mm-1024 === Subject: cracks in Euclidean Geometry and why Reals are fake Re: Does Euclidean geometry exist as a physical entity, or does only Riemannian and Lobachevskian > I am going to have to revamp File 103 on FLT, and File 120 of 3 and > only 3 geometries and File 125 of two proofs of the Riemann > Hypothesis in my website of www.iw.net/~a_plutonium/ > > I did not do much mathematics after 1997 and recently when I reviewed > my Riemann Hypothesis proof I realized that it is the p-adics that are > on the 1/2 Real line which means that lines are curved when out at > infinity. There are no straightlines. I have to change and revise my > Poincare Conjecture proof also. > > Physics is duality and not triality. If I go by that presumption then > I have to concede that there are only really 2 geometries and not 3. > Of the three known geometries of Riem, Loba, and Eucl, I would bet > that Euclidean is the nonexistant one. The P-adics create a nice point > by point Riemannian geometry and they naturally curve back around. But > I have trouble forming Lobachevskian geometry with a one to one > correspondence with algebraic numbers. Perhaps the Doubly-Infinites? > Perhaps the negative Reals. > But then I have trouble with the Reals for they seem to be Euclidean > geometry. But are they really? Euclidean geometry is zero curvature. > But the number 0 exists in p-adics and doubly-infinites. The > complex-numbers I can rule out as just a gimmick that gives added > dimensions. > So I am faced with 3 number sets of P-adics, Reals, and > Doubly-Infinites. If physics is the final word on this that duality > exists but triality is nonsense, then I am going to have to find out > what 2 and only 2 geometries exist and that one of them is a mental > illusion for human minds. Just as Newtonian absolute space and > absolute time was just a mental illusion. > The most perfect match is P-adics to Riemannian Geometry because the > P-adics are all positive numbers which Riem geometry deals with only > positive quantities and they naturally curve back around such as in > 10-adics the number ....99998 is equivalent of -2 and then ....99999 > is equivalent to -1. So it is a beautiful matchup. > But I have trouble in finding a number set to matchup with > Lobachevskian Geometry. It cannot be the Reals because they have both > negative and positive but Loba requires only negative. So are the > Doubly-Infinites intrinsically negative quantities similar to the fact > that P-adics are intrinsically positive quantities? Another major crack in Euclidean Geometry is that Real Numbers really do not coincide with Euclidean Geometry, do they? Perhaps I am making a statement or perhaps I am asking a question, for I am only at the beginning of this inquiry. If we look at the P-adics, they are all positive numbers, even the zero point can be said to be positive and so they are ideal for coinciding with Riemannian Geometry with its *positive curvature*. P-adics are ideal for representing Riemannian Geometry, or, making a 1-to-1-correspondence. So one can say that P-adics are the points of Riemannian Geometry and ideally such because the p-adics have a natural curvature to them for as we start in the 10-adics with 0 and then next is .000001 and next is .000002 and going way out it comes back to ...99997 which we can conceive of as -3 then ....999998 which is -2 and excitedly ....999 which is -1 and finally back to our starting point of 0. So the P-adics can be said to be the actual algebraic points of Riemannian Geometry. But now inspecting Reals with the geometry of Euclid, it just simply does not fit together does it. Because Euclid geometry is 0 curvature and the only number in the Reals that obeys curvature is a single point which is zero itself. The positive Reals disobey Euclidean Geometry because they are positive signifying Riem geom and not Euclidean and likewise the reverse for negative Reals for they signify Lobachevskian geom. So one is left with the conclusion that the Reals never represented the geometry entailed by Euclidean Geometry. Do the Real Number system represent any geometry??? I suspect not. I suspect the Reals are as mythical or imaginary as was Newtonian Mechanics of absolute space and absolute time. Humans have minds that can dream up things which really have no physical reality such as ghosts, witches, and Newtonian absolute space and absolute time. Are the Real Numbers another dream-up thing which has no physics reality? I suspect so. But I am troubled with what numbers correspond to Lobachevskian geometry. I would like to think that the negative REals suit the Loba geometry, but that leaves the nasty question of the positive Reals. I think I can draw some clues as what the numbers that make up Lobachevskian Geometry from the P-adics making up Riemannian Geometry. If I start with this claim: P-adics == Riemannian Geometry and accept it as fully true, then there are some other numbers called Doubly-Infinites. Doubly-Infinites are what the name implies. You see, p-adics are infinite leftward strings. Doubly-Infinites would then be numbers that are both infinite leftward but also infinite rightwards. In some sense, the Real Numbers should be doubly-infinite. Perhaps that is the reason the Reals are fake and nonphysical just as the NaturalNumbers = finite-integers was a fake and dreamed-up illusions. TEST: the test of the above would be to show that the Doubly-Infinites are numbers that are all negative in sign value. What I mean is that the P-adics are all positive (even ....9999 is positive but it can represent -1). So, if I can show that Doubly-Infinites are all negative, then I will have shown a vast amount of knowledge and understanding. Because if I can show the Doubly- Infinites are all negative in sign value then these numbers are what compose Lobachevskian geometry. And that where I have: All-P-adics == Riemannian Geometry I will also have All-Doubly-Infinites == Lobachevskian Geometry This would then conclude that REals as a number system were a fake entity and would imply that the centuries of gaps and holes found in Reals such as the myriad types of differentiation and integration Lebesgue integral to name one is because the Reals are a hodge-podge-mess just as Newtonian absolute space and absolute time was a hodge-podge-mess that saddled Quantum Mechanics. Driver Motivation for the above: what drives me to many of these conclusions is that Quantum Mechanics is duality based and not threesome. Because Physics is twosome, then geometries should be twosome and not threesome. Therefore, geometry should have 2 and only 2 indepedent geometries. So, I have a choice of 2 of these geometries that really exist (1) Euclid (2) Riemannian (3) Lobachevskian. My choice is Riem and Loba. And since there are only two entails that Algebraically in Mathematics there exists only two real Number Systems. I am fully confident that P-adics are one true number system. I suspect Doubly-Infinites is the second truly existing number-system. Hence I suspect the Reals to be a fake system just as Newtonian Absolute Space and Absolute Time was a fake concept. Research: All I need is hard-core evidence that Doubly-Infinites are all negative numbers. If I can get that evidence, then all of my above thoughts would be confirmed. Archimedes Plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies (www.iw.net/~a_plutonium) website of the science of AP under revision what used to be my old science website www.newphys.se/elektromagnum/physics/LudwigPlutonium from years 1993 === Subject: Re: a linear algebra thought > All, I have a question on linear algebra. Let V be a finite dimensianl > vector space, S be a noninvertable linear transformation on V. Let V be 2 dimensional with basis b_1, b_2. Let S(b_1) = 0 and S(b_2) = b_1, then Ker(S) = Range(S). >I wonder if the Ker(S) is perpendicular to Range(S), or >ker(S)+Range(S) is direct sum and Ker(S)+Range(S)=V. === Subject: Re: a linear algebra thought > All, I have a question on linear algebra. Let V be a finite dimensianl > vector space, S be a noninvertable linear transformation on V. I > wonder if the Ker(S) is perpendicular to Range(S), or ker(S)+Range(S) > is direct sum and Ker(S)+Range(S)=V. Perhaps that I misunderstood your questions, but consider V = R^2 and S(x,y) = (y,0). Then ker(S) = Range(S). I suppose that this answers your questions. Jose Carlos Santos === Subject: a linear algebra thought All, I have a question on linear algebra. Let V be a finite dimensianl vector space, S be a noninvertable linear transformation on V. I wonder if the Ker(S) is perpendicular to Range(S), or ker(S)+Range(S) is direct sum and Ker(S)+Range(S)=V. === Subject: solution Ok, I have this solution... AS someone say, it is weel know that the number of end-zeros of n! is f(n) = [n/5] + [n/5^2] + [n/5^3]... If n in base 5 is n = a0 + a1*5 + a2*5^2 + ... + ak*5^k = (a0 a1 a2 ... ak)_5 Then using the fact that [n/5^s] = as + a(s+1)*5 + ... =(as...ak)_5 and adding f(n) = (n - sum_i ai) / 5-1 < = n-1/4 like this, f(5) = 1, f(24) = (24-8)/4 = 4 , f(25) = (25 - 1)/4 = 6 ... usint the baund f(n) < = (n-1)/ 4 and solving n-1 / 4 = 153 we faund that 151 = f(617) < = 153 and we writte 617 in base 5 we have 617 = (4432)_5 which let us know that (4444)_5 = 624 have f(624) = (624-16)/4 = 152 end zeros and that 625 = (10000)_5 has f(625) = 624/4 = 156 zeros ... so there is no number integer n such that n! ends with 153 or 154 or 155 zeros... Bonito no? This formula and metoth let us know what are the numbers wich never would be the end-zero number of n! (like 5, 28, 29, 30,153, 154, 155...) Rogelio buedia@matem.unam.mx >... >> Anyhow counting 0s lets begin with n=5: >> 1*2*3*4*5 sequence will result with 1 0s : 5!= 120 >> 6*7*8*9*10 sequence will result with 1 0s >> 11*12*13*14*15 sequence will result with 1 0s too >> and so on but sequences >> 21*22*23*24*25 >> 46*47*48*49*50 >> 71*72*73*74*75 will result with 2 0s >> acc. to 4*25=100 6*50=300 and 75*8=600 >> and so on it will be till >> 91*92*93*94*95 sequence =95/5 + 3 = 21 0s totally >> 96*97*98*99*100 seguence will result with 2 0s >> 101*102*103*104*105 sequence will result with 1 0s >> and so on till >> 196*197*198*199*200 will result with 2 0s and totally 200/5 + 8 >> also 48 0s. >> now what with 153 0s ? it is like 625/5 + 25 = 150 0s >> then 630! turns to 151 0s >> 635! to 152 0s >> could 640! contain 153 0s or I am mistaken ? >> simply it looks for some extra 0s in some inside sequences: >> Not so easy to be correct in short time ! >No. 624!=152 zeros at the end, 625!=156 zeros, thats because >625!=624!*625 = 624!*5^4, so you have 4 more 5s in the factors of >625! than in those of 624!. Combined with the 2s in the factors of >624! (you need equal amount of each because 10=2*5), thats why you go >from 152 zeros to 156. === Subject: Re: plotting hyperelliptic curves!! > could someone tell how to plot hyperelliptic curves using MAPLE, > Mathematica, Matlab..anything > for example..something like > v^2=u^5 - 5*(u^3) + 4*u + 3, over real numbers > thanx in advance for all the help.. > OP. Another way in maple that works well with many algebraic curves is: with(algcurves): f:=v^2-(u^5-5*u^3+4*u+3); plot_real_curve(f,u,v); Jim Buddenhagen === Subject: Re: plotting hyperelliptic curves!! > could someone tell how to plot hyperelliptic curves using MAPLE, > Mathematica, Matlab..anything > for example..something like > v^2=u^5 - 5*(u^3) + 4*u + 3, over real numbers > thanx in advance for all the help.. With Maple, load the plots package and then do: implicitplot(v^2=u^5-5*u^3+4*u+3,u=-1..3,v=-4..4); Mathematica also has an ImplicitPlot command. Jose Carlos Santos === Subject: plotting hyperelliptic curves!! could someone tell how to plot hyperelliptic curves using MAPLE, Mathematica, Matlab..anything for example..something like v^2=u^5 - 5*(u^3) + 4*u + 3, over real numbers thanx in advance for all the help.. OP. === Subject: Re: Polynomial solutions to Pell eqn >The Pell equation and the obvious cubic generalisation >X^2 - DY^2 = ±1 and >X^3 + DY^3 +(D^2)Z^3 - 3DXYZ = 1 >have simple polynomial solutions >(n,1) for D = n^2 ±1 >(n^2, n, 1) for D = n^3±1 >In the square case, you can use continued >fractions to expand expression of the form, say, >SQRT(an^2 + bn + c), but in the cubic case >no such option is available. >Simple cases can be guessed >for example >D = n^3 ± 3 >X = n^6 ± 3n^3 ± 1 >Y = n^5 ±2n^2 >X = n^4 ± n >D = n^3 +2, n even >X = (9n^6)/4 + (9n^3)/2 +1 >Y = (9n^5)/4 + 3n^2 >Z = 3n(3n^3 +2)/4 >I was wondering if there is any >systematic approach to obtaining >solutions ? > Not real sure about a systematic approach, but looking at a few data points for > specific D may be useful. Consider the following: > D=10^3+5*10 > X=62591931727331611501 > D=100^3+5*100 > X=4910017588770392230083131681058781350001 > D=1000^3+5*1000 > X=489788270129887199471510415004171147924480921387800135000001 > D=10000^3+5*10000 > 00000001 > D=100000^3+5*100000 > X= 48977602682444006529246451274826892825824096005404164480666792 0000448092 > 0000138780000001350000000001 > D=1000000^3+5*1000000 > X= 48977602561224440064012924645120074826892800258240960000540416 4480006667 > 920000004480920000001387800000000135000000000001 > D=10000000^3+5*10000000 > X= 48977602560012244400640001292464512000074826892800002582409600 0000540416 > 44800000666792000000004480920000000013878000000000013500000000 000001 > D=100000000^3+5*100000000 > X= 48977602560000122444006400000129246451200000074826892800000025 8240960000 > 00005404164480000000666792000000000044809200000000001387800000 000000013500 > 00000000000001 > There is a pattern here that seems to indicate that a subset of {n^3+5*n} has a > polynomial solution to P3. Im not sure what the exact answer is, but it could > probably be worked out, given sufficient time and interest. There are lots of > other curiosities that can be found this way. Yes n^3 +3n and 3n^2 +3n +1 seem easy to approximate. n^ + Dn^3 where (n,D) >1 also seems to work. I did try expanding cubrt(n^6 +2n^3 +4) and cubrt(n^3 +2),say in steps cubrt(n^6 +2n^3 +4) = n^2 + y, etc. cubrt(n^3 +2) = n + z, etc and truncating the two expression so that they both have the same denominator. This sems to work if, say, y = 1/kn where k is an integer. This suggests continued fractions but simulataneous approximation of polys seems specially difficult. > Rich === Subject: Re: Simple counterexample needed (ODEs) > Im looking for a function f(t, x, y) from R x R^2 to R^2 such that > df/dt (at some time t0) is well defined for all (x, y), but not > differentiable w.r.t. to (x, y). The grammar is questionable here. Do you want f not differentiable w.r.t. to (x, y), or df/dt not differentiable w.r.t. to (x, y)? In either case you could consider f(t,x,y) = e^t*(g(x), g(y)), where g : R -> R is as bad as you want it to be. === Subject: Re: Simple counterexample needed (ODEs) > Im looking for a function f(t, x, y) from R x R^2 to R^2 such that > df/dt (at some time t0) is well defined for all (x, y), but not > differentiable w.r.t. to (x, y). > (Basically, Im after a case that fails to meet the conditions of > the existence and uniqueness theorem for differential equations.) > jill a classical example of an ODE with branching is y = 3*y^(2/3) and the other variables can be dummy variables for an example you are looking for: f(t,x,y) = 3*x^(2/3) Here the partial df/dt is constant 0, the best of all differentiable functions. The branching: take a<0b This is a legitimate solution, besides the obvious constant zero solution, and the less obvious strictly increasing solution y=t^3. Hope it helps, ZVK(Slavek). === Subject: Simple counterexample needed (ODEs) Im looking for a function f(t, x, y) from R x R^2 to R^2 such that df/dt (at some time t0) is well defined for all (x, y), but not differentiable w.r.t. to (x, y). (Basically, Im after a case that fails to meet the conditions of the existence and uniqueness theorem for differential equations.) jill === Subject: Re: Request for comments on antiquated algebraic topology online-book > Well Ive pretty much decided not to invest too much time with Lefschetz. > However, initially, I only need to learn a small subset of algebraic > topology (AT). Specifically, the same topics as found in the first chapter > of Munkres ÔElements of AT. The application is to distributed computing > where the evolution of finite asynchronous distributed protocols are modeled > as high dimensional simplicial complexes (if thats the correct terminology). > combinatorial. The seminal explanatory paper on the approach can be found > at: > http://www.cs.brown.edu/people/mph/HerlihyR96/sv.pdf > Thats an interesting paper. I suppose the explanations are too brief > for you? If you read that papers explanations on topology and combine > it with looking at things online like Wikipedia, I bet youll be > alright. See below also. > I have taken another look at Hatcher, and youre right - it doesnt seem > nearly as intimidating as it did before I bulked up on general topology a > bit. Hopefully I wont need to shell out the cash for Munkres book. > I think for you, it would be worth looking at the first chapter of > Munkres. Just check it out from your local university/college library. > There is no need to pay up all that money for something you only need a > piece of. Hi Chan-Ho. I started reading Hatchers book. I had been reading chapter 0 even though he notes it may be skipped intitially, and was having trouble with it. Chapter 1 was far more instructive and have been successfully making my way through it. l8r, Mike N. Christoff === Subject: Re: Cardinality of pi(S^1) > Im reading Hatchers book on Algebraic topology and he proves the > following: > The map v : Z -> pi(S^1) sending an integer n to the homotopy class of the > loop w_n(s) = (cos 2*pi*n*s, sin 2*pi*n*s) based at (1,0) is an isomorphism. > --- > But isnt w_n hotomopic to w_m for all m,n in N, using a linear homotopy? > ie: [w_n] = [w_m] meaning there is only a single homotopy class to map Z > into. Actually, linear homotopies are only gauranteed in convex sets which S^1 is not. l8r, Mike N. Christoff === Subject: Cardinality of pi(S^1) Im reading Hatchers book on Algebraic topology and he proves the following: The map v : Z -> pi(S^1) sending an integer n to the homotopy class of the loop w_n(s) = (cos 2*pi*n*s, sin 2*pi*n*s) based at (1,0) is an isomorphism. --- But isnt w_n hotomopic to w_m for all m,n in N, using a linear homotopy? ie: [w_n] = [w_m] meaning there is only a single homotopy class to map Z into. l8r, Mike N. Christoff === Subject: Re: Book series Lecture Notes in Mathematics > Are these books generally directed to people with advanced interests > and degrees? So it seemed to me when I frequented my college library. It depends on the editorial house (there are many Lecture Notes series!) but by and large, yes. They cover stuff that would be covered in an advanced course or seminar for which a textbook is not available. The Editorial Policy of the LMN series from Springer (the yellow books that I suspect is what you are referring to) states in part (taken from the overleaf front cover of the 1994 printing of Mumfords Red Book, LNM 1358): 1. Lecture Notes aim to report new developments - quickly, informally, and at a high level. The texts should be reasonably self-contained and rounded-off. Thus they may, and often will, present not only results of the author but also related work by other people. The back cover states: This series reports on new developments in mathematical research and teaching - quickly, informally, and at a high level. The type of material considered for publication includes 1. Research monographs 2. Lectures on a new field or presentations of a new angle in a classical field. 3. Seminars on topics of current research. Texts which are out of print but still in demand may also be considered if they fall within these categories. The London Mathematical Society Lecture Notes series does not seem to have an editorial policy printed on their books (at least the ones I have) but they seem to be of about the same scope, except that they also often publish conference proceedings. Arturo Magidin, sans .sig === Subject: Re: Book series Lecture Notes in Mathematics > Are these books generally directed to people with advanced interests > and degrees? Yes. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Book series Lecture Notes in Mathematics > Are these books generally directed to people with advanced interests > and degrees? Yes. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Book series Lecture Notes in Mathematics Are these books generally directed to people with advanced interests and degrees? So it seemed to me when I frequented my college library. David Ames === Subject: Re: dummy variable / integration help OK...I think also Youngs inequality / Minkowskis integral inequality is directly on-point here, but...is there a way to do it directly? > Let p(x) = 1/sqrt(4 pi t) e^-(x^2/4t). > Let T f (x) = INT p(x-y) f(y) dy (the integral, and all integrals > henceforth, are assumed from -oo to oo). > I have to show that T is a bounded operator from L_2 to L_2, so (let || || > be the L^2 norm) > ||T_t f||^2 = INT ( INT p(y) f(x-y) dy ) ^2 dx > Hint: Use Jensens Inequality on (INT p(y) f(x-y) dy)^2. === Subject: Re: dummy variable / integration help > Let p(x) = 1/sqrt(4 pi t) e^-(x^2/4t). > Let T f (x) = INT p(x-y) f(y) dy (the integral, and all integrals > henceforth, are assumed from -oo to oo). > I have to show that T is a bounded operator from L_2 to L_2, so (let || || > be the L^2 norm) > ||T_t f||^2 = INT ( INT p(y) f(x-y) dy ) ^2 dx Hint: Use Jensens Inequality on (INT p(y) f(x-y) dy)^2. === Subject: dummy variable / integration help Alright, I clearly cant remember introductory calculus. Let p(x) = 1/sqrt(4 pi t) e^-(x^2/4t). Let T f (x) = INT p(x-y) f(y) dy (the integral, and all integrals henceforth, are assumed from -oo to oo). I have to show that T is a bounded operator from L_2 to L_2, so (let || || be the L^2 norm) ||T_t f||^2 = INT ( INT p(y) f(x-y) dy ) ^2 dx = INT (INT p(y)f(x-y) dy ) (INT p(z) f(x-z) dz) dz = INT INT INT p(y) p(z) f(x-y) f(x-z) dx dy dz = INT INT (INT f(x-y) f(x-z) dx ) p(y) p(z) dy dz. Now Id *like* to get ||f||^2 on the inside, but INT f(x-y) f(x-z) dx doesnt seem to be quite that, because of the (x-y) and (x-z), and any linear change of variable on x wouldnt solve the problem....where am I screwing up? === Subject: eigenvectors of the laplacian I am looking for information on the properties of the Laplacian on a bounded open set of R^n. In particular, if one consider a bounded open set O with a regular boundary, I would like to have a complete study of the properties of the eigenvectors {v1,v2,...} of the laplacian with Neumann boundary conditions. In fact, I am looking forward some approximation results. For example, given a C^1 regular function f on O, which is the convergence speed of the approximation sum{i vi} when N-->infinity ? This is in fact the study of the spectrum {} of a regular function f. At last (but not least), I would like to have some insight about discrete approximation results. For example, when I have a grid that approximates the domain O, and the laplacian is replaced by a finite difference scheme. Gabriel === Subject: Re: Question about Incompleteness (Godel) [...] > 2. What results are there about minimum complexity requirements for > a statement to be potentially undecidable in a consistent formal > system. (For instance, trivial statements like x+y > x for all x>0 > are certainly not undecidable in PA, and this is because (as I > understand) such a statement is expressable in a much weaker formal > system which is consistent and complete.) [...] Gregory Chaitin defined the number Omega, 01. Is it a trivial corollary that the set of undecidable statements, >is not only non-empty, but infinite (or even uncountable) (In some >interesting way - obviously ÔG & n=n for all n is an infinite >undecidable set) Let A_0 be the original axiom set. Let G_0 be the Goedel statement for the theory with that axiom set. For each i>0, let G_i be the Goedel statement for the theory with axiom set A_i = A_{i-1} U {G_i-1}. Then G_i is expressible in the language of the original theory, makes a statement essentially distinct from G_0,...,G_{i-1}, and being formally undecidable from A_i, it is also formally undecidable from A_0. So the set {G_0,G_1,G_2,...} is a countable set of undecidable sentences for the original theory. Thus the set is infinite. We usually do not consider uncountable languages, because there is no point: any well-formed-formula can use only finitely many variables, so a countable set of variables will suffice. If the language is at most countable, then the entire set of WFFs is countable, so the set of undecidable statements must be at most countable. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Question about Incompleteness (Godel) ... > Huh? No, there are models in which the given Goedel statement is true, > and models in which it is false; one may adjoin it or its negation to > the axioms (provided certain conditions are met) and obtain a > consistent system. Yes, the new system has a ->new<- Goedel statement > which is formally undecidable in that system, but the old one is not > true but not provable. I looked at your response and didnt believe it for a minute, but during that minute I remembered the completeness theorem: if the Goedel statement were true in every model of T it would be deducible. Post in haste, repent at leisure ... === Subject: Re: Question about Incompleteness (Godel) Adjunct Assistant Professor at the University of Montana. >>Goedels proof shows that there are statements that are formally >>undecidable. It says nothing about what form they will take, only >that >>they exist. If the set of formally undecidable statements is >nonempty, >>that means that statements may be there. >Ok, great, so a minimal existence result. How about a response to my >more interesting supplementary questions: My post was not a reply to that post, was it? So why should my reply contain answers to questions that you had not asked yet? >1. Is it a trivial corollary that this set has infinite or even >uncountable cardinality? It is trivial that it cannot be uncountable unless the (formal) language itself is uncountable: every formula is a finite sequence of symbols from the alphabet, hence there are at most countably many formulas, well formed or not. I dont see why it would be a trivial corollary (unless you accept trivial conjunctions of undecidable statements and other statements, which you explicitly excluded in your previous post). [.snip.] >And btw, since this is a fairly informal newsgroup, you should all try >to post more interesting replies than Ive seen to my question. Limosnero, y con garrote...[1] >To be >honest, I provided the best post by mentioning the Goodstein theorem, Oh, well, then you should not have posted; you could have accomplished everything just by talking to yourself in the mirror, no? You gave the best reply, after all. >which admittedly followed from researching A. Magidins initial reply. >But as far as I can tell, Goodsteins theorem is the only really >compelling demonstration of Godel as applied to Peano arithmetic. >Restating results about AC and CH is lazy, boring, uninteresting and a >pretty good demonstration of why mathematics is losing its appeal in >democracies worldwide (big drop in quality of university applicants in >the UK anyway). Sigh. You must be a real hit at office hours... You asked a question. The level of your prior knowledge was not given by the question, and could only be guessed. You asked explicitly if there were any Ôordinary mathematical statements that were undecidable. If you were so familiar with the independence results about AC and CH, then your question was, to paraphrase you, lazy, boring, uninteresting, and a pretty good demonstration of why the big drop in quality of applicants: they dont think before asking, or they do not know how to ask the question they really mean. You were already aware of statements that were independent, if you wanted to exclude those as answers, you did a poor job of phrasing your question. To now complain because people answered the question you asked rather than the question you ->meant<- is rather annoying. As for restating results, unless you were expecting someone to present you with a brand new proof of a brand new statements independence, that was all you were goin to get in a fairly informal newsgroup. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu [1] Beggar with a bat; asking for favors, and then complaining because those who freely give their time to help do not do so to the degree you would wish them to. === Subject: Re: Question about Incompleteness (Godel) >Goedels proof shows that there are statements that are formally >undecidable. It says nothing about what form they will take, only that >they exist. If the set of formally undecidable statements is nonempty, >that means that statements may be there. Ok, great, so a minimal existence result. How about a response to my more interesting supplementary questions: 1. Is it a trivial corollary that this set has infinite or even uncountable cardinality? 2. What are the known results regarding minimal complexity for a sentence to be undecidable? (If there are none, then I have a new theorem, an undecidable sentence requires >2 symbols to state it, since the minimal meaningful sentence using 2 symbols is ~A, where A is an axiom of the system,and ~A is false, hence decidable) And btw, since this is a fairly informal newsgroup, you should all try to post more interesting replies than Ive seen to my question. To be honest, I provided the best post by mentioning the Goodstein theorem, which admittedly followed from researching A. Magidins initial reply. But as far as I can tell, Goodsteins theorem is the only really compelling demonstration of Godel as applied to Peano arithmetic. Restating results about AC and CH is lazy, boring, uninteresting and a pretty good demonstration of why mathematics is losing its appeal in democracies worldwide (big drop in quality of university applicants in the UK anyway) JG === Subject: Re: Question about Incompleteness (Godel) > Yes, the new system has a ->new<- Goedel statement > which is formally undecidable in that system, but the old one is not > true but not provable. For the usual cases, the Godel sentence for T is true, given that T is consistent, but unprovable in T. Or in other words, given the equivalence in T of G and (a formalization of) T is consistent, that T is consistent is true, if T is consistent, but unprovable in T. === Subject: Re: Question about Incompleteness (Godel) Adjunct Assistant Professor at the University of Montana. [.snip.] >So, all true ... >... but there is still a _big_ difference between the situations >wrt AC, CH, GCH (and, on more elementary ground, the Euclidean >parallel postulate) and the Goedel statement G in some theory T: >The well-known undecidable (relative to [case-specific mumble]) >statements of set theory and geometry are undecidable because >the base theories have models in which these statemenets are >true, and (other) models in which they are false. Goedel-type >statements are true but not provable in _every_ model of the >theory in which they are expressed. Huh? No, there are models in which the given Goedel statement is true, and models in which it is false; one may adjoin it or its negation to the axioms (provided certain conditions are met) and obtain a consistent system. Yes, the new system has a ->new<- Goedel statement which is formally undecidable in that system, but the old one is not true but not provable. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Question about Incompleteness (Godel) >How do we then conclude >that Ôstandard mathematical statements, like ÔEvery even number is >prime or the sum of two primes might be undecidable? Doesnt Godel >only show that Ômeta-statements like G exist whch are undecidable? As Goedel did it, the final undecidable statement is not a Ômeta-statement. We now know it can have the form: Here is an explicit polynomial in many variables with integer coefficients: [long-ish polynomial written here] When natural numbers are plugged in for the variables, the result is never zero. Concrete but (as stated) uninteresting. === Subject: Re: Question about Incompleteness (Godel) > Ah. I was thinking of something like that. Would you happen to have a > refernce (or if its just a simple illumination insight, sketch a proof > or hint at the right direction)? The arithmetic part is simple: for any arithmetical theorem A of ZF+V=L, the relativization of A to L is provable in ZF, but this relativization is equivalent in ZF to A itself, since the hereditarily finite sets in L are the hereditarily finite sets. Indeed, Godels relative consistency result concerning GCH and AC is much better formulated as a proof that GCH and AC are conservative over ZF with respect to theorems in finite mathematics. In an old FOM posting, Steve Simpson lists various results about conservative extensions, including ZFC + V=L is conservative over ZF for Pi^1_2 sentences. ZFC + GCH (actually ZFC + V=L(r) for some real r) is conservative over ZF for Pi^1_3 sentences. ZFC + GCH is conservative over ZFC (actually over ZF + a well ordering of the reals) for Pi^2_1 sentences. === Subject: Re: Question about Incompleteness (Godel) > Goedel-type statements are true but not provable in _every_ model of the > theory in which they are expressed. What does it mean for a statement to be unprovable in every model of the theory in which they are expressed? By completeness theorem, if a sentence G is not provable from theory a (consistent) T there is a model in which T holds but G does not and if G is independent, there is a model in which both T and G hold as well as a model in which T holds but G does not. The G.9adel statement says that there is no natural number which codes the proof of the statement itself, and since its independent of the theory T (provided T satisfies certain conditions, which you can slightly weaken by using the Rosser sentence) there is a model of T in which there is a non-standard natural number coding a proof of G and in addition there are models in which no such standard or non-standard number exists. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Question about Incompleteness (Godel) > G.9adels interpretation of GCH and ZFC in L tells you (as first > pointed out, I believe, by Kreisel) that they have no new arithmetical > consequences. Ah. I was thinking of something like that. Would you happen to have a refernce (or if its just a simple illumination insight, sketch a proof or hint at the right direction)? -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Question about Incompleteness (Godel) (Note: Im replying after 5 posts in this thread even though Google Groups is only showing 2 at the moment) that answers things pretty clearly. Although to be absolutely sure I understand the situation I need clarification of these two points: 1. Is it a trivial corollary that the set of undecidable statements, is not only non-empty, but infinite (or even uncountable) (In some Ôinteresting way - obviously ÔG & n=n for all n is an infinite undecidable set) 2. What results are there about minimum complexity requirements for a statement to be potentially undecidable in a consistent formal system. (For instance, trivial statements like x+y > x for all x>0 are certainly not undecidable in PA, and this is because (as I understand) such a statement is expressable in a much weaker formal system which is consistent and complete.) I was really looking for. It led to me to the incredible ÔGoodstein theorem and its proof (which is impossible in PA) === Subject: Re: Question about Incompleteness (Godel) > Yes. Its also known that these imply no new Pi_1-truths, i.e. truths of > form Ax(P(x)=0), where P(x) is prim. rec. I dont know if this is best > we can do, i.e. whether GCH and AC are actually conservative over ZFC > for more than just Pi_1-sentences. G.9adels interpretation of GCH and ZFC in L tells you (as first pointed out, I believe, by Kreisel) that they have no new arithmetical consequences. Much stronger (optimal) results along these lines were proved in the sixties - unfortunately I forget the reference. === Subject: Re: Question about Incompleteness (Godel) > There are plenty of natural/ordinary statements that are known to > be formally undecidable in given systems. Most famously, perhaps, the > Axiom of Choice is undecidable from the usual axioms of > Zermelo-Fraenkel, as are the Continuum Hypothesis (from both ZF and > ZF+Choice) and the Generalized Continuum Hypothesis (also from both ZF > and ZF+Choice). Yes. Its also known that these imply no new Pi_1-truths, i.e. truths of form Ax(P(x)=0), where P(x) is prim. rec. I dont know if this is best we can do, i.e. whether GCH and AC are actually conservative over ZFC for more than just Pi_1-sentences. Although both AC and GCH are interesting examples of mathematically undecidable sentences, they are not undecidable because of G.9adels theorem - in any sense Im aware of. > I believe certain results from Ramsey Theory are > formally undecidable in Peano Arithmetic. Yes. There is the Paris-Harrington theorem, which is a slightly tweaked finite form of the infinite Ramseys Theorem (about existence of homogenous colourings). There are various variations, such as the Kirby-Paris theorem and what you have. There is also Friedmans Finite Form of the Kruskal Tree theorem, which is unprovable in theories much stronger than PA - in fact, its provably beyond means of predicative mathematics. > But the important point is that while the Goedel statement G may seem > contrived, it is a perfectly valid Ôordinary mathematical statement > about numbers. Is it? There seems to be a very clear sense in which it is not: it would not have arisen in the course of the study of number theory or related fields. If you just mean that theres nothing magically metamathematical about the statement, youre right, of course. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Question about Incompleteness (Godel) >>Something that has confused me for a while and I would appreciate any >>clarification: >>Godels incompleteness proof essentially demonstrates that if you have >>a consistent formal system defining at least simple arithmetic then >>some statements will be undecidable in the axioms of that system. Now, >>in his proof, the undecidable statement constructed is the >>meta-statement G = ÔG is not demonstrable. > The undecidable statement constructed is a statement about numbers; it > states there there does not exist a number which stands in a > well-defined relation with a given number y. This statement, which is > a statement purely about numbers, ->may be interpreted<- in the > metalanguage as stating that the formal statement with Goedel number y > has no proof in the system. But the statement itself is a statement in > the object language. >>How do we then conclude >>that Ôstandard mathematical statements, like ÔEvery even number is >>prime or the sum of two primes might be undecidable? Doesnt Godel >>only show that Ômeta-statements like G exist whch are undecidable? > Goedels proof shows that there are statements that are formally > undecidable. It says nothing about what form they will take, only that > they exist. If the set of formally undecidable statements is nonempty, > that means that statements may be there. >>Are there are proofs of incompleteness which construct (or exhibit >>directly) an Ôordinary undecidable mathematical statement, rather >>than one like G, which seems like a meta-statement, of little >>practical interest. > There are plenty of natural/ordinary statements that are known to > be formally undecidable in given systems. Most famously, perhaps, the > Axiom of Choice is undecidable from the usual axioms of > Zermelo-Fraenkel, as are the Continuum Hypothesis (from both ZF and > ZF+Choice) and the Generalized Continuum Hypothesis (also from both ZF > and ZF+Choice). I believe certain results from Ramsey Theory are > formally undecidable in Peano Arithmetic. > But the important point is that while the Goedel statement G may seem > contrived, it is a perfectly valid Ôordinary mathematical statement > about numbers. In particular, it asserts that a certain Diaphontine system has no solutions. So, all true ... ... but there is still a _big_ difference between the situations wrt AC, CH, GCH (and, on more elementary ground, the Euclidean parallel postulate) and the Goedel statement G in some theory T: The well-known undecidable (relative to [case-specific mumble]) statements of set theory and geometry are undecidable because the base theories have models in which these statemenets are true, and (other) models in which they are false. Goedel-type statements are true but not provable in _every_ model of the theory in which they are expressed. === Subject: Re: Question about Incompleteness (Godel) Adjunct Assistant Professor at the University of Montana. >Something that has confused me for a while and I would appreciate any >clarification: >Godels incompleteness proof essentially demonstrates that if you have >a consistent formal system defining at least simple arithmetic then >some statements will be undecidable in the axioms of that system. Now, >in his proof, the undecidable statement constructed is the >meta-statement G = ÔG is not demonstrable. The undecidable statement constructed is a statement about numbers; it states there there does not exist a number which stands in a well-defined relation with a given number y. This statement, which is a statement purely about numbers, ->may be interpreted<- in the metalanguage as stating that the formal statement with Goedel number y has no proof in the system. But the statement itself is a statement in the object language. >How do we then conclude >that Ôstandard mathematical statements, like ÔEvery even number is >prime or the sum of two primes might be undecidable? Doesnt Godel >only show that Ômeta-statements like G exist whch are undecidable? Goedels proof shows that there are statements that are formally undecidable. It says nothing about what form they will take, only that they exist. If the set of formally undecidable statements is nonempty, that means that statements may be there. >Are there are proofs of incompleteness which construct (or exhibit >directly) an Ôordinary undecidable mathematical statement, rather >than one like G, which seems like a meta-statement, of little >practical interest. There are plenty of natural/ordinary statements that are known to be formally undecidable in given systems. Most famously, perhaps, the Axiom of Choice is undecidable from the usual axioms of Zermelo-Fraenkel, as are the Continuum Hypothesis (from both ZF and ZF+Choice) and the Generalized Continuum Hypothesis (also from both ZF and ZF+Choice). I believe certain results from Ramsey Theory are formally undecidable in Peano Arithmetic. But the important point is that while the Goedel statement G may seem contrived, it is a perfectly valid Ôordinary mathematical statement about numbers. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Question about Incompleteness (Godel) Something that has confused me for a while and I would appreciate any clarification: Godels incompleteness proof essentially demonstrates that if you have a consistent formal system defining at least simple arithmetic then some statements will be undecidable in the axioms of that system. Now, in his proof, the undecidable statement constructed is the meta-statement G = ÔG is not demonstrable. How do we then conclude that Ôstandard mathematical statements, like ÔEvery even number is prime or the sum of two primes might be undecidable? Doesnt Godel only show that Ômeta-statements like G exist whch are undecidable? Are there are proofs of incompleteness which construct (or exhibit directly) an Ôordinary undecidable mathematical statement, rather than one like G, which seems like a meta-statement, of little Ôpractical interest. (Apologies for the slightly imprecise language, I hope some of you will not be too pedantic) === Subject: Re: linear difference equations by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1Q1IGF07985; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1Q02Ii01553 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i1Q02Ip21160; >I want to find x(n) >x(n+2) + 2*x(n+1) + c(n)*x(n) = 0 >were c(n) is a given sequence and x(0) and x(1) are given. Try transform methods, analogous to solving a 2nd order ODE using Laplace transforms. phil === Subject: Questions I am beginner in algebraic topology and I want to understand the classifying spaces,so I read from Steenrods book and Milnors paper then I compared them with Husemollers book and I could understand by it, but I could not solve some exercises (Ex.11 and 13, page 58,Husemollersbook Fiber Bundles), so would you be so kind as to solve them ? These exercises are: 11.If u: G ---> H is a continuous group morphism, define a map B(u): BG ->BH. Under what circumstances do you have a functor ? 13. Let $=(X,P,B) be a numerable principal G-bundle. Prove that $ is universal if and only if X is contractible Best wishes Rabab, === Subject: Re: divergent series >S_1^inf [1-(ln(n))/n]^n = inf >how to prove this? >I assume you mean > oo > --- ln(n) n > > ( 1 - ----- ) [1] > --- n > n=1 >-Yes, thats exactly what I mean. > When a < 1/2, log(1-a) > -a-a^2. Therefore, letting a = ln(n)/n, > multiplying by n, and exponentiating, we get > ln(n) n > ( 1 - ----- ) > exp(-ln(n)-ln(n)^2/n) > n > 1 1 > > - - [2] > n 2 >For all n >= 1, ln(n)/n < 1/2, therefore, we have [2] for all n >= 1. >Since each term of [1] is greater than 1/(2n), compare with the harmonic >series. >your inequalities? For the inequality, log(1-a) > -a-a^2 when a < 1/2, I looked at the Maclaurin series for log(1-a) = -a - a^2/2 + O(a^3), and figured it had to be bigger than -a - a^2 for some values of a around 0. Using the derivative of log(1-a) + a + a^2, I found that it is greater than 0 when a < .6838, so definitely when a < 1/2. For the inequality ln(n)/n < 1/2, I needed that inequality, so I just used the derivative of ln(n)/n to find the extrema of ln(n)/n and found that it is always less than 1/e which is less than 1/2. Of course n must be greater than 0. Rob Johnson take out the trash before replying === Subject: Re: divergent series > > Let me keep the thread going and pose: does the series > > sum_n 1/2^sqrt(n) > > converge or diverge? Curiously, I dont recall ever seeing this on a > calculus exam. > > --Ron Bruck > > converges, by the integral test >>Sounds good, but offhand I dont see how to integrate 1/2^sqrt(x). Is >>it really integrable by elementary functions? > | sqrt(x) > | 2 dx > | > | > = | 2^y d(y^2) > | > | > = | 2y 1/log(2) d(2^y) > | > | > = 2/log(2) y 2^y - | 2/log(2) 2^y dy > | > = 2/log(2) y 2^y - 2/log(2)^2 2^y > = 2/log(2)^2 2^y (log(2) y - 1) > sqrt(x) > = 2/log(2)^2 2 (log(2) sqrt(x) - 1) Oops; while the integration above is correct, it does not converge as | -sqrt(x) | 2 dx | | = | 2^-y d(y^2) | | = | 2y -1/log(2) d(2^-y) | | = -2/log(2) y 2^-y + | 2/log(2) 2^-y dy | = -2/log(2) y 2^-y - 2/log(2)^2 2^-y = -2/log(2)^2 2^-y (log(2) y + 1) -sqrt(x) = -2/log(2)^2 2 (log(2) sqrt(x) + 1) This does converge as x->oo. >>On the other hand, its clear that 2^sqrt(n) is larger than, say, n^2, >>for all sufficiently large n. (Take logs, so need to know sqrt(n) >>grows faster than log(n).) Hence the series converges by the >>comparison test with sum of 1/n^2. >>More generally, if C > 1 and e > 0 are constants, then the sum of >>1/C^(n^e) converges. Proof is the same. Rob Johnson take out the trash before replying === Subject: Re: divergent series >> Let me keep the thread going and pose: does the series >> >> sum_n 1/2^sqrt(n) >> >> converge or diverge? Curiously, I dont recall ever seeing this on a >> calculus exam. >> >> --Ron Bruck >> converges, by the integral test >Sounds good, but offhand I dont see how to integrate 1/2^sqrt(x). Is >it really integrable by elementary functions? | sqrt(x) | 2 dx | | = | 2^y d(y^2) | | = | 2y 1/log(2) d(2^y) | | = 2/log(2) y 2^y - | 2/log(2) 2^y dy | = 2/log(2) y 2^y - 2/log(2)^2 2^y = 2/log(2)^2 2^y (log(2) y - 1) sqrt(x) = 2/log(2)^2 2 (log(2) sqrt(x) - 1) >On the other hand, its clear that 2^sqrt(n) is larger than, say, n^2, >for all sufficiently large n. (Take logs, so need to know sqrt(n) >grows faster than log(n).) Hence the series converges by the >comparison test with sum of 1/n^2. >More generally, if C > 1 and e > 0 are constants, then the sum of >1/C^(n^e) converges. Proof is the same. Rob Johnson take out the trash before replying === Subject: Re: divergent series > > Let me keep the thread going and pose: does the series > > sum_n 1/2^sqrt(n) > > converge or diverge? Curiously, I dont recall ever seeing this on a > calculus exam. > > --Ron Bruck > > converges, by the integral test > Sounds good, but offhand I dont see how to integrate 1/2^sqrt(x). Is > it really integrable by elementary functions? > On the other hand, its clear that 2^sqrt(n) is larger than, say, n^2, > for all sufficiently large n. (Take logs, so need to know sqrt(n) > grows faster than log(n).) Hence the series converges by the > comparison test with sum of 1/n^2. > More generally, if C > 1 and e > 0 are constants, then the sum of > 1/C^(n^e) converges. Proof is the same. Yes, its integrable. Make the change of variable u = sqrt(x) in int 2^(-sqrt(x))dx to get int 2^(-u) 2u du and then integrate by parts. I confess this wasnt the way I solved the problem. That was just an order-of-magnitude calculation (show its < 1/n^2 for n sufficiently large). And also the condensation test, that sum a_n is summable (given a_n decreasing) iff sum 2^n a_{2^n} is summable. --Ron Bruck === Subject: Re: divergent series > Let me keep the thread going and pose: does the series > > sum_n 1/2^sqrt(n) > > converge or diverge? Curiously, I dont recall ever seeing this on a > calculus exam. > > --Ron Bruck > converges, by the integral test Sounds good, but offhand I dont see how to integrate 1/2^sqrt(x). Is it really integrable by elementary functions? On the other hand, its clear that 2^sqrt(n) is larger than, say, n^2, for all sufficiently large n. (Take logs, so need to know sqrt(n) grows faster than log(n).) Hence the series converges by the comparison test with sum of 1/n^2. More generally, if C > 1 and e > 0 are constants, then the sum of 1/C^(n^e) converges. Proof is the same. JoeS === Subject: Re: divergent series > Let me keep the thread going and pose: does the series > sum_n 1/2^sqrt(n) > converge or diverge? Curiously, I dont recall ever seeing this on a > calculus exam. > --Ron Bruck It converges. In fact, sum_n 1/a^(n^b) converges for any a>1, b>0, by comparison with sum_n 1/n^2, for example. This follows from the fact that n^b is eventually greater than 2 log n for any b>0. === Subject: Re: divergent series S_1^inf [1-(ln(n))/n]^n = inf how to prove this? I assume you mean oo --- ln(n) n > ( 1 - ----- ) [1] --- n n=1 -Yes, thats exactly what I mean. When a < 1/2, log(1-a) > -a-a^2. Therefore, letting a = ln(n)/n, multiplying by n, and exponentiating, we get ln(n) n ( 1 - ----- ) > exp(-ln(n)-ln(n)^2/n) n 1 1 > - - [2] n 2 For all n >= 1, ln(n)/n < 1/2, therefore, we have [2] for all n >= 1. Since each term of [1] is greater than 1/(2n), compare with the harmonic series. your inequalities? === Subject: Re: divergent series > Let me keep the thread going and pose: does the series > sum_n 1/2^sqrt(n) > converge or diverge? Curiously, I dont recall ever seeing this on a > calculus exam. > --Ron Bruck converges, by the integral test -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: divergent series >S_1^inf [1-(ln(n))/n]^n = inf >how to prove this? > I assume you mean > oo > --- ln(n) n > > ( 1 - ----- ) [1] > --- n > n=1 > When a < 1/2, log(1-a) > -a-a^2. Therefore, letting a = ln(n)/n, > multiplying by n, and exponentiating, we get > ln(n) n > ( 1 - ----- ) > exp(-ln(n)-ln(n)^2/n) > n > 1 1 > > - - [2] > n 2 > For all n >= 1, ln(n)/n < 1/2, therefore, we have [2] for all n >= 1. > Since each term of [1] is greater than 1/(2n), compare with the harmonic > series. Let me keep the thread going and pose: does the series sum_n 1/2^sqrt(n) converge or diverge? Curiously, I dont recall ever seeing this on a calculus exam. --Ron Bruck === Subject: Re: divergent series >S_1^inf [1-(ln(n))/n]^n = inf >how to prove this? I assume you mean oo --- ln(n) n > ( 1 - ----- ) [1] --- n n=1 When a < 1/2, log(1-a) > -a-a^2. Therefore, letting a = ln(n)/n, multiplying by n, and exponentiating, we get ln(n) n ( 1 - ----- ) > exp(-ln(n)-ln(n)^2/n) n 1 1 > - - [2] n 2 For all n >= 1, ln(n)/n < 1/2, therefore, we have [2] for all n >= 1. Since each term of [1] is greater than 1/(2n), compare with the harmonic series. Rob Johnson take out the trash before replying === Subject: Re: divergent series > S_1^inf [1-(ln(n))/n]^n = inf > how to prove this? ln(1-ln(n)/n) ~ -ln(n)/n , etc ... the general term of your series is equivalent to 1/n (and sum(1/n) diverges) === Subject: Re: divergent series > S_1^inf [1-(ln(n))/n]^n = inf > how to prove this? S = sum, I guess. Limit comparison with 1/n === Subject: divergent series S_1^inf [1-(ln(n))/n]^n = inf how to prove this? === Subject: Re: How to prove this probability Inequality >X,Y,Z are random variables >Try to prove >E[EX-E(X|Z)]^2>=E[EX-E(X|Y,Z)]^2 > When is your homework due? It is a homework, it is a problem my friend asked me he is preparing qualifying exam Weve already solved that I dont like this kind of words When is your homework due? please dont waste your time on typing this.sorry for the offence though maybe you are a professor === Subject: Re: How to prove this probability Inequality by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1PIt0802061; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1PIpTi01822 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i1PIpT523699; >X,Y,Z are random variables >Try to prove >E[EX-E(X|Z)]^2>=E[EX-E(X|Y,Z)]^2 When is your homework due? === Subject: Re: Cubic Formula > So how would you go about solving x^3 + 9x^2 + 9x + 8 = 0? > Long ago from self-study of an old College Algebra book (by C.H > Lehmann, published in 1962) I found, I learned the following method. > By Descartes Rule of Signs, the given equation has no positive root and > either three or one negative root. > If it has a rational root, then this is of the form p/q where p is a > divisor of the constant term and q is a divisor of the leading > coefficient. Thus, in this case, the only possible rational roots are > -1, 2, -4, and -8. It just occurred to me that finding any rational number solutions to polynomial equations using the method described above. Using the quadaratic, cubic, or quartic formulas would only be necessary to find roots which are not rational numbers. Michael === Subject: Re: Cubic Formula > So how would you go about solving x^3 + 9x^2 + 9x + 8 = 0? Long ago from self-study of an old College Algebra book (by C.H Lehmann, published in 1962) I found, I learned the following method. By Descartes Rule of Signs, the given equation has no positive root and either three or one negative root. If it has a rational root, then this is of the form p/q where p is a divisor of the constant term and q is a divisor of the leading coefficient. Thus, in this case, the only possible rational roots are -1, 2, -4, and -8. Now check (e.g., by synthetic division) each of these values to find that -8 is a root and the remainder on division by x + 8 is x^2 + x + 1. More generally, one rarely has need for precise closed-form solutions of such equations and one uses approximation algorithms. Some, like Newtons method, use calculus (which I assume you have not studied yet). Yet there are others, e.g., Horners method, which do not. Also, one can often make qualitative statements about the roots of a polynomial without explicitly finding them. I use calculus in the present example: Let f(x) be the given polynomial. Since f(x) approaches +/- infinity as x does, f has at least one real root (as do all real polynomials of odd degree). Since f maps nonnegative numbers to positive numbers, all its real roots are negative. It must have either one or three real roots, since if non-real z is a root, then so is the conjugate of z. (and f has exactly three complex roots, counting multiplicities). Now consider the derivative f(x) = 3x^2 + 18x + 9 = 3(x^2 + 6x + 3). Since 6^2 - 4*3 is positive, this has two real roots, whence f has two relative extrema. The larger x value, -3 + sqrt(6), must be a local minimum. f(-3+sqrt(6)) > 0, so f has exactly one real root, and this is less than -3 - sqrt(6). I think you might enjoy looking at such an old-style College Algebra book. The discussion here comes under the topic, Theory of Equations. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Cubic Formula > What is the method of finding the cube roots of any complex number? > Michael Write the complex number in polar form: z = r * exp(i*theta), where r and theta are real and r >= 0. Let s be the real cube root of the real number r then the cube roots of z are are s*exp(i*theta/3) s*exp(i*(theta+2*pi)/3) s*exp(i*(theta+4*pi)/3) === Subject: Re: Cubic Formula solutions to most cubic equations, is not included in most high school > algebra textbooks? > Michael (1) Not enough time. > (2) Not enough patience. > (3) The formula is messy and hard to memorize. > (4) It is the method, rather than the final formula, that is applied; this > goes against the spirit of high-school math teaching. > (5) The famous obstacle: real cubics with real distinct roots lead to cube > roots of complex (non-real) numbers (casus irreducibilis). > What is the method of finding the cube roots of any complex number? > Michael Stepping out of elementary algebra and trisecting an angle: if w = r*(cos(t) + i*sin(t) with r>0 then the cube roots of w are r^(1/3) * (cos((t+2*k*pi)/3) + i*sin((t+2*k*pi)/3), k=0, 1, -1} To improve an approximate cube root z of w, one can use a cubically convergent version of Newtons Method: z_new = z - (z^3-w) * z/(2*z^3+w) === Subject: Re: Cubic Formula Michael Ejercito escribi.97: > Is there any reason why this formula, which is crucial to finding > solutions to most cubic equations, is not included in most high > school algebra textbooks? > Michael >> Yes, first of all it is *NOT* crucial to finding such roots, more >> often than not you can easyly solve the equation by other means. >> Refer to the same textbook which does not include the formula for >> more explanation on this . > So how would you go about solving x^3 + 9x^2 + 9x + 8 = 0? It is > not like you can factor it into two binomials. What about synthetic division of x^3 + 9x^2 + 9x + 8 by (x - a), where a is a integr divisor of 8? -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Cubic Formula > So how would you go about solving x^3 + 9x^2 + 9x + 8 = 0? It is >not like you can factor it into two binomials. x^3 + 9x^2 + 9x + 8 = 0 <=> x^3 + x^2 + x + 8x^2 + 8x + 8 = 0 <=> x(x^2 + x + 1) + 8(x^2 + x + 1) = 0 <=> (8 + x)(x^2 + x + 1) = 0 -- Im not interested in mathematics that might have anything to do with reality. -- Russell Easterly, in sci.math === Subject: Re: Cubic Formula > Is there any reason why this formula, which is crucial to finding > solutions to most cubic equations, is not included in most high school > algebra textbooks? > Michael > (1) Not enough time. > (2) Not enough patience. > (3) The formula is messy and hard to memorize. > (4) It is the method, rather than the final formula, that is applied; this > goes against the spirit of high-school math teaching. > (5) The famous obstacle: real cubics with real distinct roots lead to cube > roots of complex (non-real) numbers (casus irreducibilis). What is the method of finding the cube roots of any complex number? Michael === Subject: Re: Cubic Formula >Is there any reason why this formula, which is crucial to finding >solutions to most cubic equations, is not included in most high school >algebra textbooks? > Michael > Yes, first of all it is *NOT* crucial to finding such roots, more > often than not you can easyly solve the equation by other means. Refer > to the same textbook which does not include the formula for more > explanation on this . So how would you go about solving x^3 + 9x^2 + 9x + 8 = 0? It is not like you can factor it into two binomials. > Also, it is messy, hard or imposible to memorize, etc etc. I have no problem memorizing it, once I found a pattern. Michael === Subject: Re: Cubic Formula >Is there any reason why this formula, which is crucial to finding >solutions to most cubic equations, is not included in most high school >algebra textbooks? A point in addition to others: The derivation of the quadratic formula is instructive and simple enough to teach in high school. The derivations of the cubic and quartic formulae are far more complicated and have less pedagogical value for the general audience. The interested student can go learn about it on his/her own. And such an exercise in research is in itself very valuable. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Cubic Formula >Is there any reason why this formula, which is crucial to finding >solutions to most cubic equations, is not included in most high school >algebra textbooks? > Michael Yes, first of all it is *NOT* crucial to finding such roots, more often than not you can easyly solve the equation by other means. Refer to the same textbook which does not include the formula for more explanation on this . Also, it is messy, hard or imposible to memorize, etc etc. === Subject: Re: Cubic Formula > Is there any reason why this formula, which is crucial to finding > solutions to most cubic equations, is not included in most high school > algebra textbooks? > Michael (1) Not enough time. (2) Not enough patience. (3) The formula is messy and hard to memorize. (4) It is the method, rather than the final formula, that is applied; this goes against the spirit of high-school math teaching. (5) The famous obstacle: real cubics with real distinct roots lead to cube roots of complex (non-real) numbers (casus irreducibilis). (6) (This may not be known to many high-school textbook authors, but I include it anyway:) the formula is unpredictable when it comes to numerical stability: see Nicholas J. Higham: Accuracy and Stability of Numerical Algorithms, Sec. 24.3.3 Solving a Cubic. (A stable method, for general polynomials: roots sought as the eigenvalues of the companion matrix using QR-iterations; that is way out of high-school curriculum.) === Subject: Cubic Formula Is there any reason why this formula, which is crucial to finding solutions to most cubic equations, is not included in most high school algebra textbooks? Michael === Subject: Re: . The hardest of all hard facts . >Surely the anisotropy decreased over the 100 years from your first citation >to your last. > The first citation considers the anisotropy of the speed of light in > air and other gases, the last the anisotropy of the speed of light in > vacuum. These are two different things. Yes and no. If the gas molecules are imaged on the aether, or are otherwise dependent on the aether for EM-moderated repulsion, which is the concensus, then they are the same thing. You still have a secular decay in the aether drift. So you can choose to believe that you have poked a hole in the bottom of the aetherists boat, or it is simply that one cannot assert certainty at the resolution limits of a given apparatus. I vote for the latter, but you can certainly claim the former if you wish. David A. Smith === Subject: Re: . The hardest of all hard facts . >Surely the anisotropy decreased over the 100 years from your first citation >to your last. The first citation considers the anisotropy of the speed of light in air and other gases, the last the anisotropy of the speed of light in vacuum. These are two different things. In theory the Lorentz contraction will ensure that the observed speed of light in vacuum is always equal to c, irrespectiver of whether or not the propagation of light occurs through an ether. This has been confirmed by experiment to a high degree of precision. In contrast the maths which Cahill worked out for gas filled Michelson-Morley type interferometers predicts that if the interferometer is in motion relative to quantum foam, the observed speed of light will vary depending on the direction of the light beam relative to the motion of the interferometer. This maths is explained in the first paper cited: R.T. Cahill http://arxiv.org/pdf/physics/0312082 This considered the Michelson-Morley experiment which in effect compared the speed of light in two orthogonal directions by using an air filled interferometer. A small anisotropy was observed. If my memory is correct, using Newtonion physics and ignoring the effect of the air, Michelson calculated that the anisotropy was equivalent to a speed through the ether of about 8 km/second. However, as he knew that the orbital velocity of the earth was 30 km / second relative to the sun, a speed of only 8 km/second relative to the ether did not make sense. Therefore he reported that the experiment had failed to detect motion relative to the ether. However when Cahill reanalysed the Michelsons data so as to take into account the effects of the Lorentz contraction and the effect of the air (which required some new maths), he found the small anisotropy was equivalent to a speed of hundreds of km/second relative to space/quantum foam (whatever one likes to call it). He was able to obtain similar results when he reanalysed data from other M-M type experiments with gas filled interferometers. The last citation was: Modern Michelson-Morley experiments and gravitationally-induced anisotropy of c M. Consoli Istituto Nazionale di Fisica Nucleare, Sezione di Catania http://arxiv.org/pdf/gr-qc/0306105 Since this paper reports on experiments using vacuum interferomenters, no-one (not even Cahill) would have expected any anisotropy to have been observed. So I thought it was interesting that a very small anisotropy has now even been seen for a vacuum M-M experiment. Peter === Subject: are Doubly-Infinites the points of Lobachevskian Geometry Re: cracks in Euclidean Geometry and why Reals are fake (most snipped) > Research: All I need is hard-core evidence that Doubly-Infinites are > all negative numbers. If I can get that evidence, then all of my above > thoughts would be confirmed. Let us look at a few P-adic of a 10-adic such as .....333333. and ....67 and .....101100111000. And the question of sign value does not become an issue for they are all of one sign. They are all of a positive number sign value even though some of them can resemble a negative value such as ....99998 which can be constrewed as -2. Not that it is -2 but has the resemblance of -2. But all the P-adics thus have this *positive value characteristic*. But now take a look at Doubly-Infinites, can we likewise unravel what sign value these numbers possess? Are they all a value opposite of that of P-adics? Are they all *negative sign value* whereas p-adics are *positive sign value*?? Let us write some Doubly-Infinites such as .....333333.67676767..... Another Doubly-Infinite is .....99998.101100111000......... The main question then is whether that infinite string rightwards upsets the sign value of the infinite string leftwards so very much so that the sign value of the entire number entity in question changes from positive to negative. P-adics are infinite strings leftward and are all positive valued specimens. But what is the sign value of Doubly-Infinites? Does that rightward infinite string force so much change upon the leftward infinite string that the number specimen in question is transformed into a negative number? I am trying to get a handle on this question by contrasting actual physical objects. And in contrast with Speed and Velocity and Acceleration in Physics. If we are allowed to speed ahead with velocity and with acceleration this would be the infinite leftward string. But in our speeding and accelerating we are forced to pay attention to little things, little distances would slow us down. In going from 1 to 2 to 3 to 4 we can keep the speed. But if forced to look at 1.3 between 1 and 2 and then 1.4 between 1 and 2 we are slowed down. Another physical model is the shape of a convex lens to a concave lens in contrast to infinite leftward strings compared to doubly-infinites. With no infinite rightward string to slow down the leftward string it becomes convex in shape. But when details of numbers between two consecutive rightward string numbers is demanded then a concave surface is traversed. What I am trying to convey is that P-adics are infinite leftward strings and they are all positive numbers. But in Doubly Infinites, because you have a infinite rightward string attached to a infinite leftward string, that the rightward string concaves the path. So that where the P-adic would form a convex lens surface, the Doubly-Infinite forms the opposite surface of concave and thus the opposite sign of negative vice positive. Perhaps there is a better physics analogy to what a rightward minutia detail string does to the macro large string that goes leftward. Or perhaps mathematics itself can decide this case without any need to refer to physics. But I doubt it because all of mathematics is a tiny subset of physics. Archimedes Plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies (www.iw.net/~a_plutonium) website of the science of AP under revision what used to be my old science website www.newphys.se/elektromagnum/physics/LudwigPlutonium from years 1993 === Subject: Re: Bayesian Class and Math/Stat Teaching... (Space shuttle o-rings) >The other was >that the technical people didnt communicate well. In particular, I >recall that they produced a scatterplot of temperature versus amount >of o-ring erosion FOR LAUNCHES WHERE THE O-RINGS WERE ERODED, which >showed no clear correlation. Not shown on the plot were the launches >with NO EROSION, which were all at warm temperatures. A more suitable >plot would have made clear that there was plenty of evidence that >erosion was greater at low temperatures than at high temperatures, and >that launching at a temperature substantially lower than any previous >launch was madness. Your summary is correct. Dropping the 0s (because they didnt add any more information) was a little more than a failure to communicate. The statistical reference for this is Dalal, Fowlkes and Hoadley, 1989. Risk analysis of the space shuttle: Pre-Challenger prediction of failure. JASA 84:945--957 -- My real email address is mcintosh ##at## research ##dot## telcordia ##dot## com === Subject: Re: Bayesian Class and Math/Stat Teaching... (Space shuttle o-rings) > ... If I recall, the point was that >just because there are many data points, one may still not have enough >information with which to proceed. All the data points they had indicated >that the o-ring was fine but the had no data points for how the o-ring might >behave in cold weather. > No. My fairly clear recollection is that they had enough data to be > positively suspicious that disaster might strike if they launched at a > temperature lower than any previous launch. This WASNT just a > general conservative view that one shouldnt push into unexplored > territory. [...] I believe you are correct. The original poster may be thinking of the more recent shuttle disaster. If I remember correctly, as part of the infamous Power Point presentation there is a graph showing some data about results of objects striking the wing, and there is a footnote on the graph that explicitly states, Estimated kinetic energy is greater than any yet measured (paraphrasing). When they did, later, carry out a simulation of the impact, it blew a hole in the wing. Robert Dodier === Subject: Re: Bayesian Class and Math/Stat Teaching... (Space shuttle o-rings) > ... If I recall, the point was that >just because there are many data points, one may still not have enough >information with which to proceed. All the data points they had indicated >that the o-ring was fine but the had no data points for how the o-ring might >behave in cold weather. No. My fairly clear recollection is that they had enough data to be positively suspicious that disaster might strike if they launched at a temperature lower than any previous launch. This WASNT just a general conservative view that one shouldnt push into unexplored territory. There were two problems. One was that the managers didnt pay enough attention to the technical peoples worries. The other was that the technical people didnt communicate well. In particular, I recall that they produced a scatterplot of temperature versus amount of o-ring erosion FOR LAUNCHES WHERE THE O-RINGS WERE ERODED, which showed no clear correlation. Not shown on the plot were the launches with NO EROSION, which were all at warm temperatures. A more suitable plot would have made clear that there was plenty of evidence that erosion was greater at low temperatures than at high temperatures, and that launching at a temperature substantially lower than any previous launch was madness. Radford Neal === Subject: Re: Saxon Math critique > I would like information regarding the Saxon Math program. How well does it meet national math standards, etc.? Having had to teach (in a semi-independent environment) Saxons Algebra I, I felt that my life was being drained away by a horde of invisible leeches. If youre trying to use the texts in a Ôhome school environment, then the amount of repetition will only bore one (or however many children you have) to a semi-comatose state. Using the texts to try to teach a dozen or so bright kids was a form of cruel and unusual punishment. I finally took an entire week-end and edited the idiot things exercises to remove the problems based on lessons about a hundred pages back from that set of problems. Now, dont mistake me: As a mathematician, I was handed a shock the like of which Id not felt since ÔNam when I first ran into that institutionalized idiocy of new new math, quite erroneously called ÔCollege Prep Math and its attendant stupidities of using calculators to avoid having the little darlings learn to do simple calculations on their own, and the relegation of math teachers to nannies wiping the intellectual drool from their charges alleged minds. Saxon was one of the first to oppose this damnfoolishness, and his texts reßect this opposition. The repetition in his system is an effort to make sure that the student will learn the methods in the text, BUT if you are dealing with one or more really bright student(s), the repetition will not only bore them, it might well convince them not to enjoy math. That would be a definite tragedy. As to the so-called national standards, the NCTM (to take an extremely disgusting example) is all in favor of the CPM rubbish. Id take a look at the requirements of your state (or province, or...) and then decide if the Saxon texts will cover the material. Here in California, the land of the Governator (Goddess help us!), Saxons two algebra books do a pretty good job. However, YMMV, this is all IMPO, beware the phase or the Moon, the Ides of March, &c. Bottom line: If youre trying to educate one or more students on your own, then Saxons books seem to be the best of the lot. If youre looking for a supplementary text for a student that is bright but is being bored silly by what passes for Ôpublic Ôeducation, then they might well serve the turn. I hope you have access to a copy, so that you might judge for yourself. HTH, Drieux PS Note that the spelling of my name in the email is part of trap. === Subject: Saxon Math critique by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1OJL5d08719; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1OJJ1i08191 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i1OJJ1520298; I would like information regarding the Saxon Math program. How well does it meet national math standards, etc.? === Subject: Re: 3 Squares Covering 1 Circle :-))) === Subject: Re: 3 Squares Covering 1 Circle > I was asked (via e-mail) to explain this here in more detail. > If a sense, this is what I did when I found that Jakes > configuration yielded a saddle point. Three of the eigenlines > were hot -- 2 slightly, and 1 much more so. So I looked at > that hot line, and it suggested the twisted configuration that > produced the bona fide local maximum. > Hi Jim, > But there is still a large gap between the general explanation, > which you gave in a vivid manner, and the (I think) awkward > looking matrix itself, full of lengthy terms. > It would be a great pleasure for me, if you showed this beast, > which eventually gave you the best solution so far. Maybe I or > some of the interested lurkers feels so strong now as to try > by myself/herself/himself to play with the matrix and get an > improvement. Or else get a better feeling for why this should > be the best solution indeed. I cant really show you this beast. Its too beastly. Even the formula for the area itself is complicated: its a 15-gon whose vertices have messy formulas depending on 9 parameters + 3 circular segements. To then write down the 45 second derivatives of this formula . . . Ugh. What I can write down is the formula for the two-parameter family which yields the optimal configuration. The input parameters are r and phi (described in a previous post). The formula for the area of the resulting figure is then given as follows (in Mathematica notation): rc = Sqrt[2] r Cos[phi]; rs = Sqrt[2] r Sin[phi]; sm = 1 - (rc - rs)/2; sp = 1 - (rc + rs)/2; cm = Sqrt[1 - sm^2]; cp = Sqrt[1 - sp^2]; qm = Sqrt[1 + 2 sm cm]; qp = Sqrt[1 + 2 sp cp]; am = ArcSin[(sm - cm)/Sqrt[2]]; ap = ArcSin[(sp - cp)/Sqrt[2]]; xtra = 12 - 9 rc - 14 sm sp - r^2 (1 + 2 Sqrt[3]); area = (3/4) ( xtra + 2 (am + ap) + (1 - rc) (cm + cp) + (1 + rs) qm + (1 - rs) qp ); > I like these geometric puzzles very much and I want to say > to be a strong tool to achieve insight as well as results. You call the Hessian a strong tool. Id say the strong tool is Taylor series: f(x) = f(x0) + (grad f)(x0).x + (grad grad f)(x0):xx / 2 + ... Because (grad f)(x0) = 0 at the critical point (by definition of a critical point), the local behavior is just f(x) = f(x0) + (grad grad f)(x0):xx / 2 + higher-order terms, i.e., a constant plus a quadratic form. It would be interesting to see a multidimensional problem in which grad f, grad grad f, and grad grad grad f were all 0, so that one had to resort to analyzing a *quartic* form. Not as simple as good old eigenvalues and eigenvectors. Hmmm. > configuration so far, > Rainer Rosenthal > r.rosenthal@web.de === Subject: Re: 3 Squares Covering 1 Circle > I was asked (via e-mail) to explain this here in more detail. > If a sense, this is what I did when I found that Jakes > configuration yielded a saddle point. Three of the eigenlines > were hot -- 2 slightly, and 1 much more so. So I looked at > that hot line, and it suggested the twisted configuration that > produced the bona fide local maximum. > Hi Jim, > But there is still a large gap between the general explanation, > which you gave in a vivid manner, and the (I think) awkward > looking matrix itself, full of lengthy terms. > It would be a great pleasure for me, if you showed this beast, > which eventually gave you the best solution so far. Maybe I or > some of the interested lurkers feels so strong now as to try > by myself/herself/himself to play with the matrix and get an > improvement. Or else get a better feeling for why this should > be the best solution indeed. > I like these geometric puzzles very much and I want to say > to be a strong tool to achieve insight as well as results. You are welcome. :) But I should thank, in turn, 3 square napkins and a paper plate for the inspiration... In any case, this is most likely a well-known problem. I know that, generally, such coverings are well-studied. (Although I know not the history of my particular problem.) Leroy Quet > configuration so far, > Rainer Rosenthal > r.rosenthal@web.de === Subject: Re: 3 Squares Covering 1 Circle > I was asked (via e-mail) to explain this here in more detail. > If a sense, this is what I did when I found that Jakes > configuration yielded a saddle point. Three of the eigenlines > were hot -- 2 slightly, and 1 much more so. So I looked at > that hot line, and it suggested the twisted configuration that > produced the bona fide local maximum. Hi Jim, But there is still a large gap between the general explanation, which you gave in a vivid manner, and the (I think) awkward looking matrix itself, full of lengthy terms. It would be a great pleasure for me, if you showed this beast, which eventually gave you the best solution so far. Maybe I or some of the interested lurkers feels so strong now as to try by myself/herself/himself to play with the matrix and get an improvement. Or else get a better feeling for why this should be the best solution indeed. I like these geometric puzzles very much and I want to say to be a strong tool to achieve insight as well as results. configuration so far, Rainer Rosenthal r.rosenthal@web.de === Subject: Re: 3 Squares Covering 1 Circle > Ac = 2.61034477 = 83.0898548% of the circle. > The value 2.60808705 for Jakes configuration is a local maximum for > a constrained 1-parameter problem, and the value above is a local > maximum for a constrained 2-parameter problem. But of greater interest > is whether the values are local maxima for the unconstrained problem, > which has 9 parameters (3 for each square). > I looked into this. Both constrained maxima are indeed critical points > for the unconstrained problem (the gradient of the area function is 0 > for each). To assess whether the critical points are local maxima, we > form the Hessian matrix and look at its eigenvalues. One eigenvalue > must be 0 (corresponding to the eigenvector that rotates the entire > system). For Jakes configuration, there are 5 negative, and 3 positive > positive eigenvalues, so it is a saddle rather than a local maximum. > For the new configuration Ive given, all 8 non-zero eigenvalues of the > Hessian are negative. The configuration is indeed a local maximum of > the unconstrained problem, and hence a viable candidate for the title > of global maximum. I was asked (via e-mail) to explain this here in more detail. Its the n-dimensional generalization of what one does in first-year calculus to determine how a function behaves at a point. For simplicity, we will assume that all functions are C^infinity, i.e., that you can take as many derivatives as you want. First, lets outline the first-year calculus procedure for determining the behavior of some function f at a point x0: ----- Set n = 1. Begin Loop Let D = the nth derivative of f at x0. If D > 0: f is increasing at x0. Stop. If D < 0: f is decreasing at x0. Stop. n++ (i.e., increase the value of n by 1) Let D = the nth derivative of f at x0. If D > 0: f has a minimum at x0. Stop. If D < 0: f has a maximum at x0. Stop. n++ Go to beginning of the loop. End Loop ----- What if the above is an infinite loop? I.e., what if all the derivatives are zero? Such functions are called C^infinity ßat. If f is analytic in such a case, then f must be a constant function. Otherwise it f is either increasing or decreasing at x0, or has a maximum or a minimum there, or has none of those behaviors. This last case is kind of weird: consider for example, the function f(x) = 0 for x = 0, f(x) = e^(-1/x^2) sin(1/x) otherwise. But I digress. In n dimensions, we first ask if the first derivative of f is zero at a point, that first derivative being the gradient of f, which is a vector. If so, then we continue to second order to look for the local behavior of f. Thus we take the second derivative of f, which is a matrix (or a second rank tensor, if you like) called the Hessian of f. The (i,j) entry in the Hessian matrix H is simply the partial derivative d^2 f / dx_i dx_j. Taylors expansion says that in the neighborhood of x0, f is approximately f(x) = x0 + (1/2) (x - x0)^T H (x - x0). H is symmetric, so all its eigenvalues are real. f(x) is greater than x0 everwhere in a neighborhood of x0 if H is positive definite (i.e., if all the eigenvalues of H are positive). That is, if all the eigenvalues of H are positive, x0 is a local minimum. Likewise, if all the eigenvalues of H are negative, x0 is a local maximum. If H has both positive and negative eigenvalues, then x0 is a saddle point. Now suppose youre at a saddle point, and you want to sense the environment around you. Lets say f represents temperature. If you compute the eigenvectors of H, the will form a set of mutually orthogonal lines (call them eigenlines) extending out from x0, each with an associated eigenvalue. If you stretch out both hands along an eigenline, they will be hotter than your body if the eigenvalue associated with the eigenline is positive, and colder if the eigenvalue is negative. If a sense, this is what I did when I found that Jakes configuration yielded a saddle point. Three of the eigenlines were hot -- 2 slightly, and 1 much more so. So I looked at that hot line, and it suggested the twisted configuration that produced the bona fide local maximum. -Jim Ferry === Subject: Re: who can >hi,who can solve my problem? > x^2*y+x*y+y=0 ,f(x)=y if x->tan(t) now write x^2*y+x*y+y=0 with t,f(t) Maple can. I get / 2 | d | | /d --- Y(t) | | 2 |-- Y(t)| tan(t) 2 | 2 | dt / dt | tan(t) |- ------------------ + -----------| /d | 2 2| |-- Y(t)| tan(t) 1 + tan(t) 1 + tan(t) / dt / -------------------------------------------- + ---------------- 2 2 1 + tan(t) 1 + tan(t) + Y(t) = 0 which simplifies somewhat: / 2 /d |d | 1/4 |-- Y(t)| sin(4 t) + (-1/8 cos(4 t) + 1/8) |--- Y(t)| + Y(t) = 0 dt / | 2 | dt / Why would you want to do this? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: who can hi,who can solve my problem? x^2*y+x*y+y=0 ,f(x)=y if x->tan(t) now write x^2*y+x*y+y=0 with t,f(t) thank you for your help hupo === Subject: re:measure of stuck-togetherness That doesnt work. ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: Re: measure of stuck-togetherness >So the earths surface is about one-quarter land. That land could be >distributed across the surface of the globe in different ways - I can >imagine at one extreme, all the land stuck together in one roundish >continent, and at the other extreme, all the land spread out evenly, in >equally-spaced tiny islands. Is there a measure, preferably just one >number, that captures these differences? If so, how is it calculated? > I dont know if theres an accepted measure, but heres one for you: > Partition the surface into many equal-sized regions. In each region, > measure the percentage of land. Take the standard deviation across > all the regions. Thats your number. I could be wrong, but that doesnt seem to work. Call each of the equal-sized regions a cell, and suppose there are n cells. Let p_k be the proportion of the kth cell-area thats land. Then the number you describe is m = s.d.{p_1,...,p_n}. If P is the overall proportion of the global area thats land, then avg{p_1,...,p_n} = P, and m**2 = (1/n) sum( (p_k - P)**2, k=1..n ). There seem to be two cases: n is either large enough to give a stable value for m (so making n even larger wont matter much), or n is too small (and m will be reßecting n rather what its supposed to be measuring). In the acceptable, stable case, it seems that n would eliminate most boundary effects, leaving a sum that comes mainly from two contributions: m**2 ~ (1/n) [ (#cells on land)(1-P)**2 + (#cells off land)(0-P)**2 ] ~ (1/n) [ n P (1-P)**2 + n (1-P) (0-P)**2 ] ~ P(1-P) In other words, either n is too small for m to measure what its supposed to, or n is large enough but then m measures only the global property P(1-P). --r.e.s. === Subject: Re: measure of stuck-togetherness >So the earths surface is about one-quarter land. That land could be >distributed across the surface of the globe in different ways - I can >imagine at one extreme, all the land stuck together in one roundish >continent, and at the other extreme, all the land spread out evenly, in >equally-spaced tiny islands. Is there a measure, preferably just one >number, that captures these differences? If so, how is it calculated? I dont know if theres an accepted measure, but heres one for you: Partition the surface into many equal-sized regions. In each region, measure the percentage of land. Take the standard deviation across all the regions. Thats your number. -- Matthew T. Russotto mrussotto@speakeasy.net Extremism in defense of liberty is no vice, and moderation in pursuit of justice is no virtue. But extreme restriction of liberty in pursuit of a modicum of security is a very expensive vice. === Subject: re:measure of stuck-togetherness Suppose that only the land on the surface has mass, and the rest of the earth (the oceans and the interior of the earth) is massless. We still suppose the earth holds together. In each point X in the interior of the earth measure the absolute value of the gravitaional pull F(X) and gravitational potential E(X). The following are possible measures of dispersion: 1) max F(X) 2) mean F(X) or mean F(X)^2 etc. 3) max (E(X) - E(Y)) 4) mean (E(X) - E(Y)) or mean (E(X) - E(Y))^2 etc. For uniform distribution those are all about zero. ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: Re: measure of stuck-togetherness > So the earths surface is about one-quarter land. That land could be > distributed across the surface of the globe in different ways - I can > imagine at one extreme, all the land stuck together in one roundish > continent, and at the other extreme, all the land spread out evenly, in > equally-spaced tiny islands. Is there a measure, preferably just one > number, that captures these differences? If so, how is it calculated? Youve received lots of suggestions, but let me just add the fractal dimension of the boundary (coastline). - Risto - === Subject: Re: measure of stuck-togetherness > So the earths surface is about one-quarter land. That land could be > distributed across the surface of the globe in different ways - I can > imagine at one extreme, all the land stuck together in one roundish > continent, and at the other extreme, all the land spread out evenly, in > equally-spaced tiny islands. Is there a measure, preferably just one > number, that captures these differences? If so, how is it calculated? Various measures of dispersion could work. For points x and y on the surface, let dist(x,y) be the length of the shortest great-circle arc connecting x and y. Then two such measures are E dist(x,y) and sqrt(E dist(x,y)**2), where x and y are iid uniform on the land-region. As approximations, you could replace the expectations by averages with x and y suitably discretized. --r.e.s. === Subject: Re: measure of stuck-togetherness Standard deviation of Statistical mathematics, measures the difference. === Subject: Re: measure of stuck-togetherness > So the earths surface is about one-quarter land. That land could be > distributed across the surface of the globe in different ways - I can > imagine at one extreme, all the land stuck together in one roundish > continent, and at the other extreme, all the land spread out evenly, in > equally-spaced tiny islands. Is there a measure, preferably just one > number, that captures these differences? If so, how is it calculated? How about how many islands there are or the area of the largest one. Depends on what you mean by differences. Bill === Subject: Re: measure of stuck-togetherness > So the earths surface is about one-quarter land. That land could be > distributed across the surface of the globe in different ways - I can > imagine at one extreme, all the land stuck together in one roundish > continent, and at the other extreme, all the land spread out evenly, in > equally-spaced tiny islands. Is there a measure, preferably just one > number, that captures these differences? If so, how is it calculated? The perimeter is one such measure. === Subject: measure of stuck-togetherness So the earths surface is about one-quarter land. That land could be distributed across the surface of the globe in different ways - I can imagine at one extreme, all the land stuck together in one roundish continent, and at the other extreme, all the land spread out evenly, in equally-spaced tiny islands. Is there a measure, preferably just one number, that captures these differences? If so, how is it calculated? === Subject: Re: Public Notice: Copyright Violations <4Ne%b.4390$Zp.3684@fed1read07> Discussion, linux) > Seems to be covered the fair use clause, even if it were marked Ôcopyright: > from: http://www.copyright.gov/title17/92chap1.html#107 > Notwithstanding the provisions of sections 106 and 106A, the fair use of a > copyrighted work, including such use by reproduction in copies or > phonorecords or by any other means specified by that section, for purposes > such as criticism, comment, news reporting, teaching (including multiple > copies for classroom use), scholarship, or research, is not an infringement > of copyright. In determining whether the use made of a work in any > particular case is a fair use the factors to be considered shall include - > (1) the purpose and character of the use, including whether such use is of a > commercial nature or is for nonprofit educational purposes; > (2) the nature of the copyrighted work; > (3) the amount and substantiality of the portion used in relation to the > copyrighted work as a whole; and You seem to be overlooking the third clause. Entire books (or not clear whether entire Usenet posts can be reproduced for the same purpose. (Note: I know that many classes involve a substantial number of practice.) > (4) the effect of the use upon the potential market for or value of the > copyrighted work. > The fact that a work is unpublished shall not itself bar a finding of fair > use if such finding is made upon consideration of all the above factors. > -- Jesse F. Hughes If anything is true in general about Usenet, its that people can go on and on about just about anything. -- James Harris speaks the truth. === Subject: Re: Public Notice: Copyright Violations >By posting communications on or through the Service, you automatically >grant Google a royalty-free, [...] >Im thinking similar stipulations apply no matter which vehicle you choose >to post to Usenet. Why on earth would that be so? Who would make these stipulations, and how would posters agree to them? Its conceivable that a Usenet provider (an ISP for example) might have a contract with its customers that allows them to do such things, but its hardly likely to allow Google to do them. And of course not all of us post through commercial providers. Weve had Usenet access for 20 years, and Im not aware of any such agreements with the sites that our news servers peer with. Furthermore, Google obtained the historical Usenet archives without any possibility of the original posters having agreed to it, since no-one considered the possibility at that time. -- Richard -- Spam filter: to mail me from a .com/.net site, put my surname in the headers. FreeBSD rules! === Subject: Re: Public Notice: Copyright Violations >Actually, complaining about copying toilet paper patterns is a way to >start some good lawsuits. For example, take Roger Penroses lawsuit >against the Kimberly-Clark corporation, for using a Penrose tiling on >their Kleenex Quilted toilet paper. Pentaplex, a company that licenses >Penroses patterns, is also a plaintiff. A spokesman for Pentaplex >says, ...when it comes to the population of Great Britain being >invited by a multi-national to wipe their bottoms on what appears to be >the work of a Knight of the Realm without his permission, then a last >stand must be made. Think how much more inappropriate it would have been for a Penrose tiling to have been used for the outer layer of a sanitary napkin, though. Lee Rudolph === Subject: Re: Public Notice: Copyright Violations > Let me make this simple for you. Complaining about copying usenet postings > is like complaining about copying toilet paper patterns. Actually, complaining about copying toilet paper patterns is a way to start some good lawsuits. For example, take Roger Penroses lawsuit against the Kimberly-Clark corporation, for using a Penrose tiling on their Kleenex Quilted toilet paper. Pentaplex, a company that licenses Penroses patterns, is also a plaintiff. A spokesman for Pentaplex says, ...when it comes to the population of Great Britain being invited by a multi-national to wipe their bottoms on what appears to be the work of a Knight of the Realm without his permission, then a last stand must be made. Googling for Penrose and toilet paper, brings up among many other interesting links. Is this a frivolous lawsuit? Well, lets just say that the mighty toilet paper empire was threatened enough to change the design of their Kleenex Quilted toilet paper. Ive stepped into this discussion unawares, but I think my example is probably not what Richard Henry was thinking. I believe the purpose of his comment was to indicate that copying toilet paper patterns is perfectly fine. Thus, I couldnt resist pointint out this Penrose toilet paper incident. While I realize posting about all sorts of craziness is the norm for sci.math, I couldnt make this same assumption about all those other groups, so Ive eliminated them from the header, even though my post can be said to be more math-related than the previous ones in the discussion ;-) === Subject: Re: Public Notice: Copyright Violations <67w_b.121$h23.35@fed1read06> <878yiradzk.fsf@phiwumbda.org> Discussion, linux) >> Well, it is as relevant to the discussion as your simple >> declaration. As Wade said, no one was complaining about copying >> Usenet postings. > Sorry, but youre wrong. He was complaining about copying Usenet postings > to a website. Yes, youre right. He was. Though it is still not at all obvious whether posting to Usenet gives an implicit right for others to present your posts on webpages -- especially webpages that are not part of Usenet propagation or archiving. So, I take it back. Richards comment, Complaining about copying usenet postings is like complaining about copying toilet paper patterns, is not irrelevant. It is relevant, but depending on what the simile means, it is likely false. -- Jesse F. Hughes C is for Cookie. Thats good enough for me. Cookie Monsters === Subject: Re: Public Notice: Copyright Violations message >> How can I make this simple for you? He mentioned websites. Websites. >> Websites. Can you read? Websites are not Usenet. So whatever may be > common this >> simple point escapes you escapes me. > Let me make this simple for you. Complaining about copying usenet postings > is like complaining about copying toilet paper patterns. > Let me make this simple for you. If you dont regularly change your > fuel filter, then your engine performance will decline. > Well, it is as relevant to the discussion as your simple > declaration. As Wade said, no one was complaining about copying > Usenet postings. Sorry, but youre wrong. He was complaining about copying Usenet postings to a website. KeithK > I am not sure what the fuel filter-changing habits are for the readers > of sci.math, but it is likely that a few readers havent given the > matter any thought in some time. Therefore, while my advice was no > more relevant than yours, it might have practical benefits for the > readers. > -- > Jesse Hughes > By definition m is a variable. By definition all then (sic) numbers > represented by letters are variables--thats algebra[,] Magidin. > -- James Harris shows deep understanding of algebra === Subject: Re: Public Notice: Copyright Violations >> I believe Google has been granted the legal right to archive Usenet >> Granted by whom? There are millions of copyright-owning authors on >usenet. >> Suppose you post through Google. Then your posts are subject to the >> following: >> By posting communications on or through the Service, you automatically >> grant Google a royalty-free, perpetual, irrevocable, non-exclusive >license >> to use, reproduce, modify, publish, edit, translate, distribute, >perform, >> and display the communication alone or as part of other works in any >form, >> media, or technology whether now known or hereafter developed, and to >> sublicense such rights through multiple tiers of sublicensees. >> It seems likely that similar stipulations apply to all Usenet posts. If >> true, the millions of copyright-owning authors on Usenet have granted >> licenses to specific parties to use their work - basically as those >parties >> see fit. >It seems likely? >>I can offer only one data point: I for one have never signed any >>contract containing terms remotely like the above. >Now read your Misoft licensing agreements. I dont think the validity of shrink-wrap licenses has been tested in this jurisdiction. > One no longer has >to sign to cede intellectual property. Another magic >incantation to make things go poof is EULA (I think Ive >spelled it correctly). Indeed. -- Richard Herring === Subject: Re: Public Notice: Copyright Violations <9OW_b.1170$h23.739@fed1read06> <45D+nCZTNIPAFwmU@baesystems.comIn message , Richard Henry message > I believe Google has been granted the legal right to archive Usenet Granted by whom? There are millions of copyright-owning authors on >>usenet. > Suppose you post through Google. Then your posts are subject to the > following: > By posting communications on or through the Service, you automatically > grant Google a royalty-free, perpetual, irrevocable, non-exclusive license > to use, reproduce, modify, publish, edit, translate, distribute, perform, > and display the communication alone or as part of other works in any form, > media, or technology whether now known or hereafter developed, and to > sublicense such rights through multiple tiers of sublicensees. > It seems likely that similar stipulations apply to all Usenet posts. If > true, the millions of copyright-owning authors on Usenet have granted > licenses to specific parties to use their work - basically as those >>parties > see fit. >>It seems likely? >I can offer only one data point: I for one have never signed any >contract containing terms remotely like the above. Now read your Misoft licensing agreements. One no longer has to sign to cede intellectual property. Another magic incantation to make things go poof is EULA (I think Ive spelled it correctly). /BAH Subtract a hundred and four for e-mail. === Subject: Re: Public Notice: Copyright Violations <9OW_b.1170$h23.739@fed1read06> In message , Richard Henry >> I believe Google has been granted the legal right to archive Usenet >> Granted by whom? There are millions of copyright-owning authors on >usenet. >> Suppose you post through Google. Then your posts are subject to the >> following: >> By posting communications on or through the Service, you automatically >> grant Google a royalty-free, perpetual, irrevocable, non-exclusive license >> to use, reproduce, modify, publish, edit, translate, distribute, perform, >> and display the communication alone or as part of other works in any form, >> media, or technology whether now known or hereafter developed, and to >> sublicense such rights through multiple tiers of sublicensees. >> It seems likely that similar stipulations apply to all Usenet posts. If >> true, the millions of copyright-owning authors on Usenet have granted >> licenses to specific parties to use their work - basically as those >parties >> see fit. >It seems likely? I can offer only one data point: I for one have never signed any contract containing terms remotely like the above. -- Richard Herring === Subject: Re: Public Notice: Copyright Violations Discussion, linux) >> The fact that you have no idea what constitutes fair use says a lot >> here. >> And this fact is based on what exactly? >> You might start with the US Code, since it seems to apply in this case. >> http://www.copyright.gov/title17/92chap1.html#107 > Once again you appear to have reading difficulty. The alleged fact is > that JSH has no idea of what constitutes fair use. I doubt that the US > Code has commented on this allegation. It was in a footnote. The US Code gives an explicit (but partial) list of persons with no clue of fair use rights. James S. Harris appears in the footnote, three lines above Jesse F. Hughes. -- [R]eality has a fascinating ability to check us when we get a little too big for our britches... Make no mistake. There isnt a mathematician alive today that I cant now touch, and not a mathematical career on the planet that I cant now affect. --James Harris, render of worlds === Subject: Re: Public Notice: Copyright Violations <871xojadfg.fsf@phiwumbda.org> Discussion, linux) > Well I admit its just a theory. Its based on the idea that when you post > to Usenet through Google, your posts are subject to the following: > By posting communications on or through the Service, you automatically > grant Google a royalty-free, perpetual, irrevocable, non-exclusive license > to use, reproduce, modify, publish, edit, translate, distribute, perform, > and display the communication alone or as part of other works in any form, > media, or technology whether now known or hereafter developed, and to > sublicense such rights through multiple tiers of sublicensees. I think that, without doubt, Google users have given Google the right you describe. > Im thinking similar stipulations apply no matter which vehicle you choose > to post to Usenet. Then Im thinking these vehicles have granted Google > the right to archive. If you go to > youll see > ----- > Google Groups - Terms and Conditions > ... > The contents within the Service are protected by copyright and other laws > in both the United States and elsewhere. The Service includes both content > owned or controlled by the Google as well as content owned or controlled by > third parties and licensed to Google (collectively, the Materials). > Google authorizes you to view and download a single copy of the Materials > solely for your personal, non-commercial use. You may not sell or modify > the Materials or reproduce, display, publicly perform, distribute, or > otherwise use the Materials in any way for any public or commercial purpose > without the written permission of Google. Special rules may apply to the > use of certain software and other items provided via the Services, and are > noted where appropriate. > ---------- > Google certainly makes it sound like they have a legal right to the > archived posts. Perhaps its all bluff. So my thought was Google has been > granted licenses by lots of third parties (ISPs and other Usenet > providers), who were granted licenses by the original posters in the manner > given above. But I admit its just a theory. I doubt that the various third parties have the ability to grant Google the right to archive and serve my posts. If Google has such a right, then it would surely have to come implicitly from the poster, who grants the right by posting. But if that argument holds water, then its unclear why Google would implicitly have the right to archive and present Usenet posts and, say, Dik Decker or Rick Winter or whoever wouldnt have the same right (actually, Dik and Rick dont quote Usenet posts, I think). Well, as Barbie *should* have said, Copyright law is hard. I havent any clue what rights one implicitly grants when posting to Usenet and what he doesnt. The situation would certainly be clearer if the X-no-archive header were replaced by an X-archive header (but then Googles archives would fit on a ßoppy). Dot sig serendipity strikes again. The quote below was selected randomly. -- Sale or rental of this disc is ILLEGAL. If you have rented or purchased this disc, please call the MPAA at 1-800-NO-COPYS. -- The MPAA begins a new anti-piracy program, found on a DVD purchased in China === Subject: Re: Public Notice: Copyright Violations > I believe Google has been granted the legal right to archive Usenet > Really? Google has been granted this right by whom? When? > If anything, Id guess that Google is relying on a presumed right to > archive, but I dont know that there is any such right explicit in the > law. I dont think Google has any more or less right to archive > Usenet postings or web sites than anyone else. Some other archivist > could do just what Google does with Usenet now (and as DejaNews did > before Google). > As far as I know, the Wayback Machine is on similarly shaky footing. > (As is Googles caching and serving of web pages.) > I dont think the legal status of these practices has been settled all > that much. But, again, I dont know diddly here. If you have some > good references, then Id be happy to see them. Popular journalism is > preferable to technical legal documents for my addled brane. Well I admit its just a theory. Its based on the idea that when you post to Usenet through Google, your posts are subject to the following: By posting communications on or through the Service, you automatically grant Google a royalty-free, perpetual, irrevocable, non-exclusive license to use, reproduce, modify, publish, edit, translate, distribute, perform, and display the communication alone or as part of other works in any form, media, or technology whether now known or hereafter developed, and to sublicense such rights through multiple tiers of sublicensees. Im thinking similar stipulations apply no matter which vehicle you choose to post to Usenet. Then Im thinking these vehicles have granted Google the right to archive. If you go to youll see ----- Google Groups - Terms and Conditions ... The contents within the Service are protected by copyright and other laws in both the United States and elsewhere. The Service includes both content owned or controlled by the Google as well as content owned or controlled by third parties and licensed to Google (collectively, the Materials). Google authorizes you to view and download a single copy of the Materials solely for your personal, non-commercial use. You may not sell or modify the Materials or reproduce, display, publicly perform, distribute, or otherwise use the Materials in any way for any public or commercial purpose without the written permission of Google. Special rules may apply to the use of certain software and other items provided via the Services, and are noted where appropriate. ---------- Google certainly makes it sound like they have a legal right to the archived posts. Perhaps its all bluff. So my thought was Google has been granted licenses by lots of third parties (ISPs and other Usenet providers), who were granted licenses by the original posters in the manner given above. But I admit its just a theory. === Subject: Re: Public Notice: Copyright Violations Discussion, linux) > I believe Google has been granted the legal right to archive Usenet Really? Google has been granted this right by whom? When? If anything, Id guess that Google is relying on a presumed right to archive, but I dont know that there is any such right explicit in the law. I dont think Google has any more or less right to archive Usenet postings or web sites than anyone else. Some other archivist could do just what Google does with Usenet now (and as DejaNews did before Google). As far as I know, the Wayback Machine is on similarly shaky footing. (As is Googles caching and serving of web pages.) I dont think the legal status of these practices has been settled all that much. But, again, I dont know diddly here. If you have some good references, then Id be happy to see them. Popular journalism is preferable to technical legal documents for my addled brane. -- You lack the ability to reason, but instead get an idea in your head and hold on to it against all evidence. I dont find you credible, and reject your claims, as coming from a ßawed source. === Subject: Re: Public Notice: Copyright Violations Discussion, linux) >> Ive noticed people with webpages extensively quoting from posts Ive >> made and am now giving public notice that there is no right to copy >> extensively, that is, to go beyond fair use, from my writings without >> my express written permission. >> The fact that you have no idea what constitutes fair use says a lot >> here. > And this fact is based on what exactly? Well, who knows what page James is complaining about? I didnt see much quoting (none?) on Ricks page, so it must be Diks page at . But, Diks page doesnt contain Jamess proof. It contains an altered proof of a different statement (a false statement). True, it does contain rather a lot of Jamess words, but it looks to me like a legitimate criticism of Jamess writing. Obviously, my opinion on these matters is worth squat, but just as obviously, Ill give it anyway. Diks use of Jamess writing looks like fair use to me. It doesnt look excessive. In fact, it looks like a page or maybe a bit more, edited and with commentary. Maybe theres some other page JSH has in mind, but he doesnt usually play his cards close to his chest. -- Jesse F. Hughes Ultimately, I can bring the entire mathematical establishment to its knees... Live in a fantasy world if you wish, but to me thats just an expression of your intellectual inferiority. --James Harris === Subject: Re: Public Notice: Copyright Violations <67w_b.121$h23.35@fed1read06> Discussion, linux) >> How can I make this simple for you? He mentioned websites. Websites. >> Websites. Can you read? Websites are not Usenet. So whatever may be > common >> simple point escapes you escapes me. > Let me make this simple for you. Complaining about copying usenet postings > is like complaining about copying toilet paper patterns. Let me make this simple for you. If you dont regularly change your fuel filter, then your engine performance will decline. Well, it is as relevant to the discussion as your simple declaration. As Wade said, no one was complaining about copying Usenet postings. I am not sure what the fuel filter-changing habits are for the readers of sci.math, but it is likely that a few readers havent given the matter any thought in some time. Therefore, while my advice was no more relevant than yours, it might have practical benefits for the readers. -- Jesse Hughes By definition m is a variable. By definition all then (sic) numbers represented by letters are variables--thats algebra[,] Magidin. -- James Harris shows deep understanding of algebra === Subject: Re: Public Notice: Copyright Violations >> Im not a copyright lawyer, but I know one. Heres some info you >> mgiht find enlightening... >> ...[F]air use is a highly fact-specific determination. Copyright >> Office document FL102 puts it this way: The distinction between >> Ôfair use and infringement may be unclear and not easily defined. >> There is no specific number of words, lines, or notes that may safely >> be taken without permission. Acknowledging the source of the >> copyrighted material does not substitute for obtaining permission. >> The document then quotes from the 1961 Report of the Register of >> Copyrights on the General Revision of the U.S. Copyright Law., >> providing the following examples of activities that courts have held >> to be fair use: >> - Quotation of excerpts in a review or criticism for purposes of >> illustration or comment; >> - Quotation of short passages in a scholarly or technical work >> for illustration or clarification of the authors observations; >> - Use in a parody of some of the content of the work parodied; >> news report; >> Document FL102 is included in Copyright Office information kit 102 >> (Fair Use), which can be ordered from the Copyright Office. >> Proponents of the implied license idea point out that Usenet postings >> are routinely copied and quoted, and anyone posting to Usenet is >> granting an implied license for others to similarly copy or quote >> that posting, too. Its not clear whether such implied license >> extends beyond Usenet, or indeed, what Usenet really means (does it >> include, for example, Internet mailing lists? Does it include >> netnews on CD- ROM?). If a posting includes an express limitation on >> the right to copy or quote, its not at all certain whether the >> express limitation or the implied license will control. No doubt it >> depends on the specific facts. For example, was the limitation >> clearly visible to the person who did the copying? Was the >> limitation placed such that it would be visible only after the person >> who did the copying invested time and money to get the posting, >> believing it to be without any limitation? >> These theories are largely speculative, because there has been little >> litigation to test them in the courts. As a practical matter, most >> postings, with a small number of notable exceptions, are not >> registered with the Copyright Office. As such, to prevail in court, >> the copyright holder would need to show actual damages (see section >> 2.5). Since most of these cases will result in little or no actual >> damage, no cases have been be brought; its simply too expensive to >> sue for negligible damages. >> Just some food for thought. >> - Tim >> permission of the author. > Why not? Hey, thats cheating. - Tim -- Timothy M. Brauch Graduate Student Department of Mathematics Wake Forest University email is: news (dot) post (at) tbrauch (dot) com === Subject: Re: Public Notice: Copyright Violations > Uh, you do know that quoting passages is fair use, presumably. And > that there are thousands of news servers copying your post without > expressed written permission? Nooooooooooooooooooo! - Tim -- Timothy M. Brauch Graduate Student Department of Mathematics Wake Forest University email is: news (dot) post (at) tbrauch (dot) com === Subject: Re: Public Notice: Copyright Violations Discussion, linux) > How can you tell whether something is going to be emvarassing > years later? Well in _your_ case its very simple: dont make any > posts on mathematical topics! This is hardly sufficient. Consider his recent foray into international copyright law. -- A set having three members is a single thing wholly constituted by its members but distinct from them. After this, the theological doctrine of the Trinity as Ôthree in one should be childs play. --Max Black, _Caveats and Critiques_ === Subject: Re: Public Notice: Copyright Violations Discussion, linux) > Google has a _web page_ where one can find _complete_ quotes of > _all_ your posts. Holy cow! I just checked! Google *does* have pages and pages devoted to old and discarded arguments of James going back *years*. They *maintain* these pages, even though James has moved on. Man, are they in for a thrashing. -- This page contains information of a type (text/html) that can only be viewed with the appropriate Plug-in. Click OK to download Plugin. --- Netscape 4.7 error message === Subject: Re: Public Notice: Copyright Violations > People putting up webpages exceeding fair usage are in violation of > international copyright law and may face prosecution to the full > extent of those laws. > I dont think that copyright infringement is a criminal matter. Good > luck with that prosecution. In order to be awarded damages in a civil suit, JSH may have to show that he has sustained damages. But a countersuit using the Google record of both sides postings might end up with JSH having to pay damages instead. No sane lawyer would take such a case on a contingent fee basis. The danger of losing is too great, and the damages that could be extracted from anyone earning only a professors pay isnt going to pay for the lawyers time, even of JSHs case were rock solid. And if JSH is still not working, he is unlikely to have enough to hire a lawyer. It would appear that JSHs threats of legal action are no more solid that his claims of mathematical discovery. === Subject: Re: Public Notice: Copyright Violations Discussion, linux) > Its always at least a civil matter (a tort). 17 U.S.C. 501(b) details > the mechanisms by which an owner of a copyright may file a civil suit, > and 28 U.S.C. 1338 expressly refers to civil actions arising under the > copyright act. > However, under certain circumstances, it may also be a federal crime. A > copyright infringement is subject to criminal prosecution if infringement > is willful and for purposes of commercial advantage or private financial > gain. 17 U.S.C. 506(a). If the offense consists of the reproduction or > distribution, during any 180-day period, of 10 or more copies having a > retail value of more than $2,500, the offense is a felony; otherwise, the > offense is a misdemeanor. 18 U.S.C. 2319. > As a side note, although 18 U.S.C. 2319 purports to prescribe the > penalties for criminal infringement, all crimes covered by Title 18 have > their penalties determined by the U.S. Sentencing Guidelines, another > part of Title 18. > permission of the author. Uh, you do know that quoting passages is fair use, presumably. And that there are thousands of news servers copying your post without expressed written permission? -- Analysis/ editors have evaluated the paper, they accepted it for publication and they have the copyright of its contents - and thus they are responsible for its correctness, she [said]. === Subject: Re: Public Notice: Copyright Violations > Im not a copyright lawyer, but I know one. Heres some info you mgiht > find enlightening... > ...[F]air use is a highly fact-specific determination. Copyright Office > document FL102 puts it this way: The distinction between Ôfair use and > infringement may be unclear and not easily defined. There is no > specific number of words, lines, or notes that may safely be taken > without permission. Acknowledging the source of the copyrighted > material does not substitute for obtaining permission. > The document then quotes from the 1961 Report of the Register of > Copyrights on the General Revision of the U.S. Copyright Law., providing > the following examples of activities that courts have held to be fair > use: > - Quotation of excerpts in a review or criticism for purposes of > illustration or comment; > - Quotation of short passages in a scholarly or technical work > for illustration or clarification of the authors observations; > - Use in a parody of some of the content of the work parodied; > news report; > Document FL102 is included in Copyright Office information kit 102 > (Fair Use), which can be ordered from the Copyright Office. > Proponents of the implied license idea point out that Usenet postings > are routinely copied and quoted, and anyone posting to Usenet is > granting an implied license for others to similarly copy or quote that > posting, too. Its not clear whether such implied license extends > beyond Usenet, or indeed, what Usenet really means (does it include, > for example, Internet mailing lists? Does it include netnews on CD- > ROM?). If a posting includes an express limitation on the right to copy > or quote, its not at all certain whether the express limitation or the > implied license will control. No doubt it depends on the specific > facts. For example, was the limitation clearly visible to the person > who did the copying? Was the limitation placed such that it would be > visible only after the person who did the copying invested time and > money to get the posting, believing it to be without any limitation? > These theories are largely speculative, because there has been little > litigation to test them in the courts. As a practical matter, most > postings, with a small number of notable exceptions, are not registered > with the Copyright Office. As such, to prevail in court, the copyright > holder would need to show actual damages (see section 2.5). Since most > of these cases will result in little or no actual damage, no cases have > been be brought; its simply too expensive to sue for negligible > damages. > Just some food for thought. > - Tim > permission of the author. Why not? === Subject: Re: Public Notice: Copyright Violations >> People putting up webpages exceeding fair usage are in violation of >> international copyright law and may face prosecution to the full >> extent of those laws. > I dont think that copyright infringement is a criminal matter. Good > luck with that prosecution. Its always at least a civil matter (a tort). 17 U.S.C. 501(b) details the mechanisms by which an owner of a copyright may file a civil suit, and 28 U.S.C. 1338 expressly refers to civil actions arising under the copyright act. However, under certain circumstances, it may also be a federal crime. A copyright infringement is subject to criminal prosecution if infringement is willful and for purposes of commercial advantage or private financial gain. 17 U.S.C. 506(a). If the offense consists of the reproduction or distribution, during any 180-day period, of 10 or more copies having a retail value of more than $2,500, the offense is a felony; otherwise, the offense is a misdemeanor. 18 U.S.C. 2319. As a side note, although 18 U.S.C. 2319 purports to prescribe the penalties for criminal infringement, all crimes covered by Title 18 have their penalties determined by the U.S. Sentencing Guidelines, another part of Title 18. - Tim permission of the author. -- Timothy M. Brauch Graduate Student Department of Mathematics Wake Forest University email is: news (dot) post (at) tbrauch (dot) com === Subject: Re: Public Notice: Copyright Violations Discussion, linux) > People putting up webpages exceeding fair usage are in violation of > international copyright law and may face prosecution to the full > extent of those laws. I dont think that copyright infringement is a criminal matter. Good luck with that prosecution. -- At the Microsoft-sponsored cocktail reception in the Galaxy Ballroom that evening, Robert Dees urges us Ôto network on behalf of the people of Iraq, -- Naomi Klein reports on Microsofts efforts to further democracy. === Subject: Re: Help proving a lemma > Given this situation: You have a circle tangent to a line. The points of > the midpoint of circle are given and the equation of the line is given. > Find the radius. > I found the radius by using the distance formula and substitution. > Good. > My teacher has given us this llama though: r = | Ax + By + C | / [ sqrt > (A^2 > + B^2) ] > To me this looks like *the* distance formula. I assume Ax+By+C=0 is the > equation for the line, and x,y is a point. Then r is the distance from the > point to the line, if I am not mistaken. > I wonder what distance formula you used... > -Michael. D = | Au + Bv + C | / [ sqrt(A^2+ B^2) ] is a fairly standard formula for the distance from point (u,v) to the line with equation Ax + By + C = 0. There are a variety of ways of validating the formula. Probably the easiest is in terms of vectors. Then vector [A, B] is a vector perpendicular to line Ax + By + C = 0, and if [x, y] is any point on the line one must have that [A,B]*[x,y] = -C, where * indicates dot product, and then the distance from [u,v] to the line is the magnitude of the projection of [u-x, v-y] perpendicular to the line (e.i., parallel to [A,B]. Filling in the details should give you a proof. === Subject: Re: Help proving a lemma > Given this situation: You have a circle tangent to a line. The points of > the midpoint of circle are given and the equation of the line is given. > Find the radius. > I found the radius by using the distance formula and substitution. Good. > My teacher has given us this llama though: r = | Ax + By + C | / [ sqrt (A^2 > + B^2) ] To me this looks like *the* distance formula. I assume Ax+By+C=0 is the equation for the line, and x,y is a point. Then r is the distance from the point to the line, if I am not mistaken. I wonder what distance formula you used... -Michael. === Subject: Re: Help proving a lemma Igor Y > Given this situation: You have a circle tangent to a line. The points of > the midpoint of circle are given and the equation of the line is given. > Find the radius. > I found the radius by using the distance formula and substitution. > My teacher has given us this llama though: r = | Ax + By + C | / [ sqrt (A^2 > + B^2) ] > I have to prove it. > I dont know where to start. Can someone show me how to prove such a lemma? If the line is Ax + By + C = 0 and the _center_ of the circle is given, say (u,v), then the point of tangency (s,t) will satisfy As + Bt + C = 0 B(s-u) - A(t-v) = 0 the distance from (u,v) to (s,t). That distance is the radius. (The formula for the radius will not involve x and y. Maybe your lemma is only part of the problem?) LH === Subject: Help proving a lemma Given this situation: You have a circle tangent to a line. The points of the midpoint of circle are given and the equation of the line is given. Find the radius. I found the radius by using the distance formula and substitution. My teacher has given us this llama though: r = | Ax + By + C | / [ sqrt (A^2 + B^2) ] I have to prove it. I dont know where to start. Can someone show me how to prove such a lemma? === Subject: Farey series= equivalent statement to Riemann Hypothesis Huh? In various recent popular mathematics books they write of the equivalent statement to RH that relates to the difference in the order of the fractions in the farey series and the ordered value of the fractions. Could someone translate this? Just what property is being tested? Wouldnt the order of the farey series be arbitrary depending on what direction(s) you want to start from. Eg up down left or right or even a tree spanning algorithm. Or a mix? Anyone? === Subject: Re: Fundamental Group of the circle > I am looking over a proof relating to this. > Let p : R -> S^1, p(s) = (cos 2 pi s, sin 2 pi s) > I = [0,1] R > They then state: > a) For each path f: I -> S^1 starting at a point x0 in S^1 and each ~x0 in > p^(-1)(x0) there is a unique lift ~f : I -> R starting at ~x0. > They then state: > c) Given a map F : Y x I -> S^1 and a map ~F : Y x {0} -> R lifting F|Y x > {0}, then there is a unique map ~F: Y x I -> R lifting F and restricting to > the given ~F on Y x {0}. > Finally they state that a) is the special case of c) where Y is a point. > Im having trouble seeing this relation. I think I see this. I will omit the Y domain, since a) only requires a single element. Then the path f:I->S^1,f(0)=x0 in S^1 from a) is an instance of the function F from c). ~F:Yx{0}->R can be a constant function since |Y|=1, so it is simply ~f:{0}->R, ~f(0) = ~x0 from a). Using p from a) we see that (p o ~f) : {0}->S^1 = f|{0}->S^1, so ~f is a lift of f|{0} as starting at ~x0 since we designed ~f(0) = ~x0 above. > Also, can I assume that lifting F and restricting to the given ~F on Y x > {0} means that the image of ~F: Y x I equals the image of ~F : Y x {0}? I think this means simply that ~F:YxI = ~F:Yx{0} on Yx{0}, but ~F:YxIs full image may be larger than ~F:Yx{0}s. l8r, Mike N. Christoff === Subject: Fundamental Group of the circle I am looking over a proof relating to this. Let p : R -> S^1, p(s) = (cos 2 pi s, sin 2 pi s) I = [0,1] R They then state: a) For each path f: I -> S^1 starting at a point x0 in S^1 and each ~x0 in p^(-1)(x0) there is a unique lift ~f : I -> R starting at ~x0. They then state: c) Given a map F : Y x I -> S^1 and a map ~F : Y x {0} -> R lifting F|Y x {0}, then there is a unique map ~F: Y x I -> R lifting F and restricting to the given ~F on Y x {0}. Finally they state that a) is the special case of c) where Y is a point. Im having trouble seeing this relation. Also, can I assume that lifting F and restricting to the given ~F on Y x {0} means that the image of ~F: Y x I equals the image of ~F : Y x {0}? l8r, Mike N. Christoff === Subject: Re: 6 Universali Domande Tranello Per Nascondere la Propria Ignoranza in Matematica > SIX ALL-PURPOSE TRICK QUESTIONS THAT WILL KEEP YOUR IGNORANCE FROM noi tutti siamo in guerra con chi cerca di sfruttarci non possiamo arrenderci senza lottare, la posta in gioco .8f troppo alta Ô j === Subject: Re: 6 Universali Domande Tranello Per Nascondere la Propria Ignoranza in Matematica ti guardo. ti assorbo. mi ipnotizzi. occhi senza parlpebre sempre fissi su di te. Ô j > ROTFL!!!! > keroppi === Subject: Re: 6 Universali Domande Tranello Per Nascondere la Propria Ignoranza in Matematica ROTFL!!!! keroppi ha scritto nel messaggio > SIX ALL-PURPOSE TRICK QUESTIONS THAT WILL KEEP YOUR IGNORANCE FROM > BEING EXPOSED IN TREACHEROUS MATHEMATICS DISCUSSION GROUPS > 1. Can that result be restated in terms of category theory? > 2. Isnt the constant in that equation suspiciously close to the > square root of pi divided by e cubed? > 3. Wasnt your theorem prefigured in the work of Euler? > 4. Same as above, but replace Euler with Poincar.8e. > 5. Is any of this applicable to linear p-adic groups, p-adic Galois > groups, or p-affine Schur algebras? > 6. Yes, but can you prove that lemma for the case of n=3? > -- > keroppi > www.geocities.com/rosacrux === Subject: Re: 6 Universali Domande Tranello Per Nascondere la Propria Ignoranza in Matematica Plonk, plonk! Scusate, avrei qualcosa da ridire anchio, dopo che keroppi ha scritto, al riguardo, le seguenti parole: > or p-affine Schur algebras? E lalgebra della sciura piu affine (cioe piu fine) che mi intriga. Assomiglia un po ai conti della serva appliocata alla finanza creativa. Ce un nesso tra le due teorie? === Subject: 6 Universali Domande Tranello Per Nascondere la Propria Ignoranza in Matematica SIX ALL-PURPOSE TRICK QUESTIONS THAT WILL KEEP YOUR IGNORANCE FROM BEING EXPOSED IN TREACHEROUS MATHEMATICS DISCUSSION GROUPS 1. Can that result be restated in terms of category theory? 2. Isnt the constant in that equation suspiciously close to the square root of pi divided by e cubed? 3. Wasnt your theorem prefigured in the work of Euler? 4. Same as above, but replace Euler with Poincar.8e. 5. Is any of this applicable to linear p-adic groups, p-adic Galois groups, or p-affine Schur algebras? 6. Yes, but can you prove that lemma for the case of n=3? -- keroppi www.geocities.com/rosacrux === Subject: Re: JSH: Apology to Ramsay, why I post |> So I apologize to Keith Ramsay for questioning his honesty here. |> suspect its because I just havent ever. |> Excuse me if I say a little about this before letting it go, though. |> Ive read that a compete canonical apology contains five parts. One, a |> statement of what was done wrong. Two, agreement that it was wrong. Three, |> a statement of the principle violated by it, i.e. why it was wrong. Four, |> an explanation of why it happened. Five, reason to believe it wont |> happen in the future. (Probably the main reason to say why one did it is |> to support ones explanation for why the other person doesnt need to |> worry about ones doing it again.) | |Are you sane? Yes. Do you see anything suggesting insanity in the above? Im not saying that every apology should be formal, by the way, if thats whats bothering you. Usually most of those components are implicit. But its not quite an apology if they arent at least implicitly there. It may seem like a paradox that a person would explain why they did something in order to help convince someone they wont do it again, but it makes perfect sense. Often the only way to be sure not to do something is to understand yourself well enough to know why you had a tendancy to do it in the first place. As far as I can see, however, you dont have this kind of self-knowledge when it comes to your accusing others of lies. Certainly Ill consider your latest accusation against me settled, but I dont have much confidence that you wouldnt do it again if similar circumstances arose. I think it usually seems quite clear to you that were lying to you when you make such an accusation. Youre just not looking at that squarely. You kid yourself. Keith Ramsay === Subject: Re: JSH: Apology to Ramsay, why I post >Are you sane? Would an insane person know that they werent sane? Hence isnt that a rather meaningless question? Unless of course it was merely rhetorical and you simple wish to be insulting. === Subject: Re: JSH: Apology to Ramsay, why I post === >Subject: Re: JSH: Apology to Ramsay, why I post serious error: thinking that because he is Special he can take a short >cut right to the finish line without having to learn or understand much >of anything. Im not saying this out of jealousy or desire to protect >the Establishment. Just look at his track record. He makes a lifetimes >worth of serious conceptual errors every month or so ... without even >trying. In the end, not one single realized accomplishment, although >I do think he has set some kind of high-water mark for being the most >incredibly successful and annoying jerk on the Internet up to the present >time. The Isaac Newton, the pioneering Daniel Boone of crank/trolls! >This alone might eventually get him a spot on Letterman. > Nora B. Top Ten Reasons Why James Harris Should Appear On Letterman 10. Mathematicians are evil. 9. There is an error in core. 8. Mathematicians are assholes. 7. No one has ever found a way to count primes using a partial difference equation. 6. Mathematicians are lapdogs. 5. It turns out that there *has* to be a ring beyond algebraic integers. 4. Mathematicians are liars. 3. His peers are Gauss and Dedekind. 2. Mathematicians are racists. 1. The chicks will dig him. -- Mensanator Ace of Clubs === Subject: Re: How big can a manifold be? > I read some of David Gualds paper and settled my conjectures about > 1-manifolds. I dont know if youre still interested in the original question, but I was, and Ive now managed to answer it: > Is there any limit to the cardinality of a connected n-manifold, that > is a connected Hausdorff space every point of which has a > neighborhood homeomorphic to R^n? Any such manifold has the cardinality of the continuum (assuming its not empty, and n is nonzero). More generally: Let k be a cardinal greater than 1. Let X be a connected sequential space such that every point of X has a neighbourhood of cardinality at most k^aleph_0, and each sequence in X has at most k^aleph_0 limits. Then |X| is at most k^aleph_0. (A sequential space is a topological space in which sequentially closed sets are closed. A sequentially closed set is a set D that contains all limits of sequences in D. Manifolds are first countable, and therefore sequential.) Sketch of proof: Put t = k^aleph_0. For each x in X, let U_x be an open neighbourhood of x of cardinality at most t. Assume X is noempty, and take some y in X. Define A_g for each ordinal g <= w_1 (where w_1 denotes the first uncountable ordinal) by transfinite recursion, as follows. Let A_0 = U_y. For each countable ordinal g, let B_g be the set of all limits of sequences in A_g, and define A_(g+1) to be the union of all U_x with x in B_g. For each limit ordinal g <= w_1, define A_g to be the union of all the A_h with h < g. Then show that A_w_1 is open, closed (because sequentially closed) and nonempty (as it contains y). So X = A_w_1, because X is connected. Then show that |A_w_1| <= t. === Subject: re:How big can a manifold be? I read some of David Gualds paper and settled my conjectures about 1-manifolds. ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: Re: How big can a manifold be? > >> I reduced this claim to the following set-theoretic statement, but >>cannot prove it from the first glance: >>Consider a linearly ordered by inclusion family of sets, all of them >>having the cardinality of the continuum. Then their union has the >>cardinality of the continuum as well. > > > Because you cant prove it at first glance. You have to > take a second glance and assume the Axiom of Choice is true. > Then the proof is more trivial than the equivalent intersection > of sets of the power of the continuum. Since in order to > prove that it is true you have to *assume* that there is another set, A, > which also has the power of the continuum, > and for every set T in the ordered sets: > > T is a subset of A. > I dont follow: > The original question may be interpreted as follows: We know thats it possible to interpret the question that way. Since the only thing Logicians are really useful for is interpreting the word cardinality any way they feel pre-disposed to interpret it as. And if you havent noticed, most of them interpret cardinality to mean ordinality. Which is where the word function comes from, not from mathematics. Since the only thing mathemticians do full-time is interpret the word function to mean functional. Since Newton invented the word functional and its simple. Its so simple that even British and Latin Philosophers can understand it. And my question isnt at all ambiguous. Given that Lambda Calculi aret derived from (<,+,*,=,0) over R. They derived from ([,U,^,P,@,.EQ.) over Z. > Given an ordinal lambda, and a functional set f with domain > lambda, alpha < beta < lambda => f(alpha) subset f(beta) > and f(alpha) is of the cardinality of the continuum for each > alpha, the union of all f(alpha) is of the cardinality of the > continuum. > wlog, we can assume that if alpha < beta, then f(alpha) is not > equal to f(beta). But that still allows lambda to be as big as > 2^c. But that observation is irrelevent to Set Theory though, since its only true because of philosophy, not science. Since its a priori true that R is sufficiently large to allow anything that is not R to be as large as R. === Subject: Re: How big can a manifold be? > Consider a linearly ordered by inclusion family of sets, all of them > having the cardinality of the continuum. Then their union has the > cardinality of the continuum as well. >> Unfortunately, this set-theoretic statement is false. Let k be any >> infinite cardinal. The ordinals of cardinality k are linearly ordered >> by inclusion, but their union has cardinality greater than k. >> That works if c is an ordinal (for example, if we assume either AC or CH). >> What if it isnt? > Of course, I was assuming AC (and various other axioms of ZFC too). > But I think you can do it in ZF, if you really want to. > Let b be the least ordinal of cardinality not less than or equal to c. > Then b is a cardinal. Let R be the reals. Consider the set > X = { R cup a | a in b }. The elements of X are linearly ordered by > inclusion, and all have cardinality c. The union of X is a superset of > b. So cardinality of the union of X is greater than or equal to b, and > therefore not equal to c. > I dont know what Im doing without AC, so I could easily have made a > mistake, but it looks OK. It was much easier in ZFC. -- Dave Seaman Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: How big can a manifold be? >> I reduced this claim to the following set-theoretic statement, but >>cannot prove it from the first glance: >>Consider a linearly ordered by inclusion family of sets, all of them >>having the cardinality of the continuum. Then their union has the >>cardinality of the continuum as well. > Because you cant prove it at first glance. You have to > take a second glance and assume the Axiom of Choice is true. > Then the proof is more trivial than the equivalent intersection > of sets of the power of the continuum. Since in order to > prove that it is true you have to *assume* that there is another set, A, > which also has the power of the continuum, > and for every set T in the ordered sets: > T is a subset of A. I dont follow: The original question may be interpreted as follows: Given an ordinal lambda, and a functional set f with domain lambda, alpha < beta < lambda => f(alpha) subset f(beta) and f(alpha) is of the cardinality of the continuum for each alpha, the union of all f(alpha) is of the cardinality of the continuum. wlog, we can assume that if alpha < beta, then f(alpha) is not equal to f(beta). But that still allows lambda to be as big as 2^c. === Subject: Re: How big can a manifold be? >> Consider a linearly ordered by inclusion family of sets, all of them >> having the cardinality of the continuum. Then their union has the >> cardinality of the continuum as well. > Unfortunately, this set-theoretic statement is false. Let k be any > infinite cardinal. The ordinals of cardinality k are linearly ordered > by inclusion, but their union has cardinality greater than k. > That works if c is an ordinal (for example, if we assume either AC or CH). > What if it isnt? Of course, I was assuming AC (and various other axioms of ZFC too). But I think you can do it in ZF, if you really want to. Let b be the least ordinal of cardinality not less than or equal to c. Then b is a cardinal. Let R be the reals. Consider the set X = { R cup a | a in b }. The elements of X are linearly ordered by inclusion, and all have cardinality c. The union of X is a superset of b. So cardinality of the union of X is greater than or equal to b, and therefore not equal to c. I dont know what Im doing without AC, so I could easily have made a mistake, but it looks OK. It was much easier in ZFC. === Subject: Re: How big can a manifold be? > > Is there any limit to the cardinality of a connected n-manifold, > that is a connected Hausdorff space every point of which has a > neighborhood homeomorphic to R^n? > > and later: > > Im mainly interested in 1-dimentional manifolds anyway. > > For 1-manifolds, the answer to your question appears to be that the > cardinality of the continuum is the limit. > I reduced this claim to the following set-theoretic statement, but > cannot prove it from the first glance: > Consider a linearly ordered by inclusion family of sets, all of them > having the cardinality of the continuum. Then their union has the > cardinality of the continuum as well. Because you cant prove it at first glance. You have to take a second glance and assume the Axiom of Choice is true. Then the proof is more trivial than the equivalent intersection of sets of the power of the continuum. Since in order to prove that it is true you have to *assume* that there is another set, A, which also has the power of the continuum, and for every set T in the ordered sets: T is a subset of A. > Simeon === Subject: Re: How big can a manifold be? >> >> Is there any limit to the cardinality of a connected n-manifold, >> that is a connected Hausdorff space every point of which has a >> neighborhood homeomorphic to R^n? >> >> and later: >> >> Im mainly interested in 1-dimentional manifolds anyway. >> >> For 1-manifolds, the answer to your question appears to be that the >> cardinality of the continuum is the limit. >> I reduced this claim to the following set-theoretic statement, but >> cannot prove it from the first glance: >> Consider a linearly ordered by inclusion family of sets, all of them >> having the cardinality of the continuum. Then their union has the >> cardinality of the continuum as well. > Unfortunately, this set-theoretic statement is false. Let k be any > infinite cardinal. The ordinals of cardinality k are linearly ordered > by inclusion, but their union has cardinality greater than k. That works if c is an ordinal (for example, if we assume either AC or CH). What if it isnt? -- Dave Seaman Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: How big can a manifold be? > > Is there any limit to the cardinality of a connected n-manifold, > that is a connected Hausdorff space every point of which has a > neighborhood homeomorphic to R^n? > > and later: > > Im mainly interested in 1-dimentional manifolds anyway. > > For 1-manifolds, the answer to your question appears to be that the > cardinality of the continuum is the limit. > I reduced this claim to the following set-theoretic statement, but > cannot prove it from the first glance: > Consider a linearly ordered by inclusion family of sets, all of them > having the cardinality of the continuum. Then their union has the > cardinality of the continuum as well. Unfortunately, this set-theoretic statement is false. Let k be any infinite cardinal. The ordinals of cardinality k are linearly ordered by inclusion, but their union has cardinality greater than k. === Subject: Re: How big can a manifold be? > Is there any limit to the cardinality of a connected n-manifold, > that is a connected Hausdorff space every point of which has a > neighborhood homeomorphic to R^n? > and later: > Im mainly interested in 1-dimentional manifolds anyway. > For 1-manifolds, the answer to your question appears to be that the > cardinality of the continuum is the limit. I reduced this claim to the following set-theoretic statement, but cannot prove it from the first glance: Consider a linearly ordered by inclusion family of sets, all of them having the cardinality of the continuum. Then their union has the cardinality of the continuum as well. Simeon === Subject: Re: How big can a manifold be? > Is there any limit to the cardinality of a connected n-manifold, > that is a connected Hausdorff space every point of which has a > neighborhood homeomorphic to R^n? and later: > Im mainly interested in 1-dimentional manifolds anyway. For 1-manifolds, the answer to your question appears to be that the cardinality of the continuum is the limit. David Gauld has a paper on his website http://www.math.auckland.ac.nz/~gauld/ giving 88 equivalent conditions for a connected n-manifold to be metrizable. In it, he mentions in passing that there are only four (nonempty) connected 1-manifolds: the circle, the line, the long ray (by which he means what is usually called the long line), and the long line (long in both directions). He doesnt give a reference for this result, but some of the references at the end may be of help. === Subject: Re: derivative in the sense of distributions >> Is it true that a function in L^2 is always differentiable in the sense of >> distributions? In other words, for any f in L^2, does it exists a function >> g in L^1_loc such that: >> - = for all v in C_0^infty >As David pointed out, every L^2 function (like every distribution) has a >distibutional derivative, but the latter need not be in L^1 locally. For >example, let f be the characteristic function of [0,1]. Then - = >v(0) - v(1). So the distributional derivative of f is the singular measure >delta_0 - delta_1, and theres no way this is going to be given by a g in >L^1_loc. Or more generally, if the distribution derivative is in L^1_loc then f must be continuous (or f = g ae where g is continuous), in fact locally absolutely continuous. ************************ David C. Ullrich === Subject: Re: derivative in the sense of distributions > Is it true that a function in L^2 is always differentiable in the sense of > distributions? In other words, for any f in L^2, does it exists a function > g in L^1_loc such that: > - = for all v in C_0^infty As David pointed out, every L^2 function (like every distribution) has a distibutional derivative, but the latter need not be in L^1 locally. For example, let f be the characteristic function of [0,1]. Then - = v(0) - v(1). So the distributional derivative of f is the singular measure delta_0 - delta_1, and theres no way this is going to be given by a g in L^1_loc. === Subject: Re: derivative in the sense of distributions >Is it true that a function in L^2 is always differentiable in the sense of >distributions? Yes. >In other words, for any f in L^2, does it exists a function >g in L^1_loc such that: > - = for all v in C_0^infty No. Your in other words is wrong. Any element of L^2 has a derivative in the sense of distribiutions, but this derivative need _not_ be a function in L^1_loc. >I could find it in any book. Is there a good reference? _Any_ reference developing the theory of distributions from a mathematical point of view contains the following two facts: (i) any L^2 function defines a distribution (ii) every distribution has a derivative, in the sense of distributions. (On the other hand, youre not going to find a proof that the derivative of an L^2 function must be in L^1_loc in most books, because its false.) >Diogo ************************ David C. Ullrich === Subject: derivative in the sense of distributions Is it true that a function in L^2 is always differentiable in the sense of distributions? In other words, for any f in L^2, does it exists a function g in L^1_loc such that: - = for all v in C_0^infty I could find it in any book. Is there a good reference? Diogo === Subject: Re: Cantor: ignorant, harmless fool or intentional liar? Discussion, linux) > Is safe to assume that this poster is just pulling our leg? Duh. Whats remarkable is the number of respondents who point out (repeatedly) that Nathan the Great should be considerably older than 11 by now. Personally, I always find a joke much funnier when its been patiently explained to me. Repeatedly. -- you, but just remember, Im the guy who proved Fermats Last Theorem in just a bit over 6 years [...] My standards are kind of high. --James Harris, founding a new mathematical school === Subject: Re: Cantor: ignorant, harmless fool or intentional liar? Considering this past posts: And particulary this one: Is safe to assume that this poster is just pulling our leg? Arraitz ----------------------------------------------- Thats free enterprise, friends: freedom to gamble, freedom to lose. And the great thing -- the truly democratic thing about it -- is that you dont even have to be a player to lose. -Barbara Ehrenreich > > I have to confess, this is homework, so I am not looking for a full > answer, just some hints where to look. > Im taking a math history class and the assignment is to write a paper > on whether Cantor was just misinformed and ultimately innocent despite > the harm he did mathematics... or whether he intentionally strove to > murder mathematics. > So far Ive read into his biography and I am leaning toward the 2nd > case, since when he was young he was sexually abused by a math > teacher. That leads me to believe he had a big grudge against > mathematics in general. > If I adopt that route, one of my main arguments in this paper will be > that it would be impossible for anyone to be SO wrong about infinite > numbers purely be coincidence. I mean, hed have to be really tripping > on some hard stuff if were to believe he REALLY believed the nonsense > he published about infinite numbers. > > Im sure most of you have written a paper to this effect some time or > other in one of your mathematics history classes :-) > > Nathaniel Deeth > Age 11 > Math Genious > Welcome back, Nathaniel. Youve been an 11-year-old math genius for > some years now, so I presume that you were born on Feb 29. I am too > lazy (see previous posts) to figure out when you were born using this > information. > The truth is that Cantor intentionally strove to murder mathematics. > He had been laughed at by mathematicians because, even as a child, he > had a lousy singing voice in spite of the fact that his last name was > Cantor. This rankled all the more as he was not Jewish. His family > had converted some years before. > He was also utterly incensed by the fools who insisted that .999999... > was acually equal to 1.0, a pseudo-fact that any fool can see is > false. He determined to go them one better. He started out trying to > show that it was not possible to represent every number as a sum of at > most n cubes (for any n), but he found the effort wearing, so he moved > on to other pursuits. He published several proofs of Goldbachs > conjecture and Fermats last theorem using only the mathematics of > Fermats and Goldbachs day, but these efforts were met with scorn and > derision. Finally he decided to get even. He invented a confusing > and mystical doctrine of the infinite that has devolved into chaos and > anarchy, just as had planned. He spent the remainder of his days > addicted to zornication, and he finally died insane as a result. > Achava === Subject: Re: Cantor: ignorant, harmless fool or intentional liar? > > Nathaniel Deeth > Age 11 > Math Genious > If you are really a math genius, you would know that 11+5 = 16. Whatever capability Deeth may declare himself to have in mathematics, he wisely does not claim for his spelling. === Subject: Re: Cantor: ignorant, harmless fool or intentional liar? > Nathaniel Deeth > Age 11 > Math Genious If you are really a math genius, you would know that 11+5 = 16. === Subject: Re: Cantor: ignorant, harmless fool or intentional liar? > I have to confess, this is homework, so I am not looking for a full > answer, just some hints where to look. > Im taking a math history class and the assignment is to write a paper > on whether Cantor was just misinformed and ultimately innocent despite > the harm he did mathematics... or whether he intentionally strove to > murder mathematics. > So far Ive read into his biography and I am leaning toward the 2nd > case, since when he was young he was sexually abused by a math > teacher. That leads me to believe he had a big grudge against > mathematics in general. > If I adopt that route, one of my main arguments in this paper will be > that it would be impossible for anyone to be SO wrong about infinite > numbers purely be coincidence. I mean, hed have to be really tripping > on some hard stuff if were to believe he REALLY believed the nonsense > he published about infinite numbers. > Im sure most of you have written a paper to this effect some time or > other in one of your mathematics history classes :-) > Nathaniel Deeth > Age 11 > Math Genious Welcome back, Nathaniel. Youve been an 11-year-old math genius for some years now, so I presume that you were born on Feb 29. I am too lazy (see previous posts) to figure out when you were born using this information. The truth is that Cantor intentionally strove to murder mathematics. He had been laughed at by mathematicians because, even as a child, he had a lousy singing voice in spite of the fact that his last name was Cantor. This rankled all the more as he was not Jewish. His family had converted some years before. He was also utterly incensed by the fools who insisted that .999999... was acually equal to 1.0, a pseudo-fact that any fool can see is false. He determined to go them one better. He started out trying to show that it was not possible to represent every number as a sum of at most n cubes (for any n), but he found the effort wearing, so he moved on to other pursuits. He published several proofs of Goldbachs conjecture and Fermats last theorem using only the mathematics of Fermats and Goldbachs day, but these efforts were met with scorn and derision. Finally he decided to get even. He invented a confusing and mystical doctrine of the infinite that has devolved into chaos and anarchy, just as had planned. He spent the remainder of his days addicted to zornication, and he finally died insane as a result. Achava === Subject: Re: Cantor: ignorant, harmless fool or intentional liar? >I conclude that the teacher is a son of a bitch. And I conclude that the teacher does not even exist. --- === Subject: re:Cantor: ignorant, harmless fool or intentional liar? I conclude that the teacher is a son of a bitch. ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: Re: Cantor: ignorant, harmless fool or intentional liar? Discussion, linux) > > I have to confess, this is homework, so I am not looking for a full > answer, just some hints where to look. > Im taking a math history class and the assignment is to write a paper > on whether Cantor was just misinformed and ultimately innocent despite > the harm he did mathematics... or whether he intentionally strove to > murder mathematics. > So far Ive read into his biography and I am leaning toward the 2nd > case, since when he was young he was sexually abused by a math > teacher. That leads me to believe he had a big grudge against > mathematics in general. > If I adopt that route, one of my main arguments in this paper will be > that it would be impossible for anyone to be SO wrong about infinite > numbers purely be coincidence. I mean, hed have to be really tripping > on some hard stuff if were to believe he REALLY believed the nonsense > he published about infinite numbers. > > Im sure most of you have written a paper to this effect some time or > other in one of your mathematics history classes :-) > > Nathaniel Deeth > Age 11 > Math Genious >>Nathaniel, >>This is supposed to be done on your own. You are not allowed to ask >>anyone else for help, as this is cheating. >>Your teacher, >>Dr. Ben Zona > And Craig Feinstein, or Dr. Ben Zona, proves that he is no genuine > mathematician, as he did not even realize that Mike Deeth was lying > when he said that it was homework. No class on the history on > mathematics would set as homework the assignment which Mike Deeth > claims to have been set, since no genuine history of mathematics > Deeth has used here. In short, Deeth is a crank and a troll who > wishes to espouse his own opinions and pretend that they are the > mainstream. Gosh, I thought that both Nathan the Great and the good Doctor were being humorous. So, Dr. Zona didnt know that Nathan was lying when he said it was homework? Does that mean Dr. Zona also forgot that hes *not* Nathans teacher? -- Jesse F. Hughes Its easy folks. Just talk about my approach to your favorite mathematician. If they cant be interested in it, theyve demonstrated a lack of mathematical skill. -- James Harris === Subject: re:Cantor: ignorant, harmless fool or intentional liar? I have one question: was it a male or a female math teacher? ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: Re: Cantor: ignorant, harmless fool or intentional liar? >> I have to confess, this is homework, so I am not looking for a full >> answer, just some hints where to look. >> Im taking a math history class and the assignment is to write a paper >> on whether Cantor was just misinformed and ultimately innocent despite >> the harm he did mathematics... or whether he intentionally strove to >> murder mathematics. >> So far Ive read into his biography and I am leaning toward the 2nd >> case, since when he was young he was sexually abused by a math >> teacher. That leads me to believe he had a big grudge against >> mathematics in general. >> If I adopt that route, one of my main arguments in this paper will be >> that it would be impossible for anyone to be SO wrong about infinite >> numbers purely be coincidence. I mean, hed have to be really tripping >> on some hard stuff if were to believe he REALLY believed the nonsense >> he published about infinite numbers. >> Im sure most of you have written a paper to this effect some time or >> other in one of your mathematics history classes :-) >> Nathaniel Deeth >> Age 11 >> Math Genious >Nathaniel, >This is supposed to be done on your own. You are not allowed to ask >anyone else for help, as this is cheating. >Your teacher, >Dr. Ben Zona And Craig Feinstein, or Dr. Ben Zona, proves that he is no genuine mathematician, as he did not even realize that Mike Deeth was lying when he said that it was homework. No class on the history on mathematics would set as homework the assignment which Mike Deeth claims to have been set, since no genuine history of mathematics Deeth has used here. In short, Deeth is a crank and a troll who wishes to espouse his own opinions and pretend that they are the mainstream. David McAnally At the moment, they (the Time Lords) are far from being all-powerful. Thats why its been left up to me and me and me. quote by: Patrick Troughton in The Three Doctors ------- === Subject: Re: Cantor: ignorant, harmless fool or intentional liar? > Im taking a math history class and the assignment is to write a paper > on whether Cantor was just misinformed and ultimately innocent despite > the harm he did mathematics... or whether he intentionally strove to > murder mathematics. As I understand it, one of the main ideas of the Jewish mystical teachings known as kaballah is that the world of the infinite is a higher, more perfect world, and that by seeking knowledge of that world, we are learning about god. It appears that Cantor was strongly inßuenced by those teachings. So the conclusion would be that he innocently tried to merge religion with mathematics. > when he was young he was sexually abused by a math teacher. Is that a fact? === Subject: Re: Cantor: ignorant, harmless fool or intentional liar? > I have to confess, this is homework, so I am not looking for a full > answer, just some hints where to look. > Im taking a math history class and the assignment is to write a paper > on whether Cantor was just misinformed and ultimately innocent despite > the harm he did mathematics... or whether he intentionally strove to > murder mathematics. > So far Ive read into his biography and I am leaning toward the 2nd > case, since when he was young he was sexually abused by a math > teacher. That leads me to believe he had a big grudge against > mathematics in general. > If I adopt that route, one of my main arguments in this paper will be > that it would be impossible for anyone to be SO wrong about infinite > numbers purely be coincidence. I mean, hed have to be really tripping > on some hard stuff if were to believe he REALLY believed the nonsense > he published about infinite numbers. > Im sure most of you have written a paper to this effect some time or > other in one of your mathematics history classes :-) > Nathaniel Deeth > Age 11 > Math Genious Nathaniel, This is supposed to be done on your own. You are not allowed to ask anyone else for help, as this is cheating. Your teacher, Dr. Ben Zona === Subject: Re: Cantor: ignorant, harmless fool or intentional liar? > I have to confess, this is homework, so I am not looking for a full > answer, just some hints where to look. > Im taking a math history class and the assignment is to write a paper > on whether Cantor was just misinformed and ultimately innocent despite > the harm he did mathematics... or whether he intentionally strove to > murder mathematics. > So far Ive read into his biography and I am leaning toward the 2nd > case, since when he was young he was sexually abused by a math > teacher. That leads me to believe he had a big grudge against > mathematics in general. > If I adopt that route, one of my main arguments in this paper will be > that it would be impossible for anyone to be SO wrong about infinite > numbers purely be coincidence. I mean, hed have to be really tripping > on some hard stuff if were to believe he REALLY believed the nonsense > he published about infinite numbers. > Im sure most of you have written a paper to this effect some time or > other in one of your mathematics history classes :-) > Nathaniel Deeth > Age 11 > Math Genious *********************************************** the fact that the message of N. Deeth shows clear signs of mature stupidity and thus hes suspicious of being quite older than he say he is, he ALREADY knows what the harm Cantor did to mathematics is, and hes only interested in finding out why (so much for intellectual honestity...). Would he REALLY be that young, Nathaniel is one of the youngest trolls Ive ever met, though not quite the dumbest: for this you need to work harder, my boy! Tonio === Subject: Re: Cantor: ignorant, harmless fool or intentional liar? >>Nathaniel Deeth >>Age 11 >>Math Genious > I will not comment on you being a genious (such a sad destiny...), but > so young and already a crank and a troll? But hes the oldest age 11 troll on usenet :-) -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Cantor: ignorant, harmless fool or intentional liar? Robert J. Kolker says... >Grow up, 11 year old genius. Thats not going to happen any time soon. Hes been 11 for many years now. -- Daryl McCullough Ithaca, NY === Subject: Re: Cantor: ignorant, harmless fool or intentional liar? Are you always going to be one year older than Bart Simpson? > I have to confess, this is homework, so I am not looking for a full > answer, just some hints where to look. > Im taking a math history class and the assignment is to write a paper > on whether Cantor was just misinformed and ultimately innocent despite > the harm he did mathematics... or whether he intentionally strove to > murder mathematics. > So far Ive read into his biography and I am leaning toward the 2nd > case, since when he was young he was sexually abused by a math > teacher. That leads me to believe he had a big grudge against > mathematics in general. > If I adopt that route, one of my main arguments in this paper will be > that it would be impossible for anyone to be SO wrong about infinite > numbers purely be coincidence. I mean, hed have to be really tripping > on some hard stuff if were to believe he REALLY believed the nonsense > he published about infinite numbers. > Im sure most of you have written a paper to this effect some time or > other in one of your mathematics history classes :-) > Nathaniel Deeth > Age 11 > Math Genious === Subject: Re: Cantor: ignorant, harmless fool or intentional liar? > I have to confess, this is homework, so I am not looking for a full > answer, just some hints where to look. > Im taking a math history class and the assignment is to write a paper > on whether Cantor was just misinformed and ultimately innocent despite > the harm he did mathematics... or whether he intentionally strove to > murder mathematics. False dichotomy. It both assumes and concludes with conditions contrary to fact. Typical of an 11 year old mind with no self discipline. > So far Ive read into his biography and I am leaning toward the 2nd > case, since when he was young he was sexually abused by a math > teacher. That leads me to believe he had a big grudge against > mathematics in general. > If I adopt that route, one of my main arguments in this paper will be > that it would be impossible for anyone to be SO wrong about infinite > numbers purely be coincidence. I mean, hed have to be really tripping > on some hard stuff if were to believe he REALLY believed the nonsense > he published about infinite numbers. What are you tripping on that leads you to think so? > Im sure most of you have written a paper to this effect some time or > other in one of your mathematics history classes :-) My best advice is to go cold turkey. > Nathaniel Deeth > Age 11 > Math Genious And spelling failure. === Subject: Re: Cantor: ignorant, harmless fool or intentional liar? > I have to confess, this is homework, so I am not looking for a full > answer, just some hints where to look. > Im taking a math history class and the assignment is to write a paper > on whether Cantor was just misinformed and ultimately innocent despite > the harm he did mathematics... or whether he intentionally strove to > murder mathematics. David Hilbert, the greatest mathematician of his day said: No one will drive us from the paradise that Cantor has created. Mathematics without a wide ranging abstract set theory is unthinkable. We would not have topology, generalized functions, measure theory or a generalized theory of computation without Cantors work. Grow up, 11 year old genius. Bob Kolker === Subject: Re: Cantor: ignorant, harmless fool or intentional liar? >I have to confess, this is homework, so I am not looking for a full >answer, just some hints where to look. >Im taking a math history class and the assignment is to write a paper >on whether Cantor was just misinformed and ultimately innocent despite >the harm he did mathematics... or whether he intentionally strove to >murder mathematics. >So far Ive read into his biography and I am leaning toward the 2nd >case, since when he was young he was sexually abused by a math >teacher. That leads me to believe he had a big grudge against >mathematics in general. >If I adopt that route, one of my main arguments in this paper will be >that it would be impossible for anyone to be SO wrong about infinite >numbers purely be coincidence. I mean, hed have to be really tripping >on some hard stuff if were to believe he REALLY believed the nonsense >he published about infinite numbers. >Im sure most of you have written a paper to this effect some time or >other in one of your mathematics history classes :-) >Nathaniel Deeth >Age 11 >Math Genious I will not comment on you being a genious (such a sad destiny...), but so young and already a crank and a troll? G. Rodrigues === Subject: Cantor: ignorant, harmless fool or intentional liar? I have to confess, this is homework, so I am not looking for a full answer, just some hints where to look. Im taking a math history class and the assignment is to write a paper on whether Cantor was just misinformed and ultimately innocent despite the harm he did mathematics... or whether he intentionally strove to murder mathematics. So far Ive read into his biography and I am leaning toward the 2nd case, since when he was young he was sexually abused by a math teacher. That leads me to believe he had a big grudge against mathematics in general. If I adopt that route, one of my main arguments in this paper will be that it would be impossible for anyone to be SO wrong about infinite numbers purely be coincidence. I mean, hed have to be really tripping on some hard stuff if were to believe he REALLY believed the nonsense he published about infinite numbers. Im sure most of you have written a paper to this effect some time or other in one of your mathematics history classes :-) Nathaniel Deeth Age 11 Math Genious === Subject: Re: missing pages from Spaniers Algebraic Topology >I ordered a copy of Spaniers Algebraic Topology, and it arrived >today, minus page 13-14. (...) >Springer. Instead, I was wondering if someone with access to a >scanner would just scan these two pages (page 13 and 14 in the 1991 >printing--basically the first two pages of chapter 1) and email me the >scans at npenton@hotmail.com. (...) > I have a scanner and the book, but its a much older edition. If this > does not make it useless for you, I can send you such a scan as > a GIF image, but tell me first with what begins page 15 of your book, > so that I can better guess what content you are missing. In my book > the half-empty first page of chapter 1 is a short introduction, that > continues on the next page before section 1; do you have this > introduction ? Page 15 begins with and it is called the identity morphism of Y. Given morphisms f:X->Y and g:Y->X such that gf=1_X, g is called a left inverse of f and f is called a right inverse of g. A two sided inverse... Page 12 is just a title page, reading Chapter one: Homotopy and the Fundamental Group So no, I do not think I have the introduction to chapter one. If you could send me whatever lies between these in your edition, I would be grateful. My understanding is that he only made minor revisions and corrections between editions. Nathan === Subject: Re: missing pages from Spaniers Algebraic Topology >I ordered a copy of Spaniers Algebraic Topology, and it arrived >today, minus page 13-14. (...) >Springer. Instead, I was wondering if someone with access to a >scanner would just scan these two pages (page 13 and 14 in the 1991 >printing--basically the first two pages of chapter 1) and email me the >scans at npenton@hotmail.com. (...) I have a scanner and the book, but its a much older edition. If this does not make it useless for you, I can send you such a scan as a GIF image, but tell me first with what begins page 15 of your book, so that I can better guess what content you are missing. In my book the half-empty first page of chapter 1 is a short introduction, that continues on the next page before section 1; do you have this introduction ? Ulysse from CH === Subject: missing pages from Spaniers Algebraic Topology I ordered a copy of Spaniers Algebraic Topology, and it arrived today, minus page 13-14. This is mildly frustrating; although I dont really need to see the definition of a category again, it feels quite incomplete, and disrupts the ßow of exposition something awful. I suppose I could complain to the bookseller, but I doubt its his fault, and I fear it will take some time to get a replacement from Springer. Instead, I was wondering if someone with access to a scanner would just scan these two pages (page 13 and 14 in the 1991 printing--basically the first two pages of chapter 1) and email me the scans at npenton@hotmail.com. Anyone who would take the few minutes to do this for me will be rewarded with my undying gratitude and possibly the soul of my first-born son. Nathan === Subject: What is freedom ? Hi Starblade Darksquall, You suggested, nature IS progress. The future is not set, but the means by which we must get to the future is essentially set. However, it also possesses degrees of freedom. . Just because we dont perfectly know what the future will be that cant make nature somehow incomplete. Nature is complete by definition. So although the future is imperfectly known, the future is sure to be perfectly set. ( i.e. Time is perfectly spatial, not directional ) Q. What is progress ? A. Control is the goal. And nature never worries about that. ( or anything else ) Q. What is freedom ? A. Its facing the same trivial choices day after day ... And then feeling a false sense of freedom because of that. === Subject: principal bundles in algebraic geometry Im looking for a nice and introductory exposition of the concept of principal bundles P-->X in algebraic geometry, that explains me in modern language e.g. why mumfords definition (in GIT) is the same as saying that it should be locally trivial in the etale topology (is it?), that every principal GL bundle is already locally trivial in the Zariski topology (I know, there is always Serre), and the constructions of associated geometric and/or algebraic (meaning locally free sheaf) vector bundle. I found some helpful stuff in Milnes Etale Cohomology, but I would appreciate a more extensive treatment especially concerning the constructions of frame bundle associated to a vector bundle and vector bundle associated to a principal bundle. Im thankful for any hints, markus === Subject: Re: intercept length when a random line intersect with ellipsoid shell or parallelepiped shell with fixed thickness Actually i want to know the intercept length when a random line intersect with a convex shell with fixed thickness. how many methods the random line can be choosed? Can you get me the details? thank you in advance! > ChenHS escribi.97: > If a random line intersects with a ellipsoid shell or parellelepiped > shell with a fixed thickness, how much is the statistical intercept > length? > You must define in a precisse way how the random line is choosed. The answer > depend on it and it can be very different. === Subject: Re: determinant > çÁ ófi[CapitalEth]zçmachava@hotmail.co m (Achava Nakhash, the Loving Snake)çnóÛé.beÂ2 64çG > This is a skew-symmetric matrix. It has the property that all eigenvalues > are purely imaginary. > As an added hint, you might take note that iA is Hermitian, and > Hermitian matrices necessarily have non-negative eigenvalues. What is > ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ > Let A be nxn Hermitian maatrix. > If c is an eigenvalue of A, => there is x != 0 such that Ax=cx. > c = = = = = = conjugate(c) c = conjugate(c). > => c is real. > Hou do you show that c is non-negative? You dont, because its false. Example: A = [-1]. I think Achava is confused with Ôpositive definite. Wilbert === Subject: Re: determinant Á¡ Ûfi[Hyphen]zÁmachava@hotmail.com (Achava Nakhash, the Loving Snake)ÁnÛ¤Èæ ¬´ÁG > This is a skew-symmetric matrix. It has the property that all eigenvalues > are purely imaginary. > As an added hint, you might take note that iA is Hermitian, and > Hermitian matrices necessarily have non-negative eigenvalues. What is ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Let A be nxn Hermitian maatrix. If c is an eigenvalue of A, => there is x != 0 such that Ax=cx. c = = = = = = conjugate(c) => c = conjugate(c). => c is real. Hou do you show that c is non-negative? > the relationship between the eigenvalues of A and those of iA? > Achava -- === Subject: Re: determinant by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1QEWU910487; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1QE2Li07574 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i1QE2LC08005; >Hello! >Please help me to solve the following determinant: >| a1+a1 b1+b1 c1+c1 d1+d1 e1+e1 | >| a2+a2 b2+b2 c2+c2 d2+d2 e2+e2 | >| a3+a3 b3+b3 c3+c3 d3+d3 e3+e3 |=? >| a4+a4 b4+b4 c4+c4 d4+d4 e4+e4 | >| a5+a5 b5+b5 c5+c5 d5+d5 e5+e5 | >1. is there some shortcut to get the full solution to this determinant, i mean rather than solve it brute force.... __________________________________________________________ Wanting to get it without brute force is good, but I hope you realize this problem is easily within the range of computer algebra systems. I did it with Fermat, it took 0.133 seconds (on a two year old Mac), and the answer has 3840 terms. I didnt try other systems. Robert H. Lewis http://www.bway.net/~lewis/