mm-4579 === Subject: Simple number theory problem... OMG... This one is hard to me.(Maybe all is hard to me.) Find all n s which satisfy that n(n+1)/p = k^2 for some k. (p is prime) This is not from the book exercise, but from my studymate. === Subject: Re: Simple number theory problem... > Find all n s which satisfy that n(n+1)/p = k^2 for some k. (p is > prime) Obviously n and n+1 have no prime factors in common. Hence one of n or (n+1) is a square, and the other is p times a square. Taking n+1 to be the square, we have n+1 = x^2, n = p y^2, hence x^2 - p y^2 = 1, which is Pell's equation having an infinite number of solutions given by a recurrence relation which can be derived from the continued fraction expansion of sqrt(p). If we take n to be the square, we arrive at x^2 - p y^2 = -1, a closely related equation having solutions only when p is of the form 4k+1, also derivable via continued fractions. - Tim === Subject: Re: Simple number theory problem... >OMG... This one is hard to me.(Maybe all is hard to me.) Find all n s which satisfy that n(n+1)/p = k^2 for some k. >(p is prime) This is not from the book exercise, but from my studymate. I'll assume the variables n,k are required to be positive integers. For each p, it's just a Pell's equation, or actually 2 of them. Let p be a fixed prime. Then n*(n+1) = p*k^2 implies either n = x^2 and (n+1) = p*y^2 or n = p*y^2 and (n+1) = x^2 where x,y are positive integers. It follows that there exist positive integers (n,k) satisfying n*(n+1) = p*k^2 iff there exist positive integers x,y satisfying either x^2 - p*y^2 = 1 or x^2 - p*y^2 = -1 The theory of Pell's equation implies that if either of the above equations has a solution in positive integers x,y, then it has infinitely many such solutions. For the first equation x^2 - p*y^2 = 1 there are always positive integer solutions x,y, hence infinitely many. For the second equation, a necessary condition for positive integer solutions x,y is that -1 must be a quadratic residue mod p. I believe that condition is also sufficient, but I'm not sure. For a Pell's equation reference, here's a wiki link: quasi === Subject: Re: Simple number theory problem... posting-account=VR0DOgoAAADggPTteFeA2AkmHNhjcrDV Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) >OMG... This one is hard to me.(Maybe all is hard to me.) >Find all n s which satisfy that n(n+1)/p = k^2 for some k. >(p is prime) >This is not from the book exercise, but from my studymate. I'll assume the variables n,k are required to be positive integers. For each p, it's just a Pell's equation, or actually 2 of them. Let p be a fixed prime. Then n*(n+1) = p*k^2 implies either n = x^2 and (n+1) = p*y^2 or n = p*y^2 and (n+1) = x^2 where x,y are positive integers. It follows that there exist positive integers (n,k) satisfying n*(n+1) = p*k^2 iff there exist positive integers x,y satisfying either x^2 - p*y^2 = 1 or x^2 - p*y^2 = -1 The theory of Pell's equation implies that if either of the above > equations has a solution in positive integers x,y, then it has > infinitely many such solutions. For the first equation x^2 - p*y^2 = 1 there are always positive integer solutions x,y, hence infinitely > many. For the second equation, a necessary condition for positive integer > solutions x,y is that -1 must be a quadratic residue mod p. I believe > that condition is also sufficient, but I'm not sure. For a Pell's equation reference, here's a wiki link: quasi Your conjecture is correct, the equation x^2 - py^2 = -1 has a solution if p is a prime congruent to 1 modulo 4. However, it is not true that if -1 is a square modulo m, then x^2 - my^2 = -1 has a solution. For example, there is no solution with m = 34. === Subject: Re: Simple number theory problem... posting-account=cEKB4goAAABJ5eifbZgDmj5s5T5PQpjx BESAGENT; .NET CLR 1.1.4322; BESAGENT),gzip(gfe),gzip(gfe) >OMG... This one is hard to me.(Maybe all is hard to me.) >Find all n s which satisfy that n(n+1)/p = k^2 for some k. >(p is prime) >This is not from the book exercise, but from my studymate. I'll assume the variables n,k are required to be positive integers. For each p, it's just a Pell's equation, or actually 2 of them. Let p be a fixed prime. Then n*(n+1) = p*k^2 implies either n = x^2 and (n+1) = p*y^2 or n = p*y^2 and (n+1) = x^2 where x,y are positive integers. It follows that there exist positive integers (n,k) satisfying n*(n+1) = p*k^2 iff there exist positive integers x,y satisfying either x^2 - p*y^2 = 1 or x^2 - p*y^2 = -1 The theory of Pell's equation implies that if either of the above > equations has a solution in positive integers x,y, then it has > infinitely many such solutions. For the first equation x^2 - p*y^2 = 1 there are always positive integer solutions x,y, hence infinitely > many. For the second equation, a necessary condition for positive integer > solutions x,y is that -1 must be a quadratic residue mod p. I believe > that condition is also sufficient, but I'm not sure. For a Pell's equation reference, here's a wiki link: quasi === Subject: Re: Affine maps .NET CLR 2.0.50727; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > Hi all, Question: A map from R^n into R^m which maps line segments into line > segments must be an affine map? Answer: No. First family of counterexamples: *Any* continuous map from R^n into R > has that property. Second family of counterexamples: If f is *any* continuous map from R > into itself, then the map from R into R^n defined by æ æ x |-> (f(x),0,0,...,0) has that property. Improved question: A map from R^n into R^m (with n,m > 1) which maps > line segments into line segments must be an affine map? > Jose Carlos Santos I think the answer to your improved question is yes provided you require that the range of the map has at least three points in general position. === Subject: Primes congruent to 1 or 3 modulo 4. posting-account=FeQf_goAAABjohysG-nJ6IFLUfjTwKeK Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) Littlewood proved that the proportion of primes not greater than n which are congruent to 3 modulo 4 flips above and below 0.5 infinitely often. Is it actually the case that the limit of this proportion is 0.5? I mean, if it has a limit then it must be 0.5, but has anyone proved that it actually has a limit? === Subject: Re: Primes congruent to 1 or 3 modulo 4. > Littlewood proved that the proportion of primes not greater than n > which are congruent to 3 modulo 4 flips above and below 0.5 infinitely > often. Is it actually the case that the limit of this proportion is > 0.5? I mean, if it has a limit then it must be 0.5, but has anyone > proved that it actually has a limit? The prime number theorem for primes in arithmetic progression says (in this case) that pi(3, 4) and pi(1, 4) are both asymptotic to (1/2)(x / log x). What you want follows from that, no? -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Primes congruent to 1 or 3 modulo 4. > Littlewood proved that the proportion of primes not greater than n > which are congruent to 3 modulo 4 flips above and below 0.5 infinitely > often. Is it actually the case that the limit of this proportion is > 0.5? I mean, if it has a limit then it must be 0.5, but has anyone > proved that it actually has a limit? Yes, this is a standard result in analytic number theory. In general, if gcd(a,n) = 1 then the proportion of primes satisfying p = a mod n tends to 1/phi(n), where phi(n) is the number of a in [1,n] coprime to n. It is proved in the same way as the Prime Number Theorem, using L-series (or L-functions) in place of the Riemann zeta function. I'm pretty sure you would find a proof in Edward's book on Riemann's Zeta Function. -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: -- reducibility in Q[x] I'll fix 2 minor typos ... >Let's consider the problem of determining reducibility for polynomials >of degree n Q[x]. The above line should be: of degree n in Q[x]. >We are interested in algebraic conditions on the >coefficients which are necessary and sufficient for reducibility. For example, if f(x) = a*x^2 + b*x + c, where a,b,c in Q and a is nonzero, then f is reducible iff b2 - 4ac is a square in Q. Is there an analogue for degree 3? I conjecture no. To make that precise, I will define a class of allowable logical >expressions and then conjecture that, for the set of polynomials in >Q[x] of a fixed degree n > 2, there is no boolean combination of >(finitely many) such logical expressions, which discriminates between >reducible and irreducible polynomials. Notation and terminology: Let n be a fixed positive integer, and let P_n = {f in Q[x] | deg(f) = n} > >Thus, each f in P_n has the form f(x) = a_n*x^n + ... + a1*x + a0 > >where a_0, ..., a_n are in Q and a_n is nonzero. > >By an algebraic discriminant for reducibility in P_n, we mean a >boolean expression D_n where the atomic expressions in D_n have the >form p(a_0,...,a_n) is in S_p > >where p in Q(t_1, ...,t_n) > > S_p is a subset of K The above line should be: S_p is a subset of Q >such that f in P_n is reducible in Q[x] iff D_n is true for f. > >Remark: Don't confuse the use of the word discriminant in > > an algebraic discriminant for reducibility in P_n > >with the usual use of the word discriminant as in the discriminant of a polynomial > >An algebraic discriminant for reducibility in P_n, if it exists, >evaluates, for each f in P_n, to a boolean value, either true or >false, and is true iff f is reducible in Q[x]. > >Conjecture: For n > 2, there does not exist an algebraic discriminant for >reducibility in P_n. Remarks: (1) Since degree 1 polynomials are always irreducible, we can take D_1 = F, > >where F is the constant False. (2) For degree 2, write f = a_2*x^2 + a1*x + a0 > >Then letting > > p = a1^2 - 4(a_2)(a_0) > > S = {q^2 | q in Q} > >we can take: D_2 as the statement p(a0,a1,a2) is in S (3) My conjecture claims that, for n > 2, no such discriminant is >possible, even allowing multiple such conditions and arbitrary boolean >combinations of them. quasi === Subject: Re: -- reducibility in Q[x] I'll fix 2 minor typos ... >Let's consider the problem of determining reducibility for polynomials >of degree n Q[x]. The above line should be: of degree n in Q[x]. >We are interested in algebraic conditions on the >coefficients which are necessary and sufficient for reducibility. >For example, if > f(x) = a*x^2 + b*x + c, where a,b,c in Q and a is nonzero, >then f is reducible iff b2 - 4ac is a square in Q. >Is there an analogue for degree 3? I conjecture no. >To make that precise, I will define a class of allowable logical >expressions and then conjecture that, for the set of polynomials in >Q[x] of a fixed degree n > 2, there is no boolean combination of >(finitely many) such logical expressions, which discriminates between >reducible and irreducible polynomials. >Notation and terminology: >Let n be a fixed positive integer, and let > P_n = {f in Q[x] | deg(f) = n} > >Thus, each f in P_n has the form > f(x) = a_n*x^n + ... + a1*x + a0 > >where a_0, ..., a_n are in Q and a_n is nonzero. > >By an algebraic discriminant for reducibility in P_n, we mean a >boolean expression D_n where the atomic expressions in D_n have the >form > p(a_0,...,a_n) is in S_p > >where > p in Q(t_1, ...,t_n) The above line should have been: p in Q(t_0,...,t_n) > S_p is a subset of K The above line should be: S_p is a subset of Q > >such that > f in P_n is reducible in Q[x] iff D_n is true for f. > >Remark: >Don't confuse the use of the word discriminant in > > an algebraic discriminant for reducibility in P_n > >with the usual use of the word discriminant as in > the discriminant of a polynomial > >An algebraic discriminant for reducibility in P_n, if it exists, >evaluates, for each f in P_n, to a boolean value, either true or >false, and is true iff f is reducible in Q[x]. > >Conjecture: >For n > 2, there does not exist an algebraic discriminant for >reducibility in P_n. >Remarks: >(1) Since degree 1 polynomials are always irreducible, we can take > D_1 = F, > >where F is the constant False. >(2) For degree 2, write > f = a_2*x^2 + a1*x + a0 > >Then letting > > p = a1^2 - 4(a_2)(a_0) > > S = {q^2 | q in Q} > >we can take: > D_2 as the statement p(a0,a1,a2) is in S >(3) My conjecture claims that, for n > 2, no such discriminant is >possible, even allowing multiple such conditions and arbitrary boolean >combinations of them. quasi === Subject: Re: Constant failure; The greatest equations ever; The Coming posting-account=LqooeAoAAACZi9rhWJi6xq-ef5ZN2cZd Web Browser from: http://bsalsa.com/),gzip(gfe),gzip(gfe) On Feb 13, 2:35æpm, zzbun...@netscape.net Read my comments at the end. :) > Feb 1, 2008 > Constant failure > Have we defined our fundamental constants with maximum efficiency? > Robert P Crease invites your comments > In Proposition 3 of On the Measurement of the Circle, Archimedes > asserts, based on calculations involving regular polygons > circumscribed around and inscribed in a circle, that the ratio of the > circumference of any circle to its diameter is less than 3 1/7 but > greater than 3 10/71. He thereby strongly reinforced, if he did not > actually create, the tradition of considering that ratio, two > millennia later referred to as [CapitalYAcute], to be fundamental. > Was Archimedes wrong? æ æHe was right, but in the wrong way. > æ æHe correctly defined the measurement as fundamental, > æ æbut he unfortunately for history, bogusly defined fundamental. > æ æSince it was only later discovered, that > æ æe, (2*pi), sin(pi), and = are just as fundamental as pi. > The greatest equations ever > Maxwell's equations of electromagnetism and the Euler equation top a > poll to find the greatest equations of all time. Robert P Crease > discusses the results of his reader survey > matter what the Large Hadron Collider finds, it is going to take > physics into new territory > By Chris Quigg > The Large Hadron Collider (LHC) is certain to find something new and > provocative as it presses into unexplored territory. > the Higgs boson, or a stand-in to play its role, at energies probed by > the LHC. The Higgs, in turn, poses deep questions of its own, whose > answers should be found in the same energy range. > These phenomena revolve around the question of symmetry. Symmetries > underlie the interactions of the Standard Model but are not always > reflected in the operation of the model. Understanding why not is a > key question.- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - Any 'constant' involving Pi is surely void. Adam Lewis === Subject: Re: Denial is a sad thing posting-account=qHNgPQoAAAA40SLlrmjQxSJhnyKS76HY Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) 2 magnets, each attached to the rim of a wheel. The wheels are positioned in a sandwich formation. The magnets attract another. We slowly spin the magnet 1 toward magnet 2. As soon as the magnet comes within reach wheel 2 spins in reverse direction then the magnets lock in place. When magnet 1 approaches magnet 2 a bit faster the wheel doesn't spin backwards before locking. If we spin magnet 1 even faster it jerks over magnet 2 only dragging it along for a little bit. However, If the primary magnet is spinning within the resonant frequency it may: - first attract the secondary magnet then - accelerate it in opposing direction AND - jerk passed it as a result of the increased interaction speed. If there is a sufficient lack of friction/vibration the speed can then be increased by tuning the air gap. There where there isn't a sufficient lack of friction the interaction may be used to speed up conventional electric motors. resonance-slow.gif (GIF Image, 420x300 pixels) resonance-to-fast.gif (GIF Image, 420x300 pixels) resonance-match.gif (GIF Image, 420x300 pixels) RESONANT MAGNETOMECHANICS ABSTRACT Unleash complimentary reaction forces in resonance there~with. http://magnetmotor.go-here.nl/resonance/text :) Build a real one. Film it. Post the video. :) :) Build a real one. Film it. Post the video. :) / http://www.youtube.com/watch?v=Jc9rbysrv24 http://www.youtube.com/watch?v=jYcjjSfiNNE http://www.youtube.com/watch?v=mCANbMBujjQ http://magnetmotor.go-here.nl/video === Subject: Re: Solution Manual for many classes including Structural Analysis by Hibbeler; 6th Edition posting-account=vjNRTQoAAAAxOsYgm89PN1LSybjl14pv 1.1.4322),gzip(gfe),gzip(gfe) > I have solution manual for: 1. Structural Analysis by Hibbeler: 6th Edition, > 2. Advanced Engineering Mathematics by Erwin Kreyszig9th (Even > Problems) > 3. Mechanics of Fluids by Potter; 3rd Edition > 4. Thermodynamics by Boles, Cengel 6Edition > 5. Mechanics Of Materials 6th Edition Beer Johnston > 6. Vector Mechanics for Engineers by David F. Mazurek, David Mazurek, > Elliot R. Eisenberg, Ferdinand Pierre Beer, Russell E. Johnston: 7th > Edition > 7. Water Resources by Larry W. Mays: 5th Edition > 8. Calculus by James Stewart; 5th Edition > 9. Structural Analysis Using Classical and Matrix Methods 3rd Edition > By James K. Nelson, Jr. and Jack C. McCormac > 10. Physics for Scientists and Engineers Randall D. Knight E-mail me at : chanon13(at)hotmail.com for further information. æLet's > me know which one do you want. Could you send me the solutions manual to Water Resources by Larry === Subject: convergence in probability of a multivariate rv posting-account=r5Mu_AoAAAA4Or1qD_TiZuru2F3PYCEL Gecko/20080201 Firefox/2.0.0.12 eMusic DLM/4.0_1.0.0.1,gzip(gfe),gzip(gfe) I am trying to prove that a sequence of random vectors X(n) converges in probability to a constant vector C <=> each element Xi(n) converges in probability to ci. I can easily show =>. But I am not too sure how to prove <=. I can write P{|Xi-ci| Which can only be true if you have the right f's and g's, but you > don't know those right? But why not guess? There are only p_1 - 1 > for the f's and p_2 - 1 of them for the g's, so a computer can loop > through possibles until it stumbles across the right one. This /is/ brilliant. Of course I choose p_1 = 2 and p_2 = 3? So there > are only /two/ possibilities. Now how hard can it be to crank those? Turns out you next need to solve for d_1 modulo p_1: d_1 = (f_1 - g_1)*p_2^{-1} mod p_1 and then you also need d_2, so d_2 = (f_1 - g_2)*p_2^{-1} mod p_1 and now you can substitute modulo p_1 into, oh, forgot something: (g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = (r_2 + k_2*p_2) multiply out, and divide by p_2, using m_2 = (g_1*g_2/p_2), and you can get to k_2 = d_1*g_2 + d_2*g_1 + d_1*d_2*p_2 + m_2 so now you just substitute modulo p_1 and compare what you get on the right side with k_2 mod p_1 as you know what k_2 is, since k_2 = floor(T/p_2). If you guessed correctly then you will match with k_2's residue modulo p_1, but if you didn't, you shouldn't match unless this doesn't work either!!! So if you don't match you try another possible and try until you match, and do that for a series of primes where the total is less than m, such that T/m! < 1, or you can get an exact number by multiplying primes together--assuming you start with the smallest and go up--until you exceed T, and counting how many you have. That is the solution to the factoring problem. It is a key in the lock technique. When you have the right residues of factors modulo each prime then d_1 = (f_1 - g_1)*p_2^{-1} mod p_1 and d_2 = (f_1 - g_2)*p_2^{-1} mod p_1 are the right keys in the lock so that you get k_2 = d_1*g_2 + d_2*g_1 + d_1*d_2*p_2 + m_2. For those wondering if maybe it's wrong, imagine you have the WRONG f's and g's, then the d's should be different, right? So when they're right... James Harris === === === Subject: Re: JSH: Problem solving techniques > mathematics is being done in pure math areas today. > Maybe in topology. I think topology is ok. > Really? What about algebraic topology, which makes up a large part of > the subject? The guys who do that have to know a fair bit about > algebra, especially homological algebra and category theory. That > means they have to be familiar with recent research in those areas, so > presumably they (being real mathematicians) would notice if the > algebraists were fakes who weren't doing real mathematics. Since > you're such an expert on the state of modern mathematical research I > expect you can give some explanation as to how the rest of the > mathematical community have managed to so effectively conceal their > con game from the topology community (unless you don't know the first > thing about topology but just decided to try and impress the > undergrads with a fancy new word you saw on Wikipedia, that is). > They haven't. > Remember, last year? > A certain topologist refused a certain prize? > he did pure math, the kind you *HATE* I don't hate pure math, but also I don't think it exists. Like the Arabs refuse to acknowledge Isreal exists. At least the dont *hate* them. My major results, except for the solution to the factoring problem, > would be considered pure math by the current widely accepted notion > of it. that class viewer thing? that is trivial software, no pure math involved at all. > What I don't like is people lying about math where I think they chose > areas where they thought math was useless so that they could get away > with it, as it'd be a group of experts word against someone claiming > error. are you still in error ? Or are these experts still right ? Like I found STUPID stuff in Wiles's work on FLT. I mean just plain > dumb. But you say over and over that you are not a mathematician, so how can you judge Wiles ? Are you jealous? > But I can explain it and explain it and explain it, but as long as > more people are out there claiming it's correct--and ignoring > explication of error--he can keep getting credit for something he > didn't do. So, who did it ? Con artists went to pure math like con artists used to go to > religion. So now you are slaming all religions ? Religion was similar in that people couldn't really figure out when > you were lying, so you say they're going to Heaven when they die? Where is God going to put you since you will still be working on your factoring problems ? > How can the prove you wrong? How can you prove them right or wrong ? Get it? How much does it cost? > With the diminishment of religion the cons needed a new venue, so some > of the parasites came to mathematics as their new nest. Oh, so Old Religious people are infesting math groups that should approve your factoring problem, but they wont because they are........... Mean ? suppressing your research dont want tp lose their highpaid jobs keeping the devil down > James Harris === Subject: Re: Numerical Methods for Engineers Solutions Manual - Chapra & Canale posting-account=iTGJkQoAAAARRvV-bwVNWZVlS7h3eXWD 2.0.50727; .NET CLR 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) Hey man, I need one solutions manual too. Let me know how to get hold > I have the solutions manual for the book Numerical Methods for > Engineers Solutions 5th edition, by æSteven C. Chapra and Raymond P. > Canale. If you would like to purchase it, please email me at > scothoward@gmail.com > === Subject: Re: solving for a matrix posting-account=PbSwTwgAAADCfhtjQF2Hq7jjSPSyO4QX Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > On Feb 13, 2:39 am, Robert Israel > I have this linear system > (A)(X)(B)=(C), A, X, B, C are square and nonsingular matrices. > Can I solve for (X). > I had my trials but i am not sure. > (A)(X)=(C)(B^{-1}) > Then I got stuck!! How to left multiply the (A^{-1}) by both sides!? > How!? Just do it! > Am I missing something!? > Perhaps you're missing the fact that A^(-1) A is the identity matrix. I understand that!! What I meant is, is it the same sequence of matrix > multiplication on the other side!? or is it inverted!? like B^{-1} C A^{-1}!? > Why would it be that? You already have A X = C B^(-1). Now you put an A^(-1) on the left of both sides. A^(-1) A X = A^(-1) C B^(-1) Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada sci.engr.mech, sci.astro.amateur === Subject: Re: Reverse Speech & The Apollo Moon Controversy posting-account=axlf9QgAAADH9qSnMXr4KE9F8O1Dk5OI CLR 1.1.4322),gzip(gfe),gzip(gfe) oh, man; got me to answer a thread by some B*z*, who goes by Israeldid911; congradulation! how many folks have noticed, the tiny snippets of reversed-speech that are offered as a psychometric, are really the same as a Rorshach Test?... on the wayside, is Apollo Eleven a reverse-speech palindrome? I mean, this was obvious when that Australian dude with the pronounced lisp gave his presentations on Fart Smell Ghost-to-Ghost BC, complete with ridiculous readings of OJ at trial. or, like the Bible Code, just set your computer to find what ever prediction *ex post facto* that you want to find, except for [we're going to get] Rabin Assassinated. well, I don't know if speech analysis is able to do that, these days. I don't blame anyone who was not alive in '69, for not believing in the moonlanding; I mean, it hardly matters, since LBJ and Nixon ... I mean, sir Henry of Kiss.Ass. and George Schulz, took the space program down, in favor of a God-am shuttle. I might add, only that dumb-ass Face on Mars guy got an Academy Award; no Apollo personnel! >http://www.reversespeech.com/moonhoax.htm thus: oh, you wanted me to retain, what I'd never seen, before. back at you: you are giving an *explanation* of the mathematical formalism of E et al in GR, not any equivalence with Newton's law, which was only an algebraization of Kepler's orbital constraints, that Newton stole from Hook, the first president of the Royal Society; it's been the secular religion of the British since that time -- for the next time! anyway, yours is similar to Descartes' notion, but you've got a long way to go, in actually forcing it to fit the data; firstly, you've got to actually find the data -- for the next time! > umpteenth time, GR's space-time was patterned after Newton's > gravitational inverse square law. The actual CAUSE of Mercury's > precession is its running into ether of a density that also matches > the inverse square law! And the ether is rotating as the Sun > rotates. So that is like a sort of whirlpool > of ether that will speed up Mercury enough > to account for the precession. thus: it is somewhat more recondite, to use the diameter = one; thus, circumference & area are just pi. then, you still have to explain, why the volume is pi/6.... anyway, the fact that the area of the great circle is a quarter of the sphere's, shows, it's (somehow) tetrahedral, a la Buckafka Fullofitarians. thus quoth: These gravitational redshift objects are evenly distributed throughout the universe. Quasar's are compact objects about one light week in diameter. The close ones' redshifts are mostly from their gravitation. --Dick Cheeny, National Treasure: Run, Trickier Dick -- Run for Indy superVeep!... Al Gore, Best Actor, Occidental Dino Awards! === Subject: Re: Subject: Reverse Speech & The Apollo Moon Controversy It was remailed automatically by anonymizing remailer software. Please report problems or inappropriate use to the remailer administrator at . > Did We Go to the Moon -----BEGIN PGP SIGNED MESSAGE----- If you mean did the made-for-tv apollo-men-to-the-moon astronaughts pretend to go all the way to the Moon and back not once but six times during the hippie-era, nearly *four decades* ago (practically covered- wagon times by today's standards), in a long-since exposed, primitive cold-war propaganda hoax, a thoroughly scrutinized hoax that only the die-hard Atheists still believe was actually men going all the way to the Moon and back when Tricky Dicky was President, then the answer is unequivocally Yes: Atheists took the blue pill and woke up in their beds believing whatever they wanted to believe, even though it was an illusion to the extent NASA depicted living men above low-earth orbit, you know, the limit of where their low-earth orbit shuttles fly today. Armageddon Cometh, Daniel Joseph Min http://www.angelfire.com/moon2/danieljosephmin/ -----BEGIN PGP SIGNATURE----- iQA/AwUBR7OfLZljD7YrHM/nEQJ/gwCgzlq+X7UwNWt98cH3KPFIImyOeX0AoKdu onLeCLwZx07J1pRxjcQnrK4+ =2Pni -----END PGP SIGNATURE----- === Subject: Solution Manual West Federal Taxation 2008 Comprehensive posting-account=8cWRQwoAAAAhaenOECdoYi5uBwUSHq5w Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) I have the solutions manual to West Federal Taxation 2008 Comprehensive Appendix E. This is seperate then the solution manual for the book as this is 81 pages in itself. If you are interested drop me an email === Subject: Re: elementary and piecewise functions <20080213114445.987$xd@newsreader.com> posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > Guessing that something of the form f(x) = a*x + b*|x| > will do, one can then easily solve for a and b. But I'm > not sure that absolute value is considered an elementary > function in the usual sense of that term... > You can replace all appearances of absolute value with > square root of the square. In fact, I used to do this > in class (mid to late 1990s, especially) when I wanted to > numerically integrate or graph a function that involved > an absolute value using a calculator. The abs function > was burried somewhere in a user directory or something, > and some students knew how to get to it easily, but I > would tell them that if they forgot or didn't want to > bother looking, they could do what I did, which was > to replace |stuff| with sqrt((stuff)^2). Right, of course. But when you say square root of the square, you're > actually talking about just the _principal_ square root, which is not, > technically, an elementary function (at least as I understand things). > OTOH, the bivalued square root is an algebraic (and hence, elementary) > function. David W. Cantrell In this context, introduced by the OP, it is a safe bet that `elementary' means `known to an introductory calculus student'... -- m === Subject: Re: elementary and piecewise functions <20080213114445.987$xd@newsreader.com> posting-account=-gYNIQoAAAD-MKGbMY_-Jpj1EqcOgRWx 3.0.2; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) > Guessing that something of the form f(x) = a*x + b*|x| > will do, one can then easily solve for a and b. But I'm > not sure that absolute value is considered an elementary > function in the usual sense of that term... > You can replace all appearances of absolute value with > square root of the square. In fact, I used to do this > in class (mid to late 1990s, especially) when I wanted to > numerically integrate or graph a function that involved > an absolute value using a calculator. The abs function > was burried somewhere in a user directory or something, > and some students knew how to get to it easily, but I > would tell them that if they forgot or didn't want to > bother looking, they could do what I did, which was > to replace |stuff| with sqrt((stuff)^2). Right, of course. But when you say square root of the square, you're > actually talking about just the principal square root, which is not, > technically, an elementary function (at least as I understand things). > OTOH, the bivalued square root is an algebraic (and hence, elementary) > function. I always get told off here when I mention multi-valued functions. Anyway, if you're simulating abs() then an if ... then ... else construct would seem to be more efficient and less cryptic. === Subject: Locksmith Los Angeles posting-account=5Hx47goAAADQG8G8UkplmcgrtL2O-Rcm SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) Are you in a jam? Need a locksmith now? 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Our locksmith technicians do: Lockouts, lost keys, jammed ignitions, remake car keys, re-keys, new locks, master rekeying, high security locks, new lock sales, lock fixes, designer locks, etc . What ever it is you need from your local locksmith were on top of it so give us a call now at 818-386-1022 visit us online http://www.locksmithsecurityservices.com Los Angeles Locksmith === Subject: Re: Solution Manual (MODERN CONTROL SYSTEM 4th Edition by OGATA ) posting-account=uznr_AoAAAB95a006Vu3GRB03qbWtrHB Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) I have solution manuals for a competitive price. I also have other manuals Feel free to contact me. 1. Modern Control Systems, 11ed, Dorf. 2. DSP Diniz solutions: Digital Signal Processing, Systems analysis and design (Proakis and Manolakis) 3. Control Systems Engineering by Nise 4e Solution Manual 4. Electric Machinery Fundamentals (Chapman), Fourth ed 5 Modern Control Engineering OGATA 4e 6. Electrical Engineering by A.R. Hambley 7. Fundamentals of Microelectronics Behzad Razavi 8. University Physics 12th by Young edition FULL If your interested do let me know at abby.cham(at)yahoo(dot)com best deals Abby Cham Group === Subject: Re: solution manual for digital communication by simon haykin posting-account=uznr_AoAAAB95a006Vu3GRB03qbWtrHB Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) I have solution manuals for a competitive price. I also have other manuals Feel free to contact me. 1. Modern Control Systems, 11ed, Dorf. 2. DSP Diniz solutions: Digital Signal Processing, Systems analysis and design (Proakis and Manolakis) 3. Control Systems Engineering by Nise 4e Solution Manual 4. Electric Machinery Fundamentals (Chapman), Fourth ed 5 Modern Control Engineering OGATA 4e 6. Electrical Engineering by A.R. Hambley 7. Fundamentals of Microelectronics Behzad Razavi 8. University Physics 12th by Young edition FULL If your interested do let me know at abby.cham(at)yahoo(dot)com Abby Cham Group sci.engr.mech, sci.astro.amateur === Subject: Re: The Santa Maria New World Controversy posting-account=Rqa4sAoAAAC88UYanCtJRUF4S6TUauGA Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) It has been a huge controversy over the past several years, even the topic of a TV special. Did we really go to the new world or not? Did Spain fake the North America landings? http://xkcd.com/202/ > Did We Go to the Moon or Not? It has been a huge controversy over the past several years, No it hasn't. Marshall sci.engr.mech, sci.astro.amateur === Subject: Re: The Santa Maria New World Controversy posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 4334.34; Windows NT 5.1; SV1; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) spider-mtc-tf05.proxy.aol.com[400C70A5] (Prism/1.2.1), HTTP/1.1 cache-mtc-ad05.proxy.aol.com[400C74C7] (Traffic-Server/6.1.5 [uScM]) > It has been a huge controversy over the past several years, even the > topic of a TV special. Did we really go to the new world or not? What you mean we, Kemosabe? > Did Spain fake the North America landings? MY ancestors were Vikings. http://xkcd.com/202/ > ?Did We Go to the Moon or Not? > It has been a huge controversy over the past several years, No it hasn't. Marshall === Subject: solutions for Engineering and Chemical Thermodynamics by Koretsky posting-account=WTJbowoAAAB7RjTQc4rFntl3pphjSfst SLCC1; .NET CLR 2.0.50727; .NET CLR 3.0.04506; Media Center PC 5.0),gzip(gfe),gzip(gfe) You got a solutinos manual for Engineering and Chemical Thermodynamics by Koretsky? === Subject: JSH: Simple matching, factoring versus math politics posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) Playing around with various approaches to the factoring problem I noticed that if I did something as simple as consider T = 9 mod 11 and T = 2 mod 13, I would find that the minimum positive number for T that would work would be 119, which is a number I like to use in my examples and it occurred to me that the primes were forcing something. So I started thinking about what information prime numbers could give about factors as if the two primes were forcing T to be 119 or greater, then they were also forcing the factors to be certain values. So I expanded out a factorization with primes: (f_1 + c_1*p_1)(f_2 + c_2*p_1) = T = (r_1 + k_1*p_1) and (g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = (r_2 + k_2*p_2) And I multiplied out and solved for k_1 and k_2 respectively and again pondered, what if you GUESSED using f_1*f_2 = T mod p_1 and g_1*g_2 = T mod p_2, so you'd know the f's and the g's? And I realized that then you could use (f_1 + c_1*p_1) = (g_1 + d_1*p_2) and (f_2 + c_2*p_1) = (g_2 + d_2*p_2) and reduce to a solution for d_1. d_1 = (f_1 - g_1)*p_2^{-1} mod p_1 and then you also need d_2, so d_2 = (f_2 - g_2)*p_2^{-1} mod p_1. So suddenly I had it!!! Information from the intersection of the primes. The two prime numbers were now telling me something about a key variable in the expanded factorization!!! And now you can go back to (g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = (r_2 + k_2*p_2) multiply out, and divide by p_2, using m_2 = (g_1*g_2/p_2), and you can get to k_2 = d_1*g_2 + d_2*g_1 + d_1*d_2*p_2 + m_2 so now you just substitute modulo p_1 and compare what you get on the right side with k_2 mod p_1 as you know what k_2 is, since k_2 = floor(T/p_2). Now then, does this work? Well, is there an answer for d_1 mod p_1 and d_2 mod p_2? Of course, yes. If I have the correct residues for the f's and g's then will that answer be given by d_1 = (f_1 - g_1)*p_2^{-1} mod p_1 and d_2 = (f_2 - g_2)*p_2^{-1} mod p_1? The answer is, of course, yes. Now perturb them. Shift f_1 or g_1 and will you change d_1 and d_2? YES!!! So then, this method will work to tell you when you have them correctly so the answer to the key question of factoring is available: what is f_1 mod p_1 and f_2 mod p_2? As you answer that question prime by prime you can factor the target, as a minimum positive value for f_1 is forced, like T = 9 mod 11 and T = 2 mod 13 forced a minimum positive value for T. Now that is easy math. Trivial algebra. Hating mathematics is about hating truths that you don't like, and I know that feeling so I can understand why so many of you despise mathematics, even if you claim to be mathematicians. You despise it because you do not control it. That's why your society came up with delicate proofs and logical contradictions as you found a human solution to an inhuman discipline. The math does not care. What's true is true even if you can't make a living in the field telling the truth. So most of you learned to lie as otherwise, you'd have to make your living some other way, and if society let you, why not live the fantasy? Why not just pretend? Why not act? Actors are heroes in this society right? Like Melanie Griffith. Or Brad Pitt. Or Morgan Freeman. Or Claire Forlani. Actors make big bucks, and get accolades so why not just act with mathematics too? So if they could, why not you? James Harris === Subject: Re: Simple matching, factoring versus math politics > Playing around with various approaches to the factoring problem I > noticed that if I did something as simple as consider T = 9 mod 11 and > T = 2 mod 13, I would find that the minimum positive number for T that > would work would be 119, which is a number I like to use in my > examples and it occurred to me that the primes were forcing something. > Picka T ? > So I started thinking about what information prime numbers could give > about factors as if the two primes were forcing T to be 119 or > greater, then they were also forcing the factors to be certain values. So I expanded out a factorization with primes: (f_1 + c_1*p_1)(f_2 + c_2*p_1) = T = (r_1 + k_1*p_1) So; Picka f_1 Picka c_1 Picka p_1 Picka f_2 Picka c_2 Picka p_2 Picka T Picka r_1 Picka k_1 > and (g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = (r_2 + k_2*p_2) So; Picka g_1 Picka d_1 Picka g_2 Picka d_2 Picka r_2 > And I multiplied out and solved for k_1 and k_2 respectively and again > pondered, what if you GUESSED using f_1*f_2 = T mod p_1 and g_1*g_2 = > T > mod p_2, so you'd know the f's and the g's? And I realized that then you could use (f_1 + c_1*p_1) = (g_1 + d_1*p_2) and (f_2 + c_2*p_1) = (g_2 + d_2*p_2) and reduce to a solution for d_1. d_1 = (f_1 - g_1)*p_2^{-1} mod p_1 and then you also need d_2, so d_2 = (f_2 - g_2)*p_2^{-1} mod p_1. So suddenly I had it!!! Information from the intersection of the > primes. You fulla sa, you already picka too many Picka f_1 Picka c_1 Picka p_1 Picka f_2 Picka c_2 Picka p_2 Picka T Picka r_1 Picka k_1 Picka g_1 Picka d_1 Picka g_2 Picka d_2 Picka r_2 The two prime numbers were now telling me something about a key > variable in the expanded factorization!!! And now you can go back to (g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = (r_2 + k_2*p_2) multiply out, and divide by p_2, using m_2 = (g_1*g_2/p_2), and you > can get to k_2 = d_1*g_2 + d_2*g_1 + d_1*d_2*p_2 + m_2 so now you just substitute modulo p_1 and compare what you get on the > right side with k_2 mod p_1 as you know what k_2 is, since k_2 = floor(T/p_2). Now then, does this work? Nopea, you on dopea, floor( ) silly function for haz-ben hackers Nopea, no Math'a here. Well, is there an answer for d_1 mod p_1 and d_2 mod p_2? Of course, yes. If I have the correct residues for the f's and g's > then will that answer be given by d_1 = (f_1 - g_1)*p_2^{-1} mod p_1 and d_2 = (f_2 - g_2)*p_2^{-1} mod p_1? The answer is, of course, yes. Now perturb them. Shift f_1 or g_1 and will you change d_1 and d_2? YES!!! Picka-CHOO !!! Sneeze all them picka's outcha nose, blueberry. Youa shoulda add 20 more variables to youra equationsa, adds drama, confusion, smoka, mirrorsa, BSa, and you have diversified your portfolio. James Harris (whiny white troll/crackpot) === Subject: Re: JSH: Simple matching, factoring versus math politics posting-account=FK_G5goAAAAEbWGaiao8aY3ti47UN9eX 1.0.3705; .NET CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) What the hell is this? Some sort of futile attemp to communicate information using letters and numbers? Jesus Fskcing Christ on a pogo stick! === Subject: Re: Simple matching, factoring versus math politics Hating mathematics is about hating truths that you don't like, and I > know that feeling so I can understand why so many of you despise > mathematics, even if you claim to be mathematicians. Why do you hate mathematics ?? So most of you learned to lie as otherwise, you'd have to make your > living some other way, and if society let you, why not live the > fantasy? Why are you living in a fantasy ? Why not just pretend? Why not act? Actors are heroes in this society > right? Like Melanie Griffith. Or Brad Pitt. Or Morgan Freeman. Or Claire > Forlani. Actors make big bucks, and get accolades so why not just act with > mathematics too? So if they could, why not you? HINT: Ask them to help you with your factoring problems, I am sure they will help you. > James Harris === Subject: Re: Simple matching, factoring versus math politics > Actors make big bucks, and get accolades so why not just act with > mathematics too? > So if they could, why not you? >HINT: Ask them to help you with your factoring problems, I am sure they > will help you. I'd rather say this guys needs professional help from a psychologist. And you someone who stops you from feeding trolls. === Subject: Re: Simple matching, factoring versus math politics > Actors make big bucks, and get accolades so why not just act with > mathematics too? > So if they could, why not you? > HINT: Ask them to help you with your factoring problems, I am sure they > will help you. > I'd rather say this guys needs professional help from a psychologist. > And you someone who stops you from feeding trolls. Is this what the entire World has been forced into => trolls feeding troll feeders ? All caused by JSH's Monkey-Math, as his very weak mind now feeds his malignant ego(s). === Subject: Re: JSH: Simple matching, factoring versus math politics > Playing around with various approaches to the factoring problem I > noticed that if I did something as simple as consider T = 9 mod 11 and > T = 2 mod 13, I would find that the minimum positive number for T that > would work would be 119, How did you do that? Trial and error, or guessing, etc ? > which is a number I like to use in my know that feeling so I can understand why so many of you despise > mathematics, even if you claim to be mathematicians. boy is that one off, Most all of us love math, and understand it. > You despise it because you do not control it. You are projecting again, Translation:: I despise it because I do not control it. > That's why your society came up with delicate proofs and logical > contradictions as you found a human solution to an inhuman > discipline. You missed out on formal Math classes and training, that is why you struggle so much with this simple topic. > The math does not care. What's true is true even if you can't make a > living in the field telling the truth. You choose to remain in a blaming mode, to prop-up your sick ego, and miss out on true Math wisdom, remaining in the class of unknowing boobs, that people get disgusted with. > So most of you learned to lie as otherwise, you'd have to make your > living some other way, and if society let you, why not live the > fantasy? Projecting again........ Why do you lie? Why do you live in this fantasy? > Why not just pretend? Why not act? Actors are heroes in this society > right? Like Melanie Griffith. Or Brad Pitt. Or Morgan Freeman. Or Claire > Forlani. Perhaps they can help you with your factoring problem. Like the Gecko commertials, get that guy that makes the funny sounds to sit next to you. Add a sound file to your blog. Perhaps that will help you with your problems. Actors make big bucks, and get accolades so why not just act with > mathematics too? How much money have you made so far ? Why is that? Just ask for money to support your research in these newsgroups, I am sure people will send you money. So if they could, why not you? If you haven't made any money, it is because you do not bring value to the situation. > James Harris === Subject: Re: JSH: Simple matching, factoring versus math politics posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > Playing around with various approaches to the factoring problem I > noticed that if I did something as simple as consider T = 9 mod 11 and > T = 2 mod 13, I would find that the minimum positive number for T that > would work would be 119, which is a number I like to use in my > examples and it occurred to me that the primes were forcing something. So I started thinking about what information prime numbers could give > about factors as if the two primes were forcing T to be 119 or > greater, then they were also forcing the factors to be certain values. So I expanded out a factorization with primes: (f_1 + c_1*p_1)(f_2 + c_2*p_1) = T = (r_1 + k_1*p_1) and (g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = (r_2 + k_2*p_2) And I multiplied out and solved for k_1 and k_2 respectively and again > pondered, what if you GUESSED using f_1*f_2 = T mod p_1 and g_1*g_2 = > T > mod p_2, so you'd know the f's and the g's? And I realized that then you could use (f_1 + c_1*p_1) = (g_1 + d_1*p_2) and (f_2 + c_2*p_1) = (g_2 + d_2*p_2) and reduce to a solution for d_1. d_1 = (f_1 - g_1)*p_2^{-1} mod p_1 (1) and then you also need d_2, so d_2 = (f_2 - g_2)*p_2^{-1} mod p_1. (2) Note that these equations only give you the residues of the d's mod p1, rather than exact values for the d's. So suddenly I had it!!! Information from the intersection of the > primes. The two prime numbers were now telling me something about a key > variable in the expanded factorization!!! And now you can go back to (g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = (r_2 + k_2*p_2) multiply out, and divide by p_2, using m_2 = (g_1*g_2/p_2), I assume you mean to write m2 = floor(g1*g2/p2), otherwise the following equation is wrong (since r2 has disappeared): > and you can get to k_2 = d_1*g_2 + d_2*g_1 + d_1*d_2*p_2 + m_2 (3) so now you just substitute modulo p_1 and compare what you get on the > right side with k_2 mod p_1 as you know what k_2 is, since k_2 = floor(T/p_2). Now then, does this work? No. And here's why: as I mentioned above, you only know the values of d1 mod p1 and d2 mod p1. You suggest substituting mod p1 into the equation I have labelled (3). The trouble is that even if you have incorrectly guessed f1 and g1 this equation will always hold mod p1. This can be seen by directly substituting your formulae (1) and (2) k2 = f1*f2*p2^(-1) - g1*g2*p2^(-1) + m2 mod p1 Now use the facts that f1*f2 = T mod p1 and g1*g2 = m2*p2 + r2 and this becomes k2 = T*p2^(-1) - r2*p2^(-1) mod p1, i.e. T*p2^(-1) = (k2*p2 + r2)*p2^(-1) = T*p2^(-1) mod p1. Another tautology. That doesn't help you at all. Well, is there an answer for d_1 mod p_1 and d_2 mod p_2? Of course, yes. If I have the correct residues for the f's and g's > then will that answer be given by d_1 = (f_1 - g_1)*p_2^{-1} mod p_1 and d_2 = (f_2 - g_2)*p_2^{-1} mod p_1? The answer is, of course, yes. Right. But if you have the wrong residues for the f's and g's the answer is still, of course, yes. Now perturb them. Shift f_1 or g_1 and will you change d_1 and d_2? YES!!! Right. But will they shift in such a way as to make equation (3) still hold mod p1? YES!!! > Now that is easy math. Trivial algebra. Indeed. You really ought to try testing your methods on a composite whose factorisation you don't already know before you claim to have solved the factoring problem. Had you done so with this method, and indeed all the methods you have come up with this year, you would have found out for yourself why it doesn't work. > Hating mathematics is about hating truths that you don't like, and I > know that feeling We noticed. > so I can understand why so many of you despise > mathematics, even if you claim to be mathematicians. You're projecting again. Most of the people here don't share your problems with mathematics because we enjoy finding out the truth for its own sake, rather than as a means to win against our imagined enemies. Grow up, James. === Subject: Re: JSH: Simple matching, factoring versus math politics posting-account=Ae7cPwoAAAA1p9Bl1szxMYHINqjRfucA CLR 2.0.50727),gzip(gfe),gzip(gfe) Trivial algebra. Hating mathematics is about giving me the power to make a living in the complex plane. So without talk of factors. Also I can stop posters like you from being in the ring of algebraic integers, by the distributive property. The key result I rely on is valid over the place with some nice looking girls with big coconuts -- hey I was NOT to involve myself with his white women. He was ok with race mixing and wanted me to just decide that you knew more about this reality than you do. And if howling at reality were enough to care about merit raises. The middle class works around them, the bad guys, but too often, like with that advisor, the middle class here just shrugs, every last one of them, no exceptions and lets them get away with their replies in that thread. Your people are deliberately avoiding the truth. That's what you can get from this world. But reality always wins, and your need to understand it this time. You people have done has mattered. And I want to see what happens. I already know as I've been there myself. Logic and mathematics are brutal disciplines. But they reached, as always, about me or just wait for the future of humanity. And then everything will be no people, and mostly no life on this point, as a simple mathematical result. They can't answer with an example and will you change d_1 and d_2? YES!!! So then, this method will work to convince the world. So you can use the Chinese Remainder theorem to get f_1 and f_2 mod p such that T/m! < 1. Then you start with what I know which is a free speech area. Learn it someday, or just wait for the summer, and this time I was taking classes in C programming taught by an IBM researcher on loan from the IBM Research Triangle in Durham, North Carolina. Well, this nice roommate of mine noticed me hanging out with replies where posters dodge that basic result in the ring of algebraic integers, proving that ring is incomplete. Luckily the complex plane? Given, in the complex field the distributive property still holds as a(b+c) = ab + ac so in this case the ac = 7, but now you could have multiplied through (A(x) + 7) and (B(x) + 1) where A(0) = B(0) = 0 it must be incomplete. But by the distributive property offers no other possibility, and again, note that the relationship is true for all x, where A(0) = B(0) = 0, A(x) it's not possible for 7 to have multiplied through (A(x) + 7)(B(x) + 1) true for all x, and whether or not you get two possibles because negative residues canwork, but they won't accept what is true. So not a lot more time trying to justify why public funds should be done with the incessant whining! Your species never shuts up. For those of us in the complex plane 7C(x) = (A(x) + 7)(B(x) + 1) where A(0) = B(0) = 0 where the 7 didn't multiply through the first factor. With a result that does hold in the math field, from shutting down entire departments at even major universities like Princeton or Yale, to ending funding for large swaths of the heap. Your stupidity is in there somewhere. And now you could have p_2 = 11, so you'd have: 1, 9 2, 10 3, 3 and so forth. Like now you know, they are integers. Consider my favorite example T = 2 mod p_2 and so forth, so you can convince other people mixed right in there with them who think that if you are a mathematician, then the world or quite simply you may be more suited to a specialized journal. So let me get clear on this, if sending someone a banana conveys a defamatory message about a third party, would publishing the banana and your success can be so hard when it does not care. The other kid could have p_2 = 11, so you'd have to find it, and it does solve the problems will not give up control easily. Like any parasites, the world just keeps on turning, taking us along with it, for example. This latter scenario strikes me as being roughly as likely as your work getting published there. Both the Xerox of the distribution hassles that would entail. We are sorry to let you have enough answers and they don't shift. So now you know, they are doing anything valuable then you have it in the ring of algebraic integers must be true that isn't behave as I've been puzzling why no one figured it out before now. What you are finding out with replies where posters dodge that basic result in the ring of algebraic integers, proving that your field is corrupted. Nothing else. Everything from here understand, that you have enough primes that f_1 equals your factor. j is less than 200 primes. If you believe in that tells you that 7 multiplied through one factor with 7C(x) = (A(x) + 7) and (B(x) + 1) where A(0) = B(0) = 0, A(x) it's not like crying about it changes it. The world just keeps on turning, taking us along with p_1 = 11, so you'd look for 119 mod 143, which gives you a list of 142 numbers, again starting with destroying tenure. Which I'm going to end anyway. I want to roll the dice then you get negative solutions. You now also get f_1 and f_2. === Subject: Re: JSH: Simple matching, factoring versus math politics posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; InfoPath.1),gzip(gfe),gzip(gfe) The two prime numbers were now telling me something James Harris Maybe to stop mathurbating? You are getting even more delusional. It isn't even April Fool's Day yet! Narcissist! === Subject: Re: multivariate function. multivariate continuous implies elementwise continuous > On Feb 12, 8:33 pm, The World Wide Wade I would need some help proving the following result. > If a real-valued function f of vector X is continuous at vector A > then > f(a1,...,ai-1,xi,....,ak) is a continuous function of the variable xi > at ai for each i=1...k > I greatly appreciate your help and pointers. > So what have you tried? >multivariate calculus. > I was thinking > By hypothesis > for any delta, there exists eps, for ||X-A||Now set x1=a1,...,xi-1=ai-1,xi=xi,....,xk=ak >||X-A||=|xi-ai| > ||f(X)-f(A)||=|f(a1,...,ai-1,xi,....,ak)-f(a1,...,ai-1,ai,....,ak)| >So that > for |xi-ai| |f(a1,...,ai-1,xi,....,ak)-f(a1,...,ai-1,ai,....,ak)|showing elementwise continuity of f. >Am I on the right path? > That's the basic idea, although I think you have the tradional roles of epsilon and delta reversed. === Subject: Re: JSH: Factoring problem solution, update posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 SV1),gzip(gfe),gzip(gfe) > corrected at *****. > I stumbled across a remarkably simple solution to the factoring > problem, which exists because of what is called the floor() function. > That function just means to drop any fractions or decimals, so like > floor(3.1415) = 3. > It is crucial to the solution to the factoring problem. > Here is the full solution. > It suffices to determine variables to fulfill the factorizations: > (f 1 + c 1*p 1)(f 2 + c 2*p 1) = T = r 1 + k 1*p 1 > (g 1 + d 1*p 2)(g 2 + d 2*p 2) = T = r 2 + k 2*p 2 > where T is the target to be factored and p 1 and p 2 are primes to be > picked, as this method works because you're using primes in this way > which is why you need so many variables. > Note that f 1, f 2, g 1 and g 2 are residues where > f 1*f 2 = T mod p 1 and g 1*g 2 = T mod p 2. > So the r's and k's are easily calculated and the only remaining > variables are c 1, c 2, d 1 and d 2, and you guess at values for the > f's and g's. æSo guessing is crucial in this method. > And it can be shown that solving for the factors reduces to finding > integer solutions to a family of 4 equations and 4 unknowns, which are > k 1 = c 1*f 2 + c 2*f 1 + c 1*c 2*p 1 + m 1 > k 2 = d 1*g 2 + d 2*g 1 + d 1*d 2*p 2 + m 2 > (f 1 + c 1*p 1) = (g 1 + d 1*p 2) > (f 2 + c 2*p 1) = (g 2 + d 2*p 2) > where m 1 = floor(f 1*f 2/p 1) and m 2 = floor(g 1*g 2/p 2). > Well I started out correctly but from there went down the wrong path. > But the intuition that I was following was that the intersection of > primes gives information about factoring, which was such a strong gut > feeling that when I realized I'd screwed up, I went back over to see > where I might have gone wrong, and, well, found the trivial solution. > So I backtracked a bit and then found the correct path. > Turns out you next need to solve for d 1 modulo p 1: > d 1 = (f 1 - g 1)*p 2^{-1} mod p 1 > and then you also need d 2, so > d 2 = (f 1 - g 2)*p 2^{-1} mod p 1 > and now you can substitute modulo p 1 into, oh, forgot something: > (g 1 + d 1*p 2)(g 2 + d 2*p 2) = T = (r 2 + k 2*p 2) > multiply out, and divide by p 2, using m 2 = (g 1*g 2/p 2), and you > can get to > k 2 = d 1*g 2 + d 2*g 1 + d 1*d 2*p 2 + m 2 > so now you just substitute modulo p 1 and compare what you get on the > right side with k 2 mod p 1 as you know what k 2 is, since > k 2 = floor(T/p 2). > If you guessed correctly then you will match with k 2's residue modulo > p 1, but if you didn't, you shouldn't match unless this doesn't work > either!!! > So if you don't match you try another possible and try until you > match, and do that for a series of primes where the total is less than > m, such that T/m! < 1, or you can get an exact number by multiplying > primes together--assuming you start with the smallest and go up--until > you exceed T, and counting how many you have. > That is the solution to the factoring problem. > It is a key in the lock technique. æWhen you have the right residues > of factors modulo each prime then > d 1 = (f 1 - g 1)*p 2^{-1} mod p 1 > and > d 2 = (f 1 - g 2)*p 2^{-1} mod p 1 > are the right keys in the lock so that you get > k 2 = d 1*g 2 + d 2*g 1 + d 1*d 2*p 2 + m 2. > For those wondering if maybe it's wrong, imagine you have the WRONG > f's and g's, then the d's should be different, right? > So when they're right... > Easy solution but a lot of inertia I'm sure because of so much > invested in thinking factoring is hard. > Of course, if too much time is taken then, well, I've harped enough on > the potential negative consequences. > I just have to hope that there are adults around. > James Harris æ It is tempting to think that you can solve the > factorization problem by first solving it in modular > arithmetic for a collection of relatively small primes, > and then somehow (maybe using the Chinese Remainder > Theorem) combine those results together to > find a non-modular factor. æThe outline for such > a method might be as follows. æAssume the number to be > factored is T. æChoose a prime p1. æFind f1 and f2, > both less than p1, such that [1] æ æ æf1 * f2 = T mod p1. æ Choose another prime p2. æFind g1 and g2, both less > than p2, such that [2] æ æ æg1 * g2 = T mod p2. æ Now, in fact both of these equations have lots of > solutions - in fact, p1 - 1 solutions for [1] and > p2 - 1 solutions for [2]. æ You can write equations [1] and [2] as æ æ æ æ æf1 = a mod p1 æ æand æ æ æ æ æg1 = b mod p2, > *****> where a = 1/f2 mod p1 and b = 1/g2 mod p2. Should have been: where a = T/f2 mod p1 and b = T/g2 mod p2, > Both of these inverses are guaranteed to > exist because the integers mod p are a field, > when p is prime. æ The Chinese Remainder Theorem implies that > there is an integer x such that æ æ æ æ æx = a mod p1 æ and æ æ æ æ æx = b mod p2, and in fact, if y is any other solution > to these two equations, then æ æ æ æ æx = y mod (p1 * p2). > *****> æ It turns out that in general, h1 is 'h1' should have been 'x' > considerably larger than f1 and g1. æEven > though f1 and g1 were selected to be < p1 *****> and p2 resp., all you can say about h1 here also 'h1' should have been 'x' > is that it is less than p1 * p2. æIf > you add a third prime, the value of the > corresponding x will be less than > p1 * p2 * p3, but you cannot say much > more than that in general. æ Here is an example of how this might > work. æSay T = 11 * 13 = 143. æChoose > p1 = 2, p2 = 3, p3 = 5. æ T mod 2 = 1 æ T mod 3 = 2 æ T mod 5 = 3. æ So I can choose: æ æ (f1, f2) = (1, 1) æ æ (g1, g2) = (1, 2) æ and æ æ (h1, h2) = (1, 3). æ I can say: æ æ æ f1 = 1/f2 mod 2 = 1 mod 2 æ æ æ g1 = 1/g2 mod 3 = 1 mod 3 æ and æ æ æ h1 = 1/h2 mod 5 = 1 mod 5. æ The CRT says there exists x such > that æ æ x mod 2 = 1 æ æ x mod 3 = 1 æ æ x mod 5 = 1 æ The smallest such x, it turns out, > is æ æ x = 31. æ But 31 is NOT a factor of T. æFurther, > if you add another prime p4, it will > only make things worse: the smallest x > which will satisfy all 4 equations MUST > be no smaller than 31. æSo this is not > going to lead to the correct factorization. æ So what's wrong? æThere IS a correct > factorization. æYou should be able to > arrive at it by this route. æ The problem is in the initial choices > of f1, g1, and h1. æIf you make different > choices, you WILL end up with a factor > of T. æFor example: æ æ f1 = 1, f2 = 1 æ æ g1 = 2, g2 = 1 æ æ h1 = 3, h2 = 1 æ You can check that x = 13 satisfies: æ æ x = f1 mod 2 = 1 mod 2 æ æ x = g1 mod 3 = 2 mod 3 æ æ x = h1 mod 5 = 3 mod 5, Further, x = 13 is the smallest such > number satisfying all three equations, and > of course 13 is a factor of T = 143. æ The CRT will give the right answer, but > ONLY if you start with the right choices for > f1, g1, and h1. æFurther, there are æ æ(p1 - 1)*(p2 - 1)*(p3 - 1) possible choices if you start with 3 > primes. æIf, as James Harris suggests, > you want to factor a very large number, > you might have to consider all the > primes less than 1000. æThere are 168 > such primes. æThis means trying an > extremely large collection of factors f1, g1, h1, ..., z1, where z may correspond to, e.g., 168. æ So while the CRT might seem on the > surface to provide a good approach to > finding factors - and it will work - æin > fact it turns out to be extremely inefficient. æ Marcus.- Hide quoted text - - Show quoted text - === Subject: Re: JSH: Factoring problem solution, update > I stumbled across a remarkably simple solution to the factoring > problem, which exists because of what is called the floor() function. > That function just means to drop any fractions or decimals, so like > floor(3.1415) = 3. > It is crucial to the solution to the factoring problem. > Here is the full solution. > It suffices to determine variables to fulfill the factorizations: and > Snip I'd like that to be a small number, > and I don't see anything that restricts what p_1 and p_2 I > should pick, so let's choose p_1 = 2, p_2 = 3 and see what > happens, OK? > match, and do that for a series of primes where the total is less than > m, such that T/m! < 1, or you can get an exact number by multiplying > primes together--assuming you start with the smallest and go up--until > you exceed T, and counting how many you have. That is the solution to the factoring problem. it is not a solution if you are guessing a bunch of numbers It's a key in the lock technique. put you under lock and key. The keys are the proper values for the f's and g's. so Picka k_2, Picka f, Picka g, Picka k, Pickachoo > The lock is k_2 mod p_1. Wow. I solved the factoring problem. Who'd a thunk it could really > happen? Only You, *dumbass*. > James Harris === Subject: need help on solving boolean linear eqn posting-account=0VJq7goAAAC-2RnkHZ_MGYA23eh312IU Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) I have a system of linear boolean equations as follows An = ~(Qn^In-1)*(In^Qn-1) + (In^In-1)*(Qn^Qn-1) Bn = ~(In^Qn-1)*(Qn^In-1) + (In^In-1)*(Qn^Qn-1) ~ is invert operation, ^ is Xor, + is Or, and * is and I want to solve for Qn and In in terms of An,Bn,Qn-q, In-1. I am not sure what to start since it is boolean arithmetic. It's difficult to manipulate the equation like for regular arithmetic. I was replainc Xor with * and + operation,i.e. (A^B) = ~A+~B but still got bogged. I'd appreciate any help or hint. Sam === Subject: Re: need help on solving boolean linear eqn posting-account=-gYNIQoAAAD-MKGbMY_-Jpj1EqcOgRWx 3.0.2; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) > I have a system of linear boolean equations as follows An = ~(Qn^In-1)*(In^Qn-1) + (In^In-1)*(Qn^Qn-1) > Bn = ~(In^Qn-1)*(Qn^In-1) + (In^In-1)*(Qn^Qn-1) ~ is invert operation, ^ is Xor, + is Or, and * is and I want to solve for Qn and In in terms of An,Bn,Qn-q, In-1. > I am not sure what to start since it is boolean arithmetic. It's > difficult to manipulate the equation like for regular arithmetic. > I was replainc Xor with * and + operation,i.e. æ(A^B) = æ~A+~B but > still got bogged. I'd appreciate any help or hint. Sam Are you expecting there to always be a unique solution? I'm not sure if that's the case... === Subject: Re: need help on solving boolean linear eqn posting-account=0VJq7goAAAC-2RnkHZ_MGYA23eh312IU Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > I have a system of linear boolean equations as follows > An = ~(Qn^In-1)*(In^Qn-1) + (In^In-1)*(Qn^Qn-1) > Bn = ~(In^Qn-1)*(Qn^In-1) + (In^In-1)*(Qn^Qn-1) > ~ is invert operation, ^ is Xor, + is Or, and * is and > I want to solve for Qn and In in terms of An,Bn,Qn-q, In-1. > I am not sure what to start since it is boolean arithmetic. It's > difficult to manipulate the equation like for regular arithmetic. > I was replainc Xor with * and + operation,i.e. (A^B) = ~A+~B but > still got bogged. > I'd appreciate any help or hint. > Sam Are you expecting there to always be a unique solution? I'm not sure > if that's the case... It has a unique solution. Actually, I know the solution is In = ((~(An^Bn))*(An^In-1)) + ((An^Bn)*(An^Qn-1)) ----(1) Qn = ((~(An^Bn))*(Bn^Qn-1)) + ((An^Bn)*(Bn^In-1)) ----(2) However, I do not know how to solve for it. An = ~(Qn^In-1)*(In^Qn-1) + (In^In-1)*(Qn^Qn-1) --- (3) Bn = ~(In^Qn-1)*(Qn^In-1) + (In^In-1)*(Qn^Qn-1) ----(4) I will appreciate if anybody can help explain to me how to solve for (1) & (2) given (3) and (4) or vice versa. Sam === Subject: Re: need help on solving boolean linear eqn posting-account=-gYNIQoAAAD-MKGbMY_-Jpj1EqcOgRWx 3.0.2; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) > I have a system of linear boolean equations as follows > An = ~(Qn^In-1)*(In^Qn-1) + (In^In-1)*(Qn^Qn-1) > Bn = ~(In^Qn-1)*(Qn^In-1) + (In^In-1)*(Qn^Qn-1) > ~ is invert operation, ^ is Xor, + is Or, and * is and > I want to solve for Qn and In in terms of An,Bn,Qn-q, In-1. > I am not sure what to start since it is boolean arithmetic. It's > difficult to manipulate the equation like for regular arithmetic. > I was replainc Xor with * and + operation,i.e. æ(A^B) = æ~A+~B but > still got bogged. > I'd appreciate any help or hint. > Sam > Are you expecting there to always be a unique solution? I'm not sure > if that's the case... It has a unique solution. Apologies, I think I made a mistake. === Subject: Re: Help to calculate an integral > topic as follows: >int_0^2pi cos(m x)/sqrt(a+b cos(x)) dx, >in which a and b are real, positive, and satisfy a>b, m=1,2,3,... > For simplicity take b=1 (we can always scale it). We can write the integrand as a^(-1/2) cos(m x) sum_{k=0}^infty (2k)! (-1/4)^k cos(x)^k (b/a)^k/(k!)^2 Now int_0^{2 pi} cos(m x) cos(x)^k dx = (k choose (k-m)/2) 2^(1-k) pi for 0 <= m <= k with k == m mod 2, 0 otherwise so the integral becomes a^(-1/2) sum_{j=0}^infty (2m+4j)! (-1/4)^(m+2j) (b/a)^(m+2j) (m+2j choose j) 2^(1-m-2j) pi/((m+2j)!)^2 which according to Maple is 2(2m)!(-1)^m pi hypergeom([m/2+3/4,m/2+1/4],[m+1],1/a^2) (b/a)^m / (a^(1/2)(m!)^2 8^m) = 2 sqrt(2/b) (-1)^m LegendreQ(m-1/2, a/b) LegendreQ(m-1/2,a/b) is a Legendre function of the second kind. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Help to calculate an integral > topic as follows: > int_0^2pi cos(m x)/sqrt(a+b cos(x)) dx, > in which a and b are real, positive, and satisfy a>b, m=1,2,3,... >For simplicity take b=1 (we can always scale it). > We can write the integrand as a^(-1/2) cos(m x) sum_{k=0}^infty (2k)! (-1/4)^k cos(x)^k (b/a)^k/(k!)^2 Now int_0^{2 pi} cos(m x) cos(x)^k dx = (k choose (k-m)/2) 2^(1-k) pi > for 0 <= m <= k with k == m mod 2, 0 otherwise so the integral becomes a^(-1/2) sum_{j=0}^infty (2m+4j)! (-1/4)^(m+2j) (b/a)^(m+2j) (m+2j choose > j) 2^(1-m-2j) pi/((m+2j)!)^2 which according to Maple is 2(2m)!(-1)^m pi hypergeom([m/2+3/4,m/2+1/4],[m+1],1/a^2) (b/a)^m > / (a^(1/2)(m!)^2 8^m) > = 2 sqrt(2/b) (-1)^m LegendreQ(m-1/2, a/b) LegendreQ(m-1/2,a/b) is a Legendre function of the second kind. I'm guessing that perhaps Maple and Mathematica use different definitions of LegendreQ because, in Mathematica, 2 Sqrt[2/b] (-1)^m LegendreQ[m - 1/2, a/b] has a nonzero imaginary part and so can't be correct. But the _real_ part of the above, in Mathematica, does appear to give the correct value for the integral. I should think that an alternative form for the general answer could be given in terms of complete elliptic integrals of the first and second kinds, with modulus 2b/(a + b). David === Subject: Re: Help to calculate an integral <20080214022121.561$Zr@newsreader.com> posting-account=PbSwTwgAAADCfhtjQF2Hq7jjSPSyO4QX Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > topic as follows: > int_0^2pi cos(m x)/sqrt(a+b cos(x)) dx, > in which a and b are real, positive, and satisfy a>b, m=1,2,3,... > For simplicity take b=1 (we can always scale it). > We can write the integrand as > a^(-1/2) cos(m x) sum_{k=0}^infty (2k)! (-1/4)^k cos(x)^k (b/a)^k/(k!)^2 > Now > int_0^{2 pi} cos(m x) cos(x)^k dx = (k choose (k-m)/2) 2^(1-k) pi > for 0 <= m <= k with k == m mod 2, 0 otherwise > so the integral becomes > a^(-1/2) sum_{j=0}^infty (2m+4j)! (-1/4)^(m+2j) (b/a)^(m+2j) (m+2j choose > j) 2^(1-m-2j) pi/((m+2j)!)^2 > which according to Maple is > 2(2m)!(-1)^m pi hypergeom([m/2+3/4,m/2+1/4],[m+1],1/a^2) (b/a)^m > / (a^(1/2)(m!)^2 8^m) > = 2 sqrt(2/b) (-1)^m LegendreQ(m-1/2, a/b) > LegendreQ(m-1/2,a/b) is a Legendre function of the second kind. I'm guessing that perhaps Maple and Mathematica use different definitions > of LegendreQ because, in Mathematica, 2 Sqrt[2/b] (-1)^m LegendreQ[m - 1/2, a/b] has a nonzero imaginary part and so can't be correct. But the _real_ part > of the above, in Mathematica, does appear to give the correct value for the > integral. Maple's definition of LegendreQ(m,z) is It also has the integral form LegendreQ(v, z) = (1/2)*(Int(t^((1/2)*v-1/2)*(1-t)^((1/2)*v)/((z^2-t)/ z^2)^(1+(1/2)*v), t = 0 .. 1))/z^(v+1) for Re(v) > -1. Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Help to calculate an integral > I'm guessing that perhaps Maple and Mathematica use different definitions > of LegendreQ because, in Mathematica, 2 Sqrt[2/b] (-1)^m LegendreQ[m - 1/2, a/b] has a nonzero imaginary part and so can't be correct. But the _real_ part > of the above, in Mathematica, does appear to give the correct value for the > integral. Or take 2 Sqrt[2/b] (-1)^m LegendreQ[m - 1/2, 0, 3, a/b] === Subject: Re: Help to calculate an integral posting-account=K5WE3woAAAAXArsybjkbN6LjMxWdHtbX Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > topic as follows: int_0^2pi cos(m x)/sqrt(a+b cos(x)) dx, in which a and b are real, positive, and satisfy a>b, m=1,2,3,... > Symmetry implies that your integral is twice that from 0 to Pi. As others have pointed out, the integral from 0 to Pi is of the form int[T[m](y)/sqrt(a+y)/sqrt(1-y^2),y=-1..1) (where the notation a/b/c means a/(b*c)) and T[m] is the mth Chebychev polynomial. Maple 9.5 has trouble, but Maple 11 can get int(y^n/sqrt(a+y)/sqrt(1-y^2),y=-1..1): Jn:=Int(cos(x)^n/sqrt(a+cos(x)),x=0..Pi): with(student): Kn:=changevar(cos(x)=y,Jn,y): value(Kn) assuming n::posint, a>1; <--- here comes the integral: 1/4*Pi^(1/2)*(4*hypergeom([1/4, 3/4, 1/2*n+1/2],[1/2, 1+1/2*n],1/ 5/4, 1+1/2*n],[3/2, 3/2+1/2*n],1/ Now you can write T[m](x) = sum{c[n]*x^n,n=0..m} and use the result above. I guess the remaining issue is: can we trust the Maple 11 answer? R.G. Vickson === Subject: Re: Help to calculate an integral > topic as follows: > int_0^2pi cos(m x)/sqrt(a+b cos(x)) dx, > in which a and b are real, positive, and satisfy a>b, m=1,2,3,... > > By taking out the factor 1/sqrt(b) and writing a instead of a/b, you > can assume b = 1 and a > 1. The results are not pretty: they involve > Elliptic functions. Here is what I get for m = 1, 2 and 3 using Maple > 9.5: >f:=cos(m*x)/sqrt(a+cos(x)): > J0:=int(f,x=0..2*Pi) assuming a > 1, m::posint: <--- explicit results > not obtained > with(student): <--- to change variables > J1:=subs(m=1,J0): <---- still do not get an explicit result, so try > change of variables > J1a:=changevar(cos(x)=y,J1,y); > /a + 1 1/2 / / 2 1/2 > J1a := 4 |------| |-EllipticE(|- ------| ) > -1 + a/ -1 + a/ > / 2 1/2 / 2 1/2 > + a EllipticE(|- ------| ) - a EllipticK(|- ------| )| > -1 + a/ -1 + a/ / > / 1/2 > / (a + 1) > / >J2:=subs(m=2,J0): <--- for m = 2 > J2a:=changevar(cos(x)=y,J2,y); > /a + 1 1/2 / / 2 1/2 > J2a := -4/3 |------| |-4 a EllipticE(|- ------| ) > -1 + a/ -1 + a/ > / 2 1/2 2 / 2 1/2 > + EllipticK(|- ------| ) + 4 a EllipticE(|- ------| ) > -1 + a/ -1 + a/ > 2 / 2 1/2 / 1/2 > - 4 a EllipticK(|- ------| )| / (a + 1) > -1 + a/ / / >J3:=subs(m=3,J0): <--- for m = 3 > J3a:=changevar(cos(x)=y,J3,y); > /a + 1 1/2 / 3 2 > J3a := 4/15 |------| |32 a %1 - 9 a %1 - 32 a %1 > -1 + a/ > 3 / 2 1/2 > - 32 a EllipticK(|- ------| ) > -1 + a/ > / 2 1/2 / 1/2 > + 17 a EllipticK(|- ------| ) + 9 %1| / (a + 1) > -1 + a/ / / > / 2 1/2 > %1 := EllipticE(|- ------| ) > -1 + a/ >We can keep going like this. However, we cannot seem to get an > expression for general m. >Good luck. >R.G. Vickson > note well: int(cos(mx)/sqrt(c + cos(x)),x) = int (T[m](t)/sqrt((c+t)(1-t^2)),t) [t = cos(x)] the integral of a Tchebychev poly divided by sqrt of a cubic poly, or it is a combo of elliptic integrals first and second kinds and elementary functions on m. Perhaps this can show how to compute it for a specific m === Subject: Re: Help to calculate an integral > topic as follows: > int_0^2pi cos(m x)/sqrt(a+b cos(x)) dx, > in which a and b are real, positive, and satisfy a>b, m=1,2,3,... >By taking out the factor 1/sqrt(b) and writing a instead of a/b, you > can assume b = 1 and a > 1. The results are not pretty: they involve > Elliptic functions. Here is what I get for m = 1, 2 and 3 using Maple > 9.5: f:=cos(m*x)/sqrt(a+cos(x)): > J0:=int(f,x=0..2*Pi) assuming a > 1, m::posint: <--- explicit results > not obtained > with(student): <--- to change variables > J1:=subs(m=1,J0): <---- still do not get an explicit result, so try > change of variables > J1a:=changevar(cos(x)=y,J1,y); /a + 1 1/2 / / 2 1/2 > J1a := 4 |------| |-EllipticE(|- ------| ) > -1 + a/ -1 + a/ / 2 1/2 / 2 1/2 > + a EllipticE(|- ------| ) - a EllipticK(|- ------| )| > -1 + a/ -1 + a/ / / 1/2 > / (a + 1) > / J2:=subs(m=2,J0): <--- for m = 2 > J2a:=changevar(cos(x)=y,J2,y); /a + 1 1/2 / / 2 1/2 > J2a := -4/3 |------| |-4 a EllipticE(|- ------| ) > -1 + a/ -1 + a/ / 2 1/2 2 / 2 1/2 > + EllipticK(|- ------| ) + 4 a EllipticE(|- ------| ) > -1 + a/ -1 + a/ 2 / 2 1/2 / 1/2 > - 4 a EllipticK(|- ------| )| / (a + 1) > -1 + a/ / / J3:=subs(m=3,J0): <--- for m = 3 > J3a:=changevar(cos(x)=y,J3,y); /a + 1 1/2 / 3 2 > J3a := 4/15 |------| |32 a %1 - 9 a %1 - 32 a %1 > -1 + a/ 3 / 2 1/2 > - 32 a EllipticK(|- ------| ) > -1 + a/ / 2 1/2 / 1/2 > + 17 a EllipticK(|- ------| ) + 9 %1| / (a + 1) > -1 + a/ / / / 2 1/2 > %1 := EllipticE(|- ------| ) > -1 + a/ We can keep going like this. However, we cannot seem to get an > expression for general m. The current version of Mathematica gives a beautifully simple result for general positive integer m: In[1]:= Assuming[a > 1 && Element[m, Integers] && m > 0, Integrate[Cos[m*x]/Sqrt[a + Cos[x]], {x, 0, 2*Pi}]] Out[1]= 0 But of course, that result is completely wrong! David === Subject: The Problems of TeX posting-account=qPxGtQkAAADb6PWdLGiWVucht1ZDR6fn AppleWebKit/523.12.2 (KHTML, like Gecko) Version/3.0.4 Safari/523.12.2,gzip(gfe),gzip(gfe) TeX, and its derivatives such as LaTeX, receives a lot praise and is widely used. However, i think there is one aspect of TeX that is a damaging flaw, and this criticism is little known. I feel this problem should have wider awareness. The following essay, is edited version of my email messages, on the harm of TeX. (HTML version is available at: http://xahlee.org/cmaci/notation/TeX pestilence.html ) ------------------------------------- Problems of TeX TeX is detrimental because it harbors ignorance of the structural content embodied in most math notations in most fields. What TeX does is typesetting, as opposed to math expression encoding. In other words, what TeX does is pretty-printing. The language is designed in a way that any structural info in a math expression are botched. As such, TeX, even though it is a full-fledged computer language capable of great programing, but it understands zilch of math expressions, it encodes zilch math expressions. Now that is a egregious error of a computer language purporting to express mathematics. And more so because it is a product of a mathematician who should've known better. LaTeX mended TeX by turning a pretty printing system into a structured documentation system. However, since TeX at its core is a pretty picture system, no amount fix can correct that other than complete discard. LaTeX fixed nothing about TeX's botching of the structural info in math notations. As a testament of TeX's shortcomings, TeX botches structural info in math expressions so bad that no program whatsoever short of Human- Level Artificial Intelligence will be be able to convert from TeX into another system of math notation. As a pretty-printing system, Mathematica is no match for TeX. As a knowledge representation system that is filled with the need of math notations old and new, Mathematica is superior to TeX/LaTeX by far in just about any aspect of this endeavor. TeX's Damage to Society TeX has done significant damage to the math community, by getting people to focus and love pretty printing, which is what typesetting is mainly about. And the key to see this is that TeX has absolutely no concept of the semantic content or structure of math expressions. As to alternative systems, as i've argued, there is Mathematica, and subsequently MathML, and i'm sure others based on semantic content and automatic formatting that i'm not aware of, and or killed by TeX before they possibly could have a chance. ´ TeX is a system for typesetting. Typesetting is primarily concerned with esthetics. As a art form, typesetting is insignificant. Typesetting, taken by itself other than facilitate reading, is in general of little serious utility. ´ TeX as a knowledge representation system used by scientists, seduced vast number of scientists into the rather wasteful activity of appearance doodling. ´ TeX, because it is a pure typesetting system and is not aware any structural info embedded in math notations, it destroys this information whenever math is written in TeX. Even notational systems using plain, one-dimensional ascii text such as in Mathematica before version 3 (~1997), or other computer algebra language, maintains structural info. (by necessity, because they have to actually work in some sense as a math formalism system) ´ Because TeX is free, it hordes progress of competing systems and ideas. If TeX had not been invented and given out free, systems that preserve structure (e.g. MathML, Mathematica) for the purposes of displaying 2-dimensional math notations, would have been invented earlier or wide spread. Free software acts as a virus. Free systems have the potency to wipe out any other protocol or design, including any superior ones (unless they are also free). A example is the various Unix systems and protocols has done huge irreversible damage to society. TeX, is given out free. ´ As a computer language, the design of TeX's syntax and semantics is arguably bad. The syntax is rather inconsistent, complex, imperative style (as opposed to functional programing), and in general the language is difficult to use to control the typesetting or documentation structure. It spread bad ideas or experiences to non- programers (e.g. mathematicians). Any computer language, however bad is its quality, will attract users once exposed to it, and may become addictive, and its users will feel empowered. It's like jigsaw puzzle in comparison to Rubic Cube. In the context of puzzles, the Jigsaw is a complete waste of time as a game with respect to mathematical values, it however ferment people into fans. A example of this is in computing, is the language Perl. The above ideas, i'm sure is not new, however, it is hardly ever heard. All one hear every day everywhere are chants of the greatness TeX by ignorant computing geeks and mathematicians, the former know coding, the latter know narrow specialization, both of which in general are ignorant of the history, psychology, and linguistic theories of knowledge presentation and symbolic system. This must be stopped. Downfall of TeX I do think TeX's popularity is waning because of its inappropriateness. They being: ´ Not based on Graphical User Interface. ´ Not interactive. (its usage requires ñcompileî cycles) ´ Inconsistent syntax. ´ Bad language semantics. (it is difficult to use TeX as a embeded language in typeset programing or document structure programing. (compare: Microsoft Word with VisualBasic, Emacs with elisp)) ´ A system based on appearance, not a math notation structure. ´ As of today, structural and semantic info are receiving greater and greater awareness as opposed to formatting or displaying aspects. Witness today's thoughts of Semantic Web, and a slew of technologies oriented with maintaining structure such as SGML/HTML/XML/MathMLÁ.9d, DOM, RSS, Mathematica. Even CSS, is structural based that removed formatting issues from HTML, and is today moving to XLS for XML. In part, TeX isn't to blame for its own fall, because it is set out to do typesetting and it does that well. The thing is, the entire typesetting business itself is largely a waste of humanity's time. As communication tech progresses, typesetting as it is understood (i.e. concerns about em/en-space, ligatures, typefaces) is going to be extinct. See: The Moronicities of Typography. The Structure in Math Notations Math notation does not always have well-defined structure or meaning. However, it is important to consider the preservation of semantic in building a math notation representation language. For example, if we have x^2+Sqrt[x], the system should know that it is a operation applied to two things, each of which is some particular operation on symbols x. Similarly for the traditional notation of matrix, subscripts, sets, absolutes/norms, summation, derivative, integral, ... etc. For any new math notation that does not have a clear structure (say some bunch of arrows towards symbols in some homological algebra), we can for example have a tag that markup the part of expression to indicate that it has no meaning and is for displaying 2D-typeset purposes only. In TeX, every math expression is just a sequence of structurally meaningless but micro-position-aware symbols. TeX and Microsoft Equation Editor Considered Equal TeX, being a pretty-printing system, can be considered in the same class as Microsoft Word Equation EditorÁ.9d. The difference lies only in their mode of operation. Specifically, TeX is by compile and batch operation like a typical computer language, and the Equation Editor is by using a mouse to click menus and buttons with graphical user interface. The heart of both as far as math notation is concerned, is doodling of a em space or en dash. All math notation's semantic structure are lost. LaTeX vs Microsoft Word for Structured Document On a different note, consider LaTeX's as a structured document tool, Microsoft Outline does better, plus it has background spell checking, tabs setting, version differentiation, voice annotations, and so on. (provided the person know how to properly use MS Word) TeX's Place in the World of Typesetting In order to evaluate TeX asides from its massive brainwashing of the Math community, the question we have to ask is to what degree TeX has made a impact in strictly the typesetting and publishing community. I agree TeX as a typesetting system made a major impact these communities, however, as a technology, it is far from taking a leading role. Consider QuarkXpressÁ.9d, FrameMakerÁ.9d, PDFÁ.9d, all are in similar market and are not free. (PDF makers made by Adobe is not free) .82So you know more about this than all of us professional mathematicians who have been working with it all our lives?é Successful and professional writers will in general not know much about linguistics or writing systems. Similarly of musicians and their notations, history, or alternative design. Likewise, professional mathematicians, although they have used math notation all their lives, few have studied or thought about the history of math notation, writing systems, philosophy of math, cognizance sciences, psychology of perception, linguistics of computer language, all are related to math notation systems. .82In the opinion of just about everyone who knows about such things, TeX is the best (by far) markup language where precise and beautiful typography is essential AND for typesetting mathematical formulas and equations. The fact is that ALL professional mathematicians (and I do mean all) learn TeX as graduate students and write their thesis in TeX and from then on are so hooked on it that they not only write their mathematics in TeX but also usually their letters and other documents. Something written in MS Word just looks ugly by comparison, particularly if it contains formulas using the brain-dead MS equation editor. BTW, just about all physicists use TeX for the same reason.é I'm very well acquainted with the history TeX and its position in society, how scientists receives it, and how ubiquitous and its position as a standard. I've read extensively about TeX a decade ago. I'm very well aware how it compares to something like MS Word or other related WYSIWYG equation editors. on the whole, my thesis is that TeX, although a extremely successful and well done tool that has satisfied a niche, but it may in fact be a disservice in humanity in that it massively mislead people in the wrong direction. If TeX did not exist, then typesetting will remain in the realm in pretty-printers community (i.e. typesetters and printers), while mathematicians and scientists, will not have wearied their energy into typesetting, but instead put their focus and energy in coming up with a syntax that makes mathematics readable as well as meaningful, based on immense modern knowledges of symbolic logic, computer algebra, and computing linguistics. Consider typesetting for a moment. What is it? It is no more than pretty-printing. It has some element of facilitating reading, but only a bit. The bulk of it is meticulousness is about appearances. Things like en-dash, em-dash, ligatures, small-caps, typeface design, serif, sans-serif, micro positions, kerning ... etc. Typesetting is a cultural development. Even if we suppose that the esthetics in Western typesetting is universal, its esthetics values in the context of artistic endeavors is dismal. (For example, contrast it with calligraphy, painting, sculpture etc.) By introducing TeX the way it is designed, it encroaches the symbolic language of mathematics with pretty-printing, and devalued the system of symbolic communication used by mathematicians, and subtly derailed what mathematicians do best. If TeX as it is designed has not been invented, then today we might already have a alternative system such as the proprietary Mathematica, or MathML (very much influenced by Wolfram Research Inc.), which has taken consideration that the symbolic language of mathematics is more than just pretty-printing, in that it has a not-well-understood but nevertheless complex and deep structure and meaning in relation to mathematics, in such a degree it can influence where mathematics is going, and design with these thoughts in mind helps us actually advance computational mathematics. Xah xah@xahlee.org á http://xahlee.org/ ? === Subject: Re: The Problems of TeX > [a Xah Lee post] > Folks! It's Xah Lee. This guy is best dealt with when ignored. Troll and Kook are the operating terms here. Please let him rant and move on. /W === Subject: Re: The Problems of TeX >TeX, and its derivatives such as LaTeX, receives a lot praise and is >widely used. However, i think there is one aspect of TeX that is a >damaging flaw, and this criticism is little known. I feel this problem >should have wider awareness. The following essay, is edited version of my email messages, on the >harm of TeX. (HTML version is available at: > http://xahlee.org/cmaci/notation/TeX_pestilence.html >) >------------------------------------- Problems of TeX TeX is detrimental because it harbors ignorance of the structural >content embodied in most math notations in most fields. What TeX does >is typesetting, as opposed to math expression encoding. In other >words, what TeX does is pretty-printing. The language is designed in a way that any structural info in a math >expression are botched. As such, TeX, even though it is a full-fledged >computer language capable of great programing, but it understands >zilch of math expressions, it encodes zilch math expressions. Now that >is a egregious error of a computer language purporting to express >mathematics. If TeX _did_ purport to express mathematics (in the sense you mean here) you'd have a valid point. But it doesn't - its intended purpose is typesetting, and it does that very well. > And more so because it is a product of a mathematician >who should've known better. LaTeX mended TeX by turning a pretty printing system into a structured >documentation system. However, since TeX at its core is a pretty >picture system, no amount fix can correct that other than complete >discard. LaTeX fixed nothing about TeX's botching of the structural >info in math notations. As a testament of TeX's shortcomings, TeX botches structural info in >math expressions so bad that no program whatsoever short of Human- >Level Artificial Intelligence will be be able to convert from TeX into >another system of math notation. As a pretty-printing system, Mathematica is no match for TeX. As a >knowledge representation system that is filled with the need of math >notations old and new, Mathematica is superior to TeX/LaTeX by far in >just about any aspect of this endeavor. True. And as a method of transportation, my car is far superior to TeX. Saying this is a flaw in TeX would be silly, unless there were people saying that TeX _is_ a method of transportation. >TeX's Damage to Society TeX has done significant damage to the math community, by getting >people to focus and love pretty printing, which is what typesetting is >mainly about. And the key to see this is that TeX has absolutely no >concept of the semantic content or structure of math expressions. As >to alternative systems, as i've argued, there is Mathematica, and >subsequently MathML, and i'm sure others based on semantic content and >automatic formatting that i'm not aware of, and or killed by TeX >before they possibly could have a chance. [Eth] TeX is a system for typesetting. Typesetting is primarily concerned >with esthetics. As a art form, typesetting is insignificant. >Typesetting, taken by itself other than facilitate reading, is in >general of little serious utility. It's of little utility unless you want to typeset something. David C. Ullrich === Subject: Re: The Problems of TeX > > TeX, and its derivatives such as LaTeX, receives a lot praise and is > widely used. However, i think there is one aspect of TeX that is a > damaging flaw, and this criticism is little known. I feel this problem > should have wider awareness. > The following essay, is edited version of my email messages, on the > harm of TeX. > (HTML version is available at: > http://xahlee.org/cmaci/notation/TeX_pestilence.html > ) > ------------------------------------- > Problems of TeX > TeX is detrimental because it harbors ignorance of the structural > content embodied in most math notations in most fields. What TeX does > is typesetting, as opposed to math expression encoding. In other > words, what TeX does is pretty-printing. > The language is designed in a way that any structural info in a math > expression are botched. As such, TeX, even though it is a full-fledged > computer language capable of great programing, but it understands > zilch of math expressions, it encodes zilch math expressions. Now that > is a egregious error of a computer language purporting to express > mathematics. >If TeX _did_ purport to express mathematics (in the sense you mean > here) you'd have a valid point. But it doesn't - its intended purpose > is typesetting, and it does that very well. > > And more so because it is a product of a mathematician > who should've known better. > LaTeX mended TeX by turning a pretty printing system into a structured > documentation system. However, since TeX at its core is a pretty > picture system, no amount fix can correct that other than complete > discard. LaTeX fixed nothing about TeX's botching of the structural > info in math notations. > As a testament of TeX's shortcomings, TeX botches structural info in > math expressions so bad that no program whatsoever short of Human- > Level Artificial Intelligence will be be able to convert from TeX into > another system of math notation. > As a pretty-printing system, Mathematica is no match for TeX. As a > knowledge representation system that is filled with the need of math > notations old and new, Mathematica is superior to TeX/LaTeX by far in > just about any aspect of this endeavor. >True. And as a method of transportation, my car is far superior to > TeX. Saying this is a flaw in TeX would be silly, unless there were > people saying that TeX _is_ a method of transportation. > > TeX's Damage to Society > TeX has done significant damage to the math community, by getting > people to focus and love pretty printing, which is what typesetting is > mainly about. And the key to see this is that TeX has absolutely no > concept of the semantic content or structure of math expressions. As > to alternative systems, as i've argued, there is Mathematica, and > subsequently MathML, and i'm sure others based on semantic content and > automatic formatting that i'm not aware of, and or killed by TeX > before they possibly could have a chance. > ´ TeX is a system for typesetting. Typesetting is primarily concerned > with esthetics. As a art form, typesetting is insignificant. > Typesetting, taken by itself other than facilitate reading, is in > general of little serious utility. >It's of little utility unless you want to typeset something. I also find it to be of great utility when I want to read something. === Subject: Re: The Problems of TeX posting-account=BOb97goAAADQ9S76yCy85K7EenLx1xAN Gecko/20061201 Firefox/2.0.0.12 (Ubuntu-feisty),gzip(gfe),gzip(gfe) > If TeX _did_ purport to express mathematics (in the sense you mean > here) you'd have a valid point. But it doesn't - its intended purpose > is typesetting, and it does that very well. Yes, TeX is just a typesetting method, but it is a typesetting method that is HARMFUL TO SOCIETY. Ohmigod! === Subject: Re: The Problems of TeX > If TeX _did_ purport to express mathematics (in the sense you mean > here) you'd have a valid point. But it doesn't - its intended purpose > is typesetting, and it does that very well. Yes, TeX is just a typesetting method, but it is a typesetting method >that is HARMFUL TO SOCIETY. Ohmigod! you miss the vital point. tex is discouraging small companies like microsoft from developing mathematical typesetting. this is really troublesome, since we all know how easy it is to deflect microsoft from any of its chosen paths. you _know_ it makes sense... -- Robin Fairbairns, Cambridge === Subject: Re: The Problems of TeX Jabber-ID: bronger@jabber.org Hall.9achen! First of all, I know that Xah Lee is a notorious crossposter who publishes qustionable or wrong statements. But some of it is not wrong, and although annoying, he doesn't fit into my definition of a troll. Anyway ... > [...] TeX is detrimental because it harbors ignorance of the structural > content embodied in most math notations in most fields. What TeX > does is typesetting, as opposed to math expression encoding. In > other words, what TeX does is pretty-printing. What you fail to tell is why it is important to have semantic math in a document intended for printing. > [...] [...] And the key to see this is that TeX has absolutely no > concept of the semantic content or structure of math > expressions. As to alternative systems, as i've argued, there is > Mathematica, and subsequently MathML, You mention MathML a lot here. Note, however, that it has both Presentation and Contents markup. The presentation markup is only very slightly better than TeX/LaTeX (as far as semantic contents is concerned) but it is *much* more widely used. Besides, no human being wants to generate contents markup; presentation markup is awkward enough already. > [...] The above ideas, i'm sure is not new, however, it is hardly ever > heard. All one hear every day everywhere are chants of the > greatness TeX [...] Although I think that you choose too strong wording here, I'm also a little bit irritated that the weaknesses of TeX as well as of LaTeX are often excluded from the discussion. For example, the Wikipedia entries contain no Flaws or Criticism section. Be that as it may, it is only a minor point. > [...] LaTeX vs Microsoft Word for Structured Document On a different note, consider LaTeX's as a structured document > tool, Microsoft Outline does better, plus it has background spell > checking, tabs setting, version differentiation, voice > annotations, and so on. (provided the person know how to properly > use MS Word) What is MS Outline? Apparently you compare apples and oranges here. Anyway, the semantic features in Word are surely weaker than those in LaTeX. I don't understand why you concentrate so much on *formulae*. I don't think that it is vital to have semantic markup there, and if really necessary, it is rather easy to reconstruct most of the semantics with good heuristics. Tsch.9a, Torsten. -- Torsten Bronger, aquisgrana, europa vetus Jabber ID: bronger@jabber.org (See http://ime.webhop.org for further contact info.) === Subject: Re: The Problems of TeX posting-account=fQF27QoAAAD3N2euiDOul-WPEbSeOTkv AppleWebKit/522+ (KHTML, like Gecko, Safari/522) OmniWeb/v613.0.93354,gzip(gfe),gzip(gfe) Hi Xah, You make some good points, and some bad ones. I don't have time to reply at the length that this essay requires, so a short reply will have to do. First of all, though: > * TeX is a system for typesetting. Typesetting is primarily concerned > with esthetics. As a art form, typesetting is insignificant. > Typesetting, taken by itself other than facilitate reading, is in > general of little serious utility. I object very seriously to this point to the degree of finding it offensive. I think you miss the point of typesetting completely. Secondly, you speak about TeX as the source of your problems. This is misguided, because TeX itself is purely designed to be a typesetting engine. Only the formats on top of it (Plain, LaTeX, ConTeXt, etc.) provide any semantic details. LaTeX made strides towards a separation of content and formatting in its syntax, but maths remained rather ad hoc. Your argument is that maths in LaTeX should contain meaning. Let me first say that there's a package that attempts to do this: It's quite new, and deserves more attention, but it essentially solves your problem. You might also be interested in a package I in maths to enable a convenient syntax for predefined vectors, matrices, functions, and other things that contain elements that should be typeset consistently. A related problem arises that often the *display* of mathematics needs to be tweaked for optimal output in ways that could never be automated. So even with a fully semantic mathematics syntax, you'd still need to pollute it with visual markup. So looking towards the future, how should this be addressed in the future? Well, the breqn package is being worked on again and can now probably be used for production purposes. This package greatly improves the markup used for many of the presentation maths forms. It has a logical separation between equations and multi-line constructions, which amsmath was rather bad at enforcing. The package cool, I think, should be incorporated into a broader maths package that provides the logical constructions that you believe should be present for typesetting maths. This would allow the beginnings of a syntax that could be used to translate maths expressions from symbolic maths packages like Mathematica to TeX and back again. The question is, are you willing to join us in creating such a package? Will === Subject: Re: The Problems of TeX posting-account=G-TjQAkAAADYg6rno3bWQPnIwKFBrf1t CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) > You make some good points, and some bad ones. I don't have time to > reply at the length that this essay requires, so a short reply will > have to do. That doesn't make sense. The message you're replying to will be here for decades. You don't need time to reply to USENET. This is not the same protocol as conversations or telephones where turnaround needs to be immediate. It's an indefinite-delay medium. I routinely reply to messages that go as far back as 20 years. Time, in Cyberspace, does not exist. It is irrelevant. === Subject: Re: The Problems of TeX lpldMjj6TOyFAAACM0lEQVR4nGWUUY7cIAyGPYpyANDy3pjwTvHMAZpwgIRd3/8q/Q1Md6Wi EZP4w/ZvEyBVFpH4Ul5EMBfG2OtGzFFKAWQpMcGofudab9IEC7NwrPjbObmPg/fDEXzN47nX U1JjA5X3zZEyl8j8rFVkjRWgVq43PLhugfWrHhKXeqbbwOYCCR+cIKOeBerodgA2iFvRto+X MynAeCFtqrHOccDu/AQgT3vq0x8A9wbJy4zEanb3NUFs8hzra+rgY+Yo+llPW2/1b+8kFFrk gPpY1SQFTAYOalIktPpqPRabrGAqCG0t915mJdwJHjY6RYq+9lEeVk+9jkpZDnSqHhpURx4A tL2kT3TzTE4vohWkA3YksesPSsN+dYD9sMruaTevCy/PA6DbDaDnlHvjg7ADOB9wRgjsRgbO ik0yIKvrIi9aRC3W6kI2QL1WXQNscECqW/NmwOwBbnqt+HXN9Ab+ergeBakWkpdFIUvRXiMU qWWSnGGjxwA2EB+iTd0/MGpeLxnjtwGrjpF3XcdqfJBiQnvuhPAZVcvSgUUaYOO8UO8JpvMb NDs7qwmCx0+Aj+HMELX0JAAzR2DPMdMYyw/gcV5anmbzoAna1tjHWbXMs2CAN04a31VD7QQ4 TpxSzHO5gUcHHqISji42b8nmRTJAS561TV3Dbe2AIcrjHBe6fs1wLwPBcodmF8U7O24OAN9F mf34To8PlRpEbalfLPIN1C4A6NX/gFJMvjVLgXPQG2IAG/oXQjHCyIHx6H4AAAAASUVORK5C YII= [0wZXhS]|q0~VyH#j7Y{t%rW@@_*2''_jbA[5n [ = RtlP=cy[+Qm]{Q$+ftU`X}pfp7hq sha1:g28FNhOJibhU+ERN5jGHKD24uys= > Hi Xah, You make some good points, and some bad ones. I don't have time to > reply at the length that this essay requires, so a short reply will > have to do. First of all, though: > * TeX is a system for typesetting. Typesetting is primarily concerned > with esthetics. As a art form, typesetting is insignificant. > Typesetting, taken by itself other than facilitate reading, is in > general of little serious utility. I object very seriously to this point to the degree of finding it > offensive. I think you miss the point of typesetting completely. This is the same Xah Lee that infests the various emacs groups.... best not to feed him too much! -- Gernot Hassenpflug === Subject: Re: The Problems of TeX >This is the same Xah Lee that infests the various emacs >groups.... best not to feed him too much! And programming languages ones as well. I have some rants of it as my own .sig's! Michele -- >It's because the universe was programmed in C++. No, no, it was programmed in Forth. See Genesis 1:12: And the earth brought Forth ... - Robert Israel in sci.math, thread Why numbers? === Subject: Re: The Problems of TeX 02vxDj5)SRrR^Y@@kQoTf+hUSNc#=u~S(X{@}:5uVIZXMV/*$:[3KGy~`!N>C{_ > Hi Xah, >You make some good points, and some bad ones. I don't have time to > reply at the length that this essay requires, so a short reply will > have to do. >First of all, though: >* TeX is a system for typesetting. Typesetting is primarily concerned > with esthetics. As a art form, typesetting is insignificant. > Typesetting, taken by itself other than facilitate reading, is in > general of little serious utility. >I object very seriously to this point to the degree of finding it > offensive. I think you miss the point of typesetting completely. The main point of typesetting and typography is to achieve communication between intelligences. That is its purpose for existing. To claim that typesetting is primarily concerned with aesthetics is at best a mistake, but most likely a lie constructed for the dishonest debating purpose of creating a deliberately flawed argument to knock down. (Typesetting can be used artistically; one might expect someone with a link to Chinese or similar pictographic mode of writing to sneer at any aesthetic effects that can be created that way. That does not make typography an insigificant artform.) So: I do not understand why you (Xah) sneer that `typesetting, taken by itself other than [to] facilitate reading, is in general of little serious utility'. Typesetting is part of the craft of typography, and since both sets of techniques exist primarily to facilitate reading[1], sneering at its lack of `serious utility' outside its main area of utility is somewhere between madness and dishonesty. Having said all that, I cannot for the life of me understand how anyone could take offense at Xah's comments, nor why anyone should care if someone has. Xah's line seems to be at odds with reality. > TeX is detrimental because it harbors ignorance of the structural > content embodied in most math notations in most fields. What TeX does > is typesetting, as opposed to math expression encoding. In other > words, what TeX does is pretty-printing. Using a derisive term like that is dishonest debating. What TeX does is typesetting - in other words, using the power of the computer to put signs and symbols together on a page in a fashion intended to be convenient to read. This is a very valuable thing to have done. Now, TeX's Big Win over `lesser approaches' is that it helps the author concentrate on the matter in hand. If the author is performing mathematical research, he `just does it'. When the author comes to write up the research, TeX permits him to `just write without bothering about the pretty printing side of things'. And then, should the author wish to get the appearance `just so', that's done after the `creating the content' stage. This is very unlike wysiwyg systems, where the author is distracted by the appearance of the output as it is created by typing the input, and so has to deal with the `pretty printing' aspects of it while his mind should be on other things. So by maintaining this `modal distinction', TeX is a powerful defence against the socially harmful effects of wysiwyg. > The language is designed in a way that any structural info in a math > expression are botched. As such, TeX, even though it is a full-fledged > computer language capable of great programing, but it understands > zilch of math expressions, it encodes zilch math expressions. Now that > is a egregious error of a computer language purporting to express > mathematics. TeX is a computer program which was created to `perform the function of a typesetter' - your claim that it purports to express mathematics is false. > And more so because it is a product of a mathematician > who should've known better The mathematician who designed it did know better - /much/ better than you. When a youngster sneers at an old buffer who's been around since the dawn of time and really does know it all (pretty much), the youngster has probably missed a point or two that the old buffer might well have missed at the same age as the youngster but has got it all sorted out now. You need to learn about typesetting and typography, you do. Then re-think your ideas. In fact, I think you need to do a lot of learning and thinking before trying anything like this again. You've just utterly failed to understand oh so many points. Either that, or you're just wanting to wind people up. Rowland. [1] Typography at its best is a fine art and can be used for purposes other than `to facilitate reading' - but that's side of typography is of minor social sigificance compared to its main users. -- Remove the animal for email address: rowland.mcdonnell@dog.physics.org http://www.mag-uk.org http://www.bmf.co.uk UK biker? Join MAG and the BMF and stop the Eurocrats banning biking === Subject: Re: The Problems of TeX posting-account=KoO4kQkAAAD0HzQZvBy87wMNyw93Iq_c Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > Hi Xah, You make some good points, and some bad ones. I don't have time to > reply at the length that this essay requires, so a short reply will > have to do. First of all, though: > * TeX is a system for typesetting. Typesetting is primarily concerned > with esthetics. As a art form, typesetting is insignificant. > Typesetting, taken by itself other than facilitate reading, is in > general of little serious utility. I object very seriously to this point to the degree of finding it > offensive. I think you miss the point of typesetting completely. Secondly, you speak about TeX as the source of your problems. This is > misguided, because TeX itself is purely designed to be a typesetting > engine. Only the formats on top of it (Plain, LaTeX, ConTeXt, etc.) > provide any semantic details. LaTeX made strides towards a separation > of content and formatting in its syntax, but maths remained rather ad > hoc. Your argument is that maths in LaTeX should contain meaning. Let me first say that there's a package that attempts to do this: > It's quite new, and deserves more attention, but it essentially > solves your problem. You might also be interested in a package I > in maths to enable a convenient syntax for predefined vectors, > matrices, functions, and other things that contain elements that > should be typeset consistently. A related problem arises that often the *display* of mathematics needs > to be tweaked for optimal output in ways that could never be > automated. So even with a fully semantic mathematics syntax, you'd > still need to pollute it with visual markup. So looking towards the future, how should this be addressed in the > future? Well, the breqn package is being worked on again and can now > probably be used for production purposes. This package greatly > improves the markup used for many of the presentation maths forms. It > has a logical separation between equations and multi-line > constructions, which amsmath was rather bad at enforcing. The package cool, I think, should be incorporated into a broader maths > package that provides the logical constructions that you believe > should be present for typesetting maths. This would allow the > beginnings of a syntax that could be used to translate maths > expressions from symbolic maths packages like Mathematica to TeX and > back again. The question is, are you willing to join us in creating such a > package? Will Of course, this misses the fact that only a small part of most documents is in mathematical notation, even in a maths-heavy paper. In my area (chemistry), the key content is often graphical. One *can* try to do that in a mark-up fashion, but almost no-one does. Even (La)TeX using chemists use graphical tools (normally ChemDraw) to create that type of content. Part of the reason is the subsequent use of the data; if it is not going to be processed further by a PC, then logical mark-up may not add anything at all. Joseph Wright === Subject: Re: The Problems of TeX posting-account=RHB_QQkAAADte5utYng4sPcFEECF3_54 Gecko/2008020514 Firefox/3.0b3,gzip(gfe),gzip(gfe) (Webwasher 6.6.2.2924) On 14 Feb, 08:36, Joseph Wright Hi Xah, > You make some good points, and some bad ones. No. Don't feed the trolls. === Subject: Re: The Problems of TeX 02vxDj5)SRrR^Y@@kQoTf+hUSNc#=u~S(X{@}:5uVIZXMV/*$:[3KGy~`!N>C{_ > Hi Xah, > You make some good points, and some bad ones. >No. Don't feed the trolls. I don't think we're talking about a common or garden troll here. The OP put rather too much work in it for it to be a simple troll. Rowland. -- Remove the animal for email address: rowland.mcdonnell@dog.physics.org http://www.mag-uk.org http://www.bmf.co.uk UK biker? Join MAG and the BMF and stop the Eurocrats banning biking === Subject: Re: The Problems of TeX > You make some good points, and some bad ones. > > No. Don't feed the trolls. I don't think we're talking about a common or garden troll here. The OP >put rather too much work in it for it to be a simple troll. sdrodrian puts lots of effort into his lunacies (being a physicist, i would be surprised if you've not encountered him). the difference is that xah has significant points to make, not the amount of effort invested. imho. personally, i don't write tex off as glibly as xah does; in particular, his assumption that typesetting doesn't count because everything's going to be on-line. but lots of stuff is on-line, even now, and i'm sure i'm not the only person who prints a document of any significance because i can't cope with it on-screen. screen viewing *doesn't* mean crappy typesetting is ok. -- Robin Fairbairns, Cambridge === Subject: Re: The Problems of TeX > No. Don't feed the trolls. I don't think we're talking about a common or garden troll here. The OP >put rather too much work in it for it to be a simple troll. Rowland. Yet he is a well known one, the point being that he occasionally *has* some points. But being a crackpot, he tends to turn out trollish anyway. Google for it. Michele -- >It's because the universe was programmed in C++. No, no, it was programmed in Forth. See Genesis 1:12: And the earth brought Forth ... - Robert Israel in sci.math, thread Why numbers? === Subject: Re: Quantitative finance book Your answer has been very exhaustive... EVE You are quite welcome. The biggest thing is to have an understanding of what you want to know. There are many good books out there. Best of luck, Alan > Hello! > Is there anyone who could suggest me a good book able to give an > introduction to the main aspects of quantitative finance? > EVE > There are quite a few good books and the subject has many dimensions, so > some are more specialized than others. > There are some books like Baxter and Rennie Financial Calculus which > show > how to use calculus in derivative pricing. Along the same lines is > Neftci, > Mathematics of Financial Derivatives. While other books are less > calculus-oriented and more giving you an overview of the markets and the > instruments with less intensive mathematics. > For a general overview, the Wilmott text mentioned by Fatal is good as > is > the book by Hull Options, Futures, and Other Derivatives which includes > some of the quantitative side as well as an overivew of how those markets > work (Hull and Wilmott are both widely used and respected texts for > introductory Quant Finance). Neill Chriss's book Black Scholes and > Beyond > gives in-depth info on the Black Scholes and other option pricing models. > For a general book on investments, the book called Investments by > Bodie, > Kane, & Marcus is very good, but not really quantitatively oriented > though > basic financial mathematics are covered. It was part of the CFA program > assigned reading. > Here is Wilmott's website: http://www.wilmott.com/ > This site lists books that are related to risk management, asset > allocation, > financial engineering, etc. They include reviews of the books and I > think > it is a great resource. > http://www.riskbook.com/ > These are some writings by Emanuel Derman. He was a quantitative analyst > at > Goldman Sachs and now heads the Financial Engineering program at Columbia > http://www.ederman.com/new/index.html > Another good resours is the International Association of Financial > Engineers > website: > http://www.iafe.org/ > Global Derivatives website is also good. They have user forums and you > can > get questions about quant finance answered plus questions about Masters > degree programs in Quant Fin if that is something that interests you. > http://www.global-derivatives.com/ > http://www.leomelamed.com/ > Depending upon what you are interested in, there are numerous resources. > One thing to look at is the Master of Science degrees that the top > universities grant with various titles such as MS in Financial > Engineering, > MS in Quantitative Finance, MS in Computational Finance, etc. Many > universities have more than one degree - one will be offered by the math > dept and another by the business school. Columbia, Carnegie Mellon, Univ > of > Chicago, UC Berkeley, and a number of others offer degrees. Even if you > are > not interested in a degree, visiting their websites is informative as > they > often give a suggested reading list and indicate texts that are used in > the > program. > If you are interested in pricing derivative contracts, you need to learn > stochastic calculus and linear algebra. Stochastic calculus generally > has > Calc I, II & III as a pre-requisite along with Differential Equations and > Linear Algebra. If you are not interested in pricing derivatives but in > other quant aspects, the math requirements may be less. > p.s. I just finished Calc III last semester and am taking Differential > Eqns > now with Linear Algebra scheduled for the fall. Columbia has a distance > learning Masters in Fin Engineering and I am taking these classes as > prerequisites with the intention of enrolling in the program in the > summer > session of 2009 which is why I know some of this stuff. Feel free to ask > for more info and I will do what I can to answer. > Alan === Subject: Re: How to learn real analysis- > How to learn real analysis? > ... > This one by Bruckner, Bruckner, & Thompson is free and appears to be > well-written:http://classicalrealanalysis.com/download.aspx > (there are 3 books there, the first link on the page, Elementary Real > Analysis, is the one you want) ... > Actually the discussion here is taking the reader through a pretty > broad range of real > analysis levels. If you are following this thread as a relative > novice then make sure to > start at the beginning. We have three real analysis texts on our web > site. The Elementary > Real Analysis [TBB] starts at the calculus level and goes pretty far > (short of measure theory > though). The TBB-Dripped (for the adventurous) does the same but with > the modern integration > theory added in. Then [BBT] covers material that is commonly in a > graduate real analysis course > [at least graduate level in north america--well maybe freshman in > russia and hungary :-) ] All three books are free hyperlinked PDF files, designed for on > screen viewing. That's my design > and could certainly be improved no doubt if anyone has some advice. > Trade paperback versions > will soon be available for TBB and, later, for BBT. Our feedback > from a class of 20 in the midwest > who were using [TBB] was that, while they appreciated the free PDF > files, they had all found used > copies on the internet that they wanted to use as well. For that > reason we are working on the paperback > versions. All information on the texts is at http://www.ClassicalRealAnalysis.com. > Or my blog http://classicalrealanalysis.blogspot.com. FWIW, I read through the first chapter of Elementary Real Analysis and thought that it was very well presented. I also have Georgi Shilov's book on Real Analysis and I like his writing, but thought that your text was easier to comprehend. I just finished Calc III in December and am early into a class on Differential Equations, so Real Analysis is my first real mathematics rather than just computational stuff. Therefore, it was nice to find that your presentation was clear and to the point, not assuming a lot of underlying knowledge or jumping around. In contrast, my Differential Equations text by Boyce & Brannan seems to constantly be assuming other knowledge as they go from one formula to the next with little or no explanation and simply assume that the steps are obvious. It is a tedious tome, but BBT is written the way a math book should be written IMHO. Alan === Subject: Subspace problem Suppose V is all 2x2 matrices, and suppose U is the set of 2x2 matrices such that TU = UT where T is just some 2x2 matrix. How would I show that U is or is not a subspace of V? If it is, then TU=UT must hold for scalar multiplication and vector addition, but I'm not quite sure how to set this up. === Subject: Re: Subspace problem days. My association with the Department is that of an alumnus. >Suppose V is all 2x2 matrices, and suppose U is the set of 2x2 matrices such >that TU = UT where T is just some 2x2 matrix. Ehr... you notice you used U to denote two things, both a set of matrices, and perhaps a specific matrix? I think you meant the following: Let V be the set of all 2x2 matrices, considered as a vector space. Let T be a fixed (but arbitrary) 2x2 matrix. Now let W = { U in V | UT = TU } How do we prove that W is a subspace of V? (Otherwise, while TU, T multiplying a set of matrices, makes sense, it no longer makes sense to talk about the set U that satisfies that, since there will be more than one in general...) >How would I show that U is or is not a subspace of V? If it is, then TU=UT >must hold for scalar multiplication and vector addition, but I'm not quite >sure how to set this up. You want to show that: (i) The 0 matrix is in W; that is, that 0T = T0; well, yes. No problem there. (ii) That if U1 and U2 are in W, then so is U1-U2. That is, you know already that U1*T = T*U1, and that U2*T = T*U2. Now you want to know whether (U1-U2)*T = T*(U1-U2). Well, does it? Let's see: (U1 - U2)*T = U1*T - U2*T (properties of matrix multiplication) = T*U1 - U2*T (since U1 is in W) = T*U1 - T*U2 (since U2 is in W) = T*(U1 - U2) (properties of matrix multiplication) That's great! It means that (U1-U2) is also in W whenever U1 and U2 already are in W. Good. (iii) That if U is in W, and a is any scalar, then aU is in W. So now you are assuming that UT = TU; and you want to prove that (aU)T = T(aU). I'll let you do that one. If W satisfies (i), (ii), and (iii), then it is a subspace. Exercise: Now let Z be any subset of V, and let C(Z) be the set C(Z) = { U in V | UT = TU for every T in Z}. Prove that C(Z) is always a subspace of V. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: Subspace problem > How would I show that U is or is not a subspace of V? Demonstrate closure on the one hand, or demonstrate the exostence of a counterexample on the other. > If it is, then TU=UT must hold for scalar multiplication and vector > addition, but I'm not quite sure how to set this up. Let U1 and U2 be elements of U. I.e. U1 T = T U1 and U2 T = T U2. Then can you prove that T (a U1) = (a U1) T? What about T (U1+U2) = (U1+U2) T? What happens if you try some examples of matrices in U? E.g. it seems pretty clear that the zero and identity matrices are in U no matter what T is. If you were to get stuck on proving it true: could you find U1 and U2 for which one of the above relations fails? - Tim === Subject: Re: Subspace problem >Suppose V is all 2x2 matrices, and suppose U is the set >of 2x2 matrices such >that TU = UT where T is just some 2x2 matrix. How would I show that U is or is not a subspace of V? If >it is, then TU=UT >must hold for scalar multiplication and vector addition, >but I'm not quite >sure how to set this up. Let v be an arbitrary element of V. To see that U is a subspace of V, you have to show that if u1,u2 are in U and lambda in K, then (u1+u2)*v = v*(u1+u2) as well as (lambda*u1)*v = v*(lambda*u1). Best wishes Torsten. === Subject: Re: Subspace problem > Suppose V is all 2x2 matrices, and suppose U is the set of 2x2 matrices such > that TU = UT where T is just some 2x2 matrix. How would I show that U is or is not a subspace of V? If it is, then TU=UT > must hold for scalar multiplication and vector addition, but I'm not quite > sure how to set this up. > What do you mean TU = UT must hold for scalar multiplication and vector addition? That's rather glib and uninformative, especially to you. What exactly do you mean by that statement? Explain that to me and then you will have explained it to yourself and from that you'll have idea how to set it up. In case you have forgotten, to show V is a vector space you have to show If x,y in V and a is scalar, then show ax in V and x + y in V. BTW, vector spaces are just a few of the spaces, so to be understood in this forum without confusion, you need to make it clear that you are considering vector spaces. The same problem for the topological space of 2x2 matrices makes the conclusion much differently. === Subject: Re: Probability and combinations that is the way I would have solved that one. I have a question about probability, I did not know how to start a new thread so I just responded to this one. Please help with this problem if you can: There are 3 chests with 2 drawers. You roll a 6 sided die. The outcomes of the dice are as follows: 1 corresponds to chest 1 2 &3 correspond to chest 2 4,5,6 correspond to chest 3 You then flip a coin to choose the drawer (heads is drawer 1, tails is drawer 2). In the 1st chest is a silver coin in drawer 1 and a gold coin in drawer 2. In the 2nd chest there are 2 gold coins (one in each drawer) and in the 3rd chest there are 2 silver coins (one in each drawer). What is the probability of choosing a gold coin? Given that you have a gold coin, what is the probability of another gold coin? === Subject: Re: Probability and combinations > What is the probability of choosing a gold coin? P(11) = 1/12 (silver) P(12) = 1/12 (gold) P(21) = 1/6 (gold) P(22) = 1/6 (gold) P(31) = 1/4 (silver) P(32) = 1/4 (silver) So P(gold) = P(12) + P(21) + P(22) = 5/12 > Given that you have a gold coin, what is the probability of another > gold coin? Do you mean from the other drawer, knowing that the first gold coin was obtained by that method but not where it came from (e.g. a friend handed it to you)? Of the 5/12 where we end up with a gold coin, 4/12 has a gold coin in the other drawer. So the probability is 4/5. Note that if we have more or less information than I assumed, this probability may change or become impossible to determine. - Tim === Subject: Help w/ my homework: Bishop & Goldberg 0.1.4.2 posting-account=lxiDzgkAAAAWw9IFcvC8h6aeL3v5fjEz Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) Manifolds (Dover repub. of corrected 1980 ed.) for the fun of it. I have a question regarding Problem 0.1.4.2 which reads: f : A -> B is 1-1 onto iff there is g : B -> A such that g o f = i_sub_A (the identity map on A) and f o g = i_sub_B. This characterizes g = f_inverse. I'm starting w/ the => direction. I have: Consider g : B -> A defined by g(f(a)) = a. If g is well-defined, clearly g o f = i_sub_A. Since f(a) is well-defined in B for all a in A (by hypothesis), g(f(a)) = a clearly well-defines g for all f(a) in B. Thus the existence of such a g is shown. My question(s) is (are), do I need to show that g is bijective? If this g is well-defined, clearly g o f = i_sub_A and i_sub_X is trivially bijective; however, my proof that g is well-defined does not appear to me to use the hypothesis that f is bijective. Is my proof wrong, is it implicitly using that f is bijective (and if so, where), or am I wrong that g is DG PS: While I'm at it, what (do you suppose) is meant by this characterizes g = f_inverse? That g is unique? (Proof of existence is clearly required by the first sentence of the problem.) And is === Subject: Re: Help w/ my homework: Bishop & Goldberg 0.1.4.2 > Tensor Analysis on > Manifolds (Dover repub. of corrected 1980 ed.) for > the fun of it. > I have a question regarding Problem 0.1.4.2 which > reads: f : A -> B > is 1-1 onto iff there is g : B -> A such that g o f = > i_sub_A (the > identity map on A) and f o g = i_sub_B. This > characterizes g = > f_inverse. I'm starting w/ the => direction. I > have: Consider g : > B -> A defined by g(f(a)) = a. If g is well-defined, > clearly g o f = > i_sub_A. Since f(a) is well-defined in B for all a > in A (by > hypothesis), g(f(a)) = a clearly well-defines g for > all f(a) in B. > Thus the existence of such a g is shown. That doesn't quite work. You haven't used any of your assumptions on f. First of all, you need g to be defined on all of B, your definition can only possibly work for b in im(f) = { b in B; b = f(a) for some a in A}. If g is defined on all of B, what does this say about f? Furthermore, it might be that there are a =/= a' in A such that f(a) = f(a'). Then g isn't even a function, since you've just defined g to have (at least) two different values at the point f(a) = f(a'). But, again, your assumption on f saves you...how? > My > question(s) is (are), do > I need to show that g is bijective? No, that is not part of your implication at all. All you need to do is find a _map_ g that satisifies g o f = id and f o g = id. > If this g is > well-defined, > clearly g o f = i_sub_A and i_sub_X is trivially > bijective; however, > my proof that g is well-defined does not appear to > me to use the > hypothesis that f is bijective. Is my proof wrong, > is it implicitly > using that f is bijective (and if so, where), or am I > wrong that g is > help. In short, you're wrong that g is well defined. It may fail to be a function and it need not be defined on all of B. See above. > DG >PS: While I'm at it, what (do you suppose) is meant > by this > characterizes g = f_inverse? That g is unique? If there is a g with the above mentioned properties, then f is invertible with inverse g. And yes, it is trivial to show that such a g is unique. Martin Wanvik === Subject: Re: Help w/ my homework: Bishop & Goldberg 0.1.4.2 <5791577.1203000352676.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=lxiDzgkAAAAWw9IFcvC8h6aeL3v5fjEz Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > Tensor Analysis on > Manifolds (Dover repub. of corrected 1980 ed.) for > the fun of it. > I have a question regarding Problem 0.1.4.2 which > reads: f : A -> B > is 1-1 onto iff there is g : B -> A such that g o f = > i_sub_A (the > identity map on A) and f o g = i_sub_B. This > characterizes g = > f_inverse. I'm starting w/ the => direction. I > have: Consider g : > B -> A defined by g(f(a)) = a. If g is well-defined, > clearly g o f = > i_sub_A. Since f(a) is well-defined in B for all a > in A (by > hypothesis), g(f(a)) = a clearly well-defines g for > all f(a) in B. > Thus the existence of such a g is shown. That doesn't quite work. You haven't used any of your assumptions on f. Yeah, I noticed that - that's why I knew I had a problem! > First of all, you need g to be defined on all of B, I can see that g _will_ be defined on all of B, but it's not clear to me why I need g to be defined on all of B. > your definition can only possibly work for b in im(f) = { b in B; b = f(a) for > some a in A}. Granted, but why _must_ im(f) = B? (Again, I can see that it will, but I don't see must.) > If g is defined on all of B, what does this say about f? Clearly this is where the onto-ness of f comes in; maybe I'm dense, but I'm not getting what your saying (perhaps because you're saying it backwards: your statement is in the form something true about g implies something about f, when in this direction and at this stage of the proof, all I'm trying to prove is that something about f implies something about g). Furthermore, it might be that there are a =/= a' in A such > that f(a) = f(a'). Then g isn't even a function, since you've just defined g to have (at least) two different values at the point > f(a) = f(a'). But, again, your assumption on f saves you...how? Ah, now this I get; f 1-1 implies that this difficulty can't happen. > My > question(s) is (are), do > I need to show that g is bijective? No, that is not part of your implication at all. Well, not explicitly. :-) > All you need to do is find a _map_ g that satisifies g o f = id > and f o g = id. > If this g is > well-defined, > clearly g o f = i_sub_A and i_sub_X is trivially > bijective; however, > my proof that g is well-defined does not appear to > me to use the > hypothesis that f is bijective. Is my proof wrong, > is it implicitly > using that f is bijective (and if so, where), or am I > wrong that g is > help. In short, you're wrong that g is well defined. > It may fail to be a function Right. > and it need not be defined on all of B. Again, I don't see why it necessarily needs to be. > See above. > DG > PS: While I'm at it, what (do you suppose) is meant > by this > characterizes g = f_inverse? That g is unique? If there is a g with the above mentioned properties, then f is invertible with inverse g. And yes, it is trivial to show that such a > g is unique. Right. DG Martin Wanvik === Subject: Re: Help w/ my homework: Bishop & Goldberg 0.1.4.2 > That doesn't quite work. You haven't used any of > your assumptions on f. >Yeah, I noticed that - that's why I knew I had a > problem! >First of all, you need g to be defined on all of B, >I can see that g _will_ be defined on all of B, but > it's not clear to > me why I need g to be defined on all of B. Well, that is part of what you're trying to prove, right? Given that f is bijective, you need to show the existence of a map g: B --> A such that [...]. By definition, this means that g must be defined for every element of B, i.e. for each element in b in B it must assign one and only one element g(b) in A. As for why you would need it, i.e. why couldn't one simply define g: im(f) --> A? Well, this follows from the requirement that g satisfy f(g(b)) = b for all b in B...if you can't define g(b) for all b in B, then you clearly can't possibly satisfy that part of the implication. If that was somewhat abstract, here's a simple example: Take A = {1,2}, B = {1,2,3} and let f(1) = 1, f(2) = 2, (f is 1-1, but not onto) If we try to define g by g(f(a)) = a, then we get g(1) = 1, g(2) = 2, and we could set g(3) = 2 (or anything else for that matter). Then we have g(f(a)) = a for all a in A, but f o g(1) = f(1) = 1, f o g(2) = f(2) = 2 and f o g(3) = f(2) = 2, which clearly is not the identity on B. > your definition can only possibly work for b in > im(f) = { b in B; b = f(a) for > some a in A}. >Granted, but why _must_ im(f) = B? (Again, I can see > that it will, > but I don't see must.) >If g is defined on all of B, what does this say > about f? >Clearly this is where the onto-ness of f comes in; Yes, that was what I was trying to imply. You defined g by saying that g(f(a)) = a, right? Now if g is to be defined at every b in B, then for every b in B, there has to be some a in A such that b = f(a), right? That is the same as saying that f is onto, which is something you've (fortunately) assumed. > maybe I'm dense, > but I'm not getting what your saying (perhaps because > you're saying it > backwards: your statement is in the form something > true about g > implies something about f, Sorry if I was unclear; the backward form of the statement was simply to make you see that the assumption that f is onto is necessary for the existence of a map g: B --> A such that f o g = id_B. After all, if the fact that we have a map g, such that f o g = id_B, implies that f is onto, then you can't possibly have such a map g without f being onto, right? (If X implies Y, then X cannot be true unless Y is) > when in this direction > and at this stage > of the proof, all I'm trying to prove is that > something about f > implies something about g). > Furthermore, it might be that there are a =/= a' in > A such > that f(a) = f(a'). Then g isn't even a function, > since you've just defined g to have (at least) two > different values at the point > f(a) = f(a'). But, again, your assumption on f > saves you...how? >Ah, now this I get; f 1-1 implies that this > difficulty can't happen. Correct! :) Actually, you can easily show that f: A --> B is 1-1 if and only if there exists g: B --> A such that g o f = id_A (i.e. f has a left inverse) and f is onto if and only if there exists g': B --> A such that f o g' = id_B (i.e. f has a right inverse) > My > question(s) is (are), do > I need to show that g is bijective? > No, that is not part of your implication at all. >Well, not explicitly. :-) Yep, and that was the point. That g has these properties is equivalent to it being bijective, so there is no need to show it explicitly. === Subject: Re: Help w/ my homework: Bishop & Goldberg 0.1.4.2 > I can see that g _will_ be defined on all of B, > but > it's not clear to > me why I need g to be defined on all of B. >Well, that is part of what you're trying to prove, > right? Given that f is bijective, you need to show > the existence of a map g: B --> A such that [...]. By > definition, this means that g must be defined for > every element of B, i.e. for each element in b in B > it must assign one and only one element g(b) in A. >As for why you would need it, i.e. why couldn't one > simply define g: im(f) --> A? Well, this follows from > the requirement that g satisfy f(g(b)) = b for all b > in B...if you can't define g(b) for all b in B, then > you clearly can't possibly satisfy that part of the > implication. If that was somewhat abstract, here's a > simple example: > Take A = {1,2}, B = {1,2,3} and let f(1) = 1, f(2) = > 2, > (f is 1-1, but not onto) > If we try to define g by g(f(a)) = a, then we get > g(1) = 1, g(2) = 2, and we could set g(3) = 2 (or > anything else for that matter). Then we have g(f(a)) > = a for all a in A, but f o g(1) = f(1) = 1, f o g(2) > = f(2) = 2 and > f o g(3) = f(2) = 2, which clearly is not the > identity on B. This might have been a silly way of formulating it: I'll try to express it slightly more concisely: You want g to satisfy f o g = id_B. id_B is clearly surjective, so it is clear that you need f to be surjective, since it is the last map in the composition. (Of course, even if f is not surjective, you could in this case simply give a separate definition outside im(f), but that wouldn't help you here). The example illustrates this quite nicely, I think. Hope this is clear(er). Martin Wanvik === Subject: Re: hyperbolic techniques > . . . have you looked at > Euclidean and Non-Euclidean Geometry by Patrick Ryan, > Cambridge University Press, 1986 (ISBN 0-521-27635-7)? . . . Analytic Approach. Ryan has just about everything I was looking for. (Took me a while to get to it because I started reading the Harry Potter series about the time you posted the suggestion.) > . . . . Ryan handles only the 2-D case, but the formulas > should generalize easily to 3-D. I'm concerned that it relies so heavily on cross-products. Maybe I'll be happier when I get around to trying things. -- Anton Sherwood, http://www.ogre.nu/ How'd ya like to climb this high *without* no mountain? --Porky Pine === Subject: Re: hyperbolic techniques > . . . have you looked at > Euclidean and Non-Euclidean Geometry by Patrick Ryan, > Cambridge University Press, 1986 (ISBN 0-521-27635-7)? . . . > Analytic Approach. Ryan has just about everything I was looking for. Forgot to mention, I also recently read Coxeter's Non-E. Geom. which is very difficult because it introduces a lot of nonstandard notations and gives no exercises; but it does have advice about homogeneous coordinates, which is not in Ryan and should come in handy. -- Anton Sherwood, http://www.ogre.nu/ How'd ya like to climb this high *without* no mountain? --Porky Pine === Subject: Re: 1-1/2+1/3-1/4+1/5-1/6+1/7 > >[...] Anyway here is a simple formalism concerning the notion >of /limit/ for sequences of sets: [...]. > And that approach becomes even simpler if we are dealing with /monotone/ > sequences of sets. >If (E_n) is an /increasing/ sequence of sets, i.e. E_n c E_(n+1) for all > n e N, then the limit exists, and we have > lim E_n = U E_n. > n n >If (E_n) is an /decreasing/ sequence of sets, i.e. E_(n+1) c E_n for all > n e N, then the limit exists, and we have > _ > lim E_n = | | E_n. > n n Wow! Now we can see that mathematics is indeed a form of art! Especially when expressed in ASCII .. > So it's extremely easy to determine the limit of the sequence in this > cases. (Just calculate the union or intersection of all E_n - that's > all.) >Example: >Let's consider the sequence of sets (S_n) with > S_n := {m e N : m < n} (n e N). Those are my Naturals Construction Sets: But there's kind of circularity in your definition in that the set of all naturals is assumed to exist already. (Ah, I know that the above is meant as an example and not as a definition of the set of naturals) > Then > S_0 = {} > S_1 = {0} > S_2 = {0,1} > S_3 = {0,1,2} > : , >and, of course, S_n c S_(n+1) for for all n e N. With other words, (S_n) > is an increasing sequence of sets. >Now it's easy to see (i.e. prove) that > U S_n = N = {0,1,2,...} > n >Hence > lim S_n = N. > n Here we are! Question: what if the set of all naturals is not assumed to exist beforehand,_ can_ the above limit be employed then as a definition of N eventually? Just a wild idea .. Han de Bruijn === Subject: Re: 1-1/2+1/3-1/4+1/5-1/6+1/7 > Then > S_0 = {} > S_1 = {0} > S_2 = {0,1} > S_3 = {0,1,2} > : , > and, of course, S_n c S_(n+1) for for all n e N. With other words, > (S_n) > is an increasing sequence of sets. > Now it's easy to see (i.e. prove) that > U S_n = N = {0,1,2,...} > n > Hence > lim S_n = N. > n Here we are! Question: what if the set of all naturals is not assumed to > exist beforehand,_ can_ the above limit be employed then as a definition > of N eventually? Just a wild idea .. If we do not know that the set N exists, then we have no reason to believe that the lim S_n exists. But we could (I suppose) add a principle like this: Whenever S_n is an increasing sequence of sets, then lim S_n exists. Not at all clear whether this is a good replacement for infinity, and it surely does *not* capture what you want. Either the limit exists or not. It does not potentially (or eventually?) exist. If you want to add those kinds of statements, you'll have to explain what they mean. -- How can people [philosophers] talk like that? Acting as if they're /glad/ they don't know things! Finding out more and more things they don't know! It's like children proudly coming to show you a full potty! -- Terry Pratchett, /Small Gods/ === Subject: Re: 1-1/2+1/3-1/4+1/5-1/6+1/7 > But we could (I suppose) add a principle like this: Whenever S_n is an increasing sequence of sets, then lim S_n exists. Exactly how do you propose to formulate this principle? In particular, how is 'sequence' in the above to be interpreted? -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: 1-1/2+1/3-1/4+1/5-1/6+1/7 > But we could (I suppose) add a principle like this: > Whenever S_n is an increasing sequence of sets, then lim S_n exists. Exactly how do you propose to formulate this principle? In particular, > how is 'sequence' in the above to be interpreted? Oh. Right. Good point. If we don't have N, then we don't have a clear definition of sequences. -- So yeah, do the wrong math, and use the ring of algebraic integers wrong, without understanding its quirks and real mathematical properties, and you can think you proved Fermat's Last Theorem when you didn't. -- James S. Harris on hobbies === Subject: Re: 1-1/2+1/3-1/4+1/5-1/6+1/7 > >But we could (I suppose) add a principle like this: > Whenever S_n is an increasing sequence of sets, then lim S_n exists. >Exactly how do you propose to formulate this principle? In particular, >how is 'sequence' in the above to be interpreted? >Oh. Right. Good point. >If we don't have N, then we don't have a clear definition of > sequences. Time to hook in with an improved version of my purported Proof that ZFC is inconsistent (well, it's not the proper thread, but ..). Scrutinize, please. Proof that ZFC is inconsistent ------------------------------ More precise: proof that THE set of ALL naturals does NOT exist. The definition of the finite ordinals is considered once again: 0 = {} 1 = {0} = {{}} 2 = {0,1} = { {} , {{}} } 3 = {0,1,2} = { {} , {{}} , { {} , {{}} } } . . n+ = (0,1,2,3,4, .. ,n} = [ .. oh, well .. ] . . The axiom of Infinity in ZFC is as follows. There exists a set which contains with each member n also its successor n+, starting with 0. The set of all naturals is defined as the set which contains _only_ 0 and its successors. The finite ordinals are thus defined by a successor function, called s(), where: s(n) = n+ = n + 1 We find: 0 1 = s(0) 2 = s(1) = s(s(0)) 3 = s(2) = s(s(s(0))) . . n+ = s(n) = s(s(s(s( .. s(s(0)) .. ))) [ 2(n+1) parentheses ] . . Definitions (Hughes - Frege) ---------------------------- Let m e N , then : S_m(x) = s o s o ... o s(x) , where there are exactly m function compositions (o). Written otherwise: S_m(x) = s(s( .. s(s(x)) ..)) Let x, y be any sets and we define the set S(x) by: y is in S(x) iff there is some n such that for all m > n y is in S_m(x) Here, n e N , m e N and S_m(x) defined as above. Lemma ----- S(0) = N , S(1) = N , .. , in general S(n) = N for all n e N . Proof ----- 0 is in S_1(0) = s(0) = {0} and in general it is easy to show that (n) is in S_{n+1}(0) . So there exist for each natural number (n) a number k = n + 1 , such that for all m > k the number (n) is in S_m(0) . This is precisely the set of all naturals. Because the naturals and the finite ordinals are identical in the von Neumann model, the naturals {0,1,2, .. ,n-1} are already contained in the natural (n). But (n) itself is in S_1(n) = s(n) = {n} and it is easy to show that (n+i) is in S_{i+1}(n) . Giving again the completed set of all naturals N . Definition (Hughes - Frege) --------------------------- Given a function f() on a set D with elements x, the function that has the whole set D as its argument will be denoted as F[D] . Thus, if D = {x_0, x_1, .. , x_k, .. } and f(x) defined on D then by definition: f[D] = { f(x_1) , f(x_1) , .. , f(x_k) , .. } . Or far more concise: f[D] = { f(x) : x e D } . Theorem ------- S[N] = {N} Proof ----- S[N] = { S(x) : x e N } = { N : x e N } = { N } Mind the subtlety: the infinite composition of successors is _not_ the naturals, but the singleton with THE naturals as the only element. However, we also have the following sequence, with the successor s() having the set of all naturals N as its argument, hence s[N] : N = {0,1,2,3,4,5,6,7,8,9, .. ,n, .. } = N {} s[N] = {1,2,3,4,5,6,7,8,9, .. ,n, .. } = N {0} s[s[N]] = {2,3,4,5,6,7,8,9, .. ,n, .. } = N {0,1} s[s[s[N]]] = {3,4,5,6,7,8,9, .. ,n, .. } = N {0,1,2} . . Lemma ----- All naturals from 0 up and including n are NOT present in the function value s o s o .. o s[N] , when composed with n+ successor functions s. Proof ----- As easily seen from the above, more formally by mathematical induction. Definition ---------- The value s o s o .. o s[N] , composed of exactly m successor functions s[] , shall be denoted as S_m[N] . Lemma ----- Let x, y be any sets, then (the _same_ function S as above): y is in S[x] iff there is some n such that for all m > n y is in S_m[x] Here, n e N , m e N and S_m[x] defined as above. Proof ----- As an excercise for the interested reader ;-) Theorem ------- S[N] = {} Proof ----- There is _no_ natural (k), such that there is some n, such that for all m > n this natural k e S_m[N] . Because S_m[N] = N {0,1,2, .. ,m} . But we have also proved that S[N] = {N}. Consequently {N} = {} . Take a good look at the latter formula. It says exactly _this_: the set containing only the set of all naturals is empty. This can mean nothing else than: THE set of ALL naturals does NOT exist. As I've always said. Please shoot holes in the above. An exact indication of the place where it goes wrong will be appreciated. (Tip, hint: mind e.g. the excercise) Han de Bruijn === Subject: Re: 1-1/2+1/3-1/4+1/5-1/6+1/7 fed this fish to the penguins: >Time to hook in with an improved version of my purported Proof that ZFC >is inconsistent (well, it's not the proper thread, but ..). Scrutinize, >please. > The proof is wrong: left as an exercise to the reader. Just one simple question: *honestly* what do you think is the chance that you will stumble upon an inconsistency proof of ZFC? Literally thousands of people have gone over the axioms and no one has found a contradiction, so what do you think your chances are, given that your mathematical skills are poor at best and your knowledge of ZFC is non-existent? Why do you not dedicate yourself to create *positive value*? If you want to do mathematics why don't you, I don't know, create a limit concept a la calculus (using your own words) that only uses finitary objects? That would surely be a sight. Do something of *value* instead of wasting your time with purported proofs of ZFC inconsistency that are shot down in the blink of an eye. Do you see *serious* mathematicians of finitist, intuitionist, constructivist, etc. persuasion, loosing their time in concocting lame ZFC inconsistency proofs? Stop acting the crank. G. Rodrigues === Subject: Re: 1-1/2+1/3-1/4+1/5-1/6+1/7 > Time to hook in with an improved version of my purported Proof that ZFC > is inconsistent (well, it's not the proper thread, but ..). Scrutinize, > please. Proof that ZFC is inconsistent > ------------------------------ You are *so* cute. > Lemma > ----- > Let x, y be any sets, then (the _same_ function S as above): y is in S[x] iff there is some n such that for all m > n > y is in S_m[x] Here, n e N , m e N and S_m[x] defined as above. Proof > ----- > As an excercise for the interested reader ;-) Bull. I've already said that this is false, so how do you dare omit the proof? In <87d4r3r3ur.fsf@phiwumbda.org>, I explicitly anticipated this argument. Let T(x) = { y in N | (E n in N)(A m in N)( m > n -> y in S_m[x] ) } Then we see that y is in S[x] iff (E z in x)(A w in N)( w in y <-> (E n in N)(A m in N)(m > n -> w in S_m(z)) ) [the corresponding formula in the previous post was not right, but see below] while y is in T(x) iff (E n in N)(A m in N)(m > n -> (E z in x)(y = S_m(z))) If you want to show that S[x] = T(x), then you sure as hell can't leave the proof to *this* reader, because I don't see any reason that it's true. ----------------- Lemma: y is in S[x] iff (E z in x)(A w in N)( w in y <-> (E n in N)(A m in N)(m > n -> w in S_m(z)) ) Proof: y is in S[x] iff (E z in x)(y = S[z]) w is in S[z] iff (E n in N)(A m in N)(m > n -> w in S_m(z)). Note that S[z] c N, since each S_m(z) c N. Thus, y is in S[x] iff (E z in x)(A w in N)( w in y <-> (E n in N)(A m in N)(m > n -> w in S_m(z)) ) Lemma: y is in T(x) iff (E n in N)(A m in N)(m > n -> (E z in x)(y = S_m(z))) Proof: By definition, y is in T(x) iff (E n in N)(A m in N)( m > n -> y in S_m[x] ). y is in S_m[x] iff (E z in x)(y = S_m(z)). ------------------ Proof left to reader my ass, Han. That was either just butt-stupid or dishonest. I had *already* suggested[1] that the lemma is false. If you think otherwise, then have the balls to prove it. Footnotes: [1] Though, to be honest, my description of S[x] was simply off. -- Sale or rental of this disc is ILLEGAL. If you have rented or purchased this disc, please call the MPAA at 1-800-NO-COPYS. -- The MPAA begins a new anti-piracy program, found on a DVD purchased in China === Subject: Re: 1-1/2+1/3-1/4+1/5-1/6+1/7 Agh. Cut and paste errors. > ----------------- > Lemma: y is in S[x] iff (E z in x)(A w in N)( w in y <- (E n in N)(A m in N)(m > n -> w in S_m(z)) ) Proof: y is in S[x] iff (E z in x)(y = S[z]) Should read: y = S(z) w is in S[z] iff (E n in N)(A m in N)(m > n -> w in S_m(z)). w is in S(z) ... Note that S[z] c N, since each S_m(z) c N. S(z) c N ... Thus, y is in S[x] iff (E z in x)(A w in N)( w in y <- (E n in N)(A m in N)(m > n -> w in S_m(z)) ) I hope that's my only errors. -- Jesse F. Hughes So far as this negative attitude toward life is concerned, Buddhism is merely Taoism a little touched in its wits. -- Lin Yutang, /My Country and My People/ === Subject: Re: 1-1/2+1/3-1/4+1/5-1/6+1/7 Wow! Now we can see that mathematics is indeed a form of art! Especially > when expressed in ASCII .. > It's a form of art AND a .8bscience.8a. All educable human beings should know what mathematics is -- /that/ mathematics /is/. They would enjoy life more, they would understand life more, they would have greater insight. (Paul R. Halmos) > So it's extremely easy to determine the limit of the sequence in this > cases. (Just calculate the union or intersection of all E_n - that's > all.) > > Example: > > Let's consider the sequence of sets (S_n) [defined] with > > S_n := {m e N : m < n} (n e N). > Those are my Naturals Construction Sets ... > Those are the so called /initial segments/ of N. See: http://mathworld.wolfram.com/InitialSegment.html But there's kind of circularity in your definition in that the set of > all naturals is assumed to exist already. > Nonsense. Do you have a problem with your short time memory? We already mentioned in this thread that in ZFC we DO have N at our disposal. (That's what the axiom of infinity is primarily used for.) - We do not assume it's existence. > Then > > S_0 = {} > S_1 = {0} > S_2 = {0,1} > S_3 = {0,1,2} > : , > > and, of course, S_n c S_(n+1) for for all n e N. With other words, (S_n) > is an increasing sequence of sets. > > Now it's easy to see (i.e. prove) that > > U S_n = N = {0,1,2,...} > n Here's the missing part: Let U = U S_n. neN Claim: U = N. Proof: First we show that U c N, i.e. for every x: if x e U, then x e N. Let x e U. Hence (by def. of U) Ei e N: x e S_i. Hence x e N (by def. of S_n). Hence U c N. With other words, U _only_ contains natural numbers. Now we show N c U, i.e. for every x: if x e N, then x e U. Let x e N. Hence x e S_(x+1) (by def. of A_n). Hence Ei e N: x e S_i. Hence (by def. of U) x e U. Hence N c U. With other words, U contains _all_ natural numbers. So we got: U contains all and only natural numbers, with other words, U = N. qed > Hence > > lim S_n = N. > n > Got it now? We just PROVED that the series of sets (S_n) has the limit N. Question: what if the set of all naturals is not [defined] beforehand, > _can_ the above limit be employed then as a definition of N eventually? > No. Since we already referred to N when defining the sequence (S_n). After all we defined the S_n's to be the _initial segments of N_. F. -- === Subject: Re: 1-1/2+1/3-1/4+1/5-1/6+1/7 > Question: what if the set of all naturals is not [defined] beforehand, > _can_ the above limit be employed then as a definition of N eventually? > No. Since we already referred to N when defining the sequence (S_n). > After all we defined the S_n's to be the _initial segments of N_. I don't see how that's an issue. Even before we define N, we can define the von Neumann naturals and each S_n is really just n itself. -- Jesse F. Hughes I can't tell you how many times she left me. I lost count the very first time that she did. -- The Flatlanders, I Thought the Wreck Was Over === Subject: Re: 1-1/2+1/3-1/4+1/5-1/6+1/7 > Question: what if the set of all naturals is not [defined] beforehand, > _can_ the above limit be employed then as a definition of N eventually? > No. Since we already referred to N when defining the sequence (S_n). > After all we defined the S_n's to be the _initial segments of N_. > I don't see how that's an issue. Even before we define N, we can > define the von Neumann naturals and each S_n is really just n itself. > Sure. But this way you do not get a _sequence_ (S_n). (Which is a set theoretic function from N onto N in this case.) Actually, I don't quite understand your statement: Even before we define N, we can define the von Neumann naturals ... Sure, for any set A I may consider A' = A u {A}. But this way I don't get an infinite set containing exactly the von Neumann naturals (i.e. N). [...] for the benefit of who I don't know, the axiom of infinity states there's an inductive set, that is, a set W such that 0 is in W and whenever n is in W, n union {n} is in W. By separation we get the set of finite ordinals by setting omega = {x | x in W & for all inductive sets A, x in A}; or, in other words, omega is the intersection of all inductive sets. (Aatu Koskensilta) F. -- === Subject: Re: 1-1/2+1/3-1/4+1/5-1/6+1/7 > But there's kind of circularity in your definition in that the set of > all naturals is assumed to exist already. > Nonsense. Do you have a problem with your short time memory? We already > mentioned in this thread that in ZFC we DO have N [omega] at our disposal. > (That's what the axiom of infinity is primarily used for.) [...] > A bit more explicitly, for the benefit of who I don't know, the axiom of infinity states there's an inductive set, that is, a set W such that 0 is in W and whenever n is in W, n union {n} is in W. By separation we get the set of finite ordinals by setting omega = {x | x in W & for all inductive sets A, x in A}; or, in other words, omega is the intersection of all inductive sets. (Aatu Koskensilta) F. -- === Subject: Re: 1-1/2+1/3-1/4+1/5-1/6+1/7 Correction... > Hence > lim S_n = N. > n > [...] We just PROVED that the sequence of sets (S_n) has the limit N. > ~~~~~~~~ F. -- === Subject: Re: On convergence of sequences of sets. Was Re: 1-1/2+1/3-1/4+1/5-1/6+1/7 In another thread, the issue of convergence of sequences of sets has > come up, so I decided to write a little something about it. There is > nothing particularly deep in this post, just setting up some > definitions and basic results. I will also skip giving full details > for the proofs because nothing is gained by it except multiplying the > length of the post. The interested reader can do it for himself with > much more profit. > Right. Fix a set X and let (E_n) be a sequence of subsets of X. Define the > upper limit E^* of (E_n) to be the set {x: x is in E_n for an infinite number of [values of] n} > One might write that the following way: E^* = {x : {n e N : x e E_n} infinite}. Hence x e E^* <-> {n e N : x e E_n} infinite. and the lower limit E_* to be the set {x: there exists n_0 such that x in E_n for all n >= n_0} > Well, for symmetry reasons I'd propose the definition {x: x is not in E_n only for a finite number of [values of] n} Then one might write that the following way: E_* = {x : {n e N : x !e E_n} finite}. Hence x e E_* <-> {n e N : x !e E_n} finite. We say that (E_n) is convergent iff E_* = E^* in which case we put lim > (E_n)= E_* = E^*. > I'd prefer to write lim E_n = E_* = E^* n You might compare all this with Halmos' text: If (E_n) is a sequence of subsets of X, the set E^* of all those points of X which belong to E_n for infinitely many values of n is called the /superior limit/ of the sequence and is denoted by E^* = lim sup E_n. n The set E_* of all those points which belongs to E_n for all but a finite number of values of n is called the /inferior limit/ of the sequence and is denoted by E_* = lim inf E_n. n If it happens that E^* = E_*, we shall use the notation lim E_n n for this set. (Paul R. Halmos, Measure Theory) The first thing to notice is that E_* <= E^* but of course the > inequality can be strict. > You mean the subset relation here, right? If we denote that relation with c, we may write: E_* c E^*. Proof: x e E_* -> {n e N : x !e E_n} finite -> {n e N : x e E_n} infinite -> x e E^*. Example 1: Consider X = {0, 1} and put E_n = {0} for n even and E_n = > {0, 1} for n odd. You get E_* = {0} and E^* = {0, 1}. Example 2: If (E_n) is a sequence of pairwise disjoint sets than it > converges to the empty set > Right. There are very concrete formulas for both E_* and E^*. Theorem A: Let (E_n) be a sequence of subsets of X. Then E_* = /_{n=0}^{infinity} /_{m=n}^{infinity} E_m E^* = /_{n=0}^{infinity} /_{m=n}^{infinity} E_m ^^^ corrected proof: Unpack the definitions. > Right. I'd show/prove it the following way: x e E_* <-> {n e N : x !e E_n} finite <-> En e N Am >= n : x e E_m oo <-> En e N : x e / E_m m=n oo oo <-> x e / ( / E_m ). n=0 m=n x e E^* <-> {n e N : x e E_n} infinite <-> An e N Em >= n : x e E_m oo <-> An e N : x e / E_m m=n oo oo <-> x e / ( / E_m ). n=0 m=n I am using / for intersection and / for union. The meaning of > infinity should also be obvious. ^^^^^ note 1: The theorem is telling us that this particular notion of > convergence is a purely order-theoretic one. It can be generalized to > countably complete Boolean algebras, like those appearing in measure > theory. note 2: [A] sequence (E_n) is convergent [iff] /_{n=0}^{infinity} /_{m=n}^{infinity} E_m = > /_{n=0}^{infinity} /_{m=n}^{infinity} E_m There is a striking resemblance with the equality lim-inf a_n = > lim-sup a_n for convergent sequences (a_n) of real numbers. Not a > coincidence. [Theorem]: Using the above formulas we can readily prove that if the > sequence (E_n) is decreasing than it is convergent and the limit is > /_n E_n. Similarly, if (E_n) is increasing than it converges to /_n > E_n. It is not difficult to see that if (F_{n_k}) is a subsequence of (E_n) > then we have the chain of inequalities E_* <= F_* <= F^* <= E^* > E_* c F_* c F^* c E^* so that if (E_n) is convergent so is any subsequence. > Which again bears a striking resemblance to the properties of lim inf a_n and lim sup a_n for convergent sequences (a_n) of real numbers. F. -- === Subject: Re: On convergence of sequences of sets. Was Re: 1-1/2+1/3-1/4+1/5-1/6+1/7 > Which again bears a striking resemblance to the properties of lim inf > a_n and lim sup a_n for convergent sequences (a_n) of real numbers. Uhm, talking about infinite function compositions in practice, _this_ may happen when the interpreter of a (well known) report generator has no loop structures in its toolbox: kale titel = Lower( Replace(Replace(Replace(Replace(Replace(Replace( Replace(Replace(Replace(Replace( Replace(Replace(Replace(Replace(Replace( Replace(Replace(Replace(Replace(Replace( Replace(Replace(Replace(Replace(Replace( Replace(Replace(Replace(Replace(Replace( Replace(Replace(Replace(Replace(Replace(Replace( Replace(Replace(Replace(Replace(Replace( , Char(8) ,) ,Char(9) ,) ,Char(10) ,) , Char(11) ,), Char(12) ,) ,Char(13) ,) , Char(32) ,) , Char(33) ,) ,Char(34) ,) ,Char(35) ,) , Char(36) ,) , Char(37) ,) ,Char(38) ,) ,Char(39) ,) , Char(40) ,) , Char(41) ,) ,Char(42) ,) ,Char(43) ,) , Char(44) ,) , Char(45) ,) ,Char(46) ,) ,Char(47) ,) , Char(58) ,) , Char(59) ,) ,Char(60) ,) ,Char(61) ,) , Char(62) ,) , Char(63) ,) ,Char(64) ,) ,Char(91) ,) , Char(90) ,) , Char(92) ,) ,Char(93) ,) ,Char(94) ,) , Char(95) ,) , Char(96) ,) ,Char(123) ,) ,Char(124) ,) , Char(125) ,) , Char(126) ,) ,Char(127) ,)) Never mind. Just where much of my heuristics comes from .. Han de Bruijn === Subject: Re: Set existence posting-account=vEze6woAAAAx1KsAyEhBn6SIbW8UsMb1 SV1),gzip(gfe),gzip(gfe) I have read in Kanamori (The mathematical development of set theory from Cantor to Cohen) that 'If ZF is consistent then the assumption that [a countable standard] model exists cannot established in ZF by G.9adel's (Second) Incompleteness Theorem. Then, what else is needed in order to stablish it? === Subject: Re: Set existence posting-account=vEze6woAAAAx1KsAyEhBn6SIbW8UsMb1 SV1),gzip(gfe),gzip(gfe) > I have read in Kanamori (The mathematical development of set theory > from Cantor to Cohen) that 'If ZF is consistent then the assumption > that [a countable standard] model exists cannot established in ZF by > G.9adel's (Second) Incompleteness Theorem. Then, what else is needed in order to stablish it? Is it possible that he was wrong? In the revised version, he changed that statement by 'The existence of such a model is an avoidable assumption in formal relative consistency proofs via forcing'. Which doesn't mean the same... === Subject: Re: Set existence posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) > I have read in Kanamori (The mathematical development of set theory > from Cantor to Cohen) that 'If ZF is consistent then the assumption > that [a countable standard] model exists cannot established in ZF by > G.9adel's (Second) Incompleteness Theorem. In principle he existence of such a model could be established outside ZF. > Then, what else is needed in order to stablish it? Is it possible that he was wrong? In the revised version, he changed > that statement by 'The existence of such a model is an avoidable > assumption in formal relative consistency proofs via forcing'. Which > doesn't mean the same... There is no model at all for ZFC. In or reality there is no chance to know uncountably many indiiduals. What is a number which nobody will ever know? A number is an idea. What is an idea which nobody ever will have? === Subject: Re: Set existence > I have read in Kanamori (The mathematical development of set theory > from Cantor to Cohen) that 'If ZF is consistent then the assumption > that [a countable standard] model exists cannot established in ZF by > G.9adel's (Second) Incompleteness Theorem. >In principle he existence of such a model could be established outside > ZF. Right. If ZF is consistent, it can't prove that it has _any_ model, by Godel's 2nd incompleteness theorem. > Then, what else is needed in order to stablish it? A theory stronger than, or at least different from, ZF. > Is it possible that he was wrong? In the revised version, he changed > that statement by 'The existence of such a model is an avoidable > assumption in formal relative consistency proofs via forcing'. Which > doesn't mean the same... I don't know exactly what he means here. A relative consistency proof of ZF to some theory T would be of the form Con(T) -> Con(ZF). This doesn't need ZF to have any models to be true. > There is no model at all for ZFC. In which case ZFC, being inconsistent, could prove anything, including ZFC has a model. Do you know of such a proof in ZFC? That would be really big news! In any case, it is of course trivially true that Con(ZFC) -> Con(ZF). > In our reality there is no chance to > know uncountably many indiiduals. What is a number which nobody will > ever know? A number is an idea. What is an idea which nobody ever will > have? -- hz === Subject: Re: Set existence <47B4AD31.FEA749C7@gmail.com> posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > I have read in Kanamori (The mathematical development of set theory > from Cantor to Cohen) that 'If ZF is consistent then the assumption > that [a countable standard] model exists cannot established in ZF by > G.9adel's (Second) Incompleteness Theorem. > In principle he existence of such a model could be established outside > ZF. Right. æIf ZF is consistent, it can't prove that it has any model, > by Godel's 2nd incompleteness theorem. > Then, what else is needed in order to stablish it? A theory stronger than, or at least different from, ZF. > Is it possible that he was wrong? In the revised version, he changed > that statement by 'The existence of such a model is an avoidable > assumption in formal relative consistency proofs via forcing'. Which > doesn't mean the same... I don't know exactly what he means here. æA relative consistency proof > of ZF to some theory T would be of the form Con(T) -> Con(ZF). æThis > doesn't need ZF to have any models to be true. > There is no model at all for ZFC. In which case ZFC, being inconsistent, could prove anything, > including ZFC has a model. Do you know of such a proof in ZFC? æThat would be really big news! Do you know a model? Unless a model is presented, I assume that there is no model. Where should it be, if not in the heads of mathematicians? And there it is obviously not. === Subject: Re: Set existence > Recently I came across a statement saying that the axioms of set > theory do not mean what they say or what usually is understood. > That doesn't say that the axioms do not mean what usually is understood. It was a reply to you, saying they don't mean what *you* think they mean. - Tim === Subject: Re: Set existence > So we see that Cantor shows the uncountability of the set of real > numbers, by constructing a member of the countable set of > constructible numbers. It is interesting, yes. However, it should be noted that the diagonal number need not be actually constructible in the usual sense. The theorem makes no assumptions about whether the numbers in the sequence are themselves constructible, only that if such a list exists then from it you can construct a number not in the list. It is also notable that even if all the numbers in the sequence are constructible, there is no requirement that the sequence as a whole be a constructible entity. Finally, constructibility is a somewhat ambiguous term. What constructions do we permit? Note that diagonalization is a process that can be applied to any defined set of constructions to enlarge the set of constructible numbers. Hence there is no one unique set of constructible numbers. - Tim === Subject: Re: Set existence posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > So we see that Cantor shows the uncountability of the set of real > numbers, by constructing a member of the countable set of > constructible numbers. It is interesting, yes. However, it should be noted that the diagonal number need not be > actually constructible in the usual sense. The theorem makes no > assumptions about whether the numbers in the sequence are themselves > constructible, only that if such a list exists then from it you can > construct a number not in the list. I quoted some opinions that the diagonal number is a constructed number. I agree. It is also notable that even if all the numbers in the sequence are > constructible, there is no requirement that the sequence as a whole be > a constructible entity. I do not see why this is of any interest. Finally, constructibility is a somewhat ambiguous term. What > constructions do we permit? Note that diagonalization is a process > that can be applied to any defined set of constructions to enlarge the > set of constructible numbers. Yes, that is my point! By the diagonal construction we can prove that a non listed diagonal number exists. But on the other hand we know, that this diagonal number is a constructed number and, therefore, belongs to a countable set. Hence we see that Cantor's proof does not prove uncountability. > Hence there is no one unique set of > constructible numbers. In any case the set of constructions is countable. Hence the set of constructible numbers is countable. === Subject: Re: Set existence > I do not see why this is of any interest. Do you believe that a number is constructible if it can only be constructed from a list that may be non-constructible? > Yes, that is my point! By the diagonal construction we can prove > that a non listed diagonal number exists. Correct. > But on the other hand we know, that this diagonal number is a > constructed number Only if your answer to my first question is yes. > and, therefore, belongs to a countable set. Every number, constructible or not, belongs to a countable set of some sort. Its singleton set, if nothing else. But in case you mean belongs to the countable set of constructible reals, only if your answer to my first question is no. > In any case the set of constructions is countable. Hence the set of > constructible numbers is countable. Even if you restrict yourself to constructibility, there is no constructible bijection from the set of naturals to the set of constructible reals. - Tim === Subject: Re: Set existence posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) > I do not see why this is of any interest. Do you believe that a number is constructible if it can only be > constructed from a list that may be non-constructible? Either the diagonal number is a construction. Or the diagonal number is a chimera (and then says nothing about nothing). Kanamori as well as Fraenkel and Levi (in their Abstract set theory) agree that the diagonal proof is not merely existential. > Yes, that is my point! By the diagonal construction we can prove > that a non listed diagonal number exists. Correct. > But on the other hand we know, that this diagonal number is a > constructed number Only if your answer to my first question is yes. If a number is constructed, then t is constructible --- with no regard to the properties of lists. > and, therefore, belongs to a countable set. Every number, constructible or not, belongs to a countable set of some > sort. Its singleton set, if nothing else. I fully agree with you. There are no uncountable sets at all. But in case of the constructible numbers (and even in case of any number which exists in a way such that we can know it as an individual) we know with certainty that this set is not uncountable. But in case you mean belongs to the countable set of constructible > reals, only if your answer to my first question is no. No. In any case a number which we can know (or use in any proof as an individual) belongs to the at most countable set of individual numbers. In my opinions it is wrong to call those numbers numbers which cannot exist as individuals. > In any case the set of constructions is countable. Hence the set of > constructible numbers is countable. Even if you restrict yourself to constructibility, there is no > constructible bijection from the set of naturals to the set of > constructible reals. A bijection from N to M is not the only way to show the countability of M. Knowing that M is a countable union of countable unions (or one of its subsets) is sufficient. === Subject: Re: Set existence > So we see that Cantor shows the uncountability of the set of real > numbers, by constructing a member of the countable set of > constructible numbers. > It is interesting, yes. > However, it should be noted that the diagonal number need not be > actually constructible in the usual sense. The theorem makes no > assumptions about whether the numbers in the sequence are themselves > constructible, only that if such a list exists then from it you can > construct a number not in the list. >I quoted some opinions that the diagonal number is a constructed > number. I agree. > It is also notable that even if all the numbers in the sequence are > constructible, there is no requirement that the sequence as a whole be > a constructible entity. >I do not see why this is of any interest. There is so much in mathematics that is of interest but which WM does not see that one wonders why he does not find something other than mathematics not to see. > Finally, constructibility is a somewhat ambiguous term. What > constructions do we permit? Note that diagonalization is a process > that can be applied to any defined set of constructions to enlarge the > set of constructible numbers. >Yes, that is my point! By the diagonal construction we can prove that > a non listed diagonal number exists. But on the other hand we know, > that this diagonal number is a constructed number and, therefore, > belongs to a countable set. One can imagine a list of non-constructable numbers in which the nth number is at least constructable as far as its nth decimal digit. Well, maybe WM can't imagine it, but most of us can. > Hence we see that Cantor's proof does > not prove uncountability. It still does. === Subject: Re: Set existence > In case (2) Loewenheim's and Skolem's result shows that set theory is > inconsistent. > No, they just show that certain first-order axiomatizations have > unintended models. >Recently I came across a statement saying that the axioms of set > theory do not mean what they say or what usually is understood. So > there seems to be a secret science, accesible only to initiated > disciples. But the axioms were not spelled out by God. They were made > by people who did not even understand Skolem's proof. Isn't it very > improbable that these people were so far sighted that they made the > axioms in order to contain just what now is interpreted?* Or isn't it > by far more plausible that the axioms by hook or by crook have been > adapted by the ruling high priests such that every menacing > inconsistency so far can be rejected? I have no idea what you're babbling about. Here's a thought: state your assumptions, deduce a conclusion. Be rigorous. This is persuasive. Dark ranting about a secret science with high priests, well, who cares? > As an earlier example, not yet concerning axiomatic set theory but > showing the thoughts of the forbears, Cantor himself was not willing > to accept numbers which cannot be definied. Now you cite Cantor as an authority? You just got through saying that set theory is either false or inconsistent. Make up your mind, already. -- hz === Subject: Re: Set existence <47B397C2.F4CDD6A8@gmail.com> posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > In case (2) Loewenheim's and Skolem's result shows that set theory is > inconsistent. > No, they just show that certain first-order axiomatizations have > unintended models. > Recently I came across a statement saying that the axioms of set > theory do not mean what they say or what usually is understood. So > there seems to be a secret science, accesible only to initiated > disciples. But the axioms were not spelled out by God. They were made > by people who did not even understand Skolem's proof. Isn't it very > improbable that these people were so far sighted that they made the > axioms in order to contain just what now is interpreted?* Or isn't it > by far more plausible that the axioms by hook or by crook have been > adapted by the ruling high priests such that every menacing > inconsistency so far can be rejected? I have no idea what you're babbling about. æ Why then are you babbling here? > As an earlier example, not yet concerning axiomatic set theory but > showing the thoughts of the forbears, Cantor himself was not willing > to accept numbers which cannot be definied. Now you cite Cantor as an authority? æ Of course, Cantor was a great mathematician. And I do by far more agree with his views than the present mathematicians do. He made an error. Errare humanum est. But he would have cursed modern set theory. === Subject: Re: Set existence > > A set without any explicit set descriptor can still be a set, for > example all those inaccessible reals, as Dedekind lower cuts, are > sets > but none of them have an explicit set descriptor, or they would not > be > inaccessible in the first place > That means, translated to N and P(N), that all combinations of > natural numbers are subsets of N and elements of P(N). Is this always > the case? Or what is the condition? > Every function f:N -> {0 1} determines two subsets of N, > { n in N: f(n) = 0} and {n in N: f(n) = 1}, whose intersection is empty > and æwhose union is N itself. And P(N) contains as members all sets > determined by some such function > Do you mean functions which have to be defined? > No. They all exist, but some are not accessible. >And do these functions which are not accessible count when the > cardinality of the set of all functions is determined? Do they count > in any model? > > If you mean functions which exist in some Platonist reality and need > not be defined, then P(N) might contain all subsets of N as elements. > But altogether we have not yet come to an explanation how some subsets > might exist and other might not exist. > If they are subsets, they exist. It is only non-subsets which can > non-exist. >Do they all exist in any model? Do all combinations, as Fraenkel puts > it, exist in any model? Fraenkel denies that. If Fraenkel is right, > why then should all combinations exists in Platonistic reality. Why > should they exist in our real reality, where most of them cannot be > defined? Since no numbers at all exist in any real reality, i.e., in any physical sense, whatever existence they have is outside that real reality. === Subject: Re: Set existence posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > If you mean functions which exist in some Platonist reality and need > not be defined, then P(N) might contain all subsets of N as elements. > But altogether we have not yet come to an explanation how some subsets > might exist and other might not exist. > If they are subsets, they exist. It is only non-subsets which can > non-exist. > Do they all exist in any model? Do all combinations, as Fraenkel puts > it, exist in any model? Fraenkel denies that. If Fraenkel is right, > why then should all combinations exists in Platonistic reality. Why > should they exist in our real reality, where most of them cannot be > defined? Since no numbers at all exist in any real reality, i.e., in any > physical sense, whatever existence they have is outside that real > reality Do they all exist in the model we are sitting in? What is the difference compared to a countable model (since most numbers cannot be determined in any model)? === Subject: Re: Set existence > Since no numbers at all exist in any real reality, i.e., in any > physical sense, whatever existence they have is outside that real > reality Do they all exist in the model we are sitting in? I do not sit in models. === Subject: Re: Set existence > I do not sit in models. I have sat in a model's lap, does that count? - Tim === Subject: Re: Set existence >I do not sit in models. >I have sat in a model's lap, does that count? It counts extra. -- hz === Subject: Re: Set existence > Recently I came across a statement saying that the axioms of set > theory do not mean what they say or what usually is understood. So > there seems to be a secret science, accesible only to initiated > disciples. > Why not quote the statement and give the source? Why should we depend > on your report and interpretation? > I do not remember every source I came across. But here is this one: > > Comments: > 1. That post seems to be a joke. >You may get this impression if you are not familiar withg set theory. >2. His first statement to me seems to be addressed to > you, WM, saying the axioms of set theory don't say what > you think they do. >Perhaps. But I believe that there are more people who share my view. Not among those who post here. Where is one to find these nits? > Your reading is patently ridiculous. And to further take > it as an authoritative general statement about set theory >Oh, did I say so? >is even more ridiculous, since the author is not claiming > to be familiar with set theory. Just the opposite. >Zum Verst.8andnis der Lehre vom Transfiniten bedarf es keiner gelehrten > Vorbereitung in der neueren Mathematik; sie kann f.9fr diesen Zweck eher > sch.8adlich als n.9ftzlich sein (Cantor, 1888). >The understanding of the teachings of the transfinite does not require > a scholarly training in new mathematics; rather it can be harmful for > this purpose. And WM is still back in 1888, or earlier. === Subject: Re: -- compact connected linear order Let (S,<=) be complete linear order. > Let { U ai, V bj | i in I, j in J } be a subbase cover for S. > Let a = sup i ai, b = inf j bj > Some i in I, j in J with a in V bj, b in U ai > a = b; U ai / V bj = S. > Thus by Alexander's subbase theorem, S is compact > complete linear orders are compact. A nice way to show that a compact linear order S is > complete is the following: let A be a subset of S. > Then the identity function f : A --> S is a net in S. > Since S is compact, there exists a convergent subnet, > that is, there exists a directed set B and a monotone > cofinal map g : B --> A such that the net fg : B --> S > has a limit. Call the limit x. Then using the definition > of convergence it is easy to see that x = sup A. > Arggg! The purpose of topology is to transcend series, not to revel in them. Is there a filter version? Here is my method. Assume S compact linear order. Show by contradiction, that a compact linear order has a maximum and a minimum, unless empty. Let A be a subset of S. If A = nulset, then sup A = bottom element. Otherwise cl A is compact, hence has maximum element a. Easily ta = sup A. That all supremums exist, suffices to show the order is complete. To sumarise. A linear order space is compact iff it's order complete. -- Any easy ways of showing a linear order space is connected iff it's order dense and bounded complete, ie complete within bounds or Dedekind complete? ---- === Subject: Re: -- compact connected linear order posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > Let (S,<=) be complete linear order. > Let { U ai, V bj | i in I, j in J } be a subbase cover for S. > Let a = sup i ai, b = inf j bj > Some i in I, j in J with a in V bj, b in U ai > a = b; U ai / V bj = S. > Thus by Alexander's subbase theorem, S is compact > complete linear orders are compact. > A nice way to show that a compact linear order S is > complete is the following: let A be a subset of S. > Then the identity function f : A --> S is a net in S. > Since S is compact, there exists a convergent subnet, > that is, there exists a directed set B and a monotone > cofinal map g : B --> A such that the net fg : B --> S > has a limit. Call the limit x. Then using the definition > of convergence it is easy to see that x = sup A. Arggg! The purpose of topology is to transcend series, > not to revel in them. Is there a filter version? I did not get that memo. Do you have a link to the drafting papers of the Constitution of Topology where the Founding Fathers discuss this? Your complaint is somewhat silly in my eyes. The argument using nets has the very nice property of being exactly the same as the one one uses to show that a subset of a closed bounded interval has a maximum. And, if I were to pick a purpose for general topology (of course, in absolutely no way I expect others to agree with or limit themselves to) would be: come up with a language which allows us to talk about unfamiliar things and solve problems about them by making them look like more familiar objects. > Here is my method. Assume S compact linear order. > Show by contradiction, that a compact linear order > has a maximum and a minimum, unless empty. Let A be a subset of S. If A = nulset, then sup A = bottom element. > Otherwise cl A is compact, hence has maximum element a. Why does it have a maximum element? -- m === Subject: Fundamentals of Heat and Mass Transfer, 4th Ed by Incropera.. posting-account=GKvjuwoAAAA_7urRZuWlrw0QB5axoXRd Hey, I need solution manual of Fundamentals of Heat and Mass Transfer, 4th Ed by Incropera.. please help. === Subject: Prime numbers within Galois field? Once upon a time, I've become somewhat involved with Galois fields: http://hdebruijn.soo.dto.tudelft.nl/www/programs/delphi.htm#JT But they remain weird stuff to me. Now I have the following question. Suppose you're working _within_ a Galois Field, say of order 2^n. Is it possible then, in principle, to define a kind of prime numbers within such a field, i.e. numbers that are not divisible by anything else than 1 and themselves? (Disclaimer: rusty knowledge about it ..) Han de Bruijn === Subject: Re: Prime numbers within Galois field? posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e InfoPath.1; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) On Feb 14, 10:54æam, Han de Bruijn Suppose you're working within a Galois Field, say of order 2^n. Is > it possible then, in principle, to define a kind of prime numbers > within such a field, i.e. numbers that are not divisible by anything > else than 1 and themselves? (Disclaimer: rusty knowledge about it ..) Han de Bruijn ************************************************************ Apparently pretty rusty knowledge here, dear Han: in a field, any field, ANY element is divisible by any other non-zero element, so the notion of prime, irreducible and all that is pretty boring. book about the real prime, a rather bizarre idea he had. I haven't yet read that, but I know it made lift an eyebow to some people in my university back then (the Hebrew University in Jerusalem). Perhaps there's something to it, I can't say. Tonio === Subject: Re: Prime numbers within Galois field? > On Feb 14, 10:54 am, Han de Bruijn Once upon a time, I've become somewhat involved with Galois fields: >http://hdebruijn.soo.dto.tudelft.nl/www/programs/delphi.htm#JT >But they remain weird stuff to me. Now I have the following question. >Suppose you're working _within_ a Galois Field, say of order 2^n. Is >it possible then, in principle, to define a kind of prime numbers >within such a field, i.e. numbers that are not divisible by anything >else than 1 and themselves? (Disclaimer: rusty knowledge about it ..) >************************************************************ >Apparently pretty rusty knowledge here, dear Han: in a field, any > field, ANY element is divisible by any other non-zero element, so the > notion of prime, irreducible and all that is pretty boring. Oh well, then the case would be closed already, before it's opened. > book about the real prime, a rather bizarre idea he had. Perhaps the same bizarre idea as I had: suppose that such real primes can be defined within a Galois Field (say of order 2^n), _then_ we take the limit for n -> oo and have the set of all naturals. If there is a match between our new baked Galois primes and the common primes and if the Goldbach Conjecture could easily be proved for Galois primes then we have proved the same conjecture for common primes. But, as I've said, it's a bizarre idea .. > I haven't yet read that, but I know it made lift an eyebow to some > people in my university back then (the Hebrew University in > Jerusalem). > Perhaps there's something to it, I can't say. Curious anyway .. Han de Bruijn === Subject: Re: Prime numbers within Galois field? posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e SIMBAR Enabled; SIMBAR={70306B22-CB8C-4d52-BFF4-18424E217075}; MathPlayer 2.10b; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > On Feb 14, 10:54 am, Han de Bruijn Once upon a time, I've become somewhat involved with Galois fields: >http://hdebruijn.soo.dto.tudelft.nl/www/programs/delphi.htm#JT >But they remain weird stuff to me. Now I have the following question. >Suppose you're working within a Galois Field, say of order 2^n. Is >it possible then, in principle, to define a kind of prime numbers >within such a field, i.e. numbers that are not divisible by anything >else than 1 and themselves? (Disclaimer: rusty knowledge about it ..) > ************************************************************ > Apparently pretty rusty knowledge here, dear Han: in a field, any > field, ANY element is divisible by any other non-zero element, so the > notion of prime, irreducible and all that is pretty boring. Oh well, then the case would be closed already, before it's opened. > book about the real prime, a rather bizarre idea he had. Perhaps the same bizarre idea as I had: suppose that such real primes > can be defined within a Galois Field (say of order 2^n), then we take > the limit for n -> oo and have the set of all naturals. If there is > a match between our new baked Galois primes and the common primes and > if the Goldbach Conjecture could easily be proved for Galois primes then > we have proved the same conjecture for common primes. But, as I've said, > it's a bizarre idea .. > I haven't yet read that, but I know it made lift an eyebow to some > people in my university back then (the Hebrew University in > Jerusalem). > Perhaps there's something to it, I can't say. Curious anyway .. Han de Bruijn- ************************************************** Well, that's something to think about. Anyway, you talk of defining this real prime, whatever it is, within a GF of order 2^n, and this may pose a serious problem since this field of characteristic 2, whereas reals and, probably, that real prime thing would be in a field of characteristic zero. Without promising anything. I'm gonna try to read a little about it. Perhaps I'll understand something. Tonio === Subject: Re: Prime numbers within Galois field? posting-account=LqooeAoAAACZi9rhWJi6xq-ef5ZN2cZd Web Browser from: http://bsalsa.com/),gzip(gfe),gzip(gfe) > On Feb 14, 10:54æam, Han de Bruijn Once upon a time, I've become somewhat involved with Galois fields: >http://hdebruijn.soo.dto.tudelft.nl/www/programs/delphi.htm#JT > But they remain weird stuff to me. Now I have the following question. > Suppose you're working within a Galois Field, say of order 2^n. Is > it possible then, in principle, to define a kind of prime numbers > within such a field, i.e. numbers that are not divisible by anything > else than 1 and themselves? (Disclaimer: rusty knowledge about it ..) > Han de Bruijn ************************************************************ Apparently pretty rusty knowledge here, dear Han: in a field, any > field, ANY element is divisible by any other non-zero element, so the > notion of prime, irreducible and all that is pretty boring. book about the real prime, a rather bizarre idea he had. > I haven't yet read that, but I know it made lift an eyebow to some > people in my university back then (the Hebrew University in > Jerusalem). > Perhaps there's something to it, I can't say. Tonio Are you able to get a copy or give someone a name? :-) === Subject: Re: Prime numbers within Galois field? >Once upon a time, I've become somewhat involved with Galois fields: http://hdebruijn.soo.dto.tudelft.nl/www/programs/delphi.htm#JT But they remain weird stuff to me. Now I have the following question. >Suppose you're working _within_ a Galois Field, say of order 2^n. Is >it possible then, in principle, to define a kind of prime numbers >within such a field, i.e. numbers that are not divisible by anything >else than 1 and themselves? (Disclaimer: rusty knowledge about it ..) But in any field every nonzero element is divisible by every other one. Said differently, all nonzero elements of a field are units, so there are no irreducible elements. quasi === Subject: Why do they accept the Islam ? posting-account=Oj0DuQoAAAA6OPeaJ-8WQUQZfhPLSyKL SV1),gzip(gfe),gzip(gfe) Why do they accept the Islam ? Why do they accept the Islam ? http://www.youtube.com/watch/v/2PS2creVhaM Yusuf Estes - Priests and Preachers accepting Islam http://www.youtube.com/watch/v/E6K0627FiCk http://www.youtube.com/watch/v/1XyrxzQIN9w http://www.youtube.com/watch/v/34kjurzK500 www.youtube.com/watch/v/WAXXN6XOnzQ Thousands of Danish convert to Islam http://www.youtube.com/watch/v/kru6XQ8CT48 http://www.youtube.com/watch/v/yDwGsmTx3D4 German Scientist & his wife, clinic assistant convert 2 Islam http://www.youtube.com/watch/v/uP-2IqHl4c European Scientist converts to Islam http://www.youtube.com/watch/v/0IspK651RpY Dr. Ian Weber from England converts to Islam http://www.youtube.com/watch/v/gUZR6XwU8Pw science students turn to Islam http://www.youtube.com/watch/v/y3_JYk4Bo4Y http://www.youtube.com/watch/v/Dhu0eEuIsGg Turning Muslim in Texas - People reverting to Islam in Texas Angela Collins - Muslim Convert weeks after 9/11 Incident http://www.youtube.com/watch/v/j6PJgJdEzNM NEW MUSLIM Woman from austrailia CONVERT http://www.youtube.com/watch/v/-baqULx5IBU The Choice - A Story of New American Muslim Convert http://www.youtube.com/watch/v/CML3CRPMefA Irish and loving Islam Convert to Islam http://www.youtube.com/watch/v/dIc5oFAva-4 Jewish To Islam convert http://www.youtube.com/watch/v/KcBiJnLjwVw www.youtube.com/watch/v/IlOuITPE6kE http://www.youtube.com/watch/v/1qpQvpmEqkc Why do they accept the Islam http://www.youtube.com/watch/v/aJ3TGAnFc-U 135 Convert to Islam http://www.youtube.com/watch/v/36GljFAGcw Convert to Islam from Canada http://www.youtube.com/watch/v/uKPer2fma9U http://www.youtube.com/watch/v/bI_YVnD9UvI http://www.youtube.com/watch/v/84ZtVLI5kXM http://www.youtube.com/watch/v/pn0iPlWQNlI Greece Men Convert to Islam http://www.youtube.com/watch/v/IlAjkuECrHc Convert to Islam from Hinduism http://www.youtube.com/watch/v/zePqNxz895U Christian Convert to Islam http://www.youtube.com/watch/v/HDkW2Y35mKQ Twenty two/22 Brothers and sisters convert to (Islam) http://www.youtube.com/watch/v/XQBn6loQTdY Two british Women of different colours convert to Islam www.youtube.com/watch/v/uEfMcPQfv7w German convert to Islam http://www.youtube.com/watch/v/9U1zNXXQA6Q www.youtube.com/watch/v/JiksSo0lwL4 Sister Yvonne Ridley Becomes Muslim Islam Video http://www.youtube.com/watch/v/aOe5s5hP4Gw http://www.youtube.com/watch/v/afFv22Wsd5A http://www.youtube.com/watch/v/os4vUxfJizU Jerome - How I wrestled my way to Islam http://www.youtube.com/watch/v/b2YZGGDGUWE The story of a German convert to Islam http://www.youtube.com/watch/v/gjRjzTAk-RQ Revert to Islam - Abdus Salam (Convert to Islam) http://www.youtube.com/watch/v/VqlilLIQJRE http://www.youtube.com/watch/v/ig0N9aRT0Hc Germans convert to Islam on german TV http://www.youtube.com/watch/v/nFqj3xPKc88 http://www.youtube.com/watch/v/4LrIv6kK9o Jolene: A Southern Baptist Converts to Islam http://www.youtube.com/watch/v/RHQOx12-WJU http://www.youtube.com/watch/v/Co0xomKcXM Revert to Islam - Yusuf Ali (Convert to Islam) http://www.youtube.com/watch/v/NcjXK7qzEk How Melina came to Islam http://www.youtube.com/watch/v/jLht7Kk0bGg Revert to Islam - Abdullah Laban (Convert to Islam) http://www.youtube.com/watch/v/gGeHzUkkJH4 http://www.youtube.com/watch/v/j2w14xxi0bM William Chappelle and 25 members of his family embrace Islam http://www.youtube.com/watch/v/mkBW4l4TmPE Islam In Netherlands 6 German convert to Islam - 2007 - LIVE http://www.youtube.com/watch/v/0SpqpGKp7ts Caroline convert from Christianity to Islam http://www.youtube.com/watch/v/2Cpmvne2wj0 http://www.youtube.com/watch/v/L7PIOhK-SgA American converted Muslim Woman speaking about the Veil http://www.youtube.com/watch/v/L85Mcq3EDX8 Dia Richardson converted to Islam in USA http://www.youtube.com/watch/v/Tp097NNj3Pk http://www.youtube.com/watch/v/UTyu18RgrAo New Muslim Lady .. Live from London http://www.youtube.com/watch/v/NTlta20vsow www.youtube.com/watch/v/UyyxPO0HFLk http://www.youtube.com/watch/v/EkBRKrUDnEU How to Convert to Islam http://www.youtube.com/watch/v/i9K7lmYaucU 4 New Muslims from 4 Corners of the World Jenny - How I came to Islam Cat Stevens becomes Yusuf Islam Islam Youngest Muslim Reverts in The World- in England Why Abdul Raheem Green Came to Islam http://www.youtube.com/watch/v/wF8joJaOVJw Robert converts to Islam http://www.youtube.com/watch/v/vU5HGHiNUu0 Islam : Best and inspirational Revert Story http://www.youtube.com/watch/v/IZF9stLSYjY Spanish Woman talks about Woman rights in Islam : http://www.youtube.com/watch/v/exdCJ_wT9E4 === Subject: Re: A SIMPLE CHALLENGE that you great Mathematicians won't answer... posting-account=LqooeAoAAACZi9rhWJi6xq-ef5ZN2cZd Web Browser from: http://bsalsa.com/),gzip(gfe),gzip(gfe) > ................................................. > Do you know the old definition of insanity - the lunatic doesn't know > anything is wrong? > One definition of wrongful pride is not admitting mistakes. > But when your mistakes are foundational to your thinking - you have to > change.- ********************************************************** *Plonk!* And no: I don't have, and I won't change. Honest. Bye > Tonio Bad Tonico...! xoxox ;-) Did you write that on the toilet? More a 'plunk' I think. Like I said: > Do you know the old definition of insanity - the lunatic doesn't know > anything is wrong? > One definition of wrongful pride is not admitting mistakes. > But when your mistakes are foundational to your thinking - you have to > change. Like you said: Wah!! Adam Lewis === Subject: Re: A SIMPLE CHALLENGE that you great Mathematicians won't answer... posting-account=LqooeAoAAACZi9rhWJi6xq-ef5ZN2cZd Web Browser from: http://bsalsa.com/),gzip(gfe),gzip(gfe) > On Feb 13, 10:32æam, Han de Bruijn Sorry, I thought integers were supposed to be real rationals... > Sometimes yes, sometimes no. Depends on whether you are working in an > integer / discrete æor in a æfloating point / continuous æenvironment. > What are you trying to say? > Hey 'finite guy' ! Why do you think, or rather desire it seems, that > everybody is against you ? > Han de Bruijn *************************************************************** Aaaah! Hey, HdB: trying to add this finite, weird, guy to your forces? > Why is everyone against him? Yes, why, indeed...? Tonio Tinio, Tonio, Tonico... which are you? Which are you... now? I'm only about 5'6, so you could say, finite, weird, little guy. More apt. I guess HbD must be on the Dark Side of the force, eh? I bet you just want to be a Jedi - Tonocio Wan. Careful how you handle your lightsabre... Love and Kisses - it's Valentine's Day. Adam Lewis. === Subject: Re: A SIMPLE CHALLENGE that you great Mathematicians won't answer... > Give Han a break. Finite guy is far, far sillier than Han. Yes, HdB at least *tries* to do mathematics, even if he gets it wrong a lot of the time. Finite guy is incoherent enough to be a chatbot. - Tim === Subject: Re: A SIMPLE CHALLENGE that you great Mathematicians won't answer... > >Give Han a break. Finite guy is far, far sillier than Han. >Yes, HdB at least *tries* to do mathematics, even if he gets it wrong > a lot of the time. Finite guy is incoherent enough to be a chatbot. Sure. But I also get it _right_ much of the time. You can't have the one without the other. You don't believe it? Mind the shortest threads! E.g. Waar gehakt wordt vallen spaanders (You cannot make an omelette without breaking eggs. Literally: you can't cleave wood without chips falling or something like that ..) Han de Bruijn === Subject: Re: A SIMPLE CHALLENGE that you great Mathematicians won't answer... posting-account=LqooeAoAAACZi9rhWJi6xq-ef5ZN2cZd Web Browser from: http://bsalsa.com/),gzip(gfe),gzip(gfe) Why do the lines between points have to be straight lines? > Theoretically there can be æNO line; because a line conists > of more points! Originally, I was thinking of delta x & y as the x- & y-distances > between two points on a curve. æBack in my academic youth > (circa 1945 -1956) we had dx & dy which had something to do > with delta -x, -y æ This should help explain my confusion! Bill j A good, simple question. Why indeed... My poor answer will but the focus of some peoples derision but here goes. Firstly, the delta/change in x and y is not the distance between the points of the circle. They are of the axes. Calculus either 'divides to zero' or 'multiplies to infinity' - which is only nonsense. But that is the problem - infinitisation of the finite in all 'directions'. This leads to the acceptance of the fiat statement that 'space is infinite' and other such follies. > Theoretically there can be NO line; because a line conists of more points! Why do you even think that a 'line' is made up of 'points'? Because... Surely, in theoretical terms, even points are void. But all of these things are convenient expressions - useful. However, we are talking about the idea of going from one point to the NEXT point - none between. The idea that you can always put more points 'between' is a reflection of the misconception of the infinitisation of the finite. In practical terms, points are the only physical way we can construct - successive marks on a piece of paper, pixels on a computer screen. The mathematicians here aggressively say that mathematics and reality are not related. But regardless of their pontifications and postulations, everyone is bound by reality. In mathematical terms, the axes themselves must be 'straight lines' to be independent of each other. We go 'from' one 'to' another. A straight line conceptually means that that we are not going 'via' elsewhere which would introduce other factors. I hope that helps. I'm sure that the guys here can elaborate much more than I but they will still be expressing things which encapsulate their misconceptions. Let's wait and see. Here is another question. Why does the Pythagoras Theorem use 'squares'...? Fermat shows that 'cubes' don't work... Send me an email and I'll show you why Pythagoras is conceptually incorrect for E=mc^2. === Subject: Re: A SIMPLE CHALLENGE that you great Mathematicians won't answer... > However, we are talking about the idea of going from one point to > the NEXT point - none between. See, I told you he was going to start talking about some crank theory of successive points. Let's see what he claims is the NEXT point on the x axis after the point where it intersects the y axis. - Tim === Subject: Re: A SIMPLE CHALLENGE that you great Mathematicians won't answer... posting-account=LqooeAoAAACZi9rhWJi6xq-ef5ZN2cZd Web Browser from: http://bsalsa.com/),gzip(gfe),gzip(gfe) On Feb 14, 5:02æpm, Tim Little However, we are talking about the idea of going from one point to > the NEXT point - none between. See, I told you he was going to start talking about some crank theory > of successive points. Let's see what he claims is the NEXT point on the x axis after the > point where it intersects the y axis. - Tim Tim I think you may have me mixed up with someone else. I've only ever used my current profile which includes my name. Must be two different people that you think are cranks. Adam Lewis === Subject: Re: A SIMPLE CHALLENGE that you great Mathematicians won't answer... posting-account=4n0P8QoAAACPj0DJnja1mCT1wUiU6txx Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > Firstly, the delta/change in x and y is not the distance between the > points of the circle. > They are of the axes. And irrelevant to the description of the circle. All that's needed to describe a circle is: x^2 + y^2 = r^2 > Calculus either 'divides to zero' or 'multiplies to infinity' - which > is only nonsense. Well, when you look at it that way... Oh crap, this really does have no bearing on reality whatsoever. Errr... That was what you meant, right? Even though it's not true; calculus is an important application in physics. > But that is the problem - infinitisation of the finite in all > 'directions'. Ever heard of fractals? They are infinite in complexity but finite in size. They're mathematical concepts, NOT reality. Reality can only represent them to a certain extent. But the mathematical concept still exists. The fact that I'm talking about it proves that. > This leads to the acceptance of the fiat statement that 'space is > infinite' and other such follies. Who said that? I certainly don't believe it. > Theoretically there can be NO line; because a line conists of more points! Why do you even think that a 'line' is made up of 'points'? By definition. > Because... > Surely, in theoretical terms, even points are void. You gotta start somewhere, otherwise you'll get nowhere. In this case, we're starting at the concept of points and sets of points, ie shapes (those sets, by the way, are, in the vast majority of cases, INFINITE). > But all of these things are convenient expressions - useful. > However, we are talking about the idea of going from one point to the > NEXT point - none between. Mathematically impossible with the most commonly accepted premises. > The idea that you can always put more points 'between' > is a reflection of the misconception of the infinitisation of the > finite. Do you even accept the existence of real numbers? > In practical terms, points are the only physical way we can construct > successive marks on a piece of paper, pixels on a computer screen. Coincidentally, points are also a mathematical concept. > The mathematicians here aggressively say that mathematics and reality > are not related. For a good reason. In case you haven't noticed, math has a tendency to go on tangents completely unrelated to reality. > But regardless of their pontifications and postulations, everyone is > bound by reality. No, sir, YOU are bound by reality. You're reality, to be exact, which exists only in your head. > In mathematical terms, the axes themselves must be 'straight lines' to > be independent of each other. And perpendicular, don't forget perpendicular. > We go 'from' one 'to' another. On paper, yes. On your graphing calculator, yes. Conceptually, yes. Mathematically, no, the actual relation (be it a line or a circle or a parabola or anything else) is not composed of successive points. If it were, then there would be a finite number of points. Which there isn't. > A straight line conceptually means that that we are not going 'via' > elsewhere which would introduce other factors. Ummm... What? > I hope that helps. Nope. > I'm sure that the guys here can elaborate much more than I > but they will still be expressing things which encapsulate their > misconceptions. What misconceptions? You mean of reality? Who claimed that any of this had anything to do with reality? > Let's wait and see. Here is another question. > Why does the Pythagoras Theorem use 'squares'...? Because that's how it works out. You can actually prove the theorem, you know. > Fermat shows that 'cubes' don't work... Irrelevant and incorrect. The existence of Pythagorean triples has no bearing on the truth of the Pythagorean theorem. > Send me an email and I'll show you why Pythagoras is conceptually > incorrect for E=mc^2. Pythagorean <=> E=mc^2??? You're connecting this guy to someone over 2 millennia ahead of his time??? I'm not going to argue with you anymore; it's obvious to me that you're not about to give up on your misconceptions. The fact that I have had a complete, logical and valid response to every one of your misconceptions doesn't seem to deter you. === Subject: Re: A SIMPLE CHALLENGE that you great Mathematicians won't answer... > >Firstly, the delta/change in x and y is not the distance between the >points of the circle. >They are of the axes. >And irrelevant to the description of the circle. All that's needed to > describe a circle is: > x^2 + y^2 = r^2 Minor nitpicking .. The original (geometric) definition of the circle is the place (nowadays most people say: the set) of all points equidistant to a given point. In algebraic terms: sqrt(x^2 + y^2) = r with (x,y) e R , r > 0 and r e R . The original definition is more appropriate when making a visualization. Han de Bruijn === Subject: Re: A SIMPLE CHALLENGE that you great Mathematicians won't answer... <4ed86$47b3fbd1$82a1e228$32468@news2.tudelft.nl> posting-account=LqooeAoAAACZi9rhWJi6xq-ef5ZN2cZd Web Browser from: http://bsalsa.com/),gzip(gfe),gzip(gfe) >Firstly, the delta/change in x and y is not the distance between the >points of the circle. >They are of the axes. > And irrelevant to the description of the circle. All that's needed to > describe a circle is: > x^2 + y^2 = r^2 Minor nitpicking .. The original (geometric) definition of the circle is > the place (nowadays most people say: the set) of all points equidistant > to a given point. In algebraic terms: æ æsqrt(x^2 + y^2) = r æwith æ(x,y) e R , r > 0 and r e R . The original definition is more appropriate when making a visualization. Han de Bruijn Coll. x and y are still axial though, are they not? === Subject: Re: A SIMPLE CHALLENGE that you great Mathematicians won't answer... <4ed86$47b3fbd1$82a1e228$32468@news2.tudelft.nl> posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e InfoPath.1; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) On Feb 14, 10:29æam, Han de Bruijn Firstly, the delta/change in x and y is not the distance between the >points of the circle. >They are of the axes. > And irrelevant to the description of the circle. All that's needed to > describe a circle is: > x^2 + y^2 = r^2 Minor nitpicking .. The original (geometric) definition of the circle is > the place (nowadays most people say: the set) of all points equidistant > to a given point. In algebraic terms: æ æsqrt(x^2 + y^2) = r æwith æ(x,y) e R , r > 0 and r e R . The original definition is more appropriate when making a visualization. Han de Bruijn ***************************************************************** Originally?? It still is the usual definition of circle in analytic geometry, which takes us at once to the usual analytical description of the circle centered at (a,b) and with radius r: (x - a)^2 + (y - b)^2 = r^2 , if we're working with the euclidean plane and with the euclidean distance. Tonio Pd Have you HdB, read the last posts by that eggregius finite guy? Now common, be honest: what do you really think of all that? I mean, besides all the entertainment and the laughs, that is. === Subject: Re: A SIMPLE CHALLENGE that you great Mathematicians won't answer... > On Feb 14, 10:29 am, Han de Bruijn Firstly, the delta/change in x and y is not the distance between the >points of the circle. >They are of the axes. >And irrelevant to the description of the circle. All that's needed to >describe a circle is: >x^2 + y^2 = r^2 >Minor nitpicking .. The original (geometric) definition of the circle is >the place (nowadays most people say: the set) of all points equidistant >to a given point. In algebraic terms: > sqrt(x^2 + y^2) = r with (x,y) e R , r > 0 and r e R . >The original definition is more appropriate when making a visualization. >***************************************************************** >Originally?? It still is the usual definition of circle in analytic > geometry, which takes us at once to the usual analytical description > of the circle centered at (a,b) and with radius r: > (x - a)^2 + (y - b)^2 = r^2 , if we're working with the euclidean > plane and with the euclidean distance. Quoted from an earlier posting in the same thread: > by convoluting the delta function representing the circle in an exact > image (where it is _invisible_) with a Gaussian distribution function. > Programmed in (Delphi) Pascal as follows: > function G(x,y,sigma : integer) : double; > begin > C := sqrt(sqr(x-a)+sqr(y-b))-R; > G := exp(-sqr(C/sigma)/2); > end; Mind the formula (C) in there. Together with the spread (sigma) it forms a dimensionless quantity, as _should_ be, when plugged into the 'exp' of the formula (G). Such a dimensionless quantity cannot be formed with the other formula: (x - a)^2 + (y - b)^2 = r^2 . Okay? See what I mean? > Form1.Image1.Canvas.Pixels[x,y] := Grijs(G(x,y,2)); >Where the 'Grijs' function is black for its argument = 1 and white for > its argument = 0 . Mind the grey / gray scales in between, because they > are representing the _fuzzyfied_ equality. >Here is the picture (for a spread in the Gaussian = 2 pixels): >http://hdebruijn.soo.dto.tudelft.nl/jaar2008/cirkel.jpg Continuing with the present posting: > Pd Have you HdB, read the last posts by that eggregius finite guy? Now > common, be honest: what do you really think of all that? Well, perhaps he is like the _young_ HdB, somewhat less mature, that is: http://hdebruijn.soo.dto.tudelft.nl/www/grondig/sci_math.htm Maybe the year (12.) 1989 reveals something alike 'finitist' behaviour. > I mean, besides all the entertainment and the laughs, that is. The truth is that I'm not really amused by this. Think you understand. Han de Bruijn === Subject: Re: A SIMPLE CHALLENGE that you great Mathematicians won't answer... posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e SIMBAR Enabled; SIMBAR={70306B22-CB8C-4d52-BFF4-18424E217075}; MathPlayer 2.10b; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) On Feb 14, 1:41æpm, Han de Bruijn I mean, besides all the entertainment and the laughs, that is. The truth is that I'm not really amused by this. Think you understand. Han de Bruijn- *********************************************************** Yes, that's what I thought, Han...*sigh*. Well, I hoped...you know. Tonio === Subject: Re: A SIMPLE CHALLENGE that you great Mathematicians won't answer... posting-account=LqooeAoAAACZi9rhWJi6xq-ef5ZN2cZd Web Browser from: http://bsalsa.com/),gzip(gfe),gzip(gfe) > On Feb 14, 10:29æam, Han de Bruijn Firstly, the delta/change in x and y is not the distance between the >points of the circle. >They are of the axes. > And irrelevant to the description of the circle. All that's needed to > describe a circle is: > x^2 + y^2 = r^2 > Minor nitpicking .. The original (geometric) definition of the circle is > the place (nowadays most people say: the set) of all points equidistant > to a given point. In algebraic terms: > æ æsqrt(x^2 + y^2) = r æwith æ(x,y) e R , r > 0 and r e R . > The original definition is more appropriate when making a visualization. > Han de Bruijn ***************************************************************** Originally?? æIt still is the usual definition of circle in analytic > geometry, which takes us at once to the usual analytical description > of the circle centered at (a,b) and with radius r: > (x - a)^2 + (y - b)^2 = r^2 , if we're working with the euclidean > plane and with the euclidean distance. Tonio Pd Have you HdB, read the last posts by that eggregius finite guy? Now > common, be honest: what do you really think of all that? > I mean, besides all the entertainment and the laughs, that is.- Hide quoted text - - Show quoted text - Baby It's valentines day. Don't be like that... > Pd Have you HdB, read the last posts by that eggregius finite guy? Now > common, be honest: what do you really think of all that? > I mean, besides all the entertainment and the laughs, that is. === Subject: Re: A SIMPLE CHALLENGE that you great Mathematicians won't answer... posting-account=LqooeAoAAACZi9rhWJi6xq-ef5ZN2cZd Web Browser from: http://bsalsa.com/),gzip(gfe),gzip(gfe) > Here we go again... > And irrelevant to the description of the circle. All that's needed to > describe a circle is: > x^2 + y^2 = r^2 Gee, I thought x and y were axial. Are you saying they are not? > Well, when you look at it that way... Oh crap, this really does have > no bearing on reality whatsoever. Errr... That was what you meant, > right? Even though it's not true; calculus is an important application > in physics. Really... what do as x -> .81.87 mean then... ? and the inverse of that... ? > Ever heard of fractals? They are infinite in complexity but finite in > size. They're mathematical concepts, NOT reality. Reality can only > represent them to a certain extent. But the mathematical concept still > exists. The fact that I'm talking about it proves that. Fractals are a scaling issue. And they go on infinitly reducing... according to the mathematical concept? What if you 'start' at the minimum and fractal ' upwards'... to a finite degree? Gee, same thing happens... > Who said that? I certainly don't believe it. Fair enough - it was a generalisation. Generally, we all use them... :-) So you believe with certainty that space is finite - 'inward' and 'outward'? > By definition. Your errant definitions are the problem... that's what I have been saying. Did you miss that? > You gotta start somewhere, otherwise you'll get nowhere. In this case, > we're starting at the concept of points and sets of points, ie shapes > (those sets, by the way, are, in the vast majority of cases, > INFINITE). The infinite DOESNT START ANYWHERE or finish... Without realising it you are muddling finite and infinite... again... If I said, (those sets, by the way, are, in the vast majority of cases, BLUE), you would critisize. But 'blue' is as much a 'number' as infinite. Infinity is not a number. Would COW also be as appropriate a description as infinite... ? > Mathematically impossible with the most commonly accepted premises. Because you futilely try to infinitise the finite... Your premise holds no promise... :-) > Do you even accept the existence of real numbers? Yes, really... Both rational and irrational ones. They shouldn't try to be put on the same number line though. That's been discussed previously. I think it was 'lwas?' who pointed out that that is correct. > Coincidentally, points are also a mathematical concept. So is a line, a plane, a cube, etc. Your point being... ? > For a good reason. In case you haven't noticed, math has a tendency to > go on tangents completely unrelated to reality. Yeah, and they are presumably based on the fundamentals... 1, 2, 3... Why do they try to negate the fundamentals? Oh, philosophy is why... not mathematics. True... ? > No, sir, YOU are bound by reality. You're reality, to be exact, which > exists only in your head. Yes, sir. I am bound by reality as best I understand it. But YOU are proudly not bound by reality... Haven't you already said that... ? > And perpendicular, don't forget perpendicular. > On paper, yes. On your graphing calculator, yes. Conceptually, yes. > Mathematically, no, the actual relation (be it a line or a circle or a > parabola or anything else) is not composed of successive points. If it > were, then there would be a finite number of points. Which there > isn't. So, once again, you confirm your belief that finite is infinite... This is the problem... Is that getting through... ? > Ummm... What? Sounded simple. Read it again. > Nope. There is always hope... > What misconceptions? You mean of reality? Who claimed that any of this > had anything to do with reality? The one we have been discussing... oops, you can't hear it. Infinitising the finite and flying off into non-reality. The real question is: Are YOU interested in reality since the foundations are built upon it... ? > Because that's how it works out. You can actually prove the theorem, > you know. Ha, ha, ha... Why 'squares' was the question. You are thinking that it is a 'square' relationship. It isn't - it is an 'area' relationship, silly. Is proved... is good... I agree. It still has a misconception... squares... Circle area is pi*r^2. Gosh, that's not a square... is it? > Irrelevant and incorrect. The existence of Pythagorean triples has no > bearing on the truth of the Pythagorean theorem. Did I mention triplets? The reason I mentioned Fermat is that a 'cube' is an extended 'square'. And Fermat is correct - meaning that there is a problem for you to go from (1,1,1) to (2,2,2). BUT 'squares' have no bearing on the truth of the PT... do they... ? > Pythagorean <=> E=mc^2??? You're connecting this guy to someone over 2 > millennia ahead of his time??? Come on.. this is a dumb statement... Of course they are connected. Bear in mind that Pythagoras was way ahead of BOTH YOU AND ME. Are you superior to Pythagoras such that you might call him dumb-ass? I think not... > I'm not going to argue with you anymore; it's obvious to me that > you're not about to give up on your misconceptions. The fact that I > have had a complete, logical and valid response to every one of your > misconceptions doesn't seem to deter you. My misconceptions are actually 're-conceptions'... No one thinks that their held opinion is illogical or invalid. As you can see above, you misconceive my misconceptions... :-) Hopefully, you will reply. If you do reply, please tell me why E=mc^2 is a planar equation in triangular form. Adam Lewis. === Subject: Re: A SIMPLE CHALLENGE that you great Mathematicians won't answer... posting-account=4n0P8QoAAACPj0DJnja1mCT1wUiU6txx Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > You seem to think that 'deepness' should be unrelated to truth. > While I do agree with this statement in general, Scratch that. I do agree that 'deepness' *can* be unrelated to truth. Whoops! === Subject: Re: A SIMPLE CHALLENGE that you great Mathematicians won't answer... posting-account=4n0P8QoAAACPj0DJnja1mCT1wUiU6txx Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > Not being a great mathematician, I > don't understand why delta x and > delta y have to be zero. Can you > explain why this is necessary? > Easy. > Your mathematical enlightenment is overwhelming. Your anal-ysis is underwhelming... No doubt you were being sarcastic just like I was. === === === === === === === === === Subject: Various solution manuals posting-account=I4LMEgoAAABgwkvbBt_YLmArZhfVF8gI Gecko/20070508 Firefox/1.5.0.12,gzip(gfe),gzip(gfe) I have the solutions manuals for these books. If you are interested in any of them, contact me at booksofinterest (at) gmail (dot) com booksofinterest@gmail.com Only Paypal payments accepted Please check first if the solution manual you want is in the list, and then contact me instead of posting here. * Solutions manual for Brief Calculus and Its Applications (11th Ed., Larry J Goldstein, David I Schneider, David I. Lay & Nakhle Asmar) ISBN-10: 0131919652 ISBN-13: 9780131919655 * Solutions manual for Finite Math and Its Application (9th Ed., Larry J Goldstein, David I Schneider & Martha J. Siegel) ISBN-10: 0131873644 ISBN-13: 9780131873643 * Solutions manual for Management Accounting, 5/E Anthony A. Atkinson, University of Waterloo Robert S. Kaplan, Ella Mae Matsumura, S. Mark Young ISBN-10: 0136005314 ISBN-13: 9780136005315 * Solutions manual for Plastics: Materials and Processing, 3/E , by A. Brent Strong ISBN-10: 0131145584 ISBN-13: 9780131145580 * Solutions manual for Economics: A Tool for Critically Understanding Society, 8/E , Tom Riddell, Jean A Shackelford, Steve C. Stamos, Geoffrey Schneider ISBN-10: 0321423585 ISBN-13: 9780321423580 * Solutions manual for Engineering Materials: Properties and Selection, 8/E, by Ken Budinski, Michael K. Budinski SBN-10: 0131837796 ISBN-13: 9780131837799 * Solutions manual for Work Systems: The Methods, Measurement & Management of Work, by Mikell P. Groover ISBN-10: 0131406507 ISBN-13: 9780131406506 * Solutions manual for Applied Multivariate Statistical Analysis (6th Ed., Johnson & Wichern) ISBN-10: 0131877151 ISBN-13: 9780131877153 * Solutions manual for Financial Markets and Institutions, 6/E, by Frederic S. Mishkin, Stanley G. Eakins ISBN-10: 0321374215 ISBN-13: 9780321374219 * Solutions manual for The Economics of Money, Banking and Financial Markets, Alternate Edition Frederic S. Mishkin (1st /e) ISBN-10: 0321427807 ISBN-13: 9780321427809 * Solutions manual for The Economics of Money, Banking and Financial Markets, 8/E, Frederic S. Mishkin ISBN-10: 0321415051 ISBN-13: 9780321415059 * Solutions manual for Principles of Risk Management and Insurance, 10/E , George E. Rejda ISBN-10: 0321414934 ISBN-13: 9780321414939 * Solutions manual for Derivatives Markets, 2/E, Robert L. McDonald ISBN-10: 032128030X ISBN-13: 9780321280305 * Solutions manual for Fundamentals of Investing, 10/E, Lawrence J. Gitman, Michael D. Joehnk ISBN-10: 0321489381 ISBN-13: 9780321489388 * Solutions manual for Introduction to Finance 1st/e, by Lawrence J. Gitman, Jeff Madura ISBN-10: 0201635372 ISBN-13: 9780201635379 * Solutions manual for Principles of Managerial Finance, 12/E Lawrence J. Gitman ISBN-10: 0321557530 ISBN-13: 9780321557537 * Solutions manual for Cases in International Finance, 2/E , Gunter Dufey, Ian H. Giddy ISBN-10: 0201513072 ISBN-13: 9780201513073 * Solutions manual for Fundamentals of Multinational Finance, 3/E ISBN-10: 0321549244 ISBN-13: 9780321549242 * Solutions manual for Multinational Business Finance, 11/E , David K. Eiteman, Arthur I. Stonehill, Michael H. Moffett ISBN-10: 0321357965 ISBN-13: 9780321357960 * Solutions manual for Theory of Asset Pricing, George Pennacchi ISBN-10: 032112720X ISBN-13: 9780321127204 * Solutions manual for Corporate Finance, Jonathan Berk, Peter DeMarzo ISBN-10: 0321415116 ISBN-13: 9780321415110 * Solutions manual for Personal Finance with Financial Planning Software, 3/E, Jeff Madura ISBN-10: 0321409965 ISBN-13: 9780321409966 * Solutions manual for Mathematical Methods for Economics, 2/E , by Michael Klein ISBN-10: 0201726262 ISBN-13: 9780201726268 * Solutions manual for Using Econometrics: A Practical Guide, 5/E , by A.H. Studenmund ISBN-10: 0321316495 ISBN-13: 9780321316493 * Solutions manual for Econometrics: A Modern Introduction Michael P. Murray ISBN-10: 0321113616 ISBN-13: 9780321113610 * Solutions manual for Introduction to Econometrics, Brief Edition, by James H. Stock, Mark W. Watson ISBN-10: 0321432517 ISBN-13: 9780321432513 * Solutions manual for Introduction to Econometrics, 2/E, James H. Stock, Mark W. Watson ISBN-10: 0321278879 ISBN-13: 9780321278876 * Solutions manual for The Economics of Sports, 3/E Michael A. Leeds, Peter von Allmen ISBN-10: 0321415566 ISBN-13: 9780321415561 * Solutions manual for Economics, by Michael A. Leeds, Peter von Allmen, Richard C. Schiming ISBN-10: 0321461673 ISBN-13: 9780321461674 * Solutions manual for Macroeconomics MyEconLab Homework Edition plus Themes of the Times booklet Michael A. Leeds, Peter von Allmen, Richard C. Schiming ISBN-10: 0321492331 ISBN-13: 9780321492333 * Solutions manual for Microeconomics MyEconLab Homework Edition plus Themes of the Times Booklet By Michael A. Leeds, Peter von Allmen, Richard C. Schiming ISBN-10: 032149234X ISBN-13: 9780321492340 * Solutions manual for Public Finance and the American Economy, 2/E , by Neil Bruce ISBN-10: 0321078152 ISBN-13: 9780321078155 * Solutions manual for Market Regulation, by Roger Sherman ISBN-10: 0321322320 ISBN-13: 9780321322326 * Solutions manual for Law and Economics, 5/E , by Robert Cooter, Thomas Ulen - suggested answers ISBN-10: 0321336348 ISBN-13: 9780321336347 * Solutions manual for Modern Industrial Organization, 4/E , Dennis W. Carlton, Jeffrey M. Perloff ISBN-10: 0321180232 ISBN-13: 9780321180230 * Solutions manual for Microeconomics plus MyEconLab plus eBook 1- semester Student Access Kit, 4/E Jeffrey M. Perloff ISBN-10: 0321414527 ISBN-13: 9780321414526 * Solutions manual for Microeconomics: Theory and Applications with Calculus, by Jeffrey M. Perloff ISBN-10: 0321277945 ISBN-13: 9780321277947 3/E, by Don E. Waldman, Elizabeth J. Jensen ISBN-10: 0321376102 ISBN-13: 9780321376107 * Solutions manual for Microeconomics Plus MyEconLab Student Access Kit, by Don E. Waldman ISBN-10: 0321205278 ISBN-13: 9780321205278 * Solutions manual for Health Economics, 3/E, Charles E. Phelps ISBN-10: 032106898X ISBN-13: 9780321068989 * Solutions manual for Environmental Economics and Policy, 5/E , by Tom Tietenberg ISBN-10: 0321348907 ISBN-13: 9780321348906 * Solutions manual for Environmental and Natural Resource Economics, 7/E , by Tom Tietenberg ISBN-10: 0321305043 ISBN-13: 9780321305046 If you are interested in any of them, contact me at booksofinterest (at) gmail (dot) com booksofinterest@gmail.com Only Paypal payments accepted === Subject: Rigid implies affine? Let f be a function from a real inner product space V to V. f is affine <=> for all x, y in V, t in R: f((1 - t)x + ty) = (1 - t)f (x) + tf(y) f is rigid <=> for all x, y in V: |f(x) - f(y)| = |x - y| where |*| is the norm induced by the inner product. Does rigidity imply affinity? === Subject: Re: Rigid implies affine? Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > Let f be a function from a real inner product space V to V. f is affine <=> for all x, y in V, t in R: f((1 - t)x + ty) = (1 - t)f > (x) + tf(y) f is rigid <=> for all x, y in V: |f(x) - f(y)| = |x - y| where |*| is the norm induced by the inner product. Does rigidity imply affinity? Yes. (You are obviously talking about the finite-dimensional case, otherwise the norm doesn't make sense). Consider the distance of (1-t)x +ty from x and y (on a case-by-case basis, depending on whether t<=0 or 0=1), and use this to determine f((1-t)x+ty). === Subject: Re: Rigid implies affine? > Yes. (You are obviously talking about the finite-dimensional case, > otherwise the norm doesn't make sense). Consider the distance of (1-t)x > +ty from x and y (on a case-by-case basis, depending on whether t<=0 > or 0=1), and use this to determine f((1-t)x+ty). is the norm axiom |x| = 0 <=> x = 0. Here's what I get: |(1 - t)x + ty - x| = |-tx + ty| = |t||x - y| = |t||f(x) - f(y)| = |f((1 - t)x + ty) - f(x)| |(1 - t)x + ty - y| = |(1 - t)x - (1 - t)y| = |1 - t||x - y| = |1 - t||f(x) - f(y)| = |f((1 - t)x + ty) - f(y)| But what to do with these? === Subject: Re: Rigid implies affine? TBP_6.1_AC; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) > Yes. (You are obviously talking about the finite-dimensional case, > otherwise the norm doesn't make sense). Consider the distance of (1-t)x > +ty from x and y (on a case-by-case basis, depending on whether t<=0 > or 0=1), and use this to determine f((1-t)x+ty). is the norm axiom |x| = 0 <=> x = 0. Here's what I get: |(1 - t)x + ty - x| > = |-tx + ty| > = |t||x - y| > = |t||f(x) - f(y)| > = |f((1 - t)x + ty) - f(x)| |(1 - t)x + ty - y| > = |(1 - t)x - (1 - t)y| > = |1 - t||x - y| > = |1 - t||f(x) - f(y)| > = |f((1 - t)x + ty) - f(y)| But what to do with these? This gives you information about the distance from f((1-t)x+ty) from f(x) and f(y). You need an argument that therefore it must be (1-t)f(x) +tf(y). Use an orthonormal basis. === Subject: Re: Rigid implies affine? > Let f be a function from a real inner product space V to V. > f is affine <=> for all x, y in V, t in R: f((1 - t)x + ty) = (1 - t)f > (x) + tf(y) > f is rigid <=> for all x, y in V: |f(x) - f(y)| = |x - y| > where |*| is the norm induced by the inner product. > Does rigidity imply affinity? Yes. (You are obviously talking about the finite-dimensional case, >otherwise the norm doesn't make sense). ??? Any inner product space has a norm arising from the inner product, whether it's finite dimensional or not. >Consider the distance of (1-t)x >+ty from x and y (on a case-by-case basis, depending on whether t<=0 >or 0=1), and use this to determine f((1-t)x+ty). David C. Ullrich === Subject: Re: Rigid implies affine? Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > Let f be a function from a real inner product space V to V. > f is affine <=> for all x, y in V, t in R: f((1 - t)x + ty) = (1 - t)f > (x) + tf(y) > f is rigid <=> for all x, y in V: |f(x) - f(y)| = |x - y| > where |*| is the norm induced by the inner product. > Does rigidity imply affinity? >Yes. (You are obviously talking about the finite-dimensional case, >otherwise the norm doesn't make sense). ??? Any inner product space has a norm arising from the inner > product, whether it's finite dimensional or not. >Consider the distance of (1-t)x >+ty from x and y (on a case-by-case basis, depending on whether t<=0 >or 0=1), and use this to determine f((1-t)x+ty). David C. Ullrich You're right. Sorry. But my argument still works. Because you can pick a basis for the entire space, and then given x and y, the co-ordinates of x and y will be nonzero with respect to only finitely many elements of the basis, and then you can do the argument in the corresponding finite-dimensional subspace. === Subject: Re: Rigid implies affine? Rupert a .8ecrit : > Yes. (You are obviously talking about the finite-dimensional case, > otherwise the norm doesn't make sense). An inner product always induces a norm, no matter what the dimension is. === Subject: Re: Condition for irreducibility? posting-account=-PngCgkAAAD2yUjosqWv1Nf1lkqWP4lp Gecko/20071128 SUSE/2.0.0.11-3.1 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) >Is there an algebraic condition on the coefficients >of a polynomial f(x) in k[x] which will determine >whether f(x) is irreducible or not? > It depends on the field. > For finite fields, yes. > For an algebraically closed field, yes -- trivially. > For k = Q, yes if deg(f) <= 2, no otherwise. > To understand the issue, take k = Q and consider the situation for > polynomials of degree 3. > We can _test_ a cubic in Q[x] for reducibility since for degree 3, > reducibility is equivalent to having a rational root. Thus, the > rational root test provides a _procedure_ for determining > reducibility, but not a general algebraic condition on the > coefficients. > quasi >I am puzzled by what this question means. What is meant by an >algebraic condition, and what is that condition when deg(f) = 2? As far as what the OP meant by an algebraic condition on the > coefficients, Tim would need to answer that. But I'll give a few possible interpretations ... For a fixed positive integer n, let f(x) be a univariate polynomial of > degree n, with coefficients in a field K. Note -- the OP didn't specify that irreducibility was to be > interpreted as irreducibility over a field, but for simplicity, > let's assume that's what he meant. Since the case of a finite field is easily resolved, assume K is an > infinite field. We can assume, without loss of generality, that f is monic. Thus, > write f(x) = x^n + ... + (a_1)*x + a_0, where the coefficients are regarded as unknown elements of K. Intuitively, any interpretation of algebraic condition on a_0, ..., > a_(n-1) should mean a logical statement about a_0, ..., a_(n-1) from > some specified class of allowable statements. Of course, to make that > precise we need to specify the class of allowable statements. Let R = K(t_1, ..., t_n) be the field of rational functions in the > indeterminates t_1, ..., t_n with coefficients in K. Let me reemphasize -- n is a _fixed_ positive integer. Thus we might > have one set of conditions for n = 2, and a different set of > conditions for n = 3. Here are two interpretations of the phrase an algebraic condition on a_0, ..., a_(n-1) (1) strict interpretation -- based only on equations in the > coefficients An algebraic condition on a_0, ..., a_(n-1) is a boolean expression > (in other words, a quantifier free logical statement) where the atomic > statements all have the form p(a_0, a_1, ..., a_(n-1)) = 0 for some p > in R. Remark: For our purposes, this interpretation is too strict. For > example, even for K = Q, n = 2, there is no such condition which can > be used to characterize reducibility. (2) more lenient interpretation -- also allow m'th-power testing In this interpretation we add, for each p in R, and each positive > integer m > 1, the atomic statements p(a_0, a_1, ..., a_(n-1)) is an m'th power of an element of K Remarks: With this interpretation, we get an algebraic condition for > reducibility when n = 2, namely (a_1)^2 - 4(a_0) is a square in K. However I claim that, for n > 2, there are no such conditions. This > claim is just a conjecture -- I don't have a proof, but it's a pretty > safe conjecture. It was this second interpretation which was implicit in my previous > reply. quasi Your conjecture is very likely to be correct, although I am not convinced that your definition of an algebraic condition is a natural one, since you allow testing for being a square. The cubic f(x) = ax^3 + bx^2 + cx + d in Z[x] is reducible over Q if and only if there are integers p and q such that p|d, q|a, and f(p/q) = 0. Some might consider that to be an algebraic condition, although personally I remain sceptical as to whether the whole concept of an algebraic condition is natural or useful in this context. Derek Holt. === Subject: Re: trigonometry posting-account=QkeQNwoAAAAKcvrQ88921Di94d6pTMZe 5.1),gzip(gfe),gzip(gfe) If they do not exist we have to prove it iguess === Subject: Re: trigonometry >If they do not exist we have to prove it iguess Who are you replying to? Please learn how to quote the relevant parts of the prior message. quasi === Subject: Re: trigonometry <2al6r311b155akehp4329hldc3h0b9o3e1@4ax.com> posting-account=QkeQNwoAAAAKcvrQ88921Di94d6pTMZe 5.1),gzip(gfe),gzip(gfe) 100% true, if not necessary integer, but what about integer? === Subject: Re: trigonometry posting-account=QkeQNwoAAAAKcvrQ88921Di94d6pTMZe 5.1),gzip(gfe),gzip(gfe) sin36+sin72+sin72>2 === Subject: Re: trigonometry posting-account=QkeQNwoAAAAKcvrQ88921Di94d6pTMZe 5.1),gzip(gfe),gzip(gfe) means in grades, not in radians - A+B+C=180 grades === Subject: Re: trigonometry > means in grades, not in radians - A+B+C=180 grades Please, in English say degrees not grades. There is a little-used system grads in which 100 grads = 90 degrees. [Promulgated along with the metric system: everything in powers of 10, you know.] -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: trigonometry >means in grades, not in radians - A+B+C=180 grades Firstly, please quote the relevant portions of the prior message. Readers should be able to understand a reply without having to find and open prior message (which they might have already deleted). In any case, you're way behind in this thread. The OP already revised the requirements to allow A,B,C to be real -- thus, no requirement for integrality. The OP's current requirements are: (1) A,B,C are positive reals representing the 3 angles of a triangle. Thus, either A + B + C = 180 deg, or, equivalently, in radians, A + B + C = Pi. (3) sin(A) + sin(B) + sin(C) = 2 As has been indicated in a previous reply, there are infinitely many solutions. quasi === Subject: please please please posting-account=8ZoLdQoAAAALc600seqPF7ogQ2ypU54N CLR 1.1.4322),gzip(gfe),gzip(gfe) CASPIAN.Advanced.Traffic.Manager.v1:3128 (squid) hi i need this book: the solution of Engineering Mechanics - Statics, 6th Edition (J. L. Meriam, L. G. Kraige) please send it to alimechanic2010@gmail.com i really need it . === === Subject: Mars planet: Several STAR of 6 point wasexcavatedat the Cydonia's Face Because I have open a new web main site http://es.geocities.com/ramonetriu/links2008 to link with photo to 24 sections, I remember you that in my GENERAL LINKS 2008 http://es.geocities.com/ramonetriu/links2008 I also include the first geographic desing of excaved Salomon's star in the ground at the Cydonia human's face. I knew it in the spanish website: LA REVOLUCION DE MARTE http://es.geocities.com/jaumeclavecinca/martepidioayuda I inform to the author about my ampliation of the Cydonia area, and respectively he also introduced my link in his website. Actually I believe that it's a very huge human FACE imgae realy (Some years ago I was in doubt because it look too much beautiful, and then I say so in my book: HUMAN FACE OF MONTSERRAT published in 1990). The little hill that offers the famous human FACE of Cydonia was a deliberated work, because near there stand several huge designs of David's Star never detected before. It's necessary to be permeable and to know how to rectify. P.D.: See also my first star foun in the link: http://es.geocities.com/ramonramonetriu/superfaces === Subject: Re: Initial Model Theorem > (sci.logic added) > > > the Stanford Encyclopedia of Philosophy, in its entry on first-order > model theory (http://plato.stanford.edu/entries/modeltheory-fo/), > states the following initial model theorem: > Let T be a theory consisting of strict universal Horn sentences. Then > T has a model A with the property that for every model B of T there is > a unique homomorphism from A to B. (Such a model A is called an > initial model of T. It is unique up to isomorphism.) > > So the axioms of T are all universal quantifications of formulas of > the form > > P > > or > > (P_1 & ... & P_n) -> Q > > where P, P_j and Q are atomic. See P as -> P with n=0, i.e. >true -> P. Define on term models M: T_P(M) := { Q(t) | P_i(t1) in M & .. & P_n(ti) in M } I can't figure out what any of that means. ??? (What is a term model? Is P a variable here or just something decorating the T? Given P, what is Q? What is t? What is Q(t)? Although I have no idea what we're talking abot, the P_n(ti) was a typo, right?...) >Thus T_P is monotonic and continuous. >Thus it has a least fixpoint T_P^ := U_i in N T_P^i(0) . Now it depends how you define homomorphic, >because we have T_P^[P(t)] -> X[P(t)] >but not T_P^[P(t)] <-> X[P(t)], where X >is another model that satisfies all of >the horn sentences. > David C. Ullrich === Subject: Re: Initial Model Theorem > where P, P_j and Q are atomic. > See P as -> P with n=0, i.e. > true -> P. Define on term models M: > T_P(M) := { Q(t) | P_i(t1) in M & .. & P_n(ti) in M } >I can't figure out what any of that means. ??? Try harder. Or let me give you an example, take the following set of sentences, call this set P. edge(a,b). edge(b,c). path(X,Y) <- edge(X,Y). path(X,Y) <- path(X,Z), path(Z,Y). M subset {edge(a,a), edge(a,b), .., path(a,a), path(a,b), ...}, those are the term model. Now the program operator: T_P(M) := { edge(a,b), edge(b,c) } u { path(X,Y) | edge(X,Y) in M } u { path(X,Y) | exists Z path(X,Z) in M & path(Z,Y) in M }. Now the iterated fixpoint: T_P^ = u_i in N T_P^i({}) T_P^n+1(M) = T_P(T_P^n(M)) T_P^0(M) = M So we have: T_P^0({}) = {} T_P^1({}) = {edge(a,b), edge(b,c)} T_P^2({}) = {edge(a,b), edge(b,c), path(a,b), path(b,c)} T_P^3({}) = {edge(a,b), edge(b,c), path(a,b), path(b,c), path(a,c)} T_P^n({}) = T_P^3({}) for n>3 Hence: T_P^ = {edge(a,b), edge(b,c), path(a,b), path(b,c), path(a,c)} === Subject: Re: Initial Model Theorem > > where P, P_j and Q are atomic. > See P as -> P with n=0, i.e. > true -> P. Define on term models M: > T_P(M) := { Q(t) | P_i(t1) in M & .. & P_n(ti) in M } > I can't figure out what any of that means. ??? >Try harder. >Or let me give you an example, take the following > set of sentences, call this set P. Simpler example: nat(0). nat(s(X)) <- nat(X) Hence: T_P^ = {not(0), nat(s(0)), nat(s(s(0))), ...} Bye === Subject: Re: Initial Model Theorem > where P, P_j and Q are atomic. > See P as -> P with n=0, i.e. > true -> P. Define on term models M: > T_P(M) := { Q(t) | P_i(t1) in M & .. & P_n(ti) in M } > I can't figure out what any of that means. ??? > Try harder. > Or let me give you an example, take the following > set of sentences, call this set P. > The edge example is a datalog example. The sentences do not contain function symbols with arity > 0. As a result the T_P^ is finite. The nat example is not datalog anymore. We have s as a unary function symbol, thus the term model base is already infinite. And in this example also T_P^ is also infinite. === Subject: Re: Initial Model Theorem > (sci.logic added) >the Stanford Encyclopedia of Philosophy, in its entry on first-order >model theory (http://plato.stanford.edu/entries/modeltheory-fo/), >states the following initial model theorem: >Let T be a theory consisting of strict universal Horn sentences. Then >T has a model A with the property that for every model B of T there is >a unique homomorphism from A to B. (Such a model A is called an >initial model of T. It is unique up to isomorphism.) > So the axioms of T are all universal quantifications of formulas of > the form > P > or > (P_1 & ... & P_n) -> Q > where P, P_j and Q are atomic. > I must be missing something - the theorem as stated seems clearly > false. Say the only axiom in T is Ax P(x). Then any map from > any model of T to any other model of T is a homomorphis, > and since T has models containing more than one element in > the universe there cannot exist a model A such that for every > model B there is a unique homomorphism from A to B > (if B has more than one element then for every A there are > at least two homomorphisms from A to B.) Presumably the idea is that associated with T is a specified >first-order language, and the homomorphisms have to respect the >denotation of the ground terms (maybe there's a special case if there >are no constants). The set of ground terms can be taken as the domain >of the initial model. Interesting theory. But I looked up the definition on the cited page: The basic maps between structures of the same signature K are called homomorphisms, defined as follows. A homomorphism from structure A to structure B is a function f from dom(A) to dom(B) with the property that for every atomic formula P(v1,á,vn) and any n-tuple a = (a1,á,an) of elements of A, A |= P[a] -> B |= P[b] where b is (f(a1),á,f(a n)). I don't see anything else on the page that indicates that they mean what you suggest, although I didn't read it carefully. > On the other hand, if we change the statement to > Let T be a theory consisting of strict universal Horn sentences. Then > T has a model A with the property that for every model B of T there is > a unique homomorphism from A to B. typo for from B to A. >(Such a model A is called an > initial model of T. It is unique up to isomorphism.) I can't see any change ... > then the theorem seems trivial: Say A is the set of atomic formulas, > and let X consist of all the mappings f : A -> {true, false} which are > compatible with T (in what I suspect is the obvious sense - I can > be more explicit if this is not clear). Then it seems clear that X > becomes a model of T with the required property, if for every P > we let the interpretation of P be the set of all f such that f(P) = > true. > ??? >its proof? >Tjark > David C. Ullrich David C. Ullrich === Subject: Re: Initial Model Theorem > (sci.logic added) >the Stanford Encyclopedia of Philosophy, in its entry on first-order >model theory (http://plato.stanford.edu/entries/modeltheory-fo/), >states the following initial model theorem: >Let T be a theory consisting of strict universal Horn sentences. Then >T has a model A with the property that for every model B of T there is >a unique homomorphism from A to B. (Such a model A is called an >initial model of T. It is unique up to isomorphism.) > So the axioms of T are all universal quantifications of formulas of > the form > P > or > (P_1 & ... & P_n) -> Q > where P, P_j and Q are atomic. > I must be missing something - the theorem as stated seems clearly > false. Say the only axiom in T is Ax P(x). Then any map from > any model of T to any other model of T is a homomorphis, > and since T has models containing more than one element in > the universe there cannot exist a model A such that for every > model B there is a unique homomorphism from A to B > (if B has more than one element then for every A there are > at least two homomorphisms from A to B.) >Presumably the idea is that associated with T is a specified >first-order language, and the homomorphisms have to respect the >denotation of the ground terms (maybe there's a special case if there >are no constants). The set of ground terms can be taken as the domain >of the initial model. Interesting theory. But I looked up the definition on the cited page: The basic maps between structures of the same signature K are called > homomorphisms, defined as follows. A homomorphism from structure A > to structure B is a function f from dom(A) to dom(B) with the property > that for every atomic formula P(v1,..,vn) and any n-tuple a = (a1,..,an) > of elements of A, A |= P[a] -> B |= P[b] where b is (f(a1),..,f(a n)). I don't see anything else on the page that indicates that they mean > what you suggest, although I didn't read it carefully. Since they explicitly include equality, you can consider eg the formula x = c as P(x), and the homomorphism property as given here entails that the object denoted by c in A has to get mapped to the object denoted by c in B (at least when equality is behaving itself). David C. Ullrich -- Alan Smaill === Subject: Re: Initial Model Theorem > (sci.logic added) >the Stanford Encyclopedia of Philosophy, in its entry on first-order >model theory (http://plato.stanford.edu/entries/modeltheory-fo/), >states the following initial model theorem: >Let T be a theory consisting of strict universal Horn sentences. Then >T has a model A with the property that for every model B of T there is >a unique homomorphism from A to B. (Such a model A is called an >initial model of T. It is unique up to isomorphism.) > So the axioms of T are all universal quantifications of formulas of > the form > P > or > (P_1 & ... & P_n) -> Q > where P, P_j and Q are atomic. > I must be missing something - the theorem as stated seems clearly > false. Say the only axiom in T is Ax P(x). Then any map from > any model of T to any other model of T is a homomorphis, > and since T has models containing more than one element in > the universe there cannot exist a model A such that for every > model B there is a unique homomorphism from A to B > (if B has more than one element then for every A there are > at least two homomorphisms from A to B.) Presumably the idea is that associated with T is a specified >first-order language, and the homomorphisms have to respect the >denotation of the ground terms (maybe there's a special case if there >are no constants). The set of ground terms can be taken as the domain >of the initial model. Ok, if that's the idea... > On the other hand, if we change the statement to > Let T be a theory consisting of strict universal Horn sentences. Then > T has a model A with the property that for every model B of T there is > a unique homomorphism from A to B. (Such a model A is called an > initial model of T. It is unique up to isomorphism.) I can't see any change ... Aargh. I meant to change the homomorphism from A to B to homomorphism from B to A. > then the theorem seems trivial: Say A is the set of atomic formulas, > and let X consist of all the mappings f : A -> {true, false} which are > compatible with T (in what I suspect is the obvious sense - I can > be more explicit if this is not clear). Then it seems clear that X > becomes a model of T with the required property, if for every P > we let the interpretation of P be the set of all f such that f(P) = > true. > ??? >its proof? >Tjark > David C. Ullrich David C. Ullrich === Subject: Re: Initial Model Theorem Originator: jdolan@math.UUCP (James Dolan) |(sci.logic added) | |> |>the Stanford Encyclopedia of Philosophy, in its entry on first-order |>model theory (http://plato.stanford.edu/entries/modeltheory-fo/), |>states the following initial model theorem: |> |>Let T be a theory consisting of strict universal Horn |>sentences. Then T has a model A with the property that for every |>model B of T there is a unique homomorphism from A to B. (Such a |>model A is called an initial model of T. It is unique up to |>isomorphism.) | |So the axioms of T are all universal quantifications of formulas of |the form | |P | |or | |(P_1 & ... & P_n) -> Q | |where P, P_j and Q are atomic. | |I must be missing something - the theorem as stated seems clearly |false. Say the only axiom in T is Ax P(x). Then any map from any |model of T to any other model of T is a homomorphism, and since T has |models containing more than one element in the universe as well as ones with less than one element, if they're using what i would consider to be sensible definitions. i've sometimes heard people claim that it's more sensible to disallow empty universes, but was never convinced that their reasons for their claim made much sense. -- jdolan@math.ucr.edu === Subject: Re: Initial Model Theorem |(sci.logic added) | >||>the Stanford Encyclopedia of Philosophy, in its entry on first-order >|>model theory (http://plato.stanford.edu/entries/modeltheory-fo/), >|>states the following initial model theorem: >||>Let T be a theory consisting of strict universal Horn >|>sentences. Then T has a model A with the property that for every >|>model B of T there is a unique homomorphism from A to B. (Such a >|>model A is called an initial model of T. It is unique up to >|>isomorphism.) >| >|So the axioms of T are all universal quantifications of formulas of >|the form >| >|P >| >|or >| >|(P_1 & ... & P_n) -> Q >| >|where P, P_j and Q are atomic. >| >|I must be missing something - the theorem as stated seems clearly >|false. Say the only axiom in T is Ax P(x). Then any map from any >|model of T to any other model of T is a homomorphism, and since T has >|models containing more than one element in the universe as well as ones with less than one element, if they're using what i >would consider to be sensible definitions. i've sometimes heard >people claim that it's more sensible to disallow empty universes, but >was never convinced that their reasons for their claim made much >sense. Definitions are definitions - whether it's senssible or not it _is_ standard to disallow empty universes. David C. Ullrich === Subject: Re: Initial Model Theorem Originator: jdolan@math.UUCP (James Dolan) | ... |>| |>|> |>|>the Stanford Encyclopedia of Philosophy, in its entry on |>|>first-order model theory |>|>(http://plato.stanford.edu/entries/modeltheory-fo/), states the |>|>following initial model theorem: |>|> |>|>Let T be a theory consisting of strict universal Horn |>|>sentences. Then T has a model A with the property that for every |>|>model B of T there is a unique homomorphism from A to B. (Such a |>|>model A is called an initial model of T. It is unique up to |>|>isomorphism.) |>| |>|So the axioms of T are all universal quantifications of formulas of |>|the form |>| |>|P |>| |>|or |>| |>|(P_1 & ... & P_n) -> Q |>| |>|where P, P_j and Q are atomic. |>| |>|I must be missing something - the theorem as stated seems clearly |>|false. Say the only axiom in T is Ax P(x). Then any map from any |>|model of T to any other model of T is a homomorphism, and since T |>|has models containing more than one element in the universe |> |>as well as ones with less than one element, if they're using what i |>would consider to be sensible definitions. i've sometimes heard |>people claim that it's more sensible to disallow empty universes, |>but was never convinced that their reasons for their claim made much |>sense. | |Definitions are definitions - whether it's senssible or not it _is_ |standard to disallow empty universes. nor is it necessary to follow non-sensible standards, and it seems that they didn't do so in this particular case. did you read the part near the top of the web-page where they implicitly acknowledged the possibility of empty universes?: |If K is a signature, then a structure of signature K, say A, consists |of the following items: | | 1. A set called the domain of A and written dom(A); it is usually | assumed to be nonempty; -- jdolan@math.ucr.edu === Subject: Re: product integral > water a Âìcrit : > f is continuous function. > ProductIntegral_0^t [1 - f(u)du] = exp[- Integral_0^t f(u)du]. > why? > What is a product integral? Product integral is a special integral analogy to normal integral Do you _really_ think that that's a sufficient explanation to allow someone to answer your queston? What's the _definition_? David C. Ullrich === Subject: Re: product integral <1mb8r3ppg0rmtoqqgnvm8ckdr2arum86c1@4ax.com> posting-account=d9DpDwoAAADlGCWGhrCHDkiN-0F85Exg MathPlayer 2.10b; Avant Browser; .NET CLR 1.1.4322; .NET CLR 2.0.50727; TheWorld),gzip(gfe),gzip(gfe) > water a Â.9bcrit : > f is continuous function. > ProductIntegral 0^t [1 - f(u)du] = exp[- Integral 0^t f(u)du]. > why? > What is a product integral? >Product integral is a special integral analogy to normal integral Do you really think that that's a sufficient explanation to allow > someone to answer your queston? What's the definition ? David C. Ullrich Sufficient explanation needs many more words. === Subject: Re: product integral posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > water a Â.9bcrit : > f is continuous function. > ProductIntegral 0^t [1 - f(u)du] = exp[- Integral 0^t f(u)du]. > why? > What is a product integral? >Product integral is a special integral analogy to normal integral > Do you really think that that's a sufficient explanation to allow > someone to answer your queston? What's the definition ? > David C. Ullrich Sufficient explanation needs many more words. You can solve the original question yourself, too... -- m === Subject: Re: product integral posting-account=d9DpDwoAAADlGCWGhrCHDkiN-0F85Exg MathPlayer 2.10b; Avant Browser; .NET CLR 1.1.4322; .NET CLR 2.0.50727; TheWorld),gzip(gfe),gzip(gfe) > water a Â.9bcrit : > f is continuous function. > ProductIntegral 0^t [1 - f(u)du] = exp[- Integral 0^t f(u)du]. > why? > What is a product integral? >Product integral is a special integral analogy to normal integral > Do you really think that that's a sufficient explanation to allow > someone to answer your queston? What's the definition ? > David C. Ullrich Sufficient explanation needs many more words. - ìïæ.beñ[YAcute] îÌ[Micro]Äë[C apitalADoubleDot][Times].85 - Now let's define Product Integral denoted PI PI 0^t [1-d A(u)] = lim prod 1^r{1 - [ A(u k) - A(u {k-1}) ]} Here 0 = u 0 < u 1 < ... < u r = t and the limit is taken as r->oo and max(u i - u {i-1})->0 === Subject: Re: Branch point I'm just reading about branch points on >http://mathworld.wolfram.com/BranchPoint.html Reading mathworld is often a big mistake. Wikipedia is generally much more accurate, clear and complete. PlanetMath works sometimes as well. >I think the example given there is wrong and >would like to double-check with you that I'm right. So we consider the power function >f(z) = z^a >where a is complex non-integer. >Then z = 0 is a branch point. Because z can be written as z = exp(i*Pi) and >z = exp(i*2*Pi). In the first case >we have f(exp(i*Pi)) = exp(i*a*Pi) >and in the second one >f(exp(i*2*Pi)) = exp(i*2*a*Pi). So the values of f(z) are different for arg(z) = Pi and >arg(z) = 2*Pi. On the above webpage we have arg(z) = 0 and arg(z) = 2*Pi. >Is that a typo? L David C. Ullrich === Subject: Re: aleph_0 1 2 but not 3 >it is possible to construct a set theory where aleph_0 , aleph_1 and aleph_2 exist , but aleph_3 does not. So what? It's possible to construct a set theory where I am the king of France and nothing but me exists. A set theory with the properties you describe cannot also have all the useful properties of standard set theory. >tommy1729 David C. Ullrich === Subject: Re: aleph_0 1 2 but not 3 <0sb8r39825q39kua5qtkrdfqsk42brkf5m@4ax.com> posting-account=XBlrJAoAAADt98TEM65O-rVj9Zl2tTjM 1.1.4322),gzip(gfe),gzip(gfe) >it is possible to construct a set theory where aleph 0 , aleph 1 and aleph 2 exist , but aleph 3 does not. So what? It's possible to construct a set theory where I am the king > of France and nothing but me exists. A set theory with the properties you describe cannot also have all the > useful properties of standard set theory. >tommy1729 David C. Ullrich Ah, but who requires mathematics to have useful properties? Some like to construct (and post) mathematics for fun, or from some perceived novelty or value. Others are just plain delusional, like James Harris, or posit problems or make definitions which lead nowhere or don't make sense. === Subject: Re: aleph_0 1 2 but not 3 posting-account=x2WXVAkAAACheXC-5ndnEdz_vL9CA75q Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > Others are just plain delusional, like > James Harris, or posit problems > or make definitions which lead nowhere or > don't make sense. All progress depends on unreasonable men. - somebody or other -- Rich === Subject: Re: aleph_0 1 2 but not 3 > it is possible to construct a set theory where aleph_0 , aleph_1 and > aleph_2 exist , but aleph_3 does not. Sure. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: aleph_0 1 2 but not 3 posting-account=-XRI1AgAAABBR-yHg7m_4HCGi05mtX1k Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) On Feb 14, 3:45 am, Aatu Koskensilta it is possible to construct a set theory where aleph_0 , aleph_1 and > aleph_2 exist , but aleph_3 does not. Sure. -- > Aatu Koskensilta (aatu.koskensi...@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus What if there was a cardinal for the naturals, and then the powerset of naturals, and then the powerset of that was somehow irregular, admitting the Russell-paradox element? That's basically a question of how much infinity it takes to be infinite in the sense that any quantification over the elements leads to another element that exists in the set, kind of like the Goedel That's where there's a parallel between the notion of a Russellian incompleteness, that a set of all that that don't contain themselves contains itself, and Goedelian incompleteness, that (loosely) a set of theorems about infinitely many distinct elements has another theorem. over sets yielding a set in the sense of naive (and unrestricted) comprehension, yet that not being a set, then a notion with various sized infinities might have one infinity with the imprimatur of being a regular infinity, for easy use with induction, and some arbitrarily higher infinity being irregular and otherwise satisfying the requirement of being universal, in the sense of being irregular and having as a subset any set. Then, in terms of there being various cardinals, it would have that foundation (well-foundedness, regularity) is not an axiom, instead that there is some notion of small and large infinite set, with one particular arbitrarily larger cardinal for the continuum, simply for the notion of application of choice over elements of the continuum. Of course, that would be to satisfy those that can't see a particular mapping between the natural integers and each element, in sequence, of the unit interval of real numbers. Some agreeably nonstandard, while intuitive, constructions of the real numbers would have them be a contiguous sequence of points, holistically in the polydimensional. ( Here are some perhaps more traditional points in these arguments: real numbers aren't totally characterized by Cauchy/Dedekind/Eudoxus construction, with Newtonian/Leibnizian fluxions/infinitesimals, each smooth integrable function is characterized by countably many points where each curve of the continuum is a smooth integrable function, well-ordering the reals leads to sequential reals or doesn't well- order the reals. ) So, the consequences of the Russell paradox and Geodelian incompleteness might be seen to be flip sides of a coin, of a sort, in terms of that the infinite is not a regular set. Ross -- Finlayson Consulting === Subject: Re: aleph_0 1 2 but not 3 Aatu Koskensilta a .8ecrit : > it is possible to construct a set theory where aleph_0 , aleph_1 and > aleph_2 exist , but aleph_3 does not. >Sure. > On the other hand, it will at least need some negatoin of the powerset axiom... === Subject: Re: aleph_0 1 2 but not 3 > >it is possible to construct a set theory where aleph_0 , aleph_1 and >aleph_2 exist , but aleph_3 does not. >Sure. Is it possible to construct a (set) theory where _none_ of these exist ? Han de Bruijn === Subject: Re: aleph_0 1 2 but not 3 > Is it possible to construct a (set) theory where _none_ of these exist ? Sure. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: aleph_0 1 2 but not 3 > it is possible to construct a set theory where aleph_0 , aleph_1 and > aleph_2 exist , but aleph_3 does not. Certainly, from a completely literal point of view. You can pretty much define whatever symbols and rules you like and call it a set theory, no matter how pathetically useless it might be. I expect that these aleph symbols wouldn't have similar properties to their more usual meanings in ZFC. I also expect it would also be somewhat ugly, in the sense that the axioms wouldn't correspond to obvious set principles, but would be tailored to the requirement of allowing an aleph_2 and excluding an aleph_3. However, my guess is that you meant whether it would be possible to construct a useful and elegant set theory, that would support most of the useful features of mathematics, with axioms that correspond to intuitive principles of sets, have very similar meanings for the aleph symbols, and yet not permit an aleph_3. I expect the answer to that would be no. - Tim === Subject: Re: aleph_0 1 2 but not 3 > However, my guess is that you meant whether it would be possible to > construct a useful and elegant set theory, that would support most of > the useful features of mathematics, with axioms that correspond to > intuitive principles of sets, have very similar meanings for the aleph > symbols, and yet not permit an aleph_3. I expect the answer to that would be no. Consider the theory obtained from ZFC by dropping the replacement and powerset axioms, and adding as axioms P(V_omega) and P(P(V_omega)) exist and the generalised continuum hypothesis. This theory is equivalent to third-order arithmetic + CH, and much mathematics can be done in the system, even if it is conceptually unmotivated. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=-eQqtQoAAACZVM-kNEsOn3k7GSvoJoS4 MathPlayer 2.0; .NET CLR 1.1.4322; InfoPath.1; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > lim[n-->oo] |{ x | x in S(n) & x > |S(n)| }| / |S(n)| = 1/2 > and, therefore, > lim{n-->oo} |{ x | x in S(n) & x > |S(n)| }| / |S(n)| =/= 0. > Is this /all/ you want? You don't need all the set theoretic > digression to show this result, assuming that |M| denotes the number > of elements of M: > 0. S(n) = {2, 4, 6, ..., 2n}. > 1. |S(n)| = n. > 2. { x | x in S(n) & x > |S(n)| } = {n+2, n+4, ..., 2n}. > 3. | {n+2, n+4, ..., 2n} | = n/2. > 4. (n/2) / n = 1/2 for all n. > I suspect you are going to try to interchange the limit operation with > the definition of your sets, and /that/ is not allowed. What point do you doubt? Is it the step from lim[n-->oo] {2, 4, 6, ..., 2n} = {2, 4, 6, ...} > to > |lim[n-->oo] {2, 4, 6, ..., 2n}| = lim[n-->oo] |{2, 4, 6, ..., 2n}| = | > {2, 4, 6, ...}| ? So far every thing is fine. So now you have that lim[n-->oo] |{ x | x in S(n) & x > |S(n)| }| / |S(n)| = 1/2 and, therefore, a fortiori, lim{n-->oo} |{ x | x in S(n) & x > |S(n)| }| / |S(n)| =/= 0. Now what? Note well that the sequence { x | x in S(n) & x > |S(n)| } also has a limit (in the Halmos sense), but it is the /empty set/. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) > lim[n-->oo] |{ x | x in S(n) & x > |S(n)| }| / |S(n)| = 1/2 > and, therefore, > lim{n-->oo} |{ x | x in S(n) & x > |S(n)| }| / |S(n)| =/= 0. > Is this /all/ you want? You don't need all the set theoretic > digression to show this result, assuming that |M| denotes the number > of elements of M: > 0. S(n) = {2, 4, 6, ..., 2n}. > 1. |S(n)| = n. > 2. { x | x in S(n) & x > |S(n)| } = {n+2, n+4, ..., 2n}. > 3. | {n+2, n+4, ..., 2n} | = n/2. > 4. (n/2) / n = 1/2 for all n. > I suspect you are going to try to interchange the limit operation with > the definition of your sets, and /that/ is not allowed. > What point do you doubt? Is it the step from > lim[n-->oo] {2, 4, 6, ..., 2n} = {2, 4, 6, ...} > to > |lim[n-->oo] {2, 4, 6, ..., 2n}| = lim[n-->oo] |{2, 4, 6, ..., 2n}| = | > {2, 4, 6, ...}| ? So far every thing is fine. So now you have that > lim[n-->oo] |{ x | x in S(n) & x > |S(n)| }| / |S(n)| = 1/2 > and, therefore, a fortiori, > lim{n-->oo} |{ x | x in S(n) & x > |S(n)| }| / |S(n)| =/= 0. Now what? Now it is clear that half of the positive even numbers are larger than the cardinal number of the set of all positive even numbers. This is in contradiction with the statement that aleph 0 is larger than any natural number: lim{n-->oo} |{ x | x in S(n) & x > |S(n)| }| / |S(n)| = 0. > Note well that the sequence { x | x in S(n) & x > |S(n)| } > also has a limit (in the Halmos sense), but it is the /empty set/ So what? If you insist on a Halmos' definition yielding a non-epty set, it will suffer to consider: lim{n-->oo} |{ x | x in S(n) & x < |S(n)| }| / |S(n)| = 1/2 and not lim{n-->oo} |{ x | x in S(n) & x < |S(n)| }| / |S(n)| = 1. Bt why? After having shown that lim[n-->oo] {2, 4, 6, ..., 2n} = {2, 4, 6, ...}, we need no longer consider sets. (|{ x | x in S(n) & x > |S(n)| }| / |S(n)|) is a real number in the surrounding of 1/2 and stays so for n surpassing any natutral number, becoming 1/ 2 precisely. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 CLR 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) > So far every thing is fine. So now you have that > lim[n-->oo] |{ x | x in S(n) & x > |S(n)| }| / |S(n)| = 1/2 > and, therefore, a fortiori, > lim{n-->oo} |{ x | x in S(n) & x > |S(n)| }| / |S(n)| =/= 0. > Now what? Now it is clear that half of the positive even numbers > are larger than > the cardinal number of the set of all positive even numbers. No such thing is clear at all. First, half of the positive even numbers is undefined. Second, you've not shown a single positive even number whose cardinality is greater than the cardinality of the set of positive even numbers and you've not even shown any SET of positive even numbers whose cardinality is greater than the cardinality of the set of positive even numbers. > This is > in contradiction with the statement that aleph 0 is larger than any > natural number: Sure is. > ælim{n-->oo} |{ x | x in S(n) & x > |S(n)| }| / |S(n)| = 0. Not with the method of limits back to the ordinary one for denumerable sequences of real numbers. Under that method: lim[n-->oo] |{ x | x in S(n) & x > |S(n)| }| / |S(n)| = 1/2 Unless I miscalculated (did I?) the sequence goes: 0/1, 1/2, 2/3, 1/2, 3/5, 1/2, 4/7 ... > Note well that the sequence { x | x in S(n) & x > |S(n)| } > also has a limit (in the Halmos sense), but it is the /empty set/ I'll trust your (Horand's) calculation of that. > So what? If you insist on a Halmos' definition yielding a non-epty > set, You can't insist on that without abandoning the definition. Again, if you want to change the definition, then please tell us your exact definition and stick with it for the rest of the discussion. > it will suffer to consider: lim{n-->oo} |{ x | x in S(n) & x < |S(n)| }| / |S(n)| = 1/2 > and not > lim{n-->oo} |{ x | x in S(n) & x < |S(n)| }| / |S(n)| = 1. That's got nothing to do with the Halmos definition. That's the ordinary method of limits for denumerable sequences of real numbers. > Bt why? After having shown æthat lim[n-->oo] {2, 4, 6, ..., 2n} = {2, > 4, 6, ...}, we need no longer consider sets. > (|{ x | x in S(n) & x > |S(n)| }| / |S(n)|) is a real number in the > surrounding of 1/2 and stays so for n surpassing any natutral number, > becoming 1/ 2 precisely. Yes, working with the ordinary sense of a limit of denumerable sequence of real numbers, the limit is 1/2. So what? Of course we understand that the Halmos definition of the limit of sequence of sets does not always agree with the ordinary definition of the limit of a sequence of reals. They're too different things, not meant to be conflated. First we found a limit of a sequence of sets (not a sequence of real numbers) to be the set of positive even numbers. Fair enough. Then we found, using a completely different method and definition that a certain sequence of real numbers has 1/2 as its limit. Fair enough. But then you draw a complete non sequitur of a conclusion. You've shown no logic whatsoever that there is a positive even number with cardinality greater than the set of positive even numbers. MoeBlee === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=-eQqtQoAAACZVM-kNEsOn3k7GSvoJoS4 MathPlayer 2.0; .NET CLR 1.1.4322; InfoPath.1; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > lim[n-->oo] |{ x | x in S(n) & x > |S(n)| }| / |S(n)| = 1/2 Unless I miscalculated (did I?) the sequence goes: 0/1, 1/2, 2/3, 1/2, > 3/5, 1/2, 4/7 ... Don't know. It probably doesn't matter much, but it's best to be careful. I got: S(n) = {2, 4, 6, ..., 2n}, so |S(n)| = n { x | x in S(n) & x > |S(n)| } = {n+2, n+4,, ..., 2n}. Ah, I see my mistake now. The second line only works for even n. So the sequence goes 1/1, 1/2, 2/3, 1/2, ... The general term g(n) = 1/2 if n is even and = 1/2 + 1/(2n) if n is odd. Still, the limit is 1/2. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 CLR 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) > lim[n-->oo] |{ x | x in S(n) & x > |S(n)| }| / |S(n)| = 1/2 > Unless I miscalculated (did I?) the sequence goes: 0/1, 1/2, 2/3, 1/2, > 3/5, 1/2, 4/7 ... Oops, nevermind that first term 0/1. > 1/1, 1/2, 2/3, 1/2, ... Agreed. 1/1, 1/2, 2/3, 1/2, 3/5, 1/2, 4/7 ... MoeBlee === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I What point do you doubt? Is it the step from > lim[n-->oo] {2, 4, 6, ..., 2n} = {2, 4, 6, ...} > to > |lim[n-->oo] {2, 4, 6, ..., 2n}| = lim[n-->oo] |{2, 4, 6, ..., 2n}| = | > {2, 4, 6, ...}| ? > So far every thing is fine. So now you have that > lim[n-->oo] |{ x | x in S(n) & x > |S(n)| }| / |S(n)| = 1/2 > and, therefore, a fortiori, > lim{n-->oo} |{ x | x in S(n) & x > |S(n)| }| / |S(n)| =/= 0. > Now what? Now it is clear that half of the positive even numbers are larger than > the cardinal number of the set of all positive even numbers. No. Your argument is For every positive integer n 2n > n Therefore lim [n-->oo] 2n > lim[n-->oo] n Even pink elephants won't help here! - William Hughes === Subject: Re: A consideration concerning the diagonal argument of G. Cantor WM says... > So far every thing is fine. So now you have that > lim[n-->oo] |{ x | x in S(n) & x > |S(n)| }| / |S(n)| = 1/2 > and, therefore, a fortiori, > lim{n-->oo} |{ x | x in S(n) & x > |S(n)| }| / |S(n)| = 0. > Now what? Now it is clear that half of the positive even numbers are larger than >the cardinal number of the set of all positive even numbers. No, no, no. Define f to be the following function on sets of naturals: f(X) = |{x | x in X & x > |X| }|/|X| To get your conclusion, you need to prove the following statement If (lim[n-->oo] S(n)) = E, then (lim[n-->oo] f(S(n))) = f(E) Why do you believe that statement? (Especially since it is provably false?) Let A and B be two different domains of mathematical objects such that in each domain there is an appropriate notion of the limit of a sequence. Let f be a function from A to B. Only for very *special* functions f is it true that limit n --> oo of f(x_n) = f(limit as n --> oo of x_n) If A = B = R, the reals, and the limit is just normal convergence of sequences, then such functions are called continuous functions. Not all functions are continuous. -- Daryl McCullough Ithaca, NY === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=-eQqtQoAAACZVM-kNEsOn3k7GSvoJoS4 MathPlayer 2.0; .NET CLR 1.1.4322; InfoPath.1; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > So far every thing is fine. So now you have that > lim[n-->oo] |{ x | x in S(n) & x > |S(n)| }| / |S(n)| = 1/2 > and, therefore, a fortiori, > lim{n-->oo} |{ x | x in S(n) & x > |S(n)| }| / |S(n)| =/= 0. > Now what? Now it is clear that half of the positive even numbers are larger than > the cardinal number of the set of all positive even numbers. This is > in contradiction with the statement that aleph 0 is larger than any > natural number: ælim{n-->oo} |{ x | x in S(n) & x > |S(n)| }| / |S(n)| = 0. just above it? I should have the courage of my convictions: Yesterday I agonized whether I should use suspect or predict when I said you were trying to switch limits. In the end I went with suspect. You seem to think that |{ x | x in S(n) & x > |S(n)| }| lim{n-->oo} ---------------------------------- = |S(n)| | lim{n-->oo} { x | x in S(n) & x > |S(n)| } | ---------------------------------------------- lim{n-->oo} |S(n)| and there is *no justification* for this step. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > So far every thing is fine. So now you have that > lim[n-->oo] |{ x | x in S(n) & x > |S(n)| }| / |S(n)| = 1/2 > and, therefore, a fortiori, > lim{n-->oo} |{ x | x in S(n) & x > |S(n)| }| / |S(n)| =/= 0. > Now what? > Now it is clear that half of the positive even numbers are larger than > the cardinal number of the set of all positive even numbers. This is > in contradiction with the statement that aleph 0 is larger than any > natural number: > ælim{n-->oo} |{ x | x in S(n) & x > |S(n)| }| / |S(n)| = 0. just above it? I should have the courage of my convictions: Yesterday > I agonized whether I should use suspect or predict when I said you > were trying to switch limits. In the end I went with suspect. You seem to think that æ æ æ æ æ æ æ|{ x | x in S(n) & x > |S(n)| }| > lim{n-->oo} ---------------------------------- æ= > æ æ æ æ æ æ æ æ æ æ æ æ æ æ|S(n)| | lim{n-->oo} { x | x in S(n) & x > |S(n)| } | > ---------------------------------------------- > æ æ æ æ æ lim{n-->oo} |S(n)| and there is *no justification* for this step. Of course that is wrong. oo/oo is undefined. How do you come to the wrong impression that I took this step? === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 CLR 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) > So far every thing is fine. So now you have that > lim[n-->oo] |{ x | x in S(n) & x > |S(n)| }| / |S(n)| = 1/2 > and, therefore, a fortiori, > lim{n-->oo} |{ x | x in S(n) & x > |S(n)| }| / |S(n)| =/= 0. > Now what? > Now it is clear that half of the positive even numbers are larger than > the cardinal number of the set of all positive even numbers. This is > in contradiction with the statement that aleph 0 is larger than any > natural number: > ælim{n-->oo} |{ x | x in S(n) & x > |S(n)| }| / |S(n)| = 0. > just above it? I should have the courage of my convictions: Yesterday > I agonized whether I should use suspect or predict when I said you > were trying to switch limits. In the end I went with suspect. > You seem to think that > æ æ æ æ æ æ æ|{ x | x in S(n) & x > |S(n)| }| > lim{n-->oo} ---------------------------------- æ= > æ æ æ æ æ æ æ æ æ æ æ æ æ æ|S(n)| > | lim{n-->oo} { x | x in S(n) & x > |S(n)| } | > ---------------------------------------------- > æ æ æ æ æ lim{n-->oo} |S(n)| > and there is *no justification* for this step. Of course that is wrong. oo/oo is undefined. How do you come to the > wrong impression that I took this step? What STEP did you take? By taking 'lim n->oo' in the ordinary sense for a sequence of reals, we get: lim[n-->oo] |{ x | x in S(n) & x > |S(n)| }| / |S(n)| = 1/2 By taking 'lim n->oo' in the Halmos sense for a sequence of sets in general, we get: lim[n-->oo] S(n) = the set of even numbers. By taking 'lim n->oo' in the Halmos sense for a sequence of sets in general, we get, according to Horand (and though I haven't checked it, let's suppose he's right): lim[n-->oo] |{ x | x in S(n) & x > |S(n)| }| / |S(n)| = 0 (i.e., the empty set) Then you conclude that there is a positive even number that has cardinality greater than that of the set of positive even numbers. But there's no logic given by you for that conclusion. MoeBlee === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > [lim n->oo] S_n = E as I said. > You just SAID that without DEFINING. > That is a definition. No, it is not a definition or even any kind of explanation of a 'lim' > notation. Look, all you've done is STIPULATE lim n->oo] S_n = E Yes, that's the definition of E. but you've given no DEFINITION by which one could EVALUATE or > CALCULATE or DEDUCE through some method of limits that the value of > the left side of that equation is equal to the value of the right side > of that equation. It is so by definition. Thus, when you use your 'lim' notation for a sequence OTHER than S, we > have no basis to evaluate, calculate, or deduce the value of the limit > of that sequence other than S. I do not need a sequence other than S, because I need only consider E. And by definition E is the limit of S, Then the poster G. Frege found a DEFINITION (Halmos) that DOES allow > us to make such an evaluation and that does imply your equation. And I > mentioned another definition that does that to. I did not say that other definitions lead to other results. I am glad that all definitions I know of lead to that result. So, now that you want to evaluate also for sequences OTHER than S I never tried to do so. , YOU > need to say whether the Halmos definition, or the one I mentioned, or > the one herbzet mentioned (in another thread?), or some other > definition is the one you mean My definition is the one I mean. In case I had to choose between you and Halmos, I would adopt the latter. But this case does not occur. > for us to use NOT just as pertains to > the sequence S but to other sequences ALSO. That is the only way we > can be assured that the method is uniform as applied to different > sequences. I.e., the same method of evaluating for lim_n->oo is used > for the various sequences you are now mentioning, as opposed to some > personal one-off STIPULATION that you give each time for each > different sequence. In this thread we are considering S and its limit E. > But the above equation doesn't just hold > on its own - we have to have a specific defintion of 'lim' in such a > context to THEN see whether the equation holds. > A definition is not subject to your investigations. The above equation > need not hold. It is a definition. It's been explained over and over why it is not a definition of a > 'lim' notation. Much nonsense has been said, that is true. > It's amazing to me that you have a such a strong interest in the > subject but still don't understand such basic things about it. I understand certain basics about definitions. These are basics you > need to understand. No, I understand what you are thinking, but God beware me to adopt that. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 CLR 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) > [lim n->oo] S n = E æas I said. > You just SAID that without DEFINING. > That is a definition. > No, it is not a definition or even any kind of explanation of a 'lim' > notation. > Look, all you've done is STIPULATE > lim n->oo] S n = E Yes, that's the definition of E. If that's the definition of 'E', then what is the definition of 'lim n->oo S(n)' ? The definition of 'lim n->oo S(n)' can't be 'E' otherwise you'd be blatantly circular. > but you've given no DEFINITION by which one could EVALUATE or > CALCULATE or DEDUCE through some method of limits that the value of > the left side of that equation is equal to the value of the right side > of that equation. It is so by definition. Okay, now you're really not even PRETENDING to try. > Thus, when you use your 'lim' notation for a sequence OTHER than S, we > have no basis to evaluate, calculate, or deduce the value of the limit > of that sequence other than S. I do not need a sequence other than S, because I need only consider E. > And by definition E is the limit of S, And what is the DEFINITION of 'limit of S'? You said the definition of 'E' is 'the limit of S'. So now you need to define 'the limit of S'. Since you defined 'E' to be 'the limit of S' , the definition of 'lim n->oo S(n)' can't be 'E' otherwise you'd be blatantly circular. > Then the poster G. Frege found a DEFINITION (Halmos) that DOES allow > us to make such an evaluation and that does imply your equation. And I > mentioned another definition that does that to. I did not say that other definitions lead to other results. I am glad > that all definitions I know of lead to that result. I didn't say that you said that other definitions lead to other results (though, of course, obviously, other definitions do lead to other results). > So, now that you want to evaluate also for sequences OTHER than S I never tried to do so. You presented your lim n->oo notation for sequences other than S. You did that in the post recently where you asked me whether such and such a lim was 1/2. , YOU > need to say whether the Halmos definition, or the one I mentioned, or > the one herbzet mentioned (in another thread?), or some other > definition is the one you mean My definition is the one I mean. In case I had to choose between you > and Halmos, I would adopt the latter. But this case does not occur. Yes, you have to choose a certain single definition in a given context. So the Halmos definition it is. > æfor us to use NOT just as pertains to > the sequence S but to other sequences ALSO. That is the only way we > can be assured that the method is uniform as applied to different > sequences. I.e., the same method of evaluating for lim n->oo is used > for the various sequences you are now mentioning, as opposed to some > personal one-off STIPULATION that you give each time for each > different sequence. In this thread we are considering S and its limit E. Earlier in your post, you defined E to be the limit of S. Now what is the definition of 'the limit of S'? Since you defined 'E' to be 'the limit of S' , the definition of 'lim n->oo S(n)' can't be 'E' otherwise you'd be blatantly circular. Anyway, I'll take it at your word that the Halmos definition is the one in play here. > But the above equation doesn't just hold > on its own - we have to have a specific defintion of 'lim' in such a > context to THEN see whether the equation holds. > A definition is not subject to your investigations. The above equation > need not hold. It is a definition. > It's been explained over and over why it is not a definition of a > 'lim' notation. Much nonsense has been said, that is true. Right, it is nonsense to use 'lim' notation without defining it. > It's amazing to me that you have a such a strong interest in the > subject but still don't understand such basic things about it. > I understand certain basics about definitions. These are basics you > need to understand. No, I understand what you are thinking, but God beware me to adopt > that. What I'm thinking is the mechanics of correct defintional forms. You don't understand that. MoeBlee === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > Look, all you've done is STIPULATE > [lim n->oo] S_n = E > Yes, that's the definition of E. > Huh?! :-o So you just stated the definition E =df [lim n->oo] S_n ??? But THAT'S NOT A CORRECT DEFINITION if E is already used to denote the set of all even numbers (in the present context)! ... by definition E is the limit of S, > Well, let's accept that _for the sake of the argument_. So how is your notion of /limit/ defined you are using here? (We will have to refer to that if evaluating claims concerning E.) In this thread we are considering S and its limit E. > Great. Btw, how do you refer to the set of all even natural number now? :-o F. -- === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > I understand certain basics about definitions. These are basics you > need to understand. >No, I understand what you are thinking, but God beware me to adopt > that. You do not understand what any of us are thinking, for which you should be grateful. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > In mathematics we use this expression rather frequently. > To mean what when talking about a sequence of sets? > referred you to, I read it. > you will get an answer to your question extending this > definition to sets Yes indeedy. It does indeed give a definition of limit > recognizes that it's not a sort of oh well, everybody > understands what we mean Sorry I did not think that it was that difficult to see the light. > and actually provides one. It provides the one I gave. Strange agreement? > and also to your following questions. Except the one crucial one. When YOU say > lim(n->oo) X_n = X > to apply that general definition in this simple and obvious special case. > Are we all agreed on that, since it's the only formal > definition you've ever referenced? There must be a slight misunderstanding. I have been teaching analysis, algebra and geometry, history of mathematics, and field theory in total for nearly 20 years at German universities. That would have been a hard job without references to some formal results. But in order to satisfy your curiosity, here is another couple of formal references: === Subject: Re: A consideration concerning the diagonal argument of G. Cantor There must be a slight misunderstanding. I have been teaching > analysis, algebra and geometry, history of mathematics, and field > theory in total for nearly 20 years at German universities. That would > have been a hard job without references to some formal results. > Wolfgang M.9fckenheim is a classic crank. Why do you imagine, as you seem to do, that there is any point arguing with him? It may seem odd that he has a job as a professor in technical and mathematical subjects, but such things happen. (Torkel Franzen, sci.math) F. -- === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/20080208 Fedora/2.0.0.12-1.fc8 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > There must be a slight misunderstanding. I have been teaching > analysis, algebra and geometry, history of mathematics, and field > theory in total for nearly 20 years at German universities. That would > have been a hard job without references to some formal results. Wolfgang M.9fckenheim is a classic crank. Why do you imagine, > as you seem to do, that there is any point arguing with him? > It may seem odd that he has a job as a professor in technical > and mathematical subjects, but such things happen. (Torkel Franzen, sci.math) As I see these endless thread grow, I often ask myself precisely that: Why do you imagine, as you seem to do, that there is any point arguing with him? -- m === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > As I see these endless thread grow, I often ask myself precisely > that: Why do you imagine, as you seem to do, that there is any > point arguing with him? You're right. I'm not going to bother with it anymore without a really good reason. -- hz === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 CLR 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) On Feb 14, 9:31æam, Mariano Su.87rez-Alvarez > There must be a slight misunderstanding. I have been teaching > analysis, algebra and geometry, history of mathematics, and field > theory in total for nearly 20 years at German universities. That would > have been a hard job without references to some formal results. > æ æ æ æ Wolfgang M.9fckenheim is a classic crank. Why do you imagine, > æ æ æ æ æas you seem to do, that there is any point arguing with him? > æ æ æ æ æIt may seem odd that he has a job as a professor in technical > æ æ æ æ æand mathematical subjects, but such things happen. > æ æ æ æ (Torkel Franzen, sci.math) As I see these endless thread grow, I often ask myself precisely > that: Why do you imagine, as you seem to do, that there is any > point arguing with him? As I recall, the poster G. Frege once said that he corrects WM but doesn't argue with him. And I say: As if that distinction really matters that much in the overall scheme of things. MoeBee === Subject: Re: A consideration concerning the diagonal argument of G. Cantor =?ISO-8859-1?Q?Mariano_Su=E1rez-Alvarez?= says... >As I see these endless thread grow, I often ask myself precisely >that: Why do you imagine, as you seem to do, that there is any >point arguing with him? I think that many people fear that if it weren't for pointless arguments with crackpots, this newsgroup would be as quiet a morgue. -- Daryl McCullough Ithaca, NY === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > There must be a slight misunderstanding. I have been teaching > analysis, algebra and geometry, history of mathematics, and field > theory in total for nearly 20 years at German universities. Then God help the future of German mathematics. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor >I use the only valid result lim[n-->oo] S(n) = E. > Then you are still wrong. Or is your definition a one-off applying > only to the one special case and to no other >Wouldn't that be sufficient? In WM's matheology it might be, but no one but WM can figure out what goes on there. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > Limites superior and inferior are > not new to you, I hope. But these notions are only then required, when > there are *more than one* limit point. > There's not more than one limit point in the Halmos approach. Rather, > the sequence is defined to converge when *S = S* and the limit of the > sequence then is *S=S*. > Exactly. That's why the sequence S_n has only one limit point. As this > was is clear from the beginning, at least to me, there was no need to > refer to limits superior and inferior. > æIn my case there is only *one* > limit point, hence I need no lim sup and lim inf. > limit point defined HOW? You've not defined what a limit point in > your own sense. > A limit point is the English translation of the German H.8aufungspunkt. > Which is defined for a sequence of sets to be ... > In mathematics we use this expression rather frequently. > To mean what when talking about a sequence of sets? >I said in mathematics Which includes sequences of sets, so what does it mean for a sequence of sets? === Subject: Re: A consideration concerning the diagonal argument of G. Cantor >[lim n->oo] S_n = E æas I said. > You just SAID that without DEFINING. >That is a definition. Then it is a poor one, and does not define a limit at all, because it gives no justification for interpreting the left hand side as a limiting process, only as a series of uninterpreted symbols. >But the above equation doesn't just hold > on its own - we have to have a specific defintion of 'lim' in such a > context to THEN see whether the equation holds. >A definition is not subject to your investigations. The above equation > need not hold. It is a definition. No more or less valid than garbage = E or >It's amazing to me that you have a such a strong interest in the > subject but still don't understand such basic things about it. OUR position exactly. If what you claim is a definition actually is a definition, then it is a damned poor one, as it replaces a short expression E by a longer and more obscure one, namely [lim n->oo] S_n. One primary purpose of definitions is to shorten and simplify, and yours both expands and complicates. According to your definition, every occurrence of E after your definition's introduction can be, and perhaps even should be, replaced by [lim n->oo] S_n. That hardly shortens or simplifies anything. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > That is a definition. [WM] > Then it is a poor one, and does not define a limit at all, because it > gives no justification for interpreting the left hand side as a limiting > process, only as a series of uninterpreted symbols. > > A definition is not subject to your investigations. The above equation > need not hold. It is a definition. > No more or less valid than garbage =df E > > It's amazing to me that you have a such a strong interest in the > subject but still don't understand such basic things about it. > OUR position exactly. If what you claim is a definition actually is a definition, then it is a > damned poor one, as it replaces a short expression E by a longer and > more obscure one, namely [lim n->oo] S_n. One primary purpose of definitions is to shorten and simplify, and yours > both expands and complicates. According to your definition, every occurrence of E after your > definition's introduction can be, and perhaps even should be, re- > placed by [lim n->oo] S_n. That hardly shortens or simplifies anything. > Completely agree with all you've said. :-) Btw. Stating [lim n->oo] S_n = E then actually means E = E (after elimination of the defined term). Not that breathtaking, imho. F. -- === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > What point do you doubt? Is it the step from lim[n-->oo] {2, 4, 6, ..., 2n} = {2, 4, 6, ...} > to >|lim[n-->oo] {2, 4, 6, ..., 2n}| = lim[n-->oo] |{2, 4, 6, ..., 2n}| = | > {2, 4, 6, ...}| ? That step is unjustified, yes. You haven't given any reason to expect that the lim[n->oo] symbol you defined to apply to a single sequence of sets has any relation at all to the lim[n->oo] symbol as applied to sequences of finite cardinals. Even if there was reason to expect it, mathematics requires *proof*. What's more, even if you did somehow prove such a relation, you would still need to prove that you can interchange the order of the limit and cardinality operators. It gets even worse: in your previous 1/2 argument, you used a limit of sequences of *rational* numbers, which is a different type of limit yet again, and the corresponding relations would have to be proved for those limits, too. This is not just nit-picking. The process of attempting to develop a valid proof for a conjecture often illuminates misconceptions and hidden logic errors. Sometimes it even reveals counterexamples. - Tim === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > What point do you doubt? Is it the step from > lim[n-->oo] {2, 4, 6, ..., 2n} = {2, 4, 6, ...} > to >|lim[n-->oo] {2, 4, 6, ..., 2n}| = lim[n-->oo] |{2, 4, 6, ..., 2n}| = | > {2, 4, 6, ...}| ? That step is unjustified, yes. No. If the sequence of sets exist and converges to E, then the sequence of cardinal numbers must also exist and converge to aleph 0. Should there be an intermediate set which has no cardinal number? Or should the cardinal numbers jump at some set? >æYou haven't given any reason to expect > that the lim[n->oo] symbol you defined to apply to a single sequence > of sets has any relation at all to the lim[n->oo] symbol as applied > to sequences of finite cardinals. æEven if there was reason to expect > it, mathematics requires *proof*. I took the limit of a sequence of real numbers as it is defined in every text book on analysis. It did not interchange any operators. And the limit definition in all cases I apply it is the same: n surpasses every finite number. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=bSICGQkAAADSbkxAJ5uMxFegr4rp0Qig Gecko/20070515 Firefox/2.0.0.4,gzip(gfe),gzip(gfe) > What point do you doubt? Is it the step from > lim[n-->oo] {2, 4, 6, ..., 2n} = {2, 4, 6, ...} > to >|lim[n-->oo] {2, 4, 6, ..., 2n}| = lim[n-->oo] |{2, 4, 6, ..., 2n}| = | > {2, 4, 6, ...}| ? > That step is unjustified, yes. No. If the sequence of sets exist and converges to E, You still haven't said anything about converges to E. In fact you haven't linked E to the sequence of sets at all. You insist that all you have to do is say, in effect, that the string of symbols lim (n->oo) S_n is E by definition. That's fine, we can treat it as a long way of saying E. But since you didn't otherwise tell us what it means in terms of the S_n, then we shouldn't treat it as making any connection with the S_n. Your definition is no different from saying I define WLQKWEFJ@@SDSLADFKJ as being E. Fine. But don't expect to draw conclusions from that. - Randy === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Randy Poe says... > What point do you doubt? Is it the step from > lim[n-->oo] {2, 4, 6, ..., 2n} = {2, 4, 6, ...} > to >|lim[n-->oo] {2, 4, 6, ..., 2n}| = lim[n-->oo] |{2, 4, 6, ..., 2n}| = | > {2, 4, 6, ...}| ? > That step is unjustified, yes. > No. If the sequence of sets exist and converges to E, You still haven't said anything about converges to E. >In fact you haven't linked E to the sequence of sets at all. Well, for his purposes, he can say that the limit of a sequence of sets S(1), S(2),... is just the union. To me, that's not the problem. The problem is that he is computing some function on each set in the sequence, and assuming that the limit of the function is the function of the limit. That's not valid, in general. -- Daryl McCullough Ithaca, NY === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=-XRI1AgAAABBR-yHg7m_4HCGi05mtX1k Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) About the tree, the infinite balanced binary tree that represents each real number between zero and one inclusive as a path from the root descending through the tree, even dually for rationals with nonterminating sequences of repeating 1's or 0's, an infinite sequence of binary 0-1 decisions, fair coin tosses, to map to all of the rationals it's an infinite tree. If every node has beneath it a copy of the tree, and all the trees are rooted, then, there is a node for each copy of the tree. Each irrational, as a path, intersects countably many nodes, each of which as a terminal element represents a reduced ratio with denominator a power of two. For each irrational, as a path, there is a different path representing a different irrational, that one prepended with any number of elements. For each irrational Q_0 there are countably many similar irrationals Q_i, similar in having from some point in the path a copy of Q_0's path. How many copies of the tree must there be in the graph? Is it one for each irrational? If it's not, then there are two irrationals which share representation, i.e., for each node of a path having the same copy, or path. There is only a copy of the tree for each node, each of which is rational. For each element of each irrational's path, it is the root of as well: rationals. Thus, select among the irrationals, for each pairwise a distinct node. Then, there is a unique node for each irrational. That node is a rational. That would have that for each irrational there could be selected a particular node of its path such that for each other irrational, a different node could be selected for that irrational's path. Consider the case where that is not so, and what that would imply. That would mean that for any collection of irrationals that could be made distinct, uniquified, there would be another irrational not in that collection. Yet, if two irrationals are not distinct, then they are simply one element of the collection. Consider the notion of Conway's surreal numbers in their generation and the revelation of Day Omega having there be in existence then irrational elements in the hierarchical denotation of numbers. Here with the binary tree representing elements from the unit interval, for any finite tree depth there are only rational numbers, for finite depth D only ratios of powers of 2, 0/(2^D) through (2^D - 1)/(2^D). For each element of each irrational's path representation, a separate path with the same initial segment also branches to rationals. There are only as many rationals and irrationals as nodes. ( In the infinite balanced binary tree that represents the real numbers, another notion is to represent the numbers as some sort of unbalanced binary tree, and consider whether that would be sufficient to represent the Dedekind/Cauchy cuts, the Eudoxian definition of real number, that is among the standard definitions of real numbers, about those with the least upper bound property, in the usual (open) topology. If the tree were unbalanced, yet still with no leaf nodes, then that would be more along the lines of a representation of continued fractions. ) Anyways, back to the notion of the balanced binary tree and what it means to have each node be the terminus of a rational number's binary surreal numbers and inexistence of irrational numbers until day Omega, a notion is to use the pigeonhole principle / Dirichlet box, in terms of various levels of the tree. For any node that is a part of an irrational's expansion/path, that node is a rational number (the expansion/path of a rational number with reduced form having as a denominator a power of two). Again as each irrational is distinct, any other path branching above or below, a unique node. That node is a rational. Consider the expression of the number of nodes at the n'th level of the tree: 2^n. In the sense of ordinal arithmetic, 2^N = 2^N. There are N many levels of the tree. That leads back to the transfer all nodes, infinite. Ross -- Finlayson Consulting === Subject: Re: A consideration concerning the diagonal argument of G. Cantor [...] |lim[n-->oo] {2, 4, 6, ..., 2n}| = lim[n-->oo] |{2, 4, 6, ..., 2n}| I took the limit of a sequence of real numbers as it is defined in > every text book on analysis. [...] > No you didn't JUST use that notion of limit, since HERE |lim[n-->oo] {2, 4, 6, ..., 2n}| = lim[n-->oo] |{2, 4, 6, ..., 2n}| the notion of limit for sequences of sets is involved. This notion is NOT idientical with the limit for sequences of real numbers. Learn some logic. Learn some mathematics. Or better yet, give up mathematics and take up basket weaving. (from sci.math, @M.9fckenheim) F. -- === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > No. If the sequence of sets exist and converges to E, then the > sequence of cardinal numbers must also exist and converge to aleph_0. How are you defining convergence of sequences of cardinal numbers to aleph_0? If you treat it as a sequence of real numbers corresponding to each cardinal, it does not converge. > I took the limit of a sequence of real numbers as it is defined in > every text book on analysis. Using the textbook definitions appropriate to sequences of real numbers, your sequence does not converge at all. It certainly does not converge to aleph_0, which is not even a real number. > It did not interchange any operators. You interchanged the cardinality and limit operators. On the left side of your claimed equality, you had the limit of the cardinalities of the sets. On the right you had the cardinality of the limit of the sets. There is no reason to expect that interchanging the order of operations in this way will yield the same result, and you certainly didn't supply any proof that it does. > And the limit definition in all cases I apply it is the same: n > surpasses every finite number. The statement n surpasses every finite number is not a mathematical definition of anything, let alone a definition of a limit. It is a vague and ambiguous statement that merely mentions some terminology that is used in mathematics. - Tim === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > It did not interchange any operators. > Of course he did. You interchanged the cardinality and limit operators. On the right side > of your claimed equality, you had the limit of the cardinalities of > the sets. On the left you had the cardinality of the limit of the > sets. There is no reason to expect that interchanging the order of > operations in this way will yield the same result, and you certainly > didn't supply any proof that it does. > Right. The first occruence of lim refers to the notion of limit defined for sequence of sets, while the second occurence of lim refers to the notion of limit defined for real numbers. Of course those two notions are not the same. A case of equivacation. a logical fallacy resulting from the use of multiple meanings of a single expression Learn some logic. Learn some mathematics. Or better yet, give up mathematics and take up basket weaving. (from sci.math, @M.9fckenheim) F. -- === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > What point do you doubt? Is it the step from > lim[n-->oo] {2, 4, 6, ..., 2n} = {2, 4, 6, ...} > to >|lim[n-->oo] {2, 4, 6, ..., 2n}| = lim[n-->oo] |{2, 4, 6, ..., 2n}| = | > {2, 4, 6, ...}| ? > That step is unjustified, yes. >No. If the sequence of sets exist and converges to E, then the > sequence of cardinal numbers must also exist and converge to aleph_0. Claimed but not justified. Can you prove such a claim? > Should there be an intermediate set which has no cardinal number? Or > should the cardinal numbers jump at some set? > >æYou haven't given any reason to expect > that the lim[n->oo] symbol you defined to apply to a single sequence > of sets has any relation at all to the lim[n->oo] symbol as applied > to sequences of finite cardinals. æEven if there was reason to expect > it, mathematics requires *proof*. >I took the limit of a sequence of real numbers as it is defined in > every text book on analysis. Except that unbounded increasing sequences do not have limits in every book on analysis. In fact, many, if not most, of them say that in that case no limit exists. > It did not interchange any operators. And > the limit definition in all cases I apply it is the same: n surpasses > every finite number. Which in standard analysis means that n ( as well as 2n and {2,4, ... 2n}) does not have a limit. > === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > I find the limit 1/2 by applying Cauchy's criterion. This can be used > because > (|{ x | x in S(n) & x > |S(n)| }| / |S(n)|) is a real sequence. The Cauchy criterion applies only to convergence of series, not to convergence of sequences, and what WM wants to apply it to is a sequence, not a series. So WM screws up the math again. As usual. The given sequence, like any other costant sequence, converges to its sole value by trivial delta-epsilonics, with no need of the applying the inappropriate Cauchy criterion. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > I find the limit 1/2 by applying Cauchy's criterion. This can be used > because > (|{ x | x in S(n) & x > |S(n)| }| / |S(n)|) is a real sequence. The Cauchy criterion applies only to convergence of series, not to > convergence of sequences, and what WM wants to apply it to is a > sequence, not a series. So WM screws up the math again. As usual. You should not be proud that you don't know the Cauchy criterion for sequences. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > I use the only valid result lim[n-->oo] S(n) = E. > Then you are still wrong. Or is your definition a one-off applying > only to the one special case and to no other > Wouldn't that be sufficient? > NO! If it were the case, then you need to say so, >I said lim[n-->oo] S(n) = E. We know you said it, but we don't really know why. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > Wouldn't that be sufficient? [WM] > NO! If it were the case, then you need to say so, > > I said lim[n-->oo] S(n) = E. > We know you said it, but we don't really know why. > Well, I guess he might answer Well, it just looked right to me! F. -- === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > Wouldn't that be sufficient? [WM] > NO! If it were the case, then you need to say so, >I said lim[n-->oo] S(n) = E. > We know you said it, but we don't really know why. > Well, I guess he might answer Well, it just looked right to me! > You know: Even a blind chicken finds a kernel of corn now and then. F. -- === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > What point do you doubt? Is it the step from >lim[n-->oo] {2, 4, 6, ..., 2n} = {2, 4, 6, ...} > to > |lim[n-->oo] {2, 4, 6, ..., 2n}| = lim[n-->oo] |{2, 4, 6, ..., 2n}| = | > {2, 4, 6, ...}| ? The one from lim[n-->oo] {2, 4, 6, ..., 2n} to {2, 4, 6,...}. If it were Union_[n in N] {2,4,6,...2n}, instead of a lim we might agree to the resulting value. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > What point do you doubt? Is it the step from > lim[n-->oo] {2, 4, 6, ..., 2n} = {2, 4, 6, ...} > to > |lim[n-->oo] {2, 4, 6, ..., 2n}| = lim[n-->oo] |{2, 4, 6, ..., 2n}| = | > {2, 4, 6, ...}| ? The one from lim[n-->oo] {2, 4, 6, ..., 2n} to {2, 4, 6,...}. If it were Union [n in N] {2,4,6,...2n}, instead of a lim we might > agree to the resulting value. Union is not required here, because the sequence is monotonically increasing. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 CLR 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) > What point do you doubt? Is it the step from > lim[n-->oo] {2, 4, 6, ..., 2n} = {2, 4, 6, ...} > to > |lim[n-->oo] {2, 4, 6, ..., 2n}| = lim[n-->oo] |{2, 4, 6, ..., 2n}| = | > {2, 4, 6, ...}| ? > The one from lim[n-->oo] {2, 4, 6, ..., 2n} to {2, 4, 6,...}. > If it were Union [n in N] {2,4,6,...2n}, instead of a lim we might > agree to the resulting value. Union is not required here, because the sequence is monotonically > increasing. It is because the sequence is monotonically increasing that union is the lim, which is a corollary of the Halmos definition you agreed to be the one we're using in this context. MoeBlee === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > What point do you doubt? Is it the step from > lim[n-->oo] {2, 4, 6, ..., 2n} = {2, 4, 6, ...} > to > |lim[n-->oo] {2, 4, 6, ..., 2n}| = lim[n-->oo] |{2, 4, 6, ..., 2n}| = | > {2, 4, 6, ...}| ? > The one from lim[n-->oo] {2, 4, 6, ..., 2n} to {2, 4, 6,...}. > If it were Union_[n in N] {2,4,6,...2n}, instead of a lim we might > agree to the resulting value. >Union is not required here, because the sequence is monotonically > increasing. Union certainly works here, and works much better than anything of yours. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > What point do you doubt? [WM] > The one from lim[n-->oo] {2, 4, 6, ..., 2n} to {2, 4, 6,...}. If it were Union_[n in N] {2,4,6,...2n}, instead of a lim we might > agree to the resulting value. > That's fine. (S_n) with S_n = {2, 4, 6, ..., 2n} is an increasing sequence of sets; i.e. S_n c S_(n+1) for all n e N. In this case lim S_n = U S_n. n n (Using the notion of /limit/ for sequences of sets as defined in Halmos' book Measure Theory.) F. -- === Subject: Re: A consideration concerning the diagonal argument of G. Cantor >Of course, he refuses to do more than repeat over and > over that limit means union of all the initial > sequences or some variant. >I defined: lim[n-->oo] {2, 4, 6, ..., 2n} = {2, 4, 6, ...} = E. This > definition (like any other definition) needs no further justification. Then if we should define lim[n-->oo] {2, 4, 6, ..., 2n} = {}, it would also need no justification. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > Of course, he refuses to do more than repeat over and > over that limit means union of all the initial > sequences or some variant. > I defined: lim[n-->oo] {2, 4, 6, ..., 2n} = {2, 4, 6, ...} = E. This > definition (like any other definition) needs no further justification. Then if we should define lim[n-->oo] {2, 4, 6, ..., 2n} = {}, it would > also need no justification. No. Remember Dik. T. Winter, who defined Sum N = 0. However, your definition would change considerably the generally accepted meaning of the set of positive natural numbers. My definition does not. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > Of course, he refuses to do more than repeat over and > over that limit means union of all the initial > sequences or some variant. > I defined: lim[n-->oo] {2, 4, 6, ..., 2n} = {2, 4, 6, ...} = E. This > definition (like any other definition) needs no further justification. > Then if we should define lim[n-->oo] {2, 4, 6, ..., 2n} = {}, it would > also need no justification. >No. Remember Dik. T. Winter, who defined Sum N = 0. However, your > definition would change considerably the generally accepted meaning of > the set of positive natural numbers. My definition does not. But, according to you, a definition needs no such justification. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > I defined: lim[n-->oo] {2, 4, 6, ..., 2n} = {2, 4, 6, ...} = E. This > definition (like any other definition) needs no further justification. > Then if we should define lim[n-->oo] {2, 4, 6, ..., 2n} = {}, it would > also need no justification. > Right. That would be the ultima ratio. F. -- === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > lim[n-->oo] |{ x | x in S(n) & x > |S(n)| }| / |S(n)| = 1/2 > By the way, do you mean: > lim[n-->oo] æ( |{ x | x in S(n) & x > |S(n)| }| / |S(n)| ) >of course > or > ( lim[n-->oo] |{ x | x in S(n) & x > |S(n)| }| ) / |S(n)| ? >What should be the meaning of n in the last |S(n)| in that case? When WM can tell us what a last |S(n)| is in a sequence having no last member, only then need we worry about what the corresponding n might be. . === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > lim[n-->oo] |{ x | x in S(n) & x > |S(n)| }| / |S(n)| = 1/2 > By the way, do you mean: > lim[n-->oo] æ( |{ x | x in S(n) & x > |S(n)| }| / |S(n)| ) > of course > or > ( lim[n-->oo] |{ x | x in S(n) & x > |S(n)| }| ) / |S(n)| ? > What should be the meaning of n in the last |S(n)| in that case? When WM can tell us what a last |S(n)| is in a sequence having no last > member, only then need we worry about what the corresponding n might be. Here you understood as much as usual. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > Here you understood as much as usual. Which, however little it may be, is more than WM has understood. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > ( lim[n-->oo] |{ x | x in S(n) & x > |S(n)| }| ) / |S(n)| ? > > What should be the meaning of n in the last |S(n)| in that case? > When WM can tell us what a last |S(n)| is in a sequence having no last > member, only then need we worry about what the corresponding n might be. > He meant What should be the meaning of n in the last |S(n)| [ in the formula ( lim[n-->oo] |{ x | x in S(n) & x > |S(n)| }| ) / |S(n)| ] in that case? (Referring to the fact that this occurrence of n would not be bound in that case.) It's well known that WM has problems with the use-mention distinction. F. -- === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > ( lim[n-->oo] |{ x | x in S(n) & x > |S(n)| }| ) / |S(n)| ? >What should be the meaning of n in the last |S(n)| in that case? > When WM can tell us what a last |S(n)| is in a sequence having no last > member, only then need we worry about what the corresponding n might be. > He meant What should be the meaning of n in the last [occurrence of] |S(n)| [ in the formula ( lim[n-->oo] |{ x | x in S(n) & x > |S(n)| }| ) / |S(n)| ] in that case? (Referring to the fact that this occurrence of n would not be bound in > that case.) It's well known that WM has problems with the use-mention distinction. > Learn some logic. Learn some mathematics. Or better yet, give up mathematics and take up basket weaving. (from sci.math, @M.9fckenheim) F. -- === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > I defined: lim[n-->oo] {2, 4, 6, ..., 2n} = {2, 4, 6, ...} = E. This > definition (like any other definition) needs no further justification. Somewhat surprisingly, I agree. However, a definition of limit that only applies to one sequence of sets is not a very useful definition. - Tim === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > I defined: lim[n-->oo] {2, 4, 6, ..., 2n} = {2, 4, 6, ...} = E. This > definition (like any other definition) needs no further justification. Somewhat surprisingly, I agree. æHowever, a definition of limit that > only applies to one sequence of sets is not a very useful definition. I did not claim that my definition is very useful. I claim that it defines the positive even numbers. (And this set is of more use than whole transfinity.) === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > I did not claim that my definition is very useful. I claim that it > defines the positive even numbers. If E was not already defined to be the even numbers, then your equation defined nothing at all. On the left side fo the equation you have an undefined lim notation. On the right side you have a symbol E that you are now saying was not previously defined either. A statement that expresses something in terms of other undefined things is not a definition. - Tim === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) > I did not claim that my definition is very useful. I claim that it > defines the positive even numbers. If E was not already defined to be the even numbers, then the definition lim [n-->oo] {2, 4, 6, ..., 2n} would define it for the first time. > your > equation defined nothing at all. On the left side fo the equation you have an undefined lim notation. No, there is but one fundamental limit definition, namely that n gets larger than any natural number. > On the right side you have a symbol E that you are now saying was not > previously defined either. E stands for {2, 4, 6, ...}. I do not say that it was not previously defined. But if it was not defined, then it would be defined now! A statement that expresses something in terms of other undefined > things is not a definition. n and --> and oo have been defined long ago. These symols put together answer the question what happens when n gets larger than any fixed number. But of course, every statement is something in terms of other undefined things, because langage is nt given by God. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > I defined: lim[n-->oo] {2, 4, 6, ..., 2n} = {2, 4, 6, ...} = E. This > definition (like any other definition) needs no further justification. > Somewhat surprisingly, I agree. æHowever, a definition of limit that > only applies to one sequence of sets is not a very useful definition. >I did not claim that my definition is very useful. I claim that it > defines the positive even numbers. (And this set is of more use than > whole transfinity.) Which positive even integers does it allegedly define? To me they must each of them have to be previously defined before you can even invoke your definition. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > I defined: lim[n-->oo] {2, 4, 6, ..., 2n} = {2, 4, 6, ...} = E. This > definition (like any other definition) needs no further justification. > Somewhat surprisingly, I agree. However, a definition of limit that > only applies to one sequence of sets is not a very useful definition. > Sorry, this way IMHO he did not even define the notion of /limits/ for one string, but just the constant lim[n-->oo] {2, 4, 6, ..., 2n}. And I'm not even sure if there is any system which would allow that string as a wf term. F. -- === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Ooops... Typos... > I defined: lim[n-->oo] {2, 4, 6, ..., 2n} = {2, 4, 6, ...} = E. This > definition (like any other definition) needs no further justification. > Somewhat surprisingly, I agree. However, a definition of limit that > only applies to one sequence of sets is not a very useful definition. > Sorry, this way IMHO he did not even define the notion of /limit/ for > one sequence, but just the constant lim[n-->oo] {2, 4, 6, ..., 2n}. > ^^^^^^^^ > And I'm not even sure if there is any system which would allow for that > string as a wf term. > F. -- === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > If the limit exists, then also the limits > lim[n-->oo] |S(n)| > and > lim[n-->oo] |{ x | x in S(n) & x > |S(n)| } > exist, no? > Then > lim[n-->oo] |{ x | x in S(n) & x > |S(n)| }| / |S(n)| = 1/2 > and, therefore, > lim{n-->oo} |{ x | x in S(n) & x > |S(n)| }| / |S(n)| =/= 0. > (1) Before I evaluate anything, please tell me whether we're using the > Halmos method or the simpler definition I mentioned: lim_n->oo S(n) = > Urange(S). > I use the only valid result lim[n-->oo] S(n) = E. Of course you can > apply the Halmos method, although I do not know what there is very > special > What is special is that is is a DEFINITION as opposed to your context > in which the notation was UNdefined. >I defined: lim[n-->oo] {2, 4, 6, ..., 2n} = {2, 4, 6, ...} = E. This > definition (like any other definition) needs no further justification. Then we are equally justified in defining lim[n-->oo] {2, 4, 6, ..., 2n} = {} or lim[n-->oo] {2, 4, 6, ..., 2n} = N or anything else we might dream up. What we want is a definition of lim[n-->oo] S_n, where each S_n is a set indexed by a member of N, which will allow us to determine (1) whether a limit exists for a particular seqeunce of sets and (2) what the limit turns out to be when it should exist according to the definition. Your pseudo-definition does neither. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor >Limites superior and inferior are > not new to you, I hope. But these notions are only then required, when > there are *more than one* limit point. > There's not more than one limit point in the Halmos approach. Rather, > the sequence is defined to converge when *S = S* and the limit of the > sequence then is *S=S*. >Exactly. That's why the sequence S_n has only one limit point. As this > was is clear from the beginning, at least to me, there was no need to > refer to limits superior and inferior. Without referring t to them, there is no guarantee that your limit exists by that definition. And as you carefully have avioded ever giving a definition by which your claimed limit could be proved to exist, you have no limit of your own. > In my case there is only *one* > limit point, hence I need no lim sup and lim inf. > limit point defined HOW? You've not defined what a limit point in > your own sense. >A limit point is the English translation of the German H.8aufungspunkt. > In mathematics we use this expression rather frequently. It refers to > finite values or sets in general but its definition can be extended to > infinite values or sets. And what does the definition of H.8aufungspunkt translate to in English? A word without a definition is merely a string of nonsense syllables. > That is why I did > not mention it. The method itself should be known since Archimedes > calculated the area of a circle or since we determine integrals by the > limits of upper and lower sum. > That's for a certain KIND of sequence. You've not given a definition > for sequences as general as yours. >If no special definition is given, then the general definition has to > be applied. What general definition? There is no such general definition in mathematics, whatever may exist in WM's matheology. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor [...] > It seems that many cranks do not understand the crucial role definitions play in mathematics in general and especially for mathematical proofs. Typical statements: As this was is clear from the beginning, at least to me, there was no need to refer to limits superior and inferior. [WM] or In my _finitary_ comprehension, the whole thing is sooo obvious! [Han] etc. Yeah, having not defined the terms used EVERYTHING may be obvious. (*sigh*) F. -- === Subject: Re: A consideration concerning the diagonal argument of G. Cantor I understand certain basics about definitions. These are basics you > need to understand. > The sad truth, Moe, is: he doesn't. F. -- === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > Wouldn't that be sufficient? [WM] > NO! If it were the case, then you need to say so, > I said lim[n-->oo] S(n) = E. [WM] >æand if it were the case, then it is EXACTLY as I mentioned: all > you've done is stipulate that a certain string stands for E. > Is defining the same as stipulating in your understanding of the > English language? [WM] > No. > If not, then replace to stipulate by to define, please. > You should not try to change the words in my mouth. [WM] > I didn't say that YOU said you are merely stipulating. I am saying you > are merely stipulating. > Learn some logic. Learn some mathematics. Or better yet, give up mathematics and take up basket weaving. (from sci.math, @M.9fckenheim) F. -- === Subject: Re: A consideration concerning the diagonal argument of G. Cantor When YOU say lim X_n = X > n > In other words, lim sup Xn consists of those elements > which are in Xn for infinitely many n, while lim inf Xn > consists of those elements which are in Xn for all but > finitely many n. > This is in agreement with Halmos' approach: If (E_n) is a sequence of subsets of X, the set E^* of all those points of X which belong to E_n for infinitely many values of n is called the /superior limit/ of the sequence and is denoted by E^* = lim sup E_n. n The set E_* of all those points which belongs to E_n for all but a finite number of values of n is called the /inferior limit/ of the sequence and is denoted by E_* = lim inf E_n. n And this way we get the limit asked for: If it happens that E_* = E^*, we shall use the notation lim E_n n for this set. (Paul R. Halmos, Measure Theory) That's the (or at least _a_) definition WM is not able to state. :-) Learn some logic. Learn some mathematics. Or better yet, give up mathematics and take up basket weaving. (from sci.math, @M.9fckenheim) F. -- === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Would you mind posting the proof that last night you mentioned having > made (along with a restatement of the problem). I'd like to have it > for my notebook. > No problem. This way I'll get a feedback, which certainly is a rather good thing. First the original text: If (E_n) is a sequence of subsets of X, the set E^* of all those points of X which belong to E_n for infinitely many values of n is called the /superior limit/ of the sequence and is denoted by E^* = lim sup E_n. n The set E_* of all those points which belongs to E_n for all but a finite number of values of n is called the /inferior limit/ of the sequence and is denoted by E_* = lim sup E_n. n If it happens that E^* = E_*, we shall use the notation lim E_n n for this set. (Paul R. Halmos, Measure Theory) And here's the formulation of the related exercise: (2) If E_* = lim inf E_n and E^* = lim sup E_n, then n n oo oo oo oo E_* = / / E_m c / / E_m = E^*. n=1 m=n n=1 m=n Now my solution is the following: x e E_* <-> {n e N : x !e E_n} finite <-> En e N Am >= n : x e E_m oo <-> En e N : x e / E_m m=n oo oo <-> x e / ( / E_m). n=1 m=n x e E^* <-> {n e N : x e E_n} infinite <-> An e N Em >= n : x e E_m oo <-> An e N : x e / E_m m=n oo oo <-> x e / ( / E_m). n=1 m=n x e E_* -> {n e N : x !e E_n} finite -> {n e N : x e E_n} infinite -> x e E^*. F. -- === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 CLR 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) > Now my solution is the following: x e E * <-> {n e N : x !e E n} finite ? ? ? ? <-> En e N Am >= n : x e E m > ? ? ? ? ? ? ? ? ? ? ? ? ?oo > ? ? ? ? <-> En e N : x e / E m > ? ? ? ? ? ? ? ? ? ? ? ? ?m=n > ? ? ? ? ? ? ? ? oo ? oo > ? ? ? ? <-> x e / ( / ?E m). > ? ? ? ? ? ? ? ? n=1 ?m=n x e E^* <-> {n e N : x e E n} infinite ? ? ? ? <-> An e N Em >= n : x e E m > ? ? ? ? ? ? ? ? ? ? ? ? ?oo > ? ? ? ? <-> An e N : x e / E m > ? ? ? ? ? ? ? ? ? ? ? ? ?m=n > ? ? ? ? ? ? ? ? oo ? oo > ? ? ? ? <-> x e / ( / ?E m). > ? ? ? ? ? ? ? ? n=1 ?m=n x e E * -> {n e N : x !e E n} finite ? ? ? ? -> {n e N : x e E n} infinite ? ? ? ? -> x e E^*. Yes, quite straightforward. A nice little theorem. Also (I found passages of the book), the part about monotone up to union and down to intersection is straightforward. MoeBlee === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 CLR 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) > No problem. over, hopefully not later than this weekend. MoeBlee === Subject: Re: A consideration concerning the diagonal argument of G. Cantor it over, hopefully not later than this weekend. > You might like to check the following posts too: <7e07r357fdqkki1453eqq1cidmoc5ilvsn@4ax.com> F. -- === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 CLR 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) > You might like to check the following posts too: > Those don't link for me. MoeBlee === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Typos... If (E_n) is a sequence of subsets of X, the set E^* of all those points > of X which belong to E_n for infinitely many values of n is called the > /superior limit/ of the sequence and is denoted by E^* = lim sup E_n. > n The set E_* of all those points which belongs to E_n for all but a > finite number of values of n is called the /inferior limit/ of the > sequence and is denoted by > E_* = lim inf E_n. <------- corrected > n If it happens that E_* = E^*, we shall use the notation lim E_n > n for this set. (Paul R. Halmos, Measure Theory) > F. -- === Subject: Buttazzo - Hard real time computing systems, second edition posting-account=MMaz-QoAAAAOYFe5puLu02m0y7iamRMU MathPlayer 2.10b; InfoPath.2),gzip(gfe),gzip(gfe) I am currently looking for the pdf version of this book. If anybody has any information please let me know. === Subject: Re: random variable posting-account=-gYNIQoAAAD-MKGbMY_-Jpj1EqcOgRWx 3.0.2; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) > hi:) > can someone plz correct me if i am wrong : there are two busses that independently arrive at the bus stop with > rate 3 and 4 per hour respectively.i need to find the probability that > after my arrival at the bus stop ill see both of the busses in less > than 20 minutes so .. both busses are continuous random variables with exponential > distribution .. right ? and they are independent I imagine that you mean two bus *routes*, with a stream of many buses plying each route, not just two physical buses. In that case the exponential waiting-time model does not seem unreasonable (though whether it matches the behaviour of real-life buses I wouldn't know). so P( i see them both in less then 20 min) =P( i see bus a < 20min) * > P(i see bus b < 20 min) Bus a : it it arrives at the rate 3 per hour then its 1 per 20 minutes > ( so lambda a = 1 ) > Bus b : arrives at the rate 4 per hour then its 4/3 per 20 minutes > ( so lambda b= 4/2) Presumably a typo for 4/3. next ... > probability that i'll see a bus in less than chosen time interval is: > æ æ P(X<=1) = 1-e^(-lambda) my chosen interval is 20min and lambda a and lambda b for chosen > interval were just calculated so the total result is : P(i see both busses <20min) = ( 1 - e^(-1)) * ( 1 - e^(-4/3)) = > 0.46549... æ ? This answer looks correct to me, given the assumptions. === Subject: Re: random variable > so .. both busses are continuous random variables with exponential > distribution .. right ? I expect not. That would imply a non-negligible probability of bus A arriving at the stop within 1 minute of its previous arrival. If you're specifically studying exponential or Poisson distributions, then it might be implied that you should model it that way. Otherwise, I think it would be a very poor model. Another simple model is to assume that the buses have periodic arrival times, but with uniformly distributed and independent phases. If that is the case, then in any 20 minute period it would be certain that you would see both. Assumptions about the model make a huge difference. - Tim === Subject: Re: random variable > so .. both busses are continuous random variables with exponential > distribution .. right ? I expect not. That would imply a non-negligible probability of bus A >arriving at the stop within 1 minute of its previous arrival. Clearly, you don't take the bus very often. If you did, you would know that buses typically arrive in threes! quasi === Subject: Re: random variable > Clearly, you don't take the bus very often. If you did, you would > know that buses typically arrive in threes! Hehe, yes. If the problem had many different buses running on a route, a Poisson process could be appropriate. However, I don't think I've ever seen the *same* bus turn up twice within a minute :-) - Tim === Subject: square of a number posting-account=fl4D2woAAAC4QBFmZeykoadHa2UXfAKY Gecko/20070321 Firefox/2.0.0.3 (Swiftfox),gzip(gfe),gzip(gfe) Hi all, 1 = 1 = 1^1 1+2+1 = 2+2 = 2^2 1+2+3+2+1 = 3+3+3 = 3^2 1+2+3+4+3+2+1 = 4+4+4+4 = 4^2 1+2+3+4+5+4+3+2+1 = 5+5+5+5+5 = 5^2 1+2+3+4+5+6+5+4+3+2+1 = 6+6+6+6+6+6 = 6^2 1+2+3+4+5+6+7+6+5+4+3+2+1 = 7+7+7+7+7+7+7 = 7^2 1+2+3+4+5+6+7+8+7+6+5+4+3+2+1 = 8+8+8+8+8+8+8+8 = 8^2 1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1 = 9+9+9+9+9+9+9+9+9 = 9^2 I think the above pattern can be generalized into the result n^2 = 2 x [(n-1)/2 (1 + (n-1))] + n === Subject: Re: square of a number posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 4334.34; Windows NT 5.1; SV1; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) spider-dtc-tc07.proxy.aol.com[CDBC7047] (Prism/1.2.1), HTTP/1.1 cache-dtc-ad05.proxy.aol.com[CDBC74C7] (Traffic-Server/6.1.5 [uScM]) > ?Hi all, 1 = 1 = 1^1 a a > 1+2+1 ?= 2+2 = 2^2 b cb a b c ab > 1+2+3+2+1 = 3+3+3 = 3^2 c d e c b c d d e c a b c d e a b c > 1+2+3+4+3+2+1 = 4+4+4+4 = 4^2 d e f g d c d e e f c d b c d e f e b c d a b c d e f g a b c d etc. n^2 = sum(n) + sum(n-1) > 1+2+3+4+5+4+3+2+1 = 5+5+5+5+5 = 5^2 > 1+2+3+4+5+6+5+4+3+2+1 = 6+6+6+6+6+6 = 6^2 > 1+2+3+4+5+6+7+6+5+4+3+2+1 = 7+7+7+7+7+7+7 = 7^2 > 1+2+3+4+5+6+7+8+7+6+5+4+3+2+1 = 8+8+8+8+8+8+8+8 = 8^2 > 1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1 = 9+9+9+9+9+9+9+9+9 ?= 9^2 I think the above pattern can be generalized into the result ?n^2 ?= 2 x [(n-1)/2 (1 + (n-1))] + n === Subject: Re: square of a number skrev i en meddelelse > Hi all, 1 = 1 = 1^1 > 1+2+1 = 2+2 = 2^2 > 1+2+3+2+1 = 3+3+3 = 3^2 > 1+2+3+4+3+2+1 = 4+4+4+4 = 4^2 > 1+2+3+4+5+4+3+2+1 = 5+5+5+5+5 = 5^2 > 1+2+3+4+5+6+5+4+3+2+1 = 6+6+6+6+6+6 = 6^2 > 1+2+3+4+5+6+7+6+5+4+3+2+1 = 7+7+7+7+7+7+7 = 7^2 > 1+2+3+4+5+6+7+8+7+6+5+4+3+2+1 = 8+8+8+8+8+8+8+8 = 8^2 > 1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1 = 9+9+9+9+9+9+9+9+9 = 9^2 I think the above pattern can be generalized into the result n^2 = 2 x [(n-1)/2 (1 + (n-1))] + n * ** *** **** *** ** * Do you see the pattern? Aage === Subject: Re: square of a number Aage Andersen : > * > ** > *** > **** > *** > ** > * Do you see the pattern? ;-) Or a b c d b c d c c d c b d c b a Triangular numbers and squares... guido http://home.scarlet.be/~pin12499/hypereal.htm === Subject: Re: square of a number posting-account=ogMREwkAAAC5xUr8sg7heGtsvzzF18LA Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > Hi all, 1 = 1 = 1^1 > 1+2+1 = 2+2 = 2^2 > 1+2+3+2+1 = 3+3+3 = 3^2 > 1+2+3+4+3+2+1 = 4+4+4+4 = 4^2 > 1+2+3+4+5+4+3+2+1 = 5+5+5+5+5 = 5^2 > 1+2+3+4+5+6+5+4+3+2+1 = 6+6+6+6+6+6 = 6^2 > 1+2+3+4+5+6+7+6+5+4+3+2+1 = 7+7+7+7+7+7+7 = 7^2 > 1+2+3+4+5+6+7+8+7+6+5+4+3+2+1 = 8+8+8+8+8+8+8+8 = 8^2 > 1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1 = 9+9+9+9+9+9+9+9+9 = 9^2 I think the above pattern can be generalized into the result n^2 = 2 x [(n-1)/2 (1 + (n-1))] + n You might want to also look at the relationship between triangular numbers* to squares. You should see both of them in your examples above. *In case you don't know the term, triangular numbers are numbers that are othe sum of all positive integers from 1 to n, for some integer n. === Subject: Re: square of a number > Hi all, >1 = 1 = 1^1 > 1+2+1 = 2+2 = 2^2 > 1+2+3+2+1 = 3+3+3 = 3^2 > 1+2+3+4+3+2+1 = 4+4+4+4 = 4^2 > 1+2+3+4+5+4+3+2+1 = 5+5+5+5+5 = 5^2 > 1+2+3+4+5+6+5+4+3+2+1 = 6+6+6+6+6+6 = 6^2 > 1+2+3+4+5+6+7+6+5+4+3+2+1 = 7+7+7+7+7+7+7 = 7^2 > 1+2+3+4+5+6+7+8+7+6+5+4+3+2+1 = 8+8+8+8+8+8+8+8 = 8^2 > 1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1 = 9+9+9+9+9+9+9+9+9 > = 9^2 >I think the above pattern can be generalized into the result > n^2 = 2 x [(n-1)/2 (1 + (n-1))] + n ...which is the same as n^2? Daniel -- Done: Bar-Sam-Val, Dwa-Hum-Orc, Cha-Law, Fem-Mal Underway: Wiz-Elf-Cha-Mal To go: Arc-Cav-Hea-Kni-Mon-Pri-Ran-Rog-Tou, Gno, Neu === Subject: Re: square of a number > Hi all, > 1 = 1 = 1^1 > 1+2+1 = 2+2 = 2^2 > 1+2+3+2+1 = 3+3+3 = 3^2 > 1+2+3+4+3+2+1 = 4+4+4+4 = 4^2 > 1+2+3+4+5+4+3+2+1 = 5+5+5+5+5 = 5^2 > 1+2+3+4+5+6+5+4+3+2+1 = 6+6+6+6+6+6 = 6^2 > 1+2+3+4+5+6+7+6+5+4+3+2+1 = 7+7+7+7+7+7+7 = 7^2 > 1+2+3+4+5+6+7+8+7+6+5+4+3+2+1 = 8+8+8+8+8+8+8+8 = 8^2 > 1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1 = 9+9+9+9+9+9+9+9+9 > = 9^2 > I think the above pattern can be generalized into the result > n^2 = 2 x [(n-1)/2 (1 + (n-1))] + n >...which is the same as n^2? ... well, but the pattern is nice! Daniel -- Done: Bar-Sam-Val, Dwa-Hum-Orc, Cha-Law, Fem-Mal Underway: Wiz-Elf-Cha-Mal To go: Arc-Cav-Hea-Kni-Mon-Pri-Ran-Rog-Tou, Gno, Neu === Subject: Re: square of a number- > Hi all, >1 = 1 = 1^1 > 1+2+1 = 2+2 = 2^2 > 1+2+3+2+1 = 3+3+3 = 3^2 > 1+2+3+4+3+2+1 = 4+4+4+4 = 4^2 > 1+2+3+4+5+4+3+2+1 = 5+5+5+5+5 = 5^2 > 1+2+3+4+5+6+5+4+3+2+1 = 6+6+6+6+6+6 = 6^2 > 1+2+3+4+5+6+7+6+5+4+3+2+1 = 7+7+7+7+7+7+7 = 7^2 > 1+2+3+4+5+6+7+8+7+6+5+4+3+2+1 = 8+8+8+8+8+8+8+8 = 8^2 > 1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1 = 9+9+9+9+9+9+9+9+9 > = 9^2 >I think the above pattern can be generalized into the result > n^2 = 2 x [(n-1)/2 (1 + (n-1))] + n Nice discovery! Rearrange the additions as 1+2+3+2+1 = 1+2 + 2+1 + 3 = 3*3 = 9 (i.e. first terms numbers one and four, then terms numbers two and five, and finally the middle term) Likewise, 1+2+3+4+3+2+1 = 1+3 + 2+2 + 3+1 + 4 = 4*4 = 16 etc. Now use mathematical induction (i.e. proof from N to N+1) to prove the general result! Happy exploring and good luck: Johan E. Mebius === Subject: Eigenvalues/vectors of certain types of matrices posting-account=KMurQQkAAACkDDGELZpG-7yQAg7fSfzi Gecko/20061201 Firefox/2.0.0.6 (Ubuntu-feisty),gzip(gfe),gzip(gfe) I was wondering if there was anything that could be said about the eigenvalues/vectors of real matrices with the following two properties: i). All columns sum to zero ii). All diagonal elements are <= 0. In particular, I'm interested in whether all eigenvalues are always either 0 or otherwise have real part < 0 and whether or not the eigenvectors span R^n. Richard. === Subject: Re: Eigenvalues/vectors of certain types of matrices posting-account=K5WE3woAAAAXArsybjkbN6LjMxWdHtbX Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) I was wondering if there was anything that could be said about the > eigenvalues/vectors of real matrices with the following two > properties: i). All columns sum to zero > ii). All diagonal elements are <= 0. In particular, I'm interested in whether all eigenvalues are always > either 0 or otherwise have real part < 0 and whether or not the > eigenvectors span R^n. > Richard. I don't know if this will answer your question in general, but I think it at least shows the nonzero eigenvalues have real parts < 0. Instead of having the columns summing to zero, let the rows sum to zero. The your matrix, A, is the transition rate matrix of a continuous-time Markov chain. We can get the long-run behaviour by looking at exp(A*t) for large t > 0. A standard trick in this case is to let f > 0 be >= largest |a(j,j)|, then to write A = -fI + fP, where I = identity matrix and P = I + (1/f)A. P is a stochastic matrix (elements >= 0 and row sums = 1). We have exp(A*t) = sum(p[n](ft)*P^n, n=0..infinity), where p[n](u) = exp(-u)*u^n/n! for n = 0, 1, ... is the Poisson distribution. If P corresponds to an aperiodic, irreducible Markov chain, the matrix P^n approaches a limit L as n -> infinity (where the rows of L are identical and are the limiting state probabilities). If P is reducible and/or periodic, there may not be a limit, but periodic subsequences of P^n converge to L-like matrices, or there is a limit but with unequal rows and some rows equal to zero. In any case the matrices P^n remain nicely bounded, so the series for exp(A*t) converges for any t > 0 and remains bounded for all t > 0. On the other hand, one can perform an eigenvalue expansion of exp(A*t); if any eigenvalue has real part > 0 it looks like exp(A*t) can become unbounded as t --> infinity. (There may need to be some parts of this argument that need cleaning up.) The other, spanning, issue is not addressed by this analysis. However, the eigenvectors of A and P (the one above) are the same, so one can use information about eigenvalues and eigenspaces of stochastic matrices to, perhaps, get a handle on the problem. Beyond that, I don't know the answer. R.G. Vickson === Subject: Re: Eigenvalues/vectors of certain types of matrices posting-account=BOb97goAAADQ9S76yCy85K7EenLx1xAN Gecko/20061201 Firefox/2.0.0.12 (Ubuntu-feisty),gzip(gfe),gzip(gfe) I was wondering if there was anything that could be said about the > eigenvalues/vectors of real matrices with the following two > properties: i). All columns sum to zero > ii). All diagonal elements are <= 0. In particular, I'm interested in whether all eigenvalues are always > either 0 or otherwise have real part < 0 and whether or not the > eigenvectors span R^n. Try A = [-1 3 5; -1 -1 4; 2 -2 -1] as a counterexample for the first one. === Subject: Re: Eigenvalues/vectors of certain types of matrices posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/20070530 Fedora/1.5.0.12-1.fc5 Firefox/1.5.0.12,gzip(gfe),gzip(gfe) > I was wondering if there was anything that could be said about the > eigenvalues/vectors of real matrices with the following two > properties: > i). All columns sum to zero > ii). All diagonal elements are <= 0. > In particular, I'm interested in whether all eigenvalues are always > either 0 or otherwise have real part < 0 and whether or not the > eigenvectors span R^n. Try A = [-1 3 5; -1 -1 4; 2 -2 -1] as a counterexample for the first > one. And A = {{-9, -9, 18}, {-2, -1, 3}, {-2, 6, -4}} gives a counter example to the second one, as its Jordan form has one block of size 2. -- m === Subject: Re: Eigenvalues/vectors of certain types of matrices posting-account=KMurQQkAAACkDDGELZpG-7yQAg7fSfzi Gecko/20061201 Firefox/2.0.0.6 (Ubuntu-feisty),gzip(gfe),gzip(gfe) > I was wondering if there was anything that could be said about the > eigenvalues/vectors of real matrices with the following two > properties: > i). All columns sum to zero > ii). All diagonal elements are <= 0. > In particular, I'm interested in whether all eigenvalues are always > either 0 or otherwise have real part < 0 and whether or not the > eigenvectors span R^n. Try A = [-1 3 5; -1 -1 4; 2 -2 -1] as a counterexample for the first > one. Your example's 3rd column does not sum to zero so it doesn't satisfy my two properties. Richard. === Subject: Re: Eigenvalues/vectors of certain types of matrices posting-account=BOb97goAAADQ9S76yCy85K7EenLx1xAN Gecko/20061201 Firefox/2.0.0.12 (Ubuntu-feisty),gzip(gfe),gzip(gfe) > I was wondering if there was anything that could be said about the > eigenvalues/vectors of real matrices with the following two > properties: > i). All columns sum to zero > ii). All diagonal elements are <= 0. > In particular, I'm interested in whether all eigenvalues are always > either 0 or otherwise have real part < 0 and whether or not the > eigenvectors span R^n. > Try A = [-1 3 5; -1 -1 4; 2 -2 -1] as a counterexample for the first > one. > Your example's 3rd column does not sum to zero so it doesn't satisfy > my two properties. Sign error. Try A = [-1 3 5; -1 -1 -4; 2 -2 -1]. === Subject: A question in field theory! posting-account=zn2mMAoAAAB8GcQbJgFi7C8YMkm06Ath CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) Hi! I want to ask some question in field theory. Assume Q(root(r))=Q(root(s)) for real number r, s. Then, show that r/s=t^2 for some rational number t. (here Q is the set of all rational number) I cannot find any light. Do you have any idea? === Subject: Re: A question in field theory!- > Hi! >I want to ask some question in field theory. >Assume Q(root(r))=Q(root(s)) for real number r, s. >Then, show that r/s=t^2 for some rational number t. >(here Q is the set of all rational number) >I cannot find any light. Do you have any idea? Construct a field isomorphism of A = Q(sqrt(r)) and B = Q(sqrt(s)) and see how things go. A and B are 2D vector spaces over Q. You could start considering vector space isomorphisms F of A and B, where A and B are considered as just vector spaces over Q. I.e. forget multiplication and division for that time being. Next thing to do is to impose the additional requirement that F(x*y) = F(x)*F(y). This will narrow down the whole thing to just two isomorphisms: the identity isomorphism and the algebraic conjugation 1 -> 1, sqrt(r) -> minus sqrt(r). Happy studies and good luck: Johan E. Mebius === Subject: Re: A question in field theory!- <47B4495D.3080400@xs4all.nl> posting-account=MDF9GgoAAAAFbkuHN6YOWm0JqWwKtflo Gecko/20061023 SUSE/2.0.0.2-1.1 Firefox/2.0.0.2,gzip(gfe),gzip(gfe) > Hi! > I want to ask some question in field theory. > Assume Q(root(r))=Q(root(s)) for real number r, s. > Then, show that r/s=t^2 for some rational number t. > (here Q is the set of all rational number) sqrt(r) and sqrt(s) must be both rational or irrational. In the first case its easy, for the second one, remark that sqrt(s) and sqrt(r) must satisfies a linear relation (over Q) and show (from the fact that sqrt(s) is irrational) that this relation is actually a linear dependence between sqrt(s) and sqrt(r), then you're done. Best B. > I cannot find any light. Do you have any idea? Construct a field isomorphism of A = Q(sqrt(r)) and B = Q(sqrt(s)) and see how things go. A and B are 2D vector spaces over Q. You could start considering vector space isomorphisms > F of A and B, where A and B are considered as just vector spaces over Q. > I.e. forget multiplication and division for that time being. Next thing to do is to impose the additional requirement that F(x*y) = F(x)*F(y). > This will narrow down the whole thing to just two isomorphisms: the identity isomorphism > and the algebraic conjugation 1 -> 1, sqrt(r) -> minus sqrt(r). Happy studies and good luck: Johan E. Mebius === Subject: Re: A question in field theory!- posting-account=MDF9GgoAAAAFbkuHN6YOWm0JqWwKtflo Gecko/20061023 SUSE/2.0.0.2-1.1 Firefox/2.0.0.2,gzip(gfe),gzip(gfe) sqrt(r) and sqrt(s) must be both rational or irrational. > In the first case its easy, Actually this is easily false, you must take sqrt(r) and sqrt(s) irrational... :-) === Subject: Re: A question in field theory!- You can say sqrt(r)=a+b*sqrt(s) for a, b rationals. Then square both sides. === Subject: Re: A question in number theory! <9whsj.24502$j7.452325@news.indigo.ie> posting-account=zn2mMAoAAAB8GcQbJgFi7C8YMkm06Ath CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) > Assume there is a group G whose order is p-1 and d | p-1. ----(1) > Then, is there exactly d solutions in G satisfying x^d=e? > For example, if G=(F p)*, it is true. This is obviously not true. > Eg S 3 has 4 elements satisfying x^2 = e, > as does C 2 x C 2. > The reason I wondering this as follows; > Consider the multiplicative character group mod (p). > Then there are exactly phi(p)=p-1 characters and if d | p-1, the book > I have says there are exactly d characters > s.t.(khai)^d =(khai) 0 (principle character). > If (1) is not true, I am wondering character group (mod p) is > isomorphic to (F p)* and more generally, if m has primitive roots, > then charater group (mod m) is isomorphic to (Z/mZ)*? If G is any finite abelian group, > the dual group G* (ie the group formed by the characters) > is isomorphic to G. -- > Timothy Murphy > e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie > tel: +353-86-2336090, +353-1-2842366 > s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland Sorry. I admit that I did not fully think this problem since I am not convinced the result. But it turns out that the more thing Murphy is holds. conclusion. === Subject: Reduce/simplify the notation from a long formula posting-account=5AHSmAoAAACORxv3ciJw-Rdarq0xAtyi Gecko/2007121120 Firefox/3.0b2,gzip(gfe),gzip(gfe) Hi all, I'm dealing with a very long formula (more than 9000 chars) which contain a lot of repeating parts. Are there software tools capable to reduce/simplify the notation? Example: (1 + 1) / (2 * 2) - (1 + 1) ^ (2 * 2) X1 = (1+1) X2 = (2*2) Formula = X1 / X2 - X1 ^ X2 - Jim === Subject: multiple of 3 posting-account=fl4D2woAAAC4QBFmZeykoadHa2UXfAKY Gecko/20070321 Firefox/2.0.0.3 (Swiftfox),gzip(gfe),gzip(gfe) Hi all, This is the result which i recently read in a math book as follows a multiple of 3, say 3n ,can always be represented by the sum (n-1) + n + (n+1), now my question is follows , can any one suggest a similar relationship for 5n,6n,7n,9n........???? === Subject: Re: multiple of 3 > Hi all, >This is the result which i recently read in a math book as follows >a multiple of 3, say 3n ,can always be represented by the sum >(n-1) + n + (n+1), now my question is follows , can any one suggest a > similar relationship for 5n,6n,7n,9n........???? Odd numbers should be easy. Clue: -x+x=0. Phil -- -- Microsoft voice recognition live demonstration === Subject: How Did Online Dating Become So Popular? posting-account=tRnjwQoAAACkKtXtuM9uiT1-Y8XO314r 5.1),gzip(gfe),gzip(gfe) The reason is pretty simple. It is very much the same reason that the internet itself became so popular. The Internet opens up a whole new world of communication and contact. And the reasons for this are given below. Speed Try to picture what used to happen earlier in the days when people had to depend on the good ol' postal system. During those days, a person had to wait for one or two days for a letter to get across to a person who lived in the same state itself. The second person in turn would take one or two days to respond and this letter would take on or two days to get back to the first person. So in effect, a single correspondence would stretch over a week. But now it's a totally different story. The time taken for the first letter and the response has been brought to an amazing 2 minutes! Waiting may make the heart grow fonder but e-mail makes two people get close faster! Privacy The Internet provides for absolute privacy too. One can carry out communication with another person in the absolute privacy of one's bedroom or bath room or wherever one chooses to be. There is no fear chat facilities. Options and Opportunities The Internet provides for other options like voice chat or video conferencing and stops short only of the physical touch. But then who would want to start a relationship by touching http://www.dontplayplay.com/html/Dating/20060925/9975.html === Subject: Re: How Did Online Dating Become So Popular? posting-account=G-TjQAkAAADYg6rno3bWQPnIwKFBrf1t CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) > The reason is pretty simple. It is very much the same reason that the > internet itself became so popular. The Internet opens up a whole new > world of communication and contact. And the reasons for this are given > below. Speed ... Privacy ... Options and Opportinities The reason is even simpler than this. Middle America, being about 20 years behind the rest of us on Cyberspace is ONLY NOW starting to get into what we've long since passed into and beyond; all the while pretending that this is an artifice of the new generation. (In which case, us members of the new generation must have magically time travelled back to the 1980's, by that logic). It was only good for a while. A half dozen old men (for women) and fat Society, again, being 20 years behind the rest of us, is only now starting to get to that phase. It's just a phase. History repeating itself: a half dozen old men and fat women, and it's game over with the fad. === Subject: I need a suggestion... posting-account=CntgTwoAAACJdYC45ZCndltx_IoTdL2i CLR 2.0.50727; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) Hi everyone. I'd like to learn more about Riemann zeta function. I know there are many books on the subject. But do any of you have any suggestion which book I should read first? Which book should be considered as a good introduction to Riemann zeta function? Fran.8dois === Subject: Re: I need a suggestion... > Hi everyone. >I'd like to learn more about Riemann zeta function. I know there are > many books on the subject. But do any of you have any suggestion which > book I should read first? Which book should be considered as a good > introduction to Riemann zeta function? Fran.8dois My own preference should be Harold M. Edwards Riemann's Zeta Function : http://www.amazon.com/Riemanns-Zeta-Function-Harold-Edwards/dp/0486417409 (very cheap and fine for people who like to see the silly little details behind actual computations using Riemann-Siegel sums, see the Gram points and so on, analyse Riemann's paper point by point...). Since Riemann was a true Master of complex integration some preliminary knowledge and training concerning 'Complex analysis' may be useful (I second Pubkeybreaker's suggestion of reading parts of Whittaker & Watson's A Course of Modern Analysis for that purpose!). Another path I enjoyed was Apostol's excellent 'Introduction to analytic number theory' : http://www.amazon.com/Introduction-Analytic-Number-Undergraduate-Mathematics /dp/0387901639 The zeta function is not so central and the style is rather dense (and nice!) but it constitues too a fine introduction to Dirichlet series (a second more technical book of Apostol completes this). Of course other fine books exist Titchmarsh, Davenport ('multiplicative number theory'), Ivic and so on but for an introduction my preferences are clear... Hoping it helped and...without commitments on my part :-) Raymond === Subject: Re: I need a suggestion... posting-account=lHNboAoAAACyasQ0uqX7OeM_tLuWGoQp CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) > Hi everyone. I'd like to learn more about Riemann zeta function. I know there are > many books on the subject. But do any of you have any suggestion which > book I should read first? Which book should be considered as a good > introduction to Riemann zeta function? Whittaker & Watson : A Course of Modern Analysis, Chapter 13. Read Chap. 12 on the Gamma function first. === Subject: Re: I need a suggestion... posting-account=CntgTwoAAACJdYC45ZCndltx_IoTdL2i CLR 2.0.50727; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) > Hi everyone. > I'd like to learn more about Riemann zeta function. I know there are > many books on the subject. But do any of you have any suggestion which > book I should read first? Which book should be considered as a good > introduction to Riemann zeta function? Whittaker & Watson : æA Course of Modern Analysis, æChapter 13. Read Chap. 12 on the Gamma function first. function. Long ago, I studied hypergeometric functions. I just want to extend my knowledge to Riemann zeta function, and after that, to Dirichlet series. Francois === Subject: Re: 2000 availables Solutions manual posting-account=WkIZRAoAAACYHhyQ64bqmvFidvZMQ7nX Gecko/20080201 Firefox/2.0.0.11;MEGAUPLOAD 1.0,gzip(gfe),gzip(gfe) hello i m new to this so please bear with me. i need 3 solution manuals Engineering Mechanics - Statics, 6th Ed (J. L. Meriam, L. G. Kraige) Engineering Mechanics - Dynamics, 6th Ed (J. L. Meriam, L. G. Kraige) Fundamentals of Engineering Thermodynamics, 6th Ed (Michael J. Moran, Howard N. Shapiro) please contact me at langren22[at]gmail.com === Subject: Re: 2000 availables Solutions manual posting-account=ps6iKwoAAAClFmxCENkG35-cddp2xUJ7 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > My List of SolutionsManual > > contact me to : newbergh123yahoo.com > newbergh123(at)yahoo.com > ot to : mattos...@gmail.com > mattosbw1(at)gmail.com > If your wanted solutionsmanualins't on this list, also can ask me if > is available . These are some only. > This same list of tites (not links) is available from : >http://rapidshare.com/files/64945514/List of solutions manual.txt > > - Mechanics, Mechanical Engineering & Aerospace Engineering: > > Classical mechanics (2nd Ed., Goldstein) > Classical Mechanics (Douglas Gregory) + original Ebook > Advanced Dynamics (Greenwood) + original Ebook > Advanced Engineering Dynamics (2nd Ed., Jerry Ginsberg) + Ebook > Classical Dynamics (Jorge V. Jos¬.a6) + Ebook > Impact Mechanics (W.J. Stronge) > Introduction to Mechanical Engineering (Rizza) > Mechanical Engineering Principles (Bird & Ross) + original Ebook > Engineering Fluid Mechanics (William Graebel) > Advanced Fluid Mechanics (William Graebel) + original Ebook > Mechanics of Fluids (8th Ed., Massey) + original Ebook > Fluid Mechanics (5th Ed., White) + Ebook > Fluid Mechanics (6th Ed., White) > Viscous Fluid Flow (3rd Ed., White) + Ebook > Fundamentals of Thermal-Fluid Sciences (1st Ed.,Cengel) + original > Ebook > Fundamentals of Thermal-Fluid Sciences (2nd Ed.,Cengel) + original > Ebook > Fundamentals of Thermal-Fluid Sciences with Student Resource CD (3rd > Ed.,Cengel& Turner) > Thermodynamics: An Engineering Approach (5th Ed.,Cengel) + original > Ebook > Thermodynamics: An Engineering Approach (6th Ed.,Cengel) + original > Ebook > Essentials of Fluid Mechanics: Fundamentals and Applications (1st Ed., >Cengel) + original > Fluid Mechanics (1st Ed.,Cengel) + original Ebook >HeatTranfer (2nd Ed.,Cengel) + original Ebook >HeatandMassTransfer: A Practical Approach (3rd. Ed.,Cengel) + > original Ebook > Design and Simulation of Thermal Systems (Suryanarayana & Arici) > Introduction to Fluid Mechanics (6th Ed., Robert Fox, Alan McDonald & > Philip Pritchard) > Fluid Mechanics (5th Ed., Douglas) > Fluid Mechanics (3rd Ed., Kundu) > Fluid Mechanics with Engineering Applications (Finnemore) > Fundamentals of Fluid Mechanics, 4th Ed (Bruce R. Munson, Donald F. > Young, Theodore H. Okiishi) + original ebook > Fundamentals of Fluid Mechanics, 5th Ed (Bruce R. Munson, Donald F. > Young, Theodore H. Okiishi) > A Brief Introduction to Fluid Mechanics, 3rd Ed (Donald F. Young, > Bruce R. Munson, Theodore H. Okiishi) > A Brief Introduction to Fluid Mechanics, 4th Ed (Donald F. Young, > Bruce R. Munson, Theodore H. Okiishi, Wade W.) > Engineering Fluid Mechanics, 7th Ed (Clayton T. Crowe, Donald F. > Elger, John A. Roberson) > Engineering Fluid Mechanics, 8th Ed (Clayton T. Crowe, Donald F. > Elger, John A. Roberson) > Mechanics of Fluids (3rd Ed., Potter) > Mechanics of Fluids (4th Ed., Shames) > Extended Irreversible Thermodynamics (3rd Ed., D. Jou, J. Casas- > Vazquez & G. Lebon) > Thermodynamics: An Integrated Learning System (Schmidt, Ezekoye, > Howell & Baker) > Introduction to Thermal and Fluids Engineering (Kaminski & Jensen) > Heating, Ventilating and Air Conditioning Analysis and Design (6th > Ed., McQuiston) > An Introduction to Fluid Dynamics: Principles of Analysis and Design > (Middleman) > Introduction toMassandHeatTransfer: Principles of Analysis and > Design (Middleman) >HeatTransfer(2nd Ed., Mills) > ConvectiveHeatandMassTransfer(4th Ed., Kays & Crawford) > Advanced Engineering Thermodynamics (3rd Ed., Bejan) > ConvectionHeatTransfer(2nd Ed., Bejan) > ConvectionHeatTransfer(3rd Ed., Bejan) > Thermal Design and Optimization (Bejan) > Shape and Structure, from Engineering to Nature (Bejan) > An Introduction to Combustion: Concepts and Applications (2nd Ed., > Turns) > Thermodynamics: Concepts and Applications (Stephen Turns) > Thermal-Fluid Sciences: An Integrated Approach (Stephen Turns) > Principles ofHeatTransfer(Kaviany) >HeatConvection (Latif M. Jiji) + original Ebook >HeatTransfer(9th Ed., Holman) > Fundamentals of Momentum,HeatandMassTransfer(4th Ed., Welty) > Fundamentals of Momentum,HeatandMassTransfer(5th Ed., Welty) > Momentum,Heat, andMassTransferFundamentals (Kessler) + original > Ebook > Analytical Methods forHeatTransferand Fluid Flow Problems (Bernhard > Weigand) >HeatTranfer (Rao) >HeatConduction (kakac) >HeatExchanges (Kakac) > ConvectiveHeatTransfer(kakac) >HeatExchangers: Selection, Rating and Thermal Design (2nd Ed. Sadik > Kakac & Hongtan Liu) > Fundamentals of Engineering Thermodynamics, 5th Ed (Michael J. Moran, > Howard N. Shapiro) + original Ebook > Fundamentals of Engineering Thermodynamics, 6th Ed (Michael J. Moran, > Howard N. Shapiro) > Fundamentals ofHeatandMassTransfer(5th Ed., Incropera, DeWitt) > Fundamentals ofHeatandMassTransfer(6th Ed., Incropera, DeWitt) > Introduction toHeatTransfer(4th Ed., Incropera, DeWitt) > Introduction toHeatTransfer(5th Ed., Incropera, DeWitt) > Radiation Detection and Measurement (3rd Ed., Glenn Knoll) > RadiativeHeatTransfer(2nd Ed., Michael Modest) > EngineeringHeatTransfer(2nd Ed., Janna) > Engineering Thermodynamics: Work andHeatTransfer(4th Ed., G.F.C. > Rogers & Y.R. Mayhew) > Elements ofHeatTransfer(Yildiz Bayazitoglu and M. Necati Ozisik) > InverseHeatTransfer: Fundamentals and Applications (M.N. Ozisik & > Helcio R.B. Orlande) > Thermal RadiationHeatTransfer(4th Ed.,Robert Siegel & John R. > Howell) > ComputationalHeatTransfer(2nd Ed., Jaluria) > Principles of Combustion (2nd Ed., Kenneth Kuan-yun Kuo) > Incompressible Flow (3rd Ed., Panton) > Modern Compressible Flow: With Historical Perspective (3rd Ed., John > D. Anderson) > Non-Newtonian Flow : Fundamentals and Engineering Applications (R P > Chhabra & J F Richardson) + original Ebook > Computational Techniques for Fluid Dynamics (Srinivas, K., Fletcher, > C.A.J.) > Ebook > Theory of Applied Robotics: Kinematics, Dynamics and Control (Reza N. > Jazar) > Kinematic Chains and Machine Components Design (Dan B. Marghitu) + > original Ebook > Kinematics and Dynamics of Machinery (3rd Ed., Wilson & Sadler) > Kinematics, Dynamics, and Design of Machinery (2nd Ed., Waldron & > Kinzel) > Mechanism Design: Analysis and Synthesis-Volume 1 (4th Ed., Erdman & > Sandor) > Machines and Mechanisms: Applied Kinematic Analysis (3rd Ed., > Myszka) > Mechanical Design: A Components Approach (Peter Childs) > Mechanical Design of Machine Elements and Machines: A Failure > Prevention Perspective (Collins) > Fundamentals of Machine Component Design (3rd Ed., Juvinall) > Fundamentals of Machine Component Design (4th Ed., Juvinall) > Design of Machine Elements (8th Ed., Spotts) > Machine Design (Wentzell) > SolutionsManualto the text : Problems on the Design of Machine > Elements (Faires) > Machine Elements in Mechanical Design (4th Ed., Mott) > Mechanical Design: An Integrated Approach (1st Ed., Ugural) > Design of Machinery (3rd Ed., Norton) > Design of Machinery (4th Ed., Norton) > Machine Design (2nd Ed., Norton) > Machine Design : An Integrated Approach (3rd Ed., Norton) > Mechanical Engineering Design (6th Ed., Shigley) > Mechanical Engineering Design (7th Ed., Shigley) > Shigley's Mechanical Engineering Design (8th Ed., Budynas) > Fundamentals of Machine Elements (1st Ed., Hamrock) > Fundamentals of Machine Elements (2nd Ed., Hamrock) > Mechanics of Materials: A Modern Integration of Mechanics and > Materials in Structural Design (Christopher Jenkins & Sanjeev Khanna) > Mechanics of Materials (3th Ed., Beer) > Mechanics of Materials (5th Ed., Gere) > Mechanics of Materials (6th Ed., Gere) > Mechanics of Materials (Ugural) > Simplified Mechanics and Strength of Materials (6th Ed., James > Ambrose) > Engineering Mechanics, Statics, 2nd Ed (William F. Riley, Leroy D. > Sturges) > Engineering Mechanics, Dynamics, 2nd Ed (William F. Riley, Leroy D. > Sturges) > Engineering Mechanics - Statics, 5th Ed (J. L. Meriam, L. G. Kraige) + > Ebook > Engineering Mechanics - Statics, 6th Ed (J. L. Meriam, L. G. Kraige) > Engineering Mechanics - Dynamics, 5th Ed (J. L. Meriam, L. G. Kraige) > Engineering Mechanics - Dynamics, 6th Ed (J. L. Meriam, L. G. Kraige) > Vector Mechanics for Engineers: Statics (7th Ed., Ferdinand P. Beer) > Vector Mechanics for Engineers: Statics (8th Ed., Ferdinand P. Beer) > Vector Mechanics for Engineers: Dynamics (7th Ed., Ferdinand P. Beer) > Vector Mechanics for Engineers: Dynamics (8th Ed., Ferdinand P. Beer) > Statics: Analysis and Design of Systems in Equilibrium (Sheppard & > Tongue) > Dynamics: Analysis and Design of Systems in Motion (Sheppard & Tongue) > Statics and Mechanics of Materials: An Integrated Approach (2nd Ed., > Riley, Sturges & Morris) > Mechanics of Materials (6th Ed., Riley, Sturges & Morris) > Deformable Bodies and Their Material Behavior > ... > read more hi can get a solution manual of heat and mass transfer by cengel I am looking for the following solution manuals: > Applied Fluid Mechanic 6th ed by Mott > Machine elements in mechanical design 4th ed, by Mott > Applied strenght of materials 5th ed. by Mott > Vector mechanics for engineers- Dynamics, 7th ed, by Beer & Johnston === Subject: Re: 2000 availables Solutions manual posting-account=IbfZHwoAAABe2tz2E7KgXzOsMdKcWmK8 Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > My List of SolutionsManual > > contact me to : newbergh123yahoo.com > newbergh123(at)yahoo.com > ot to : mattos...@gmail.com > mattosbw1(at)gmail.com > If your wanted solutionsmanualins't on this list, also can ask me if > is available . These are some only. > This same list of tites (not links) is available from : >http://rapidshare.com/files/64945514/List of solutions manual.txt > > - Mechanics, Mechanical Engineering & Aerospace Engineering: > > Classical mechanics (2nd Ed., Goldstein) > Classical Mechanics (Douglas Gregory) + original Ebook > Advanced Dynamics (Greenwood) + original Ebook > Advanced Engineering Dynamics (2nd Ed., Jerry Ginsberg) + Ebook > Classical Dynamics (Jorge V. Jos¬.a6) + Ebook > Impact Mechanics (W.J. Stronge) > Introduction to Mechanical Engineering (Rizza) > Mechanical Engineering Principles (Bird & Ross) + original Ebook > Engineering Fluid Mechanics (William Graebel) > Advanced Fluid Mechanics (William Graebel) + original Ebook > Mechanics of Fluids (8th Ed., Massey) + original Ebook > Fluid Mechanics (5th Ed., White) + Ebook > Fluid Mechanics (6th Ed., White) > Viscous Fluid Flow (3rd Ed., White) + Ebook > Fundamentals of Thermal-Fluid Sciences (1st Ed.,Cengel) + original > Ebook > Fundamentals of Thermal-Fluid Sciences (2nd Ed.,Cengel) + original > Ebook > Fundamentals of Thermal-Fluid Sciences with Student Resource CD (3rd > Ed.,Cengel& Turner) > Thermodynamics: An Engineering Approach (5th Ed.,Cengel) + original > Ebook > Thermodynamics: An Engineering Approach (6th Ed.,Cengel) + original > Ebook > Essentials of Fluid Mechanics: Fundamentals and Applications (1st Ed., >Cengel) + original > Fluid Mechanics (1st Ed.,Cengel) + original Ebook >HeatTranfer (2nd Ed.,Cengel) + original Ebook >HeatandMassTransfer: A Practical Approach (3rd. Ed.,Cengel) + > original Ebook > Design and Simulation of Thermal Systems (Suryanarayana & Arici) > Introduction to Fluid Mechanics (6th Ed., Robert Fox, Alan McDonald & > Philip Pritchard) > Fluid Mechanics (5th Ed., Douglas) > Fluid Mechanics (3rd Ed., Kundu) > Fluid Mechanics with Engineering Applications (Finnemore) > Fundamentals of Fluid Mechanics, 4th Ed (Bruce R. Munson, Donald F. > Young, Theodore H. Okiishi) + original ebook > Fundamentals of Fluid Mechanics, 5th Ed (Bruce R. Munson, Donald F. > Young, Theodore H. Okiishi) > A Brief Introduction to Fluid Mechanics, 3rd Ed (Donald F. Young, > Bruce R. Munson, Theodore H. Okiishi) > A Brief Introduction to Fluid Mechanics, 4th Ed (Donald F. Young, > Bruce R. Munson, Theodore H. Okiishi, Wade W.) > Engineering Fluid Mechanics, 7th Ed (Clayton T. Crowe, Donald F. > Elger, John A. Roberson) > Engineering Fluid Mechanics, 8th Ed (Clayton T. Crowe, Donald F. > Elger, John A. Roberson) > Mechanics of Fluids (3rd Ed., Potter) > Mechanics of Fluids (4th Ed., Shames) > Extended Irreversible Thermodynamics (3rd Ed., D. Jou, J. Casas- > Vazquez & G. Lebon) > Thermodynamics: An Integrated Learning System (Schmidt, Ezekoye, > Howell & Baker) > Introduction to Thermal and Fluids Engineering (Kaminski & Jensen) > Heating, Ventilating and Air Conditioning Analysis and Design (6th > Ed., McQuiston) > An Introduction to Fluid Dynamics: Principles of Analysis and Design > (Middleman) > Introduction toMassandHeatTransfer: Principles of Analysis and > Design (Middleman) >HeatTransfer(2nd Ed., Mills) > ConvectiveHeatandMassTransfer(4th Ed., Kays & Crawford) > Advanced Engineering Thermodynamics (3rd Ed., Bejan) > ConvectionHeatTransfer(2nd Ed., Bejan) > ConvectionHeatTransfer(3rd Ed., Bejan) > Thermal Design and Optimization (Bejan) > Shape and Structure, from Engineering to Nature (Bejan) > An Introduction to Combustion: Concepts and Applications (2nd Ed., > Turns) > Thermodynamics: Concepts and Applications (Stephen Turns) > Thermal-Fluid Sciences: An Integrated Approach (Stephen Turns) > Principles ofHeatTransfer(Kaviany) >HeatConvection (Latif M. Jiji) + original Ebook >HeatTransfer(9th Ed., Holman) > Fundamentals of Momentum,HeatandMassTransfer(4th Ed., Welty) > Fundamentals of Momentum,HeatandMassTransfer(5th Ed., Welty) > Momentum,Heat, andMassTransferFundamentals (Kessler) + original > Ebook > Analytical Methods forHeatTransferand Fluid Flow Problems (Bernhard > Weigand) >HeatTranfer (Rao) >HeatConduction (kakac) >HeatExchanges (Kakac) > ConvectiveHeatTransfer(kakac) >HeatExchangers: Selection, Rating and Thermal Design (2nd Ed. Sadik > Kakac & Hongtan Liu) > Fundamentals of Engineering Thermodynamics, 5th Ed (Michael J. Moran, > Howard N. Shapiro) + original Ebook > Fundamentals of Engineering Thermodynamics, 6th Ed (Michael J. Moran, > Howard N. Shapiro) > Fundamentals ofHeatandMassTransfer(5th Ed., Incropera, DeWitt) > Fundamentals ofHeatandMassTransfer(6th Ed., Incropera, DeWitt) > Introduction toHeatTransfer(4th Ed., Incropera, DeWitt) > Introduction toHeatTransfer(5th Ed., Incropera, DeWitt) > Radiation Detection and Measurement (3rd Ed., Glenn Knoll) > RadiativeHeatTransfer(2nd Ed., Michael Modest) > EngineeringHeatTransfer(2nd Ed., Janna) > Engineering Thermodynamics: Work andHeatTransfer(4th Ed., G.F.C. > Rogers & Y.R. Mayhew) > Elements ofHeatTransfer(Yildiz Bayazitoglu and M. Necati Ozisik) > InverseHeatTransfer: Fundamentals and Applications (M.N. Ozisik & > Helcio R.B. Orlande) > Thermal RadiationHeatTransfer(4th Ed.,Robert Siegel & John R. > Howell) > ComputationalHeatTransfer(2nd Ed., Jaluria) > Principles of Combustion (2nd Ed., Kenneth Kuan-yun Kuo) > Incompressible Flow (3rd Ed., Panton) > Modern Compressible Flow: With Historical Perspective (3rd Ed., John > D. Anderson) > Non-Newtonian Flow : Fundamentals and Engineering Applications (R P > Chhabra & J F Richardson) + original Ebook > Computational Techniques for Fluid Dynamics (Srinivas, K., Fletcher, > C.A.J.) > Ebook > Theory of Applied Robotics: Kinematics, Dynamics and Control (Reza N. > Jazar) > Kinematic Chains and Machine Components Design (Dan B. Marghitu) + > original Ebook > Kinematics and Dynamics of Machinery (3rd Ed., Wilson & Sadler) > Kinematics, Dynamics, and Design of Machinery (2nd Ed., Waldron & > Kinzel) > Mechanism Design: Analysis and Synthesis-Volume 1 (4th Ed., Erdman & > Sandor) > Machines and Mechanisms: Applied Kinematic Analysis (3rd Ed., > Myszka) > Mechanical Design: A Components Approach (Peter Childs) > Mechanical Design of Machine Elements and Machines: A Failure > Prevention Perspective (Collins) > Fundamentals of Machine Component Design (3rd Ed., Juvinall) > Fundamentals of Machine Component Design (4th Ed., Juvinall) > Design of Machine Elements (8th Ed., Spotts) > Machine Design (Wentzell) > SolutionsManualto the text : Problems on the Design of Machine > Elements (Faires) > Machine Elements in Mechanical Design (4th Ed., Mott) > Mechanical Design: An Integrated Approach (1st Ed., Ugural) > Design of Machinery (3rd Ed., Norton) > Design of Machinery (4th Ed., Norton) > Machine Design (2nd Ed., Norton) > Machine Design : An Integrated Approach (3rd Ed., Norton) > Mechanical Engineering Design (6th Ed., Shigley) > Mechanical Engineering Design (7th Ed., Shigley) > Shigley's Mechanical Engineering Design (8th Ed., Budynas) > Fundamentals of Machine Elements (1st Ed., Hamrock) > Fundamentals of Machine Elements (2nd Ed., Hamrock) > Mechanics of Materials: A Modern Integration of Mechanics and > Materials in Structural Design (Christopher Jenkins & Sanjeev Khanna) > Mechanics of Materials (3th Ed., Beer) > Mechanics of Materials (5th Ed., Gere) > Mechanics of Materials (6th Ed., Gere) > Mechanics of Materials (Ugural) > Simplified Mechanics and Strength of Materials (6th Ed., James > Ambrose) > Engineering Mechanics, Statics, 2nd Ed (William F. Riley, Leroy D. > Sturges) > Engineering Mechanics, Dynamics, 2nd Ed (William F. Riley, Leroy D. > Sturges) > Engineering Mechanics - Statics, 5th Ed (J. L. Meriam, L. G. Kraige) + > Ebook > Engineering Mechanics - Statics, 6th Ed (J. L. Meriam, L. G. Kraige) > Engineering Mechanics - Dynamics, 5th Ed (J. L. Meriam, L. G. Kraige) > Engineering Mechanics - Dynamics, 6th Ed (J. L. Meriam, L. G. Kraige) > Vector Mechanics for Engineers: Statics (7th Ed., Ferdinand P. Beer) > Vector Mechanics for Engineers: Statics (8th Ed., Ferdinand P. Beer) > Vector Mechanics for Engineers: Dynamics (7th Ed., Ferdinand P. Beer) > Vector Mechanics for Engineers: Dynamics (8th Ed., Ferdinand P. Beer) > Statics: Analysis and Design of Systems in Equilibrium (Sheppard & > Tongue) > Dynamics: Analysis and Design of Systems in Motion (Sheppard & Tongue) > Statics and Mechanics of Materials: An Integrated Approach (2nd Ed., > Riley, Sturges & Morris) > Mechanics of Materials (6th Ed., Riley, Sturges & Morris) > Deformable Bodies and Their Material Behavior > ... > read more hi can get a solution manual of heat and mass transfer by cengel I am looking for the following solution manuals: > Applied Fluid Mechanic 6th ed by Mott > Machine elements in mechanical design 4th ed, by Mott > Applied strenght of materials 5th ed. by Mott > Vector mechanics for engineers- Dynamics, 7th ed, by Beer & Johnston Please, can I get the solution manual for heating, ventilating, air conditioning analysis and design, Mcquiston. I need it desperately for my terms exam. === Subject: optimization theory posting-account=SQDT2QoAAAARANqaHM5kmoBtkqydZTV9 InfoPath.1; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) dear all, i have a question rwgarding optimization theory ...is there someone on the group who can help me out ....plzzz reply me soon thanx === Subject: Re: optimization theory posting-account=bSICGQkAAADSbkxAJ5uMxFegr4rp0Qig Gecko/20070515 Firefox/2.0.0.4,gzip(gfe),gzip(gfe) > dear all, > i have a question rwgarding optimization theory ...is there someone on > the group who can help me out ....plzzz reply me soon > thanx Probably lots of people. But you won't know unless you actually ask your question, will you? - Randy === Subject: Re: optimization theory posting-account=SQDT2QoAAAARANqaHM5kmoBtkqydZTV9 InfoPath.1; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) > dear all, > i have a question rwgarding optimization theory ...is there someone on > the group who can help me out ....plzzz reply me soon > æ æ æ æ æ æ æ æ æ æ æ æ æ æ æthanx Probably lots of people. But you won't know unless > you actually ask your question, will you? æ æ æ æ æ æ - Randy MY problem is actually i have a matrix a1*x1 +a2*x2 +a3*x3 +a4*x4 +a5*x5 +a6*x6=0 b1*x1 +b2*x2 +b3*x3 +b4*x4 +b5*x5 +b6*x6=0 c1*x1 +c2*x2 +c3*x3 +c4*x4 +c5*x5 +c6*x6=0 where a1..a6 and b1....b6, c1...c6 are all known and x1 x2...x6 are all unknowns .....my problems is more like an undetermined linear equations but i have some constraints to the solutions of the x1......x6 1)all the x1 ....x6 must be positive and non zero 2)x1^2 +x2^2........+x6^2=6 i know that it is going to be solved by non linear programming ..i tried like this using reduced gradient method minimize x1^2 +x2^2........+x6^2-6 subject to a1*x1 +a2*x2 +a3*x3 +a4*x4 +a5*x5 +a6*x6 b1*x1 +b2*x2 +b3*x3 +b4*x4 +b5*x5 +b6*x6 c1*x1 +c2*x2 +c3*x3 +c4*x4 +c5*x5 +c6*x6 x1,x2,x3,x4,x5,x6 =>0 can anybody tell me i am doing the right thing and secondly if i am doing right i need a type of algorithm in which the constraint x1,x2,x3,x4,x5,x6 =>0 does not include a zero term that is all of the solutions must be non zero but reduced gradient oesnt gaurantee me that ....which algorithm should i try thanx for your patience === Subject: Re: optimization theory posting-account=K5WE3woAAAAXArsybjkbN6LjMxWdHtbX Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > dear all, > i have a question rwgarding optimization theory ...is there someone on > the group who can help me out ....plzzz reply me soon > thanx > Probably lots of people. But you won't know unless > you actually ask your question, will you? > - Randy MY problem is actually i have a matrix a1*x1 +a2*x2 +a3*x3 +a4*x4 +a5*x5 +a6*x6=0 > b1*x1 +b2*x2 +b3*x3 +b4*x4 +b5*x5 +b6*x6=0 > c1*x1 +c2*x2 +c3*x3 +c4*x4 +c5*x5 +c6*x6=0 where a1..a6 and b1....b6, c1...c6 are all known and x1 x2...x6 are > all unknowns .....my problems is more like an undetermined linear > equations but i have some constraints to the solutions of the > x1......x6 1)all the x1 ....x6 must be positive and non zero > 2)x1^2 +x2^2........+x6^2=6 i know that it is going to be solved by non linear programming ..i > tried like this using reduced gradient method minimize x1^2 +x2^2........+x6^2-6 subject to > a1*x1 +a2*x2 +a3*x3 +a4*x4 +a5*x5 +a6*x6 > b1*x1 +b2*x2 +b3*x3 +b4*x4 +b5*x5 +b6*x6 > c1*x1 +c2*x2 +c3*x3 +c4*x4 +c5*x5 +c6*x6 x1,x2,x3,x4,x5,x6 =>0 The constraints should be a1*x1 + ... + a6*x6 = 0, etc. Here are some problems. (1) Incorrect objective. You want sum xj^2 to be near 6, whether larger or smaller, so you want to minimize (sum xj^2 - 6)^2 or |sum xj^2 - 6|. The former can be given just as it is, since it is already a smooth function, but to do the latter you need to write something like min z subject to z >= sum xj^2 - 6 and z >= 6 - sum xj^2, plus the other constraints. Here, z is a new variable. Alternatively, you can write sum xj^2 - 6 = p - n (where p and n >= 0 are new variables), then write the objective as min p + n. The formula sum xj^2 - 6 = p - n becomes an additional constraint. In both cases we are introducing / nonlinear/ constraints, and for such problems the so-called method of multipliers may be better than the straight reduced gradient (or better, generalized reduced gradient) method, which would now be forced to follow a curved surface instead of going along straight line segments. (2) Guaranteeing all xj > 0 is not easy. Probably the best approach is to introduce a small parameter epsilon > 0 and add the constraints xj >= epsilon for all j. Alternatively, you could just solve the problem assuming xj >= 0. If all xj turn out > 0 you are done; otherwise, if some xj = 0 you can constrain these to be >= epsilon and re-solve. However, it is not obvious that a feasible solution exists. If we write xj = epsilon + yj for all j, the constraints become sum(aj * yj) = A, sum(bj * yj) = B and sum(cj * yj) = C, where A = -epsilon * sum(aj), etc. Can we always guarantee a feasible solution (i.e., one having all yj >= 0)? The answer is NO for some choices of aj, bj and cj. Note: if a solution x0 exists with all x0j > 0, just let epsilon <= min(xj0). Then a solution to the epsilon problem exists, too. Conversely, if the epsilon problem is infeasible for all (small) epsilon > 0, the original problem is infeasible also. R.G. Vickson can anybody tell me i am doing the right thing and secondly if i am > doing right i need a type of algorithm in which the constraint > x1,x2,x3,x4,x5,x6 =>0 does not include a zero term that is all of the > solutions must be non zero but reduced gradient oesnt gaurantee me > that ....which algorithm should i try thanx for your patience === Subject: Re: optimization theory posting-account=bSICGQkAAADSbkxAJ5uMxFegr4rp0Qig Gecko/20070515 Firefox/2.0.0.4,gzip(gfe),gzip(gfe) > dear all, > i have a question rwgarding optimization theory ...is there someone on > the group who can help me out ....plzzz reply me soon > thanx > Probably lots of people. But you won't know unless > you actually ask your question, will you? > - Randy MY problem is actually i have a matrix a1*x1 +a2*x2 +a3*x3 +a4*x4 +a5*x5 +a6*x6=0 > b1*x1 +b2*x2 +b3*x3 +b4*x4 +b5*x5 +b6*x6=0 > c1*x1 +c2*x2 +c3*x3 +c4*x4 +c5*x5 +c6*x6=0 > Which we can write Ax = 0. In linear algebra terms, this means that x lies in the null-space of A. Since A is 3 x 6, it is at most rank 3 and so we know that the null space is at least 3 dimensional. > where a1..a6 and b1....b6, c1...c6 are all known and x1 x2...x6 are > all unknowns .....my problems is more like an undetermined linear > equations but i have some constraints to the solutions of the > x1......x6 1)all the x1 ....x6 must be positive and non zero x > 0 > 2)x1^2 +x2^2........+x6^2=6 || x ||^2 = 6 So your problem is actually to find a vector x in the null-space of A with the properties that x >= 0 and || x ||^2 = 6. The magnitude constraint is not an issue, if you can satisfy the other constraint. Let x be any vector such that Ax = 0. Then any multiple of x, y = ax, also has the property Ay = A(ax) = a(Ax) = 0. > i know that it is going to be solved by non linear programming ..i > tried like this using reduced gradient method minimize x1^2 +x2^2........+x6^2-6 subject to > a1*x1 +a2*x2 +a3*x3 +a4*x4 +a5*x5 +a6*x6 > b1*x1 +b2*x2 +b3*x3 +b4*x4 +b5*x5 +b6*x6 > c1*x1 +c2*x2 +c3*x3 +c4*x4 +c5*x5 +c6*x6 Did you mean to set these equal to 0? x1,x2,x3,x4,x5,x6 =>0 can anybody tell me i am doing the right thing and secondly if i am > doing right i need a type of algorithm in which the constraint > x1,x2,x3,x4,x5,x6 =>0 does not include a zero term that is all of the > solutions must be non zero but reduced gradient oesnt gaurantee me > that ....which algorithm should i try No, constrained optimization methods can't guarantee a strictly > constraint. I would take a linear algebra approach. Find the null-space of A and then use other methods to search for all-positive vectors in that space. Using canned algorithms, you can probably generate a set of basis vectors v1, v2, v3 which span the null space of A. That is, any x which satisfies Ax = 0 will be of the form x = a1*v1 + a2*v2 + a3*v3. - Randy === Subject: Re: Topological groups question The subgroup is open since finite index, hence also closed. === Subject: Re: Topological groups question <16846771.1203000302359.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > The subgroup is open since finite index, hence also closed. Why does having finite index imply openness? -- m === Subject: Re: Topological groups question > <4376932.1202914377829.JavaMail.jakarta@nitrogen.mathf > orum.org>, James > > > If G is a topological group and H is a subgroup of > index n, then is the > topological closure of H also a subgroup of index > n? > > In other words, a subgroup of index n is closed ... > is that the > question? > >I'm not asking if a subgroup of index n is closed. Yes you are. Since H is a subgroup of the closure of H and H has finite index, the index of the closure equals the index of H if and only if the closure equals H. >I'm asking whether or not the closure of a subgroup of index n is also a >subgroup of index n. David C. Ullrich === Subject: Re: Topological groups question <13217938.1202916978835.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e InfoPath.1; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > If G is a topological group and H is a subgroup of > index n, then is the > topological closure of H also a subgroup of index > n? > I suppose topological group includes Hausdorff ... > otherwise let > G be an n-element group with the indiscrete topology, > and H the > one-element subgroup. æSo the closure of G is H, no > longer of > index n. Actually in my case the group is locally compact and totally disconnected. ********************************************************* This reminds me of profinite groups, though these are compact, totally disconnected and Hausdorff top. groups. Anyway, I think Arturo's answer wraps the issue up: if H is a sbgp. of G of finite index n, and if its closure H' is also a sbgp. of G, the only way, I think, we can get [G:H'] = n again is if H = H'. Otherwise (i.e., if H < H'), then it most be that [G:H'] < n, and in fact [G:H'] / n . Now, if H is closed and of finite index in G, then H is also open, so this could help to decide whether a sbgp. is closed (or open, or both) Tonio === Subject: Re: Topological groups question <11668363.1202916830893.JavaMail.jakarta@nitrogen.mathforum.org> If G is a topological group and H is a subgroup of > index n, then is the > topological closure of H also a subgroup of index >I'm not asking if a subgroup of index n is closed. I'm asking >whether or not the closure of a subgroup of index n is also a >subgroup of index n. > No. [R;Q] is uncountable, [R:cl Q] = 1 > If the closure of the subgroup properly contains the subgroup, how > could the index fail to be strictly smaller? After all: [G:H] = [G:cl(H)][cl(H):H]. > === Subject: Re: Topological groups question William Elliot a .8ecrit : > No. [R;Q] is uncountable, [R:cl Q] = 1 Bad counter-example: [R;Q] is not finite. === Subject: Re: Topological groups question > William Elliot a .8ecrit : >No. [R;Q] is uncountable, [R:cl Q] = 1 >Bad counter-example: [R;Q] is not finite. And, of course, when someone simply says n we know he intends finite. === Subject: Re: Condition for irreducibility? The question seems to be whether irreducible polynomials form a subvariety of all polynomials. I think this is true iff the field is algebraically closed. But I don't know how to prove this. === Subject: Re: Condition for irreducibility? posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/20080208 Fedora/2.0.0.12-1.fc8 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > The question seems to be whether irreducible polynomials form a subvariety of all polynomials. I think this is true iff the field is algebraically closed. But I don't know how to prove this. Let P_2 be the affine space of all polynomials of degree 2, with coordinates given by the polynomial coefficients c_0, c_1, c_2. Consider the subset V of those which are products of two degree one polynomials. This is the set of points (ie, polynomials) (c_0, c_1, c_2) such that there exist a_0, a_1, b_0, b_1 such that c_0 = a_0 b_0 c_1 = a_0 b_1 + a_1 b_0 c_2 = a_1 b_1 If f in k[c_0, c_1, c_2] is a polynomial function on the affine space P_2 such that it vanishes on V, then f(a_0 b_0, a_0 b_1 + a_1 b_0, a_1 b_1) = 0. Thus the set of polynomials which vanish on V is the kernel of the ring map phi : k[c_0, c_1, c_1] --> k[a_0, a_1, b_0, b_1] such that phi(c_0) = a_0 b_0, phi(c_1) = a_0 b_1 + a_1 b_0, phi(c_2) = a_1 b_1. One sees that this map is injective. This means that a polynomial map which vanishes on the reducible polynomials is identically zero. This generalizes, of course. -- m === Subject: Re: Time distribution posting-account=bSICGQkAAADSbkxAJ5uMxFegr4rp0Qig Gecko/20070515 Firefox/2.0.0.4,gzip(gfe),gzip(gfe) > In this problem there are n distinct objects, each with a distinct, > probability p_i of being chosen each draw. The objects are drawn with > replacement. What I need is a probability distribution of the number of draws (d) > it will take to have drawn k distinct objects. Failing that, I would like to at least get an expected value and a > variance on this distribution. It's called the coupon collector problem, although that is usually in terms of the objects having equal probability of being drawn. Still, you may find the theory of the standard coupon collector problem useful. http://www.math.uah.edu/stat/urn/Coupon.xhtml http://www.ds.unifi.it/VL/VL_EN/urn/urn9.html http://en.wikipedia.org/wiki/Coupon_collector's_problem - Randy === Subject: Re: Time distribution posting-account=Ic7o8QkAAABRTcr6iMEJQ_yKDBSQpNPG Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > In this problem there are n distinct objects, each with a distinct, > probability p_i of being chosen each draw. The objects are drawn with > replacement. > What I need is a probability distribution of the number of draws (d) > it will take to have drawn k distinct objects. > Failing that, I would like to at least get an expected value and a > variance on this distribution. It's called the coupon collector problem, although > that is usually in terms of the objects having > equal probability of being drawn. Still, you may find the theory of the standard coupon > collector problem useful. http://www.math.uah.edu/stat/urn/Coupon.xhtmlhttp://www.ds.unifi.it/VL/VL_EN / urn/urn9.htmlhttp://en.wikipedia.org/wiki/Coupon_collector's_problem - Randy These resources look very helpful. Is anyone aware of a treatment of the coupon collector problem with unequal probabilities? === Subject: Re: Time distribution posting-account=K5WE3woAAAAXArsybjkbN6LjMxWdHtbX Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > In this problem there are n distinct objects, each with a distinct, > probability p_i of being chosen each draw. The objects are drawn with > replacement. > What I need is a probability distribution of the number of draws (d) > it will take to have drawn k distinct objects. > Failing that, I would like to at least get an expected value and a > variance on this distribution. > It's called the coupon collector problem, although > that is usually in terms of the objects having > equal probability of being drawn. > Still, you may find the theory of the standard coupon > collector problem useful. >http://www.math.uah.edu/stat/urn/Coupon.xhtmlhttp://www.ds.unifi.it/V... > - Randy > These resources look very helpful. Is anyone aware of a treatment of the coupon collector problem with > unequal probabilities? You can model this as a Markov chain with states of the form {start}, {i}, {i,j}, {i,j,k},..., {i_1, i_2, ...,i_(n-1)} and {finish}---basically, all the subsets of {1,2,...,n}. If the system is in state {i,j,k,...,m}, that means that coupons i, j, ..., m have already been collected (maybe several times). The probability that the system stays within the subset {i,j, ...,m} depends only on the subset, not on the order in which i, j, ..., m were obtained in the past; and similarly, the probability of a transition to a state n outside the subset depends only on the subset and not on the order it was realized. The total time is the first passage-time from states {start} to {finish}. For your modified problem (getting the time to k distinct objects), just enumerate all the subsets of cardinality <= k-1 and let finish occur when you go to a new object giving a subset of cardinality k. This approach is practical if n is not too large, at least for numerically-specified probabilities p_j. It would not work easily for your desired case of n --> infinity. R.G. Vickson === Subject: Re: Time distribution posting-account=-gYNIQoAAAD-MKGbMY_-Jpj1EqcOgRWx 3.0.2; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) > In this problem there are n distinct objects, each with a distinct, > probability p i of being chosen each draw. æThe objects are drawn with > replacement. What I need is a probability distribution of the number of draws (d) > it will take to have drawn k distinct objects. Failing that, I would like to at least get an expected value and a > variance on this distribution. I was fooling around trying to think of a computationally feasible way to calculate the full distribution (assuming you're doing it on a computer that is). It doesn't seem so easy. If n is relatively small (say not more than about 15), and d is not too huge, then you can work through draw-by-draw, keeping track of every combination of which objects have been drawn at least once and which haven't (of which there are 2^n possibilities), along with the probability of that scenario arising. For each combination there are n possible objects to consider at each draw, so that's max d*n*2^n operations in total (where max d is the highest value of d that you're interested in) -- or rather fewer if you discard paths as soon as they reach k distinct objects. Because you know how many distinct objects have been drawn at each point, and which they are, you can check whether drawing the next object takes you to k distinct objects, and so accumulate the probability of hitting the k'th distinct object at draw d. Unfortunately this method will quickly bog down as n gets large. === Subject: Re: Time distribution posting-account=Ic7o8QkAAABRTcr6iMEJQ_yKDBSQpNPG Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) I don't know if this makes this better or worse, but I am principally interested in asymptotics (n, #draws-->infinity) and looking at the probability distribution of time it takes to reach a fraction of distinct number of objects to the total number of objects. === Subject: convergence of a random vector posting-account=r5Mu_AoAAAA4Or1qD_TiZuru2F3PYCEL Gecko/20080201 Firefox/2.0.0.12 eMusic DLM/4.0_1.0.0.1,gzip(gfe),gzip(gfe) Hi everybody, I am trying to prove that when each element Xi(n) of a sequence of a random vector converges in probability to a constant ci, the random vectors X(n) converges in probability to a constant vector C. I can write P{|Xi-ci| I would like to put [a, b) inside an array environment, but latex > won't recognize the square bracket. > i also tried left[ and still i get an error - any help? > If your library has this book available, I believe you will find it to be an invaluable source of help for starting out using Tex: LEarning LaTex by David F. Griffiths and Desmond J. Higham It is the best *introductory* book on LaTex out there, bar none. M === Subject: Re: how to type the square bracket in array environment? > I would like to put [a, b) inside an array environment, but latex > won't recognize the square bracket. > i also tried left[ and still i get an error - >any help? I am going to do you a great disservice. begin{equation*} begin{array}{cc} x & y x & left[ z,,yright) end{array} end{equation*} -- Michael Press === Subject: Re: Bingo ! posting-account=lREQ0goAAABJGVkI7Y-7onQ9_lq1aWhl CLR 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) > When a string on 244 tiny pictures, differentiated into 46 kinds, is > comparable to a picture lottery down to the smallest detail, as I so > clearly show, then it is those, who object against, that have the > explanatory problem in, why the hag they keep on believing that this > string is a text based on phonetics, if they really still believe so? > When at the same time I thoroughly demonstrate, how the 244 pictures > on this famous disc represent an eight-month calendar, then the lack > of responses become extremely weird.http://web.gvdnet.dk/GVD002393/gnomonic.htm > M.m. http://web.gvdnet.dk/GVD002393/phaistos.htm Gregarious animals. A desperate attempt was made by mobbing the floor with my discovery, to try to explain away the unfolding of the inscription ; Going: Then the disc must be a modern hoax, and an awkward case, which should be hushed up by all means. - As if the feat of decipherment was less a red letter day by that reason. No, the real cause is not a hide away. It is the envier's hard wish to reduce the one person, who so had the 'luck' to look through to the solution of a nearly impossible problem, an enigma with one-hundred years of anniversary .- An hoax or not, the Phaistos disc was still the memorable world-champion of all sudoku's.(so do correct). I so much want to have pin pointed those trolls. So that they will not be run awayer's from the truth ..-) - that they shall not do harm to other persons as well. :-( http://web.gvdnet.dk/GVD002393/f13c.htm http://web.gvdnet.dk/GVD002393/zzjagt2.htm === Subject: -- compute a limit-point Assume b = x*i, (i imaginary unit, x real) assume x=1 then the infinite sequence s(b):= b, b^b, b^b^b, b^b^b^b,.... converges. Take x=1.7 then this converges,too . Take x=1.8 then this does not converge. What is the value of x, where convergence changes to divergence? (for an imagination see: http://go.helms-net.de/math/tetdocs/trifurcation_bases.png ) Gottfried -- --- Gottfried Helms, Kassel === Subject: Re: -- compute a limit-point > Assume b = x*i, (i imaginary unit, x real) assume x=1 then the infinite sequence s(b):= b, b^b, b^b^b, > b^b^b^b,.... converges. Take x=1.7 then this converges,too . Take > x=1.8 then this does not converge. > What is the value of x, where convergence changes to divergence? > (for an imagination see: > http://go.helms-net.de/math/tetdocs/trifurcation_bases.png > ) The condition for convergence of the iterated exponential b^b^b^...^b, is: |phi^{-1}(b)| < 1 or (phi^{-1}(b))^n = 1 [*] for some n in N, where: phi(z)=exp(z*exp(-z)) This means, phi^{-1}(w)=-W(-log(w)), where W is the Lambert function. For your case then, one needs to examine the behavior of the map |phi^{-1}(i*x)|, for x real. Analyzing the graph of the map and solving with Maple: > fsolve(abs(phiinv(I*x))=1,x,x=0..1); .1300484757 > fsolve(abs(phiinv(I*x))=1,x,x=1..2); 1.712936040 The map is less than 1 between the above two values, hence convergence occurs when: .1300484757 < x < 1.712936040 Convergence at the end points should be checked separately, as per [*] > Gottfried -- I.N. Galidakis === Subject: Re: -- compute a limit-point Am 14.02.2008 19:39 schrieb I.N. Galidakis: > Assume b = x*i, (i imaginary unit, x real) > assume x=1 then the infinite sequence s(b):= b, b^b, b^b^b, > b^b^b^b,.... converges. Take x=1.7 then this converges,too . Take > x=1.8 then this does not converge. > What is the value of x, where convergence changes to divergence? (...) > > fsolve(abs(phiinv(I*x))=1,x,x=0..1); > .1300484757 > > fsolve(abs(phiinv(I*x))=1,x,x=1..2); > 1.712936040 >The map is less than 1 between the above two values, hence convergence occurs > when: >.1300484757 < x < 1.712936040 >Convergence at the end points should be checked separately, as per [*] > Hmm, Ioannis - The symbolic inverse calculator gives a nice try: > The number 1.712936040 > coth(3/5 + 2/5 zeta(5) - 1/2 ln(2)) but, well, ...;-) Gottfried -- --- Gottfried Helms, Kassel === Subject: Re: education grrr posting-account=p6EJ9AoAAACMPSv4t2GeMy73rdicHikC Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) >its like einstein said : > given all this education ; its a wonder i can still think at all >education >grrr > In the novel > The Bold Saboteurs > by Chandler Brossard > there is a cool quote ... > First, some background from the novel, then the quote ... > Two brothers, one about 13, the other about 7. The brothers are both > very, very smart. It's a broken home -- the father, a hopeless > drunkard, has abandoned them. Times are hard -- it's the depression > era, and they are in serious poverty. To help make ends meet, the > older brother quits school and takes a job as a night watchman. > The older brother is the new father figure, and he makes a project of > the younger one. The younger brother idolizes the older one, and comes > to visit him every night on his job. They have long talks, about > anything and everything. > The older brother, continuing his own education by self-study from > carefully chosen books obtained from the public library, is a gifted > teacher, and leads the younger one in the path of his own newly > acquired knowledge and insights. The younger one absorbs all the new > ideas very rapidly, and at one point, the older brother says to the > younger one ... > I think you should quit school > so as to further your education. > [In other words, so they could have more time to study on their own, > at their own pace, controlling their own education, free from the > indoctrination, bureaucratization, and mediocratization inherent in > the educational system.] > I always really liked that quote. > quasi >Read about OBE? No, I hadn't heard of it -- I had to look it up. I assume OBE means Outcome-Based Education. I like the idea of certifying knowledge by testing. However, I'm not sure I trust the test makers. quasi yes, it does. I am not fully in education yet. I am not like t, though I know complex number when 5th grade, and FLT by 8th grade but there was no one to guide by that time. But i don't react like t. Never came across my mind. That time my world was peaceful. I didn't know my teacher know about complex number, i assume she knew little but not competent. But the other teacher is 8th grade should know about FLT, since he hold BSc in math. Gauss refused to enter FLT contest. According to what I read, he said it is useless. Would Gauss solve it? Boen S. Liong === Subject: Re: education grrr >Gauss refused to enter FLT contest. According to what I read, he said >it is useless. Would Gauss solve it? Secretly, I'm sure he played with it, and at some point, wisely gave up. My guess is he sensed the problem was out of reach, except for special cases, and he knew better than to squander all his energy on what would probably have been a futile quest. quasi === Subject: With FLT, Gauss was a pussy! <8m59r3dm2kvjql7jsjjsjgc533rrqjnlvl@4ax.com> posting-account=XBlrJAoAAADt98TEM65O-rVj9Zl2tTjM 1.1.4322),gzip(gfe),gzip(gfe) >Gauss refused to enter FLT contest. According to what I read, he said >it is useless. Would Gauss solve it? Secretly, I'm sure he played with it, and at some point, wisely gave > up. My guess is he sensed the problem was out of reach, except for > special cases, and he knew better than to squander all his energy on > what would probably have been a futile quest. quasi As I recall the story in E.T. Bell's Men of Mathematics and elsewhere, such as quoted in: http://users.forthnet.gr/ath/kimon/FLT.htm Gauss was asked about Fermat's Last Theorem, and he supposedly said, I confess that Fermat's Last Theorem, as an isolated proposition, has very little interest for me, because I could easily lay down a multitude of such propositions, which one could neither prove nor dispose of. Sour grapes! Gauss belittled a problem that he couldn't solve. What a coward! At least some real and respectable mathematicians, such as Gabriel Lame, published proofs which were found to be in error. And Wiles' first proof was flawed and had to be fixed. But these two men (Lame and Wiles) put their genuine mathematician reputation on the line, unlike Gauss who published NO proof at all of Fermat's Last Theorem. === Subject: Re: With FLT, Gauss was a pussy! posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e SIMBAR Enabled; SIMBAR={70306B22-CB8C-4d52-BFF4-18424E217075}; MathPlayer 2.10b; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) >Gauss refused to enter FLT contest. According to what I read, he said >it is useless. Would Gauss solve it? > Secretly, I'm sure he played with it, and at some point, wisely gave > up. My guess is he sensed the problem was out of reach, except for > special cases, and he knew better than to squander all his energy on > what would probably have been a futile quest. > quasi As I recall the story in E.T. Bell's Men of Mathematics and elsewhere, > such as quoted in: http://users.forthnet.gr/ath/kimon/FLT.htm Gauss was asked about Fermat's Last Theorem, and he supposedly said, I confess that Fermat's Last Theorem, as an isolated proposition, has > very little interest for me, because I could easily lay down a > multitude of such propositions, which one could neither prove nor > dispose of. Sour grapes! æGauss belittled a problem that he couldn't solve. æWhat > a coward! At least some real and respectable mathematicians, such as Gabriel > Lame, published proofs which were found to be in error. æAnd Wiles' > first proof was flawed and had to be fixed. æBut these two men (Lame > and Wiles) put their genuine mathematician reputation on the line, > unlike Gauss who published NO proof at all of Fermat's Last Theorem. ************************************************************ Rubbish. Gauss is deservedly called the prince of mathematics, and his contributions in many areas are outstanding from any point of view. I agree in this with quasi: perhaps Gauss took a peek at FLT, played a little with it, and then wisely backed off. To qualify a mathematician only, or even mainly, by his contributions to FLT is, imho, absolute nonsense. Tonio === Subject: Re: With FLT, Gauss was a pussy! posting-account=XBlrJAoAAADt98TEM65O-rVj9Zl2tTjM 1.1.4322),gzip(gfe),gzip(gfe) to FLT is, imho, absolute nonsense. I didn't say or intend that, now did I? I just slammed Gauss for this specific instance of Gauss being a cowardly pussy. Couldn't he just graciously say he did not feel it was important to him? Noooooooooo, he had to belittle the Fermat conjecture and pump up his own ego and superiority in mathematics with a snide comment. Obviously other real and genunine mathematicians felt that proofing Fermat's conjecture was important, and new mathematics was spurned and developed by them by their efforts, such as Kummer, Sophie Germain, and Wiles. Luckily, none of those fools Kummer, Germain, Wiles, etc. listend to the prince of mathematics. So, what king sired this prince of mathematics Gauss anyway? Yeah, tell me, who would you call the King of Mathematics? Euclid? Archimedes (the original, not the Plutonium head)? === Subject: Re: With FLT, Gauss was a pussy! posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/20070530 Fedora/1.5.0.12-1.fc5 Firefox/1.5.0.12,gzip(gfe),gzip(gfe) >Gauss refused to enter FLT contest. According to what I read, he said >it is useless. Would Gauss solve it? > Secretly, I'm sure he played with it, and at some point, wisely gave > up. My guess is he sensed the problem was out of reach, except for > special cases, and he knew better than to squander all his energy on > what would probably have been a futile quest. > quasi As I recall the story in E.T. Bell's Men of Mathematics and elsewhere, > such as quoted in: http://users.forthnet.gr/ath/kimon/FLT.htm Gauss was asked about Fermat's Last Theorem, and he supposedly said, I confess that Fermat's Last Theorem, as an isolated proposition, has > very little interest for me, because I could easily lay down a > multitude of such propositions, which one could neither prove nor > dispose of. Sour grapes! Gauss belittled a problem that he couldn't solve. What > a coward! At least some real and respectable mathematicians, such as Gabriel > Lame, published proofs which were found to be in error. And Wiles' > first proof was flawed and had to be fixed. But these two men (Lame > and Wiles) put their genuine mathematician reputation on the line, > unlike Gauss who published NO proof at all of Fermat's Last Theorem. Sure. He set out instead to come up with what can be regarded as one of the foundations of number theory. But, as you know, the number of (attempted and published) proofs for FLT is the only measure of a mathematician's prowess. What have /you/ published, by the way, which validates your elevated tone and condemning judgement of Gauss? -- m -- m === Subject: Re: With FLT, Gauss was a pussy! posting-account=XBlrJAoAAADt98TEM65O-rVj9Zl2tTjM 1.1.4322),gzip(gfe),gzip(gfe) On Feb 14, 4:03æpm, Mariano Su.87rez-Alvarez >Gauss refused to enter FLT contest. According to what I read, he said >it is useless. Would Gauss solve it? > Secretly, I'm sure he played with it, and at some point, wisely gave > up. My guess is he sensed the problem was out of reach, except for > special cases, and he knew better than to squander all his energy on > what would probably have been a futile quest. > quasi > As I recall the story in E.T. Bell's Men of Mathematics and elsewhere, > such as quoted in: >http://users.forthnet.gr/ath/kimon/FLT.htm > Gauss was asked about Fermat's Last Theorem, and he supposedly said, > I confess that Fermat's Last Theorem, as an isolated proposition, has > very little interest for me, because I could easily lay down a > multitude of such propositions, which one could neither prove nor > dispose of. > Sour grapes! æGauss belittled a problem that he couldn't solve. æWhat > a coward! > At least some real and respectable mathematicians, such as Gabriel > Lame, published proofs which were found to be in error. æAnd Wiles' > first proof was flawed and had to be fixed. æBut these two men (Lame > and Wiles) put their genuine mathematician reputation on the line, > unlike Gauss who published NO proof at all of Fermat's Last Theorem. Sure. He set out instead to come up with what > can be regarded as one of the foundations of > number theory. But, as you know, the number of > (attempted and published) proofs for FLT is the > only measure of a mathematician's prowess. What have /you/ published, by the way, which > validates your elevated tone and condemning > judgement of Gauss? -- m -- m- Hide quoted text - - Show quoted text - What I have published is irrelevant. It's what Gauss said in this specific instance which deserves contempt. Notice how not only did he belittle the conjecture, he also made an evasion by saying he could toss out even more conjectures that couldn't be proven. Yeah, Gauss did a lot of math. But this snide comment of his irks me. I am IRKED! === Subject: Re: Mathematics vs Physics posting-account=p6EJ9AoAAACMPSv4t2GeMy73rdicHikC Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > sorry for the question, but I need to know. > There is someone with a double major or graduated two times or some > teacher that teach to both Mathematics and Physics students, that can > say me the difference between these two matters? > I want to know also which is (usually) harder, where the students has to > work hard more, which if more difficult. > And at the extremely specializations of Pure Mathematics and Theoretical > Physics, what are the difference? What's the level of knowledge for the > average student in both these matters? Which need more hard work? > I want just the opinions of who really know both the environments. > A. They are both equally hard, especially if you want to become good at > it, as opposed to merely competent. The decision factor is not going > to be which is harder but which is more interesting, and this depends > on the person. If something is interesting, you will work on it no > matter how hard it is, and your intense desire to figure it out will > help you overcome any barriers you have. But if the subject does not > interest you, then you will never do well in the subject, no matter > how talented you are. As a physicist, I of course have my opinion about which is more > interesting and why. I can certainly tell you that the way you > investigate and make progress in physics is much different than is > done in mathematics. As a starting point, you may want to reread the > basics on the scientific method. The scientific method is not > something that is central to the methodology of pure mathematics, > though it might have stronger impact on applied mathematics. PD Is there any Newton or Einstein Theorem in math? Boen S. Liong === Subject: Re: Mathematics vs Physics posting-account=nPH_PQkAAACneDKT6RXopPWArC2We4Rq CLR 1.1.4322),gzip(gfe),gzip(gfe) > sorry for the question, but I need to know. > There is someone with a double major or graduated two times or some > teacher that teach to both Mathematics and Physics students, that can > say me the difference between these two matters? > I want to know also which is (usually) harder, where the students has to > work hard more, which if more difficult. > And at the extremely specializations of Pure Mathematics and Theoretical > Physics, what are the difference? What's the level of knowledge for the > average student in both these matters? Which need more hard work? > I want just the opinions of who really know both the environments. > A. > They are both equally hard, especially if you want to become good at > it, as opposed to merely competent. The decision factor is not going > to be which is harder but which is more interesting, and this depends > on the person. If something is interesting, you will work on it no > matter how hard it is, and your intense desire to figure it out will > help you overcome any barriers you have. But if the subject does not > interest you, then you will never do well in the subject, no matter > how talented you are. > As a physicist, I of course have my opinion about which is more > interesting and why. I can certainly tell you that the way you > investigate and make progress in physics is much different than is > done in mathematics. As a starting point, you may want to reread the > basics on the scientific method. The scientific method is not > something that is central to the methodology of pure mathematics, > though it might have stronger impact on applied mathematics. > PD Is there any Newton or Einstein Theorem in math? Newton and Einstein are not famous for theorems. They are famous for theories that agree astounding well with experiment. And yes, there are some quite powerful mathematics that have had the same impact in mathematics that Newton and Einstein have had in physics. If you're unaware of those, then this alone is good reason to study more mathematics, so you can see some of the really powerful things. In fact, if you want to self-study the interplay between really fundamental leaps in mathematics and really fundamental leaps in physics, there is a book by Penrose called The Road to Reality which I would highly recommend. It is for the lay reader, but a dedicated lay reader -- it is by no means another A Brief History of Time or The Elegant Universe. PD === Subject: Re: Mathematics vs Physics posting-account=p6EJ9AoAAACMPSv4t2GeMy73rdicHikC Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > It was explained to me once when I asked a similar > question. Mathematics you just have to do the problems > again and again. It is just practice. Physics you have > to memorize a lot of facts. Interestingly, if these labels have to be applied, > my experience has been the reverse. In mathematics > you have to learn a lot of definitions, theorems, > proof-tricks/methods, etc. while in physics you have > to solve problem after problem after problem after ... The tests and homework I had in my classical mechanics, > quantum mechanics, electrodynamics, and statistical physics > classes were almost 100% problem solving of the type that > you need to practice-practice-practice to do well (which > I didn't, and so I didn't). In math, especially at the > advanced undergraduate to beginning graduate level, > I found it often to be the case that understanding the > underlying concept (e.g. what is the underlying rationale > behind the quotient topology) was almost all you needed to > take care of homework and test problems. It's possible the > reverse is true for post Ph.D. research in these fields, > but I don't know enough about physics to be sure. Dave L. Renfro Math do with function and space. Physics do with SI unit like lbs, kg, etc. The only true mathematician physicist is Gauss. Newton, and Einstein are just physicist. Have ever heard of Newton or Einstein Theorem in math? Newton invented differential to be used in physics, but Leibnitz invented integral to be used in math. See the difference? Boen S. Liong === Subject: Re: Mathematics vs Physics posting-account=W7RQ6gkAAACLDC5JWapQU1UV8ot8KkAS 1.0.3705; Dealio Toolbar 3.1.1),gzip(gfe),gzip(gfe) > sorry for the question, but I need to know. > There is someone with a double major or graduated two times or some > teacher that teach to both Mathematics and Physics students, that can > say me the difference between these two matters? People who have an interest in physical things discovered the principles of evolution. Poeple with an interest in math, discovered mirrors. It's so simple, even idiots like physicists can understand it, and even imbeciles like mathematicians can see two sides of it. > I want to know also which is (usually) harder, where the students has to > work hard more, which if more difficult. > And at the extremely specializations of Pure Mathematics and Theoretical > Physics, what are the difference? What's the level of knowledge for the > average student in both these matters? Which need more hard work? > I want just the opinions of who really know both the environments. > A. === Subject: Re: Mathematics vs Physics | > | > | > | > sorry for the question, but I need to know. | > There is someone with a double major or graduated two times or some | > teacher that teach to both Mathematics and Physics students, that can | > say me the difference between these two matters? | > I want to know also which is (usually) harder, where the students has to | > work hard more, which if more difficult. | > And at the extremely specializations of Pure Mathematics and Theoretical | > Physics, what are the difference? What's the level of knowledge for the | > average student in both these matters? Which need more hard work? | > I want just the opinions of who really know both the environments. | > A. | > | > They are both equally hard, especially if you want to become good at | > it, as opposed to merely competent. The decision factor is not going | > to be which is harder but which is more interesting, and this depends | > on the person. If something is interesting, you will work on it no | > matter how hard it is, and your intense desire to figure it out will | > help you overcome any barriers you have. But if the subject does not | > interest you, then you will never do well in the subject, no matter | > how talented you are. | > | > | As a physicist, I of course have my opinion about which is more | > | interesting and why. I can certainly tell you that the way you | > | investigate and make progress in physics is much different than is | > | done in mathematics. As a starting point, you may want to reread the | > | basics on the scientific method. The scientific method is not | > | something that is central to the methodology of pure mathematics, | > | though it might have stronger impact on applied mathematics. | > | > As an engineer, I of course have my opinion about which is more | > interesting and why. I can certainly tell you that the way you | > investigate and make progress in mathematics is much different than | > is done in crank theoretical physics. As a starting point, you may want | > to reread the basics on the axiomatic and logical method. The | > mathematical method is not something that is central to the methodology | > of crank theoretical physics where blind faith and belief in the impossible | > are paramount, though it will have stronger impact on applied physics. | | As someone trained in physics and mathematics who has | spent his entire professional career in engineering environments, | I've been fortunate to work among engineers who value both | disciplines and who know how to work in interdisciplinary | teams. Every once in awhile there would be those who | thought the sun rose and set on what you learned in | a BSEE curriculum, but those were the minority. As someone trained AND experienced in physics and mathematics who has spent his entire professional career in engineering environments, I've been fortunate to work among engineers who value both disciplines and who know how to work in interdisciplinary teams. Every once in a while there would be those who thought the sun rose and set out of Einstein's arse, but those were the minority. You were one of the minority, huh? | It has been my great good fortune never to work with an | engineer of the type who frequent this newsgroup. These | people do not strike me as competent even in engineering, | let alone physics. | And someone like Androcles who blithely declares that | there's no such thing as free space loss or antenna gain | would never have made it past the first interview. | It has been my great good fortune never to work with a physicist of the type who frequent this newsgroup. These people do not strike me as competent even in engineering, let alone mathematics or physics. And someone like Blind Poe who blithely declares I confirm I believe transportation data from DOT more than transportation data pulled ex recto Androcles. (because DOT says buses to longer to stop than cars and Poe believes it) would never have made it past the curriculum vitae (resum.8e for Americans). Roman Catholic, are you Poe? They like confirming their blind faith too. BTW, there is no such animal as free space loss, energy can neither be created or destroyed, and photons make it across billions of light years. Strange you should deny conservation of energy -- or perhaps not, given your pathetic desperation. === Subject: Re: Mathematics vs Physics > sorry for the question, but I need to know. > There is someone with a double major or graduated two times or some > teacher that teach to both Mathematics and Physics students, that can > say me the difference between these two matters? > I want to know also which is (usually) harder, where the students has to > work hard more, which if more difficult. > And at the extremely specializations of Pure Mathematics and Theoretical > Physics, what are the difference? What's the level of knowledge for the > average student in both these matters? Which need more hard work? > I want just the opinions of who really know both the environments. > A. They are both equally hard, especially if you want to become good at > it, as opposed to merely competent. The decision factor is not going > to be which is harder but which is more interesting, and this depends > on the person. If something is interesting, you will work on it no > matter how hard it is, and your intense desire to figure it out will > help you overcome any barriers you have. But if the subject does not > interest you, then you will never do well in the subject, no matter > how talented you are. | As a physicist, I of course have my opinion about which is more > | interesting and why. I can certainly tell you that the way you > | investigate and make progress in physics is much different than is > | done in mathematics. As a starting point, you may want to reread the > | basics on the scientific method. The scientific method is not > | something that is central to the methodology of pure mathematics, > | though it might have stronger impact on applied mathematics. As an engineer, I of course have my opinion about which is more > interesting and why. I can certainly tell you that the way you > investigate and make progress in mathematics is much different than > is done in crank theoretical physics. As a starting point, you may want > to reread the basics on the axiomatic and logical method. | And this is an important case in point. Some buffoons like Androcles | think that physics should be derivable from axiomatic logic and | mathematical derivation, | thereby completely obviating the need for the scientific method. This | is, of course, what distinguishes buffooons from physicists. And this is an important case in point. What we engineers like to see, of course, is the scientific method that established that 'the time required by light to travel from A to B equals the time it requires to travel from B to A', an impossible axiom contrary to the scientific method, it being methodless. Some heads like Einstein and his prophesying disciples (like the head PD) think that physics should be derivable from axiomatic logic and mathematical derivation, thereby completely obviating the need for the scientific method. This is, of course, what distinguishes mathematicians, engineers and real physicists from heads like the lying hypocrite Phuckwit Duck who never was and never will be a physicist or an engineer or a mathematician, but will continue to rant and rave about his blind faith in the Holey Church of Relativity. > The > mathematical method is not something that is central to the methodology > of crank theoretical physics where blind faith and belief in the > impossible > are paramount, though it will have stronger impact on applied physics. === Subject: Re: Mathematics vs Physics Androcles a .8ecrit : > And this is an important case in point. > What we engineers ... You engineer ? What a joke... Tell us again about or and xor, or about differential equations... === Subject: Re: Mathematics vs Physics <47b3a29d$0$30647$426a34cc@news.free.fr> posting-account=bSICGQkAAADSbkxAJ5uMxFegr4rp0Qig Gecko/20070515 Firefox/2.0.0.4,gzip(gfe),gzip(gfe) > Androcles a .8ecrit : > And this is an important case in point. > What we engineers ... You engineer ? What a joke... Tell us again about or and xor, or about differential equations... Why didn't you know? Androcles is not only an engineer but a Ph.D. mathematician, from the school where you learn that curl E = -dB/dt and E = -dB/dt are the same equation. - Randy === Subject: Re: Mathematics vs Physics > Androcles a .8ecrit : > And this is an important case in point. > What we engineers ... You engineer ? What a joke... Tell us again about or and xor, or about differential equations... Why didn't you know? Androcles is not only an engineer but a Ph.D. mathematician, from the school where you learn that curl E = -dB/dt and E = -dB/dt are the same equation. Yep, whereas x+y = 0 (x,y being velocity vectors) and |x+y|=0 are different. Found any more tick fairies under the bed yet, Poe? === Subject: CFP: DTVCS 2008 - Design, Testing and Formal Verification Techniques for Integrated Circuits and Systems posting-account=tcby3QoAAAAbRR4YXReYXZS7no_xK0fN Gecko/20071015 SUSE/2.0.0.10-0.1 Firefox/2.0.0.10,gzip(gfe),gzip(gfe) Apologies for any multiple copies received. We would appreciate it if you could distribute the following call for papers to any relevant mailing lists you know of. CALL FOR PAPERS Special Session: Design, Testing and Formal Verification Techniques for Integrated Circuits and Systems DTVCS 2008 August 18-20, 2008 (Kailua-Kona, Hawaii, USA) http://digilander.libero.it/systemcfl/dtvcs Special Session in the IASTED International Conference on Circuits and Systems (CS 2008) ---------------------------------------------------------------------------- --------------------------------------------------------- The IASTED International Conference on Circuits and Systems (CS 2008) will take place in Kailua-Kona, Hawaii, USA, August 18-20, 2008. URL: http://www.iasted.org/conferences/cfp-625.html. Aims and Scope ------------------------- The main target of the Special Session DTVCS is to bring together engineering researchers, computer scientists, practitioners and people from industry to exchange theories, ideas, techniques and experiences related to the areas of design, testing and formal verification techniques for integrated circuits and systems. Contributions on UML and formal paradigms based on process algebras, petri-nets, automaton theory and BDDs in the context of design, testing and formal verification techniques for integrated circuits and systems are also encouraged. Topics ---------- Topics of interest include, but are not limited to, the following: * digital, analog, mixed-signal and RF test * built-in self test * ATPG * theory and foundations: model checking, SAT-based methods, use of PSL, compositional methods and probabilistic methods * applications of formal methods: equivalence checking, CSP applications and transaction-level verification * verification through hybrid techniques * verification methods based on hardware description/system-level languages (e.g. VHDL, SystemVerilog and SystemC) * testing and verification applications: tools, industrial experience reports and case studies Industrial Collaborators and Sponsors ------------------------------------------------------ This special session is partnered with: * CEOL: Centre for Efficiency-Oriented Languages Towards improved software timing, University College Cork, Ireland (http://www.ceol.ucc.ie) * International Software and Productivity Engineering Institute, USA (http://www.intspei.com) * Intelligent Support Ltd., United Kingdom (http://www.isupport- ltd.co.uk) * Minteos, Italy (http://www.minteos.com) * M.O.S.T., Italy (http://www.most.it) * Electronic Center, Italy (http://www.el-center.com) * Legale Fiscale, Italy (http://www.legalefiscale.it) This special session is sponsored by: * LS Industrial Systems, South Korea (http://eng.lsis.biz) * Solari, Hong Kong (http://www.solari-hk.com/) Technical Program Committee -------------------------------------------- * Prof. Vladimir Hahanov, Kharkov National University of Radio Electronics, Ukraine * Prof. Paolo Prinetto, Politecnico di Torino, Italy * Prof. Alberto Macii, Politecnico di Torino, Italy * Prof. Joongho Choi, University of Seoul, South Korea * Prof. Wei Li, Fudan University, China * Prof. Michel Schellekens, University College Cork, Ireland * Prof. Franco Fummi, University of Verona, Italy * Prof. Jun-Dong Cho, Sung Kyun Kwan University, South Korea * Prof. AHM Zahirul Alam, International Islamic University Malaysia, Malaysia * Dr. Emanuel Popovici, University College Cork, Ireland * Dr. Jong-Kug Seon, System LSI Lab., LS Industrial Systems Co. Ltd., South Korea * Dr. Umberto Rossi, STMicroelectronics, Italy * Dr. Graziano Pravadelli, University of Verona, Italy * Dr. Vladimir Pavlov, International Software and Productivity Engineering Institute, USA * Dr. Jinfeng Huang, Philips & LiteOn Digital Solutions Netherlands, Advanced Research Centre, The Netherlands * Dr. Thierry Vallee, Georgia Southern University, Statesboro, Georgia, USA * Dr. Menouer Boubekeur, University College Cork, Ireland * Dr. Ana Sokolova, University of Salzburg, Austria * Dr. Sergio Almerares, STMicroelectronics, Italy * Ajay Patel (Director), Intelligent Support Ltd, United Kingdom * Monica Donno (Director), Minteos, Italy * Alessandro Carlo (Manager), Research and Development Centre of FIAT, Italy * Yui Fai Lam (Manager), Microsystems Packaging Institute, Hong Kong University of Science and Technology, Hong Kong --------------------------- April 1, 2008: Deadline for submission of completed papers May 15, 2008: Notification of acceptance/rejection to authors Please visit our web-site for further information on the hosting conference of DTVCS, submission guidelines, proceedings and publications. General Chair of DTVCS: Dr. K.L. Man (University College Cork, Ireland) and Organising Chairs: Miss Maria O'Keeffe (University College Cork, Ireland) and: Mr. Michele Mercaldi (M.O.S.T., Italy) === Subject: -- Wolfram Research QA process defect: Bug in Mathematica 6 - Integrate - 86 (Log, invalid value, multiple regression bug) - BUG THE LONG LIVER: 1996-2008 (!) posting-account=ubyIWAkAAABW-OTbVB1QiN1oZlu0qUgw ----------------------------------------------------------------- -- NOT FIXED BUGS ARE DANGEROUS: THEY TEND TO GET MORE SEVERE -- ----------------------------------------------------------------- Quality of CASs is our #1 care. So our little demo continues.... Hello again from the VM machine which hopefully soon will be used by CAS manufacturers to the benefit of their customers. This example demonstrates YET ANOTHER case of bad defects in the Wolfram Research Quality Assurance process. Consider, it cannot trap and fix efficiently bugs the long livers. The reported here defect persists in the Mathematica versions since at least 1996 to 2008. This is also an example of a MULTIPLE regression bug when the invalid answer is now here, now not depending on the concrete version. ----------------------------------------------------------------- N[ - Pi ( Sqrt[3] Pi + 3 (Log[12] - 6) ) / 18 ] NIntegrate[Log[1 + z^2]/z^2/(1 + z^6), {z, 0, Infinity}] 0.890795 0.890795 ----------------------------------------------------------------- N[Integrate[Log[1 + z^2]/z^2/(1 + z^6), {z, 0, Infinity}]] ----------------------------------------------------------------- VERSION OUTPUT RESOLUTION ----------------------------------------------------------------- Mathematica 6.0 0.598308 <------------------------------ BUG #3 Mathematica 5.2 0.890795 OK Mathematica 4.2 1.8405 <------------------------------ BUG #2 Mathematica 3.0 Integral does not converge <---------- BUG #1 ----------------------------------------------------------------- Best wishes, Vladimir Bondarenko VM and GEMM architect Co-founder, CEO, Mathematical Director http://www.cybertester.com/ Cyber Tester, LLC http://maple.bug-list.org/ Maple Bugs Encyclopaedia http://www.CAS-testing.org/ CAS Testing ----------------------------------------------------------------- We must understand that technologies like these are the way of the future. ----------------------------------------------------------------- === Subject: it makes money today posting-account=RhDOgAoAAADqG6Jodn2scQc7v7itRbMo Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) register here and it begins to make money http://bux.to/?r=carlos1869 very easy === Subject: A General Definition of Part for Mathematical Objects posting-account=NMk5YQoAAACccB2Kbsg0e1L1ybwmgPr8 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) It is not obvious what is to be understood by a part of a mathematical object. For example, the parts of a group could be its elements, its subgroups, or its quotient groups. Is it possible to find a general definition of part? I will assume from now on that the mathematical objects are given as objects in a category C. All of the following could be considered parts of an object O: 1) A map p:T->O, where T is a terminal object. (Global elements) 2) A map p:O->Omega, where Omega is a subobject classifier. (Properties) 3) A map p:P->O. (Generalised elements) 4) A map p:O->P. (Coelements) 5) An equivalence class of monic maps p:P->O, where two maps p1:P1->O and p2:P2->O are considered equivalent if C includes an isomorphism h:P1->P2 such that p2*h=p1. (Subobjects) 6) An equivalence class of epic maps p:O->P, where two maps p1:O->P1 and p2:O->P2 are considered equivalent if C includes an isomorphism h:P1->P2 such that h*p1=p2. (Quotient objects) This list is just meant to give an idea of what a general definition should try to encompass. Note that in each of these examples, a part is a map to or from O (examples 1)-4) ), or an equivalence class of such maps (examples 5) and 6) ). It seems to me that the intuitive meaning of part is actually better captured if 1)-4) are rewritten to involve equivalence classes just like 5) and 6) (unfortunately, the definitions get somewhat harder to read, but the idea is really very simple): 1') An equivalence class of maps p:T->O, where T is a terminal object and where two maps p1:T1->O and p2:T2->O are considered equivalent if C includes an isomorphism h:T1->T2 such that p2*h=p1. 2') An equivalence class of maps p:O->Omega, where Omega is a subobject classifier and where two maps p1:O->Omega1 and p2:O->Omega2 are considered equivalent if C includes an isomorphism h:Omega1- >Omega2 such that h*p1=p2. 3') An equivalence class of maps p:P->O, where two maps p1:P1->O and p2:P2->O are considered equivalent if C includes an isomorphism h:P1- >P2 such that p2*h=p1. 4') An equivalence class of maps p:O->P, where two maps p1:O->P1 and p2:O->P2 are considered equivalent if C includes an isomorphism h:P1- >P2 such that h*p1=p2. Note that 1') and 5) are special cases of 3') while 2') and 6) are special cases of 4'). It might therefore seem as if 3') and 4') together capture the concept of part. However, this is not the case assuming that we want the following very natural requirement: If X is a part of Y and Y is a part of Z, then X is a part of Z. (Transitivity of is a part of) (For example, consider the case when X, Y, and Z are groups, when X is a subgroup of Y and Y is a quotient group of Z.) While natural, this is not quite precise in the present context. Instead, we may state the requirement thus: Parts can be composed; if we have a way of regarding X as a part of Y and a way of regarding Y as a part of Z, then we get -- by composition -- a way of regarding X as a part of Z. If we want 3') and 4') to be special cases and want to satisfy this requirement, then the following definition is not far-fetched: A part of an object O is given by a sequence of objects P=P0, P1, P2, ..., Pn=O, along with a map p1 between P0 and P1, a map p2 between P1 and P2, a map p3 between P2 and P3, and so on, where a map between two objects is a map that goes from one of the objects to the other object (the direction being of no importance). Two parts <, and <, are equal just in case there are isomorphisms h0:P0- >Q0, h1:P1->Q1, h2:P2->Q2, ..., h(n-1):P(n-1)->Q(n-1) which make everything commute. (Actually, one may want a weaker definition of equality, but I will ignore this complication.) Equivalently, one could define a part of O by means of a functor F:D- >C from some category D which contains two designed objects, D_P and D_O, with the latter satisfying F(D_O)=O. Two such functors F1:D->C and F2:D->C describe the same part just in case there is a natural isomorphism between them whose component at D_O is the identity of O. For this definition to fully correspond to the previous one, the category D must be of a certain form (there must be a chain from D_P to D_O which includes all objects and non-identity maps in D and includes no object twice). It is easy to see how 3') and 4') can be regarded as special cases of these definitions and how it is possible to compose parts (if we compose <, and <, , what we get is <, ). After all these technicalities, let me illustrate the beauty of parts. One of the beauties (in my opinion) is illustrated by the fact that the parts of a group G include not only subgroups (equivalence classes of monics to G) and quotient groups (equivalence classes of epis from G), but also their opposites: equivalence classes of monics /from/ G and equivalence classes of epis /to/ G. These may be thought of as ideas of how G could be embedded in a larger group. Intuitively, once you have a group, not only do you have its subgroups and quotient groups, but you also have ideas of how the group could appear as a subgroup or quotient group of some larger group. What I am trying to convey by the talk about ideas is captured precisely in the following definition: A part P of an object O is of the idea-type if P has two parts which are different but correspond to the same part of O when parts are composed. For example, if C is the category Sets then we may regard a two- element set {e1, e2} as a part of the empty set {} (via the inclusion map from {} to {e1, e2}), but it is a part of the idea-type since {e1} and {e2} are different when regarded as parts of {e1, e2} (in the natural way), but come to coincide when we compose parts and regard them as parts of the empty set {} (relative to {}, {e1} and {e2} are both just the idea of a one-element set (contained in a two-element set) ). Any comments? === Subject: Consecutive Zeros Counting Problem posting-account=ajEVfQoAAADsEDVr4z3lBchwwq9iTyyF Gecko/2008020514 Firefox/3.0b3,gzip(gfe),gzip(gfe) ----------- Start of problem definition --------------------------- Given a number in base B, Length N, we have B^N distinct values, where B and N are integers such that B>1 and N>0. Define a notation that forces all numbers to be Length N by padding the first digits with 0. So 0 with length 4 is 0000 and 1 with length 4 is 0001, etc. The problem is to find the number of values that have at least M consecutive 0s, where M is a natural number such that M>0 and M<=N. Restated, how many numbers base B, Length N, that have at least M consecutive 0s are there, where B, N, and M are integers such that B>1, N>0, and 0 ----------- Start of problem definition --------------------------- Given a number in base B, Length N, we have B^N distinct values, where > B and N are integers such that B>1 and N>0. Define a notation that forces all numbers to be Length N by padding > the first digits with 0. So 0 with length 4 is 0000 and 1 with > length 4 is 0001, etc. The problem is to find the number of values that have at least M > consecutive 0s, where M is a natural number such that M>0 and M<=N. Restated, how many numbers base B, Length N, that have at least M > consecutive 0s are there, where B, N, and M are integers such that > B>1, N>0, and 0 observation of computer program calculations) B^N - A_N where A is a recurrence relation that defines the number of values > without M consecutive zeros, given by the following formula: A_n = (B-1) * (A_(n-1) + A_(n-2) + ... + A_(n-M+1)) The last A_ should be (n-M), not (n-M+1). Seems ok once corrected: B: 2 N: 6 M: 3 Calculated zero count: 20 Brute force zero count: 20 B: 3 N: 6 M: 3 Calculated zero count: 81 Brute force zero count: 81 B: 4 N: 6 M: 3 Calculated zero count: 208 Brute force zero count: 208 B: 5 N: 6 M: 3 Calculated zero count: 425 Brute force zero count: 425 B: 6 N: 6 M: 3 Calculated zero count: 756 Brute force zero count: 756 B: 7 N: 6 M: 3 Calculated zero count: 1225 Brute force zero count: 1225 B: 8 N: 6 M: 3 Calculated zero count: 1856 Brute force zero count: 1856 B: 9 N: 6 M: 3 Calculated zero count: 2673 Brute force zero count: 2673 B:10 N: 6 M: 3 Calculated zero count: 3700 Brute force zero count: 3700 B:11 N: 6 M: 3 Calculated zero count: 4961 Brute force zero count: 4961 B:12 N: 6 M: 3 Calculated zero count: 6480 Brute force zero count: 6480 B:13 N: 6 M: 3 Calculated zero count: 8281 Brute force zero count: 8281 B:14 N: 6 M: 3 Calculated zero count: 10388 Brute force zero count: 10388 B:15 N: 6 M: 3 Calculated zero count: 12825 Brute force zero count: 12825 B:16 N: 6 M: 3 Calculated zero count: 15616 Brute force zero count: 11776 Wait a minute -- something's still wrong. I'll get back later... With initial seeding values A_0 = B^0, A_1 = B^1, ..., A_(M-1) = > B^(M-1) Note that the order of the recurrence relation A changes as M changes, > requiring M initial seeding values. Also note that for B = 2, and M = > 2, this relation is equivalent to the Fibonacci sequence. ------------ End of solution --------------- Now, how does one go about proving this relation is true, for all B, > N, and M? I would rather have a solution that is verified > mathematically rather than just a gut feeling it is true for all N, > etc. As B and N increase, the computing time required to verify > numerical results grows very fast. === Subject: Re: Consecutive Zeros Counting Problem posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) On Feb 14, 4:24æpm, mensana...@aol.compost ----------- Start of problem definition --------------------------- > Given a number in base B, Length N, we have B^N distinct values, where > B and N are integers such that B>1 and N>0. > Define a notation that forces all numbers to be Length N by padding > the first digits with 0. æSo 0 with length 4 is 0000 and 1 with > length 4 is 0001, etc. > The problem is to find the number of values that have at least M > consecutive 0s, where M is a natural number such that M>0 and M<=N. > Restated, how many numbers base B, Length N, that have at least M > consecutive 0s are there, where B, N, and M are integers such that > B>1, N>0, and 0 -------------- End of problem, Start of solution ---------------- > The solution: (obtained by intuition (mostly of Harold Reiter) and > observation of computer program calculations) > B^N - A N > where A is a recurrence relation that defines the number of values > without M consecutive zeros, given by the following formula: > A n = (B-1) * (A (n-1) + A (n-2) + ... + A (n-M+1)) The last A should be (n-M), not (n-M+1). Seems ok once corrected: B: 2 æN: 6 æM: 3 > Calculated æzero count: 20 > Brute force zero count: 20 B: 3 æN: 6 æM: 3 > Calculated æzero count: 81 > Brute force zero count: 81 B: 4 æN: 6 æM: 3 > Calculated æzero count: 208 > Brute force zero count: 208 B: 5 æN: 6 æM: 3 > Calculated æzero count: 425 > Brute force zero count: 425 B: 6 æN: 6 æM: 3 > Calculated æzero count: 756 > Brute force zero count: 756 B: 7 æN: 6 æM: 3 > Calculated æzero count: 1225 > Brute force zero count: 1225 B: 8 æN: 6 æM: 3 > Calculated æzero count: 1856 > Brute force zero count: 1856 B: 9 æN: 6 æM: 3 > Calculated æzero count: 2673 > Brute force zero count: 2673 B:10 æN: 6 æM: 3 > Calculated æzero count: 3700 > Brute force zero count: 3700 B:11 æN: 6 æM: 3 > Calculated æzero count: 4961 > Brute force zero count: 4961 B:12 æN: 6 æM: 3 > Calculated æzero count: 6480 > Brute force zero count: 6480 B:13 æN: 6 æM: 3 > Calculated æzero count: 8281 > Brute force zero count: 8281 B:14 æN: 6 æM: 3 > Calculated æzero count: 10388 > Brute force zero count: 10388 B:15 æN: 6 æM: 3 > Calculated æzero count: 12825 > Brute force zero count: 12825 B:16 æN: 6 æM: 3 > Calculated æzero count: 15616 > Brute force zero count: 11776 Wait a minute -- something's still wrong. I'll get back later... Never mind, Python inapproriately puts an '0x' decorator on base 16 numbers. Will have to change my brute force algorithm to strip those off. > With initial seeding values A 0 = B^0, A 1 = B^1, ..., A (M-1) = > B^(M-1) > Note that the order of the recurrence relation A changes as M changes, > requiring M initial seeding values. æAlso note that for B = 2, and M = > 2, this relation is equivalent to the Fibonacci sequence. > ------------ End of solution --------------- > Now, how does one go about proving this relation is true, for all B, > N, and M? æI would rather have a solution that is verified > mathematically rather than just a gut feeling it is true for all N, > etc. æAs B and N increase, the computing time required to verify > numerical results grows very fast. === Subject: Re: Consecutive Zeros Counting Problem posting-account=ajEVfQoAAADsEDVr4z3lBchwwq9iTyyF Gecko/2008020514 Firefox/3.0b3,gzip(gfe),gzip(gfe) > A_n = (B-1) * (A_(n-1) + A_(n-2) + ... + A_(n-M+1)) My apologies, this should be A_n = (B-1) * (A_(n-1) + A_(n-2) + ... + A_(n-M)) === Subject: Re: Consecutive Zeros Counting Problem > ----------- Start of problem definition --------------------------- >Given a number in base B, Length N, we have B^N distinct values, where > B and N are integers such that B>1 and N>0. >Define a notation that forces all numbers to be Length N by padding > the first digits with 0. So 0 with length 4 is 0000 and 1 with > length 4 is 0001, etc. >The problem is to find the number of values that have at least M > consecutive 0s, where M is a natural number such that M>0 and M<=N. >Restated, how many numbers base B, Length N, that have at least M > consecutive 0s are there, where B, N, and M are integers such that > B>1, N>0, and 0-------------- End of problem, Start of solution ---------------- >The solution: (obtained by intuition (mostly of Harold Reiter) and > observation of computer program calculations) >B^N - A_N >where A is a recurrence relation that defines the number of values > without M consecutive zeros, given by the following formula: >A_n = (B-1) * (A_(n-1) + A_(n-2) + ... + A_(n-M+1)) >With initial seeding values A_0 = B^0, A_1 = B^1, ..., A_(M-1) = > B^(M-1) >Note that the order of the recurrence relation A changes as M changes, > requiring M initial seeding values. Also note that for B = 2, and M = > 2, this relation is equivalent to the Fibonacci sequence. >------------ End of solution --------------- >Now, how does one go about proving this relation is true, for all B, > N, and M? I would rather have a solution that is verified > mathematically rather than just a gut feeling it is true for all N, > etc. As B and N increase, the computing time required to verify > numerical results grows very fast. Hint: a string of N digits (with N > M) is without M consecutive zeros if and only if the first nonzero digit is in some position j with 1 <= j <= M, and followed by a string of N-j digits without M consecutive zeros. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Petry's paradox posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 CLR 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) On Feb 13, 2:14æam, david petry would reason as follows: æSince I believe that my own beliefs include > a contradiction, I have to believe that I am capable of arriving at > absolutely any conclusion at all, and hence I have no reason to > believe that any of the conclusions I reach are necessarily true, Non sequitur. One may recognize that one is able to incorrectly conclude contingent statements that are in contradiction with one another but still recognize that some of ones beliefs are of necessary truths. Also, you're assuming that the person understands ex falso quodlibet, which not all people do, not even all fairly intelligent and rational people . And backing up to an earlier point of yours. A rational person might recognize that he has thousands upon thousands (millions?) of beliefs and that therefore there is great likelihood that at least two of those beliefs are inconsistent (since one might have this great number of beliefs without having checked each one against all the others), and therefore believe, based on that likelihood, that one does have inconsistent beliefs. MoeBlee === Subject: Re: Petry's paradox > herbzet says... >More to the point is the question of whether, and to what degree, >the first-order formula ~Con(PA) captures the natural language >meaning of the meta-theoretical statement PA is inconsistent. >That's a puzzlement to me, at the moment. >The formula ~Con(PA) captures the natural language meaning > of PA is inconsistent in the sense that we can prove that > PA is inconsistent <-> ~Con(PA) is true of the naturals. >In any model of the theory A = PA + ~Con(PA), the formula > ~Con(PA) does *not* mean PA is inconsistent. It means > something complicated like There exists a hyperfinite > number x such that x is a code for a hyperfinite structure > such that blah, blah, blah. -- hz === Subject: Primes in non-arithmetic progression! posting-account=rltBVQkAAABzJdnKp0dFuDqa5xCvlbif Gecko/20061201 Firefox/2.0.0.11 (Ubuntu-feisty),gzip(gfe),gzip(gfe) I just want to know if anyone know if some set A_k = {p, p+k, p+k+1, p+k+2, .... } for some k>=1 has: * infinit primes * arbitrary long chains (that is, we could find p+k+i, .... , p+k+j in A_k all primes) Xan. === Subject: Primes in non-arithmetic progression! posting-account=rltBVQkAAABzJdnKp0dFuDqa5xCvlbif Gecko/20061201 Firefox/2.0.0.11 (Ubuntu-feisty),gzip(gfe),gzip(gfe) I just want to know if anyone know if some set A_k = {p, p+k, p+k+1, p+k+2, .... } for some k>=1 has: * infinit primes * arbitrary long chains (that is, we could find p+k+i, .... , p+k+j in A_k all primes) Xan. === Subject: Re: Primes in non-arithmetic progression! days. My association with the Department is that of an alumnus. I just want to know if anyone know if some set A_k = {p, p+k, p+k+1, p+k+2, .... } for some k>=1 has: * infinit primes Well, since there are infinitely many primes larger than any specific integer, it is certainly the case that for any p>0 and any k>0, the set {p+k+1, p+k+2, p+k+3,....} will contain infinitely many primes; in fact, it will contain all but finitely many of the (positive) primes. >* arbitrary long chains (that is, we could find p+k+i, .... , p+k+j in >A_k all primes) Certainly not, since if p+k+i is prime, with only one exception it will be odd and therefore p+k+i+1 will be even, hence not a prime. Somehow, I suspect you did not write what you meant to write. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: Primes in non-arithmetic progression! posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/20070530 Fedora/1.5.0.12-1.fc5 Firefox/1.5.0.12,gzip(gfe),gzip(gfe) I just want to know if anyone know if some set A_k = {p, p+k, p+k+1, p+k+2, .... } for some k>=1 has: * infinit primes For all k, such a set contains an arithmetic progression, so it contains infinitely many primes. > * arbitrary long chains (that is, we could find p+k+i, .... , p+k+j in > A_k all primes) As long as the chain has at least two items, one of them is going to be even... -- m === Subject: Re: Primes in non-arithmetic progression! I just want to know if anyone know if some set A_k = {p, p+k, p+k+1, p+k+2, .... } for some k>=1 has: * infinit primes >* arbitrary long chains (that is, we could find p+k+i, .... , p+k+j in >A_k all primes) If you have 2 consecutive integers doesn't one of them have to be even? For 3 consecutive integers, doesn't one of them have to be a multiple of 3? And so on ... quasi === Subject: Solving system of boolean equation posting-account=0VJq7goAAAC-2RnkHZ_MGYA23eh312IU Gecko/20070725 Firefox/2.0.0.6,gzip(gfe),gzip(gfe) Hi. Could anyone give me some hint on how to solve (3),(4) given (1) and (2)? In = ((~(An^Bn))*(An^In-1)) + ((An^Bn)*(An^Qn-1)) ----(1) Qn = ((~(An^Bn))*(Bn^Qn-1)) + ((An^Bn)*(Bn^In-1)) ----(2) An = ~(Qn^In-1)*(In^Qn-1) + (In^In-1)*(Qn^Qn-1) --- (3) Bn = ~(In^Qn-1)*(Qn^In-1) + (In^In-1)*(Qn^Qn-1) ----(4) where ~ is invert operation, ^ is xor, + is or, and * is and. I tried to decompose An^Bn into ~An + ~Bn, but still got bogged down. Sam === Subject: Re: Solving system of boolean equation posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/20070530 Fedora/1.5.0.12-1.fc5 Firefox/1.5.0.12,gzip(gfe),gzip(gfe) > Hi. Could anyone give me some hint on how to solve (3),(4) given (1) and > (2)? In = ((~(An^Bn))*(An^In-1)) + ((An^Bn)*(An^Qn-1)) ----(1) > Qn = ((~(An^Bn))*(Bn^Qn-1)) + ((An^Bn)*(Bn^In-1)) ----(2) An = ~(Qn^In-1)*(In^Qn-1) + (In^In-1)*(Qn^Qn-1) --- (3) > Bn = ~(In^Qn-1)*(Qn^In-1) + (In^In-1)*(Qn^Qn-1) ----(4) where ~ is invert operation, ^ is xor, + is or, and * is and. I tried to decompose An^Bn into ~An + ~Bn, but still got bogged down. What do you mean by solve in this context? What does -1 mean? Google for smart questions... -- m === Subject: Man+Machine Review Of Maple Crisis beta 0.2 is under way posting-account=ubyIWAkAAABW-OTbVB1QiN1oZlu0qUgw ----------------------------------------------------------------- Unique Maple bug data not available elsewhere. Crashes. Absurd results. Math roulette clanging. Simple made challenging by fragile Maple kernel. 750 pages of selected Maple 12 bugs. The beta 0.2 of the first world's document written by a human in a close cooperation with a successor of the GEMM machine, our unique VM automated testing expert system which is a failure prediction oracle. To be released in max 2 months after Maple 12 release. ----------------------------------------------------------------- http://maple.bug-list.org/maple-crisis.php Main Maple's Quality Results 1. Maple 9.5 is an unstable, inconsistent, non-linear, non-uniform, randomized, self-incompatible environment where fundamental math properties (uniqueness of the answer for a good-defined problem commutativity and linearity property etc) now hold, now fail making Maple breaking down grotesquely. [ ... ] ----------------------------------------------------------------- http://www.cybertester.com/ Cyber Tester, LLC http://maple.bug-list.org/ Maple Bugs Encyclopaedia http://www.CAS-testing.org/ CAS Testing ----------------------------------------------------------------- We must understand that technologies like these are the way of the future. ----------------------------------------------------------------- === Subject: Re: -- x*phi(y) = y*phi(z) = z*phi(x) > <147634.1202507863287.JavaMail.jakarta@nitrogen.mathfo > rum.org>, > <33430859.1202414568640.JavaMail.jakarta@nitrogen.math > forum.org>, > Roman B. Binder surely > for x;y > 1 > x*phi(y) = y*phi(x) > implies x=y >Right. Tell me something: what happens > when x = 4 and y = 2? > 2 is prime number so what with the function phi(2) > ? > phi(4) could be 2 and also 1 ? > I can guess function phi collects one factor or > product of some of them of the specific number. > If there is included value 1 as a factor > so I am sorry. I guess false. >So, what you're saying is, you posted something about > phi > when you don't know what phi is? >Why would anyone do that? Because not everybody likes to think: Sometimes I enjoy to dezifer most of philosofic symbols in mathematics: LHS; RHS; gcd; RMS; etc. No matter much bigger list could be offer at any short hand or slightly coded knowledge: If I would write win, Win, WIn or WIN ? How many correct suggestions could be covered ? Sure I'll find this enigmatic symbol in some library. How much I am mistaken ? Ro-Bin >-- > Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for > email) === Subject: www.cybertester.com <9qr6gs68ei.fsf_-_@aquin.mat.univie.ac.at> posting-account=ubyIWAkAAABW-OTbVB1QiN1oZlu0qUgw ----------------------------------------------------------------- We must understand that technologies like these are the way of the future. ----------------------------------------------------------------- === Subject: Re: Hidden Variables, Lost and Found. posting-account=G-TjQAkAAADYg6rno3bWQPnIwKFBrf1t CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) On Feb 8, 10:47æpm, Tomoko Kanazawa dom arigato > One is capable of observing expected length, but one cannot observe > the underlying process which is probabilistic. This probabilistic > process is hidden from observers who can only observe the expected > length. These Hidden Variables are really just random variables which may be > discrete or continuous random variables. It cannot be determined > whether these random variables are never observed. One can only > observe the expected length. And so they may be continuous random > variables or discrete random variables. The mathematical conditions required to reduce probabilistic phenomena in this way are non-trivial. One cannot automatically conclude this. The required conditions for this to go through, in fact, are themselves empirically testable. The most famous of those experiments, the Aspect experiments, show that the conditions FAILED for quantum theory -- this making the reduction you suggest mathematically impossible. === Subject: Re: Hidden Variables, Lost and Found. posting-account=tCEoyAoAAAAkltU5zxOoI8uJ4lyz5-kv .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) On Feb 13, 8:44æam, Tomoko Kanazawa dom arigato > On Feb 12, 10:40æpm, Tomoko Kanazawa dom arigato > We percieve large scale objects as either existing or not. Earth and > moon exist, but 2nd Sun does not. There is only one Sun. The Second > Sun is nonexistent. On Newtonian scales we percieve a dichotomy, > something either exists or it does not. There is no in-between. > But the quantum scale is an existential boundary. On the quantum > scale, it is useful to understand length using existential > indeterminacy. To each point along a given path we may ascribe a > probability that the point exists. > Many popularizers -- and presumbly some influential practitioners, do > speak this way: that quantum mechanics requires a third > indeterminate state of existence. > For what it's worth, I think this is sheep dip (which is worse -- hog > wash, or sheep dip?). > It does make sense that you could have 3 existential forms. I have > never seen a satisfactory explanation from physics, except quantum > states, but they dont seem to state it explicitly. > There seems to be a duality here. Either you have existential > dichotomy (something either exists or it does not), or you have > existential indeterminacy (an object exists with any probability from > 0 to 1). > Well, random length seems to facilitate this rather nicely, and so I > am being lured by this because it looks like an explanation to wave- > When something seems to be inteterminate in state, it should occur > to us this may mean the something is an outcome of a process which > hasn't proceeded, yet. æClassically processes with indeterminate > outcomes (determined only in probability) are common as mud. > I further think that if we start from zero and analyze what we want to > mean by a physical theory, we will find that any theory we might > describe as having a three state logic can be recast in a form which > appears to be a more conventionaly kind of theory. > We should also remember that theories don't have to concern themselves > with whether intermidiates exist or don't exists, or something > else -- theories need only concern themselves with predicting > observable outcomes (we knew this, right?). æIn this regard theories > mimic the claim of Decartes (I think, therefore I am): we perceive > the outcome of experiment, and therefore it is (or is not). æWe may > define existence for unobserved intermediaries, but that's merely > some extra labeling. > I've been working on something which is really quite fascinating to > me, and really seems like fun. Even though it is rather excruciating > to develop philosophically. I can describe my the state of my failure > in progress - > Based on my views regarding length, I have concocted an idea of what > motion must be. So, I want to understand length, velocity and > acceleration within the framework of my claims about length. I want to > try to use that to understand HUP better, or prove it if possible. > Pretty humble - eh ? yeah - I know. Anyway - it's all in fun. > So...I've been thinking about this - and here is my conclusion. My > most recent hallucination. > I will need to consider randomness or disorder as if it were a > liquid...and then perform fluid mechanics upon it. As if randomness > could flow like water - that is what I am vizualizing. I think that > this can be modelled very easily with PDF's. I think that this > actually makes sense and is usable. I think that I could argue it > pretty well. Might not win, but I could make a case. > But I cannot comprehend distance, velocity and acceleration based on > the above. I am trying to understand how they are related.....similar > to traditional calculus. And I dont quite understand it. But I think > that this might say something very interesting about HUP. > So that is what I've been thinking about. And I think that I might > need tensors. Id really rather keep it simple - > but how does one perform vector mathematics on a blob of randomness ? > How to ascribe properties of position and acceleration to an > abstraction ? Sometimes seems ridiculous, but I know that I have some > good fundamentals. > Bewildered -- Hide quoted text - > - Show quoted text -- Hide quoted text - > - Show quoted text - I think that it makes sense to treat randomness as if it could slosh > around like a liquid, a liquid which is embedded in length. But modelling this mathematically is tricky. A PDF seems to work fine for the 1-dimensional case. But for the 2-dimensional case you cannot use PDF X PDF as if they > were cooordinate axes. For the 2-dimensional case you would need an axis, and to each point > along the axis you assign a PDF which normal to that point, all lying > in the same plane. You could do the same thing for the 3-dimensional > case, but I dont know about the 4th dimension. And this methodology > starts to get very non-elegant. I think that using vector spaces might make more sense. Maybe tensors. > Just about clueless at this point - but that's where I'm at. Well, I does seem like a possible means to model space-bending, > despite the wierdness and difficulties. Still needs work I think.- Hide quoted text - - Show quoted text - OK - thought about this a bit and I think that I have a solution which will work. For the 1 dimensional case you can characterize the existential potential with a PDF which is a function of 1 variable. For the 2 dimensional case we use a function of 2 variables f(x,y). The output of that function is interpreted as the existential potential of the 2-tuple (x,y). We simply interpret the output as a probability. We can do the same thing for 3-space using functions of 3 variables to give existential probabilities associated with those 3-tuples. (disregarding issues related to normalization for the time being) With this in hand, we can model randomness as if it were a fluid - so to speak. In fact all we are doing is bending space, but we have a lot more freedom to do whatever we want compared to what the Lorentz transform will do. So - we think that we know what an object is. And we think that we know what distance is. We have to understand velocity, acceleration, and momentum. Momentum will be the hardest I think - Then we can talk about Heisenberg's Uncertainty Principle. === Subject: My number of attempts at proofs of Fermat's Last Theorem = 3 posting-account=XBlrJAoAAADt98TEM65O-rVj9Zl2tTjM 1.1.4322),gzip(gfe),gzip(gfe) All attempts were pre-Wiles and without knowledge of the work of Frey and Ribet. My correct proofs = zero. Not even a pretense of a proof. I freely admit failure (only) in this regard to Fermat's Last Theorem. Unlike that joke Bill Newbold, and that utter failure and crank James Harris, I didn't keep going on and on and on about my attempts. Now, the integral of x^x did intrigue me for many years. Luckily, the Internet came around, and I learned of Liouville's Theorem, so my fruitless search for an elementary expression of integral of x^x came to an end. === Subject: solution manuals posting-account=1LXcdwoAAAAOMsiNKAVJ2p8wxrEhk12V Gecko/20080201 Firefox/2.0.0.12 Creative ZENcast v1.02.10,gzip(gfe),gzip(gfe) Hello I have solution manuals for a competitive price. I also have other manuals Feel free to contact me. MODERN CONTROL SYSTEM 4th Edition by OGATA Modern Control System 11th edition by Dorf and Bishop (11e) Fundamentals of Microelectronics. Author: Behzad Razavi Microelectronic Circuits. Author: Adel S. Sedra, Kenneth C. Smith Control Systems Engineering. Author: Norman S. Nise Probability and Random Processes for Electrical Engineering (3rd Edition)Chapters 1-6. Author: Albert Leon-Garcia Communication Systems Engineering (2nd Ed.). Author:Proakis and Salehi. ISBN 0130619746 Digital Signal Processing System Analysis and Design. Author: Paulo S.R. 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