact, factor[2340]={{2,2},{3,2},{5,1},{13,1}} 1/2548, exact, factor[2548]={{2,2},{7,2},{13,1}} 1/2600, exact, factor[2600]={{2,3},{5,2},{13,1}} 1/2640, exact, factor[2640]={{2,4},{3,1},{5,1},{11,1}} 1/2730, exact, factor[2730]={{2,1},{3,1},{5,1},{7,1},{13,1}} 1/2860, exact, factor[2860]={{2,2},{5,1},{11,1},{13,1}} 1/3120, exact, factor[3120]={{2,4},{3,1},{5,1},{13,1}} 1/3220, exact, factor[3220]={{2,2},{5,1},{7,1},{23,1}} 1/3245, exact, factor[3245]={{5,1},{11,1},{59,1}} 1/3480, exact, factor[3480]={{2,3},{3,1},{5,1},{29,1}} 1/3612, exact, factor[3612]={{2,2},{3,1},{7,1},{43,1}} 1/3843, exact, factor[3843]={{3,2},{7,1},{61,1}} 1/4060, exact, factor[4060]={{2,2},{5,1},{7,1},{29,1}} 1/4224, exact, factor[4224]={{2,7},{3,1},{11,1}} 1/4340, exact, factor[4340]={{2,2},{5,1},{7,1},{31,1}} 1/4620, exact, factor[4620]={{2,2},{3,1},{5,1},{7,1},{11,1}} 1/4914, exact, factor[4914]={{2,1},{3,3},{7,1},{13,1}} 1/5040, exact, factor[5040]={{2,4},{3,2},{5,1},{7,1}} 1/5060, exact, factor[5060]={{2,2},{5,1},{11,1},{23,1}} 1/5100, exact, factor[5100]={{2,2},{3,1},{5,2},{17,1}} 1/5475, exact, factor[5475]={{3,1},{5,2},{73,1}} 1/5544, exact, factor[5544]={{2,3},{3,2},{7,1},{11,1}} 1/6162, exact, factor[6162]={{2,1},{3,1},{13,1},{79,1}} 1/6300, exact, factor[6300]={{2,2},{3,2},{5,2},{7,1}} 1/6844, exact, factor[6844]={{2,2},{29,1},{59,1}} 1/6970, exact, factor[6970]={{2,1},{5,1},{17,1},{41,1}} 1/7140, exact, factor[7140]={{2,2},{3,1},{5,1},{7,1},{17,1}} 1/7564, exact, factor[7564]={{2,2},{31,1},{61,1}} 1/8320, exact, factor[8320]={{2,7},{5,1},{13,1}} 1/8424, exact, factor[8424]={{2,3},{3,4},{13,1}} 1/8463, exact, factor[8463]={{3,1},{7,1},{13,1},{31,1}} 1/9030, exact, factor[9030]={{2,1},{3,1},{5,1},{7,1},{43,1}} 1/9940, exact, factor[9940]={{2,2},{5,1},{7,1},{71,1}} > Your best-fit technique invites incredibly small slivers, and thus > discardings. Did you think about other less greedy allocation > strategies? I have not experimented much in that direction. Choosing the _longest_ short side every time causes the program to fail to find a fit on inserting 1/95. -- Clive Tooth http://www.clivetooth.dk === Subject: : Sum of Product = Product of Sum (number-theoretical) I will post here a generalization of the result in Sum Of Powers Of Divisor-Function. (http://groups.google.com/groups?dq=&start=50&hl=en&lr=&ie=UTF -8&group=sci.m ath&safe=off&selm=b4be2fdf.0403131607.46772492% 40posting.google.com) Let a(p,k) be a nonnegative integer such that p^a(p,k) is the highest power of the prime p which divides k. Let {b(k)} be any sequence defined for all positive integers k. Then, for every positive integer m: --- --- | | / ( | | b(a(p,k)) ) --- p|k k|m = a(p,m) ---- -- ! | ! | (1 + / b(k) ) p|m -- k=1 (All p-products in this post are over primes p which divide k or m, depending.) In linear-mode: sum{k|m} (product{p|k} b(a(p,k)) ) = product{p|m} (1 +sum{k=1 to a(p,m)} b(k) ). And from this we can get, for any positive integer n: --- --- --- --- | | / / ... / ( | | b(a(p,k_n)) ) --- --- --- p|k_n k_1|m k_2|k_1 k_n|k_{n-1} = a(p,m) k_1 k_{n-1} ---- -- -- -- ! | /a(p,m)+n-1 ! | (| | + / / ... / b(k_n) ) p|m n - 1 / -- -- -- k_1=1 k_2=1 k_n=1 In linear-mode: sum{k_1|m}sum{k_2|k_1}... sum{k_n|k_{n-1}} (product{p|k_n} b(a(p,k_n)) ) = product{p|m} (binomial(a(p,m)+n-1,n-1) + +sum{k_1=1 to a(p,m)}sum{k_2=1 to k_1}... sum{k_n=1 to k_{n-1}} b(k_n) ) For n = 2 we also have simply --- --- | | / d(m/k) ( | | b(a(p,k)) ) --- p|k k|m = a(p,m) k_1 ---- -- -- ! | ! | (a(p,m) + 1 + / / b(k_2) ) p|m -- -- k_1=1 k_2=1 d() is the number-of-divisors function. In linear-mode: sum{k|m} d(m/k) (product{p|k} b(a(p,k)) ) = product{p|m} (a(p,m) +1 + +sum{k_1=1 to a(p,m)}sum{k_2=1 to k_1} b(k_2) ) So, an interesting specific case of the above: sum{k|m} d(m/k) /d(k) = product{p|m} ((a(p,m) + 2)(H(a(p,m)+2) -1)), where H(j) = sum{k=1 to j} 1/k, the j_th harmonic number. A specific example of first identity: sum{k|m} product{p|k} F(a(p,k)) = product{p|m} F(a(p,m)+2), where F(j) is the j_th Fibonacci number. Leroy Quet === Subject: : Re: Why are there 63360 inches in a mile? > === Subject: : Re: Why are there 63360 inches in a mile? >Message-id: Gallon? I thought you folks used cubic inches? (If _we_ had a gallon engine, >it >>would have been 4.5 litres >Not everyone realizes that some of these traditional units were already >decimal based. Most try to reconcile a base two system of half >and double when defining their sub-units, and this spoils their >simplicity. >But in this case: >One cubic foot is six and a quarter gallons That's not what the http://www.onlineconversion.com/volume.htm says: 1 cubic foot = 6.2288355 gallon [UK] >One gallon is eight pints >One pint is 20 fluid oz >-------------------------- >Hence One cubic foot is: > 6.25 x 8 x 20 oz >= 1000 oz 1 cubic foot = 996.6136777 ounce [UK, liquid] > ... which looks pretty decimal to me >1 fluid oz is (0.1 ft)^3 >The conversion factor 'six and a quarter' comes up >elsewhere. Don't dismiss it as being silly and >arbitrary. Ok, it's not silly and arvitrary, it's wrong. >- - >Ken, __O > -<,_ > (_)/ (_) >Virtuale Saluton. -- Mensanator Ace of Clubs === Subject: : Connected Graphs Question, Part II Originator: fab@soda.csua.berkeley.edu (Fabio Rojas) A few days ago, I asked about the percentage of all graphs that are connected. Klaus hoffman directed me to the onlince encyclopedia of integere sequences. So I calculate the ratio and I find that up till N=20, about 99% of graphs are connected. I calculated the ratio of sequences A001187 vs. A006125 and the ratio of A001349 vs. A000088. Have I screwed up? Small graphs should be connected but graphs with 20 nodes? It's seems that as N gets bigger, connectedness should be an extremely rare property. At some point, singletons, dyads and other little groups should start breaking off... Someone, please enlighten me. Fabio Rojas === Subject: : theoreme de Rademacher .8enonc.8e du th.8eor.8fme : une fonction lipschitzienne sur un ouvert est presque partout diferrentiable sur cet ouvert. d.8emonstration(s?) en fran.8dais, ouvrages,... Merci. === Subject: : Re: A brief note on pens and coathangers > I was recently compelled by the university to turn my thoughts > for a moment away from the pure realm of mathematics, and to suffer > myself for a while to consider the crude and clumsy applied sciences. > Recognizing me as a genious of rare calibre, when the school was > asked on short notice to provide an original exhibition for a science > fair, naturally the physics department turned in their desperation to > me for help. > It has been observed many times, and used to illustrate many > subtle and unintuitive propositions of quantum mechanics, that (left > to themselves and in the absense of an observer) coathangers will > multiply, while pens will decay at a rate which closely resembles > radioactive decay. > And so I devised the following experiment: into a lead box, > airtight, I inserted a large sample of ordinary pens, a sample of > ordinary coathangers of equal size, and an equally large sample of > long, thin pens, carefully twisted so as to also serve as coathangers. > These items were carefully made up of disparate chemical > substances, so that they might be distinguished from one another no > matter what permutations or transformations they would endure. > At the end of a month's time, the contents of the box were > analyzed: whence, by carefully examining the chemical compounds, it > was discovered that many of the ordinary pens had decayed: but, being > confounded by the confines of the box, these suffered themselves to > join the pen-coathangers, twisting themselves appropriately so as to > fit in; simultaneously, the ordinary coathangers, seeking to multiply > but being also confounded by the same box, balanced this effect by > drawing upon the pen-coathangers. > Using a mathematical model, it was demonstrated that if this > experiment were repeated with arbitrarily large samples, and allowed > to sit unobserved for arbitrarily large periods of time, the limiting > process would be an unchanged number of pen-coathangers, a complete > absense of pens, and an exact doubling of coathangers. This limit > would never actually be obtained by any finite amount of time, > however, lest the theorem of conservation of rate of change of pens > and coathangers be violated. > These principles were put to practical use for the sake of > transporting matter from one location to another: the items at their > source were enclosed within a large container made entirely of > interwoven pens, next to a giant pile of coathangers; while at the > destination, an enormous pile of pen-coathangers was gathered. As you > can guess, the the container made of pens was eventually teleported > into the remote pile of pen-coathangers, an event which was balanced > out, of course, by an equal number of the pen-coathangers untwisting > themselves and teleporting to the pile of ordinary coathangers. In > this way, the university was able to transport a small-sized > automobile from New York, NY to Beijing, China, with almost no energy > exerted. > My good colleagues from the physics department are working > closely with the department of defense to determine whether this new > technology might prove useful for national defense; I assure you that > I, however, am glad to be rid of the project, that I might return to > considerations of mathematics. > Your friend and correspondent, > Nathan the Great > Age 11 :-) Nathan, that was brilliant. But I don't think you are eleven. You have to be something like twelve or thirteen to come up with something like that. Your friend, Skip === Subject: : Re: Permutational Polynomial > I heard once, but do not have references, that the permutational polynomials > in GF(2^n) have been completely classified, so that case may be solved. === Subject: : Re: Closure property question >> Let F be a field (of sets, not necessarily a sigma-field). For each >> countable collection b = {B_j: j in N} of countable collections of >> sets in F, let c(b) = union(j=1..infty, intersection(B_j)). Let C >> be the image of c. Is C necessarily closed under countable unions >> and/or intersections? > Or to express it another way, C is the collection of all sets of the > form union(j, intesection(k, B_jk)), where the indices range over > the naturals and the B_jk are in F. > Agapito's argument about countable unions is correct: Let f: N - > N^2 be a surjection. Then > union(n, union(j, intersection(k, B_njk))) = union(n, intersection(k, > B_(f(n),k))). > It is true that in general C is not closed under countable > intersection: If C were closed under countable intersection, it > would be the sigma-field generated by F. To get a sigma-field, one > needs to repeat the alternations of union and intersection through > the ordinals, as Henry Helson told my analysis class years ago. > So let F be the field generated by the intervals of the form > [a,b), where a and b are rational. There exists a collection > (B_njk) such the intersection of unions of intersections is not in > C. Can anyone display a specific such collection? But now I am confused. Since a field is closed under complementation, every set in C can also be expressed (via De Morgan's laws) as the intersection of unions. By the same argument as above, these must be closed under countable intersection as well. Am I misquoting Helson? Somehow, it doesn't seem right that you can generate the sigma-field with countable unions of countable intersections of sets in the field. Where am I going wrong? -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: : Re: Closure property question <4de76b8.0403131417.7d8051b8@posting.google.com> <40549EE9.30706@rutcor.rutgers.edu> <4054F334.30405@rutcor.rutgers.edu But now I am confused. Since a field is closed under complementation, > every set in C can also be expressed (via De Morgan's laws) as the > intersection of unions. You mean: the *complement* of every set in C can be expressed as the intersection of unions. C is not closed under complementation. If C *is* closed under complementation as well as countable unions, then of course it is a sigma-fiels. === Subject: : Re: Primary Decomposition of Modules so there is a direct counterpart to chinese rem thm in modules... how great. This sort of result is not valid if N is non-torsion then, because the idea of p-primary components would not hold for a general module with some free-component. Anyway, the proof is very clear! thanks. > I am trying to learn about modules from Dummit and Foote. A problem > there (chapter 12.1 prob 10), asks to show that in a P.I.D. R a > torsion module N is the direct sum of its p-primary components (there > need not be finitely many of them). > I have tried to tackle this problem by using the fact that the > annihilator of N is of the form (a) in R, and that > a=p1^i1.p2^i2......pn^in (here pk^ik is pk raised to ik power). So > any n in N is annihilated by some product of the prime powers. How > does one show that the sum is direct, i.e. no every n can be > decomposed as a direct sum of p-annihilated components (for finitely > generated modules over a P.I.D this follows from applying chinese > remainder theorem to the Fundamental Theorem of Finitely Generated > Modules over a P.I.D) > It's OK by induction to split N as a direct sum of a bit N_1 > annihliated by p_1^{i_1} (I'll write this as p^i) and > a bit N_2 annihilated by b = a/p^i. Now p^i and b are coprime > and since we're in a PID there are s and t in R with > s p^i + t b = 1. Let N_1 = b N and N_2 = p^i N. Then for x in N > x = t b x + s p^i x is in N_1 + N_2. Also if y is in N_1 intersect N_2 > then p^i y = 0 (as y is in N_1) and b y = 0 (as y is in N_2). Thus > y = t b y + s p^i y = 0. So N is the direct sum of N_1 and N_2. === Subject: : Re: Calculus of Variations [Q] >I'm reading Structure and Interpretation of Classical Mechanics by >Sussman and Wisdom. On p. 93, during their discussion of how to >derive Lagrange's equations for the case of constrained motions, >the author's make a statement that doesn't make sense to me. They >start out by discussing the case in which the permissible configuration >paths q must satisfy a constraint of the form > phi(t, q(t), (D q)(t)) = 0, I assume you mean that the integral of phi vanishes, not phi. For most phi, this constraint alone would define q from an initial q(t_1). In any case, variations of q would be rather limited if the constraint were as you have stated above. >where t is time and D denotes derivative. They claim that the >proof in this constrained case is very similar to the proof (sketch) >they gave for the unconstrained case. In this earlier proof, a >key step was arguing that since the integral of > ({ (@_1 L o Gamma[q]) - D (@_2 L o Gamma[q]) }*eta)(t) >from t_1 to t_2 is 0 for all permissible variations eta, the >integrand above must be identically 0. >[NOTATIONAL ASIDE: I use @ for the partial derivative symbol and >the letter o to indicate functional composition; L is a Lagrangian >function for the physical system under consideration; Gamma is a >path dependent map that takes a path q: R -> R^n and returns the >map t |-> (t, q(t), (D q)(t)); eta is a perturbating map with the >same range and domain as q, and with the additional property that >eta(t_1) = eta(t_2) = 0; the * before the eta denotes the product >of a row/covariant vector and a column/contravariant vector.] >For the constrained motion case, the difference is that eta, in >addition to satisfying eta(t_1) = eta(t_2) = 0, must also be such >that the perturbed path q + eta satisfies the constraint on the >motion. The authors claim that the expression above must be >identically zero in this case as well, because, as they write in >the footnote on p. 93, > [g]iven any acceptable variation [eta], we may make another > acceptable variation by multiplying the given one by a bump > function that emphasizes any particular time interval. If I understand everything here correctly, and the assumption I made above is correct, it would seem to me that rather than being 0, what should be true is that (@_1 L o Gamma[q]) - D (@_2 L o Gamma[q]) should be a constant multiple of (@_1 phi o Gamma[q]) - D (@_2 phi o Gamma[q]). >I've seen this kind of argument before; it is essentially the >reasoning the authors used when they proved the unconstrained motion >case. But in this case I just don't see what justifies the assertion >that by multiplying an acceptable variation by a bump function one >ends up with another acceptable variation. After all, at this >point in the argument, the constraint is pretty abstract (nothing >more than the (phi o Gamma[q])(t) = 0 expression above). With >so little information about the constraint, what justifies the >assumption that a suitable bump function exists, that when multiplied >by the origianl variation will yield a path that satisfies the >constraint? Usually, I see these arguments handled by orthogonality rather than bump functions. One orthogonality lemma that I use in most, if not all, variational problems is http://www.whim.org/nebula/math/orthovar.html I apologize if I am wholly off-base and have caused more confusion than clarification. Rob Johnson take out the trash before replying