mm-221 I'm intrigued by the questions raised by your recent posts, >>as for>> years you were ts guy who came up with rather creative >>ways to>> insult me, and now I find it hard not to figure you're just >>doing so>> again.Is it okay to ask you questions, talk in any familiar way, or >>in any way>> act as if I wish you to reply or am addressing you?Those proscriptions apply only to Ullrich, Virgil, and >>anyone who>> posts under a palindromic pseudonym, right?Okay then. You ask a good question. Just what is my intent? >> I'm not>> sure precisely why I'm doing what I'm doing, but I'll try to >>answer you>> earnestly and respectfully.Now I'll embarrass you a bit as from what I've read on the >>web you>> have one of the ghest IQ's out there, so it seems to me >>there's>> probably some reason to what you're doing, and possibly I'm >>wrong>> about what it is.Ts does embarrass me because I'm certainly not one of the >>big fish on>> sci.math. You must be basing ts statement on the fact that >>I once>> joined sometng called Mega, wch purports to be a gh-IQ >>society for>> those of 1-in-a-million intelligence. I now realize that by >>joining, I>> was implicitly making ts arrogant claim about myself, but I >>reject that>> claim. The fact that I was able to ace the math part of >>their test just>> indicates that it wasn't hard enough, because it's easy to >>find people>> better at math than I. Indeed, you can find lots of them on >>sci.math.>> And some of them even take the time to analyze your work, >>James.You're not one of the big fish on sci.math, and I'm less >curious about>your past experiences with gh IQ societies than you probably >tnk.JSH in s best Rosanne Rosannadanna voice: Nevermind. >However, I had a theory, and testing it involved mentioning >that facet>of your public persona.JSH's attempt at regaining control.> Therefore, I'm going to give you the opportunity that I >>feel *I* don't>> get, wch is the benefit of the doubt.You're asking me to clarify me position, wch I am about to >>do. I tnk>> it's fair to say, however, that others have given you the >>same opportunity,>> i.e., that they've asked you to clarify your position.Tell me succinctly and in a way that will minimize potential>> embarrassment for both of us, what it is that your up to, >>and no, none>> of ts wild stuff about how great I supposedly am, or how >>I've proven>> FLT or any of that, as I just want you to say sometng >>that fits into>> a worldview that makes sense.What's your intent?You tnk that I'm mocking you, but you're not entirely sure. >>> Ullrich tnks it's incredible that it's not obvious to you >>that I'm>> mocking you.Well let me clear tngs up. Yes, I'm mocking you. I've >>made a series>> of posts over the last six months in wch I've appeared to >>be converted>> to a religion in wch you are the Messiah of Mathematical >>Truth being>> crucified my the benighted masses. Most people consider that >>absurd and>> therefore conclude that I must have been being sarcastic. >>You, however,>> do not consider the idea that you are the Messiah of >>Mathematical Truth>> to be absurd. You consider it to be essentially correct -- a >>little off>> somehow: a little over the top, or emotionally overblown, but >>basically>> the correct attitude to take.>So you were lying. I just needed to make sure.The problem is that I'd concluded that very intelligent people >see a>much gher value in telling the truth than others.JSH shrinking s world a little bit more.>That lead me to consider the possibility that you were in fact>sincere, but deluded and confused, possibly dealing with a lot >of>emotional pain from a difficult position--considering that I >might be>right--against tremendous social forces.But you weren't being brave. You were === JSH: Open letter to Jim Ferry> The problem is that I'd concluded that very intelligent people see a> much gher value in telling the truth than others.Another problem is that you seem to place such a low value on telling the truth.What does your own conclusion imply about your own === The problem is that I'd concluded that very intelligent people see a> much gher value in telling the truth than others.That places *your* intelligence below zero.> JSH Motto: No matter how much I succeed in lowering the integrity bar, I can still slither under it.--There are two tngs you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over === letter to Jim Ferry|The problem is that I'd concluded that very intelligent people see a|much gher value in telling the truth than others.do you consider yourself to be very intelligent? and do you consideryourself to be very honest? in particular, do you consider yourselfto have a very good record of telling the truth in your posts tosci.math?-- [e-mail === general, equals a MazeAs for representing a series of connected math-results as a maze,I would assume a more apt analogy would be connected math results as amore general graph, not necessarily as a maze (a maze with only onesolution, in any case).Anyway, I wonder if there is anytng interesting known about themetamathematical topology of connected math-results.(aside from Godel's theorem, of course.)(I mean, there HAS to be many interesting results regarding resultsin-general.):),> In > selm=78kbmti88zn8%40forum.mathforum.com>I know mathematics is a maze.>But it is a maze without end.>It is a maze where goals sft>And directions change.>It is almost impossible,>If it is not impossible,>To climb upon its walls>And view everytng whole.I would add that mathematical proof isA maze of infinite complexity,> of infinite exits and entrances> (each valid),> and of infinite dead-ends> (each invalid),> a maze of opaqueness and transparency,> of obviousness and mystery,> of unappreciated beauty> only seen as it finally becomes known> to esoteric eyes...> But seriously,...> When I sometimes try to prove a math result, I will often get> sidetracked, following other paths, paths not originally intended, to> some other related results instead. (I might as well prove> *sometng*!...)Does anyone else have an excellent example of ts?(I know the maze-metaphor applies to Wiles' {et al} proof Fermats Last> theorem, since it turned out that a once-abandoned path was actually> the way to Wiles' {et al} ultimate goal.)And of course, in mathematics, if mathematicians fail to solve> sometng as originally planned, still their work need not be> completely for notng, since other related results can still come> from their partial successes.> And more ambitiously, as far as the maze metaphor is concerned, can> viewing mathematics and proofs as a maze or as a graph (one result> leads to another leads to another), lead to any new ways to solve> math-problems in general?(I bet the Ôright-hand-rule' algorithm for solving mazes has no> obvious analog that can be applied to proving math results, as === algebra question, permutation groupsI need to find three elements o in S9 with the property thato^3=(157)(283)(469).The answer is (124586739), (142568793), (214856379)I have been trying to figure out how to work ts out for a couple hours buthaven't gotten anywhere. Can someone help me????? I must not be understandingsome part of the theory because I don't === algebra question, permutation groups> I need to find three elements o in S9 with the property that> o^3=(157)(283)(469).The answer is (124586739), (142568793), (214856379)Well, that's *one* answer. Another answer would involve (189534726), (193542768), and (126589734). And there are many more. Do you understand why (124586739)^3 = (157)(283)(469)? If you do, you should be well on your way to understanding where (124586739) came from (and if you don't, === question, permutation groups: I need to find three elements o in S9 with the property that: o^3=(157)(283)(469).:: The answer is (124586739), (142568793), (214856379):: I have been trying to figure out how to work ts out for a couple hoursbut: haven't gotten anywhere. Can someone help me????? I must not beunderstanding: some part of the theory because I don't know a trick that will work forts: one.Since the 3 of o^3 divides the 9 of S9, the o will be a permutation oßength 9 to get the three length 3 permutations you are looking for. The1st will map to the 4th, the 4th to the 7th, and the 7th back to the 1st,along with the other cycles starting at 2 and 3 (add the rigor).Now, since you know the answers, look at their (1,4,7) (2,5,8) (3,6,9)members. Now show any others must be identical === Matrix Representation of Irreps of SnI was wondering if there is anyone who has read Hamermesh's bookGroup Theory and its Applications to Physical Problems and has agood understanding of and is able to explain in explicit detail howHammermesh derived the matrix representations of the irreps of Sndisplayed from pages 224 to 230, and their relations to the Yamanoucsymbols displayed beside them. I don't understand what exactly theindices s, r, lamda_s, and, lamda_r refer to, and the formulas thatHamermesh labels (7-102), (7-104), (7-105), and (7-111) do not appearto have any relation to the matrix entries, or at least I can't figureout what that relation is. === Re: reformulated problem: how to solve for integer matrix equations? !Using Maple V R4 I find :{a1 = 4*a3, u4 = -4/a4^2, a4 = a4, u2 = -4/a3/a4, a2 = -a4, u3 = 4/a3/a4, u1 =1/a3^2, a3 = a3}is a solution.ChrisHere is a simplified reformulation of my previous problem:How to solve ts integer matrix equation?[ a1^2*u1, a1*a2*u2, a1*a2*u3, a2^2*u4]> [ a1*a3*u1, a1*a4*u2, a3*a2*u3, a2*a4*u4]> [ a1*a3*u1, a3*a2*u2, a1*a4*u3, a2*a4*u4]> [ a3^2*u1, a3*a4*u2, a3*a4*u3, a4^2*u4]> => [ 16 16 -16 -4]> [ 4 -16 -4 4]> [ 4 4 16 4]> [ 1 -4 4 -4]a1, a2, a3, a4, u1, u2, u3, u4 should be 2's positive integer power, such> as -2, +1, etc.I considered to factor the right hand side(the constant matrix) first:a1^2*u1 =16=4*4*1> a1*a2*u2=16=4*4*1> a1*a2*u3=-16=-4*4*1> a2^2*u4=-4=-2*2*1The try to allocate/assign these 2's factors to the variables a1, a2, a3,> a4, u1, u2, u3, u4...Since there are more equations than unknowns, sometimes there are> conßicting assignment to the variables, so only approximated assignment can> be made to minimize the mean square error between the left hand side and> right hand side in the above equation...It is doable by exhaustive search, for example, since there are 16 equations> and 8 unknowns, I can do 8-level iteration like the following:for a1=[-4, -2, -1, 1, 2, 4] do> for a2=[-4, -2, -1, 1, 2, 4] do> for a3=[-4, -2, -1, 1, 2, 4] do> for a4=[-4, -2, -1, 1, 2, 4] do> for u1=[-4, -2, -1, 1, 2, 4] do> for u2=[-4, -2, -1, 1, 2, 4] do> for u3=[-4, -2, -1, 1, 2, 4] do> for u4=[-4, -2, -1, 1, 2, 4] do compare MSE between LHSvsRHS for all cases> keep the best one end> end> end> end> end> end> end> endThe only problem is that one the matrix gets larger, with hundreds of> unknowns, the above method cannot be used any more...Can anybody give me more ideas and === Equations> !Is there some (huge) positive integer M with the following> property: for any z>M, there exist positive integers > x,y_1,y_2,...,y_z such thatx^x=(y_1)^(y_1)+(y_2)^(y_2)+...+(y_z)^(y_z) ?(Please remark that the y's are >=1 and need not to be> necessarily distinct). in advance.> Ignorantly Yours,> Ady.A partial result??:... x^x/(x-1)^(x-1) ~ x*e.So, (for x not y; z >1)z has to be *at least* in the _vicinity_ of x*e,if all y's = === there some (huge) positive integer M with the following> property: for any z>M, there exist positive integers > x,y_1,y_2,...,y_z such thatx^x=(y_1)^(y_1)+(y_2)^(y_2)+...+(y_z)^(y_z) ?(Please remark that the y's are >=1 and need not to be> necessarily distinct).I'm afraid you'll have to put more conditions on it than that. Otherwise, just let z=x^x and y_1 = y_2 = y_3 = y_4 = ... = 1. Thenyou get x^x = 1 + 1 + 1 + 1 + ... 1. Therefore, the answer is yes, M =0.You could put more conditions on it, like requiring y_n > 1 (not y_n>= 1) orrequiring that y_n are distinct. Did ts arise as part of a solutionto a bigger problem, or were you just looking at ts for fun? in advance. === some (huge) positive integer M with the following> property: for any z>M, there exist positive integers > x,y_1,y_2,...,y_z such thatx^x=(y_1)^(y_1)+(y_2)^(y_2)+...+(y_z)^(y_z) ?(Please remark that the y's are >=1 and need not to be> necessarily distinct).I'm afraid you'll have to put more conditions on it than that. > Otherwise, just let z=x^x and y_1 = y_2 = y_3 = y_4 = ... = 1. Then> you get x^x = 1 + 1 + 1 + 1 + ... 1. Therefore, the answer is yes, M => 0.That doesn't answer the question, not the way I read it. Take your M = 0. Take z = 2, thus satisfying the hypothesis z > M. Now try to find x, y_1, and y_2 such that x^x = (y_1)^(y_1) + (y_2)^(y_2). Give up? OP === .......genius teacher........let function f :E -> R is differentiable on open set E in R1. let any a in E , f''(a) exist.show that f''(a) = lim {f(a+h) - 2f(a) + f(a-h) / (h^2)} , h->02. let any a in E , n-derivative f^(n) (a) exist.express f^(n) (a) by limit of f(x)-um.....i solved first problem.lim {f(a+h) - 2f(a) + f(a-h) / (h^2)}> h->0= lim [{f(a+h)-f(a)} - {f(a-h+h)-f(a-h)} / (h^2)]= lim {f'(a) - f'(a-h)} / h , let h=(-t)= lim {f'(a) - f'(a+t)} / (-t)> t->0= lim {f'(a+t) - f'(a)} / t= f''(a)> it's right??but i can't solve second problem....You haven't solved the FIRST problem, either. Your passage> = lim [{f(a+h)-f(a)} - {f(a-h+h)-f(a-h)} / (h^2)]= lim {f'(a) - f'(a-h)} / h , let h=(-t)is ßawed.nt: apply L'Hopital's Rule to the original quotient, once. Then usethe === easy...analysis problem...........genius teacher........let function f :E -> R is differentiable on open set E in R1. let any a in E , f''(a) exist.show that f''(a) = lim {f(a+h) - 2f(a) + f(a-h) / (h^2)} , h->02. let any a in E , n-derivative f^(n) (a) exist.express f^(n) (a) by limit of f(x)-um.....i solved first problem.lim {f(a+h) - 2f(a) + f(a-h) / (h^2)}> h->0= lim [{f(a+h)-f(a)} - {f(a-h+h)-f(a-h)} / (h^2)]= lim {f'(a) - f'(a-h)} / h , let h=(-t)= lim {f'(a) - f'(a+t)} / (-t)> t->0= lim {f'(a+t) - f'(a)} / t= f''(a)> it's right??but i can't solve second problem....You haven't solved the FIRST problem, either. Your passage= lim [{f(a+h)-f(a)} - {f(a-h+h)-f(a-h)} / (h^2)]= lim {f'(a) - f'(a-h)} / h , let h=(-t)is ßawed.nt: apply L'Hopital's Rule to the original quotient, once. Then use> the definition of the second derivative.> Another nt for the first problem: since f''(a) exists, f'(a) existsand f(a+h) = f(a) + h f'(a) + (h2/2)f''(a) === teacher........let function f :E -> R is differentiable on open set E in R1. let any a in E , f''(a) exist.show that f''(a) = lim {f(a+h) - 2f(a) + f(a-h) / (h^2)} , h->02. let any a in E , n-derivative f^(n) (a) exist.express f^(n) (a) by limit of f(x)-um.....i solved first problem.lim {f(a+h) - 2f(a) + f(a-h) / (h^2)}> h->0= lim [{f(a+h)-f(a)} - {f(a-h+h)-f(a-h)} / (h^2)]= lim {f'(a) - f'(a-h)} / h , let h=(-t)= lim {f'(a) - f'(a+t)} / (-t)> t->0= lim {f'(a+t) - f'(a)} / t= f''(a)> it's right??but i can't solve second problem....You haven't solved the FIRST problem, either. Your passage= lim [{f(a+h)-f(a)} - {f(a-h+h)-f(a-h)} / (h^2)]= lim {f'(a) - f'(a-h)} / h , let h=(-t)is ßawed.nt: apply L'Hopital's Rule to the original quotient, once. Then use> the definition of the second derivative.Another nt for the first problem: since f''(a) exists, f'(a) exists> and f(a+h) = f(a) + h f'(a) + (h2/2)f''(a) + o(h^2).Is that true? Taylor's theorem only guarantees that f(a+h) = f(a) + h f'(a) + h^2/2 f''(z)for some z between a and a+h; but we're NOT told that f'' iscontinuous, only that it exists throughout the open === easy...analysis problem....> .......genius teacher........let function f :E -> R is differentiable on open set E in R1. let any a in E , f''(a) exist.show that f''(a) = lim {f(a+h) - 2f(a) + f(a-h) / (h^2)} , h->02. let any a in E , n-derivative f^(n) (a) exist.express f^(n) (a) by limit of f(x)-um.....i solved first problem.lim {f(a+h) - 2f(a) + f(a-h) / (h^2)}> h->0= lim [{f(a+h)-f(a)} - {f(a-h+h)-f(a-h)} / (h^2)]= lim {f'(a) - f'(a-h)} / h , let h=(-t)= lim {f'(a) - f'(a+t)} / (-t)> t->0= lim {f'(a+t) - f'(a)} / t= f''(a)> it's right??but i can't solve second problem....You haven't solved the FIRST problem, either. Your passage= lim [{f(a+h)-f(a)} - {f(a-h+h)-f(a-h)} / (h^2)]= lim {f'(a) - f'(a-h)} / h , let h=(-t)is ßawed.nt: apply L'Hopital's Rule to the original quotient, once. Then use> the definition of the second derivative.Another nt for the first problem: since f''(a) exists, f'(a) exists> and f(a+h) = f(a) + h f'(a) + (h2/2)f''(a) + o(h^2).Is that true? Taylor's theorem only guarantees that f(a+h) = f(a) + h f'(a) + h^2/2 f''(z)for some z between a and a+h; but we're NOT told that f'' is> continuous, only that it exists throughout the open set.That was why I suggested L'H.--Ron BruckUh, the hypotheses state that f is differentiable on E. So f'(a)exists for any a in E. (You're technically right, Ron, but notparticularly helpful. === a 2d screenOk.. ts question may have been asked hundreds of anytng goodso...I'm looking for the canonical formulas for plotting n-dimensionalpoints. I know how to plot a 3d graph:screenx = d * x / zscreeny = d * y / z(where d is the distance of the observer from the 3d object)I assume that to plot 4d, 5d, 6d, nd would just be an extensionof these equations somehow. I assume I could play around andfigure it out, but I figure I might just as well ask. I assumethat others there any canonical way to plot n-dimensional points?(Note: I know there are some elaborate multidimensional grapngtechniques for data-mining and such... that's not what I'mlooking for. I'm just looking for the simple and easy way atts point-- it doesn't have to even be that *usable* beyond4d or so, but you should at least be able to see === a 2d screen> I assume that to plot 4d, 5d, 6d, nd would just be an extension> of these equations somehow. I assume I could play around and> figure it out, but I figure I might just as well ask. I assume> that others might find it helpful too for ts answer n-dimensional points?Neil Sloane has (among other tngs) worked on the problem offinding representative 2D cross-sections of 4D (and gherdata). The idea is that you would show several projectionsof the gh-D data into different 2D-planes. The task at handis to somehow optimize the chosen collection of planes. Hecooked up s own version of a metric on Grassmannian === define M as a mxm complex matrix> and M^n = I> is there any properties related with ts kind of matrix? a lot!Yeah, there are a lot of them. Among the obvious are that it hasinverse M^(n-1), that all the eigenvalues are complex roots of === Helen,Well, if M is not (trivially) the identity matrix, then any eigenvalues of M will be the complex n-th roots of unity.--Eric> here we define M as a mxm complex matrix> and M^n = I> is there any properties related with ts kind of matrix? a === define M as a mxm complex matrix> and M^n = I> is there any properties related with ts kind of matrix?It's diagonalizable, for a start.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes s penis for a square root, 88-9Francis Wheen, _How === to prove that (1+X)^n >= (1+nX), n is a natural #, X is a real number>= -1.Here is what I have (and don't have). I know that (1+X)^n and (1+nX)intersect at (0,1). Since [ (1+X)^n ] Ô > n for all X >0 and since (1 + nX)' = n for all X we know that (1+X)^n > (1+nX) ( but we already knew tsfrom the binomial theorem).The real question is whether or not (1 + X) ^n and (1 === to prove that (1+X)^n >= (1+nX), n is a natural #, X is a real number>= -1.> Here is what I have (and don't have). I know that (1+X)^n and (1+nX)> intersect at (0,1). Since [ (1+X)^n ] Ô > n for all X >0 and since (1 + nX)> Ô = n for all X we know that (1+X)^n > (1+nX) ( but we already knew ts> from the binomial theorem).> The real question is whether or not (1 + X) ^n and (1 + nX ) intersect below> X =0 ? For n = 0 or 1, (1+x)^n = (1+n*x).For integers n > 1, let f(x) = (1+x)^n - (1+n*x), then showf(0) = 0, and f'(x) = 0 at x = 0 (and only === helpI need to prove that (1+X)^n >= (1+nX), n is a natural #, X is a real number>= -1.> Here is what I have (and don't have). I know that (1+X)^n and (1+nX)> intersect at (0,1). Since [ (1+X)^n ] Ô > n for all X >0 and since (1 + nX)> Ô = n for all X we know that (1+X)^n > (1+nX) ( but we already knew ts> from the binomial theorem).> The real question is whether or not (1 + X) ^n and (1 + nX ) intersect below> X =0 ?> For n = 0 or 1, (1+x)^n = (1+n*x).For integers n > 1, let f(x) = (1+x)^n - (1+n*x), then show> f(0) = 0, and > f'(x) = 0 at x = 0 (and only at x = 0), and > f''(x) > 0 at x = 0. dear virgil i tnk that i could if f(x)=(1+x)^n - (1+nx) f'(x)=n(1+X)^[n-1] -n > f'(0)=n-n=0f(x)=n(n-1)(1+x)^[n-2] >f(0)=n(n-1) we know that n>1 at result n(n-1)>0do you thonk that it is true ? === (1+nX), n is a natural #, X is a real > number>= -1.> Here is what I have (and don't have). I know that (1+X)^n and (1+nX)> intersect at (0,1). Since [ (1+X)^n ] Ô > n for all X >0 and since (1 + > nX)> Ô = n for all X we know that (1+X)^n > (1+nX) ( but we already knew ts> from the binomial theorem).> The real question is whether or not (1 + X) ^n and (1 + nX ) intersect > below> X =0 ?> For n = 0 or 1, (1+x)^n = (1+n*x).For integers n > 1, let f(x) = (1+x)^n - (1+n*x), then show> f(0) = 0, and > f'(x) = 0 at x = 0 (and only at x = 0), and > f''(x) > 0 at x = 0. dear virgil > i tnk that i could > if f(x)=(1+x)^n - (1+nx) > f'(x)=n(1+X)^[n-1] -n > f'(0)=n-n=0> f(x)=n(n-1)(1+x)^[n-2] >f(0)=n(n-1) we know that n>1 at result n(n-1)>0> do you thonk that it is true ?> === help> I need to prove that (1+X)^n >= (1+nX), n is a natural #, X is a real number>= -1.> Here is what I have (and don't have). I know that (1+X)^n and (1+nX)> intersect at (0,1). Since [ (1+X)^n ] Ô > n for all X >0 and since (1 + nX)> Ô = n for all X we know that (1+X)^n > (1+nX) ( but we already knew ts> from the binomial theorem).> The real question is whether or not (1 + X) ^n and (1 + nX ) intersect below> X =0 ?3 ways to see it: 1. (1+X)^n is strictly concave up on (-1,oo) and so stays above all of its tangent lines, including y = 1 + nX, on that interval. 2. Let f(x) = (1+X)^n - (1+nX). Then f' > 0 to the right of 0 and f' < 0 to the left of 0. By the mean value theorem, f(x) > 0 for nonzero x. 3. As Gerry Myerson === with proof of distribution like function: analysisi will first state the problem, and then show you what i have thus far.I am looking at the kernel of the fundamental solution of the heat diffusionequation i.e.K(x,t) = Exp(-x/(4*t))/sqrt(4*pi*t) t>0The problem is ts: doesLimit (x+-->0) -2*Integral(0,t) dK(x,t-s)/dx * (g(s) - g(t)) ds = 0 ?[1]I tnk it does - at least i remember that ts result is true from when ifirst studied the heat equation - but i dont remember the details, wch iswhat i am trying to reconstruct here.note:-integral(0,t) dK(x,t-s)/dx ds = 1/sqrt(pi)*integral(x/(2sqrt(t)),oo)exp(-p^2) dp =1 for x>0for x<0 -integral(0,t) dK(x,t-s)/dx ds = -1 [2]thus noting ts property what we are really trying to show isLimit (x+-->0) -2*Integral(0,t) dK(x,t-s)/dx * g(s) ds = g(t) [3]we have written ts in the form of [1] because i tnk thats moreappropriate for a proof. Now i am particularly bad at proof's (my brainjust isnt wired that way i suppose), however ts is what i have so far.assume g(t) is continuous at t. Since g(t) is continuous at t, for eache>0 there exists a d>0 such that |g(t) - g(s)| < e/2 for all |t-s|<=dok. consider the integralI = -2*Integral(0,t) dK(x,t-s)/dx * (g(s) - g(t)) dswch i choose to write asI = -2*Integral(0,t-d) dK(x,t-s)/dx * (g(s) - g(t)) ds -2*Integral(t-d,t)dK(x,t-s)/dx * (g(s) - g(t)) ds :=I1 + I2look at I2 first. from the assumption |g(t) - g(s)| < e/2 for all |t-s|<=di tnk we can write|I2| <= -e*integral(t-d,d)dK/dx ds < e/2using the above property of [2]. So thats good (or is it?). im quite surethat i am missing many important points that would make all ts rigorous -but i am more interested in the end result at the moment. Next we have toshow that |I1| i will first state the problem, and then show you what i have thus far. I am looking at the kernel of the fundamental solution of the heatdiffusion> equation i.e. K(x,t) = Exp(-x/(4*t))/sqrt(4*pi*t) t>0 The problem is ts: does Limit (x+-->0) -2*Integral(0,t) dK(x,t-s)/dx * (g(s) - g(t)) ds = 0 ?> [1] I tnk it does - at least i remember that ts result is true from when i> first studied the heat equation - but i dont remember the details, wchis> what i am trying to reconstruct here. note: -integral(0,t) dK(x,t-s)/dx ds = 1/sqrt(pi)*integral(x/(2sqrt(t)),oo)> exp(-p^2) dp =1 for x>0as x-->0. for x<0 -integral(0,t) dK(x,t-s)/dx ds = -1 [2]as show is Limit (x+-->0) -2*Integral(0,t) dK(x,t-s)/dx * g(s) ds = g(t) [3] we have written ts in the form of [1] because i tnk thats more> appropriate for a proof. Now i am particularly bad at proof's (my brain> just isnt wired that way i suppose), however ts is what i have so far. assume g(t) is continuous at t. Since g(t) is continuous at t, for each> e>0 there exists a d>0 such that |g(t) - g(s)| < e/2 for all |t-s|<=d ok. consider the integral I = -2*Integral(0,t) dK(x,t-s)/dx * (g(s) - g(t)) ds wch i choose to write as I = -2*Integral(0,t-d) dK(x,t-s)/dx * (g(s) - g(t)) ds -2*Integral(t-d,t)> dK(x,t-s)/dx * (g(s) - g(t)) ds :=I1 + I2 look at I2 first. from the assumption |g(t) - g(s)| < e/2 for all|t-s|<=d> i tnk we can write |I2| <= -e*integral(t-d,d)dK/dx ds < e/2as x-->0. using the above property of [2]. So thats good (or is it?). im quitesure> that i am missing many important points that would make all tsrigorous -> but i am more interested in the end result at the moment. Next we have to> show that |I1| x/(2*sqrt(pi))|Integral(0,t-d) exp(-x^2/(4*(t-s))/(t-s)^(3/2) (g(s) -g(t))> ds | ts i am not sure about obtaining. i *tnk* i need to use an integral> inequality say, |integral f*g| <= Sqrt((integral f^2 )*(integral g^2)) but i cant see how to get e/2. the idea being that if we can show|I2| then |I| showing classic signs of there being some sort of delta function involved> perhaps there is another way though all ts === how I wonder...> To me, integration makes perfectly sense but,Where is the irrasionality of irrasional numbers?> They seem to be fully rational.Where is the inductive logic in mathematical induction?> It seems to be deduction.Whats primary with prime numbers?> 1 seems to be more primary.Well, if you define primary as1 a : first in orderthen 1 seems more primary. But primary also can be defined as2 b : not derivable from other colors, odors or tastesSo if one adds numbers to that list, primes are not derivable fromother numbers.Whats imaginary with imaginary numbers?> -1 seems to be the only imaginary number.It's a joke. Imaginary means not real, where ts use of realrefers to existence. The mathematical use of real doesn't refer toexistence just as the mathematical use of primary doesn't refer toorder. -1 happens to be a real number mathematically. But thesquare root of -1 is not a real number in the mathematical sense.So it's called imaginary even though it does exist. Try explaining that to a bunch of drunks at a party, it's a barrel of laughs.They could have called the numbers that are not prime the SecondaryNumbers, but instead the name composite was chosen.What is platonic with platonic figures?> They seem to be of ts world.Whats defferentiated during differensiation?> No difference is attempted === Imaginary means not real, where ts use of real> refers to existence. The mathematical use of real doesn't refer to> existence just as the mathematical use of primary doesn't refer to> order. -1 happens to be a real number mathematically. But the> square root of -1 is not a real number in the mathematical sense.> So it's called imaginary even though it does exist. Try explaining > that to a bunch of drunks at a party, it's a barrel of laughs.Better yet, explain to an AC Circuits student that the charge on acapacitor is === Oh, how I wonder...Where is the irrasionality of irrasional numbers?>They seem to be fully rational.They're not ratios (of whole numbers). Hence ir-ratio-nal.Where is the inductive logic in mathematical induction?>It seems to be deduction.Whats primary with prime numbers?>1 seems to be more primary.Whats imaginary with imaginary numbers?>-1 seems to be the only imaginary number.What is platonic with platonic figures?>They seem to be of ts world.Whats defferentiated during differensiation?>No difference is attempted found.It is a mistake to tnk that the common definitions of the> words used in mathematics will tell us sometng about the> structure of the mathematical objects. They have a structure,> independent of our language(s). We make clear what structure> we wish to study by giving a formal definition. Now, we have> the OPTION of using a long name for these tngs; for example> we can refer to convex,face-congruent,face-regular polyhedron> if we wish. (That would be the chemists' approach.) But we> don't like that mouthful, and we have a record that Plato > mentioned them, so we OPT to call them Platonic solids.> That doesn't imbue them with any additional property beyond> what is already a consequence of their definition.Sometimes the choices we have collectively made turn out to> be good ones, sometimes not so good. Imaginary number was> probably a poor choice, in retrospect. Likewise transcendental.> Regular topological space was uninspired. Perfect number was > great PR for a pretty unimportant idea; I might put chaos in > ts category too. Spectral sequence would play well in the > popular press if they could only get a sense of what to do with it.But they're all just names. The trick is to learn the actual> definition and see the consequences of that definition. > A simple group by any other name would be as cool.daveDude, I === the worst mnemonic for pi i've ever seen.-- [e-mail address === to denote a right angle in diagrams? I am looking atthe Mathematical Methods book by Arfken. On page 15 in the dot productsection he has a figure that has the angle measure symbol with a dot inside.Is ts just another way to grapcally === another way to denote a right angle in diagrams? I am looking at> the Mathematical Methods book by Arfken. On page 15 in the dot product> section he has a figure that has the angle measure symbol with a dot inside.> Is ts just another way to grapcally say a right angle?Like so: http://osj.cjb.net ? That's the usual notation for a german right === another way to denote a right angle in diagrams? I am looking at> the Mathematical Methods book by Arfken. On page 15 in the dot product> section he has a figure that has the angle measure symbol with a dotinside.> Is ts just another way to grapcally say a right angle?The first figure on the page:http://mathworld.wolfram.com/RightAngle.htmlTom === notation. I am asking is there another way of denoting aright angle in diagrams. Is there another way to denote a right angle in diagrams? I am lookingat> the Mathematical Methods book by Arfken. On page 15 in the dot product> section he has a figure that has the angle measure symbol with a dot> inside.> Is ts just another way to grapcally say a right angle? The first figure on the page:> http://mathworld.wolfram.com/RightAngle.html> Tom son> === notation. I am asking is there another way of denoting a> right angle in diagrams.I understand. I sure can't draw a little square in a cornerGo Figure...L(90)(= R.A.)+Or you could make sometng up...Use the * symbol and explain it === notationIMHO, you could use the perpendicular symbol if you === 3 valued logic by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1D3GA411021;ts is an attempt at axiomatizing a universal set using three valuedlogic to avoid russell's paradox.i have changed the subsets axiom so that if binary logic is applied,then it remains the same axiom.i have changed the foundation axiom so that U can be an element of Uyet if U did not exist, the change to the foundation axiom would notexist (it is formulated as IF there is a universal set then it MAYcontain itself).some results:cantor's diagonal argument fails to prove that U does not map ontoP(U).U=P(U), where = means equals, not in bijection with.if U<=x for any x, where ts means domination, then U~x where ~ meansis bijection with. thus U is the largest set.it immediately follows that the set of all functions from U to U is inbijection with U and so OMEGA to the OMEGA is OMEGA.ts also shows that the OMEGA product of OMEGA is OMEGA.i then dabble into an investigation of U as a boolean ring and discusssome more elementarily derivable properties. i discuss the concept ofa universal limit and convergence without reference to a topology ormetric.ts is still a rough draft and all feedback is appreciated so that === with 3 valued logic(addition)> ts is an attempt at axiomatizing a universal set using three valued> logic to avoid russell's paradox.>Isn't ts a bit drastic? Why do you need a formalized universal set anyway?I have implemented a set theory (in PC-based software, downloadable athttp://www.dcproof.com ) in wch you can prove that the Russell Set and,consequently, the Universal Set do not exist.I don't see a problem witheither result. It also takes care the of paradoxical nature of Cantor'spower set theorem since is === Re: universal set with 3 valued logic> ts is an attempt at axiomatizing a universal set using three valued> logic to avoid russell's paradox.I noticed that you mention that MP is not a tautology.I am curious what you tnk about formulas that arenot false, but still aren't true.Russell- 2 many 2 countps: I was able to read your page earlier, but now I am having === data points on a 2d screen by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1D3GAD11010;For 4D, I have always plotted the 3D function and then looked at a moving picture for the 4D object and then just imagine that all the pictures I am seeing happend all at === 2d screen> For 4D, I have always plotted the 3D function and then> looked at a moving picture for the 4D object and then> just imagine that all the pictures I am seeing happend all> at once.Unfortunately, that's not an option here... the plot willactually be tracking a realtime process!I have seen pictures of n-dimensional plots like I'm talkingabout before, === plotting n-dimensional data points on a 2d screen> I have seen pictures of n-dimensional plots like I'm talking> about before, but never with a description of the algorithm.The formula you described in the 3D-case has some kind of perspective projection in it. Is ts really necessary? Why not just use a projection onto a 2D plane? That's quite easy in any-D.And about the problem of multiple different points projecting onto the same pixel of your screen, that's uncircumventable when reducing the dimensionality of your problem.-- M.vr.gr. Dave === n-dimensional data points on a 2d screen> I have seen pictures of n-dimensional plots like I'm talking> about before, but never with a description of the algorithm. The formula you described in the 3D-case has some kind of perspective> projection in it. Is ts really necessary? Why not just use a> projection onto a 2D plane? That's quite easy in any-D.> And about the problem of multiple different points projecting onto the> same pixel of your screen, that's uncircumventable when reducing the> dimensionality of your problem.Not necessarily. Look up star plot or radar plot. You lose allgeometrical aspects of the data, but at least you can plot the pointsunambiguously, provided you have a lower bound on each of yourdimensions. Ts is best done in polar coordinates, but there areformulas to transform back and fourth between those and rectangularcoordinates. Google should be able to help.Perhaps the OP should start by listing wch properties of the dataset are important to m. (such as distances between images must beproportional to distances in reality, === for your nts. Unfortunately my bipartite graphs are not >regular.Well, the same upper bound will work in the case where the maximum degree is k (since removing edges can only === Theory: Number of maximal Matcngs> for your nts. Unfortunately my bipartite graphs are not >regular.Well, the same upper bound will work in the case where > the maximum degree is k (since removing edges can only decrease the> number of possible matcngs).I'm not sure I understood the underlying argument:Are you assuming that a graph with maximum degree k can be alwaysidentified as a subgraph of a k-regular graph on the same vertices?Because that's not true.One _can_ identify every graph with maximum degree k as a subgraph ofa k-regular graph but the one might have to add vertices (wch wouldmake the bound in questions inapplicable). But I recall seeing a paperwith some estimates of number of vertices in the smallest regularsupergraph so that might help. If anybody cares, I can === Graph Theory: Number of maximal Matcngs> Well, the same upper bound will work in the case where > the maximum degree is k (since removing edges can only decrease the> number of possible matcngs).I'm not sure I understood the underlying argument:Are you assuming that a graph with maximum degree k can be always> identified as a subgraph of a k-regular graph on the same vertices?> Because that's not true.One _can_ identify every graph with maximum degree k as a subgraph of> a k-regular graph but the one might have to add vertices (wch would> make the bound in questions inapplicable). But I recall seeing a paper> with some estimates of number of vertices in the smallest regular> supergraph so that might help. If anybody cares, I can try to dig up> the paper.One can identify every graph with maximum degree k as a subgraph of a k-regular multigraph on the same vertex set. So if the upper bounds work on multigraphs and not just simple graphs (I have no idea whether they do) then you can still use them.-- Eppstein http://www.ics.uci.edu/~eppstein/Univ. of California, Irvine, === about Kripke semantics and physics (Was: Re: Study groups in science)> ,> if you are interested in scientific online workshops, please visit> site> http://de.geocities.com/scienceworkshops/> Its goal is to organize study groups on scientific topics like quantum> field theory,> probabilistic inference or neural nets, to name a few topics I am> personally interested in> (of course, arbitray topics may be suggested). What positive precautions are you taking to prevent the idiots morons and> kooks from taking it over, as has happened in sci.physics?> Franz!Let's just defuse the reference to morons and kooks by deferring toShannon's discussion concerning the statistical character of language.I want to ask you a question about the theories physicists are promulgatingthese days. So, once again, let me start with a quick Google search toestablish context.If I use the search string Ô26 dimensions string physics' I get about 959ts.For my part, I could care less about the details. It is the mathematicsthrough wch your colleagues express their explanations that is important tome. The 26 dimensions are particularly interesting here because theirsymmetries and invariants involved offer me an opportunity to ask you how totnk about truth and falsity in physical theory.Now, Galathaea has been wanting to talk about Heyting algebras and quantumlogic. Section 6.4 of the paper http://www.illc.uva.nl/Publications/ResearchReports/MoL-2001- 09.text.pdfis entitled Finite projective formulas in two variables Strangely enough,there are precisely 26 Heyting algebras associated with 2-universal modelsdiscussed here. Moreover, there is not a single mention of quantum logic.Now, isn't ts just an amazing coincidence? Logicians tell mathematiciansabout ÔT' and ÔF,' the physicists are talking about 26 dimensions, and mitchknows just where to find a paper specifying the 26 2-generated Heytingalgebras.Actually, I do not tnk it is coincidental. Unfortunately, whereas I wouldlove to offer an explanation, you would simply engage in more vulgarity.But, here is a little sometng to consider. If I am interpreting KipThorne's account correctly, Einstein applied a Kantian notion of space andtime to s formulation of special relativity. By formulating an absolutemetric, Minkowski dragged your precious theories of everytng into thefoundational arguments of mathematics. Einstein thought that was a terribletng--until he recognized that he could enhance s personal reputationfurther with general relativity. How convenient!The key combinatorial concept that ties it all together is that of a nervedefined by Eduard Cech. Apparently, these associate simplicial complexes toopen coverings. You do understand the significance of open coverings incalculus and intuitionistic semantics, don't you?But, Cech was not a physicist. Rather, he was just someone else whose ideasyour colleagues could use without proper citation because, after all, it wasmathematics. Here is a little sometng about the man in s professionalcapacity:Whenever he was doing sometng in mathematics,he always strove to aceve a thorough understandingof the subject. The result was that even outside s fieldsof research he had an extensive knowledge and deepinsight into many other areas of mathematics. Ts featureof s personality also had some other consequences.Wle he was not conceited and talked easily to peoplewith little formal education, he expected in s fellowprofessors the same qualities that he mself possessed.Ts did not contribute to smooth relations with somepeople as he was not diplomatic but, on the contrary,quite forthright in expressing s opinions. http://www-gap.dcs.st-and.ac.uk/~story/Mathematicians/ Cech.htmlI'm going to guess that he had few problems with statistical entropy andsentence construction. :-)Unlike you, I am fairly certain that I am looking at a situation that derivesfrom the fundamental theorem of algebra. But, physicists are in the businessof theories of everytng and plosopcal logicians are in the businessof telling mathematicians about truth.My only concern, as always, is the foundation of mathematics. If you couldoffer some insight--being a physicist whose education has served m well fora long career--it would be greatly appreciated. I had been under theimpression that physical theories corresponded with physical fact. But,then, facticity does not seem to be very important to distinguishedprofessors who spend more time at faculty tea than the library.By the way, Franz. The vocabulary word for the day is impudence.If anyone else has any explanations to offer or papers that might clarifyts apparent correspondence, I would appreciate that as === and physics (Was: Re: Study groups in science)> By the way, Franz. The vocabulary word for the day is impudence.Quite. It was impudent of you to waste such a long space in ts thread asthat wch I have just snipped without === EconomistsRV> I have written up a demonstration that wages and employment RV> need not be determined by the intersection of well-behaved RV> supply and demand curves for labor:RV> MW> So, what serious model does ts work improve upon, MW> and what does it tell us about the observed world that MW> we did not already an idiot: MW> Responses come, yet no answer to the original question, so again IMW> ask: MW> Does ts model show sometng of interest to any MW> serious researcher? What do we observe that it MW> explains better than competing approaches? MW> The latest response contains the following: MW> I remain of the opinion that discarding theories about the MW> world that are self-contradictory improves one's understanding MW> of the world.Poor Mark Witte continues to insult the intelligence of s readers.He presents that quote as if that is a new answer on my part. Herefuses to acknowledge that he has been ignoring an answer to squestion for some time. Personally, I'm of the opinion that one's ability to understand the world is improved by throwing out internally inconsistent theories. Personally, I'm of the opinion that one's ability to understand the world is improved by throwing out internally inconsistent theories. I remain of the opinion that discarding theories about the world that are self-contradictory improves one's understanding of the world. I remain of the opinion that discarding theories about the world that are self-contradictory improves one's understanding of the world.MW> ...the problem is that no post on ts thread indicates whatMW> self-contradictory theories we should be discarding. Is itMW> supply and demand models for labor? I have explained in tsMW> thread how the linked post is not related to simple supply andMW> demand curve analysis of labor markets and only a misunderstandingMW> of the supply and demand approach would make anyone tnk so, suchMW> as with the misuse in the context of tools like IRR. That tsMW> confusion exists on the part of the original author is made clearMW> by ts quote from the linked webpage: So much for the theory that wages and employment are determined by the intersection of well-behaved supply and demand curves in the labor market. Poor Mark Witte continues to insult your intelligence, dear reader.Note that poor Mark Witte makes no attempt whatsoever to demonstrateany misuse of tools like IRR. He tnks s mere statement willdistract you from s previous error in asserting that the Internal Rateof Return will not generally be different at different levels of costs,when revenues are unchanged.And poor Mark Witte seems to tnk readers, if any, are not awarethat, in mainstream economics, labor demand curves are supposed tobe derived from optimizing behavior of the firm. I provide acorrect analysis of such optimizing behavior. I do not end up witha labor demand curve. In fact, I do not even end up with a firmthat will necessarily adopt a more labor-intensive techninue orre more workers at a lower wage. Somehow Mark Witte tnks youwill be confused enough to believe s echo of my point, thatsuch analysis does not yield well-behaved labor demand curves,into tnking that I have not shown the foundation for well-behavedlabor demand curves is lacking.If poor Mark Witte thought otherwise, he could always try to presentan argument. For example, he might construct an equilibrium of thefirm at each possible level of the wage in my example. And thenhe might show how to draw a labor demand function. Or he couldshow how to find the equilibrium of the firm under some otherdescription of technology.I do not here address poor Mark Witte's pretence that empiricalevidence relevant to my example has not been cited on tsthread.> Question: Does the originator of ts thread believe that a> well-behaved supply and demand model is never appropriate for> explaining how labor markets function?No such claim has been advanced here. Perhaps poor Mark Wittecould outline under what special case conditions one canderive a labor demand function.By the way, I also haven't suggested firms will always want toadopt less labor-intensive techniques at lower wages. But if onethought optimizing competitive firms always adopt more laborintensive techniques at lower wages, one might outline somespecial case assumptions that yield ts conclusion.But the empirical evidence suggests that poor Mark Witte willalways beg all arguments rather than put forth a statement thatis not fallacious. It is better to be left with an empty mind than one filled with nonsense - with deductive inconsistencies and fanciful empirical hypotheses. -- Paul Samuelson since both groups of versions of marginalist equilibrium theory - the long-period versions and the neo-Walrasian versions - encounter what appear to be radical and insurmountable difficulties, one must conclude that at present there is no defensible neoclassical theory (in the sense of explanation) of prices and distribution. The onus is on the neoclassicals to show that ts is not so. Unless and until they succeed, it seems reasonable to turn to different, non-neoclassical approaches to value and distribution (and employment and growth). -- Fabio Petri, Professor Hahn on the Ôneo-Ricardian' Criticism of Neoclassical Economics, in _Value, Distribution, and Capital: Essays in Honour of Pierangelo Garegnani. Routledge, 1999.-- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.htmlr c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but ghly @ r c m unbecoming to reasonable and free men in === Advanced For Mainstream EconomistsMy last post contained the following, it remains unanswered andevidently always will. Anyone interested in understanding thesupply-and-demand framework is recommended to read an intermediatetext on microeconomics.__________________L| So much for the theory that wages and employment are determined L| by the intersection of well-behaved supply and demand curves in L| the labor market. Ts calls for answering two questions.Question 1: Does the originator of ts thread believe that awell-behaved supply and demand model is never appropriate forexplaining how labor markets function?Question 2: Give than there are many types of models for explaininglabor markets, what observed aspect of labor market behavior does tsapproach explain better than other established models?RV> I have written up a demonstration that wages and employment > RV> need not be determined by the intersection of well-behaved > RV> supply and demand curves for labor:> RV> MW> So, what serious model does ts work improve upon, > MW> and what does it tell us about the observed world that > MW> we that he tnks you an idiot:> MW> Responses come, yet no answer to the original question, so again I> MW> ask:> MW> Does ts model show sometng of interest to any > MW> serious researcher? What do we observe that it > MW> explains better than competing approaches?> MW> The latest response contains the following:> MW> I remain of the opinion that discarding theories about the > MW> world that are self-contradictory improves one's understanding > MW> of the world.Poor Mark Witte continues to insult the intelligence of s readers.> He presents that quote as if that is a new answer on my part. He> refuses to acknowledge that he has been ignoring an answer to s> question for some time. Personally, I'm of the opinion that one's ability to understand> the world is improved by throwing out internally inconsistent> theories. Personally, I'm of the opinion that one's ability to understand the> world is improved by throwing out internally inconsistent theories. I remain of the opinion that discarding theories about the world> that are self-contradictory improves one's understanding of the world. I remain of the opinion that discarding theories about the world> that are self-contradictory improves one's understanding of the world.MW> ...the problem is that no post on ts thread indicates what> MW> self-contradictory theories we should be discarding. Is it> MW> supply and demand models for labor? I have explained in ts> MW> thread how the linked post is not related to simple supply and> MW> demand curve analysis of labor markets and only a misunderstanding> MW> of the supply and demand approach would make anyone tnk so, such> MW> as with the misuse in the context of tools like IRR. That ts> MW> confusion exists on the part of the original author is made clear> MW> by ts quote from the linked webpage: So much for the theory that wages and employment are determined > by the intersection of well-behaved supply and demand curves in > the labor market. Poor Mark Witte continues to insult your intelligence, dear reader.> Note that poor Mark Witte makes no attempt whatsoever to demonstrate> any misuse of tools like IRR. He tnks s mere statement will> distract you from s previous error in asserting that the Internal Rate> of Return will not generally be different at different levels of costs,> when revenues are unchanged.And poor Mark Witte seems to tnk readers, if any, are not aware> that, in mainstream economics, labor demand curves are supposed to> be derived from optimizing behavior of the firm. I provide a> correct analysis of such optimizing behavior. I do not end up with> a labor demand curve. In fact, I do not even end up with a firm> that will necessarily adopt a more labor-intensive techninue or> re more workers at a lower wage. Somehow Mark Witte tnks you> will be confused enough to believe s echo of my point, that> such analysis does not yield well-behaved labor demand curves,> into tnking that I have not shown the foundation for well-behaved> labor demand curves is lacking.If poor Mark Witte thought otherwise, he could always try to present> an argument. For example, he might construct an equilibrium of the> firm at each possible level of the wage in my example. And then> he might show how to draw a labor demand function. Or he could> show how to find the equilibrium of the firm under some other> description of technology.I do not here address poor Mark Witte's pretence that empirical> evidence relevant to my example has not been cited on ts> thread.Question: Does the originator of ts thread believe that a> well-behaved supply and demand model is never appropriate for> explaining how labor markets function?No such claim has been advanced here. Perhaps poor Mark Witte> could outline under what special case conditions one can> derive a labor demand function.By the way, I also haven't suggested firms will always want to> adopt less labor-intensive techniques at lower wages. But if one> thought optimizing competitive firms always adopt more labor> intensive techniques at lower wages, one might outline some> special case assumptions that yield ts conclusion.But the empirical evidence suggests that poor Mark Witte will> always beg all arguments rather than put forth a statement that> is not fallacious. It is better to be left with an empty mind than one filled> with nonsense - with deductive inconsistencies and fanciful> empirical hypotheses.> -- Paul Samuelson since both groups of versions of marginalist equilibrium theory -> the long-period versions and the neo-Walrasian versions - encounter> what appear to be radical and insurmountable difficulties, one> must conclude that at present there is no defensible neoclassical> theory (in the sense of explanation) of prices and distribution.> The onus is on the neoclassicals to show that ts is not so.> Unless and until they succeed, it seems reasonable to turn to> different, non-neoclassical approaches to value and distribution> (and employment and growth).> -- Fabio Petri, Professor Hahn on the Ôneo-Ricardian' Criticism> of Neoclassical Economics, in _Value, Distribution, and Capital:> Essays in Honour of Pierangelo Garegnani. Routledge, 1999.-- > Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html> To solve Linear Programs: .../LPSolver.html> r c A game: .../Keynes.html> v s a Whether strength of body or of mind, or wisdom, or> i m p virtue, are found in proportion to the power or wealth> e a e of a man is a question fit perhaps to be discussed by> n e . slaves in the hearing of their masters, but ghly> @ r c m unbecoming to reasonable and === Problem of the Week Trig Question, I saw an interesting problem, I'm sure some of you are familiarwith it, it's the Purdue (I'm a Boilermaker fan, btw :) Problem of theWeek #5 for the Spring Series [http://www.math.purdue.edu/academics/pow/ ]. And the only reason Ithought I could tackle a problem like ts is because we've donetngs similar in our Trig class before, that is, longitude/latitudetype problems and I thought with some reading and investigating, Icould solve ts. I asked my math teacher, he said I probably wouldnot need spherical trig but mentioned the azimuth (distance to thenorthern point from the horizon). So, my question is, given thatproblem, what aspects should I read up on in Trig to help me solve it?I'm not asking for a solution, just some opinions and what inparticular I === an interesting problem, I'm sure some of you are familiar> with it, it's the Purdue (I'm a Boilermaker fan, btw :) Problem of the> Week #5 for the Spring Series [> http://www.math.purdue.edu/academics/pow/ ]. And the only reason I> thought I could tackle a problem like ts is because we've done> tngs similar in our Trig class before, that is, longitude/latitude> type problems and I thought with some reading and investigating, I> could solve ts. I asked my math teacher, he said I probably would> not need spherical trig but mentioned the azimuth (distance to the> northern point from the horizon). So, my question is, given that> problem, what aspects should I read up on in Trig to help me solve it?> I'm not asking for a solution, just some opinions and what in> particular I should read about to solve it. in advance.We are told that the observer is at latitude 42 deg. N. and that thesun's rays make an angle of 12 degrees with the plane of the equator.Are we supposed to assume that the problem statement means the sun is 12degrees *south* of the equator? It doesn't say that, and it surely makesa difference.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. We are told that the observer is at latitude 42 deg. N. and that the> sun's rays make an angle of 12 degrees with the plane of the equator.Are we supposed to assume that the problem statement means the sun is 12> degrees *south* of the equator? It doesn't say that, and it surely makes> a difference.Good question :) Well, I initially thought the suns rays would be beaming down parallel(as in North of the equator). But, I'm not certain, so I'm going toassume they are from the North and will try and work the problem outthat way, === Problem of the Week Trig Questioncircle_k> ...> Problem of the> Week #5 for the Spring Series [> http://www.math.purdue.edu/academics/pow/ ].> ... I asked my math teacher, he said I probably would> not need spherical trig but mentioned the azimuth (distance to the> northern point from the horizon).You won't need the azimuth either. The question comes down to, whatpercentage of the 42nd parallel is sunlit, when the sun's rays make an angleof 12 degrees at the equator. If it were, say 60% or 3/5, then the length ofdaylight would be 60% of 24 hours, i.e. 14 hours and 24 minutes.I would start by drawing a diagram of some === of a linear operator is th following: T(x+y) = T(x) +T(y) and T(cx) = cT(x). Now in most cases the first rule is logicallyindependent from the second. Yet if T:V->W and both v and w are vectorspaces over the rationals, the second rule follows from the first. Cananyone please show me why? I can' t === operatorsure,the first rule imply that cT(x)=T(cx) for c any integer.let a, b be integers then (a/b)*bT((1/b)x)= T((a/b)x)but (a/b)*bT(x/b)= (a/b)*T(x)here you go> Hey everyone, The definition of a linear operator is th following: T(x+y) = T(x) +> T(y) and T(cx) = cT(x). Now in most cases the first rule is logically> independent from the second. Yet if T:V->W and both v and w are vector> spaces over the rationals, the second rule follows from the first. Can> anyone please show me why? I can' t seem to find the relation. .> -Greg === has a graph wch is everywhere dense overe its plane in the sense that for every circle in the plane of its graph there exists points on the graph that belong to the interior of the circle. An explicit example will be ghly === function wch has a graph wch is everywhere dense > overe its plane in > the sense that for every circle in the plane of its graph there exists points> on the graph that belong to the interior of the circle.> An explicit example will be ghly appreciated.Let E = {sqrt(2)/m : m = 1,2,3 ...}. Let r_1, r_2, ... be an enumeration of the rationals. Then the sets E_n = r_n + E are pairwise disjoint, and each is countably infinite. So for each n there exists a function f_n from E_n onto the rationals. Now define f : R -> R by setting f = f_n on each E_n, and anytng you like on the rest of R. The graph of ts f === {sqrt(2)/m : m = 1,2,3 ...}. Let r_1, r_2, ... be an enumeration of > the rationals. Then the sets E_n = r_n + E are pairwise disjoint, and each > is countably infinite. So for each n there exists a function f_n from E_n > onto the rationals. Now define f : R -> R by setting f = f_n on each E_n, > and anytng you like on the rest of R. The graph of ts f will be dense > in the plane.That gives a function f such that f(I) is dense in R for every open interval I. Actually we can design an f such that f(I) = R for every open interval I. Let I1, I2, ... be an enumeration of the open intervals with rational endpoints. Choose a Cantor set K1 contained in I1. Because K1 has no interior, we can choose a Cantor set K2 inside I1 K1. Because K1 U K2 has no interior (we don't need Baire for ts) we can choose a Cantor set K3 inside I3 (K1 U K2), and so forth. The Kn are pairwise disjoint and each has the cardinality of R. So for each n there exists an fn mapping Kn onto R. So define f to be fn on Kn and anytng you like on R === bMh4TyZLYeY0IXsdrh5RO3W7c0hXhCNYVDu9bzC0ATjkDB111Uen6u The Does there exist a real function wch has a graph wch is everywhere dense >> overe its plane in >> the sense that for every circle in the plane of its graph there exists points>> on the graph that belong to the interior of the circle.>> An explicit example will be ghly appreciated.Let E = {sqrt(2)/m : m = 1,2,3 ...}. Let r_1, r_2, ... be an enumeration of >the rationals. Then the sets E_n = r_n + E are pairwise disjoint, and each >is countably infinite. So for each n there exists a function f_n from E_n >onto the rationals.For the proof to work, you have to find first a g on E that is ontothe Rationals. That clearly exists. Then you define f_n(x) = g(x -r_n)on E_n. I tnk you will run into troubles when you try to define thef_n independently.Thomas> Now define f : R -> R by setting f = f_n on each E_n, >and anytng you like on the rest of R. The graph === problem>Let E = {sqrt(2)/m : m = 1,2,3 ...}. Let r_1, r_2, ... be an enumeration of >the rationals. Then the sets E_n = r_n + E are pairwise disjoint, and each >is countably infinite. So for each n there exists a function f_n from E_n >onto the rationals.For the proof to work, you have to find first a g on E that is onto> the Rationals. That clearly exists. Then you define f_n(x) = g(x -r_n)> on E_n.You can define it that way if you want, but it's not necessary.> I tnk you will run into troubles when you try to define the> f_n independently.I tnk not. For the f I specified, f(I) is dense in R for every === OptimalStrategy via VonNeumann Gametheory on Playing the Stock MarketPortfolio of PAF as of 12FEB04:BCE 9,720 22.00 $213,840.00 BMY 50 29.89 $1,494.50 SBC 12,200 25.64 $312,808.00realestate land 3APR03 of 3 lots $19,000.science-art of pictures,porcelain etc starting JAN03 for $14,143.realestate land 30JUL03 another lot $11,500.Finally, finally I hope I have some clear sailing for the remainder ofts year. What I was trying to aceve for the past 8 months was toget rid of competing Strategies. For the Optimal Strategy is a uniqueand in addition I was also employing a sideshow strategy or asecondary strategy. The OS is Crossover but the secondary strategy Iwas employing with Schering Plough and with Qwest was a buyoutto get rid of the last and final Qwest. And even though Qwest or SGPhave a good and reasonable chance of being bought out in the futuremonths or years, I simply do not care. Good on them if they get boughtout. The OS of StockMarket should not be diversified for a strategyof the OS is unique. And now that I have gotten rid of SGP and Qwestthe above 3 companies form the basis for the OS of Crossovertechnique.I would like to mention that when ts portfolio started in Autumn of2002 with total funds of $461,000 that if I had simply bought SBC in2002 that it would have come to about 21,000 shares of SBC at thattime and if I look at the above portfolio now it is close to 22,000shares of the three companies of SBC, BCE, and BMY. Even though thepaper money value of the above is $571,000, that if I had just boughtSBC back in October 3, 2002 that it would have been about 21,000shares of SBC. Wch is a fine example of where people tnk they arerobustly gaining and forging ahead whereas in real wealth terms theyare mostly just crawling ahead.But today I sold my last 14,200 shares of Qwest at 4.67 and with theproceeds bought 3,000 more shares of BCE. I do not know why Q was up9% on the day with its 30 cent rise. Is it perhaps that if Vodafonedoes not get AT&T wireless they will go after Qwest? Perhaps so. But Ido not care because the OS of the Stockmarket is not a strategy thatseeks laggards and downtrodden companies. The OS of Stockmarket seekspairs of companies that are healthy and vibrant where the playerswitches out of one company into the second company and then switcngout of the second back into the first and the process is continuedover and over again.One of the troubles with SGP and Qwest was their little to no dividendwhereas SBC, BCE, BMY have lovely dividends. So if no switcng due toCrossover is realized then at least nice dividends ßow in.I like to compare the OS of Chess with the OS of StockMarket. Andwhere a pawn gain is like a Crossover of 3 instances. Because if youhave masters playing chess, a sizable gain is a pawn wch can win.But if you have nonmasters playing chess that in their sloppiness oneloses a knight or a bishop or more. So my having SGP or Qwest in theportfolio was like expecting a bishop or rook capture in chess whereasplaying Crossover is expecting a pawn advantage and a slow but steadyand assured win.My days of buying shares in companies that are downtrodden with thehopes of being bought-out are finished with, for I now seek middle ofthe road companies wch are nice dividend payers and whose priceßuctuates so that I can gain more shares by switcng back and forth.A recent example is that in my last post on ts subject I had sold100 shares of Merck and bought more BCE, but I had mentioned that aMerck to Medtronics would be a very nice switcng campaign withtrading at least a dollar gher than MRK, but today I see MRK isgher by a dollar to MDT. And so that a Crossover had been made injust 2 weeks time and that if in a month time more, if MDT goes gherthan MRK is a time to unload a sizable number of MDT shares and buyback MRK getting some of those shares as freebies.So, I am happy that I can focus and concentrate solely on the OptimalStrategy via Crossover of the StockMarket. There is one tng badabout the above portfolio in that it is concentrated in only 2companies. It would be better if I divided it out into 4 companieswith at least two switcng campaigns. One such plan would be to have1/4 in BCE, 1/4 in SBC, 1/4 in MRK and 1/4 in MDT. So now I feel Ishould diversify somewhat. The tng I do not like about the drugsrecently is that they are overpriced and that companies like MDT giveso small a dividend.Arcmedes Plutoniumwhole entire Universe is just one big atom where dotsof the === ts is my first post here, and I hope ts is the rightng, and that I'm not asking a rtfm kind of question...Here's my problem.I've got a coordinate system, and need to draw a triangle given anorigin. The origin will be the Ôcenter' (I use the term loosely) ofthe triangle. It won't be an equilateral triangle, but the trianglewill Ôpoint' in a given direction.So I need a function that will return me 3 verticies (3 sets ofpoints) given 4 inputs. The inputs are x_origin, y_origin, direction(vector?), and size. The size would be the distance from the nose ofthe pointing triangle (it's furthest point along the vector), and theback of itso if I gave inputs of 100, 100, 0 (facing right?), and 10 ...Assuming a coordinate system where 0,0 is in the upper-lefthandcorner, and x grows to the right, and y grows downward ...I would expect my verticies to be SOMETNG like:105, 100 <-- The nose of it pointing right (0 degrees)95, 97 <-- The first Ôback' vertex slightly above the origin andwell bend it95, 103 <-- The second Ôback' vertex being slightly below theorigin and bend itI know I'm presenting all ts in a very naive fasion (I never didwell in math), but for bearing with my explanation.The above computations were done just for example's sake, and they getVERY difficult for me once you start rotating the direction thetriangle points. I'm assuming I need to use trig functions to figurets out, but I have no Idea how to do it.Is ts stuff easy to compute? in advance for any help.Extra credit:I am eventually going to need to find the coordinate points at anarbitrary distance along the vector originating from the origin of thetriangle. For example, using the above triangle, if I needed to findthe coords of a point 100 along the vector, I would need a functionlike so:input: x_origin, y_origin, direction, distanceoutput: x, yand it would return 200, 100 ...I'm assuming that whatever equations will give me the desired numbers,that they won't be whole numbers... That's OK ... I'll be able to dosome rounding. again for any help... it's greatly === my first post here, and I hope ts is the right> ng, and that I'm not asking a rtfm kind of question... Here's my problem. I've got a coordinate system, and need to draw a triangle given an> origin. The origin will be the Ôcenter' (I use the term loosely) of> the triangle. It won't be an equilateral triangle, but the triangle> will Ôpoint' in a given direction. So I need a function that will return me 3 verticies (3 sets of> points) given 4 inputs. The inputs are x_origin, y_origin, direction> (vector?), and size. The size would be the distance from the nose of> the pointing triangle (it's furthest point along the vector), and the> back of it so if I gave inputs of 100, 100, 0 (facing right?), and 10 ... Assuming a coordinate system where 0,0 is in the upper-lefthand> corner, and x grows to the right, and y grows downward ... I would expect my verticies to be SOMETNG like: 105, 100 <-- The nose of it pointing right (0 degrees)> 95, 97 <-- The first Ôback' vertex slightly above the origin and> well bend it> 95, 103 <-- The second Ôback' vertex being slightly below the> origin and bend it> I know I'm presenting all ts in a very naive fasion (I never did> well in math), but for bearing with my explanation. The above computations were done just for example's sake, and they get> VERY difficult for me once you start rotating the direction the> triangle points. I'm assuming I need to use trig functions to figure> ts out, but I have no Idea how to do it. Is ts stuff easy to compute? in advance for any help. Extra credit: I am eventually going to need to find the coordinate points at an> arbitrary distance along the vector originating from the origin of the> triangle. For example, using the above triangle, if I needed to find> the coords of a point 100 along the vector, I would need a function> like so: input: x_origin, y_origin, direction, distance> output: x, y and it would return 200, 100 ... I'm assuming that whatever equations will give me the desired numbers,> that they won't be whole numbers... That's OK ... I'll be able to do> some rounding. again for any help... it's greatly appreciated!Here MakeArrow( x_origin, y_origin: FloatingPoint; direction: FloatingPoint; size: FloatingPoint; distance: FloatingPoint; var x1, y1: FloatingPoint; var x2, y2: FloatingPoint; var x3, y3: FloatingPoint; var x_point, y_point: FloatingPoint);const HeightFactor=0.6; // The aspect ratio of the arrowvar dx, dy: FloatingPoint; Theta: FloatingPoint; procedure RotateAndTranslate(var x, y: FloatingPoint); var xx, yy: FloatingPoint; begin xx:= x*Cos(Theta)-y*Sin(Theta); yy:= x*Sin(Theta)+y*Cos(Theta); x:= xx+x_origin; y:= yy+y_origin end; { RotateAndTranslate }begin { MakeArrow } dx:= size/2; dy:= dx*HeightFactor; Theta:= direction*Pi/180; x1:= dx; y1:= 0; x2:= -dx; y2:= -dy; x3:= -dx; y3:= dy; x_point:= distance; y_point:= 0; RotateAndTranslate(x1, y1); RotateAndTranslate(x2, y2); RotateAndTranslate(x3, y3); parameters to MakeArrow: x_origin, y_origin: your definition direction: degrees of clockwise rotation [if y axis is pointing down] size: your definition distance: the distance away of your test point x1, y1: vertex 1 x2, y2: vertex 2 x3, y3: vertex 3 x_point, y_point: the test pointHere are some test results: MakeArrow( 100, 100, 0, 10, 100, x1, y1, x2, y2, x3, y3, x4, y4 );105.000 100.000 95.000 97.000 95.000 103.000200.000 100.0007 degree rotation: MakeArrow( 100, 100, 7, 10, 100, x1, y1, x2, y2, x3, y3, x4, y4 );104.963 100.609 95.403 96.413 94.672 102.368199.255 112.18790 degree rotation (arrow pointing down): MakeArrow( 100, 100, 90, 10, 100, x1, y1, x2, y2, x3, y3, x4, y4 );100.000 105.000103.000 95.000 97.000 95.000100.000 200.000-- Clive === what can you say about the inf-norm(E*A^(-1)) and inf-norm(A^(-1)*E)?>Here inf-norm is the maximal absolute row sum.>A is a square matrix, E is a Ô'small'' matrix of the same order.>I wonder if there exists some theorem/conclusion/hypothesis on the>relationsp between>INF-NORM(E*A^(-1)) and INF-NORM(A^(-1)*E) ?Small doesn't mean much here because both sides are positivelyhomogeneous in E. And if A is arbitrary (other than being invertible)you might as well replace A^(-1) by another arbitrary matrix B.There really isn't any relation, e.g. consider [ a 0 ] [ 0 c ]A^(-1) = [ 0 b ], E = [ 0 0 ]where INF-NORM(E A^(-1)) = |bc| wle INF-NORM(A^(-1) E) = |ac|.The ratio of the two norms is |b|:|a|, wch can be any === rational points of known measure>Is anyone aware of a construction (or a reference to a construction)>of a (nontrivial) open set containing all rational points on an>interval with a proof of its exact measure. For example, a>construction of an open set on [0,1] with measure exactly 1/2,>containing every rational point in the interval. (Of course, it's an>easy exercise to construct a set with these properties with a measure>witn an arbitrarily good approximation of a desired value.)Let {r_j: j=1,2,...} be an enumeration of the rationals in [0,1].Start with U_1 = emptyset.At stage n, U_n will be a union of finitely many open intervals with total length 1/2 - 2^{-n}, and its complement will be a union offinitely many disjoint closed intervals C_j of total length 1/2 + 2^{-n} (of wch some may have length 0), and we will defineU_{n+1} to be the union of U_n with subintervals V_j of each C_j, where length(V_j) = 2^{-n-1} length(C_j)/(1/2 + 2^{-n}) (call ts m_j),chosen as follows. Let C_j = [a_j, b_j].1) if C_j is a single point, V_j is that point.2) otherwise, if a_j < r_j - m_j/2 < r_j + m_j/2 < b_j,V_j = (r_j - m_j/2, r_j + m_j/2).3) otherwise, if r_j - m_j/2 < a_j <= r_j, V_j = [a_j, a_j + m_j).4) otherwise, V_j = (b_j - m_j, b_j].Notice that if r_j is in C_j we'll always have === points of known measure> Is anyone aware of a construction (or a reference to a construction)> of a (nontrivial) open set containing all rational points on an> interval with a proof of its exact measure. For example, a> construction of an open set on [0,1] with measure exactly 1/2,> containing every rational point in the interval. (Of course, it's an> easy exercise to construct a set with these properties with a measure> witn an arbitrarily good approximation of a desired value.)Suppose {r_1, r_2, ...} are the rationals in (0,1). For each r in (0,1), let d(r) = min(r,1-r). For x >= 1, let f(x) denote the measure of the union of [0,1/x), (1-1/x, 1] and the open intervals centered at r_n of radius d(r_n)/x^n. Then f : [1,oo) -> (0,1], f is continuous and decreasing, f(1) = 1, and f(x) -> 0 as x -> oo. By the intermediate value theorem, f takes on all values in (0,1]. So for each m in (0,1] there is an open set in [0,1] of measure m containing all the rationals === all diagonals drawn, how many regions are there? In particular, how many triple (or N-tuple) concurrences of diagonals are there? (Kok)Pl-- Unpatched IE vulnerability: WMP local file bounceDescription: Switcng security zone, arbitrary command execution, automatic email-borne command executionReference: http://www.ntbugtraq.com/default.asp?pid=36&sid=1&A2=ind0307&L =ntbugtraq&F=P&S=&P=6783Exploit: === Geometry poserPl Carmody> Given a regular n-gon with all diagonals drawn, how many> regions are there? In particular, how many triple (or N-tuple)> concurrences of diagonals are there? (Kok)I followed the link given by N.8estor ():http://citeseer.nj.nec.com/cache/papers/cs/14168/http: zSzzSzwww.math.berkeley.eduzSz~poonenzSzpaperszSzngon.pdf/ poonen98number.pdfOn page 2 there is a most beautiful picture showing the 30-gonwith all its diagonals drawn. Sort of fractal - really lovely.The remarkable numbers for various n-gons can be found inthe OEIS http://www.research.att.com/~njas/sequences/A007678 the number of regionsA006561 the number of intersectionsIn the references you will see our most noble sci.math regularGerry Myerson (Rational Products of sines of rational angles).For n=30 your questions are answered (on page 2) as follows:N=2 #concurrences = 13800N=3 #concurrences = 2250N=4 #concurrences = 420N=5 #concurrences = 180N=6 #concurrences = 120N=7 #concurrences = 30N=15 #concurrences = 1---Sum is A006561(30) = 16801 (see OEIS)There are certainly many interesting observations witn tswonderful paper. Let me just cite the following from page 1:~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~It will result from our analysis that for n > 4, the maximumnumber of diagonals of the regular n-gon that meet at a pointother than the center is 2 if n is odd 3 if n is even but not divisible by 6 5 if n is divisible by 6 but not 30, and, 7 if n is divisible by 30with two exceptions: ts number is 2 if n=6, and 4 if n=12.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ === escribi.97:> Given a regular n-gon with all diagonals drawn, how many> regions are there? In particular, how many triple (or N-tuple)> concurrences of diagonals are there? (Kok)>Pl, go tohttp://www.math.uwaterloo.ca/JIS/sommars/ newtriangle.htmlandhttp://math.berkeley.edu/~poonen/papers/ === Geometry poser| Pl Carmody asked:| Given a regular n-gon with all diagonals drawn, how many| regions are there? In particular, how many triple (or N-tuple)| concurrences of diagonals are there? (Kok)See A007678 in the On-line Encyclopedia of Integer Sequences www.research.alt.com/~mjas/sequences/index.html ________Gerard === a regular n-gon with all diagonals drawn, how many> | regions are there? In particular, how many triple (or N-tuple)> | concurrences of diagonals are there? (Kok)See A007678 in the On-line Encyclopedia of Integer Sequences www.research.alt.com/~mjas/sequences/index.html ________Gerard S. A far more interesting, if less direct, route to the answer than the other two replies ;-)I suspect you need a new keyboard.Pl-- Unpatched IE vulnerability: NavigateAndFind file proxyDescription: cross-domain scripting, cookie/data/identity theft, command executionReference: http://safecenter.net/liudieyu/NAFfileJPU/ NAFfileJPU-Content.HTMExploit: http://safecenter.net/liudieyu/NAFfileJPU/NAFfileJ PU-MyPage.htm= === ==Subject: Re: Geometry poserhttp://citeseer.nj.nec.com/cache/papers/cs/14168/http: zSzzSzwww.math.berkeley.eduzSz~poonenzSzpaperszSzngon.pdf/ poonen98number.pdf:)Pl Carmody diagonals drawn, how many> regions are there? In particular, how many triple (or N-tuple)> concurrences of diagonals are there? (Kok)> Pl -- > Unpatched IE vulnerability: WMP local file bounce> Description: Switcng security zone, arbitrary command execution,> automatic email-borne command execution> Reference:http://www.ntbugtraq.com/default.asp?pid=36&sid=1&A2 =ind0307&L=ntbugtraq&F=P&S=&P=6783> Exploit: === Geometry poserGiven a regular n-gon with all diagonals drawn, how many> regions are there? In particular, how many triple (or N-tuple)> concurrences of diagonals are there? (Kok)SeeMR1612877 (98k:52027)Poonen, Bjorn(1-BELL); Rubinstein, Michael(1-BELL)The number of intersection points made by the diagonals of a regularpolygon. (English. English summary)-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes s penis for a square root, 88-9Francis Wheen, _How === solution> Group,> I hope someone will give me advice, or point me in the right direction> so that I can solve the following type of problem. Given:> N TNGS that produce GOOD stuff, but at a COST as follows; COST(i) = exp(slope(i)*(GOOD(i)-offset(i))); i=1,N> wch implies GOOD(i) = log(COST(i))/slope(i) + offset(i); i=1,N I have a target amount of GOOD I need to get out of my TNGS, but I wantto> minimize the COST> so> SUM(GOOD(i)) = Target> SUM(COST(i)) is minimum It seems to me that the minimum is reached when the slopes of the GOOD vs> COST curves of each TNG are equal, or 1/(COST(i) * slope(i)) =SOME_CONST. Whats the best way to go about solving ts sort of problem? ,> Chris for the useful === optimal solutionFenchel's Theorem. That, or Lagrange === someone will give me advice, or point me in the right direction> so that I can solve the following type of problem. Given:> N TNGS that produce GOOD stuff, but at a COST as follows; COST(i) = exp(slope(i)*(GOOD(i)-offset(i))); i=1,N> wch implies GOOD(i) = log(COST(i))/slope(i) + offset(i); i=1,N I have a target amount of GOOD I need to get out of my TNGS, but I wantto> minimize the COST> so> SUM(GOOD(i)) = Target> SUM(COST(i)) is minimum It seems to me that the minimum is reached when the slopes of the GOOD vs> COST curves of each TNG are equal, or 1/(COST(i) * slope(i)) =SOME_CONST. Whats the best way to go about solving ts sort of problem? ,> ChrisThe method of Lagrange multipliers yields your result as the unique criticalpoint. You can find the value of SOME_CONST === Re: there is no such tng as infinity> The program in FORTRAN is simple:00001 n=1>> 00002 1 n=n+1>> 00003 print(3,4)n>> 00004 if(n.eq.M) then print(3,4)M>> 00005 else go to 1>> 00006 end if>> 00007 endIt has currently reached about 2.0 x 10^18.You LIE! No output device in the world could keep up with the pace>you imply. If it output a million lines per second, your current>value would have taken about 60,000 years.He's seeing a burnt-in screen. Thus, he doesn't have todo any TTY output in order to see a display of numbers.I'm still trying to figure out how he adds bits ahead of theaccumulator. I'm also not questioning wch compiler languagehe's using. None of the above looks like my beloved FORTRAN.nt: FORTRAN is spelled correctly and implies wch FORTRANsI'm referencing./BAHSubtract a hundred === infinity> I've thought really hard about ts one and came to the conclusion> that there is no scientific evidence of infinity existing. The ghest> number that anyone has ever measured to according to Isaac Asimov in> s book Science and Human Thought is only about 5.0 x 10^48. No one> has ever gotten past that number. Doesn't ts sound weird?> What's to say that eventually there is a number where it is impossible> to count gher than? If someone were to find ts number and prove> that it is in fact the ghest number, then that person would> undoubtably be rich and famous.I am currently running a computer program that will eventually find> ts magic number (I hope and pray) that I call M for short. It> counts and counts and counts and my theory is that it will eventually> stop at M. I am looking for collaborators in ts experiment so that I> can use their computer time. The program in FORTRAN is simple:00001 n=1> 00002 1 n=n+1> 00003 print(3,4)n> 00004 if(n.eq.M) then print(3,4)M> 00005 else go to 1> 00006 end if> 00007 endIt has currently reached about 2.0 x 10^18. Just as Einstein proved> that there is no aether, I am convinced that I will prove that there> is no infinity and then write a book or two.Dr. Ben ZonaFirst of all infinity is not a number or any such tng. On thecontrary to the claim, infinity alone exists. As such, its appearance,the universe is only as observed. And so, wle infinity exists, theobservers' universe is simply observed.Now, ts observers' universe forms the foundation for the observerentangled quantum theory and observer related relativity theory, themodern theories of physics.S S === Fortran says positive infinity = 2147483647 and negative infinity => -2147483648. Weird tng is 2*(positive infinity) = 0. Where should I publish my findings?2*(positive infinity) = -22*(negative infinity) = 0$ create test.for integer *4 i, j i = 2147483647 j = 2 * i type *, j i = 2147483648 j = 2 * i type *, j end$ fort test$ link test$ r test -2 0Weird or not, that's === Re: there is no such tng as infinity>> Fortran says positive infinity = 2147483647 and negative infinity =>> -2147483648. Weird tng is 2*(positive infinity) = 0. Where should I publish my findings?2*(positive infinity) = -2>2*(negative infinity) = 0$ create test.for> integer *4 i, j> i = 2147483647> j = 2 * i> type *, j> i = 2147483648> j = 2 * i> type *, j> end>$ fort test>$ link test>$ r test> -2> 0Weird or not, that's two's complement without overßow detection for you.That's how your FORT compiler worked at handling numbers. Youreally have to look under the hood to see if there's a mess.I usually DDTed and stuffed a word with 36 1-bits (we had 36-bitword size) and then fed it out of our OTS (object time system)wch is the downstream code that really does the I/O conversionsinto human-human readable forms wch has no relationsp (otherthan being in the same address space) with the arithmetic macnations of the code./BAHSubtract a hundred and four for === wibbled:You tnk you can reach the largest number with a FORTAN program? How> crude! Obviously, you need to write ts program in C++. Anytng worth doing can be done in Perl, in a fraction of the memory space sometng like C++ would use.-- Next on list: === wibbled:You tnk you can reach the largest number with a FORTAN program? How>> crude! Obviously, you need to write ts program in C++. Anytng worth doing can be done in Perl, in a fraction of the memory >space sometng like C++ would use.You mean occupy, not use./BAHSubtract a hundred and four for === is no such tng as infinity>> You tnk you can reach the largest number with a FORTAN program? How>> crude! Obviously, you need to write ts program in C++.Anytng worth doing can be done in Perl, in a fraction of the memory> space sometng like C++ would use.And can be written in exponentially more obfuscated a manner than even C orassembler... :) -- Darryl L. Pierce Visit the Infobahn Offramp - What do you care what other people === infinity <4f1e888502f84d511af633bad4d7158d@news.teranews.com>In message <4f1e888502f84d511af633bad4d7158d@news.teranews.com>, Darryl You tnk you can reach the largest number with a FORTAN program? How> crude! Obviously, you need to write ts program in C++.>> Anytng worth doing can be done in Perl, in a fraction of the memory>> space sometng like C++ would use.And can be written in exponentially more obfuscated a manner than even C or>assembler... :)Do you have sometng against APL?-- === infinity> In message <4f1e888502f84d511af633bad4d7158d@news.teranews.com>, Darryl You tnk you can reach the largest number with a FORTAN program? How> crude! Obviously, you need to write ts program in C++.>> Anytng worth doing can be done in Perl, in a fraction of the memory>> space sometng like C++ would use.And can be written in exponentially more obfuscated a manner than even Cor>assembler... :) Do you have sometng against APL?APL has the benefit of being obvious to anyone who understands the === infinity>> You tnk you can reach the largest number with a FORTAN program? How>> crude! Obviously, you need to write ts program in C++. Anytng worth doing can be done in Perl, in a fraction of the memory> space sometng like C++ would use.>>And can be written in exponentially more obfuscated a manner than even C>>or assembler... :)Do you have sometng against APL?Just a court ruling in my favor and some outstanding payments for emotionaldamage.... ;) -- Darryl L. Pierce Visit the Infobahn Offramp - What do you care what other people === support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1DDdVC22805;Got any opinions on whether 2 + 2 = 4?>, 2 + 2 = 4.Good luck on your further contemplations of mathematics.Does the evaluation of the integration of f(x)=1 defined on the naturals equal 4? Please === by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1DDddr23036;>I need to prove that (1+X)^n >= (1+nX), n is a natural #, X is a real number>>= -1.>Here is what I have (and don't have). I know that (1+X)^n and (1+nX)>intersect at (0,1). Since [ (1+X)^n ] Ô > n for all X >0 and since (1 + nX)>' = n for all X we know that (1+X)^n > (1+nX) ( but we already knew ts>from the binomial theorem).>The real question is whether or not (1 + X) ^n and (1 + nX ) intersect below>X =0 ?> No, the real question is to prove that (1+X)^n>= (1+ nX) and that can be done by induction without worrying about where (if at all) two graphs intersect. When n= 1, (1+X)^1= 1+ X= 1+ (1)X so the statement is true. Assume the statement is true for some natural number N. Then (1+X)^(N+1)= (1+X)(1+X)^N<= (1+X)(1+NX) <= 1+ NX+ X+ NX^2 Since NX^2>= 0, <= 1+ (N+1)X Therefore, (1+X)^n<= 1+ nX for any natural number n and ANY real number X (the condition that === valued logic by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1DDdXA22880;yes, MP fails in 3 valued logic. so statements that aren't true and not false have the trd truth value. also, proofs by contradiction fail.intuitively yet in a clearly spelled out way, i'm really using binary logic, though, except in a few spots. thus, MP applies most of the time.i'm actually finding it difficult to find any fuzzy sets wch is to say a set such that the truth value of whether an element is in that set is the trd truth value. that is, besides the ones associated with paradoxes. the general idea is that russell's paradox fails to prove there is no universal set because M o M is T, where o is a particular extension of the iff biconditional. hence when one asks if russell's set S is a member of itself, or if the barber shaves mself, the is M rather than T or F.the link should now work. for letting me know the file was === appears from inspection that the diagonals of Pascal's Trianglerepresent the sequences associated with the N-dimensional(hyper)triangular numbers. Is ts guess correct?Please post responses, if any, to the newsgroup and === diagonals of Pascal's Triangle> It appears from inspection that the diagonals of Pascal's Triangle> represent the sequences associated with the N-dimensional> (hyper)triangular numbers. Is ts guess correct?Yes. > Please post responses, if any, to the newsgroup and not to my email.Done.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes s penis for a square root, 88-9Francis Wheen, === Accumulation and Limit PointI'm not sure if ts is standard terminology, but I have seen the>> following definitions for limit and accumulation points of a set S:>> x is a limit point of S if every neighborhood of x contains at least>> one element of S distinct from x. (I tnk ts is standard.) Yes standard> x is an accumulation point of S if every neighborhood of x contains>> infinitely many elements of S. (Not sure if ts is standard)> seems true according to mathworld> In metric spaces the 2 definitions turn out to mean the same tng.>> But in general topological spaces they are not the same concept.> In any T1-space, it's also true. In any *infinite* T1 space, you must mean. Yes, but if in a T1 space a point is a limit point of some set, thenthe space is necessarily infinite. If a T1 space is finite, then noneof its === significance of left relative residual and right relative residual matrix? >Suppose a matrix A has an appoximation B. Define left relative residual to >be L=(A-B)*A^(-1), right relative residual to be R=A^(-1)*(A-B). What are >the significance and relationsp of their norms? Would one norm always >larger than another?-Joenyimts question makes sense only in combination with a numerical method for computing A^{-1}. the error bound ||E||||A^{-1}|| is the same for both, but the true values can be quite different. typically,if you compute A^{-1} using lu decomposition of A and n right hand sides, then A*(A^{-1}_computed} -I might be much smaller than A^{-1}_computed*A-I.see the discussion in wilkinson: rounding === left relative residual and right relative residual matrix?> Suppose a matrix A has an appoximation B. Define left relative residual to> be L=(A-B)*A^(-1), right relative residual to be R=A^(-1)*(A-B). What are> the significance and relationsp of their norms? Would one norm always> larger than another?-JoenyimI tnk that examples will show that they are not the same. Certainly one cannot have that, for example, the left one is always larger than the right one, for then, by replacing A and B by their transposes, one could deduce that the right one is always larger than the left, and hence they are equal.However, given that A is invertible (wch you are implicitly assuming), you can show that L is invertible if and only if R is invertible (and more generally they will have the same eigenvalues except possibly only one of them might have === relative residual and right relative residual matrix?>> Suppose a matrix A has an appoximation B. Define left relative >> residual to>> be L=(A-B)*A^(-1), right relative residual to be R=A^(-1)*(A-B). What are>> the significance and relationsp of their norms? Would one norm always>> larger than another?-Joenyim> I tnk that examples will show that they are not the same. Certainly > one cannot have that, for example, the left one is always larger than > the right one, for then, by replacing A and B by their transposes, one > could deduce that the right one is always larger than the left, and > hence they are equal.However, given that A is invertible (wch you are implicitly assuming), > you can show that L is invertible if and only if R is invertible (and > more generally they will have the same eigenvalues except possibly only > one of them might have eigenvalue 1).> Actually tnking about it, I tnk that they will simply have all the same eigenvalues (the exception about the case 1 only applies when A and B are operators on infinite dimensional === transcendental numbers ?>>e^[ipi]=-1 >My comments>--->Since e^[iPi]=cosPi+isinPi>or , e^[iPi]=-1+i[0]>then there are two solutions here, to the given equatio:>A) e^[ipi]=-1 the real part solution and >B) e^[ipi]=i[0] , or e^[ipi]=0 the imaginary part solutio.>>The reasoning bend A and B is not evident to me. Please explain>>where B comes from.> e^[ipi]=0 is The solution if the imaginary part.No. If you are taking imaginary parts, then the conclusion is thatsin(pi) = 0, not e^{i pi] = 0. Similarly, if you take real partsof e^[i pi] = -1, you get cos(pi) = -1.> Just tnk that angle is 179 deg>the real component is cos(179deg) = -0999847695approximately.>and the imaginary is i*sin(179deg)=i*0.017452406approximately.>e^[iPi179/180]= -0999847695 +i*0.017452406approximately.>in polar form ts will be e^[iPi179/180]=mod 1, arg 179e^[iPi179/180]=mod 1, arg 179 degreesSo what does all ts have to do with your extraordinary claimthat e^[i pi] = 0?> e^[i pi] = -1 is right of course, but the last >>step that you mentioned in your derivation looks completely >>unjustified.The last step SHOULD have yielded the result that cos(pi) = -1.>My question :>What is the implication of ts second value of e^[ipi]=0 ?>>No implication. It is not true.> e^[iPi]=mod 1, arg 180e^[iPi]=mod 1, arg 180 degrees> Ts is correct in polar form.But it has absolutely notng to do with your earlier extraordinaryclaim that exp(i pi) = 0. That particular claim of yours is stillunjustified, and it is still false. McAnally Despite anytng you may have heard to the contrary, the rain in Spain stays almost === looking for a numerical (analytical if available) method totackle expectation functions like the one below:find r = [0,inf] forE( (exp(Z)-1)/(r*exp(Z)+1) ) = 0with Z(mu, sigma)Any comment on exact or approximation methods very === logarithm equationHavent had ts in years and Im reading a databaes book that uses itfor calculating trees.L= log_(n/2)(k)That means L= log sub n/2 to the base k. How do I solve for n? That equation is the inverse of (n/2)^L = k === had ts in years and Im reading a databaes book that uses it>for calculating trees.L= log_(n/2)(k)That means L= log sub n/2 to the base k. How do I solve for n? That equation is the inverse of (n/2)^L = k right?If the base is really k, you would write it L = log_k(n/2) and read itL = log base k of n/2. The === of algebraic integers, comparison checkIt turns out that there's an incredibly simple way to check the ringof algebraic integers and see that there's a problem with how it'scurrently taught.You can take two quadratics:x^2 - x + 42 = 0andy^2 - by - 7 = 0where by how algebraic integers are currently taught, you'd tnk thatthere exists an algebraic integer b, such that a root of the secondquadratic is a factor of a root of the first. Intriguingly, theredoes not, and it's easy to show.Now x^2 - x + 42 = 0 has (1 + sqrt(-167))/2 as on of its roots, andy^2 - by - 7 = 0, has as one of its roots (b + sqrt(b^2 + 28))/2.So I can simply introduce z, where(1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2wch isz = (1 + sqrt(-167))/(b + sqrt(b^2 + 28))where now the question is, can an algebraic b exist such that z is analgebraic integer?The square roots don't tell me much, but working to eliminate themadds solutions, as a first step consider:28z^2 + 2(1+sqrt(-167))bz - (1+sqrt(-167))^2 = 0where I have *two* solutions, where they arez_1 = (1 + sqrt(-167))/(b + sqrt(b^2 + 28))and z_2 = (1 + sqrt(-167))/(b - sqrt(b^2 + 28))and working still further I get196 z^4 + 28b z^3 + (186b^2 + 2324) z^2 - 168bz + 7056 = 0and I can divide both sides by 28 to finally get7z^4 + 5z^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0where I have four solutions, wch arez_1 = (1 + sqrt(-167))/(b + sqrt(b^2 + 28))z_2 = (1 - sqrt(-167))/(b + sqrt(b^2 + 28))z_3 = (1 + sqrt(-167))/(b - sqrt(b^2 + 28))z_4 = (1 - sqrt(-167))/(b - sqrt(b^2 + 28)).(Those not sure can simply multiply out(Z - z_1)(Z - z_2)(Z - z_3)(Z - z_4) to verify.)Now then our question was, could what is now z_1 be an algebraicinteger for an algebraic integer b?And now an important theorem comes into play, as an algebraic integercannot be the root of a non-monic primitive with integer coefficientsthat is irreducible over Q.So integer b's are eliminated as a possibility immediately.There are several ways to show that none of the the z's can be afraction for an integer b, but some of you may wonder why they'd needto be, like why couldn't they be irrational algebraic integers?Well, unless 7z^4 + 5z^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0is reducible over Q with that integer b, the theorem I mentionedrequires that *none* of its roots can be an algebraic integer, as it'snon-monic and primitive.Therefore, even if you had an irrational algebraic integer root, thepolynomial would have to be reducible over Q, meaning that you'd needat least one rational root.If b is irrational--remember it's an algebraic integer--then you get apolynomial of degree equal to the order of the monic polynomial withinteger coefficients for wch b is a root times 4.However, for any such polynomial, you run into the same problem thatit'd need at least one root that's rational.But you can characterize every possible root byz_j = (1 + sqrt(-167))/W_jz_k = (1 - sqrt(-167))/W_kwhere j and k are counting numbers, and W_j and W_k are algebraicintegers.The problem is that (1+sqrt(-167)/U where U is an algebraic integercannot be rational unless U has (1+sqrt(-167) itself as a factor.Therefore, in the ring of algebraic integers, the roots ofx^2 - x + 42 = 0andy^2 - by - 7 = 0never share non-unit factors in the ring of algebraic integers.Notice that the key theorem here is that one that prevents analgebraic integer from being the root of a non-monic primitive withinteger coefficients irreducible over Q.Take away the condition that you have an algebraic integer factor, andyou're ok, wch proves that there must be a ring beyond algebraicintegers, where they *can* share non-unit factors.I have called the more inclusive ring, the ring of objects, or theObject Ring.The Object Ring is a commutative ring that includes all numbers suchthat -1 and 1 are the only members that are both a unit and aninteger, where no non-unit member is a factor of any two integers thatare coprime.One of the most surprising results that follows quickly fromunderstanding the limitations of the ring of algebraic integers, andhaving the fuller definition, is a proof of === Re: Ring of algebraic integers, comparison checkhttp://www.giganews.com/info/dmca.html>[...]I have called the more inclusive ring, the ring of objects, or the>Object Ring.And if you ever gave a coherent _definition_ of the Object Ringpeople could decide whether the tngs you say about it aretrue or false - until then whatever you say about it is not evenwrong.(_Other_ people have posted _guesses_ what your definitionmight mean. But they're just guessing, and with most versionsit's not clear that there is exactly one ring satisfying the givenconditions...)>The Object Ring is a commutative ring that includes all numbers such>that -1 and 1 are the only members that are both a unit and an>integer, where no non-unit member is a factor of any two integers that>are coprime.Ok, I'll explain why ts is incoherent. To make sense, all numberssuch that must be followed by a condition that is true or falseof a _number_. Like all numbers n such that n > 5 orall numbers n such that n is green would make sense ints sense. You say all numbers such that -1 and 1 are the onlymembers... - you're following all numbers such that bya condition on the _ring_, not a condition on the elementsof the ring.Simply makes no sense.>One of the most surprising results that follows quickly from>understanding the limitations of the ring of algebraic integers, and>having the fuller definition, is a proof of Fermat's Last Theorem of>around two to three pages.Oops! Last we heard you'd disclaimed ts proof, in orderto get people off your back about it. Now you're back toclaiming you've proved FLT. People === integers, comparison check adds solutions, as a first step consider:28z^2 + 2(1+sqrt(-167))bz - (1+sqrt(-167))^2 = 0where I have *two* solutions, where they arez_1 = (1 + sqrt(-167))/(b + sqrt(b^2 + 28))and z_2 = (1 + sqrt(-167))/(b - sqrt(b^2 + 28))and working still further I get196 z^4 + 28b z^3 + (186b^2 + 2324) z^2 - 168bz + 7056 = 0As KeithK noted, that is wrong as it should be196 z^4 + 28b === Ring of algebraic integers, comparison check> It turns out that there's an incredibly simple way to check the ring> of algebraic integers and see that there's a problem with how it's> currently taught. You can take two quadratics: x^2 - x + 42 = 0 and y^2 - by - 7 = 0 where by how algebraic integers are currently taught, you'd tnk that> there exists an algebraic integer b, such that a root of the second> quadratic is a factor of a root of the first. Intriguingly, there> does not, and it's easy to show. Now x^2 - x + 42 = 0 has (1 + sqrt(-167))/2 as on of its roots, and y^2 - by - 7 = 0, has as one of its roots (b + sqrt(b^2 + 28))/2. So I can simply introduce z, where (1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2 wch is z = (1 + sqrt(-167))/(b + sqrt(b^2 + 28)) where now the question is, can an algebraic b exist such that z is an> algebraic integer? The square roots don't tell me much, but working to eliminate them> adds solutions, as a first step consider: 28z^2 + 2(1+sqrt(-167))bz - (1+sqrt(-167))^2 = 0 where I have *two* solutions, where they are z_1 = (1 + sqrt(-167))/(b + sqrt(b^2 + 28)) and z_2 = (1 + sqrt(-167))/(b - sqrt(b^2 + 28)) and working still further I get 196 z^4 + 28b z^3 + (186b^2 + 2324) z^2 - 168bz + 7056 = 0> Should ts be 168b^2 ?> and I can divide both sides by 28 to finally get 7z^4 + 5z^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0No, you get:7z^4 + bz^3 + (6.64b^2 + 83)z^2 - 6bz + 252 = 0obviously just a typo on your 5b^3, but the b^2 coefficient is suspect.KeithK where I have four solutions, wch are z_1 = (1 + sqrt(-167))/(b + sqrt(b^2 + 28)) z_2 = (1 - sqrt(-167))/(b + sqrt(b^2 + 28)) z_3 = (1 + sqrt(-167))/(b - sqrt(b^2 + 28)) z_4 = (1 - sqrt(-167))/(b - sqrt(b^2 + 28)). (Those not sure can simply multiply out (Z - z_1)(Z - z_2)(Z - z_3)(Z - z_4) to verify.) Now then our question was, could what is now z_1 be an algebraic> integer for an algebraic integer b? And now an important theorem comes into play, as an algebraic integer> cannot be the root of a non-monic primitive with integer coefficients> that is irreducible over Q. So integer b's are eliminated as a possibility immediately. There are several ways to show that none of the the z's can be a> fraction for an integer b, but some of you may wonder why they'd need> to be, like why couldn't they be irrational algebraic integers? Well, unless 7z^4 + 5z^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0 is reducible over Q with that integer b, the theorem I mentioned> requires that *none* of its roots can be an algebraic integer, as it's> non-monic and primitive. Therefore, even if you had an irrational algebraic integer root, the> polynomial would have to be reducible over Q, meaning that you'd need> at least one rational root. If b is irrational--remember it's an algebraic integer--then you get a> polynomial of degree equal to the order of the monic polynomial with> integer coefficients for wch b is a root times 4. However, for any such polynomial, you run into the same problem that> it'd need at least one root that's rational. But you can characterize every possible root by z_j = (1 + sqrt(-167))/W_j z_k = (1 - sqrt(-167))/W_k where j and k are counting numbers, and W_j and W_k are algebraic> integers. The problem is that (1+sqrt(-167)/U where U is an algebraic integer> cannot be rational unless U has (1+sqrt(-167) itself as a factor. Therefore, in the ring of algebraic integers, the roots of x^2 - x + 42 = 0 and y^2 - by - 7 = 0 never share non-unit factors in the ring of algebraic integers. Notice that the key theorem here is that one that prevents an> algebraic integer from being the root of a non-monic primitive with> integer coefficients irreducible over Q. Take away the condition that you have an algebraic integer factor, and> you're ok, wch proves that there must be a ring beyond algebraic> integers, where they *can* share non-unit factors. I have called the more inclusive ring, the ring of objects, or the> Object Ring. The Object Ring is a commutative ring that includes all numbers such> that -1 and 1 are the only members that are both a unit and an> integer, where no non-unit member is a factor of any two integers that> are coprime. One of the most surprising results that follows quickly from> understanding the limitations of the ring of algebraic integers, and> having the fuller definition, is a proof of Fermat's Last Theorem of> around two to three pages.> === check>You can take two quadratics:x^2 - x + 42 = 0andy^2 - by - 7 = 0where by how algebraic integers are currently taught, you'd tnk that>there exists an algebraic integer b, such that a root of the second>quadratic is a factor of a root of the first.Can you explain why? It doesn't seem obvious to me.-- Richard-- Spam filter: to mail me from a .com/.net site, put my === algebraic integers, comparison check Adjunct Assistant Professor at the University of Montana.>>You can take two quadratics:>>x^2 - x + 42 = 0>>and>>y^2 - by - 7 = 0>>where by how algebraic integers are currently taught, you'd tnk that>>there exists an algebraic integer b, such that a root of the second>>quadratic is a factor of a root of the first.Can you explain why? It doesn't seem obvious to me.Let r1 and r2 be the two roots of x^2-x+42 = 0. Since 7|r1*r2=42,there is a factorization of 7 into algebraic integers, 7=s*t, suchthat s|r1 and t|r2.The polynomial y^2 - (s-t)y - 7 = (y-s)(y+t) has roots -t and s. sdivides the first root of x^2-x+42, t divides the second root ofx^2-x+42. Setting b = s-t gives the what I accept as reality. --- Calvin (Calvin and === comparison check>You can take two quadratics:>>x^2 - x + 42 = 0>>and>>y^2 - by - 7 = 0>>where by how algebraic integers are currently taught, you'd tnk that>>there exists an algebraic integer b, such that a root of the second>>quadratic is a factor of a root of the first.Can you explain why? It doesn't seem obvious to me. Let r1 and r2 be the two roots of x^2-x+42 = 0. Since 7|r1*r2=42,> there is a factorization of 7 into algebraic integers, 7=s*t, such> that s|r1 and t|r2. The polynomial y^2 - (s-t)y - 7 = (y-s)(y+t) has roots -t and s. s> divides the first root of x^2-x+42, t divides the second root of> x^2-x+42. Setting b = s-t gives the result.So where does that leave James' argument?--There are two tngs you must never attempt to prove: the unprovable -- andthe obvious.--Democracy: The triumph of popularity over === algebraic integers, comparison check> It turns out that there's an incredibly simple way to check the ring> of algebraic integers and see that there's a problem with how it's> currently taught.It turns out that there's an incredibly simple way to check whethersometng posts is worth reading or not. If it startswith the words It turns out or anytng remotely similar to thesentence quoted above then you know he's just repeating the sameoldalready-disproven junk and there's no need to read any further.-- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvisefwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- === algebraic integers, comparison check> It turns out that there's an incredibly simple way to check the ring> of algebraic integers and see that there's a problem with how it's> currently taught. You can take two quadratics: x^2 - x + 42 = 0 and y^2 - by - 7 = 0 where by how algebraic integers are currently taught, you'd tnk that> there exists an algebraic integer b, such that a root of the second> quadratic is a factor of a root of the first. Intriguingly, there> does not, and it's easy to show.Please provide a citation showing that ts existence theorem is taughtanywhere or appears in any text. It appears that you just made it up basedon an observation and an unsupported conclusion. Tsk, tsk.> James If I can't find it, I'll make it up. Harris--There are two tngs you must never attempt to prove: the unprovable --and the obvious.--Democracy: The triumph of popularity over === ConjectureI have recently placed a one-page file on the proof of Beal'sConjecture on the Web. It may be viewed at the address:geocities.com/kerrymerry2000/ Let me know of your === points might have been addressed before but I wouldreallyappreciate some clarifications in your proof :- In Def. 2.2, what does it mean Ôwe define that each covering has atleast twomembers' ? Does it simply mean that, if J has got one or no member wedo notcall it a covering ?- In Def. 2.3, what is the meaning of indexing the left subscript ofJ' ? Thatis, what is the meaning of r_i ? You explain that p_i is the i-th oddprime,but it is unclear what the subscript Ôi' means on r.- Same one, regarding the definition of P' as an intersection of P andN. You appear to have defined tngs such that P(3,pl) is a subset ofN(3,q)_odd (since pl <= q), so why take the intersection or introduceP' for that matter ?- A bit later, in the definition of A B, you appear to beputting as conditions the equivalence of A and B simultaneously withthe existence of unpaired points of B ! Doesn't the equivalence implythe nonexistence of such points ?At ts point I stopped reading. Could you please clarify or === The Clearest by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1CMOQa19499; for the reply. First, let me make a request to you for clarification of your point. Then, I will tell you a counterargumentto your argument based on the hypothesis of the lemma.> The proposed proof of Goldbach's conjecture does not> work.The heart of the proposed proof is the assertion> (Lemma 2.5, second part) > that for each large odd q,> there is no setting of residues r_i corresponding to> the primes p_i < q/2,Ts should be p_i < q, not q/2.> satisfying the following two conditions:> (1) for each i, there are at least two odd integers> less than q in the > residue class (x = r_i mod p_i);> (2) each odd prime p p_zeta) is in some > residue class (r_i mod p_i).The following counting argument shows that ts> assertion is false.Letting pi(x) denote the number of primes less than> x, we know > pi(x)=(x/log x)(1+o(1)).> Ts implies that for all q sufficiently large we> have> pi(q/4 - 2) > 13 + 0.245 * pi(q).Could you tell me why q/4 - 2 appeared? > Consider the first six odd primes: p_1=3, p_2=5,> p_3=7, p_4=11, p_5=13, > p_6=17, wch I will call the small primes.> Each odd prime p_i between 18 and q is not divisible> by any of these six > small pimes; these p_i will be the large primes.> The primes p_i between 18 and (q/4 -2) will be> called medium primes; > some primes are both medium and large. Consider a choice of residue classes r_i unequal to > 0 modulo p_i for > these six small primes.> The number of choices of such residue classes is > (3-1)*(5-1)*(7-1)*(11-1)*(13-1)*(17-1)=92160.> For each such choice of (r_1,...,r_6), some fraction> of the large primes > p_i fall outside all six residue classes.Here, does fraction of large primes mean some large prime?> The average value of these fractions (averaged over> the 92160 choices) is > (3-2)*(5-2)*(7-2)*(11-2)*(13-2)*(17-2)/92160 => 22275/92160 < 0.2417.It seems to me that ts decimal is related to the first eqaution inyour argument, but it is hard for me to see how they are related.> So we can select one setting of (r_1,...,r_6) for> wch the fraction of > large primes p_i falling outside all six residue> classes is less than > 0.2417.> (In fact for q sufficiently large, each choice of> (r_1,...,r_6) gives a > fraction sufficiently close to 22275/92160.)So these six odd primes, and their associated> residue classes r_i mod p_i, > t a fraction at least 1-0.2417=0.7583 of the large> primes p_i.> For each of the remaining large primes p_k (fewer> than 0.245(q/log q) of > them), and the six small primes,> assign one of the medium primes p_i, and select r_i> so that p_k = r_i mod > p_i.> Since p_i p_i) ts at least > two odd numbers in the range (3,q).> The number of medium primes is pi(q/4)-7 6+0.245*pi(q), so there are > enough medium primes to assign to the> leftover large primes and the six small primes.The upshot is that we have chosen residue classes> (r_i mod p_i) for the > small and medium primes, such that each (small or> large) > prime falls into one of the residue classes; ts> is the F whose > existence Lemma 2.5 (part 2) denies.Don Coppersmith Now, the reason I could reply to your argument without completeunderstanding of your argument is the last paragraph:> The upshot is that we have chosen residue classes> (r_i mod p_i) for the > small and medium primes, such that each (small or> large) > prime falls into one of the residue classes; ts> is the F whose > existence Lemma 2.5 (part 2) denies.Recall the hypothesis of the lemma that the sequence of primesfor moduli of F are defined as all odd primes less than or equalto p_x_, wherep_x_ + 1 =< q.Now, in paricular, is your p_i_ for a modulo one of small primes?More precisely, are your p_i_'s related to induction hypothesis? Itmight be meaningful to point out that there does exist a layeringoutside G_E_.More directly, how is === pentiamond rep-tilesSeveral web pages imply that the rep-tility of pentiamonds is not known.There are only four pentiamonds, not counting mirror images, and two areeasy to dispose of: // __/Ts one obviously won't fit into a horn of a larger replica of itself. /__ /___/Ts one isn't obvious, but with a little experimenting you'll find thatyou cannot fill a 60-degree angle with copies of it.That leaves these two: ____ __/ /____ /___/I see no reason that these should not be rep-tilable. Neither do I seea rep-tiling of either! Does anybody?-:- ÔPATAGEOMETRY: The study of those mathematical properties wch remain invariant under brain transplants.-- Col. G. L. Sichermanhome: colonel@mail.monmouth.comwork: sicherman@att.comweb: === you can prove to the mathematicians you are right> Nobody can come as close as he has to understanding our arguments> without actually understanding them. Hence s refusal to EVER even> quote Decker's main point. If he really believed he is right, the> motivation to have s proof validated by macne would be enormous.> It is quite possible that he would become famous [Letterman, Leno,> King, etc.], even rich.And that's just the letter L! When he gets to O, there's Oprah, and> at R he gets Rose (or does he get Charlie Rose at C?). In going from O to R, you missed a more relevant show: Dr Pl... I'dpay to see that encounter...and I'm not necessarily betting on thegood === mathematicians you are rightBased on previous posts, I tnk what you want to prove is:> The expression (f_1(x) + 7)(f_2(x) + 1) = 7P(x) where f_1(x), and f_2(x) are algebraic integer functions, meaning that> for an algebraic integer x they give an algebraic integer, where f_1(0) = f_2(0) = 0, and P(x) is a polynomial that is also an algebraic integer function,> and P(0)=1 can only be true in the ring of algebraic integers if f_1(x)/7> is an algebraic integer for all x.Unfortunately, a proof of ts will be hard, because it is false. >(it is true if we add the condition that f_1 and f_2 are polynomials [1]) - William Hughes[1] Actually, though I have made ts claim before, I recently> tried to prove it and failed. The claim is true for the ordinary> integers, but there the proof involves prime factorization of> the polynomial coefficients (ts is not going to work in the> algebraic integers). I thought I had a way of substituting > GCD's, but ts fell apart when I tried to fill in the details.> I guess the statement it is true if we add the condition that> f_1 and f_2 are polynomials should be classed as a conjecture> (HnFnLC [2]). Can anyone provide a proof?[2] Hughes' neither First nor Last ConjectureYes. It follows from Dedekind's Prague Theorem, though I am sure Bill> Dubuque will shortly post explaining that I am doing tngs the hard> way (as usual):DEDEKIND'S PRAGUE THEOREM (Lemma 2 in Chapter 2, Section 5, of > lbert's Zahlbericht) If the coefficients a1,a2,....,b1,b2,... of two> polynomials in one variable F(x) = a1*x^r + a2*x^{r-1} + ...> G(x) = b1*x^s + b2*x^{s-2} + ...are algebraic integers, and the coefficients of the product F(x)*G(x) = c1*x^{r+s} + c2*x^{r+s-1} + ...are all divisible by the rational integer w, then each of the numbers> a1*b1, a1*b2, a1*b3,...,a2*b1, a2*b2,... is also divisible by w.(Also proven Kronocker, Mertens, and Hurwitz).So assume we have (f_1(x) + 7)(f_2(x) + 1) = 7P(x)and f_1, f_2, and P are polynomials with algebraic integer> coefficients. Since 7 divides the coefficients of the product, and> f_2(x)=0, then for each coefficient of f_1(x), 1 times that> coefficient must be a multiple of 7. I am not sure if there is an easier route, butmy approach to proving tngs in the integercase was essentially to prove Dedekind's Prauge Theorem for the case F(x) and G(x) in Z[x]. (I note that if p|ab then p|a or p|b then doa lot of messy bookkeeping). On the other hand, I do havea tendency to do tngs the hard way so I would notbe at all suprised to find there is an easier way. I was interested to note that the statment of the theoremrequires w to be a rational integer. (are there any knowncounterexamples if w is an algebraic integer but notat rational integer?) James' examples have always useda rational integer (indeed a prime), however some of sformulations allow for the use of an === JSH: how you can prove to the mathematicians you are right Adjunct Assistant Professor at the University of Montana. [.snip.]> I am not sure if there is an easier route, but>my approach to proving tngs in the integer>case was essentially to prove Dedekind's Prauge Theorem >for the case F(x) and G(x) in Z[x]. (I >note that if p|ab then p|a or p|b then do>a lot of messy bookkeeping). Well, ->that's<- certainly too hard. Just use the version of Gauss'sLemma that says that the content of the product of two polynomials isthe product of their contents. If F(x)*G(x) has a content wch is amultiple of 7, then 7 must divide either the content of F(x) OR thecontent of G(x), wch means one of the two polynomials hascoefficients all divisible by 7.The Prague Theorem can be interpreted as a generalization of thatresult on the contents.>I was interested to note that the statment of the theorem>requires w to be a rational integer. (are there any known>counterexamples if w is an algebraic integer but not>at rational integer?)I'm afraid I selective about what I accept as reality. --- Calvin (Calvin === denominator Kerry likes to say ``I don't tnk Presidents aren't supposed togo for the lowest common denominator, I tnk they're supposed to lookfor the ghest common denominator.''eg. http://home.att.net/~rhhardin6/imuscut.kerrydenominator.ra (32kb)Is ts a ``potatoe'' tng?-- Ron Hardinrhhardin@mindspring.comOn the internet, nobody knows === denominator> Kerry likes to say ``I don't tnk Presidents aren't supposed to> go for the lowest common denominator, I tnk they're supposed to look> for the ghest common denominator.''eg. http://home.att.net/~rhhardin6/imuscut.kerrydenominator.ra (32kb)Is ts a ``potatoe'' tng?*Lowest* common denominator is just one of those commonly usedphrases wch, in spite of (because?) the phrase's popularity, is usedanyway in an ironic manner. Another well-known example:I COULD care less!...when what is actually meant isI could *not* care less!...Our language is full of such unintentional irony, I would bet.In any case, presidents probably are not suppose to go for the literal*lowest* common denominator either!...(Hmmm...perhaps what Kerry literally meant by lowest was not 1, butmeant literally instead the *negative* of the *ghest* === lowest/ghest common denominator> Kerry likes to say ``I don't tnk Presidents aren't...''He really says it with a double === denominator> Kerry likes to say ``I don't tnk Presidents aren't...''He really says it with a double negative like ts?typo s/aren't/are/-- Ron Hardinrhhardin@mindspring.comOn === how you can prove to the mathematicians you are right by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1DGfRn06130;If w is an algebraic integer (not necessarily a rational integer), and each coefficient of F*G is divisible by w.Let v be the product of the conjugates of w,and Norm(w)=v*w is an algebraic integer.Then apply the Prague theorem to (v*F)*G:Norm(w) divides each coefficient of (v*F)*G,so it divides each (v*ai)*(bj),and w divides each ai*bj. Does that work?... >> DEDEKIND'S PRAGUE THEOREM (Lemma 2 in Chapter 2, Section 5, of >> lbert's Zahlbericht) If the coefficients a1,a2,....,b1,b2,... of two>> polynomials in one variable>>> F(x) = a1*x^r + a2*x^{r-1} + ...>> G(x) = b1*x^s + b2*x^{s-2} + ...are algebraic integers, and the coefficients of the product F(x)*G(x) = c1*x^{r+s} + c2*x^{r+s-1} + ...are all divisible by the rational integer w, then each of the numbers>> a1*b1, a1*b2, a1*b3,...,a2*b1, a2*b2,... is also divisible by w.(Also proven Kronocker, Mertens, and Hurwitz).>> ...>I was interested to note that the statment of the theorem>requires w to be a rational integer. (are there any known>counterexamples if w is an algebraic integer but not>at rational integer?) ... -William === following definitions: 1)The hyperderivative of a function f at a point x is defined as the limit as h approaches 0,of (f(x+h)/f(x))^(1/h),provided it exists. 2)The hyperintegral(or more properly the productal) of a function f over an> interval [a,b] over wch it is positive is defined as the limit as n approaches infinity,of the continued product as k varies > from 0 to n-1, of [f(a+(b-a)k/n)^1/n],provided it exists. It turns out that the two operations are related through following theorem:> Suppose that F has f as its hyperderivative over some interval (a,b) and> continuous over [a,b] then> the hyperintegral of f over [a,b] is F(b)/F(a). The calculus thus cerated has a structure analogous to the known calculus> and many theorems have their analogues. It remains true,however, that the> hyperderivative and the hyperintegral can be expressed in terms of the> known derivatives and known integrals. If ts idea is so much beautiful> why is not it popular in mathematical circles?It's one of a family of calculi already investigated - see the reviews below.-Bill Dubuque------80m:26005 26A06 (00-01)Grossman, MichaelThe first nonlinear system of differential and integral calculus.MATHCO, Rockport, Mass., 1979. xi+85 pp.------The system in question is based on a derivative of f equal to exp(f'/f),where the prime indicates the conventional derivative. It is one of afamily of calculi introduced by the author and M. Katz in their earlier book[Non-Newtonian calculus, Lee, Pigeon Cove, Mass., 1972; MR 55 #3180].The author gives intrinsic and geometric definitions of the derivative anddevelops a detailed theory paralleling ordinary calculus but adapted toexponential rather than linear phenomena. It is not yet clear whether the newcalculus provides enough additional insight to justify its use on a large scale. Reviewed by R. P. Boas, Jr.------55 #3180 26A06Grossman, Michael; Katz, RobertNon-Newtonian calculus. Lee Press, Pigeon Cove, Mass., 1972. viii+94 pp.------family of calculi, wch includes the classical calculus as well as aninfinite subfamily of non-Newtonian calculi that we had constructed threeyears earlier.Each calculus possesses the following: (1) a distinctive method of measuringchanges in function arguments; (2) a distinctive method of measuring changesin function values; (3) four operators: a gradient (i.e., an average rate ofchange), a derivative, a natural average, and an integral; (4) acharacteristic class of functions having a constant derivative; (5) a basictheorem involving the gradient, derivative, and natural average; (6) a basicproblem whose solution motivates a simple definition of the integral in termsof the natural average; and (7) two fundamental theorems wch reveal that thederivative and integral are `inversely' related in an appropriate sense.A popular method of creating a new mathematical system is to vary the axiomsof a known system. Although it may be possible to axiomatize the classicalcalculus, we did not and shall not pursue that course. Instead, in Chapter 1we formulate the classical calculus in a novel manner that leads naturally tothe subsequent construction of the non-Newtonian calculi.In Chapters 2--4, we construct three specific non-Newtonian calculi: thegeometric, anageometric, and bigeometric. Chapter 5 is devoted to arithmetics(slightly specialized complete ordered fields), wch are used in subsequentchapters. In Chapter 6 we simultaneously construct all the calculi, indicatethe uniform relationsps between the corresponding operators of any twocalculi, and provide suitable grapc interpretations. Chapters 7 and 8contain brief developments of the quadratic, anaquadratic, biquadratic,harmonic, anaharmonic, and biharmonic calculi. (The harmonic calculus is notto be confused with harmonic analysis.) Chapter 9 includes a variety ofheuristic guides for selecting gradients, derivatives, averages, andintegrals. In Chapter 10 we discuss certain generalized spaces, vectors, andleast squares methods, wch were suggested by our work with the non-Newtoniancalculi; we introduce a trend concept that may be considered as a kind ofglobal derivative; and we indicate some connections between the non-Newtoniancalculi and calculus in Banach spaces. Sundry digressions and comments havebeen placed in the Notes at the rear of the book; for example, in Note 2 weexplain how a simple algebraic identity suggested the possibility ofconstructing the geometric calculus.Since ts book is intended for a wide audience, including students,engineers, scientists, as well as mathematicians, we have presented manydetails that would not appear in a research report and we have excluded allproofs. (All results stated can be proved in a straightforward way). It isassumed, of course, that the reader has a working knowledge of the rudimentsof classical calculus.Table of Contents: Preliminaries (pp. 1--2); Chapter 1, The classical calculus(pp. 3--8); Chapter 2, The geometric calculus (pp. 9--17); Chapter 3, Theanageometric calculus (pp. 18--24); Chapter 4, The bigeometric calculus(pp. 25--31); Chapter 5, Systems of arithmetic (pp. 32--37); Chapter 6, The*-calculus (pp. 38--51); Chapter 7, The quadratic family of calculi(pp. 52--58); Chapter 8, The harmonic family of calculi (pp. 59--65); Chapter9, Heuristics (pp. 66--74); Chapter 10, Collateral issues (pp. 75--83); Notes(pp. 84--88); List of symbols (pp. 89--90); Index (pp. 91--94).------81k:26003 26A06 (00A05)Grossman, Jane; Grossman, Michael; Katz, RobertThe first systems of weighted differential and integral calculus. Arcmedes Foundation, Rockport, Mass., 1980. vi+55 pp.------In M. Grossman and R. Katz's Non-Newtonian calculus [Lee Press, Pigeon Cove,Mass., 1972; MR 55 #3180] ordinary calculus was generalized by measuringchanges in the domain by a{a^{-1}(y)-a^{-1}(z)} and changes in the rangecorrespondingly by b{b^{-1}(y)-b^{-1}(z)} , where a and b are variousone-to-one functions. In the present monograph the ordinary derivative andintegral are generalized by a procedure that is equivalent to selecting apositive continuous weight function w and a primitive W of w , andreplacing derivatives by derivatives with respect to W and int f(x) dx byint f dW (or equivalently int fw ). The authors apply ts procedure to thecalculi of the previous monograph. Reviewed by R. P. Boas, Jr.------84h:26002 26A06Grossman, Jane; Grossman, Michael; Katz, RobertAverages. A new approach.Arcmedes Foundation, Rockport, Mass., 1983. vi+61 pp.------Ts booklet introduces families of averages wch in conventional notationcorrespond to two one-to-one functions a and b via 1 / A(s) b( -------- | B(f(a(t)) dt) where A = a^-1, B = b^-1 A(s)-A(r) / A(r)Special cases have been considered in the literature or have been introducedpreviously by the authors. The results are expressed in terms of the fields in wch y + z = a[A(y) + A(z)], and so on, using a special notation.It seems possible that the notation may facilitate the study of the averages,but no new theorems about means are === Involutionary Calculus>Consider the following definitions: 1)The hyperderivative of a function f at a point x is defined as the limit as h approaches 0,of (f(x+h)/f(x))^(1/h),provided it exists. 2)The hyperintegral(or more properly the productal) of a function f over an > interval [a,b] over wch it is positive is defined as the limit as n approaches infinity,of the continued product as k varies from 0 to > n-1,of [f(a+(b-a)k/n)^1/n],provided it exists. It turns out to be that the two operations are related through following > theorem:> Suppose that F has f as its hyperderivative over some interval (a,b) and > continuous over [a,b] then> then the hyperintegral of f over [a,b] is F(b)/F(a). The calculus thus cerated has a structure analogous to the known calculus > and many theorems have their analogues.> It remains true,however,that the hyperderivative and the hyperintegral can > be expressed in terms of the known derivatives and known integrals.> If ts idea is so much beautiful why is not it popular in mathematical > circles?Does it have any use? What good is it? What is the use of knowing the Fermat's Last theorem,of Euler'sformula for the sum of the reciprocals of the squares of all natural numbers andsimilar results. The fact that a mathematical theory has no obvious use at some momentof time doesn't mean that its useless. What problems can it solve that are not otherwise solveable, if any? In what ways may it offer simpler explanations of phenomena explainedby more convoluted expression?Solve a problem with it, and challenge others to solve the problem any other way.> Rise above the === Calculus> What problems can it solve that are not otherwise solveable, if any? > In what ways may it offer simpler explanations of phenomena explained> by more convoluted expression? Solve a problem with it, and challenge others to solve the problem any other way. Rise above the cldish problem solving mentality.Easy Ashurosh, problems solving capability enhancement/new skillacquisition is the aim of almost all of further learning. Euler wasmentally calculating till the last. Practical utility or immediateapplication is a tng apart, new techniques and tools are central tofurther mathematical === the following definitions:>>> 1)The hyperderivative of a function f at a point x is defined as the >>> limit as h approaches 0,of (f(x+h)/f(x))^(1/h),provided it exists. 2)The hyperintegral(or more properly the productal) of a function f over an >> interval [a,b] over wch it is positive is defined as the limit as n approaches infinity,of the continued product as k varies from 0 >> to >> n-1,of [f(a+(b-a)k/n)^1/n],provided it exists. It turns out to be that the two operations are related through following >> theorem:>> Suppose that F has f as its hyperderivative over some interval (a,b) and >> continuous over [a,b] then>> then the hyperintegral of f over [a,b] is F(b)/F(a). The calculus thus cerated has a structure analogous to the known calculus >> and many theorems have their analogues.>> It remains true,however,that the hyperderivative and the hyperintegral can >> be expressed in terms of the known derivatives and known integrals.>> If ts idea is so much beautiful why is not it popular in mathematical >> circles?We call it taking logs. I give a damn to what you call it or in what manner can the result be > established.If you cannot see the inherent aesthetic appeal bend the> results you better keep your comments with you.Uh, right. All hail Ashurosh, the discoverer of the amazing> hypercalculus.The point is that your assumption that hypercalculus is not> popular is simply incorrect. It's just calculus plus logarithms,> and logarithms are _very_ popular... The theory exists independent of logarithms.Only the connection between them becomes obvious === Calculushttp://www.giganews.com/info/dmca.html Consider the following definitions: 1)The hyperderivative of a function f at a point x is defined as the > limit as h approaches 0,of (f(x+h)/f(x))^(1/h),provided it exists. 2)The hyperintegral(or more properly the productal) of a function f over an > interval [a,b] over wch it is positive is defined as the limit as n approaches infinity,of the continued product as k varies from 0 > to > n-1,of [f(a+(b-a)k/n)^1/n],provided it exists. It turns out to be that the two operations are related through following > theorem:> Suppose that F has f as its hyperderivative over some interval (a,b) and > continuous over [a,b] then> then the hyperintegral of f over [a,b] is F(b)/F(a). The calculus thus cerated has a structure analogous to the known calculus > and many theorems have their analogues.> It remains true,however,that the hyperderivative and the hyperintegral can > be expressed in terms of the known derivatives and known integrals.> If ts idea is so much beautiful why is not it popular in mathematical > circles?We call it taking logs.>> I give a damn to what you call it or in what manner can the result be >> established.If you cannot see the inherent aesthetic appeal bend the>> results you better keep your comments with you.Uh, right. All hail Ashurosh, the discoverer of the amazing>> hypercalculus.The point is that your assumption that hypercalculus is not>> popular is simply incorrect. It's just calculus plus logarithms,>> and logarithms are _very_ popular... The theory exists independent of logarithms.Only the connection between them> becomes obvious through the use of logarithms.That's not the only tng that becomes obvious through the use oßogarithms - one other tng is === Involutionary Calculus> The theory exists independent of logarithms.Only the connection between them> becomes obvious through the use of logarithms.> That's not the only tng that becomes obvious through the use of> logarithms - one other tng is that the theory is really no big deal.The idea seems new, interesting and promising. At least for monotonicfunctions it may provide speed of computation. An application may bein Fractals.Firstly, one might ask if a table of hyper derivativesand hyper integrals has been compiled for some functions like x^n,sin(x), log(x), exp(x), cosh(x) etc. Are there references? One may askmore questions. How are the results related to normal derivatives andintegrals? [ Surely one is not the log of the other always.] Have theybeen grapcally interpreted ? Any work about existence, uniquenessetc.? Any hyper differential equations have been set up?How can one be so judgmental so soon about the deal size? When thereis not enough clarity on these or similar matters, avoidance ofsuggestion to general condemnatives could perhaps better serve theaims of the newsgroup.Narasimham === independent of logarithms.Only the connection between them> becomes obvious through the use of logarithms.> That's not the only tng that becomes obvious through the use of> logarithms - one other tng is that the theory is really no big deal.The idea seems new, interesting and promising. At least for monotonic> functions it may provide speed of computation. An application may be> in Fractals.Firstly, one might ask if a table of hyper derivatives> and hyper integrals has been compiled for some functions like x^n,> sin(x), log(x), exp(x), cosh(x) etc. Are there references? One may ask> more questions. How are the results related to normal derivatives and> integrals? [ Surely one is not the log of the other always.] Have they> been grapcally interpreted ? Any work about existence, uniqueness> etc.? Any hyper differential equations have been set up?How can one be so judgmental so soon about the deal size? When there> is not enough clarity on these or similar matters, avoidance of> suggestion to general condemnatives could perhaps better serve the> aims of the newsgroup.Narasimham G.L. I want to add few remarks. We know that any infinitely differentiable (differentiable in the usual sense of the word) can be written as an infinite polynomial. We have an analogous theorem for the hyper derivative case. It runs as follows: Any function whose hyper derivatives exist upto all orders can be represented as an infinite product of the following form: f(x)= Product as r varies from 0 to infinity of (A(r)^((x-a)^r)) where A(r) is determined by a simple expression involving hyper derivatives of f at a and is independent of x. Ts theorem is completely analogous to the taylor's theorem. It remains true that ts theorem has a trivial proof through the use of logarithms. Perhaps I need to add further remarks as for the usefulness of a mathematical theory as I understand it. There are certain mathematical theories that have a kind of aesthetic appeal in them.These theories need not only be studied for the sake of their immediate usefulness but also for the sake of ts appeal.The question of applicability is of no doubt primary importance but the fact that no immediate application is obvious doesn't === Calculushttp://www.giganews.com/info/dmca.htmlThe theory exists independent of logarithms.Only the connection between them>> becomes obvious through the use of logarithms.>> That's not the only tng that becomes obvious through the use of>> logarithms - one other tng is that the theory is really no big deal.The idea seems new, interesting and promising. At least for monotonic>> functions it may provide speed of computation. An application may be>> in Fractals.Firstly, one might ask if a table of hyper derivatives>> and hyper integrals has been compiled for some functions like x^n,>> sin(x), log(x), exp(x), cosh(x) etc. Are there references? One may ask>> more questions. How are the results related to normal derivatives and>> integrals? [ Surely one is not the log of the other always.] Have they>> been grapcally interpreted ? Any work about existence, uniqueness>> etc.? Any hyper differential equations have been set up?How can one be so judgmental so soon about the deal size? When there>> is not enough clarity on these or similar matters, avoidance of>> suggestion to general condemnatives could perhaps better serve the>> aims of the newsgroup.Narasimham G.L.> I want to add few remarks.> We know that any infinitely differentiable (differentiable in the usual > sense of the word) can be written as an infinite polynomial.That's wrong for two reasons. First, you mean to say that anyinfinitely differentiable function can be written as a _powerseries_ - there's no such tng as an infinite polynomial.More important, it's _not_ _true_ that any infinitely differentiablefunction can be written as a power series.> We have an analogous theorem for the hyper derivative case.> It runs as follows:> Any function whose hyper derivatives exist upto all orders can be represented> as an infinite product of the following form:f(x)= Product as r varies from 0 to infinity of (A(r)^((x-a)^r))> where A(r) is determined by a simple expression involving hyper derivatives > of f at a and is independent of x.> Ts theorem is completely analogous to the taylor's theorem.> It remains true that ts theorem has a trivial proof through the use of > logarithms. Really? I'd like to see the proof. (Make certain to include a proofof the _false_ version of the standard Taylor's theorem thatyou're evidently using.)> Perhaps I need to add further remarks as for the usefulness of a mathematical > theory as I understand it.> There are certain mathematical theories that have a kind of aesthetic> appeal in them.These theories need not only be studied for the sake of > their immediate usefulness but also for the sake of ts appeal.The question > of applicability is of no doubt primary importance but the fact that no > immediate application is obvious doesn't imply === Calculushttp://www.giganews.com/info/dmca.html> The theory exists independent of logarithms.Only the connection between them>> becomes obvious through the use of logarithms.>> That's not the only tng that becomes obvious through the use of>> logarithms - one other tng is that the theory is really no big deal.The idea seems new, interesting and promising. At least for monotonic>functions it may provide speed of computation. That's an interesting concept. It may be that computing a quotientis faster than computing a difference, that taking a 1/h power isfaster than dividing by h. I would have guessed just the opposite.>An application may be>in Fractals.Firstly, one might ask if a table of hyper derivatives>and hyper integrals has been compiled for some functions like x^n,>sin(x), log(x), exp(x), cosh(x) etc. Are there references? One may ask>more questions. How are the results related to normal derivatives and>integrals? [ Surely one is not the log of the other always.] Huh? Who said one was the log of the other? If you don't even understand what the connection _is_ then you're really not ina position to comment on whether the connection means thathyperderivatives are no big deal.Let's say Hf is the hyperderivative of f and Df is the derivative.Then for f > 0 Hf = exp(D(log(f))).It's true that one may ask what the connection between Hand D is, especially if one doesn't know gh-school algebra...>Have they>been grapcally interpreted ? Any work about existence, uniqueness>etc.? Any hyper differential equations have been set up?How can one be so judgmental so soon about the deal size?Because the connection between H and D makes the answerto all these questions clear. For example hyperdifferentialequations: It's clear that Hf = f if and only if f = exp(g),where Dg = g. So understanding hyperdifferentialequations is exactly the same tng as understandingordinary differential equations.> When there>is not enough clarity on these or similar matters, avoidance of>suggestion to general condemnatives could perhaps better serve the>aims of the newsgroup.Nobody was condemming anytng, just pointing out that there'snotng new here, because we already know about log. Yousay the idea seems new. It's === radarsignal> the spectum of a radarsignal.> A radar sends out a pulsed sinosoidial waveform.> The carrier cos(wt-kz) wch is a cosine that propagates in the> z-direction> is modulated by a pulsetrain wch is a typical square-wave wich is a> sum of cosine with its odd harmonics cos(wt)+(1/3)*cos(wt)+ ...If I do the amplitude modulation (multiplication) of the pulse train> and the carrier. I get a discrete spectrum...But somebody told me that the following rule exists:> A continous signal has a discrete spectrum and a discrete signal has a> continous spectrum.> By standard meanings of the words continuous and discrete, theabove statement is ßat out wrong. I suggest you forget it as quicklyas possible as it seems to cause great confusion. ( Tnk ofcontinuous wte or colored noise for a quick example of a continuoussignal with a continuous spectrum)I have ts feeling that you may mean infinite duration signals andfinite duration signals, but one still has to be careful.> The radarsignal is a discrete signal. The emitter is on and off, on> and off,...> it should according to the rule of thumb have a continous spectrum.> Again, confusion in terminology exits. A pulsed continuous wave radarsignal is not discrete. Because of its periodicity, its spectrumappproches a discrete line spectrum. If the duration of the radarsignal were infinite, then its spectrum would indeed be discrete.Nonrigorous, on the ßy, made-up definition of discrete signals - signals whose energy exists in pulses approacng zero duration.Ts is not the case for the radar you === radarsignalClarification and erratum:A sampled input produces a periodic output.To clarify, the ratios between time coordinates of the samples needto be rational. (Ts is usually the case in engineering.)A non-periodic input produces a continuous-spectrum output.Ts statement is ßawed because the input could also be composedof discrete spectral components that are not harmonically related(having an irrational frequency ratio) and therefore not periodic.A signal composed of a number of discrete frequency components thatare not harmonically related will be quasi-periodic, meaning thatthere will be regular recurrences of a waveform that is similarto any given fragment. The gher the degree of accuracy of match,the === Fourier analysis of a radarsignal >>A Fourier transform swaps large-scale and small-scale structure.>>A periodic input produces a sampled output.>>A sampled input produces a periodic output.> Undiluted crap.> He's basically right though.Please point to the line wch is basically right, and === analysis of a radarsignal>>A Fourier transform swaps large-scale and small-scale structure.>>A periodic input produces a sampled output.>>A sampled input produces a periodic output.>Undiluted crap.>He's basically right though.> Please point to the line wch is basically right, and explain why it is> basically right.The word Ôsample' is used a broader.A dirac in time or frequency transforms to unity.A wide function in time has a narrow spectrum and anarrow function of time has a wide spectrum.The eigenfunction(s) transforming to themselves are the gaussian, BTW.A periodic signal has a set of lines in the spectrum.A periodic dirac in time has a periodic dirac in the spectrum.Rene-- Ing.Buero R.Tschaggelar - http://www.ibrtses.com& commercial newsgroups === radarsignal>>A Fourier transform swaps large-scale and small-scale structure.>>A periodic input produces a sampled output.>>A sampled input produces a periodic output.>Undiluted crap.>He's basically right though.> Please point to the line wch is basically right, and explain why it is> basically right. The word Ôsample' is used a broader. A dirac in time or frequency transforms to unity.> A wide function in time has a narrow spectrum and a> narrow function of time has a wide spectrum.> The eigenfunction(s) transforming to themselves are the gaussian, BTW.> A periodic signal has a set of lines in the spectrum.> A periodic dirac in time has a periodic dirac in the spectrum.>After all that nonsense, I suggest we call it a === The word Ôsample' is used a broader.A dirac in time or frequency transforms to unity.> A wide function in time has a narrow spectrum and a> narrow function of time has a wide spectrum.The eigenfunction(s) transforming to themselves are the gaussian, BTW.> Are you sure they are gaussian? Most electromagnet distributions have a Lorentzian distribution with a Gaussian distribution imposed on top for the noise in the measurement system. All resonance filters give a Lorentzian distribution.> A periodic signal has a set of lines in the spectrum.> A periodic dirac in time has a periodic dirac in === radarsignal The word Ôsample' is used a broader. A dirac in time or frequency transforms to unity.> A wide function in time has a narrow spectrum and a> narrow function of time has a wide spectrum. The eigenfunction(s) transforming to themselves are the gaussian, BTW.> Are you sure they are gaussian? Most electromagnet distributions have a> Lorentzian distribution with a Gaussian distribution imposed on top for> the noise in the measurement system. All resonance filters give a> Lorentzian distribution.Yes, Rene has that right. The FT of === analysis of a radarsignal> The word Ôsample' is used a broader.>> A dirac in time or frequency transforms to unity.>> A wide function in time has a narrow spectrum and a>> narrow function of time has a wide spectrum.>> The eigenfunction(s) transforming to themselves are the gaussian, BTW.>>Are you sure they are gaussian? Most electromagnet distributions have a > Lorentzian distribution with a Gaussian distribution imposed on top for > the noise in the measurement system. All resonance filters give a > Lorentzian distribution.Yes, I am.The equation is as follows : f(w)= n*FT(f(t)) = Eigenvalue equationI forgot whether n is integer, real, or complex.And ts is true for the gaussians.Just one of them for example :f(t) = exp(-alpha*t^2)gives morelessf(w) = whatever*exp(-sometng*w^2)I'd had to look up the constants.Rene-- Ing.Buero R.Tschaggelar - http://www.ibrtses.com& commercial newsgroups - === nowhere differentiable on R^1.Here's a theorem from Principles of Mathematical Analisys (W. Rudin3rd edition). Theorem 7.18: There exists a real continuous function on the real linewch is nowhere differentiable. I am trying to understand the proof from the book and ts is what Iunderstand so far: (my comments are in brackets)Proof: define p(x) = |x| (-1 <= x <= 1) and extend the definitionof p(x) to all real x by requiriring that p(x+2) = p(x). [i understand that p(x+2) = p(x) makes p(x) a periodic function]Then, for all s and t, |p(s) - p(t)| <= |s - t|. [i dont the see the point here, but i agree with the statement. ibelieve |s + 2 - (t + 2)| = |s - t| because p(x+2)=p(x). and i<= when just a = would suffice].In particular, p() is continuous on R^1. Define oo f(x) = Sum( ((3/4)^n) * p( (4^n) * x) ) n=0Since 0 <= p <= 1, theorem 7.10 shows that the series aboveconverges uniformly on R^1. By theorem 7.12, f is continuous on R^1.Now fix a real number x and a positive integer m. Put delta_m = +|- 1/2 * 4^(-m)where the sign is so chosen that no integer lies between (4^m) * x and(4^m) * (x + delta_m). Ts can be done, since 4^m * |delta_m| = 1/2. [I see that. Good so far.]Define p(4^n (x + delta_m)) - p(4^n * x)gamma_n = - delta_mWhen n > m, then 4^n * delta_m is an even integer, so that gamma_n=0.[i dont see ts. why if 4^n * delta_m is an even integer,gamma_n=0 ? since p(x+2) = p(x), i thought any delta_m would work]When 0 <= n <= m, the equation |p(s) - p(t)| <= |s - t| impliesthat |gamma_n| <= 4^n. Since |gamma_m| = 4^m, we conclude that[i dont ts either, not to mention the equation below] | f(x + delta_m) - f(x) | | oo | | --------- | = | Sum( (3/4)^n * gamma_n) | | delta_m | | n=0 | (m - 1) <= 3^m - Sum ( 3^n ) n=0 = 1/2 (3^m + 1).As m -> oo, delta_m -> 0. It follows that f is not differentiable atx. [Now how is it possible for a function be continuous on the real lineand not be differentiable at === continuous function nowhere differentiable on R^1.> Here's a theorem from Principles of Mathematical Analisys (W. Rudin> 3rd edition). Theorem 7.18: There exists a real continuous function on the real line> wch is nowhere differentiable. I am trying to understand the proof from the book and ts is what I> understand so far: (my comments are in brackets)Proof: define p(x) = |x| (-1 <= x <= 1) and extend the definition> of p(x) to all real x by requiriring that p(x+2) = p(x). [i understand that p(x+2) = p(x) makes p(x) a periodic function]with period 2Then, for all s and t, |p(s) - p(t)| <= |s - t|. [i dont the see the point here, but i agree with the statement. i> believe |s + 2 - (t + 2)| = |s - t| because p(x+2)=p(x). and i> <= when just a = would suffice].Sometimes < holds: say s=1/10, t=-1/10, then|p(s) - p(t)| = |1/10 - 1/10| = 0 < 1/5 = |s-t|In particular, p() is continuous on R^1. Define oo> f(x) = Sum( ((3/4)^n) * p( (4^n) * x) )> n=0Since 0 <= p <= 1, theorem 7.10 shows that the series above> converges uniformly on R^1. By theorem 7.12, f is continuous on R^1.Now fix a real number x and a positive integer m. Put delta_m = +|- 1/2 * 4^(-m)where the sign is so chosen that no integer lies between (4^m) * x and> (4^m) * (x + delta_m). Ts can be done, since 4^m * |delta_m| = 1/2. [I see that. Good so far.]Define p(4^n (x + delta_m)) - p(4^n * x)> gamma_n = -> delta_mWhen n > m, then 4^n * delta_m is an even integer, so that gamma_n=0.[i dont see ts. why if 4^n * delta_m is an even integer,> gamma_n=0 ? since p(x+2) = p(x), i thought any delta_m would work]No, only delta_m that is a multiple of 2. That is the pointof being periodic with period 2.When 0 <= n <= m, the equation |p(s) - p(t)| <= |s - t| implies> that |gamma_n| <= 4^n. Since |gamma_m| = 4^m, we conclude that[i dont ts either, not to mention the equation below]> | f(x + delta_m) - f(x) | | oo |> | --------- | = | Sum( (3/4)^n * gamma_n) |> | delta_m | | n=0 | (m - 1)> <= 3^m - Sum ( 3^n )> n=0> = 1/2 (3^m + 1).As m -> oo, delta_m -> 0. It follows that f is not differentiable at> x. [Now how is it possible for a function be continuous on the real line> and not be differentiable at anywhere? Sounds a little strange.]That is why there has to be a proof. Because === OrderingCountable dense orderings w/o endpoints are all isomorpc.Are all such orderings of a given cardinality isomorpc?No, the theory of dense linear orderings without endpointsis indeed countably categorical but not uncountably so, e.g.consider q * w_1 vs. q * w_1~, q = rationals, ~ = reverseCategoricity is studied in any good textbook on model === What I meant to say is that any countable dense linear orderingswithout endpoints are isomorpc> I'm curious, I know how to show dense linear orderings without> endpoints are isomorpc. However, how do you show that dense linear> orderings without endpoints admits elimination of quantifiers?? for any advice.Are you aware that the rationals and the reals have dense linear>> orderings without endpoints but are not isomorpc?Countable dense orderings w/o endpoints are all isomorpc.Are all such orderings of a given cardinality isomorpc?Surely not...For example R and R with an interval removed and the rationals> in that interval stuck back === say is that any countable dense linear orderings> without endpoints are isomorpcWch specifically excludes the === confident that the rational and reals ARE isomorpc. You are overconfident. Is the Stainless Steel Rat rusting?Not only is there no order preserving bijection, f: Q -> R, from the rationals, Q, to the reals, R, wch is required for such an isomorpsm to exist, to there is no such bijection at all of any sort whatsoever from Q to R.Georg Cantor proved ts several different ways. Others have come up with several other proofs since Cantor's. Ao the issue is settled beyond all reasonable === Dense Linear Orderinghttp://www.giganews.com/info/dmca.html>I'm pretty confident that the rational and reals ARE isomorpc.Good for you. Got any opinions on whether 2 + 2 = 4?> I'm curious, I know how to show dense linear orderings without>> endpoints are isomorpc. However, how do you show that dense linear>> orderings without endpoints admits elimination of quantifiers?? for any advice.Are you aware that the rationals and the reals have dense linear> orderings without endpoints but are not isomorpc?>>Countable dense orderings w/o endpoints are all isomorpc.>>Are all such orderings of a given cardinality isomorpc?Surely not...For example R and R with an interval removed and the rationals>> in that interval stuck back === Gallian's book and there is a lemma (I'll post in bracketswhen I am lost) that saysIf e=B1B2****************Br, where the B's are 2-cycles, then r iseven.The Proof is as follows:Clearly r does not equal 1 since a 2-cycle is not the identity. Ifr=2 we are done. SO, we suppose that r>2 and proceed by induction. Since (ij)=(ji), the product Br-1*Br can be expressed in one of thefollowing forms shown on the right:e=(ab)(ab)(ab)(bc)=(ac)(ab)(ac)(cb)=(bc)(ab)(ab)(cd)=(cd )(ab).[Ok, how the heck do you know that Br-1 Br can be expressed in one ofthose forms? What the heck do they mean (besides the identity(ab)(ab))? Why only 4 and not more?]If the first case occurs we may delete Br-1*Br from the originalproduct to obain e=B1B2*******Br-2 and by the second principle ofmathematical induction, r-2 is even. [Wouldn't you have to assume r iseven for r-2 to be even?!??!]In the other 3 cases, we replace the form of Br-1*Br on the right byits counterpart on the left to obtain a new product of r 2-cycles thatis still the identity but where the rightmost occurrence of hteinteger a is in the second-from-the-rightmost 2-cycleof the productinstead of the rightmost 2-cycle. We now repeat teh producure justdescribed with Br-2*Br-1, and, ans before, we obtain a product of(R-2) 2-cycles equal to the identity or a new product of r 2-cycles.,where hte rightmost occurance of a is in the trd 2-cycle form theright. Contignuing ts process, we must obtain a product of (r-2)2-cycles equal to the identity, because otherwise we have a productequal ot the identity in wch the only occurence of hte integer a isin the leftmost 2-cycle and such a product does not fix a, whereas theidentity does. Hence, by induction, r-2 is even and r is even aswell.End of proof. [Now I don't understand how the continual working tothe left with a ensures that it's even. I'm just totally lost withts proof. Please === using Gallian's book and there is a lemma (I'll post in brackets>when I am lost) that says>If e=B1B2****************Br, where the B's are 2-cycles, then r is>even.>The Proof is as follows:>Clearly r does not equal 1 since a 2-cycle is not the identity. If>r=2 we are done. SO, we suppose that r>2 and proceed by induction. >Since (ij)=(ji), the product Br-1*Br can be expressed in one of the>following forms shown on the right:>e=(ab)(ab)>(ab)(bc)=(ac)(ab)>(ac)(cb)=(bc)(ab)>(ab)(cd) =(cd)(ab).>[Ok, how the heck do you know that Br-1 Br can be expressed in one of>those forms? What the heck do they mean (besides the identity>(ab)(ab))? Why only 4 and not more?]>We are considering a product of two 2-cycles; each of the elements a,bmoved by the 2nd of these cycles may occur or not in the 1st of themi.e. may be moved or not by the 1st; ts makes for the 4 cases(of course, a,b,c,d are supposed to be different elements)Does ts make the tng clear for you ?If the first case occurs we may delete Br-1*Br from the original>product to obain e=B1B2*******Br-2 and by the second principle of>mathematical induction, r-2 is even. [Wouldn't you have to assume r is>even for r-2 to be even?!??!]>It is the other way round: from the induction hyp., we deduce that r-2is even, therefore we may conclude that r is evenIn the other 3 cases, we replace the form of Br-1*Br on the right by>its counterpart on the left to obtain a new product of r 2-cycles that>is still the identity but where the rightmost occurrence of hte>integer a is in the second-from-the-rightmost 2-cycleof the product>instead of the rightmost 2-cycle. We now repeat teh producure just>described with Br-2*Br-1, and, ans before, we obtain a product of>(R-2) 2-cycles equal to the identity or a new product of r 2-cycles.,>where hte rightmost occurance of a is in the trd 2-cycle form the>right. Contignuing ts process, we must obtain a product of (r-2)>2-cycles equal to the identity, because otherwise we have a product>equal ot the identity in wch the only occurence of hte integer a is>in the leftmost 2-cycle and such a product does not fix a, whereas the>identity does. Hence, by induction, r-2 is even and r is even as>well.>End of proof. [Now I don't understand how the continual working to>the left with a ensures that it's even. I'm just totally lost with>ts proof. Please help....]>Ts sounds like a is even - but in ts proof, the permutationsneed not be integers (as often supposed - at least to begin with)and if they are, there is no reason why Ôa' should be even ... ;-)BTW as done here by myself, it might help to better understandthe whole to put the name Ôa' of an element in apostrophes when usingit in an english sentence (instead of in a formula) because it will beIn the book, either the font or the style (like italic) will make thedifference ...I hope you in fact mean by it the number r ... The idea is ts:we can push the element Ôa' more and more to the left, leaving onthe right of the 2-cycle containing ts Ôa' only 2-cycles that moveonly elements different from Ôa' (i.e. >not< moving Ôa'), withoutchanging the number of cycles in the product nor the value e ofts product. Ts will stop only when one of the following tngsoccur: (1) the cycle on the left of the one we consider containingthe element Ôa' is identical to it -> then we can suppress theduplicated cycle, so that we are left with r-2 cycles and with theinduct. hyp. we are done (see above) (2) there is no more cycleto the left of the one with Ôa' in it, i.e. we have reached the leftend of the (variable) sequence of r 2-cycles, but ts leads toa contradiction because the product thus defined cannot be = e,because it does not leave Ôa' unchanged (as explained in yourquoted text) -> therefore we can discard ts case. QEDI find ts an interesting foundation to the theory of even & oddpermutations without beginning with a (totally) ordered set wherethe permutations take place (often one uses {1,2,...,n}) - an orderheavily used in the theory as I studied it ... where one begins withthe set {1,2,...,n} and an expression containing two products ofdifferences of elements of ts set, indexed by the pairs (i,j) ofintegers such that 1<=iI'm using Gallian's book and there is a lemma (I'll post in brackets>when I am lost) that saysIf e=B1B2****************Br, where the B's are 2-cycles, then r is>even.The Proof is as follows:Clearly r does not equal 1 since a 2-cycle is not the identity. If>r=2 we are done. SO, we suppose that r>2 and proceed by induction. >Since (ij)=(ji), the product Br-1*Br can be expressed in one of the>following forms shown on the right:>e=(ab)(ab)>(ab)(bc)=(ac)(ab)>(ac)(cb)=(bc)(ab)>(ab)(cd) =(cd)(ab).[Ok, how the heck do you know that Br-1 Br can be expressed in one of>those forms? What the heck do they mean (besides the identity>(ab)(ab))? Why only 4 and not more?]>A 2-cycle moves 2 points. *two* 2-cycles move at most 4 points.If the pair of 2-cycles moves 3 points, call them a, b, c, and make alist of all the possibilities. I'll start you off:(a b) (a c)(a b) (b c)(a c) (a b)etc.If the pair of 2-cycles moves 4 points, call them a, b, c, d, and makelist of all the possibilities.(a b) (c d)(a c) (b d)etc.Use the fact that (a b) = (b a), (b c) = (c b), etc. to eliminate theduplicates from your list.What's left?>If the first case occurs we may delete Br-1*Br from the original>product to obain e=B1B2*******Br-2 and by the second principle of>mathematical induction, r-2 is even. [Wouldn't you have to assume r is>even for r-2 to be even?!??!]By induction, e=B1B2*****Bn where n < r means that n is even. Notethat we only know ts in the first case, e = (a b)(a b), since itreduces to a previously known case.>In the other 3 cases, we replace the form of Br-1*Br on the right by>its counterpart on the left to obtain a new product of r 2-cycles that>is still the identity but where the rightmost occurrence of hte>integer a is in the second-from-the-rightmost 2-cycleof the product>instead of the rightmost 2-cycle. We now repeat teh producure just>described with Br-2*Br-1, and, ans before, we obtain a product of>(R-2) 2-cycles equal to the identity or a new product of r 2-cycles.,>where hte rightmost occurance of a is in the trd 2-cycle form the>right. Contignuing ts process, we must obtain a product of (r-2)>2-cycles equal to the identity, because otherwise we have a product>equal ot the identity in wch the only occurence of hte integer a is>in the leftmost 2-cycle and such a product does not fix a, whereas the>identity does. Hence, by induction, r-2 is even and r is even as>well.>End of proof. [Now I don't understand how the continual working to>the left with a ensures that it's even. I'm just totally lost with>ts proof. Please help....]Let's suppose you follow the steps listed, and never find a case whereB(n-1)*Bn is of the form (a x)(a x) = e. What does the permutationlook like after the final step?e = (a z)B2*B3 ... *BrWhere a does not appear in any of the B2...Br.So the permutation move point a to point z (the B1 term), and noother Bi term moves point a So point a is not fixed --- it'smoved to zBut e fixes point aand yet e = (a z)B2*B3...*BrHow can that === transcendental numbers ? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1DKh3x25898;>>In s proof makes use of the relationsp e^[ipi]+1=0 , or>>e^[ipi]=-1 >My comments>>--->>Since e^[iPi]=cosPi+isinPi>>or , e^[iPi]=-1+i[0]>>then there are two solutions here, to the given equatio:>A) e^[ipi]=-1 the real part solution and >B) e^[ipi]=i[0] , or e^[ipi]=0 the imaginary part solutio.The reasoning bend A and B is not evident to me. Please explain>where B comes from.> e^[ipi]=0 is The solution if the imaginary part.No. If you are taking imaginary parts, then the conclusion is that>sin(pi) = 0, not e^{i pi] = 0. Similarly, if you take real parts>of e^[i pi] = -1, you get cos(pi) = -1.> Just tnk that angle is 179 deg>>the real component is cos(179deg) = -0999847695approximately.>and the imaginary is i*sin(179deg)=i*0.017452406approximately.>e^[iPi179/180]= -0999847695 +i*0.017452406approximately.>in polar form ts will be e^[iPi179/180]=mod 1, arg 179e^[iPi179/180]=mod 1, arg 179 degreesSo what does all ts have to do with your extraordinary claim>that e^[i pi] = 0?> e^[i pi] = -1 is right of course, but the last >step that you mentioned in your derivation looks completely >unjustified.The last step SHOULD have yielded the result that cos(pi) = -1.>My question :>>What is the implication of ts second value of e^[ipi]=0 ?No implication. It is not true.> e^[iPi]=mod 1, arg 180e^[iPi]=mod 1, arg 180 degrees> Ts is correct in polar form.But it has absolutely notng to do with your earlier extraordinary>claim that exp(i pi) = 0. That particular claim of yours is still>unjustified, and it is still false.It has to do.cos pi + i sin pi=-1+i*0 =-1 is correctHowever the phasor e^[ipi] =cos pi + i sin pi , is properly defined as MOD 1 , ARG [THETA] If I state that a phasor A=-1 IS TS CORRECT?No it must be accompanied by its length and its direction.A=-1 shows a length of unity directed towards the [0,0] point [of an X-Y coordinates system crossing]from any angular direction around a unit circle,from the positive X-axis (taken as zero with the usual notation. e^[ipi] cannot loose its value as 1 (unity),and with the usualnotation it is positive ,leaving away the [0,0] point e^[ipi] = -1 it is driven towards the [0,0] point since e^[ipi] =-1+i*{0) is a technicality and SELECTIVELY one may choose either of the two componentsaccording to its interest, to give the real ,or the imaginarysolution. Simply taking care of the angle ,it is fair for one to statetha unity (mod 1 of e^[ipi]) at 180 deg ,anticlockwise from thepositive X-axis ,has two components ,one of wch the real(by convention)has the value of one, and the imaginary (by convention)Zero. === mis-interpreted the OP's question. There are plenty of> non-finite tngs of interest in mathematics: ordinals and cardinals,> sequences and series, non-terminating algorithms, etc., and if> that's what the question was about I apologize. I thought the person> was talking like a movie producer. (Oh, those mathematicians are > so different from us; they can study Ôinfinity'!)> That's why I don't bother with answering those kinds of questions. It's hard to tell if it's a movie producer or not ;-) I basically read your post without tnking of the context of thepossible movie producer mindset of the OP. [BTW, ts is DaveRusin's terminology; I have only the ghest regard for movieproducers] Your answer is undoubtedly appropriate in that context;unfortunately, many people, including me, misunderstood what you weretrying to say. There are a lot of consenting adults -- to use yourphrase from a === related post -- here on sci.math!Subject: Re: The role of accepted answer, what do you>tnk?>>Huh? Ts question doesn't make sense.> You are right, it doesn't. I'll try again:These are questions wch seem not to have an accepted answer, what do you> tnk?As you've probably seen, there tends to be a majority opinion, with a few dissenters. Ts is a lot like most other fields. We also get the occasional crackpot. Nobody tends to get kicked out unless they are doing sometng that is logically inconsistent.Personally, I would have issues with some of the Ultraintuitionist stuff just because I like mathematical induction. As long as they can make it self-consistent, they === infinity in math>How many integers are there?How would you possibly answer ts question?A. There's an infinity of them. I would never say that.B. There's an infinite number of them. Don't say that; it makes people tnk there are infinite (natural) numbers.C. There are infinitely many of them, or There's an infinite set of them. That's OK. I don't see a bit of difference between saying that and saying there's not a finite number of them (But there is at least one!)D. There are aleph-1 of them or the set of them is countably infinite. Look up the definition of countably infinite and tell me ts isn't nearly a tautology!So what's your point? I thought I made it clear that mathematics encountersinfinite sets and limits and all sorts of situations where we mightbandy about the term infinity, but it's just a shorthand for sometngmore precise, or else just a synonym for not finite. Or are you sayinginfinity is the set of integers? I tnk that's awfully limiting...>> Mathematicians (and teachers) would probably do everyone a great>> service if they never made reference to infinity; leave the>> mysticism to someone else (Buzz Lightyear, perhaps?)Modern set theory deals a great deal with infinity as a core concept.I tnk you mean it deals primarily with infinite sets and compares andcontrasts them. OK. But I tnk the emphasis is on other structures (suchas the existence of functions of various types between sets, or possiblewell-orderings on sets). The fact that most of the sets are infinite is agiven, and not itself the subject of much attention. (Disclaimer: I am not a set theorist!) Some people do wrestle with the precise question of howwe define the predicate S is infinite (usual choice: existence of abijection S->T where T is a proper subset of S) but I don't tnk that'sconsidered a core issue of modern set theory! I believe you're reading much too much into the OP's question. Ts isa secondary-school student who links mathematics with measurement andprecision. Perhaps someone has wspered about a tng called infinityand he wants to know what it is and why we care. Not infinite sets;not ordinals and cardinals, just a tng called infinity, wch isseen as some kind of number (I guess). I don't believe that's a usefulconcept in itself.I've given a talk like My infinity can beat your infinity! to ghschool students. First tng I do is to ask them to stop tnking aboutany such tngs, and instead to tnk about infinite _sets_ instead.Much more mathematically sound and it still seems to intrigue them.>> When you have a set that's not finite, and people ask you how>> many tngs you've got, you can say infinitely many, but they're>> still just ordinary tngs (points, numbers, whatever); there's>> just a lot of them -- not a finite number of them at all, so, well,>> an in-finite number of them.You are focusing on the individual elements, but when people talk about >how many elements there are in a set, the desire is to get a meaningful >value. In fact, simply saying that it is infinite is not always enough >detail. Thus the cardinals to measure levels of infinity.You mean, Thus the cardinals to represent different answers to Ôhow many'.There are finite cardinals, too, and I tnk you're reinforcing my point:infinity, per se, is pretty useless: if you want to answer how many,you point to a cardinal. If that cardinal is finite, great. If not, youcan say not finite. If you want to give more information, you specifythe appropriate cardinal, thus making it clear that the answer infinityis at best a shorthand for not finite. It's not really a useful conceptin and of itself.>The Riemann sphere has a point on it that *is* infinity. It is a value, >not a concept, in that model of the complex numbers.Um, right. I said it was a point _called_ infinity, but if now you wantto tell me that ts is, in fact, the definition of infinity, then Idon't object, exactly; but if infinity is _really_ a part of theRiemann sphere, then I wonder what the heck you meant by trotting out infinite sets before? Again I tnk you're sort of making my point for me:as much as we use the term infinity in mathematics, there really isno single tng that encompasses a synonym for not finite the whole panoply of infinite sets a point on the Riemann sphere etc.not to mention the pop-science connections to Big Bangs or whatever.It's a subtle tng, but careful attention to subtleties can helpnewcomers. Use infinite, the definable adjective, whenever possibleinstead of an undefined noun === many integers are there?> How would you possibly answer ts question?A. There's an infinity of them. I would never say that.> B. There's an infinite number of them. Don't say that; it makes people> tnk there are infinite (natural) numbers.> C. There are infinitely many of them, or There's an infinite set of them.> That's OK. I don't see a bit of difference between saying that and> saying there's not a finite number of them (But there is at least one!)> D. There are aleph-1 of them or the set of them is countably infinite.> Look up the definition of countably infinite and tell me ts isn't> nearly a tautology!I'd say C, or There's aleph-0 of themProbably C.So what's your point? I thought I made it clear that mathematics encounters> infinite sets and limits and all sorts of situations where we might> bandy about the term infinity, but it's just a shorthand for sometng> more precise, or else just a synonym for not finite. Or are you saying> infinity is the set of integers? I tnk that's awfully limiting...Probably none. At ts point we seem to be splitting hairs of one of the angels dancing on the head of a pin.>Mathematicians (and teachers) would probably do everyone a great>service if they never made reference to infinity; leave the>mysticism to someone else (Buzz Lightyear, perhaps?)>>Modern set theory deals a great deal with infinity as a core concept.> I tnk you mean it deals primarily with infinite sets and compares and> contrasts them. OK. But I tnk the emphasis is on other structures (such> as the existence of functions of various types between sets, or possible> well-orderings on sets). The fact that most of the sets are infinite is a> given, and not itself the subject of much attention. (Disclaimer: I am not > a set theorist!) Some people do wrestle with the precise question of how> we define the predicate S is infinite (usual choice: existence of a> bijection S->T where T is a proper subset of S) but I don't tnk that's> considered a core issue of modern set theory! I believe you're reading much too much into the OP's question. See my reply to the OP. I just found your response odd. Perhaps it's just me. > Ts is> a secondary-school student who links mathematics with measurement and> precision. Perhaps someone has wspered about a tng called infinity> and he wants to know what it is and why we care. Not infinite sets;> not ordinals and cardinals, just a tng called infinity, wch is> seen as some kind of number (I guess). I don't believe that's a useful> concept in itself.I've given a talk like My infinity can beat your infinity! to gh> school students. First tng I do is to ask them to stop tnking about> any such tngs, and instead to tnk about infinite _sets_ instead.> Much more mathematically sound and it still seems to intrigue them.When you have a set that's not finite, and people ask you how>many tngs you've got, you can say infinitely many, but they're>still just ordinary tngs (points, numbers, whatever); there's>just a lot of them -- not a finite number of them at all, so, well,>an in-finite number of them.>>You are focusing on the individual elements, but when people talk about >>how many elements there are in a set, the desire is to get a meaningful >>value. In fact, simply saying that it is infinite is not always enough >>detail. Thus the cardinals to measure levels of infinity.> You mean, Thus the cardinals to represent different answers to Ôhow many'.> There are finite cardinals, too, and I tnk you're reinforcing my point:> infinity, per se, is pretty useless: if you want to answer how many,> you point to a cardinal. If that cardinal is finite, great. If not, you> can say not finite. If you want to give more information, you specify> the appropriate cardinal, thus making it clear that the answer infinity> is at best a shorthand for not finite. It's not really a useful concept> in and of itself.>The Riemann sphere has a point on it that *is* infinity. It is a value, >>not a concept, in that model of the complex numbers.> Um, right. I said it was a point _called_ infinity, but if now you want> to tell me that ts is, in fact, the definition of infinity, then I> don't object, exactly; but if infinity is _really_ a part of the> Riemann sphere, then I wonder what the heck you meant by trotting out > infinite sets before? Again I tnk you're sort of making my point for me:> as much as we use the term infinity in mathematics, there really is> no single tng that encompasses> a synonym for not finite> the whole panoply of infinite sets> a point on the Riemann sphere> etc.> not to mention the pop-science connections to Big Bangs or whatever.It's a subtle tng, but careful attention to subtleties can help> newcomers. Use infinite, the definable adjective, whenever possible> instead of an undefined noun infinity.It is, IMHO, a concept best used carefully and not introduced too soon. === The role of infinity in math>How many integers are there?> How would you possibly answer ts question? A. There's an infinity of them. I would never say that.> B. There's an infinite number of them. Don't say that; it makes people> tnk there are infinite (natural) numbers.> C. There are infinitely many of them, or There's an infinite set ofthem.> That's OK. I don't see a bit of difference between saying that and> saying there's not a finite number of them (But there is at leastone!)> D. There are aleph-1 of them or the set of them is countablyinfinite.> Look up the definition of countably infinite and tell me ts isn't> nearly a tautology! I'd say C, or There's aleph-0 of them Probably C.> So what's your point? I thought I made it clear that mathematicsencounters> infinite sets and limits and all sorts of situations where we might> bandy about the term infinity, but it's just a shorthand for sometng> more precise, or else just a synonym for not finite. Or are you saying> infinity is the set of integers? I tnk that's awfully limiting... Probably none. At ts point we seem to be splitting hairs of one of> the angels dancing on the head of a pin.Mathematicians (and teachers) would probably do everyone a great>service if they never made reference to infinity; leave the>mysticism to someone else (Buzz Lightyear, perhaps?)>>Modern set theory deals a great deal with infinity as a core concept.> I tnk you mean it deals primarily with infinite sets and compares and> contrasts them. OK. But I tnk the emphasis is on other structures(such> as the existence of functions of various types between sets, or possible> well-orderings on sets). The fact that most of the sets are infinite isa> given, and not itself the subject of much attention. (Disclaimer: I amnot> a set theorist!) Some people do wrestle with the precise question of how> we define the predicate S is infinite (usual choice: existence of a> bijection S->T where T is a proper subset of S) but I don't tnk that's> considered a core issue of modern set theory! I believe you're reading much too much into the OP's question. See my reply to the OP. I just found your response odd. Perhaps it's> just me.>For what its worth, I tend to agree with you. I'm not a mathematician butRusin's comments don't seem to jibe with the way I have learned the conceptof infinity. Especially comments that seem to reduce the concept of'infinite set' to merely Ônon-finite set'. But perhaps it is appropriatefor the OP's level of familiarity with math in === infinity in mathHow many integers are there?>> How would you possibly answer ts question?>> A. There's an infinity of them. I would never say that.>> B. There's an infinite number of them. Don't say that; it makes>> people>> tnk there are infinite (natural) numbers.>> C. There are infinitely many of them, or There's an infinite set of> them.>> That's OK. I don't see a bit of difference between saying that and>> saying there's not a finite number of them (But there is at least> one!)>> D. There are aleph-1 of them or the set of them is countably> infinite.>> Look up the definition of countably infinite and tell me ts>> isn't nearly a tautology!>> I'd say C, or There's aleph-0 of them>> Probably C.>>For what its worth, I tend to agree with you. I'm not a mathematician but> Rusin's comments don't seem to jibe with the way I have learned the> concept> of infinity. As you said: you are not a mathematician.Dave's comments above nicely epitomize the mathematician's concept ofinfinite sets. Note B and the avoidance of the infinite numbers belovedof cranks.> Especially comments that seem to reduce the concept of> Ôinfinite set' to merely Ônon-finite set'. Simple etymology: infinite means not finite.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes s penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the === do wrestle with the precise question of how> we define the predicate S is infinite (usual choice: existence of a> bijection S->T where T is a proper subset of S) but I don't tnk that's> considered a core issue of modern set theory!I came up with a simple definition of an infinite set in another thread.Let S be a set of natural numbers and let x and y be members of S.Consider these two statements:1) ExAy(x>=y)2) AxEy(x we define the predicate S is infinite (usual choice: existence of a> bijection S->T where T is a proper subset of S) but I don't tnk that's> considered a core issue of modern set theory!I came up with a simple definition of an infinite set in another thread.> Let S be a set of natural numbers and let x and y be members of S.> Consider these two statements:1) ExAy(x>=y)> 2) AxEy(x If statement (2) is true then S is infinite.But (1) only works for well-ordered sets, and not all sets are well-ordered, e.g., the set of negative integers satisfies (1) so must be finite by your reckoning.How do you tell if an ordered-but-not-well-ordered set is finite?In fact sets need not be ordered at all. How do you tell whether an unordered set is finite?Your definitions are === math Some people do wrestle with the precise question of how> we define the predicate S is infinite (usual choice: existence of a> bijection S->T where T is a proper subset of S) but I don't tnkthat's> considered a core issue of modern set theory! I came up with a simple definition of an infinite set in another thread.> Let S be a set of natural numbers and let x and y be members of S.> Consider these two statements: 1) ExAy(x>=y)> 2) AxEy(x If statement (2) is true then S is infinite. But (1) only works for well-ordered sets, and not all sets are> well-ordered, e.g., the set of negative integers satisfies (1) so must> be finite by your reckoning.I said S is a set of natural numbers.That wouldn't include negative integers.> How do you tell if an ordered-but-not-well-ordered set is finite?That would be more complicated.Any set of natural numbers can be well ordered.> In fact sets need not be ordered at all. How do you tell whether an> unordered set is finite?My method won't tell you.Can't all finite sets be ordered?> Your definitions are simple enough, but wrong.My definition works for sets of natural numbers.That is all I claimed.Russell- 2 many 2 === do wrestle with the precise question of how> we define the predicate S is infinite (usual choice: existence of a> bijection S->T where T is a proper subset of S) but I don't tnk> that's> considered a core issue of modern set theory! I came up with a simple definition of an infinite set in another thread.> Let S be a set of natural numbers and let x and y be members of S.> Consider these two statements: 1) ExAy(x>=y)> 2) AxEy(x If statement (2) is true then S is infinite. But (1) only works for well-ordered sets, and not all sets are> well-ordered, e.g., the set of negative integers satisfies (1) so must> be finite by your reckoning.I said S is a set of natural numbers.> That wouldn't include negative integers.It wouldn't work for negatives either, wch is what I said.How do you tell if an ordered-but-not-well-ordered set is finite?That would be more complicated.> Any set of natural numbers can be well ordered.In fact sets need not be ordered at all. How do you tell whether an> unordered set is finite?My method won't tell you.> Can't all finite sets be ordered?Your definitions are simple enough, but wrong.My definition works for sets of natural numbers.> That is all I claimed.By implication you claimed more. You said that you had simple definition in another thread, then stated ts one, without making clear that it was not your other definition.And I complained that ts one === infinity in math> The classic plosophy (Platonism) says that all of ts stuff> (infinity) is in fact real. And just because we cannot perceive it> does not mean that it does not exist. Ts is the majority opinion,> not necessarily because the majority believe it, but more because it> is the most practical way of tnking of mathematicsWhat?!!!! (read that as I'm stunned in disagreement and intrigue)> would be much more difficult to write and describe. The derivative> of x^2 would not be 2x but would be 2x+1/M, where M is the largest> number. In practice, ts number 1/M would be so small that it> wouldn't even matter, so why write it down on paper? Limits and derivatives and in general differential calculus makessense and is 100% tanle without having to resort to theexistence of such largest number, wch can so easily be showndoes not exist. The limit of x^2 as x approaches 0 is *exactly*0 -- it's not 0 + 1/M. It's not as close to 0 as we want -- no,it is 0, because the definition directly implies so (a definitionthat is tanle and expressed in tanle terms, unambiguous, andvery indisputable, I tnk).> Pythagorean theorem would not hold, and it would be much more messy to> describe ts relationsp. But ts way of tnking still avoids> paradoxes.How does it avoid the paradox that there is the same number ofreal numbers in the interval (0,1) as in the interval (0,2) ??If you state that there are M real numbers in the interval (0,1),where M is the largest number that exist, how do you explainthat there is exactly M numbers in the interval (0,2)? (butthere is also exactly M + M numbers... What? M + M is M??I don't tnk that's === Re: The role of infinity in mathThe classic plosophy (Platonism) says that all of ts stuff> (infinity) is in fact real. And just because we cannot perceive it> does not mean that it does not exist. Ts is the majority opinion,> not necessarily because the majority believe it, but more because it> is the most practical way of tnking of mathematicsWhat?!!!! (read that as I'm stunned in disagreement and intrigue) I'm curious as to why you are stunned? Most mathematicians today are acombination of formalists and Platonists. When they congregatetogether, they only believe in the theorems that they prove becausethey follow from axioms that are standardly accepted and assumed toget rid of paradoxes.However, whenever they relate to natural scientists, it is then whenthey suddenly become Platonists and are the first to proclaim that notonly what they are doing is formally right but also is reality.The reason that the intuitionist plosophy is not mainstream is notbecause it is wrong, but because it would nder the power ofmathematics to describe the natural === reason that the intuitionist plosophy is not mainstream is not> because it is wrong, but because it would nder the power of> mathematics to describe the natural world.Really? And that part about where intuitionism avers that mathematicsand proofs cannot be communicated? That doesn't conßict with anymathematicians' experiences?-- Now I'm informing all of you that the people arguing against me are EVIL,yes they are real, live EVIL people as mathematics is that important, soit's important enough for Evil itself to send minions like them. -- === The role of infinity in math> The reason that the intuitionist plosophy is not mainstream is not> because it is wrong, but because it would nder the power of> mathematics to describe === infinity in math The classic plosophy (Platonism) says that all of ts stuff> (infinity) is in fact real. And just because we cannot perceive it> does not mean that it does not exist. Ts is the majority opinion,> not necessarily because the majority believe it, but more because it> is the most practical way of tnking of mathematics What?!!!! (read that as I'm stunned in disagreement and intrigue) would be much more difficult to write and describe. The derivative> of x^2 would not be 2x but would be 2x+1/M, where M is the largest> number. In practice, ts number 1/M would be so small that it> wouldn't even matter, so why write it down on paper? Limits and derivatives and in general differential calculus makes> sense and is 100% tanle without having to resort to the> existence of such largest number, wch can so easily be shown> does not exist.I tnk you've missed the point.http://en.wikipedia.org/wiki/Mathematical_ === Matroids and non Matroids by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1DLS5A29553;I am trying to create a non matroid that satisfies rank axioms 1 and 2 and not 3 but at a lower rank than the === universal set with 3 valued logic by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1DLp8X31615;yes, in ZF you can prove the universal set does not exist and you don't need a computer program to do that (i would be interested in what happens if your program is equipped with three logical values and the o logical connective wch is an extension of the iff biconditional). but ts is an extension of ZF and an enlargement of perspective. read the paper before you disagree on ts point, especially the first couple of pages.why does one NEED to do ts? well, the short answer is that one doesn't. one doesn't need to not do it, either. i find it satisfying on a plosopcal level that the universal set no longer need to be of a type different than a === find equation for pianist's possibilities of expert but I'd like a valid number for the possibilities ofsound from a piano. I started with ts:P = 10 x 88 x V x T--P = possibilities10 = pianisit's 10 fingers88 = number of keys on the pianoV = Velocity or volume (one could assign a number of 100 differentdiscernable volume levels)T = tempo or time between notes (one could assign a number of 1000 differentdiscernable tempos)--I don't know if my equation needs exponents though. And if so, where? Onecan play 10 notes all at once or one at a time. And, by holding the sustainpedal a pianist can make all 88 keys ring at once.The discussion that explains my tnking is below:-----------Just wanted to throw sometng your way. A discussion I had after a recenttrip to see Gonzalo Rubalcaba, Cuban piano genius, I thought might sparkyour interest a bit. I was tnking about what it means to create art andthe term was troubling for me. It seemed to emphasize the ego. It turns theartist into creator. Rather, I found it better to view art making as adiscovery process, much like astronomers and geologists to some extent. Letme start at the beginning of my tnking.Driving home from an amazing and inspiring jazz piano trio concert my motherthe hauntingly gorgeous etched melodies carved out of the slow moving spaceof s ballads to the ferocious explosions of s rhythmic up-tempo tunes. Iwondered why s improvisations and compositions seemed so fresh and new,unlike any other jazz music I'd heard. It was even fresh compared to thelast 30 times I'd seen m play.I thought about all the choices he made that night. All the possibilities hehad before m on the 88 key piano. I saw it best as the algebra equation:P = 10 x 88 x V x TWhere P is the playing field or all the possibilities of sound from thepiano. 10 represents the artist's 10 fingers. 88 is the number of keys onthe piano. V represents the velocity or volume at wch the keys are struck.And T is time or rather tempo.Even if I were to assign a value of only 100 possible Velocity levels andsay 1000 tempos (or time values between two notes), that still makes ourequation 10 x 88 x 100 x 1000 = 88,000,000 unique possibilities. Now mylocal mathematics professor would say I need to alter my equation to includeexponents. Sometng like 88vt^10 (8.8 million to the power of 10 fingers)might be closer to what my equation should look like. Or maybe 10^88vt?Regardless of my feeble attempts at algebra we know it would take millionsof years and millions of piano players to play all the possibilities of thepiano's sound. Ok, so now let's calculate possibility values for theacoustic bass and the drum set? .... Just kidding.The equations I thought about reminded me of equations for 3-D space. Itseems attractive to tnk about the piano as a 3 dimensional space so vastmy mind can barely grasp it's expansion. Hence, the pianist becomes anexplorer on a journey through a God created world, first traveling throughthe space others have trodden, and then finding new paths.The pianist's possibilities might better be portrayed as an enormous solidsubstance that stretches throughout the universe. It would be suitable forcseling away at like a sculpture artist might work with marble. Or duringa pianist's jazz improvisation it might become soft clay that is more easilyformed into shapes as spontaneously as the artist imagines them. Alwaysdiscovering new shapes. Always finding new paths.The focus is not on creating. Rather, finding and discovering. When youfinish a wood carving do you say Ts is a wood carving I created? Or doyou say Ts is a piece of a tree God created? Though it is your workindeed, the substance, the essence, the material itself is God's.The weight and pressure of creating sometng from notng is lifted. Now Ican just play with and explore throughout God's already created materialuntil I find tngs I like.Please email if you have any === number?/What is not a number?What is a number?/What is not a number? Numbers seem to be simple tngs (objects) or a property of tngs.Yet the deeper you study numbers the more complex they become.As cldren understanding of numbers starts with the Naturals then the Intergers, the Rationals, the Reals, the Complex and beyond.Mathamatics later turns ts all on its headAll Computer Data & programs that currently Exist now can be considered as on very large interger, ts Interger is unknowen, and tecnicaly impossable to calculateCarl === is a number?/What is not a number? > Numbers seem to be simple tngs (objects) or a property of tngs.> Yet the deeper you study numbers the more complex they become.As cldren understanding of numbers starts with the Naturals then the > Intergers, the Rationals, the Reals, the Complex and beyond.Mathamatics later turns ts all on its head> All Computer Data & programs that currently Exist now can be considered as > on very large interger, ts Interger is unknowen, and tecnicaly > impossable === Re: sqrt(-1)=0/0> 0/0 is every number, 0 is their placeholder.0/0=undefined, so 0 cannot be a place holder for all x in N, I or R. > 0> denotes the absence of quantity wch is instrinsically a quantity> (ie. notng is sometng).Notng has no constituent parts, and therefore cannot be divided by> or multiplied against...*> Do not equate Ônotng' with Ôzero'.earle> *Zero denotes the absence of quantity wch is intrisically a quantity,> since plosopcally, notng is sometng.*> ÔNotng' is the opposite of Ôsometng'.The opposite of sometng is sometng, the opposite of everytng isnotng.> ÔZero' is a point on the number line, like 6, -23, pi, and 1,931.Zero is a number, it is not a point. A point can be described atcoordinate zero relative to a frame of reference, but you don't needreference frames or geometry to define === everybody.Consider a finite set A with n elements and the group G ofpermutations on A. Then the order of G is n!. I was wondering, howdoes ts extend to the countably infinite case? Is G countable oruncountable? I would imagine the latter, since n! > 2^n for n > 3,and, proceeding by analogy with the power sets of finite sets, |P(A)|= 2^|A| and the naturals, but I realize of wch order of infinity a setbelongs to.Any suggestions, === Group Theory>(...)>Consider a finite set A with n elements and the group G of>permutations on A. Then the order of G is n!. I was wondering, how>does ts extend to the countably infinite case? Is G countable or>uncountable? I would imagine the latter, since n! > 2^n for n > 3,>and, proceeding by analogy with the power sets of finite sets, |P(A)|>= 2^|A| and the naturals, but I realize how tricky ts issue can be,>especially to.>Any suggestions, answers, nts?>(...)>I found a simple injection f:P(A)->G' with G'=group of permutationsof Ax{1,2} so that |G|=|G'|>=2^|A| is uncountable if A is infinite -because then |A|*2=|A|. In fact, if A is countable-infinite, then |G|=2^|A| (the power of continuum) because G is part of A^A andwith a=|A| (=aleph_0) we have a^a <= (2^a)^a = 2^(a*a)=2^a <= a^athus a^a=2^a.Define for B any subset of A: f(B) is the permutation p of Ax{1,2}such that p(a , k)=(a , k) for k=1,2 when a is not in B, andp(a , 1)=(a , 2) and p(a , === 2)=(a , 1) when a is in B.Subject: Information Theory - Mutual conditional mutualinformation of 4 random variables X1,X2,X3 and X4 ? And also the formual for the chain rule for conditional mutualinformation. I(X3;X4/X1,X2) = ????? I(Xn-1;Xn/X1,X2,.....Xn-2) = ?????? Hoping for a reply in two Unlimited-Uncensored-Secure Usenet News==----http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption === === Partial-Sum -> Some PrimesLet a(1) = 1;Let a(m) be the lowest yet unpicked positive integer such that:sum{k=1 to m} k* a(k)is a prime.Figured by hand, so very well could be wrong, I get:a(k) : 1, 2, 4, 3, 6, 5, 12, 7,...Question:Is ts a permutation of the positive integers?If so, its inverse-permutation is:1, 2, 4, 3, 6, 5, 8,...(I get some EIS matches for the few terms I give for the inverse permutation, === Difference Divides Sum Of 2 TermsLet b(1) = 1;Let b(m) = lowest positive unpicked integers such that:(m-k) divides evenly into (b(m) + b(k))for EACH k, 1 <= k <= m-1.I get (again, figured by hand, so not completely believable):b(m) : 1, 2, 3, 8, 7, 54,...(That last IS 54, not 5, 4.)Is ts sequence a permutation of the integers >= 1?(I am not even sure it is infinite, without looking harder at the sequence's mathematics.), Leroy === my functions of a complex variable class:verify that the function g(z)= ln(r) + i*t (r>0, 00)with derivative:G'(z)= (2z)/(z^2 + 1)There is a nt that Im(z^2 + 1)>0 when x,y>0I really don't understand ts question so any === w.r.t. to Random Variables ?Let's say I have two rvs, RV1 and RV2. Assume for now they are normalwith known means and variances. Let's say I have a function F of RVs(such as sum or difference, though my function is more complex). Iknow an analytical expression to calculate the mean and variance ofF(RV1,RV2) though I don't necessarily have an analytical closed formexpression for F as a whole.Now, let's say I'm doing an optimization with some objective functionFO = Constant1*mean+Constant2*stdvariationI want to look at F(RV1,RV2), and knowing FO(F) I want to determinewch RV contributes more towards FO. If these had been regularvariables, I would just take derivative of FO relative to my freeparameters and pick the largest or smallest one depending on what I'moptimizing. How do I do a comparable operation with RVs RV1 and RV2 ?Are derivatives the way to go or is there some other concept I can use? I can settle for doing ts numerically once I can at least describeit analytically. Any and all pointers much === Integer Pairs(I am posting ts for fun, for it is probably easily solved by thosewith some background in number theory.)Let us say we have 2 spherical dice wch somehow each represent allpositive integers, each integer represented exactly once per die, ontheir countably-infinite number of sides. And each integer is equallylikely to be rolled on both dice.So, the dice-pair is rolled an infinite number of times.What is the closed-form representation of the probability thatfor *at least one* roll of the dice-pairthe m_th rolling of the pair produces integers j and k whereGCD(j,k) = m?(ie. there is at least one m, where m = GCD(j,k) on the m_th rollingof the dice-pair.)Hopefully the === Volume of Astroid Help!!The problem is to get volume of solid formed by rotation of x^2/3 +y^2/3=a^2/3 around the x-axis and y-axis with a>0.1) I have never heard of an astroid or hypocycloid, so had to do somedigging.2) I found that x^2/3 + y^2/3=a^2/3 is the formula for an astroid (a foursided diamond shaped tng with rounded sides.The points of the diamondfall on the axes. It is formed from rotating a circle witn an circle.3) I also found some other formulas : x=a cos ^3 theta and y= a sin ^3theta.4) That's all I got. I was ßying through disks and washers and shells, butwhen it came to ts I t a wall because 1) I had never heard of an astroidor the shape of the graph and 2) ts formula has three variables (x,y,a)vice the normal x and y functions.5) Any help on how to work ts === problem is to get volume of solid formed by rotation of x^2/3 +> y^2/3=a^2/3 around the x-axis and y-axis with a>0.Do you mean (x^2)/3 + (y^2)/3=(a^2)/3, wch is what your notation suggests, or (x^2)^(1/3) + (y^2)^(1/3)=(a^2)^(1/3), wch is what your description suggests?And what do you mean by rotating it about _both_ axes? Do you mean around one axis OR the other, or sometng like the solid bounded bythe 3D surface |x|^(2/3) + |y|^(2/3) + |z|^(2/3) = |a|^(2/3) ?> 1) I have never heard of an astroid or hypocycloid, so had to do some> digging.2) I found that x^2/3 + y^2/3=a^2/3 is the formula for an astroid (a four> sided diamond shaped tng with rounded sides.The points of the diamond> fall on the axes. It is formed from rotating a circle witn an circle.3) I also found some other formulas : x=a cos ^3 theta and y= a sin ^3> theta.4) That's all I got. I was ßying through disks and washers and shells, but> when it came to ts I t a wall because 1) I had never heard of an astroid> or the shape of the graph and 2) ts formula has three variables (x,y,a)> vice the normal x and y functions.> 5) Any help on how to work ts === earlier but it doesn't seem to have got there, so i will postit again.I am trying to understand the Zeta function.I understand that it is defined as Zeta(s) = sum(1/n^s), n=1..infinity.Therefore, Zeta(2) = Pi^2/6, Zeta(4) = Pi^4/90, and so on.Also Zeta(1) = infinity. These make sense.However, the books also tell me that when s = 0, Zeta(0) = -1/2Also when s = -2k for k = 1,2,3,...,then Zeta(-2k) = 0When i substitute s = 0 or s = -2k into the sum function above, i getdivergence to infinity, not the specific values above.Is the sum function for Zeta(s) only valid for s = positive numbers, sgreater than or equal to 1?If so, how is the function for Zeta(s) defined when s is less than 1? in advance for any help === coverings-problem...(I myself might have already asked ts, but I do not believe that I,at least, have asked ts specific question before.)If we have a unit circle (radius = 1),and we have 3 unit squares (sides =1),then what is the maximum area of the circle we can ever cover with thesquaresa) if overlapping of the squares is not allowed?b) if overlap is allowed?I may be way off,but I conjecture that the (a) case issquares arranged like: ____ ! ! ! !---------! ! !! ! !---------But I bet tilting the squares works better.As for the (b) case,if we start with the (a) solution I give, then raise the bottomsquares (or lower the top square by the same amount), then tilt thesquares somewhat somehow,PERHAPS we might have an arrangement wch maximally covers a unitcircle placed properly on top of the squares.Any better ideas or arrangements? (or more precise descriptions ofpossible/actual === uses False Arrest, Document Destruction!Notng.Columbia University is so Liberal far to the Left that it will admit> anytng even vaguely human (short of a credentialed Wte male) and> give it a PhD for emulating Bush the Lesser's National Guard> attendence record. Columbia Emeritus Professor Barton Sholod is my> cousin. He routinely taught Spanish to spics who could not learn> Spanish, and they graduated on time fully degreed without having> suffered the hate burden of acquiring detectable education or of> paying tuition out of pocket. They had compassionate scholarsps.Anybody who cannot succeed in such a milieu is a phenomenon.Try CUNY instead - the standards are lower.But the whole situation is competely unrelated to Mathematics. Tsmust not be a subject of discussion in Sci.math === University uses False Arrest, Document Destruction!> For explanation of the message title please see wesbites> http://www.cuspeech.org and http://www.infiniteseriestheorem.org.your e-mails do not seem user-friendly. to top it off, you offer nosubstatiation to your accussations.don't get me wrong, i generally agree with the idea thatgovernment-run schools are source of waste and corruption, but i justdon't see that here. could you offer === uses False Arrest, Document Destruction!> Can you formulate in a few sentences what exactly is the problem?> That would be helpful for those who don't take the effort of reading whole> sites.Wayne Brown summarized some of the facts for you in s msg.The wider issue is what a healthy University is? I suppose the facultyis working to seek truths (in their fields) and the students areinterested in learning these truths. Then there is the administration-- suppose they are going about destroying documents, falsifyingnotes, and calling in rogue police detectives to arrest and intimidatecritics -- how does that fit in with the purpose of a University?Is there no way to hold administrators accountable for intentionalmisbehavior against students and faculty? When the President of amajor university is out of control, what can one do?Could it not be argued that destroying the truth is incompatible witha University's mission? Or is there no mission?Just what is a University? === Arrest, Document Destruction!> Can you formulate in a few sentences what exactly is the problem?> That would be helpful for those who don't take the effort of reading whole> sites.By snipping all headers and context, you have lost the bulk of your possiblereaders.Franz