mm-2949 ==== Subject: Re: http://arxiv.org/abs/gr-qc/0602022 GARBAGE Bad as that strewn by Brian Stuck Less. Vergon ==== Subject: Re: Terminology so, are you just a Wolframite, relying on math newbies to buy the package? speak the Hell **! thus quoth: Don't be a square or a blockhead; see: http://bfi.org/node/574 thus: so, what numbers are known to be normal, aside from artificial (cantorian?) cconstructions?... I prefer to take the circumference as the unit, thus making the diameter to be pith; why wouldn' pith be normal, as well -- what possible *raison* could be, for pi not being normal, and having an infinite number of the God-am message in Contact?... as a student of what's-his-name, The Weaponeer(s), Sagan had *always* been a little whack-o!... oh, yes; asking for money to make the a-bomb. > Yes, IF pi is normal, it contains all sequences of digits. But that's not known. And, besides any normal number (many of which ARE known) does that- and almost all numbers ARE normal! thus: what an exquisite invocation of apsension, taking it apart along one axis -- dood! >(http://hexdome.com/bridges/graphics/bridge_6.jpg) only comes through thus: he's not god ... just the God-am MEssiah.... with faster-than-light *drive*, he'll be popping out of the continuum and into your face in less than no time !moooorV -- alas, life is just one, long QI test. >it wasn't until then, early 1990s, when I had unlimited access to >SLAC and Stanford's voluminous library systems -- and the vast >literature of a global community of institutional, alternative and >dissident physics -- that the ubiquity of subluminal and superluminal >lightwave and plasma velocities (and hence starship speeds) became >overwhelming and trivial. it was then that I showed the obvious, that >'Black Holes' do not exist, that rather the supermassive superstellar >phenomena that populate the centers of galaxies and denser >plasma-cosmos regions are TOROIDAL SUPERSTARS, with 'horizon' >(the AVERAGE velocity of light). [and, of course, it was 1994 when >the hundreds of pages of my math and illustration and scribbling led >to the discovery of the 3-frequency superluminal lissajous topology >of the proton. [a standing, self-reflected, or re-connected >electromagnetic wave. just look at the Lissajous figures on your >oscilloscope!] >though I have had many many teachers over the decades in the areas of >propulsion and velocity, I still look to the day when I have even ONE >colleague in the world who has (honestly, ethically) studied, >reflected, and written on the superluminal electrodynamic (OBVIOUSLY >dynamic, non-static) topology of the atomic nucleus. in other words, >nucleon-group topological dynamics. (which ARE indeed >electromagnetic, so-called 'weak' and 'strong' (QED and QCD) fields >being transparent pure red-herring misdirection to take the cattle, >clones, away from looking at the ACTUAL electromagnetic wave >(picoscopic and femtoscopic variable wavelength, frequency AND >velocity) helicoidal topologies of protons, neutrons and their >assemblages. --Friends don't let friends be/do Yahoo!s ... you can login to the googolplex, but you can't logout! http://www.benfranklinbooks.com/ http://members.tripod.com/~american_almanac http://www.wlym.com/pdf/iclc/howthenation.pdf http://www.rand.org/publications/randreview/issues/rr.12.00/ http://www.rwgrayprojects.com/synergetics/plates/figs/plate02.html ==== Subject: Re: Terminology > so, are you just a Wolframite, > relying on math newbies to buy the package? > speak the Hell **! > thus quoth: > Don't be a square or a blockhead; see: > http://bfi.org/node/574 I showed the Synergetics coordinate system in Wolfram's Mathematica because it's so easy to read, write, and translate into other programming languages. Go kick yourself. Cliff Nelson Dry your tears, there's more fun for your ears, Forward Into The Past 2 PM to 5 PM, Sundays, California time, at: http://www.kspc.org/ Don't be a square or a blockhead; see: http://bfi.org/node/574 ==== Subject: Re: Trying to figure out what this summation does. helpful. ==== Subject: Re: Pi as the Mother Number <32019501.1139759545412.JavaMail.jakarta@nitrogen.mathforum.org> so, what numbers are known to be normal, aside from artificial (cantorian?) cconstructions? I prefer to take the circumference as the unit, thus making the diameter to be pith; why wouldn' pith be normal, as well -- what possible *raison* could be, for pi not being normal, and having an infinite number of the God-am message in Contact? as a student of what's-his-name, The Weaponeer(s), Sagan had *always* been a little whack-o!... oh, yes; asking for money to make the a-bomb. > Yes, IF pi is normal, it contains all sequences of digits. But that's not known. And, besides any normal number (many of which ARE known) does that- and almost all numbers ARE normal! thus: what an exquisite invocation of apsension, taking it apart along one axis -- dood! >(http://hexdome.com/bridges/graphics/bridge_6.jpg) only comes through thus: he's not god ... just the God-am MEssiah.... with faster-than-light *drive*, he'll be popping out of the continuum and into your face in less than no time !moooorV -- alas, life is just one, long QI test. >it wasn't until then, early 1990s, when I had unlimited access to >SLAC and Stanford's voluminous library systems -- and the vast >literature of a global community of institutional, alternative and >dissident physics -- that the ubiquity of subluminal and superluminal >lightwave and plasma velocities (and hence starship speeds) became >overwhelming and trivial. it was then that I showed the obvious, that >'Black Holes' do not exist, that rather the supermassive superstellar >phenomena that populate the centers of galaxies and denser >plasma-cosmos regions are TOROIDAL SUPERSTARS, with 'horizon' >(the AVERAGE velocity of light). [and, of course, it was 1994 when >the hundreds of pages of my math and illustration and scribbling led >to the discovery of the 3-frequency superluminal lissajous topology >of the proton. [a standing, self-reflected, or re-connected >electromagnetic wave. just look at the Lissajous figures on your >oscilloscope!] >though I have had many many teachers over the decades in the areas of >propulsion and velocity, I still look to the day when I have even ONE >colleague in the world who has (honestly, ethically) studied, >reflected, and written on the superluminal electrodynamic (OBVIOUSLY >dynamic, non-static) topology of the atomic nucleus. in other words, >nucleon-group topological dynamics. (which ARE indeed >electromagnetic, so-called 'weak' and 'strong' (QED and QCD) fields >being transparent pure red-herring misdirection to take the cattle, >clones, away from looking at the ACTUAL electromagnetic wave >(picoscopic and femtoscopic variable wavelength, frequency AND >velocity) helicoidal topologies of protons, neutrons and their >assemblages. --Friends don't let friends be/do Yahoo!s ... you can login to the googolplex, but you can't logout! http://www.benfranklinbooks.com/ http://members.tripod.com/~american_almanac http://www.wlym.com/pdf/iclc/howthenation.pdf http://www.rand.org/publications/randreview/issues/rr.12.00/ http://www.rwgrayprojects.com/synergetics/plates/figs/plate02.html ==== Subject: Re: limits , The > On Sun, 19 Feb 2006 19:35:32 -0800, The World Wide Wade Just to add an elementary spin y = sqrt(x + 1) is an increasing function dy / dx = (1 / 2) (x + 1)^(-1/2) >= 0 for every x >= 0. > That sqrt(x + 1) is increasing is immediate without calculus. (I > didn't write the above, by the way). But how does this help? We > need to know that if x is one of the iterates, then x <= > sqrt(1+x). Well, but: if f is increasing, and x <= f(x), then by induction x <= f(x) <= f^2(x) <= ... while if f(x) <= x then ... <= f^2(x) <= f(x) <= x. So the sequence of iterates is monotone. --Ron Bruck ==== Subject: Re: limits > , The On Sun, 19 Feb 2006 19:35:32 -0800, The World Wide Wade Just to add an elementary spin y = sqrt(x + 1) is an increasing function dy / dx = (1 / 2) (x + 1)^(-1/2) >= 0 for every x >= 0. That sqrt(x + 1) is increasing is immediate without calculus. (I > didn't write the above, by the way). But how does this help? We > need to know that if x is one of the iterates, then x <= > sqrt(1+x). > Well, but: if f is increasing, and x <= f(x), then by induction > x <= f(x) <= f^2(x) <= ... > while if f(x) <= x then > ... <= f^2(x) <= f(x) <= x. > So the sequence of iterates is monotone. So it is. This iteration stuff always confuses me. ==== Subject: Interesting strategy problem This problem was suggested to me by the post at though it may be a well-known one. Two players both pay one unit into a pot. Each is then dealt a random number uniformly distributed between zero and one. Each player sees his own number but not his opponent's. Player A then bets an additional k units, where k is some number of his choice. k can be any non-negative real number - not necessarily a whole number of units or an exact fraction. In response, player B can choose to call or fold. If B calls then he matches A's k units and there is a show of numbers. The player with the higher number takes the pot, realising a net gain of 1 + k units; the player with the lower number sees a net loss of 1 + k units. If B folds then A takes the pot whatever the numbers, resulting in a net gain of 1 unit for A and a net loss of 1 unit for B. k = 0 is equivalent to an immediate show of numbers since it is never disadvantageous for B to match the zero bet and call. Two questions then: 1. What is A's optimal strategy, and what is the corresponding expected payoff for A? 2. What is B's optimal strategy, and what is the corresponding expected payoff for B? The optimal strategy is the best worst case strategy. For example, for every possible A-strategy there is a best counter-strategy available to B (or possibly many equally good ones), which holds A to an expected payoff of x units. A's optimal strategy is the one that maximises x. B's optimal strategy is the same but the other way around. In the most general terms (the most general that I can think of anyway), A's strategy is defined by a function p(a,k) which gives the probability that A will bet k given that his number is a. To simplify matters we could let the p's take only the value zero or one (in other words, player A always does the same thing with each a), as in the example below. My guess is that the optimal strategy might be of this nature, but I am not sure. B's strategy can be defined by a function q(b,k) which gives the probability that B will call given that his number is b and A has bet k units. However, for question 1 it is fairly obvious that B's best counter-strategy is to call only if b is higher than some cutoff point which depends on k (there may also be a region where it makes no difference what B does). Regarding question 1, the best A-strategy I have so far found guarantees A an expected payoff of 1/7, if I haven't made a mistake. The strategy is as follows: If a < 1/7 then k is the unique positive solution to a = (k^3 + 6*k^2) / (7*(2 + k)^3). If 1/7 <= a < 4/7 then k = 0. If 4/7 <= a < 1 then k = Sqrt(12) / Sqrt(7 - 7*a) - 2. where, as before, a is A's number and k is the amount that A bets. A's bet rises from zero at a = 0 to become indefinitely large as a approaches 1/7. It then plummets to zero over the centre section, and rises again from a = 4/7 to become indefinitely large as a approaches 1. B's best counter-strategy (holding A's expected payoff to 1/7) is: Trivially call for k = 0. Otherwise fold if b < a_1 and call if b > a_2, where a_1 and a_2 are the two possible values of a, with a_1 < a_2. Between these values B can do either - it makes no difference. When I started looking at this I assumed that A's optimal strategy would not involve k anywhere taking a continuous range of values (dependent on a), since the value of k would then give B very specific information about the value of a. I thought that k would take on just a small number of discrete levels. What I found, though, is that a certain pattern of discrete levels gives an ever-increasing expected payoff for A as the number of levels increases, with the limit being the continuous curves given above. I have a slight conceptual problem with this continuous result though. When A bets k units (k <> 0), he reveals to B that a is one of two numbers: a_1 or a_2. The result above is calculated assuming that the probability of a being a_1 or a_2, as far as it is known to B, is in the same proportion as the corresponding values of da/dk. This works in practice because a and k can only be given to a finite number of decimal places and so there is necessarily some discretisation. But if a and k really are real numbers then I am not sure this reasoning is valid. I am not convinced that B has *any* information about the relative probabilities of a_1 and a_2. To work around this the above answer can be discretised to any degree of fineness, and A's expectation can thus be made as close to 1/7 as we want. I haven't yet looked at question 2. I don't know whether we would expect the expectation in question 2 to be the negative of the expectation in question 1. I would be interested to hear if anyone can improve on my answer of 1/7 (assuming it is correct), or make any progress with question 2. ==== Subject: Re: f(x)=f(2x)+f(2x+1) > -- > VP Cheney Burr-ed his gun as a bird flew past The nation responds > burr > as we await bird flu shots and fight a real cold war. > pmont...@cwi.nl Microsoft Research and CWI Home: Bellevue, WA > Please leave politics out of this site. *sigh* Just when I start thinking that we're approaching October, someone has to come along and remind me that we're still right in the middle of September. LuckyOne: Usenet isn't a site, and politics is fair game in .sigs (unlike lawyers, who are only fair game while hunting quail). Colin Percival ==== Subject: Re: f(x)=f(2x)+f(2x+1) > -- > VP Cheney Burr-ed his gun as a bird flew past The nation responds > burr > as we await bird flu shots and fight a real cold war. > pmont...@cwi.nl Microsoft Research and CWI Home: Bellevue, WA > Please leave politics out of this site. This isn't a site--Google copied it from sci.math. This is a newsgroup. What you object to is called a .sig. It was carefully crafted to fit within the official requirements for a .sig-- it begins with a line consisting only of --, and then has five lines of commentary--and it is WELL WITHIN the customs of usenet, and standard netiquette, to say what you like in your .sig. That said, some employers wouldn't be happy having their name mentioned in the .sig, and some will require a disclaimer. That doesn't seem to be the case here. --Ron Bruck ==== Subject: Re: JSH is manipulating you <43f7440e@dnews.tpgi.com.au> <87vevbg0hv.fsf@phiwumbda.org> <43fa126d$1@news2.actrix.gen.nz> I guess, there's some kind of minimum desiderata for readablity. thus: I always recommend starting with the 14th book of Euclid, which is actually by Hypsicles, which happens to go-along with the dictum of Bucky Fuller: Why start with the flat stuff?... as a simple example, you can prove Desargues' theorem on perspective trigona. the idea is to actually see how proofs are done, so that someone at Wikipedia could *possibly* parse your typography, or English as a secondary language. (of course, I myself am quite monolingual, alas .-) thus: so, are you just a Wolframite, relying on math newbies to buy the package?... speak the Hell **! thus quoth: Don't be a square or a blockhead; see: http://bfi.org/node/574 thus: so, what numbers are known to be normal, aside from artificial (cantorian?) cconstructions?... I prefer to take the circumference as the unit, thus making the diameter to be pith; why wouldn't pith be normal, as well -- what possible *raison* could be, for pi not being normal, and having an infinite number of the God-am message in Contact?... as a student of what's-his-name, The Weaponeer(s), Sagan had *always* been a little whack-o!... oh, yes; asking for money to make the a-bomb. > Yes, IF pi is normal, it contains all sequences of digits. But that's not known. And, besides any normal number (many of which ARE known) does that- and almost all numbers ARE normal! thus: what an exquisite invocation of apsension, taking it apart along one axis -- dood! >(http://hexdome.com/bridges/graphics/bridge_6.jpg) only comes through thus: he's not god ... just the God-am MEssiah.... with faster-than-light *drive*, he'll be popping out of the continuum and into your face in less than no time !moooorV -- alas, life is just one, long QI test. >it wasn't until then, early 1990s, when I had unlimited access to >SLAC and Stanford's voluminous library systems -- and the vast >literature of a global community of institutional, alternative and >dissident physics -- that the ubiquity of subluminal and superluminal >lightwave and plasma velocities (and hence starship speeds) became >overwhelming and trivial. it was then that I showed the obvious, that >'Black Holes' do not exist, that rather the supermassive superstellar >phenomena that populate the centers of galaxies and denser >plasma-cosmos regions are TOROIDAL SUPERSTARS, with 'horizon' >(the AVERAGE velocity of light). [and, of course, it was 1994 when >the hundreds of pages of my math and illustration and scribbling led >to the discovery of the 3-frequency superluminal lissajous topology >of the proton. [a standing, self-reflected, or re-connected >electromagnetic wave. just look at the Lissajous figures on your >oscilloscope!] >though I have had many many teachers over the decades in the areas of >propulsion and velocity, I still look to the day when I have even ONE >colleague in the world who has (honestly, ethically) studied, >reflected, and written on the superluminal electrodynamic (OBVIOUSLY >dynamic, non-static) topology of the atomic nucleus. in other words, >nucleon-group topological dynamics. (which ARE indeed >electromagnetic, so-called 'weak' and 'strong' (QED and QCD) fields >being transparent pure red-herring misdirection to take the cattle, >clones, away from looking at the ACTUAL electromagnetic wave >(picoscopic and femtoscopic variable wavelength, frequency AND >velocity) helicoidal topologies of protons, neutrons and their >assemblages. --Friends don't let friends be/do Yahoo!s ... you can login to the googolplex, but you can't logout! http://www.benfranklinbooks.com/ http://members.tripod.com/~american_almanac http://www.wlym.com/pdf/iclc/howthenation.pdf http://www.rand.org/publications/randreview/issues/rr.12.00/ http://www.rwgrayprojects.com/synergetics/plates/figs/plate02.html ==== Subject: Re: Hamiltonian Cycle on Cubic Graph > exiting on cubic graph or not in the worst case? > It's NP complete. > I haven't found a reference, but this fact is described as well-known > in this paper: > M. R. Garey, D. S. Johnson, and R. E. Tarjan. The planar hamiltonian > (They reduce the planar hamiltonian problem to the cubic hamiltonian > problem.) I found an abstract for the Garey - Johnson - Tarjan paper mentioned above. The authors show that the Hamiltonian Cycle problem remains NP-complete, even if you only consider graphs which are planar, cubic, triangle-free, and 3-connected. --- Christopher Heckman ==== Subject: ? Fourier expansion and KL expansion Hi: For a given function, Fourier expansion is derived to minimize the 2-norm of truncation error and the expansion is based on a set of mutully orthogonal functions. This seems to be the same as for Karhunen-Loeve transformation. Are these two transformation identical or what are the differences? by Cheng Cosine Feb/20/2k6 NC ==== Subject: Simple disproof of standard Galois Theory interpretation With a modification of equations originally given by a Rick Decker, head of the computer science department at Hamilton College, I will directly show a contradiction with the standard interpretations of Galois Theory in terms of factor distribution by directly forcing a key variable to be coprime to 3, when it is a root of a monic polynomial with integer coefficients that is irreducible over Q with a last coefficient that has 3 as a factor. In the ring of algebraic integers, modifying the equations given by a Rick Decker of Hamilton College, I have f Q(x) = f((x^2 + fx)(5^2) + (-1 + fx)(5) + f) = f(25 x^2 + 30fx + (f - 5)) and f Q(x) = (5a_1(x) + f)(5a_2(x) + f) where the a's are defined by a^2 - (fx - 1)a + f(x^2 + fx) = 0 and the direct solution is to pick f=-3 and solve for the a's using the quadratic formula, which gives a = ((-3x - 1) +/- sqrt((-3x - 1)^2 + 12(x^2 - 3x)))/2 and focusing inside the square root I have (3x + 1)^2 + 12(x^2 - 3x) = 9x^2 + 6x + 1 + 12x^2 - 36x which is (3x + 1)^2 + 12(x^2 - 3x) = 21x^2 - 30x + 1 and completing the square gives (3x + 1)^2 + 12(x^2 - 3x) = 21x^2 - 30x + 75/7 - 75/7+ 1 so (3x + 1)^2 + 12(x^2 - 3x) = (3(7x - 5)^2 - 68)/7 and now let g_1 g_2 = 17, and g_1 + g_2 = sqrt(3) (7x - 5) then x = ((g_1 + g_2)/sqrt(3) + 5)/7 and substituting back now I have 2a = (-3(((g_1 + g_2)/sqrt(3) + 5)/7) - 1) +/- (g_1 - g_2)/sqrt(7) multiplying both sides by 7, I have 14a = (-sqrt(3)(g_1 + g_2) + 15 - 7) +/- sqrt(7) (g_1 - g_2) The question is divisibilty of the a's by 3, and designating k_1 as the factor shared with sqrt(3) 3 of one of the a's and k_2 as the other, with k representing the unresolved case I have: - 7 +/- sqrt(7) (g_1 - g_2) = 0 mod k and I can divide off sqrt(7) since 7 is coprime to 3, to get -sqrt(7) +/- (g_1 - g_2) = 0 mod k where all of the terms with sqrt(3) as a factor, and 14a on the left side go to 0. But g_1 = - g_2 + sqrt(3)(7x - 5), so I can make that substitution and get -sqrt(7) +/- (-2g_2 + sqrt(3)(7x - 5)) = 0 mod k and again zeroing out terms multiplied by sqrt(3), I have -sqrt(7) +/- -2g_2 = 0 mod k, which is equivalent to sqrt(7) +/- 2g_2 = 0 mod k clearing out the negative signs. Since g_1 g_2 = 17 and g_1 + g_2 = sqrt(3) (7x - 5) substituting out for g_1, I have g_2^2 - sqrt(3) (7x - 5) g_2 + 17 = 0 so I can solve for g_2 and have g_2 = (sqrt(3)(7x - 5) +/- sqrt(3(7x-5)^2 - 4(17)))/2 and making that substitution, and again zeroing out terms that have sqrt(3) as a factor, except those in the radical, I have sqrt(7) +/- sqrt(3(7x-5)^2 - 68) = 0 mod k so sqrt(7) +/- sqrt(3(49x^2 -5 (7) x + 25) - 68) = 0 mod k which is sqrt(7) +/- sqrt(7) sqrt(21x^2 -15x + 1) = 0 mod k and now I can just let 21x^2 - 15x + 1 = 4 so 21x^2 - 15x - 3 = 0 so 7x^2 - 5x - 1 = 0 so that I have sqrt(7)(1 +/- 2) = 0 mod k which is 1 +/- 2 = 0 mod k and the key result 1 + 2 = 0 mod k_1 and 1 -2 = 0 mod k_2 which forces one of g_1 or g_2 coprime to 3. Now then from before I have 7x^2 - 5x - 1 = 0 so I have that x = (5 +/- sqrt(53))/14 and I can now substitute back into the equation defining the a's which again is a^2 - (fx - 1)a + f(x^2 + fx) = 0 with f=-3, and x as found, and solve for the a's to find. 9604 a^4 + 39788 a^3 + 56644 a^2 + 6559a - 18513 = 0 and using a = c/14, I can get the monic polynomial c^4 + 58c^3 + 1156c^2 + 1874c - 74052 = 0 and its irreducibility over Q completes the direct proof that Galois Theory as taught is false, hoping I got the algebra right! Note the fundamental limitation that the a's cannot be coprime to 3 in the ring of algebraic integers prevents them from being algebraic integers, but in this case, they are roots of a non-monic polynomial with a leading coefficient coprime to 3, so I can still get to a monic polynomial with integer coefficients, where it has now been proven that one of its roots is coprime to 3, but it can be shown that NONE of its roots can be coprime to 3 in the ring of algebraic integers! That completes the proof of the fundamental problem I have been talking about for years, with this being the second proof of that problem, which circumvents specious objections that rely on calls to mystery functions and doubts about the applicability of the distributive property. It is incumbent upon mathematicians to tell the truth as rapidly as possible in this situation, as each day trusting students are learning flawed information, and each day with the wrong mathematical ideas, is a day lost to humanity. In the quest for truth, necessarily, we need correct mathematical ideas! EACH DAY lost is a day that cannot be gained back, and is a day we take from ourselves and future generations. James Harris ==== Subject: Re: Simple disproof of standard Galois Theory interpretation > With a modification of equations originally given by a Rick Decker, > head of the computer science department at Hamilton College, I will > directly show a contradiction with the standard interpretations of > Galois Theory in terms of factor distribution by directly forcing a key > variable to be coprime to 3, when it is a root of a monic polynomial > with integer coefficients that is irreducible over Q with a last > coefficient that has 3 as a factor. > In the ring of algebraic integers, modifying the equations given by a > Rick Decker of Hamilton College, I have > f Q(x) = f((x^2 + fx)(5^2) + (-1 + fx)(5) + f) > = f(25 x^2 + 30fx + (f - 5)) > and > f Q(x) = (5a_1(x) + f)(5a_2(x) + f) > where the a's are defined by > a^2 - (fx - 1)a + f(x^2 + fx) = 0 > and the direct solution is to pick f=-3 and solve for the a's using the > quadratic formula, which gives > a = ((-3x - 1) +/- sqrt((-3x - 1)^2 + 12(x^2 - 3x)))/2 > and focusing inside the square root I have > (3x + 1)^2 + 12(x^2 - 3x) = 9x^2 + 6x + 1 + 12x^2 - 36x > which is > (3x + 1)^2 + 12(x^2 - 3x) = 21x^2 - 30x + 1 > and completing the square gives > (3x + 1)^2 + 12(x^2 - 3x) = 21x^2 - 30x + 75/7 - 75/7+ 1 > so > (3x + 1)^2 + 12(x^2 - 3x) = (3(7x - 5)^2 - 68)/7 > and now let g_1 g_2 = 17, and > g_1 + g_2 = sqrt(3) (7x - 5) > then > x = ((g_1 + g_2)/sqrt(3) + 5)/7 > and substituting back now I have > 2a = (-3(((g_1 + g_2)/sqrt(3) + 5)/7) - 1) +/- (g_1 - g_2)/sqrt(7) > multiplying both sides by 7, I have > 14a = (-sqrt(3)(g_1 + g_2) + 15 - 7) +/- sqrt(7) (g_1 - g_2) > The question is divisibilty of the a's by 3, and designating k_1 as the > factor shared with sqrt(3) 3 of one of the a's and k_2 as the other, > with k representing the unresolved case I have: > - 7 +/- sqrt(7) (g_1 - g_2) = 0 mod k > and I can divide off sqrt(7) since 7 is coprime to 3, to get > -sqrt(7) +/- (g_1 - g_2) = 0 mod k > where all of the terms with sqrt(3) as a factor, and 14a on the left > side go to 0. > But g_1 = - g_2 + sqrt(3)(7x - 5), so I can make that substitution and > get > -sqrt(7) +/- (-2g_2 + sqrt(3)(7x - 5)) = 0 mod k > and again zeroing out terms multiplied by sqrt(3), I have > -sqrt(7) +/- -2g_2 = 0 mod k, > which is equivalent to > sqrt(7) +/- 2g_2 = 0 mod k > clearing out the negative signs. > Since > g_1 g_2 = 17 and > g_1 + g_2 = sqrt(3) (7x - 5) > substituting out for g_1, I have > g_2^2 - sqrt(3) (7x - 5) g_2 + 17 = 0 > so I can solve for g_2 and have > g_2 = (sqrt(3)(7x - 5) +/- sqrt(3(7x-5)^2 - 4(17)))/2 > and making that substitution, and again zeroing out terms that have > sqrt(3) as a factor, except those in the radical, I have > sqrt(7) +/- sqrt(3(7x-5)^2 - 68) = 0 mod k > so > sqrt(7) +/- sqrt(3(49x^2 -5 (7) x + 25) - 68) = 0 mod k This should be sqrt(7) +/- sqrt(3(49x^2 -(2) (5) (7) x + 25) - 68) = 0 mod k You forgot a factor of 2. Increment the Ooops counter. > which is > sqrt(7) +/- sqrt(7) sqrt(21x^2 -15x + 1) = 0 mod k > and now I can just let > 21x^2 - 15x + 1 = 4 > so > 21x^2 - 15x - 3 = 0 > so > 7x^2 - 5x - 1 = 0 > so that I have > sqrt(7)(1 +/- 2) = 0 mod k which is 1 +/- 2 = 0 mod k > and the key result > 1 + 2 = 0 mod k_1 and > 1 -2 = 0 mod k_2 > which forces one of g_1 or g_2 coprime to 3. > Now then from before I have > 7x^2 - 5x - 1 = 0 This should be 7x^2 - 10x - 1 = 0 > so I have that > x = (5 +/- sqrt(53))/14 No, x = (10 +/- sqrt(128))/14 Redo the algebra and check reducibity of the quartic. -William Hughes ==== Subject: Re: Simple disproof of standard Galois Theory interpretation In sci.math, jstevh@msn.com on 20 Feb 2006 17:57:43 -0800 > With a modification of equations originally given by a Rick Decker, > head of the computer science department at Hamilton College, I will > directly show a contradiction with the standard interpretations of > Galois Theory in terms of factor distribution by directly forcing a key > variable to be coprime to 3, when it is a root of a monic polynomial > with integer coefficients that is irreducible over Q with a last > coefficient that has 3 as a factor. > In the ring of algebraic integers, modifying the equations given by a > Rick Decker of Hamilton College, I have > f Q(x) = f((x^2 + fx)(5^2) + (-1 + fx)(5) + f) > = f(25 x^2 + 30fx + (f - 5)) I'm assuming f*Q(x) = f*((x^2+f*x)*(5^2) + (fx - 1)*(5) + f) = f*(25*x^2 + 30*f*x + (f-5)) > and > f Q(x) = (5a_1(x) + f)(5a_2(x) + f) > where the a's are defined by > a^2 - (fx - 1)a + f(x^2 + fx) = 0 and can be explicitly solved for, yielding a_1 = (1/2)*( (f*x - 1) + sqrt( (f*x - 1)^2 - 4*f*(x^2+f*x)) = (1/2)*( (f*x - 1) + sqrt( f^2*x^2 - 2*f*x + 1 - 4*f*x^2 - 4*f^2*x)) a_2 = (1/2)*( (f*x - 1) - sqrt( f^2*x^2 - 2*f*x + 1 - 4*f*x^2 - 4*f^2*x)) > and the direct solution is to pick f=-3 and solve for the a's using the > quadratic formula, which gives a_1 = (1/2) * (3*x - 1) + sqrt( 9*x^2 - 6*x + 1 - 12*x^2 - 36*x) = (1/2) * (3*x - 1) + sqrt( -3*x^2 - 42*x + 1 ) a_2 = (1/2) * (3*x - 1) - sqrt( -3*x^2 - 42*x + 1 ) > a = ((-3x - 1) +/- sqrt((-3x - 1)^2 + 12(x^2 - 3x)))/2 > and focusing inside the square root I have > (3x + 1)^2 + 12(x^2 - 3x) = 9x^2 + 6x + 1 + 12x^2 - 36x > which is > (3x + 1)^2 + 12(x^2 - 3x) = 21x^2 - 30x + 1 > and completing the square gives > (3x + 1)^2 + 12(x^2 - 3x) = 21x^2 - 30x + 75/7 - 75/7+ 1 > so > (3x + 1)^2 + 12(x^2 - 3x) = (3(7x - 5)^2 - 68)/7 > and now let g_1 g_2 = 17, and > g_1 + g_2 = sqrt(3) (7x - 5) which means g are the roots of g^2 - sqrt(3) * (7*x - 5) * g + 17 = 0 and explicitably solvable again: g_1 = (1/2) (sqrt(3) * (7*x - 5) + sqrt(3*(7*x-5)^2 - 68)) = (1/2) (sqrt(3) * (7*x - 5) + sqrt(147*x^2 - 210*x + 7) ) g_2 = (1/2) (sqrt(3) * (7*x - 5) - sqrt(147*x^2 - 210*x + 7) ) One can manipulate this equation thusly: g^2 + 17 = - sqrt(3) * (7*x - 5) * g g^4 + 34*g^2 + 17 = 3 * (7*x - 5)^2 * g^2 which proves that the g's are algebraic integers if 7*x is an integer. > then > x = ((g_1 + g_2)/sqrt(3) + 5)/7 > and substituting back now I have > 2a = (-3(((g_1 + g_2)/sqrt(3) + 5)/7) - 1) +/- (g_1 - g_2)/sqrt(7) > multiplying both sides by 7, I have > 14a = (-sqrt(3)(g_1 + g_2) + 15 - 7) +/- sqrt(7) (g_1 - g_2) > The question is divisibilty of the a's by 3, x = 0: a_1 = (1/2) * (-1 + 1) = 0; divisible by 3 a_2 = (1/2) * (-1 - 1) = 1; not divisible by 3 g_1 = (1/2) * (-5 * sqrt(3) + sqrt(-68)) = -5/2*sqrt(3) + sqrt(-17) g_2 = -5/2*sqrt(3) - sqrt(-17) x = +1: a_1 = (1/2) * (2 + sqrt(-44)) = 1 + sqrt(-11); not divisible by 3 a_2 = (1/2) * (2 - sqrt(-44)) = 1 - sqrt(-11); not divisible by 3 g_1 = (1/2) * (2 * sqrt(3) + sqrt(-56) = sqrt(3) + sqrt(-14) g_2 = sqrt(3) - sqrt(-14) > and designating k_1 as the > factor shared with sqrt(3) 3 of one of the a's and k_2 as the other, > with k representing the unresolved case I have: > - 7 +/- sqrt(7) (g_1 - g_2) = 0 mod k > and I can divide off sqrt(7) since 7 is coprime to 3, to get One must also note that sqrt(7) is not a unit. > -sqrt(7) +/- (g_1 - g_2) = 0 mod k Unless k is a nonnegative integer 'mod k' is a formalism that frankly I have no idea how to interpret. Its definition with k a positive integer is of course m = n (mod k) <=> (n - m) is divisible by k and this might be extensible for any nonzero algebraic integer k, but it's not ground I'm all that comfortable with, personally. In any event, I already have a counterexample so the rest of all this muck is more or less moot. > where all of the terms with sqrt(3) as a factor, and 14a on the left > side go to 0. > But g_1 = - g_2 + sqrt(3)(7x - 5), so I can make that substitution and > get > -sqrt(7) +/- (-2g_2 + sqrt(3)(7x - 5)) = 0 mod k > and again zeroing out terms multiplied by sqrt(3), I have > -sqrt(7) +/- -2g_2 = 0 mod k, > which is equivalent to > sqrt(7) +/- 2g_2 = 0 mod k > clearing out the negative signs. > Since > g_1 g_2 = 17 and > g_1 + g_2 = sqrt(3) (7x - 5) > substituting out for g_1, I have > g_2^2 - sqrt(3) (7x - 5) g_2 + 17 = 0 > so I can solve for g_2 and have > g_2 = (sqrt(3)(7x - 5) +/- sqrt(3(7x-5)^2 - 4(17)))/2 > and making that substitution, and again zeroing out terms that have > sqrt(3) as a factor, except those in the radical, I have > sqrt(7) +/- sqrt(3(7x-5)^2 - 68) = 0 mod k > so > sqrt(7) +/- sqrt(3(49x^2 -5 (7) x + 25) - 68) = 0 mod k > which is > sqrt(7) +/- sqrt(7) sqrt(21x^2 -15x + 1) = 0 mod k > and now I can just let > 21x^2 - 15x + 1 = 4 > so > 21x^2 - 15x - 3 = 0 > so > 7x^2 - 5x - 1 = 0 > so that I have > sqrt(7)(1 +/- 2) = 0 mod k which is 1 +/- 2 = 0 mod k > and the key result > 1 + 2 = 0 mod k_1 and > 1 -2 = 0 mod k_2 > which forces one of g_1 or g_2 coprime to 3. I'm not at all sure how to determine whether an algebraic integer is coprime to 3. In the examples I have above *both* g1 and g2 are not divisible by 3 (in the sense that g1/3 and g2/3 are not algebraic integers). > Now then from before I have > 7x^2 - 5x - 1 = 0 > so I have that > x = (5 +/- sqrt(53))/14 > and I can now substitute back into the equation defining the a's which > again is > a^2 - (fx - 1)a + f(x^2 + fx) = 0 > with f=-3, and x as found, and solve for the a's to find. > 9604 a^4 + 39788 a^3 + 56644 a^2 + 6559a - 18513 = 0 > and using a = c/14, I can get the monic polynomial > c^4 + 58c^3 + 1156c^2 + 1874c - 74052 = 0 > and its irreducibility over Q completes the direct proof that Galois > Theory as taught is false, hoping I got the algebra right! > Note the fundamental limitation that the a's cannot be coprime to 3 in > the ring of algebraic integers prevents them from being algebraic > integers, but in this case, they are roots of a non-monic polynomial > with a leading coefficient coprime to 3, so I can still get to a monic > polynomial with integer coefficients, where it has now been proven that > one of its roots is coprime to 3, but it can be shown that NONE of its > roots can be coprime to 3 in the ring of algebraic integers! > That completes the proof of the fundamental problem I have been talking > about for years, with this being the second proof of that problem, > which circumvents specious objections that rely on calls to mystery > functions and doubts about the applicability of the distributive > property. > It is incumbent upon mathematicians to tell the truth as rapidly as > possible in this situation, as each day trusting students are learning > flawed information, and each day with the wrong mathematical ideas, is > a day lost to humanity. > In the quest for truth, necessarily, we need correct mathematical > ideas! > EACH DAY lost is a day that cannot be gained back, and is a day we take > from ourselves and future generations. > James Harris -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== Subject: Re: Simple disproof of standard Galois Theory interpretation > With a modification of equations originally given by a Rick Decker, > head of the computer science department at Hamilton College, I will > directly show a contradiction with the standard interpretations of > Galois Theory in terms of factor distribution by directly forcing a key > variable to be coprime to 3, when it is a root of a monic polynomial > with integer coefficients that is irreducible over Q with a last > coefficient that has 3 as a factor. Now to understand what you're saying, I'm trying to visualize an example. So, what's a monic poly, irreducible over Q, with its last coef having 3 as a factor? Well, the poly x + 3 comes to mind. Just for the purpose of helping me to understand your terminology. Now, to understand what you're saying, I have to ask myself, what do you mean by key variable? I do not know. The only variable is x, so maybe that's what you mean by key variable. Now, how can x -- or any variable, for that matter -- be coprime to anything? For some values of x, it's coprime to 47; and for other values, it's not. That's the essence of being a variable ... it takes on many values. All I'm trying to do here is understand your lead paragraph. ==== Subject: Re: Simple disproof of standard Galois Theory interpretation > With a modification of equations originally given by a Rick Decker, > head of the computer science department at Hamilton College, I will > directly show a contradiction with the standard interpretations of > Galois Theory in terms of factor distribution by directly forcing a key > variable to be coprime to 3, when it is a root of a monic polynomial > with integer coefficients that is irreducible over Q with a last > coefficient that has 3 as a factor. ... leading to the alleged killer polynomial: > c^4 + 58c^3 + 1156c^2 + 1874c - 74052 = 0 > and its irreducibility over Q completes the direct proof that Galois > Theory as taught is false, hoping I got the algebra right! > Note the fundamental limitation that the a's cannot be coprime to 3 in > the ring of algebraic integers prevents them from being algebraic > integers, but in this case, they are roots of a non-monic polynomial > with a leading coefficient coprime to 3, so I can still get to a monic > polynomial with integer coefficients, where it has now been proven that > one of its roots is coprime to 3, but it can be shown that NONE of its > roots can be coprime to 3 in the ring of algebraic integers! You have shown nothing of the sort. The polynomial above (irrespective if whether it's the one you think it is, in terms of your argument: you may wish to check what Rick Decker has done) has roots that share non- unit factors with 3, in the ring of algebraic integers. Your argument notwithstanding, this fact can be verified using ordinary arithmetic. I have found one such common factor, and in the matter of a few days will have found its expression as a cubic polynomial in the root with integer coefficients. What that means, what it has always meant, is that your arguments violate ordinary arithmetic. They produce results that are at variance with ordinary arithmetic, i.e., that produce *different answers* than one gets by doing arithmetic, and so are unsound. > That completes the proof of the fundamental problem I have been talking > about for years, with this being the second proof of that problem, > which circumvents specious objections that rely on calls to mystery > functions and doubts about the applicability of the distributive > property. No, read carefully: it can be shown *with ordinary arithmetic*, that the roots of the quartic you've presented all share non-unit factors with 3. Your argument, if it shows otherwise, simply violates the arithmetic that we all know. I'll note that it is actually not very difficult to violate ordinary arithmetic: many poorly-prepared students in elementary school do it every day. > It is incumbent upon mathematicians to tell the truth as rapidly as > possible in this situation, as each day trusting students are learning > flawed information, and each day with the wrong mathematical ideas, is > a day lost to humanity. But apparently not incumbent on you to tell the truth. Instead, you continue to claim that your flawed arguments actually prove something. This is so, despite repeated demonstration of your errors. I've said so before, and I'll likely say it again: if only you would actually read and understand some algebra, you could save yourself a whole lot of embarassment in the future, and while it would make sci.math a less interesting place, we could all use something more valuable to do than respond to your bigotry and libelous remarks regarding mathematicians. The Jeremiads can continue; they are like a large 'Kick Me' sign on someone's bottomnal region. Everyone needs someone to make the world seem just a little brighter. > In the quest for truth, necessarily, we need correct mathematical > ideas! Which, precisely, is why you must be opposed at every turn. Every idea you spawn is yet another floater that needs to be flushed away. > EACH DAY lost is a day that cannot be gained back, and is a day we take > from ourselves and future generations. Just keep saying that to yourself. The sense of self-importance is surely more comforting than the notion that you've spent a decade farting into a paper bag only to have it pop, and are bound to spend yet another decade frantically looking for the tape to mend it. > James Harris Dale. ==== Subject: Re: Simple disproof of standard Galois Theory interpretation head of the computer science department at Hamilton College, I will > directly show a contradiction with the standard interpretations of > Galois Theory in terms of factor distribution by directly forcing a key > variable to be coprime to 3, when it is a root of a monic polynomial > with integer coefficients that is irreducible over Q with a last > coefficient that has 3 as a factor. > ... leading to the alleged killer polynomial: > c^4 + 58c^3 + 1156c^2 + 1874c - 74052 = 0 > and its irreducibility over Q completes the direct proof that Galois > Theory as taught is false, hoping I got the algebra right! > Note the fundamental limitation that the a's cannot be coprime to 3 in > the ring of algebraic integers prevents them from being algebraic > integers, but in this case, they are roots of a non-monic polynomial > with a leading coefficient coprime to 3, so I can still get to a monic > polynomial with integer coefficients, where it has now been proven that > one of its roots is coprime to 3, but it can be shown that NONE of its > roots can be coprime to 3 in the ring of algebraic integers! > You have shown nothing of the sort. The polynomial above (irrespective > if whether it's the one you think it is, in terms of your argument: you > may wish to check what Rick Decker has done) has roots that share non- > unit factors with 3, in the ring of algebraic integers. Your argument > notwithstanding, this fact can be verified using ordinary arithmetic. > I have found one such common factor, and in the matter of a few days > will have found its expression as a cubic polynomial in the root with > integer coefficients. You are showing delusion. I PROVE by directly forcing coprimeness with 3 in a very direct way that the numerator of one of the a's, with the a's written as a ratio of algebraic integers, must be coprime to 3. That means that MY EXPLANATION must be the correct one and proving that IN THE RING OF ALGEBRAIC INTEGERS none of the numerators of the a's can be coprime to 3, proves that the RING IS INCOMPLETE, as I've pointed out for years now. What this method of proof does is remove the mystery function objection as my previous proof relying on the distributive property was in fact correct, and formally peer reviewed I might add. In denying mathematical truths posters like yourself relied on people's confusions about functions, and the belief that mystery functions could do the impossible. But I did what geniuses do, I found another way. If you wish to show more complete contempt for mathematics DENY THE PROOF yet again. Or, show some ing respect for mathematics and the human race, and start telling the goddamn truth!!! James Harris ==== Subject: Re: Simple disproof of standard Galois Theory interpretation Discussion, linux) > That means that MY EXPLANATION must be the correct one and proving that > IN THE RING OF ALGEBRAIC INTEGERS none of the numerators of the a's can > be coprime to 3, proves that the RING IS INCOMPLETE, as I've pointed > out for years now. Sorry, what does that mean again? -- Jesse F. Hughes Had you told it like it was, it wouldn't be like it is. -- Albert King ==== Subject: Re: Simple disproof of standard Galois Theory interpretation > Or, show some ing respect for mathematics and the human race, and > start telling the goddamn truth!!! You tell'm James! Respect! ==== Subject: Re: Simple disproof of standard Galois Theory interpretation >With a modification of equations originally given by a Rick Decker, >head of the computer science department at Hamilton College, I will >directly show a contradiction with the standard interpretations of >Galois Theory in terms of factor distribution by directly forcing a key >variable to be coprime to 3, when it is a root of a monic polynomial >with integer coefficients that is irreducible over Q with a last >coefficient that has 3 as a factor. >> ... leading to the alleged killer polynomial: >c^4 + 58c^3 + 1156c^2 + 1874c - 74052 = 0 >and its irreducibility over Q completes the direct proof that Galois >Theory as taught is false, hoping I got the algebra right! >Note the fundamental limitation that the a's cannot be coprime to 3 in >the ring of algebraic integers prevents them from being algebraic >integers, but in this case, they are roots of a non-monic polynomial >with a leading coefficient coprime to 3, so I can still get to a monic >polynomial with integer coefficients, where it has now been proven that >one of its roots is coprime to 3, but it can be shown that NONE of its >roots can be coprime to 3 in the ring of algebraic integers! >>You have shown nothing of the sort. The polynomial above (irrespective >>if whether it's the one you think it is, in terms of your argument: you >>may wish to check what Rick Decker has done) has roots that share non- >>unit factors with 3, in the ring of algebraic integers. Your argument >>notwithstanding, this fact can be verified using ordinary arithmetic. >>I have found one such common factor, and in the matter of a few days >>will have found its expression as a cubic polynomial in the root with >>integer coefficients. > You are showing delusion. Nonsense. In these several years, it has only been you who has been deluded, and the pattern continues. Wake up. > I PROVE by directly forcing coprimeness with 3 in a very direct way > that the numerator of one of the a's, with the a's written as a ratio > of algebraic integers, must be coprime to 3. You prove nothing of the sort. If you insist you have a proof, than do what I'll do: produce numbers u,v in whatever the appropriate ring is, for which uA + v3 = 1, where A is the root in question. I said it'll take a few days to do this, but I can do that much. You do the same, and we'll have something to discuss. > That means that MY EXPLANATION must be the correct one and proving that > IN THE RING OF ALGEBRAIC INTEGERS none of the numerators of the a's can > be coprime to 3, proves that the RING IS INCOMPLETE, as I've pointed > out for years now. There is no mathematical definition of incomplete ring that I am familiar with. Give a mathematical definition of complete in the context you're using. Until then, you are just babbling. > What this method of proof does is remove the mystery function objection > as my previous proof relying on the distributive property was in fact > correct, and formally peer reviewed I might add. If it was formally peer reviewed, then the peers who reviewed it were your peers in the sense of being rank incompetents. Either that, or they didn't follow the argument and just assumed someone else would catch you should know as much. By the way, how did the formal peer review of the version you sent to the Annals of Mathematics work out? Any comments? Anything at all? our publication at this time. Please accept our best wishes for a successful and fulfilling life. > In denying mathematical truths posters like yourself relied on people's > confusions about functions, and the belief that mystery functions could > do the impossible. I have not relied on anyone's confusions. I will produce arithmetic that anyone with the facility for working with large (>500 digit) integers will be able to verify. No confusion. No indirect because [mumble] is a constant mumbo-jumbo. Arithmetic. Your argument violates arithmetic. It's that simple. I can show the common factor, as a root of a quadric polynomial having integer coefficients, and a leading coefficient of 1. I will shortly be able to use ordinary arithmetic to verify the fact that this factor is indeed a common factor of the root A and 3. > But I did what geniuses do, I found another way. Oh, geniuses. People who don't have to know what they're doing, since every word they write is the very mark of genius. Got it. > If you wish to show more complete contempt for mathematics DENY THE > PROOF yet again. You have no proof. This constitutes respect for mathematics. You have no understanding of why your result is bogus, and until you learn the basics of the field, you will never have such an understanding. > Or, show some ing respect for mathematics and the human race, and > start telling the goddamn truth!!! I see. It's ing respect I need. Goddamn truth that needs to be told. Just because you can't get me to agree to your bogus argument as valid, you need to resort to the sort of language you'd use in talking to your parents at their upcoming wedding. You are of course welcome to your own opinion. As far as no one accepts your argument, you will continue to be stymied. If you actually cared to get past this particular hurdle towards acceptance, you would do what many have suggested: learn something that you don't know yet. > James Harris Dale. ==== Subject: Re: Simple disproof of standard Galois Theory interpretation *********CORRECTION*********** ... stuff deleted ... > If you insist you have a proof, than do what I'll do: produce numbers > u,v in whatever the appropriate ring is, for which > uA + v3 = 1, > where A is the root in question. Oh, boy. I didn't even remember what it was I had done. What I *didn't* do was to show that any root A was coprime to 3. Duh, indeed. What I should have said was that I'd post an argument displaying a common factor between A and 3. > I said it'll take a few days to do this, but I can do that much. That's it. Dale ==== Subject: Re: Simple disproof of standard Galois Theory interpretation >> What this method of proof does is remove the mystery function objection >> as my previous proof relying on the distributive property was in fact >> correct, and formally peer reviewed I might add. > If it was formally peer reviewed, then the peers who reviewed it were > your peers in the sense of being rank incompetents. Either that, or they > didn't follow the argument and just assumed someone else would catch > you should know as much. His paper *was* peer-reviewed... and got rejected by the reviewers. > By the way, how did the formal peer review of the version you sent to > the Annals of Mathematics work out? Any comments? Anything at all? Don't hold your breadth while waiting for an answer. Jose Carlos Santos ==== Subject: Re: Simple disproof of standard Galois Theory interpretation days. My association with the Department is that of an alumnus. > What this method of proof does is remove the mystery function objection > as my previous proof relying on the distributive property was in fact > correct, and formally peer reviewed I might add. >> If it was formally peer reviewed, then the peers who reviewed it were >> your peers in the sense of being rank incompetents. Either that, or they >> didn't follow the argument and just assumed someone else would catch >> you should know as much. >His paper *was* peer-reviewed... and got rejected by the reviewers. I disagree. The editor quite clearly screwed up by the numbers and then attempted to cover himself. He very much did James wrong. The journal claimed that every paper published was reviewed by at least two referees prior to acceptance and publication. It is clear that this was not the case with James's paper, whether out of a mishap, as the editor claimed (that the paper had been incorrectly placed in the accepted pile rather than the to be reviewed pile), or because the journal did not meet its stated policy on a regular basis. The plethora of simple grammatical and typographical errors on that paper is proof positive that it did not undergo any review whatsoever. I posted a list of purely grammatical and typographical followed by a list of mathematical comments. Quite simply, this paper was not even cursorily read before it was placed in that issue. That said, if as James contends the editor replied to him by forwarding e-mails from sci.math readers and labeling them referee Corrigendum which I submitted to them, and which was summarily rejected on the grounds that they were aware of the problems and have dealt with them. E-mails indicating errors in the paper are simply ->not<- referee reports and should not be taken or presented as such, even if they may be grounds for rejecting a paper. And the editor yanked the paper without informing or explaining to the author either in advance or contemporaneously with the action, after having informed him of acceptance and publication. That is quite simply unacceptable. While there was some debate in sci.math on whether the editor acted properly or improperly in yanking the paper, I believe there was a consensus that the editor was being disingenious and misleading (if not actively dishonest) in labeling the paper as withdrawn. Withdrawn suggests that the author requested that the paper be taken out; it gives no indication that the editors removed it without consulting or informing the author in advance. It should have been marked, at best, Removed, not withdrawn. There is absolutely no doubt in my mind that James was treated unfairly and unprofessionally by the journal. Just not in the way and not for the reasons that James seems to think. That said, I do have to wonder: Dedekind's work which he is now tearing down was peer-reviewed and published in no less than FOUR forms, and in journals and editorial houses of extremely high repute. It appeared in two different forms as supplements to Dirichlet's book, Vorlesungen uber Zahlentheorie, published by Vieweg; it appeared yet again in the Bulletin des sciences mathematiques et astronomiques, under the aegis of Rudolf Lipschitz who requested the material; it then appeared a fourth time in book form as Sur la Theorie des Nombres Entiers Algebriques, which is the one translated by Stillwell and published by Cambridge Mathematical Library; and it finally appeared in more generality, applied to function fields, in his memoir with Weber in the extremely prestigious J. reine und angew. Math. (Journal for pure and applied Mathematics), that did not even remark on the fact that the author used different fonts for the same variable is supposed to count more than the one undergone by Dedekind. Not to mention that while I see a lot of claims that the only way in which it is valid to claim an argument is wrong is to go through it and point out the mistake, I see absolutely no attempt to do this with any of the versions of Dedekind's work, or for that matter the very standard baby-application of Galois theory that has been invoked. I know Galois is not being attacked, but I can only imagine that this is because James has always had a bit of a thing for Galois and views himself as the tragic heir of Galois: the young brilliant unappreciated genius. He cannot bring himself to speak ill of Galois, even though all the overinterpretations that he claims come from Galois himself and nothing beyond. -- ' ==== Subject: Re: Simple disproof of standard Galois Theory interpretation <2DyKf.10904$rL5.9921@newssvr27.news.prodigy.net> <460qnlF8ic5qU1@individual.net> as my previous proof relying on the distributive property was in fact > correct, and formally peer reviewed I might add. > If it was formally peer reviewed, then the peers who reviewed it were >> your peers in the sense of being rank incompetents. Either that, or they >> didn't follow the argument and just assumed someone else would catch >> you should know as much. >His paper *was* peer-reviewed... and got rejected by the reviewers. > I disagree. The editor quite clearly screwed up by the numbers and > then attempted to cover himself. He very much did James wrong. > The journal claimed that every paper published was reviewed by at > least two referees prior to acceptance and publication. It is clear > that this was not the case with James's paper, whether out of a > mishap, as the editor claimed (that the paper had been incorrectly > placed in the accepted pile rather than the to be reviewed pile), > or because the journal did not meet its stated policy on a regular > basis. The plethora of simple grammatical and typographical errors on > that paper is proof positive that it did not undergo any review > whatsoever. I posted a list of purely grammatical and typographical > followed by a list of mathematical comments. Quite simply, this paper > was not even cursorily read before it was placed in that issue. > That said, if as James contends the editor replied to him by > forwarding e-mails from sci.math readers and labeling them referee > Corrigendum which I submitted to them, and which was summarily > rejected on the grounds that they were aware of the problems and have > dealt with them. E-mails indicating errors in the paper are simply > ->not<- referee reports and should not be taken or presented as such, > even if they may be grounds for rejecting a paper. > And the editor yanked the paper without informing or explaining to the > author either in advance or contemporaneously with the action, after > having informed him of acceptance and publication. That is quite > simply unacceptable. But isn't a fraudulent submission also unacceptable? Why should JSH be given the same consideration a legitimate author gets? > While there was some debate in sci.math on whether the editor acted > properly or improperly in yanking the paper, I believe there was a > consensus that the editor was being disingenious and misleading (if > not actively dishonest) in labeling the paper as withdrawn. > Withdrawn suggests that the author requested that the paper be taken > out; it gives no indication that the editors removed it without > consulting or informing the author in advance. It should have been > marked, at best, Removed, not withdrawn. That I agree with, the editor should have laid out the true facts and not tried to save face by engaging in the same fraudulent activities that JSH did. > There is absolutely no doubt in my mind that James was treated > unfairly and unprofessionally by the journal. Just not in the way and > not for the reasons that James seems to think. But I think the JSH _did_ contribute to the journal's dying. Sure the official reason is they cut the funding, but who would continue to fund a journal whose editors are unfair and unprofessional? If they really thought the journal's reputation was worth saving, the editors could have been replaced. > That said, I do have to wonder: Dedekind's work which he is now > tearing down was peer-reviewed and published in no less than FOUR > forms, and in journals and editorial houses of extremely high > repute. It appeared in two different forms as supplements to > Dirichlet's book, Vorlesungen uber Zahlentheorie, published by > Vieweg; it appeared yet again in the Bulletin des sciences > mathematiques et astronomiques, under the aegis of Rudolf Lipschitz > who requested the material; it then appeared a fourth time in book > form as Sur la Theorie des Nombres Entiers Algebriques, which is the > one translated by Stillwell and published by Cambridge Mathematical > Library; and it finally appeared in more generality, applied to > function fields, in his memoir with Weber in the extremely prestigious > J. reine und angew. Math. (Journal for pure and applied Mathematics), > that did not even remark on the fact that the author used different > fonts for the same variable is supposed to count more than the one > undergone by Dedekind. Not to mention that while I see a lot of claims > that the only way in which it is valid to claim an argument is wrong > is to go through it and point out the mistake, I see absolutely no > attempt to do this with any of the versions of Dedekind's work, or for > that matter the very standard baby-application of Galois theory that > has been invoked. I know Galois is not being attacked, but I can only > imagine that this is because James has always had a bit of a thing for > Galois and views himself as the tragic heir of Galois: the young > brilliant unappreciated genius. He cannot bring himself to speak ill > of Galois, even though all the overinterpretations that he claims > come from Galois himself and nothing beyond. > -- > It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) > Arturo Magidin > magidin@math.berkeley.edu ==== Subject: Re: Simple disproof of standard Galois Theory interpretation days. My association with the Department is that of an alumnus. [.snip.] >> And the editor yanked the paper without informing or explaining to the >> author either in advance or contemporaneously with the action, after >> having informed him of acceptance and publication. That is quite >> simply unacceptable. >But isn't a fraudulent submission also unacceptable? Why should >JSH be given the same consideration a legitimate author gets? The editor should treat all authors professionally. Even in the case of fraudulent material (say, discovery of plagiarism), the editor should at least notify the author that a decision has been made the fundamentally changes the status of the paper, given that he had already notified the author of acceptance in the first placd. I am not saying that the editor should have given James the opportunity to reply or answer; the editor was within his rights to make a unilateral decision to yank the paper, whether or not I personally approve (I said at the time I did not; although I have come to recognize that those who argue otherwise do have some valid points, I am still inclined towards thinking the editor's choice of action was at the very least unfortunate). But he should not have waited until the author contacted him to inquire in order to let him know of the change of status: he should have notified the author of the change of status on his own initiative. This is what I would expect if an editorial house decides to yank a book because of allegations against the book: whether or not the allegations are true, they should tell the author Although we had said we would publish the book, we have now decided that we will not do so. I don't demand that the editor give explanations or justify himself. But since he had made a professional promise to publish, then he should notify that this professional promise would not be kept. You may be justified in spitting at someone who has defecated on you; that does not, however, make your spitting good manners. Now, was the editor required to submit James's paper to peer review prior to rejection? No; he could have, all within the bounds of professionalism and fairness, rejected to do so on any number of valid grounds even without addressing the mathematical content. The extremely poor grammar and the inconsistent use of typographical types would be sufficient, in my mind, to reject that paper for publication anywhere without any further inquire. Was the editor required to keep the paper in the journal after he had accepted it for publication? No, he could have removed it, yanked it, whatever, all within the bounds of professionalism and fairness (even without giving James an opportunity to reply to the criticisms received). Was the editor required to explain himself to James? No, he need not explain himself and yet remain both professional and fair. But he failed in his obligations to the profession as an editor (when the paper was published, at the very least), and assuming James's account of the subsequent exchange between himself and the editor is accurate, then he was certainly both unprofessional and unfair in his response to James, by misrepresenting an e-mail as a referee report, and by misrepresenting his actions as a withdrawal of the paper. Being an abusing, dishonest, lowlife does not mean that any action of others automatically becomes fair. The editor was unfair to James, no matter how much we might feel there is some cosmic justice (or some sauce for the gander at last) in James getting slapped like that. >> There is absolutely no doubt in my mind that James was treated >> unfairly and unprofessionally by the journal. Just not in the way and >> not for the reasons that James seems to think. >But I think the JSH _did_ contribute to the journal's dying. I doubt it. If you go look through issues prior to the one in question, you will see that the quality of the journal was pretty dismal; the publishing schedule was seldom kept; it is obvious they were not adhering to their stated policy of refereeing. James's paper and the events that surrounded it were a symptom of all the problems that journal had, not a contributing cause of its terminal illness (in my humble opinion, of course). -- ' ==== Subject: Re: Simple disproof of standard Galois Theory interpretation <2DyKf.10904$rL5.9921@newssvr27.news.prodigy.netWith a modification of equations originally given by a Rick Decker, >head of the computer science department at Hamilton College, I will >directly show a contradiction with the standard interpretations of >Galois Theory in terms of factor distribution by directly forcing a key >variable to be coprime to 3, when it is a root of a monic polynomial >with integer coefficients that is irreducible over Q with a last >coefficient that has 3 as a factor. > ... leading to the alleged killer polynomial: >c^4 + 58c^3 + 1156c^2 + 1874c - 74052 = 0 and its irreducibility over Q completes the direct proof that Galois >Theory as taught is false, hoping I got the algebra right! Note the fundamental limitation that the a's cannot be coprime to 3 in >the ring of algebraic integers prevents them from being algebraic >integers, but in this case, they are roots of a non-monic polynomial >with a leading coefficient coprime to 3, so I can still get to a monic >polynomial with integer coefficients, where it has now been proven that >one of its roots is coprime to 3, but it can be shown that NONE of its >roots can be coprime to 3 in the ring of algebraic integers! >>You have shown nothing of the sort. The polynomial above (irrespective >>if whether it's the one you think it is, in terms of your argument: you >>may wish to check what Rick Decker has done) has roots that share non- >>unit factors with 3, in the ring of algebraic integers. Your argument >>notwithstanding, this fact can be verified using ordinary arithmetic. >>I have found one such common factor, and in the matter of a few days >>will have found its expression as a cubic polynomial in the root with >>integer coefficients. > You are showing delusion. > Nonsense. In these several years, it has only been you who has > been deluded, and the pattern continues. Wake up. > I PROVE by directly forcing coprimeness with 3 in a very direct way > that the numerator of one of the a's, with the a's written as a ratio > of algebraic integers, must be coprime to 3. > You prove nothing of the sort. Well then step through the argument, and show an error. That's how you are supposed to criticize a mathematical argument you claim is false. The outline of the proof is easy: I directly force the numerator of one of the a's, with the a's as a ratio of algebraic integers, to be coprime to 3. The a's are then roots of a non-monic polynomial with integer coefficients BUT it's leading coefficient is coprime to 3 while it's last is not, so I can simply move to a monic polynomial with integer coefficients without any possible shift of factors in common with 3. That done the lack of rational solutions completes the proof of the flaw. Show a problem with those logical steps if you wish to maintain there is an error. James Harris ==== Subject: Re: Simple disproof of standard Galois Theory interpretation <2DyKf.10904$rL5.9921@newssvr27.news.prodigy.net >With a modification of equations originally given by a Rick Decker, >head of the computer science department at Hamilton College, I will >directly show a contradiction with the standard interpretations of >Galois Theory in terms of factor distribution by directly forcing a key >variable to be coprime to 3, when it is a root of a monic polynomial >with integer coefficients that is irreducible over Q with a last >coefficient that has 3 as a factor. > ... leading to the alleged killer polynomial: >c^4 + 58c^3 + 1156c^2 + 1874c - 74052 = 0 and its irreducibility over Q completes the direct proof that Galois >Theory as taught is false, hoping I got the algebra right! Note the fundamental limitation that the a's cannot be coprime to 3 in >the ring of algebraic integers prevents them from being algebraic >integers, but in this case, they are roots of a non-monic polynomial >with a leading coefficient coprime to 3, so I can still get to a monic >polynomial with integer coefficients, where it has now been proven that >one of its roots is coprime to 3, but it can be shown that NONE of its >roots can be coprime to 3 in the ring of algebraic integers! >>You have shown nothing of the sort. The polynomial above (irrespective >>if whether it's the one you think it is, in terms of your argument: you >>may wish to check what Rick Decker has done) has roots that share non- >>unit factors with 3, in the ring of algebraic integers. Your argument >>notwithstanding, this fact can be verified using ordinary arithmetic. >>I have found one such common factor, and in the matter of a few days >>will have found its expression as a cubic polynomial in the root with >>integer coefficients. > You are showing delusion. Nonsense. In these several years, it has only been you who has > been deluded, and the pattern continues. Wake up. > I PROVE by directly forcing coprimeness with 3 in a very direct way > that the numerator of one of the a's, with the a's written as a ratio > of algebraic integers, must be coprime to 3. You prove nothing of the sort. > Well then step through the argument, and show an error. > That's how you are supposed to criticize a mathematical argument you > claim is false. > The outline of the proof is easy: I directly force the numerator of one > of the a's, with the a's as a ratio of algebraic integers, to be > coprime to 3. > The a's are then roots of a non-monic polynomial with integer > coefficients BUT it's leading coefficient is coprime to 3 while it's > last is not, so I can simply move to a monic polynomial with integer > coefficients without any possible shift of factors in common with 3. > That done the lack of rational solutions completes the proof of the > flaw. > Show a problem with those logical steps if you wish to maintain there > is an error. You can only demonstrate the lack of rational solutions if you get your algebra correct. Increment the Ooops counter. -William Hughes ==== Subject: Quick Quiz Was the following written by (a) someone who has found an air-tight refutation of a mathematical result and presented a clear argument, or (b) someone who has seen their refutation blown into dust and is frustrated about it? > Or, show some ing respect for mathematics and the human race, and > start telling the goddamn truth!!! --- Christopher Heckman ==== Subject: Re: Simple disproof of standard Galois Theory interpretation > With a modification of equations originally given by a Rick Decker, > head of the computer science department at Hamilton College, I will > directly show a contradiction with the standard interpretations of > Galois Theory in terms of factor distribution by directly forcing a key > variable to be coprime to 3, when it is a root of a monic polynomial > with integer coefficients that is irreducible over Q with a last > coefficient that has 3 as a factor. > In the ring of algebraic integers, modifying the equations given by a > Rick Decker of Hamilton College, I have > f Q(x) = f((x^2 + fx)(5^2) + (-1 + fx)(5) + f) > = f(25 x^2 + 30fx + (f - 5)) > and > f Q(x) = (5a_1(x) + f)(5a_2(x) + f) > where the a's are defined by > a^2 - (fx - 1)a + f(x^2 + fx) = 0 > and the direct solution is to pick f=-3 and solve for the a's using the > quadratic formula, which gives > a = ((-3x - 1) +/- sqrt((-3x - 1)^2 + 12(x^2 - 3x)))/2 > and focusing inside the square root I have > (3x + 1)^2 + 12(x^2 - 3x) = 9x^2 + 6x + 1 + 12x^2 - 36x > which is > (3x + 1)^2 + 12(x^2 - 3x) = 21x^2 - 30x + 1 > and completing the square gives > (3x + 1)^2 + 12(x^2 - 3x) = 21x^2 - 30x + 75/7 - 75/7+ 1 > so > (3x + 1)^2 + 12(x^2 - 3x) = (3(7x - 5)^2 - 68)/7 > and now let g_1 g_2 = 17, and > g_1 + g_2 = sqrt(3) (7x - 5) > then > x = ((g_1 + g_2)/sqrt(3) + 5)/7 > and substituting back now I have > 2a = (-3(((g_1 + g_2)/sqrt(3) + 5)/7) - 1) +/- (g_1 - g_2)/sqrt(7) > multiplying both sides by 7, I have > 14a = (-sqrt(3)(g_1 + g_2) + 15 - 7) +/- sqrt(7) (g_1 - g_2) > The question is divisibilty of the a's by 3, and designating k_1 as the > factor shared with sqrt(3) 3 of one of the a's and k_2 as the other, > with k representing the unresolved case I have: > - 7 +/- sqrt(7) (g_1 - g_2) = 0 mod k > and I can divide off sqrt(7) since 7 is coprime to 3, to get > -sqrt(7) +/- (g_1 - g_2) = 0 mod k > where all of the terms with sqrt(3) as a factor, and 14a on the left > side go to 0. > But g_1 = - g_2 + sqrt(3)(7x - 5), so I can make that substitution and > get > -sqrt(7) +/- (-2g_2 + sqrt(3)(7x - 5)) = 0 mod k > and again zeroing out terms multiplied by sqrt(3), I have > -sqrt(7) +/- -2g_2 = 0 mod k, > which is equivalent to > sqrt(7) +/- 2g_2 = 0 mod k > clearing out the negative signs. > Since > g_1 g_2 = 17 and > g_1 + g_2 = sqrt(3) (7x - 5) > substituting out for g_1, I have > g_2^2 - sqrt(3) (7x - 5) g_2 + 17 = 0 > so I can solve for g_2 and have > g_2 = (sqrt(3)(7x - 5) +/- sqrt(3(7x-5)^2 - 4(17)))/2 > and making that substitution, and again zeroing out terms that have > sqrt(3) as a factor, except those in the radical, I have > sqrt(7) +/- sqrt(3(7x-5)^2 - 68) = 0 mod k > so > sqrt(7) +/- sqrt(3(49x^2 -5 (7) x + 25) - 68) = 0 mod k No. Check your arithmetic. None of what follows is right. and I can now substitute back into the equation defining the a's which > again is > a^2 - (fx - 1)a + f(x^2 + fx) = 0 > with f=-3, and x as found, and solve for the a's to find. > 9604 a^4 + 39788 a^3 + 56644 a^2 + 6559a - 18513 = 0 This is wrong. The correct polynomial is the product of two irreducible quadratics, neither of which has algebraic integer roots. Sorry, James, but mathematics is safe for today. Rick ==== Subject: Re: Simple disproof of standard Galois Theory interpretation head of the computer science department at Hamilton College, I will > directly show a contradiction with the standard interpretations of > Galois Theory in terms of factor distribution by directly forcing a key > variable to be coprime to 3, when it is a root of a monic polynomial > with integer coefficients that is irreducible over Q with a last > coefficient that has 3 as a factor. > In the ring of algebraic integers, modifying the equations given by a > Rick Decker of Hamilton College, I have > f Q(x) = f((x^2 + fx)(5^2) + (-1 + fx)(5) + f) > = f(25 x^2 + 30fx + (f - 5)) > and > f Q(x) = (5a_1(x) + f)(5a_2(x) + f) > where the a's are defined by > a^2 - (fx - 1)a + f(x^2 + fx) = 0 > and the direct solution is to pick f=-3 and solve for the a's using the > quadratic formula, which gives > a = ((-3x - 1) +/- sqrt((-3x - 1)^2 + 12(x^2 - 3x)))/2 > and focusing inside the square root I have > (3x + 1)^2 + 12(x^2 - 3x) = 9x^2 + 6x + 1 + 12x^2 - 36x > which is > (3x + 1)^2 + 12(x^2 - 3x) = 21x^2 - 30x + 1 > and completing the square gives > (3x + 1)^2 + 12(x^2 - 3x) = 21x^2 - 30x + 75/7 - 75/7+ 1 > so > (3x + 1)^2 + 12(x^2 - 3x) = (3(7x - 5)^2 - 68)/7 > and now let g_1 g_2 = 17, and > g_1 + g_2 = sqrt(3) (7x - 5) > then > x = ((g_1 + g_2)/sqrt(3) + 5)/7 > and substituting back now I have > 2a = (-3(((g_1 + g_2)/sqrt(3) + 5)/7) - 1) +/- (g_1 - g_2)/sqrt(7) > multiplying both sides by 7, I have > 14a = (-sqrt(3)(g_1 + g_2) + 15 - 7) +/- sqrt(7) (g_1 - g_2) > The question is divisibilty of the a's by 3, and designating k_1 as the > factor shared with sqrt(3) 3 of one of the a's and k_2 as the other, > with k representing the unresolved case I have: > - 7 +/- sqrt(7) (g_1 - g_2) = 0 mod k > and I can divide off sqrt(7) since 7 is coprime to 3, to get > -sqrt(7) +/- (g_1 - g_2) = 0 mod k > where all of the terms with sqrt(3) as a factor, and 14a on the left > side go to 0. > But g_1 = - g_2 + sqrt(3)(7x - 5), so I can make that substitution and > get > -sqrt(7) +/- (-2g_2 + sqrt(3)(7x - 5)) = 0 mod k > and again zeroing out terms multiplied by sqrt(3), I have > -sqrt(7) +/- -2g_2 = 0 mod k, > which is equivalent to > sqrt(7) +/- 2g_2 = 0 mod k > clearing out the negative signs. > Since > g_1 g_2 = 17 and > g_1 + g_2 = sqrt(3) (7x - 5) > substituting out for g_1, I have > g_2^2 - sqrt(3) (7x - 5) g_2 + 17 = 0 > so I can solve for g_2 and have > g_2 = (sqrt(3)(7x - 5) +/- sqrt(3(7x-5)^2 - 4(17)))/2 > and making that substitution, and again zeroing out terms that have > sqrt(3) as a factor, except those in the radical, I have > sqrt(7) +/- sqrt(3(7x-5)^2 - 68) = 0 mod k > so > sqrt(7) +/- sqrt(3(49x^2 -5 (7) x + 25) - 68) = 0 mod k > No. Check your arithmetic. None of what follows is right. > and I can now substitute back into the equation defining the a's which > again is > a^2 - (fx - 1)a + f(x^2 + fx) = 0 > with f=-3, and x as found, and solve for the a's to find. > 9604 a^4 + 39788 a^3 + 56644 a^2 + 6559a - 18513 = 0 > This is wrong. The correct polynomial is the product of > two irreducible quadratics, neither of which has algebraic > integer roots. > Sorry, James, but mathematics is safe for today. > Rick Hopefully you're not that dense. The only way to refute the proof is to find a rational root. Actually, you need two rational roots. Think about it. You have a non-monic polynomial that has a leading coefficient coprime to 3. But in the proof considering the a's as a ratio of algebraic integers, I have that the numerator of one of the a's MUST be coprime to 3. Get it? This time, a simple algebra mistake doesn't block the proof. You need rational solutions Decker. Surely you do not expect me to believe that simple mathematical reality escapes you! It's over. The proof is solid. Minor algebra mistakes don't sink it. It's over Decker. I was right. Follow the math, please. James Harris ==== Subject: Re: Simple disproof of standard Galois Theory interpretation head of the computer science department at Hamilton College, I will > directly show a contradiction with the standard interpretations of > Galois Theory in terms of factor distribution by directly forcing a key > variable to be coprime to 3, when it is a root of a monic polynomial > with integer coefficients that is irreducible over Q with a last > coefficient that has 3 as a factor. In the ring of algebraic integers, modifying the equations given by a > Rick Decker of Hamilton College, I have f Q(x) = f((x^2 + fx)(5^2) + (-1 + fx)(5) + f) > = f(25 x^2 + 30fx + (f - 5)) and f Q(x) = (5a_1(x) + f)(5a_2(x) + f) where the a's are defined by a^2 - (fx - 1)a + f(x^2 + fx) = 0 and the direct solution is to pick f=-3 and solve for the a's using the > quadratic formula, which gives a = ((-3x - 1) +/- sqrt((-3x - 1)^2 + 12(x^2 - 3x)))/2 and focusing inside the square root I have (3x + 1)^2 + 12(x^2 - 3x) = 9x^2 + 6x + 1 + 12x^2 - 36x which is (3x + 1)^2 + 12(x^2 - 3x) = 21x^2 - 30x + 1 and completing the square gives (3x + 1)^2 + 12(x^2 - 3x) = 21x^2 - 30x + 75/7 - 75/7+ 1 so (3x + 1)^2 + 12(x^2 - 3x) = (3(7x - 5)^2 - 68)/7 and now let g_1 g_2 = 17, and g_1 + g_2 = sqrt(3) (7x - 5) then x = ((g_1 + g_2)/sqrt(3) + 5)/7 and substituting back now I have 2a = (-3(((g_1 + g_2)/sqrt(3) + 5)/7) - 1) +/- (g_1 - g_2)/sqrt(7) multiplying both sides by 7, I have 14a = (-sqrt(3)(g_1 + g_2) + 15 - 7) +/- sqrt(7) (g_1 - g_2) The question is divisibilty of the a's by 3, and designating k_1 as the > factor shared with sqrt(3) 3 of one of the a's and k_2 as the other, > with k representing the unresolved case I have: - 7 +/- sqrt(7) (g_1 - g_2) = 0 mod k and I can divide off sqrt(7) since 7 is coprime to 3, to get -sqrt(7) +/- (g_1 - g_2) = 0 mod k where all of the terms with sqrt(3) as a factor, and 14a on the left > side go to 0. But g_1 = - g_2 + sqrt(3)(7x - 5), so I can make that substitution and > get -sqrt(7) +/- (-2g_2 + sqrt(3)(7x - 5)) = 0 mod k and again zeroing out terms multiplied by sqrt(3), I have -sqrt(7) +/- -2g_2 = 0 mod k, which is equivalent to sqrt(7) +/- 2g_2 = 0 mod k clearing out the negative signs. Since g_1 g_2 = 17 and g_1 + g_2 = sqrt(3) (7x - 5) substituting out for g_1, I have g_2^2 - sqrt(3) (7x - 5) g_2 + 17 = 0 so I can solve for g_2 and have g_2 = (sqrt(3)(7x - 5) +/- sqrt(3(7x-5)^2 - 4(17)))/2 and making that substitution, and again zeroing out terms that have > sqrt(3) as a factor, except those in the radical, I have sqrt(7) +/- sqrt(3(7x-5)^2 - 68) = 0 mod k so sqrt(7) +/- sqrt(3(49x^2 -5 (7) x + 25) - 68) = 0 mod k > No. Check your arithmetic. None of what follows is right. x = (5 +/- sqrt(53))/14 and I can now substitute back into the equation defining the a's which > again is a^2 - (fx - 1)a + f(x^2 + fx) = 0 with f=-3, and x as found, and solve for the a's to find. 9604 a^4 + 39788 a^3 + 56644 a^2 + 6559a - 18513 = 0 > This is wrong. The correct polynomial is the product of > two irreducible quadratics, neither of which has algebraic > integer roots. > Sorry, James, but mathematics is safe for today. Rick > Hopefully you're not that dense. The only way to refute the proof is > to find a rational root. > Actually, you need two rational roots. > Think about it. > You have a non-monic polynomial that has a leading coefficient coprime > to 3. > But in the proof considering the a's as a ratio of algebraic integers, > I have that the numerator of one of the a's MUST be coprime to 3. Consider (a-1/2)*(a-3/5)=0 This gives you 10 a^2 -11 a + 3 = 0 The roots are (1/2) and (3/5), both have denominators coprime to 3 and only one has a numerator coprime to 3. Both roots are solutions of a non-monic polynomial that has a leading coefficient coprime to 3. Set y = 10 a. Then y^2 -11y +30 =0 The roots are 6 and 5, only one of these is divisible by 3. Galois theory does not collapse. -William Hughes ==== Subject: Re: Well Ordering the Reals I have told you often that a well ordering of the reals is impossible to > give explicitly. Such a well ordering must exist in ZFC, but there are > models of ZF in which a well ordering does not exist. Is your > comprehension really so bad that you can be told this many times, but it > has still never entered your brain? >ZF is inconsistent. > This is still claimed by Ross, but he has never proven it to the > satisfaction of anybody but himself. > ----- Again perhaps you refer to Feferman's result, that only applies if V =/= L. Otherwise if you care for well-ordering of the reals you might consider the natural ordering and what that means in terms of what the real numbers are in their completeness. For some, division by zero is undefined. For others, it's not. About ZF and its putative consistency, set-build X: x E X, true. There is no universe in ZF, nor ZF with class etcetera for any regular (well-founded) theory, and people along the lines of say, Cantor, argue there is a universe. Do not you agree that there is a universe or domain of discourse? Besides that ZF is inconsistent, say, it's incomplete, for the basically the same reason as it has no universe in the false axiomatization of regularity. There is a universe, the universe, it's infinite, and infinite sets are equivalent. So, V = L, well-order the reals. Do you present some other reasoning why that is not so possible? In ZFC, with the well-ordering principle explicitly by choice, the reals are a set, and well-orderable, except there is no universe in ZFC either. for your reply. Ross ==== Subject: JSH Summary #6 Part 2 Further analysis of a 48,156 word segment indicates strange patterns in JSH postings. I ran the JSH Summary #6 through a Word Frequency counter and the following was generated. 1. YOU is mentioned more than I indicating pointing finger effect. 2. 1.5% of the words is I indicate self-focus effect. 3. The number 7 makes up almost 1% of the text and twice as often as the next number indicating an extreme interest and outright Bias for that number. 4. The three words GET SOME THEORY all appear with a frequency of 121, indicating a search for finding or getting some kind of theory. 5. Same thing with GALOIS HERE WRONG at 71, indicate intense interest in proving Mr. Galois wrong here. 6. The sum of the top 22 words, which have no content, is 19,895 or 41% of the total, indicate finite verbal space. Further analysis is in work to examine and reassemble the information value/content of these JSH postings. The following is the output from http://www.georgetown.edu/faculty/ballc/webtools/web_freqs.html Text name: JSH Summary #6 Word count: 48156 Unique words: 3052 Sort order: descending 4759 > 2013 THE 1105 OF 1077 A 1067 THAT 1038 AND 958 + 949 TO 831 IS 753 YOU 713 I 573 = 509 IN 442 IT 437 7 412 AS 399 X 390 WITH 375 FOR 373 HAVE 369 NOT 353 BE 338 ALGEBRAIC 323 ARE 302 BUT 294 SO 276 INTEGER 246 CAN 246 WHERE 245 IF 235 BY 234 AN 230 RING 230 WHICH 203 0 202 INTEGERS 199 FACTOR 199 ONE 197 ON 180 MY 174 JUST 152 OUT 148 THIS 147 WILL 146 AT 146 HAS 143 THEY 142 2 138 1 134 IT'S 130 NOW 130 WHAT 129 PROOF 128 A^2 128 WAS 125 THEN 123 FROM 121 GET 121 SOME 121 THEORY 112 WROTE 110 OR 106 ABOUT 104 A'S 102 NO 101 LIKE 98 >> 98 OVER 98 POLYNOMIAL 97 COPRIME 97 F 95 THERE 93 MUST 92 ME 91 DO 91 WHEN 90 PEOPLE 89 HARRIS 87 BECAUSE 87 FACTORS 86 ALL 86 DOES 85 THEM 84 YOUR 82 SEE 81 THAT'S 80 DECKER 79 EQUATIONS 78 ANY 75 WHILE 75 X^2 72 Q(X 72 WHY 71 GALOIS 71 HERE 71 WRONG 70 HOW 70 SOLUTION 70 WHO 69 DISTRIBUTIVE 68 PROPERTY 67 1)A 67 B 67 OTHER 65 7(X^2 65 GO 65 MORE 65 WORK 64 I'VE 64 THINK 64 THOSE 63 IDEAS 62 CONSTANT 62 DON'T 62 HAD 61 T 60 ONLY 59 BEEN 59 COEFFICIENTS 59 I'M 59 KNOW 58 MATHEMATICAL 58 MATHEMATICS 58 USE 57 3 57 SIMPLE 57 WOULD 56 MONIC 55 > 55 FUNCTIONS 55 UP 54 ROOT 54 THEIR 53 MAKE 52 6 52 CANNOT 51 CASE 51 CLAIM 51 POINT 50 GIVEN ==== Subject: JSH: It is finally over, definitely So after so many long years of searching and arguing, and so many doubts and failures, it is finally over and I have the proof that leaves no room at all for any kind of reasonable doubt that I am right, and there is this problem with standard usage of Galois Theory and with the theory of ideals. That information SHOULD propagate rapidly. Some of you have a duty now to tell the truth. There were minor mistakes with my algebra but they don't change the argument when corrected. The proof is solid, standard usage of Galois Theory is refuted, the theory of ideals is refuted. It's finally over. Make no mistake, history is being made here. If nothing else, celebrate that you are a part of some of the most historic findings in the history of the world. You are now a part of history. We all are. James Harris ==== Subject: Re: JSH: It is finally over, definitely >So after so many long years of searching and arguing, and so many >doubts and failures, it is finally over and I have the proof that >leaves no room at all for any kind of reasonable doubt that I am right, >and there is this problem with standard usage of Galois Theory and with >the theory of ideals. Um. When you refer to your previous failures you should make a few things clear: (i) There have been really really really a lot of previous failures - you've been spouting nonsense in various directions on and off for ten years, at least. And every single thing you've said in all that time has turned out to be wrong (or trivial). (ii) _Every_ time you announced one of your breakthroughs that turned out to be wrong you announced it with no uncertainty whatever. _Every_ time you've explained that there's no room for doubt - _every_ time you've gone on about how nobody could possibly deny the Truth anymore, anyone who claimed to be able to show you were wrong was lying to protect his job, etc. >That information SHOULD propagate rapidly. >Some of you have a duty now to tell the truth. >There were minor mistakes with my algebra but they don't change the >argument when corrected. >The proof is solid, standard usage of Galois Theory is refuted, the >theory of ideals is refuted. >It's finally over. But _this_ time you're right, eh? Got it. >Make no mistake, history is being made here. If nothing else, >celebrate that you are a part of some of the most historic findings in >the history of the world. Make no mistake, you're accomplishing nothing but making a fool of yourself. Again. And again and again. >You are now a part of history. >We all are. >James Harris David C. Ullrich ==== Subject: Re: JSH: It is finally over, definitely > So after so many long years of searching and arguing, and so many > doubts and failures, it is finally over and I have the proof that > leaves no room at all for any kind of reasonable doubt that I am right, > and there is this problem with standard usage of Galois Theory and with > the theory of ideals. > That information SHOULD propagate rapidly. > Some of you have a duty now to tell the truth. > There were minor mistakes with my algebra but they don't change the > argument when corrected. > The proof is solid, standard usage of Galois Theory is refuted, the > theory of ideals is refuted. > It's finally over. > Make no mistake, history is being made here. If nothing else, > celebrate that you are a part of some of the most historic findings in > the history of the world. > You are now a part of history. > We all are. You, Harris, have got nothing. > James Harris ==== Subject: Re: JSH: It is finally over, definitely Discussion, linux) > That information SHOULD propagate rapidly. Yes, but won't it be slowed by the destruction of the internet? Remember? You caused that last week. > It's finally over. Again. -- Jesse F. Hughes Everybody has a heart, except some people. -- All About Eve ==== Subject: Re: JSH: It is finally over, definitely <873bidjl7l.fsf@phiwumbda.org That information SHOULD propagate rapidly. > Yes, but won't it be slowed by the destruction of the internet? > Remember? You caused that last week. > That solution is to release fully the information on solving the > factoring problem, and post the algorithm that leads to ways to crack > RSA encryption, thus ending the Internet as it currently operates, and > impacting nations and businesses around the world. > The Hammer! At last! > I'm working on it now, and luckily it should only take a few days for > the impact and this will all be over. --- Christopher Heckman ==== Subject: Re: JSH: It is finally over, definitely Great stuff, James. Up to your highest standard. Keep it up! ==== Subject: Re: JSH: It is finally over, definitely <43faaf63$0$49957$892e7fe2@authen.yellow.readfreenews.net> Lotsa wheat needing to be raised in Kansas. ==== Subject: Re: JSH: It is finally over, definitely > So after so many long years of searching and arguing, and so many > doubts and failures, it is finally over and I have the proof that > leaves no room at all for any kind of reasonable doubt that I am right, > and there is this problem with standard usage of Galois Theory and with > the theory of ideals. > That information SHOULD propagate rapidly. I can hardly wait to see your shock and incredulity when, in a week or two, you realize that the world once again has ignored your uninspired drivel. Surprise, surprise, Gomer. -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) + 1 = 0 -- Euler | -- John Myers Myers, Silverlock ==== Subject: Re: It is finally over, definitely > The proof is solid, standard usage of Galois Theory is refuted, the > theory of ideals is refuted. You can't refute anything you obviously don't know anything about. Since it's over, does this mean you're going away? Dave ==== Subject: Re: Where Is My Headline? (Was Re: Continuing Failure Of Neoclassical Theory Of Value) to be his claim that in order to have the possibility of correspondence > between theory and observation, something he calls persistence of > prices -which in his conception is different from stationarity of > prices, and it means something like current 'normal' prices should > not depend on future 'normal' prices--must be something that prices > should have. > Garegnani does not offer an argument as to why this > persistence property of prices must be present in order to have a > correspondence between theory and observation. In fact, there is some > evidence that he is ignorant of the theme in Chiappori et al (2002), > [blah, blah, blah] > Once again, Alex just does not understand what is being discussed, > in this case by Garegnani. Since it was you that claimed that you don't follow Garegnani, you're not a competent judge of that. At any rate, it remains the case that Garegnani does not offer an argument as to why his persistence property of prices is needed in order to have a correspondence between theory or observation. Nor is it clear that real world prices do have that property. It also remains true that Schefold questioned Garegnani's methodology. Alex ==== Subject: Re: Where Is My Headline? (Was Re: Continuing Failure Of Neoclassical Theory Of Value) what represents rigorous neoclassical theory - GE simply does not > make those claims about scarcity in the general case... > [blah, blah, blah] > I thank Alex for his agreement. Just because you agree with me does not make me wrong. It remains the case that you are not attacking the rigorous neoclassical price theory. You've said yourself that your work does not attack GE. What's more, the rudiments of this rigorous theory have their roots in 1870, and even though the increase in its popularity, its development and even fully grasping its implications took some time, it was recognized as the rigorous neoclassical theory much earlier than you say. Consider: Walras is in my opinion the greatest of all economists. His system of economic equilibrium, uniting, as it does, the quality of a revolutionary creativeness with the quality of classic synthesis, is the only work by an economist that will stand comparison with the achievements of theoretical physics. Compared with it, most of the theoretical writings of that period - and beyond - however valuable in themselves and however original subjectively, look like boats beside a liner, like inadequate attempts to catch some particular aspects of Walrasian truth. (J. A. Schumpeter, History of Economic Analysis, 1954) > o My assumptions are canonical in neoclassical economics. > o Yet the numerical examples I build on those assumptions are > hardly consistent with the neoclassical vision of prices as > scarcity indices. > o Hence the neoclassical theory of value and distribution, as taught > over the last century from mainstream economics textbooks, is > incorrect. Actually I was saying that your titles are misleading, as you are not attacking rigurous neoclassical theory, which is a point that I thought needed emphasis -- and with which you apparently agreed. [...] > As one can see, > Hayek does not fully understand his theory. He thinks of prices > as scarcity indices, just as I have said. I can agree with the first sentence -- it is a complex theory after all -- but this still does not imply that his rigorous model becomes incorrect. [...] > Have theories that cannot be justified by GE models continued > to be taught and applied? Consider: > Suppose the number of carpenters suddenly increases, due to the > immigration of thousands of new carpenters from Mexico. Both before > and after the change, carpenters receive their marginal revenue > product... But the wage after the migration is lower than the wage > before. Since the supply of carpenters is higher than before, the > equilibrium wage is lower. > -- David D. Friedman (1990). _Price Theory: An Intermediate > Text_, Second Edition. > If Alex were serious, he would acknowledge that David > Friedman's text is used to teach students mistaken theories that > cannot be justified by rigorous economics. David Friedman is actually clear that he is talking about PE and not GE. See for instance here (Chapter 8): It is not so simple. The intersection of supply and demand curves gives us prices. Prices (of the goods the individuals produce and sell) give us incomes. But we needed incomes to start with, since they were one of the things that determined demand curves! The same problem appears if we stop talking about prices in general and talk instead about particular prices. We run through our supply and demand argument to get the price of widgets. We then do the same to get the price of cookies. But one of the things affecting the demand for widgets is the price of cookies (if cookies are inexpensive, you spend your money on them instead of on widgets). We could solve that problem by solving for cookies first--but one of the things affecting the demand for cookies might well be the price of widgets. The fact that he gave those warnings means that he is aware that he is exposing a simplified model which might in some cases give wrong results -- and he warns the reader. However, he still thinks that PE has empirical usefulness; hence he continues to teach it. As I said, with proper warnings, I don't think that teaching PE is all that outrageous. Alex ==== Subject: Re: Where Is My Headline? (Was Re: Continuing Failure Of Neoclassical Theory Of Value) production make it unlikely that capital-theoretic paradoxes will > arise? But as I've said, it is not necessary to give the conditions as special cases of GE in which capital-theoretic paradoxes do not appear. It would be nice to be able to do this --although this kind of thing does not seem to be on many people's research agenda-- but it's not needed. The empirical irrelevance of capital theoretic paradoxes can be used just fine as a justification for ignoring them in applied work. Schefold himself notes --in a different context, but still relevant here-- that: if RCD turns out to be rare and the likelihood of a case of RCD leading to a significant deviation from equilibrium could be shown to be low, the combined probability of both events would be very low, and the pragmatic use of neoclassical theory then (but only then) could not be questioned on the basis of the arguments used in this exchange. So if capital-theoretic paradoxes are empirically irrelevant for the circumstances under study, then they can be ignored. Alex ==== Subject: Re: Where Is My Headline? (Was Re: Continuing Failure Of Neoclassical Theory Of Value) consumer side of the economy (i.e. well behaved subjective preferences) > there will always be stability. > I doubt Mandler proves any of what Alex says he does. > He proves certain conclusions in a model where he makes specific > assumptions about the modeling of consumption. Like all theorems, Mandler's theorem needs some assumptions. In this particular case, his theorem does not have to say something general about the real world -- though it might do that -- rather it was mostly useful as a critique of the points raised by Schefold and Garegnani. In particular, the assumptions under which Schefold worked in a part of his paper were the same as the assumptions of Mandler's theorem -- and yet Mandler *proved* that the instability which Schefold mentions does not exist. Hence Schefold was provably wrong as far as the stability issue is concerned. [...] > Under some assumptions about subjective > preferences, he proves the uniqueness of equilibrium. > This is not what I understand Mandler to claim in a series of papers > and a book. He has previously shown that, in certain models steady-state > equilibria are determinate. A determinate equilibrium is not one > equilibrium in a continuum of equilibria. It is locally unique, but > not necessarily globally unique. In asmuch as the conclusions of > Mandler 2005 contrast with these previous results, I need to think > more about them. > On the other hand, he claims that equilibria are indeterminate in > a model of intertemporal equilibrum. Indeterminate equilibrium are > arguably not consistent with atomistic competitive behavior. MasCollel et al. prove that indeterminacy --in the strong sense of a continuum of equilibria, or even an infinite number of equilibria-- while it can not be excluded, is in a sense a pathological result. More specifically, they prove that if an economy is regular -- i.e its matrix of price effects is nonsingular -- then it has a finite number of equilibria. (The number of equlibria is in fact an odd number). They also make precise the sense in which a nonregular economy is pathological -- the set of endowments that result in a nonregular economy is a set of measure zero. Borissov also quotes Mandler to the effect that indeterminacy occurs only at a measure zero set of endowments and does not occur when relative prices are constant through time. [...] I've read your discussion of what you were and are doing and what you hope to do in your research but I feel that I don't have any useful comment on it other that Good Luck [...] > There is another paper about sensitivity of stability results on > consumption modeling choices which I have not yet integrated into > my thinking on these matters: > Borissov, Kirill (2002). Indeterminacy of Distribution in a General > Equilibrium Model, European University at St. Petersburg, > Working Paper 2002/02 > As I understand it, Borissov assumes representative agents for each > of a number of groups of households. (One might think of these > as endogenously determined social classes.) In constrast to Mandler, > Borissov finds indeterminate steady state equilibria. Perhaps > Mandler would say that Borissov is not assuming well-behaved > consumers, in Mandler's sense. (Why anybody interested in empirically > understanding capitalist economies would prefer Mandler to > Borissov on this point, if such a contrast exists, is not clear > to me.) Mandler's assumptions are not necessary, just sufficient for his results -- see above. Borissov's results seem interesting and it might be worthwhile to investigate the precise ways in which his assumptions violate the regularity assumptions above -- but I am not going to do this for a usenet post; certainly not at this time. Alex ==== Subject: A question about Caontor's proof of the uncountability of the reals What's wrong with this argument? This is an honest question, not a troll - I've never studied mathematical logic. 1) The proposition P that the set of reals is countable (to be proved FALSE) is a statement in the language of aximatic set theory. 2) Axiomatic set theory, by Godel's incompleteness theorem, contains propositions that are undecidale. 3) Therefore the law of the excluded middle (every proposition is either true or false) does not hold, and proof by contradiction does not apply. 4) Assume P, the reals are countable, is true. This leads to a contradiction, (diagonalization argument:all reals are on the list and not all reals are on the list) and therefore by the law of non contradiction (a proposition cannot be both true and false) P is not true. 5) This leaves two alternatives: P is false, P is undecidable. 6) To complete the proof that P is false (and therefore the reals are not countable) we must show that P is not undecidable. 7) Cantor did not do this, therefore the theorem remains unproved. What am I missing? -- Michael ==== Subject: Re: A question about Caontor's proof of the uncountability of the reals > What's wrong with this argument? [...] > 7) Cantor did not do this, therefore the theorem remains unproved. ...classical logic was abstracted from the mathematics of finite sets and their subsets...Forgetful of this limited origin, one afterwards mistook that logic for something above and prior to all mathematics, and finally applied it, without justification, to the mathematics of infinite sets. This is the Fall and original sin of [Cantor's] set theory ... (Weyl, 1946) ==== Subject: Re: A question about Caontor's proof of the uncountability of the reals Discussion, linux) >> What's wrong with this argument? [...] >> 7) Cantor did not do this, therefore the theorem remains unproved. > ...classical logic was abstracted from the mathematics of finite sets > and their subsets...Forgetful of this limited origin, one afterwards > mistook that logic for something above and prior to all mathematics, > and finally applied it, without justification, to the mathematics of > infinite sets. This is the Fall and original sin of [Cantor's] set > theory ... (Weyl, 1946) Utterly unhelpful. I know, I know, this criticism is just another instance of the Cantorian cultists inflicting their religious beliefs on you. It's worse than, well, gosh, *anything*. -- Jesse F. Hughes If you are a consumer that's taking advantage of the technologies that exist ... then the spam problem for you is solved. --MS spokesman verifying that the spam problem has been solved. ==== Subject: Re: A question about Caontor's proof of the uncountability of the reals > What's wrong with this argument? This is an honest question, not a troll - > I've never studied mathematical logic. > 1) The proposition P that the set of reals is countable (to be proved FALSE) > is a statement in the language of aximatic set theory. > 2) Axiomatic set theory, by Godel's incompleteness theorem, contains > propositions that are undecidale. > 3) Therefore the law of the excluded middle (every proposition is either > true or false) does not hold, and proof by contradiction does not apply. Step 3 does not follow. Although some propositions are undecidable, so that we cannot say that they are true or false, it does remain that they are true *or* false. Thus, while G is neither provable nor disprovable, G v ~G is provable, and so is CH v ~CH. ==== Subject: Re: A question about Caontor's proof of the uncountability of the reals >What's wrong with this argument? This is an honest question, not a troll - >I've never studied mathematical logic. >1) The proposition P that the set of reals is countable (to be proved FALSE) >is a statement in the language of aximatic set theory. >2) Axiomatic set theory, by Godel's incompleteness theorem, contains >propositions that are undecidale. Meaning that there is no _proof_ withing whatever system that the proposition is true, and no proof that it's false. >3) Therefore the law of the excluded middle (every proposition is either >true or false) does not hold, and proof by contradiction does not apply. No, that doesn't follow. Because true is not the same as provable. >4) Assume P, the reals are countable, is true. This leads to a >contradiction, (diagonalization argument:all reals are on the list and not >all reals are on the list) and therefore by the law of non contradiction (a >proposition cannot be both true and false) P is not true. >5) This leaves two alternatives: P is false, P is undecidable. >6) To complete the proof that P is false (and therefore the reals are not >countable) we must show that P is not undecidable. >7) Cantor did not do this, therefore the theorem remains unproved. >What am I missing? >-- Michael David C. Ullrich ==== Subject: Re: A question about Caontor's proof of the uncountability of the reals > What's wrong with this argument? This is an honest question, not a troll - > I've never studied mathematical logic. > 1) The proposition P that the set of reals is countable (to be proved FALSE) > is a statement in the language of aximatic set theory. > 2) Axiomatic set theory, by Godel's incompleteness theorem, contains > propositions that are undecidale. Actually, it says there are true propositions which are not provable. > 3) Therefore the law of the excluded middle (every proposition is either > true or false) does not hold, and proof by contradiction does not apply. You're mixing true/false with provable/unprovable/negation is provable. --- Christopher Heckman > 4) Assume P, the reals are countable, is true. This leads to a > contradiction, (diagonalization argument:all reals are on the list and not > all reals are on the list) and therefore by the law of non contradiction (a > proposition cannot be both true and false) P is not true. > 5) This leaves two alternatives: P is false, P is undecidable. > 6) To complete the proof that P is false (and therefore the reals are not > countable) we must show that P is not undecidable. > 7) Cantor did not do this, therefore the theorem remains unproved. > What am I missing? > -- Michael ==== Subject: Re: A question about Caontor's proof of the uncountability of the reals Cybernetics espouses that there are just 2 infinities: the potential infinity and the actualized infinity. Cantor and his infinity of infinities is nice for mathematicians to play around with as a fantasy, but when you want to do something constructivistic in the real world, stick to cybernetics After all, it's good for launching rockets and guiding bombs in Iraq and steering around icebergs (cybernetics means helmsperson). See: http://pespmc1.vub.ac.be/INFINITY.html - Don ==== Subject: Re: Looking for a simple function with known values >do you see a simple increasing function with the following constraints: >f(0) = 0 >f(1) = 1/2 >f'(0) = sqrt(2)/2 >f'(1) = log(2)/2 f(x) = min(x/sqrt(2), 2^(x-2)) -- VP Cheney Burr-ed his gun as a bird flew past The nation responds burr as we await bird flu shots and fight a real cold war. pmontgom@cwi.nl Microsoft Research and CWI Home: Bellevue, WA ==== Subject: Re: Looking for a simple function with known values <38ojv1h86nl3r0i4loqppsrns649pd7sol@no.spam> do you see a simple increasing function with the following constraints: >> f(0) = 0 >> f(1) = 1/2 >> f'(0) = sqrt(2)/2 >> f'(1) = log(2)/2 >No. As log 2 and ln 2 are negative, f cannot be increasing at 1. > Huh? What's wrong with: Nothing, once one has adequate sleep. ;-) > f(x) = x/2 * [sqrt(2) * (x-1)^2 + x * (3 - 2x + log(2) * (x - 1))] > It's everywhere non-decreasing. ==== Subject: Simple Field Extension I don't understand this problem. Is there anyone can help me solve it? In the field K(x), let u= x^3 /(x+1) . Then show that K(x) is a simple ==== Subject: Re: Simple Field Extension days. My association with the Department is that of an alumnus. >I don't understand this problem. Is there anyone can help me solve it? >In the field K(x), let u= x^3 /(x+1) . Then show that K(x) is a simple You are to show that there is a single element a in K(x) such that K(x) = K(u)(a), i.e., the smallest subfield of K(x) that contains both K(u) and a is K(x). That's what it means to be a simple extension. Note that K(u) is certainly contained in K(x). But K(x) is not contained in K(u), since x is not equal to a rational function on u. -- ' ==== Subject: Representation theory question. It is well known that if a finite group has only 1 dim irreducible representationss then is is abelian. However, if it has one or two dimensional representations (irreducibe) is it tre that G must contain a normal subgroup that has index 2 in it? ==== Subject: Re: Representation theory question. >It is well known that if a finite group has only 1 dim irreducible representationss then is is abelian. >However, if it has one or two dimensional representations (irreducibe) is it tre that G must contain a normal subgroup that has index 2 in it? Presumably you mean *only* one or two-dimensional representations. Yes, this follows from a theorem of Isaacs (in his book The Character Theory of Finite Groups, p. 201), that, if the set of character degrees of a finite group G is {1,m}, then either a) G has an abelian normal subgroup of index m, or b) m=p^e (p prime) and G is a direct product of a p-group and an abelian group. There is probably a more elementary proof for the case m=2. Derek Holt. ==== Subject: Re: Representation theory question. I like mathematics. ==== Subject: List of Folks Who SAW First Plane Strike WTC -- NO, NOT Naudets' Fraudulent Video < < < < < < < < < < < < < < < < < < < < < Hmmm! It must be true! Nobody saw a plane BEFORE the explosion in the North Tower. < Ed Conrad > http://www.edconrad.com < < < If there's indeed a Smoking Gun that proves beyond ALL reasonable doubt that there was a government conspiracy responsible for the horror of 911 -- and the horrific aftermath of the on-going war in Iraq -- the following URL is it (below). < It'ss the second Naudet Brothers' video taken when the first WTC tower exploded (you'll notice, I'm not referring to their first video when the first plane commandeered by terrorists allegedly flew into the North Tower. < You see, the Naudet Brothers' video of the plane striking the North Tower is as phony as a $4 bill. < I maintain there was NO plane-- therefore no plane visible -- when it exploded. I'm convinced the plane was EDITED INTO THE VIDEO in the hours after the explosion first occurred and was only released to the networks after the plane was superimposed to make it seem like the real thing. This explains the plane's peculiar appearance and behavior when the video -- ghost plane included -- was shown to gullible Americans and people around the world. The fact that no plane was found proved that 911 is a fraud.I believed > they hired David Copperfield and David Blaine for the disappearing act ==== Subject: Re: List of Folks Who SAW First Plane Strike WTC -- NO, NOT Naudets' Fraudulent Video 9-11 is, to date, the biggest stunt of the 21st Century. More 9-11s will come. Just wait for they have determined that it was possible to create a massive tragedy in order to justify wars, invasions and even a massive nuclear strike. Yes, it will happen. -- Your Friendly Neighbourhood Spiderman. > <> Hmmm! It must be true! Nobody saw a plane BEFORE > the explosion in the North Tower. > Ed Conrad >> http://www.edconrad.com > If there's indeed a Smoking Gun that proves beyond ALL > reasonable doubt that there was a government conspiracy > responsible for the horror of 911 -- and the horrific aftermath > of the on-going war in Iraq -- the following URL is it (below). > It'ss the second Naudet Brothers' video taken when the first WTC > tower exploded (you'll notice, I'm not referring to their first video > when the first plane commandeered by terrorists allegedly flew > into the North Tower. > You see, the Naudet Brothers' video of the plane striking the North > Tower is as phony as a $4 bill. > I maintain there was NO plane-- therefore no plane visible -- > when it exploded. I'm convinced the plane was EDITED INTO > THE VIDEO in the hours after the explosion first occurred and > was only released to the networks after the plane was superimposed > to make it seem like the real thing. > This explains the plane's peculiar appearance and behavior when > the video -- ghost plane included -- was shown to gullible Americans > and people around the world. > was like a knife slicing through melted butter. > If there is any New Yorker who actually saw American Airlines Flight > 11 IN FLIGHT before striking the first tower, stand up and be heard. > All anyone can honestly say -- in all honesty -- is that they HEARD > the explosion, looked up and ASSUMED a plane had hit the tower. > No one actually saw Flight 11, or any other plane, hit the North Tower > because there was none, neither a missile. The explosion was caused > by explosives stockpiled inside. > The Naudet Brothers' video, showing a plane striking the tower, > was supposed to convince us that this IS what happened. > However, that wasn't the ONLY Naudet video taken when the > explosion occurred -- and the other video is, beyond the > question of a doubt, THE SMOKING GUN. > You see, the Naudet Brothers had taken ANOTHER video at the very > same time, showing pedestrians casually hurrying to wherever they > were going precisely when the explosion occurred high in the North > Tower. > This second video is self-explanatory (read the accompanying URL > text), and there was no reason in the world to have taken it unless it > was a feeble attempt to add insult to injury and try and provide more > fraudulent evidence that a plane had hit the tower. > This SECOND Naudet video was just one of a series of comedic errors > in the whole 911 shenanigans -- the phony cell phone calls, > the Pentagon fiasco, the bogus Osama bin Laden video in which he > confesses to masterminding 911, etc../ They were all propaganda to > convince the American public that 19 terrorists (at least a half-dozen > still alive) were responsible for 911, thus opening wide the door for > the invasion of Afghanistan and Iraq. > If the conspirators had any common sense, the second Naudet video > of the pedestrians would NEVER have seen the light of day. > THE SMOKING GUN: > http://www.911foreknowledge.com/bravenewworld.htm > I don't know who deserves credit for this URL. But, whomever > it was -- his first name is Ray -- deserves a sincere thank you. > (cc) Bush Cheney Rove Libby Barbara Olson Mark Bingham > Reuters ABC NBC CBS MSNBC Fox News CNN Larry King > New York Times Daily News Post Newsday Newsweek Wall > Street Journal Newark Star-Ledger Washington Post Times > Los Angeles Times National Enquirer Star White House Iraq > Iran Newspaper Islam History Channel Discovery People Terrorists > Freedom Fighters Baghdad Islam Osama bin Laden Halliburton > Rumsfield Rice Meet the Press Good Morning America Today > Oval Office Press Corps David Iain Greig moderator talk.origins ==== Subject: Re: List of Folks Who SAW First Plane Strike WTC -- NO, NOT Naudets' Fraudulent Video <1140499857.849448@ftpsrv1> Worse than 911 is the total Chaos these usergroups have become. ==== Subject: Re: List of Folks Who SAW First Plane Strike WTC -- NO, NOT Naudets' Fraudulent Video <1140499857.849448@ftpsrv1 Worse than 911 is the total Chaos these usergroups have become. ...completely. what the is this stuff? this is a japanese language group, not alt.paranoid.conspirists.... matt -- http://www.sushi-review.com ==== Subject: Re: List of Folks Who SAW First Plane Strike WTC -- NO, NOT Naudets' Fraudulent Video Why so? Because they challenge the lies of mainstream media controlled by the big capitalists? -- Your Friendly Neighbourhood Spiderman. > Worse than 911 is the total Chaos these usergroups have become. ==== Subject: JSH: My patience is over Some of you seem intent on destroying even hope of my goodwill at this point. If you show contempt for mathematics at THIS point with a proof that removes any confusion that might have come from all that talk about mystery functions then I don't think there is any hope for you, and I think your betrayal of humanity is complete. If you wish to know what hell on earth is in your lifetime then you fight me now and you will regret it for the rest of your lives. Mathematics is important. The truth is important. You fight the truth till you are crushed and you prove your enmity for humanity itself. You prove you are the enemies of mankind. I suggest you think carefully. Look carefully over the proof and my explanations. THINK before you step down a path you will never get to turn back from. Make no mistake. After this night, and this proof, I will feel no need to protect you from your own stupidity, nor from the consequences of proving to everyone else that you hate them, hate the truth, and despise mathematics. James Harris ==== Subject: Re: JSH: My patience is over >Some of you seem intent on destroying even hope of my goodwill at this >point. >If you show contempt for mathematics at THIS point with a proof that >removes any confusion that might have come from all that talk about >mystery functions then I don't think there is any hope for you, and I >think your betrayal of humanity is complete. >If you wish to know what hell on earth is in your lifetime then you >fight me now and you will regret it for the rest of your lives. >Mathematics is important. The truth is important. What's important is that you stop posting while drunk. Or go back on your meds, or whatever. > You fight the truth >till you are crushed and you prove your enmity for humanity itself. >You prove you are the enemies of mankind. >I suggest you think carefully. Look carefully over the proof and my >explanations. >THINK before you step down a path you will never get to turn back from. >Make no mistake. After this night, and this proof, I will feel no need >to protect you from your own stupidity, nor from the consequences of >proving to everyone else that you hate them, hate the truth, and >despise mathematics. >James Harris David C. Ullrich ==== Subject: Re: JSH: My patience is over The General system is a lot bigger than a local system. There are more things in heaven and earth than are dreamed of in your philosophy. ==== Subject: Re: JSH: My patience is over Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >Make no mistake. After this night, and this proof, I will feel no need >to protect you from your own stupidity, nor from the consequences of >proving to everyone else that you hate them, hate the truth, and >despise mathematics. And if you turn out to be mistaken, what then? Will you retract your accusations of contempt for mathematics, betrayal of humanity, and stupidity? And will you think twice before doing the same thing again in a few days, weeks, or months? Or will you just come up with a slightly modified version of your proof, and make the same accusations again? -- Richard ==== Subject: Re: JSH: My patience is over > Some of you seem intent on destroying even hope of my goodwill at this > point. > If you show contempt for mathematics at THIS point with a proof that > removes any confusion that might have come from all that talk about > mystery functions then I don't think there is any hope for you, and I > think your betrayal of humanity is complete. > If you wish to know what hell on earth is in your lifetime then you > fight me now and you will regret it for the rest of your lives. > Mathematics is important. The truth is important. You fight the truth > till you are crushed and you prove your enmity for humanity itself. > You prove you are the enemies of mankind. Yes! Tell the truth, James! Crush them! ==== Subject: Re: JSH: My patience is over > Some of you seem intent on destroying even hope of my goodwill at this > point. YOUR goodwill? We'll destroy even HOPE of YOUR GOODWILL? Are you by any chance referring to the goodwill that has continually referred to the mathematics community as being incorrigably corrupt, to the claims that we have knowingly conducted our affairs fraudulently, to the threats, too numerous to mention, of legal action, of military action, of the fomenting of mob violence, all against those who would dare to fail to accept what you write as being valid? How about the sort of goodwill that regularly refers to anyone who disagrees with your arguments as being a liar? How about your apparent belief that everyone accepts your argument as being valid, yet does not do so publicly for corrupt motives? Who on earth needs such goodwill? Whatever distinguishes this type of goodwill from active malice? > If you show contempt for mathematics at THIS point with a proof that > removes any confusion that might have come from all that talk about > mystery functions then I don't think there is any hope for you, and I > think your betrayal of humanity is complete. Really. Please show where I have used any so-called mystery functions. Your argument style is obscure, and produces results that are not in agreement with ordinary arithmetic. You use undefined terminology, and are clearly not competent in the area you pretend to have some dissatisfaction with. This is not unlike the failing calculus student who thinks that that field is not valid, or the physics know-nothing who believes he has invented the perpetual motion machine. > If you wish to know what hell on earth is in your lifetime then you > fight me now and you will regret it for the rest of your lives. I'm thinking you believe that this sort of threat actually has any effect other than the inevitable chuckles and outright derision that I'm sure will ensue. Is this how supposedly adult humans behave? Is this how you treated your host at your alma mater when he patiently listened to your discussion? > Mathematics is important. The truth is important. You fight the truth > till you are crushed and you prove your enmity for humanity itself. > You prove you are the enemies of mankind. At least I don't threaten people who have done me no wrong. I treat people with the greatest amount of respect that is reasonable. > I suggest you think carefully. Look carefully over the proof and my > explanations. > THINK before you step down a path you will never get to turn back from. Anyone who changes his mind by reading your argument here is not acting as a mathematician. He is acting out of cowardice, and not the ordinary cowardice in the face of any realistic threat: instead, it would be knuckling under to the likes of you, a posturing buffoon. > Make no mistake. After this night, and this proof, I will feel no need > to protect you from your own stupidity, nor from the consequences of > proving to everyone else that you hate them, hate the truth, and > despise mathematics. Hey, big guy. Do what you can. > James Harris Dale ==== Subject: Re: JSH: My patience is over > Some of you seem intent on destroying even hope of my goodwill at this > point. What makes you say that, tard? > If you show contempt for mathematics at THIS point with a proof that > removes any confusion that might have come from all that talk about > mystery functions then I don't think there is any hope for you, and I > think your betrayal of humanity is complete. Not complete yet, someone still owes me 30 pieces of silver. > If you wish to know what hell on earth is in your lifetime then you > fight me now and you will regret it for the rest of your lives. Are you crying yet? > Mathematics is important. The truth is important. You fight the truth > till you are crushed and you prove your enmity for humanity itself. > You prove you are the enemies of mankind. > I suggest you think carefully. Look carefully over the proof and my > explanations. > THINK before you step down a path you will never get to turn back from. Dance, monkey, dance! > Make no mistake. After this night, and this proof, I will feel no need > to protect you from your own stupidity, nor from the consequences of > proving to everyone else that you hate them, hate the truth, and > despise mathematics. tard. > James Harris ==== Subject: Re:JSH My patience is over DECODED!! The JSHCode - DECODED!! put it in the decoder at http://www.georgetown.edu/faculty/ballc/webtools/web_freqs.html and read vertically down. I THAT THINK THIS TO MATHEMATICS PROOF TRUTH WILL ANY ARE AT CAREFUL FLIGHT HATE HOPE, HUMANITY IF IMPORTANT MY NO ON POINT PROVE THEN ABOUT, AFTER, ALL BACK BEFORE BETRAYAL COME COMPLETE CONFUSION, CONSEQUENCES ,CONTEMPT, CRUSHED, DESPISE Text name: My patience is over Word count: 191 Unique words: 106 Sort order: descending 16 YOU 7 THE 6 AND 6 OF 5 IS 5 YOUR 4 FOR 4 FROM 4 I 4 THAT 4 THINK 4 THIS 4 TO 3 MATHEMATICS 3 PROOF 3 TRUTH 3 WILL 2 A 2 ANY 2 ARE 2 AT 2 CAREFULLY 2 FIGHT 2 HATE 2 HOPE 2 HUMANITY 2 IF 2 IMPORTANT 2 MY 2 NO 2 ON 2 POINT 2 PROVE 2 THEN 1 ABOUT 1 AFTER 1 ALL 1 BACK 1 BEFORE 1 BETRAYAL 1 COME 1 COMPLETE 1 CONFUSION 1 CONSEQUENCES 1 CONTEMPT 1 CRUSHED 1 DESPISE 1 DESTROYING 1 DON'T 1 DOWN 1 EARTH 1 ELSE 1 ENEMIES 1 ENMITY 1 EVEN 1 EVERYONE 1 EXPLANATIONS 1 FEEL 1 FUNCTIONS 1 GET 1 GOODWILL 1 HARRIS 1 HAVE 1 HELL 1 IN 1 INTENT 1 IT 1 ITSELF 1 KNOW 1 LIFETIME 1 LIVES 1 LOOK 1 MAKE 1 MANKIND 1 ME 1 MIGHT 1 MISTAKE 1 MYSTERY 1 NEED 1 NEVER 1 NIGHT 1 NOR 1 NOW 1 OVER 1 OWN 1 PATH 1 PROTECT 1 PROVING 1 REGRET 1 REMOVES 1 REST 1 SEEM 1 SHOW 1 SOME 1 STEP 1 STUPIDITY 1 SUGGEST 1 TALK 1 THEM 1 THERE 1 TILL 1 TURN 1 WHAT 1 WISH 1 WITH ==== Subject: Re: JSH My patience is over DECODED!! On Tue, 21 Feb 2006 00:22:15 -0600, Sulayman Your posts make less sense than Harris'. >The JSHCode - DECODED!! >put it in the decoder at >http://www.georgetown.edu/faculty/ballc/webtools/web_freqs.html and read >vertically down. >I THAT THINK THIS TO MATHEMATICS PROOF TRUTH WILL >ANY ARE AT CAREFUL FLIGHT HATE HOPE, HUMANITY >IF IMPORTANT MY NO ON POINT PROVE >THEN ABOUT, AFTER, ALL BACK BEFORE BETRAYAL COME COMPLETE CONFUSION, >CONSEQUENCES ,CONTEMPT, CRUSHED, DESPISE >Text name: My patience is over >Word count: 191 >Unique words: 106 >Sort order: descending >16 YOU >7 THE >6 AND >6 OF >5 IS >5 YOUR >4 FOR >4 FROM >4 I >4 THAT >4 THINK >4 THIS >4 TO >3 MATHEMATICS >3 PROOF >3 TRUTH >3 WILL >2 A >2 ANY >2 ARE >2 AT >2 CAREFULLY >2 FIGHT >2 HATE >2 HOPE >2 HUMANITY >2 IF >2 IMPORTANT >2 MY >2 NO >2 ON >2 POINT >2 PROVE >2 THEN >1 ABOUT >1 AFTER >1 ALL >1 BACK >1 BEFORE >1 BETRAYAL >1 COME >1 COMPLETE >1 CONFUSION >1 CONSEQUENCES >1 CONTEMPT >1 CRUSHED >1 DESPISE >1 DESTROYING >1 DON'T >1 DOWN >1 EARTH >1 ELSE >1 ENEMIES >1 ENMITY >1 EVEN >1 EVERYONE >1 EXPLANATIONS >1 FEEL >1 FUNCTIONS >1 GET >1 GOODWILL >1 HARRIS >1 HAVE >1 HELL >1 IN >1 INTENT >1 IT >1 ITSELF >1 KNOW >1 LIFETIME >1 LIVES >1 LOOK >1 MAKE >1 MANKIND >1 ME >1 MIGHT >1 MISTAKE >1 MYSTERY >1 NEED >1 NEVER >1 NIGHT >1 NOR >1 NOW >1 OVER >1 OWN >1 PATH >1 PROTECT >1 PROVING >1 REGRET >1 REMOVES >1 REST >1 SEEM >1 SHOW >1 SOME >1 STEP >1 STUPIDITY >1 SUGGEST >1 TALK >1 THEM >1 THERE >1 TILL >1 TURN >1 WHAT >1 WISH >1 WITH ==== Subject: Please provide me statistical help Iam Sriram.R, Iam from India pursuing a PhD in economics. My research involves water resources study in the rural areas of India. Iam not good at Statistics. I have come to the final stage in my research where I need to prove my research empirically. Though I have the data I do not know which statistical tool I need to use to prove my research. My research involces more of qualitative analysis. Can anyone help me with my research. You can email me at sriram.mbe@gmail.com or sriram.phd@gmail.com. Sriram.R ==== Subject: Re: Find three points P,Q,R on sides AB,BC,CA ) > Yes PQR *is* equilateral of course... > Hint added. > Find three points P,Q,R on sides AB,BC,CA > of a scalene triangle ABC so that PQ = QR = RP. > There are infinitely many solutions. > PQR being equilateral : > Choose P nearly anywhere on AB, then construct equilateral triangle > with other vertices lying on two given straight lines is a well known > problem. > hint : rotate 60 degree from P > philippe Indeed yes Philippe, it is possible to draw infinitely many lines passing through P,Q and R that meet respectively in B,C and A for sure. Actually, the query in other words is: Find an analytic function relation F needed between dividing fractions u,v,w ( where u = AP/AB, v = BQ/BC, w = CR/CA ) as obeying the required function F(u,v,w) = 0. For example: For an ex-scribed isosceles triangle u = v, w = 1 ; for an ex-scribed 60 degree right-angled triangle u = v = 1/3, w =2/3 &C... It may have appeared in several interesting de.sci.mathematik discussions earlier also in which you participated, when an inside square also was discussed so much in detail, but I couldn't find this one exactly again. In the problem of finding an equilateral triangle between a given point and two given lines I supplied a numerical, but not an analytic solution that is now sought here. Is an analytical solution possible using the hint ? ==== Subject: Re: Find three points P,Q,R on sides AB,BC,CA >> (supersedes ) >> Yes PQR *is* equilateral of course... >> Hint added. > Find three points P,Q,R on sides AB,BC,CA > of a scalene triangle ABC so that PQ = QR = RP. >> There are infinitely many solutions. >> PQR being equilateral : >> Choose P nearly anywhere on AB, then construct equilateral triangle >> with other vertices lying on two given straight lines is a well known >> problem. >> hint : rotate 60 degree from P >> philippe > Indeed yes Philippe, it is possible to draw infinitely many lines > passing through P,Q and R that meet respectively in B,C and A for sure. I understood the problem in the opposite way : *given* ABC, find PQR... (not given PQR, find ABC, as this one seems much less interesting, without some constraints on ABC, any three lines will be suitable) I specifically searched the variant : find the *smallest* PQR. Because the largest PQR is trivial. I found obvious solutions when ABC isosceles (as you note R midpoint of AC), but also when one ABC angle is 120 degree say B, ABC still being scalene : RC/RA = BC/AB. Please note that there are asymetric solutions (not minimal) with ABC isosceles, infinitely many choices of R, given ABC. > Actually, the query in other words is: Find an analytic function > relation F needed between dividing fractions u,v,w ( where u = AP/AB, v > = BQ/BC, w = CR/CA ) as obeying the required function F(u,v,w) = 0. > For example: For an ex-scribed isosceles triangle u = v, w = 1 ; Hum.. w=1/2 with your notations. w = 1 should be using w = CR/RA > for an ex-scribed 60 degree right-angled triangle u = v = 1/3, w =2/3 &C... Didn't check that case. I was (as usual) inclined in a pure geometric point of view ! But the idea of using the ratios u,v,w in a 'Ceva like' manner may be a good idea, if it really doesn't depend on the sides themselves... > It may have appeared in several interesting de.sci.mathematik > discussions earlier also in which you participated, when an inside > square also was discussed so much in detail, but I couldn't find this > one exactly again. I remember of finding a square with sides going through 4 given points, but I don't remember of finding a square inscribed in a triangle. > In the problem of finding an equilateral triangle between a given point > and two given lines I supplied a numerical, but not an analytic > solution that is now sought here. I have two very simple pure geometric constructions. One corresponding to my hint (60 degree rotations), the traditional method : (Given P, find Q on BC and R on AC) Locus of Q is line BC But R is transformed of Q by a rotation with center P and angle 60 deg (PQR equilateral) So locus of R is transformed of line BC by that rotation. But locus of R is line AC, so R is intersection point of line AC and {line BC rotated} The two solutions result from rotating +/- 60 deg. Only one of them might be suitable (Q and R inside the sides) Another one uses just inscribed angles in a suitable circle : Given ABC and any point M on AC, draw two lines from M with 60 deg angles from AC, intersecting sides AB and BC in P and Q. Draw the circumscribed circle to MPQ, this circle intersects AC in another point R. It is then easy to proove that PQR is equilateral. This gives easily the ranges where P and Q must lie on AB and BC, and reversing the construction, the range for R. I didn't see that thread. I'll look and answer also there. > Is an analytical solution possible using the hint ? seems Norm Dresner idea, corrected, might give a solution. (doesn't use my hint). For an analytic method, I think at the following (for the original question : given ABC find PQR) In complex plane, write the conditions for P,Q,R on the sides that is P = u*A + (1-u)*B etc... Then write the condition for PQR being equilateral : P + j*Q + j^2*R = 0 (if P,Q,R anticlockwise). j being e^(i*pi/3). This should result into the F(u,v,w) relation. -- philippe mail : chephip at free dot fr site : http://chephip.free.fr/ ==== Subject: Re: Find three points P,Q,R on sides AB,BC,CA contd... i.e., we need the function F( AB,BC,CA,u,v,w ) = 0 ,or otherwise,angles <) APR, <) BQP &c. Narasimham ==== Subject: Re: Basic proof? - Hamel basis > Discontinuous linear functions from R to R can be constructed with the > help of a so-called Hamel basis of the reals over the rational numbers. > To do so one needs the Axiom of Choice. The result is always a > non-measurable function in the sense of Lebesgue. The results are also linear only WRT the /rational/ vector space structure. With regard to the /real/ vector space structure, the only linear maps from R to R are (as an earlier response noted) given by f: x -> ax for some a in R -- namely a = f(1). ==== Subject: Enigma (WWII) cipher-breaking project I don't run the project, but if you've got some spare mip's on your computer, you might like to join: http://www.bytereef.org/m4_project.html It's a distributed-computing attack on three undeciphered German Navy cryptograms from 1943 or so. I joined in, and the software (which runs in the background) is causing no problems on this Windows XP box. It is available for various platforms and is open-source. ==== Subject: Re: Enigma (WWII) cipher-breaking project Why? ==== Subject: Re: Enigma (WWII) cipher-breaking project > Why? So you can pretend you're as smart as Alan Turing for a day. ==== Subject: integer least squares problem? We have an equation |y-(a1*x1+a2*x2+a3*x3...+ak*xk)|^2 where a1, a2...ak are complex constants, y, x1, x2,...,xk are complex variables. Is it possible to be approximated by an equation like y^2+b1*x1^2+ b2*x2^2+,...+bk*xk^2 + c1*x1 +c2*x2+...+ck*xk +d ? ==== Subject: Re: integer least squares problem? > We have an equation |y-(a1*x1+a2*x2+a3*x3...+ak*xk)|^2 > where a1, a2...ak are complex constants, y, x1, x2,...,xk > are complex variables. Is it possible to be approximated by > an equation like y^2+b1*x1^2+ b2*x2^2+,...+bk*xk^2 + c1*x1 > +c2*x2+...+ck*xk +d ? I do not see any equal sign,=, in what you call an equation. When I was brought up, we were not allowed to call things equations which did not contain equal signs, but society seems to be becoming less and less careful about these things with the passing of the years... ==== Subject: Re: Standard Deviation of PSIA > {...} >> This is why there IS a separate category for >> Whites and Hispanic Whites, and why each race ends up at opposite >> ends of a birthday party, glaring at each other, no matter how friendly >> they pretend to be. > Nope. Not here in Tucson. > Both races often do end up becoming quite brisk with > the pinata. And I suspect that if you showed up spewing > your racist drivel, they might, with great group spirit and > as unified as a band of brothers, treat you in the same manner. > -- cary Well, now that we're mentioning our little pasty brother, let's remember what he has had to say: The phrase German refers to a RACE of people, not just a nation. You can't be a German if you're not of that RACE. My sister, and thus I, are of the German RACE. Uh, do you think that Simple John is familiar with the word in German, Inzest? Of course, this was after Simple John had exclaimed: My sister was born in Germany, and at age 21, after they'd completed the genealogical research, she was offered German citizenship. Do you remember Simple John ever answering the question of what year his sister was offered German citizenship? Speaking of genealogical research, there was a show on History Ch yesterday titled, High Hitler. It's premise is that Hitler was a druggie and that - offhandedly - 88 stands for drugs. (Well, the latter is more /my/ theory.) One of the other History Ch shows on the bugger illustrated that he was never married because he never passed the genealogical research to show that he was a pure Aryan. Of course, there's no proof that Simple John was ever married in Germany during His epoch so it's quite possible that Simple John is totally unrelated to his sister. It's been noted that Simple John has never replied to your kind author to week whose work is the DNA of both tame animals and wild, so I can go back and get her address (ah! UC - Davis - Simple John may well be able to get a California grant to spend the time at Davis necessary to attempt to type his species and gender). -------------------------------------------- Animal Forensics KXTV-TV (Sacramento): 6/8/04 A forensic science lab at UC Davis deals with animal samples. The UCD Veterinary Genetics Lab helps criminal investigations across the globe by analyzing non-human evidence found at crime scenes. The Lab once even got a call from Scotland Yard: blood samples from a murder scene were found to be from something other than a human, and they needed help identifying the mystery witness. For years, the Davis lab specialized in horse DNA analysis to certify parentage for registries like the American Quarter Horse Association. But in the last seven years the lab has increasingly used its expertise to analyze animal DNA as part of criminal investigations. The first year they worked on six forensic cases and by 2001 their caseload was five times heavier. Now, about half their cases are criminal cases. Cases have included a murder in Indiana, dog and llama abuse cases in Florida, cattle rustling in Arizona and big horn sheep poaching in Montana. Several have involved the analysis of blood samples taken off a road to match the DNA profile of an injured or dead animal that was dragged behind a car. Dr. Sree Kanthaswamy, a primate population geneticist, is the lab's director Forensics is a science that applies medical facts to legal problems. Often, police have a good idea who committed a crime but they haven't been able to establish a clear connection between that person and the crime scene. Forensics labs the one at UC Davis can help them establish the connection by medically proving someone was at a crime scene by using DNA. DNA can be obtained from blood, urine, saliva, hair, semen, bone, teeth, muscle, internal organs, and feces. Animal forensics applies these same principles. After years of experience, the Veterinary Genetics Lab forensics unit has become expert at reliably extracting animal DNA from peculiar sources, such as a single drop of blood in a puddle of mud, or a minute spot of dried blood on a piece of wood that has been left in the sun, or fecal matter from a shoe, even dried saliva from torn clothing -- sometimes saliva appears as a faint silver shimmer. To get these samples, forensics specialists swab or scrape until they have enough material for testing. One thing that really sets the lab apart from a typical forensics lab that deals with human evidence is its large database of DNA data on a variety of animal species. Most forensic labs only have DNA data on one species: human. This data can make or break a case. For example, with a dog mauling case someone may accuse a certain dog. It is the lab's responsibility to test the accused dog against some of the dog's DNA, from hair or saliva that was left on the mauling victim. -------------------------------------------- Quite obviously, this is the place to have Simple John's DNA analysed and, perhaps, he could, again, seek to see if he still has that unusual re-action to the diploid cell Rabies Vaccine. As well, perhaps, he could get colonic therapy there as he is known to have a severe problem with waste material in his colon which he can't eliminate. Do you remember Plain John ever answering what year his sister was offered German citizenship? Gray Shockley ------------------ Yellow Bushie, Yellow Bushie; You're lying through your tushie. ==== Subject: Re: A Definition of an Algorithm > The set of all programs is partitioned into equivalence classes. > Two programs are equivalent if > they are ``essentially'' the same program. > At least one problem with your analysis is that you first require that > the algorithms be written explicitly as primitive recursive functions > before you can decide whether the algorithms are the same. This > doesn't match the usually meaning of algorithm, in which it is possible > for one algorithm (insertion sort, say) to be written in many different > programming langauges: LISP, C, Java, etc. That's why the Turing Machine standard is used. > These implementations of > the same algorhtm would be different: some might have explicit loops, > some might have no loops at all, using only recursive function calls, > but they would still be the same algorithm in the mind of the > programmer. After all, the programmer would probably say that she > So a robust definition of algorithm must be neutral to the choice of > implementation language. This is a significant obstacle, which is why > the authors you cite in your paper don't attempt to do it. > To see what you are up against, look at the page > http://www.roesler-ac.de/wolfram/hello.htm . All the programs there are > the same algorithm: display ``Hello World'' on the output. This shows > how different programs implementing the same algorithm can be. > Proginoskes replies: >Maybe this should be: Two programs P1 and P2 are equivalent if for all >inputs, P1 and P2 produce the same output ? > This is usually described by saying P1 and P2 compute the same > function. This extensional version of equality is usually viewed as > too coarse to define algorithms. For example, insertion sort and > quicksort are different algorithms, but they both sort. > You can do more devious things. Are the following the same algorithm? > They take a natural number as input and return a natural number as > output. > 1) On input x, return x. > 2) On input x, factor x into a primes. Then return x. > 3) On input x, see whether there is a counterexample to Fermat's Last > Theorem a^n + b^n = c^n with a,b,c, and n >2 all less than x by testing > all possibilities. If you find a counterexample, go into an infinite > loop. If not, return x. (Note that because there is no counterexample > to Fermat's Last Theorem, this will always return x.) > 4) On input x, send a post to sci.math containing x, then return x. > Most computer scientists would say that those descriptions are > ``different algorithms'' but that they compute the same function. So are 1) - 4) essentially the same program? You didn't answer my question about what that key word means. --- Christopher Heckman ==== Subject: Re: A Definition of an Algorithm The problem of giving a rigourous definition of algorithm that in some sense agrees with the use of the word algorithm is informal English is very difficult, and there has been little progress on it. I emphasize that it is not a mathematical question but a philosophical one to decide if a certain definition agrees with the informal use. Several approaches to defining the word algorithm are in disfavor. These approaches are formally consistant, but they do not correctly capture the informal meaning of the word algorithm. First, requiring the algorithm to be written in a specific language or computing formalism (LISP, Turing machines, etc). Informally, we know that the same algorithm can be implemented in different languages, so it isn't satisfactory for a general definition of algorithm to consider only one language or formalization. Another approach that is not favored is to define two programs to have the same algorithm if they compute the same function. There are often many different algorithms (in the informal sense of the word) that compute the same function. For example, there are at least a dozen ways of sorting a list of numbers, all of which will compute the same function. But nobody usually says, for example, that insertion sort and quicksort are the same algorithm. I don't have a reference handy, but I wager that any good introduction to computability theory for computer scientists (the type of book that includes discussion of regular and context-free languages as well as Turing machines) will have at least a short discussion of the difficulty of giving a formal definition of the word algorithm that agrees with informal use. They will go on to discuss the Church-Turing thesis, cite this as authority that the formalization they use at least includes all possible algorithms, and then formally develop things using that one formalism. I bet Knuth has good comments in the Art of Computer Programming as well. ==== Subject: Re: Graph Theory and Combinatorics Forum > Those who are interested in Graph Theory and Combinatorics, interesting > problems, brain teasers, papers, etc, Please join: Or you could just post to alt.sci.math.combinatorics ... --- Christopher Heckman ==== Subject: Hamming weight Is there book which well describe hamming weight and hamming weight properties? ==== Subject: Re: Hamming weight > Is there book which well describe hamming weight and hamming weight > properties? Yes, most books on codes. Google for it, lots on line too. The theory of error correcting codes is a good one, perhaps too advanced. ==== Subject: Re: how to inverse a highly symmetric matrix <43EB1684.4050107@yahoo.com> Your matrix is the quadratic form of just two vectors (0,1,1,1,1,1,0) (0,i,0,0,0,i,0). Therefore, it has just two nonzero eigenvalues, and it is hopelessly degenerated to have any inverse. Kunz > It's not clear exactly what the form is. Do you just mean you have a > 2n x 2n matrix A whose entries are 0's and 1's and it has the symmetries > A[i,j] = A[2n+1-i,j] and A[i,j] = A[i,2n+1-j], or is there more to it? > no, that is all > The fact that the entries are 0's and 1's is not likely to be particularly > useful for the pseudo-inverse. > ok > On the other hand, the symmetry should > help. By reversing the order of the second half of the indices for rows > and columns, you get a matrix of the form > [ A A ] > [ A A ] > and if I'm not mistaken the pseudo-inverse of this should be > [ B/4 B/4 ] > [ B/4 B/4 ] > where B is the pseudo-inverse of A. > Reverse the order of the second half of the indices for rows and > columns of this, and you get the pseudo-inverse of your original > matrix. > great, I'll try this. It reduces the size of the original matrix by a > factor 4 and I may be able to go a bit further working only on A. > Borg. ==== Subject: Re: orthogonal transformation changes eigenvectors of covariance? This is fantastic!! Please allow me to ask another question about this problem. What if we do not know the exact value of A? If our purpose is to find out what A is, is it possible to construct an appropriate W and V such that W = AV and we can conclude the A is W V^{-1} ? Roy ==== Subject: Re: orthogonal transformation changes eigenvectors of covariance? >This is fantastic!! >Please allow me to ask another question about this problem. >What if we do not know the exact value of A? If our purpose is to find >out what A is, is it possible to construct an appropriate W and V such >that W = AV and we can conclude the A is W V^{-1} ? For any orthogonal matrices W and V, A = W V^(-1) is an orthogonal matrix. If the columns of V are eigenvectors of C_X, i.e. C_X = V D V' where D is a diagonal matrix, then the columns of A V = W are eigenvectors of C_Y = A C_X A', with the same eigenvalues, i.e. A C_X A' = A V D V' A' = W D W'. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada ==== Subject: Question: Linear Mappings Don't know what I'm doing wrong here ........... ------------------------------ Let L: R3-->R3 be the linear mapping defined by : L{(a_1, a_2, a_3)} = A* [(a_1, a_2, a_3)], where the matrix A is given by: A=[(5,-12,3),(3,-10,3),(6,-24,8 )] 1) Show that L(y)=2y is a subspace of R3. 2)Find a basis {h_1, h_2} for S. 3) Find the matrix which represents L when the basis {h_1, h_2, (1,1,2)} is used in BOTH the domain and codomain. ------------------------------- OK, 1) done. 2)done. we have (A-2I)y=0, so after some green-belt kung-fu, we get,{h_1, h_2}={(4,1,0), (-1,0,1)}. 3) L(a_1, a_2, a_3)=(2a_1, 2a_2, 2a_3) So, L(4,1,0)=(8,2,0) L(-1,0,1)=(-2,0,2) L(1,1,2)=(2,2,4) So, we form the partitioned matrix [(4,-1,1|8,-2,2),(1,0,1|2,0,2),(0,1,2|0,2,4)] we turn the left matrix to the Identity matrix and obtain the matrix rep of L in the right matrix. What I get is a diagonal matrix with identical entry of 2; this is diag(2,2,2) matrix. The answer though is diag (2,2,-1) !! ==== Subject: Re: Question: Linear Mappings > Don't know what I'm doing wrong here ........... > 3) L(a_1, a_2, a_3)=(2a_1, 2a_2, 2a_3) > So, L(4,1,0)=(8,2,0) > L(-1,0,1)=(-2,0,2) The following is not true, if you compute A*(1,1,2)^T, you get (-1,-1,-2), which coresponds to the answer you mentioned as correct. > The answer though is diag (2,2,-1) !! ==== Subject: Re: Question: Linear Mappings guys, would really appreciate some help here. V > Don't know what I'm doing wrong here ........... > ------------------------------ > Let L: R3-->R3 be the linear mapping defined by : > L{(a_1, a_2, a_3)} = A* [(a_1, a_2, a_3)], where the matrix A is given > by: > A=[(5,-12,3),(3,-10,3),(6,-24,8 )] > 1) Show that L(y)=2y is a subspace of R3. > 2)Find a basis {h_1, h_2} for S. > 3) Find the matrix which represents L when the basis {h_1, h_2, > (1,1,2)} is used in BOTH the > domain and codomain. > ------------------------------- > OK, > 1) done. > 2)done. we have (A-2I)y=0, so after some green-belt kung-fu, we > get,{h_1, h_2}={(4,1,0), (-1,0,1)}. > 3) L(a_1, a_2, a_3)=(2a_1, 2a_2, 2a_3) > So, L(4,1,0)=(8,2,0) > L(-1,0,1)=(-2,0,2) > L(1,1,2)=(2,2,4) > So, we form the partitioned matrix > [(4,-1,1|8,-2,2),(1,0,1|2,0,2),(0,1,2|0,2,4)] > we turn the left matrix to the Identity matrix and obtain the matrix > rep of L in the > right matrix. What I get is a diagonal matrix with identical entry of > 2; > this is diag(2,2,2) matrix. > The answer though is diag (2,2,-1) !! ==== Subject: Region enclosing regions - mathematical proof Hi all, Is there a mathematical proof that can determine if a region (composed of 3 or more vertices, and not necessarily a regular shape) encloses another region? I'd like to program an algorithm but realised I didn't know how to do it! With best wishes Paul ==== Subject: Re: infinity sci.math_20050525_e: The set of all sets is its own powerset. There are some relatively basically show a variety of things, eg via well-ordering, domain, and transfer principles, that infinite sets are equivalent. Cantor's powerset or antidiagonal results, or separately the proof about complete sets dense in the reals such as the reals, are simple and accessible, and largely they stand by themselves. ==== Subject: Re: Looking for student solution manual > Hello Group, > Does anyone have the student solution manual for the book Methods of > Real Analysis by Goldberg, 1964 or 1976 Editions? I'll pay for any > pointers. At www.loc.gov a search on the title Methods of Real Analysis turns up the two editions of the text, but no other materials. Perhaps the manual you want does not exist. David Ames ==== Subject: FREE join We Don't Leave Anything Up to Chance. Just plug into our proven 3-Step Wealth Generation System. 1. No Selling! 2. No Meetings! 3. No Pressure! FREE join than see the website http://favorjean.probuilderplus.com/ ==== Subject: Re: Geometric Transformation Question > Given a point P in the plane, and lines l and m not passing through P, > is there some simple way of constructing points Q and R (on l and m, > respectively) such that P, Q and R are the vertices of an equilateral > triangle? I have two methods with straightedge and compass. One uses the required Geometric Transformation method, so in case of being homework, I just give you hints : In an equilateral triangle PQR, what kind of transform changes Q into R, keeping P unchanged, and being specific to PQR being *equilateral*, that is PQ=PR and angle = 60 deg ? Then consider (L) as being the locus of Q, then the locus of R is the transformed of (L) by the above transformation. But the locus of R is (M). Hence you can construct R as being the intersection point of (M) and of the transformed of (L). Then apply the reverse transform to get Q. And carefully discuss for specific cases (parallel lines, lines being at 60 deg etc). This method could as well be used analytically, but seems quite tedious. The other one uses an auxiliary line from P to change the problem into the following : Given a triangle ABC (BC = line l, AC = line m) and a point P on AB (the auxiliary line), find two points Q and R on the sides AC and BC with PQR equilateral. Draw some lines at 60 deg and one circle, and use the property of inscribed angles in a circle. But this method not using any geometric transform is off topic. -- philippe mail : chephip at free dot fr site : http://chephip.free.fr/ ==== Subject: Re: Geometric Transformation Question >> Given a point P in the plane, and lines l and m not passing through P, >> is there some simple way of constructing points Q and R (on l and m, >> respectively) such that P, Q and R are the vertices of an equilateral >> triangle? > Geometrically, take a compass and draw a circle centered at P of radius r. > This circle will intersect > (a) neither lines l nor m if the radius r is smaller than the distance > from P to these two lines > (b) one of the lines, say l, if the r is larger than the distance from P > to l but smaller than distance P to m > (c) both lines if r is larger than the larger of the distances P to l > and P to m. Call the minimum such radius r_0. > Any circle of radius > r_0 centered at P intersects the lines l and m at two > points each. Choose one of each pair and you have your three points P, Q, > and R. Of course not. Your method gives just *isosceles* triangles PQ=PR (infinitely many) Only two specific radii will result into equilateral triangles PQ=PR=QR. > Analytically, you can compute d_l and d_m as the distances from the point P > to the two lines l and m respectively. Chose any r > max( d_l , d_m ) and > you have a circle radius that satisfies the geometric conditions above. You > can then form the quadratic equations that define the 2 points on each line > that are the distance r from the point P. > Norm That is analytically you must go on with computing the distance between these two points F(r, data) and solve equation F(r, data) = r. (two tedious for me, without symbolic math software) Another method would be to draw any auxiliary line from P, intersecting the given lines in B and C to build a (scalene) triangle ABC, A being intersection point of the two given lines. Then write in the complex plane P,Q,R on the three lines : P = u*B + (1-u)*C Q = v*A + (1-v)*B R = w*A + (1-w)*C and PQR equilateral : P + j*Q + j^2*R = 0 (direct) or P + j*R + j^2*Q = 0 (retrograde) with j = e^(i*pi/3) -- philippe mail : chephip at free dot fr site : http://chephip.free.fr/ ==== Subject: Re: Convergence for fourier and taylor series On 20 Feb 2006 12:11:37 -0800, David Kanter On 20 Feb 2006 02:09:27 -0800, David Kanter Hey folks, >>I have an interesting sort of problem I've been thinking about. >>For a given class of functions, is it possible to say whether a fourier >>transform or a taylor series converges quicker? >> That would depend on what class of functions you're >> talking about... >Does this mean that there is no generalized formula or theorem about >the error of a fourier approximation? Huh? Of course this doesn't mean that. There are many results talking about the error in various sorts of fourier approximations. And whether or not it's possible to say whether a fourier approximation converges faster than a taylor series for functions in some class depends on what class you're talking about. >I certainly remember quite a >few theorems to calculate the error of a taylor approx. David C. Ullrich ==== Subject: Re: Cantorian pseudomathematics <271d5$43f19e0c$82a1e228$5729@news1.tudelft.nl> you sir, show us poor dumb folk another one. How about something like: > f(x) := (x^2) * sin (1/x) > and let f(0) := 0 (you'd agree I'm sure by your calculus methods that > this is the sensible value). Later. > Now tell me what f'(x) is, and no cheating now, then tell me what f'(0) > is. And none of this insensible discontinuous nonsense. BTW, yes f is > a differentiable function everywhere and has an everywhere existing > derivative and yes, the value at 0 is very sensible. And no. I can NOT see that f'(x) = 2.x.sin(1/x) - cos(1/x) has a sensible value for x = 0 , because of -1 < cos(1/x->0) < +1 . Han de Bruijn ==== Subject: Re: Cantorian pseudomathematics > We are all in awe of your mathematical capabilities. Please I beg of > you sir, show us poor dumb folk another one. How about something like: > f(x) := (x^2) * sin (1/x) > and let f(0) := 0 (you'd agree I'm sure by your calculus methods that > this is the sensible value). > Later. > Now tell me what f'(x) is, and no cheating now, then tell me what f'(0) > is. And none of this insensible discontinuous nonsense. BTW, yes f is > a differentiable function everywhere and has an everywhere existing > derivative and yes, the value at 0 is very sensible. > And no. I can NOT see that f'(x) = 2.x.sin(1/x) - cos(1/x) has > a sensible value for x = 0 , because of -1 < cos(1/x->0) < +1 . > Han de Bruijn But 0 <= | [f(0+h) - f(0)]/h | < |h| for all h, so f'(0) = 0. ==== Subject: Re: Cantorian pseudomathematics >> We are all in awe of your mathematical capabilities. Please I beg of >> you sir, show us poor dumb folk another one. How about something like: >> f(x) := (x^2) * sin (1/x) >> and let f(0) := 0 (you'd agree I'm sure by your calculus methods that >> this is the sensible value). >Later. >> Now tell me what f'(x) is, and no cheating now, then tell me what f'(0) >> is. And none of this insensible discontinuous nonsense. BTW, yes f is >> a differentiable function everywhere and has an everywhere existing >> derivative and yes, the value at 0 is very sensible. >And no. I can NOT see that f'(x) = 2.x.sin(1/x) - cos(1/x) has >a sensible value for x = 0 , because of -1 < cos(1/x->0) < +1 . He didn't ask for the value of 2 x sin(1/x) - cos(1/x) at x = 0. He asked for the value of f'(0). And that is very simple to work out: f'(0) = 0. The expression that you obtained is the value of f' at any nonzero x, but the method is invalid at x = 0. To calculate f'(0), you need to return to first principles. ----- ==== Subject: Re: Cantorian pseudomathematics >> We are all in awe of your mathematical capabilities. Please I beg of >> you sir, show us poor dumb folk another one. How about something like: >> f(x) := (x^2) * sin (1/x) >> and let f(0) := 0 (you'd agree I'm sure by your calculus methods that >> this is the sensible value). >Later. >> Now tell me what f'(x) is, and no cheating now, then tell me what f'(0) >> is. And none of this insensible discontinuous nonsense. BTW, yes f is >> a differentiable function everywhere and has an everywhere existing >> derivative and yes, the value at 0 is very sensible. >And no. I can NOT see that f'(x) = 2.x.sin(1/x) - cos(1/x) has >a sensible value for x = 0 , because of -1 < cos(1/x->0) < +1 . Nobody said that 2.x.sin(1/x) - cos(1/x) has a sensible value for x = 0. It's not true that f'(x) = 2.x.sin(1/x) - cos(1/x). That equation for f'(x) is valid only for x <> 0. It's very easy to show from the definition that f'(0) = 0; this has nothing at all to do with 2.x.sin(1/x) - cos(1/x). >Han de Bruijn David C. Ullrich ==== Subject: Re: Online poker RNG... <5ojhv118jvd2f6frtvhrt94h3op1401ohm@4ax.com> So, in short, yeah, I see your point...but I don't know if there's a >> direct link between the fact that the software is advertised as a >> cheating aide and the fact that the authors are necessarily lying >> about the software's ability to deliver as promised... >The programmers who come up with ways to generate random numbers will >use very advanced methods to come up with an outcome. The main fact >that they fail to realize is, they always have to instruct the computer >on how to come to a result. The computer will never pick its own result >at random. >Seems like a lizard with no legs to me. You don't really need a programmer to generate a random number. It is quite possible to attach a simple electronic device to your computer which will generate genuine random numbers very rapidly. -- Jeremy Boden ==== Subject: Re: Online poker RNG... On Mon, 20 Feb 2006 06:25:00 -0600, David C. Ullrich >On 19 Feb 2006 12:36:20 -0800, Gerry For me, David, collusion-detection is part scientific (i.e. I own >>software which tracks the number of times particular players are >>involved in hands together, and looks for markers of collusion -- >>raises and reraises in an effort to isolate a mark and drive him to >>fold, etc., etc.) and part from-the-gut (based on years of playing B&M >>and online poker). Certainly, incontrovertibly, your statement (I'll >>paraphrase it as: Good colluders know, as part of their colluding, how >>to minimize the risk of detecion.) is true. But it isn't my particular >>worry, wrong or right. >>My particular worry is that, as was shown possible in 1999 >>(http://www.cigital.com/papers/abstracts/developer_gambling.html), >>someone can know my cards... >>Again, the authors of pokerrng (at pokerrng.com) make this claim, and I >>have no way of knowing whether it's true that their product is snake >>oil or that my fear of it is well-grounded. >>None of you (and I mean this as a complimentary term, and one of >>respect!) math geeks are curious? >Could be just that nobody knows anything about the question. >I've already said that it seems to me that yes, a person could >predict things after a certain number of cards, _if_ the person >knew exactly what algorithm was being used. Reverse-engineering >the algorithm from records of past hands seems like it would >take a _lot_ of hands - do they really publish records of >_billions_ of previous hands? >Looking at that url for the first time I see that sure enough, >the problem started when a site published the shuffling algorithm - >that seems like a bad idea to me. I haven't looked at the pdf - >if I think of it when I get to the office I will. For some reason I thought it was a large pdf. I have to say the first part, about the algorithm that site was using, was just hilarious. They mention two big errors, one of which I spotted immediately - the shuffling algorithm simply does not give a uniform shuffle, even if we assume a perfect rng. There's another error that was obvious to me that the authors don't mention explicitly: the code calls randomize much too often! People get the idea from the name of the function that calling randomize as often as possible will give more random results, but in fact the opposite is true. You've said a few things here that I've been skeptical about - the paper does not support those assertions, as far as I can see. In particular, the attack they describe is _not_ based on reverse-engineering the shuffling algorithm, it depends on _knowing_ what algorithm is being used! The idea that an attack as they describe could work if the attacker knows the details of the algorithm is something I haven't disputed. (They give a few versions of the attack; the extremely efficient one depends on the fact that the seed is based on the current time, and we can determine what the current time was within a few seconds. Note that if the algorithm didn't call randomize so often then this aspect would not work, we'd be back to hundreds of millions of seeds to check instead of thousands.) Anyway. They describe well-known things that a poker site could do to prevent this problem. You seem to want someone here to tell you whether these attacks are possible on currect poker sites. It's impossible for anyone here to answer that: the answer depends on whether the site is doing things right, and that's not a mathematical question. David C. Ullrich ==== Subject: Re: Online poker RNG... [Gerry] > Just for the sake of argument (I'm unskilled in math, not in logic.), > let's assume hypothetically that it's possible, through (a) creative > coding and (b) distributed processing, a real-time way to crack the > code of the deck. Yes, the author would likely use it to make money. > And...then what? The author decides to not use other avenues to make > money as well, such as *selling* the product? Seems to me that, just as > with any business which cross-promotes or sells bundled customer info > to ad agencies or whatever, the author of the software would seek to > maximize his gain on his great product. Now, if we take those two > assumptions (The first assumption -- i.e. that it can be done -- is my > original question, right? The second assumption -- that the author > would seek to make money through sales -- is certainly debatable, but > can't be dismissed as ridiculous.) as true, for the moment, then the > question is: HOW does the author sell it? First of all, while a working > system like that could make unlimited money, and therefore might be > worth say, $1M to a customer, that doesn't mean there are customers > able to afford the original $1M investment, right? So perhaps the > author hires McKinsey to find out what people would pay for software, > gets the answer as $475, and offers for sale a VERSION of his perfect > system which is substantiall less than perfect but still worth the > $475. Remember that pokerrng purports to give only three of the five > community cards (not all five), that it doesn't give the suits of those > cards, that it gives only two opponent hands (not all 9 at a full > table) and that it doesn't even identify which players hold those two > hands! The information it gives provides an extraordinay advantage, but > there's still a substantial element of chance. So, that might be one > reason that the version would sell for $475. > Another reason for selling a stripped-down version (at a stripped-down > price) might be the fact that, if the full system relies on distributed > processing, the author might recognize that only a much less > sophisticated system can be run on the standard computers folks have at > their homes! > How about another explanation, if the two above aren't enough? How > about if (Think, here, of a bookstore which legally sells The > Anarchist's Cookbook) it is illegal to ACT in a certain way, but not > illegal to provide the INFORMATION necessary to act in that way! After > all, I can legally teach martial arts, including maneuvers designed to > produce lethal results, and, while I'd be held responsible if I > performed such maneuvers on, say, my mother-in-law, I'd be within the > bounds of the law in teaching you said maneuvers, even if you > subsequently used them on YOUR mother-in-law. (PLEASE NOTE: Before > deconstructing the previous sentence and telling me all of the ways > that the law holds certain people responsible for the actions of > others, recognize (a) that I'm a lawyer and that your lawyerly argument > will get a lawyerly response, and (b) that the example was simply for > illustrative purposes of the larger point.) I guess you're responding to my message. If so, take care you don't end up a case study in how someone who thinks they're too smart to be scammed talks themself into it ;-) > So, while I take your sarcasm in the good-natured spirit in which it > was intended, I hope that you see that it's not really well-founded. > There are plenty of reasons why a working product would be sold. Just noting that such pitches (and whether aimed at gambling, stocks, or real estate) usually come with an entertaining explanation of how the system actually works _better_ when more people use it. I missed that bit here. It's usually included because too good to be true rings warning bells. > Sure, I could spend $500 and a few hundred hours of data entry to test > the theory, but the money isn't refundable You didn't see the 100% Satisfaction Guaranteed logo on the website, or you did but didn't believe it, or you did believe it but noted that the absence of a contact address might make it hard to _ask_ for a refund, or ...? The made-up Verified Merchant graphic on the site should erase all doubt :-) Pretend you're a judge? There's all kinds of evidence on that site. > and neither is my time. I don't know what value you place on your time, but, seriously, if I had spent as much time typing about this as you have so far on sci.math and sci.crypt, I'd consider a few thousand dollars a bargain to cut it short. > I could also test whether a ladder conducts electricity by standing on > it while grabbing live wires, but I'd rather ask the expert in the store > what material comprises the ladder; similarly, I had a question about > math, and I came to a math forum to ask it. As far as I can tell, you've been answered here, and on sci.crypt, consistently wrt your technical questions: if a site uses good practice for their RNG generation, then no, the RNG can't be cracked. _Best_ practice would be to use a physically random source, which can't be predicted regardless of processing power (unless modern physics is fundamentally wrong). It's quite feasible to do that -- you can buy special HW to generate truly random bits, and you can even grab some for free (albeit in small quantities) from web services; e.g., http://www.fourmilab.ch/hotbits/generate.html A reaonable way to use truly random bits (like those) is for seeding a crytographically secure software RNG, where crytographically secure means it's computationally intractable (but not theoretically impossible, as for truly random bits) to predict future bits (Google on the phrase for more about that, but forewarned that it's a highly technical topic -- analogies with household repair break badly against this rock). > And as for your statement that [the question of] whether the owners of > the sites [I] like to frequent took [the lesson of the 1999 study] to > heart isn't really a technical question, you've made an entirely > straw-man argument. I did NOT come here and ask: Can someone on > PlanetPoker take their published algorithm, synch to their system clock > and derive a good prediction of the cards to come (which is EXACTLY > what happened in the 1999 case. Rather, what I asked, in essence, is: > In 1999, as reported in a published study, there were vulnerabilities > in some online poker sites, and cheats could have exploited them. Are > there similar ways to do that now, which is what is touted by the > authors of pokerrng (for example), and is the possibility significant > enough to be a point of concern? I believe these have answered. The answers remain no and no _provided that_ a site uses good RNG practice. The answers remain yes and yes if a site does not use good RNG practice. Since I believe these answers have been given repeatedly, I assumed you were asking some other question(s) too, and took a guess at what they might be. > Is my question REALLY not a question of mathematics??? I mean, sure, it > also comprises issues of programming and cryptography, but isn't the > methodology behind PRNGs a branch of your science? And are you REALLY > suggesting that, just because online poker rooms might have fixed > their methodology so as to not be crackable by that 1999 process, I > should assume that there's been no progression in programming, in > statistical analysis, in cryptography or in the validation of > mathematical proofs such that even the better systems can STILL be > cracked??? Yes, you should assume that a crytographically secure PRNG cannot be cracked, unless people know things about it you've said you don't want to consider (e.g., that they've somehow tapped into the system and are able to read the random bits with which it was seeded). The definition of cryptographically secure comes from computational complexity theory, and _means_ nope, there's no computationally tractable way to predict the next bit at better than chance level. It's dimly possible that a pokerrng programmer made a theoretical breakthough that the best minds in the field don't expect, like it's also dimly possible that all sites still use the RNG described in the 1999 > I'm astounded -- even confounded -- by the reaction that my question > has received. For one thing, I, personally, find it admirable when a > person admits to a level of ignorance and seeks tutelage so as to shore > up his base of knowledge, and that's all I did by coming to your forum! > For another thing, I, personally, find it gratifying to be able to > impart some of my knowledge (in the areas in which I have it) to those > who seek it, and find it hard to grasp why my request for help should > receive ANY level of sarcasm or derision. > If you review the answers to my question, you see all kinds of > irrelevancy, ranging from variations on: You shouldn't gamble to > THAT isn't your biggest concern; your biggest concern is collusion {or > whatever}. If it's any consolation, the last time I asked a lawyer a simple question, I was subjected to several thousand dollars' worth of hypothetical worst-case scenarios :-) She thought they were things I needed to know, and I was happy to pay her for not giving a literal response. > But if I'd come to this math forum and posed the following question: > Say, my neighbor is about to reshingle his roof, and wants to know > whether there's an easy way to figure out how long of a ladder he needs > in order to reach the edge of his roof, given that he'll be placing the > base of the ladder exactly 12 feet, on a level surface, from the base > of the house, and given that the roof's edge begins exactly 13 feet up > the perfectly plumb exterior wall of his house?, my supposition is > that someone would've given me a quick answer, possibly with a basic > lesson on the Pythagorean Theorem. I *highly* doubt that I would have > gotten answers like People shouldn't reshingle their own roofs or > Why worry about his roof? It's far less likely that he'll have a leaky > roof than that he'll lose energy through inefficient windows, so I'd > have him focus on that. I'm sure you can figure out the difference. > I just don't understand why some of this forum's members behave as they > do. By definition, my question is either answerable or not. If it is, > then, by definition, there are either members on here who know and can > explain the answer, or there aren't. If it IS answerable, and if people > CAN answer it, then the only two action which are acceptable, I think, > within the framework of human interaction are (1) Answer the question > while treating me with basic respect and (b) Don't answer the question. > Anything else is just...rude. Do you believe that your question is still unanswered? ==== Subject: Re: Online poker RNG... <5ojhv118jvd2f6frtvhrt94h3op1401ohm@4ax.com> <-v6dnSRKl4YFr2fenZ2dnUVZ_vydnZ2d@comcast.com> Got my answer now, certainly, and I genuinely thank those who -- like you -- took THEIR time to give the answer. (And not only to give a technically correct answer, which is nice enough to begin with, but to translate from math-speak to lay terms.) But -- and I won't ramble with this, because stuff is already there and said -- I think a review of a lot of the posts from sci.math and sci.crypt have included, as I mentioned, either jabs at me for the question itself (as though it were inherently stupid, which it wasn't. It came from a point of *ignorance*, but that's different from being stupid.), or for having been concerned with the topic -- as if a mathematician necessarily had an ability to tell a professional poker player what the player's actual concern ought to be, especially when the suggestion had nothing to do with math!! (If someone wanted to give unsolicited advice to help a poker player, the mathematician might discuss the reasons to play at different limits, or at sites with different rake arrangements, or whatever.) In essence, I just felt -- and, again, there were ABSOLUTELY several people who shared their knowledge and expertise, and to whom I'm grateful for having done so -- that there was a lot of static which wasn't warranted, andI guess I'm just sensitive to the issue of trying to help people because I'm attempting to raise a son who will have that attitude, and I'd be pretty bothered if he grows up to have a font of knowledge but refuses to share it when someone asks for help. who might've overcharged. You know that we're all scumbags, don't you? As for my time, incidentally: I type more than 75 wpm, so even my long posts are short. For you? For you, I charge $250/hr. :) ==== Subject: Re: Online poker RNG... > Is it possible that, as I'm playing online, any of the other players > are cheating by way of predicting the order of the cards in the > shuffled deck? It's highly unlikely, given the wake-up call potential investors received years ago from Paradise. Methods now are secret. And, at, at least, one site, that I know of, the code is checked and certified as generating pseudorandom deals by paid, qualified, outside agencies. > A few (some math, some poker) truths to keep in mind as you answer: > 1. Many sites use an RNG algorithm based on 2^32. Isn't that pretty > easily susceptible to brute-force attacks from someone who has a > program relying on distributed computing? This aspect of running a site is crucial. Realize that in brick-and-mortar games, the (hand-shuffled) deals are almost always biased (if the table is not using a Shufflemaster). In fact, true random deals are much more likely to be encountered on-line. So, the casino player will feel a difference. Hence, it is critically important for on-line casinos to get it right, since it will be different. > 2. In the most popular type of online poker (Texas Hold 'Em), there are > four rounds of betting, and the most important three rounds come AFTER > (a) each player gets two (face-down) cards and (b) three face-up cards > are dealt on the table -- SO, if the cheateer is, say, the third player > to be dealt cards at a table of ten players, then he will know the 3rd > and 13th cards (i.e. the two cards he's been dealt) as well as the > 21st, 22nd and 23rd cards (the three dealt face-up immediately > following the two distributed to each player). Give it a rest. There is always the possibility of cheating by two or more players, which is more likely in a brick-and-mortar setting, since casinos don't give much of a darn about anything except the rake. Collusion does not immediately affect the bottom line. Someone raised the specter of cheaters who are so good, their actions would be undetectable at the table. Well an on-line site such as Poker Stars, has ways of detecting such collusion forensically, which is impossible in a B&M setting. Another inhibiting factor, keeping good players from cheating, is that, if exposed, they become outcasts in the community. Few partnerships last forever. In my experience, by the way, in Atlantic City, there was collusion. > So, I guess my question is: how easily would someone be able -- either > by brute-force run-through of all possible combinations of deck > arrangements with those 5 cards in those specific spots, or by some > sort of reverse-engineering, or whatever -- to conclude the sequence of > cards in the deck (which, of course, would give him knowledge of > everyone's cards, as well as the two cards left to be dealt)? > And, for that matter, a related question: there's a piece of software > currently on the market (Check out pokerrng.com.) which claims that, if > a player enters into the software the details of a few thousands > recently dealt hands, the software will synch with the poker system's > RNG well enough to predict three of the five cards (not suits, but > values, which is enough to be dangerous!) left to be dealt! Is this > even plausible? Do I need to be afraid of this type of thing? > Anything online has inherent pitfalls, and cheating is -- both online > and offline -- a fact of life, but there's a certain level of security > I'd like, and I wonder if I'm jumping at shadows or rightly > concerned... You are jumping at shadows. ==== Subject: Re: Online poker RNG... Hey, N. Silver, Did you really write Give it a rest, in response to part of my -- Jeez, I've written this line so many times in the past three days, my fingers are numb -- simple request for an answer from someone with a better understanding of math than I have? WTF is your problem? Seriously. What in the WORLD (I think you can sense my growing frustration here.) can possess someone to be insulting toward anyone who comes to a gathering of scientific experts and asks with honesty and humility for a bit of the insight from their shared knowledge? Who raised you to speak to strangers like this? My god, WHAT IS WRONG WITH MOST OF THE PEOPLE WHO RESPONDED TO MY POST????? Yes, there were a few people who, out of kindness or generosity or the spirit of shared education, gave a mathematical analysis to my question. But YOU, and the others LIKE you, who deign to speak to another human being that way? You should be ashamed. And here's the sad part: while it's a generalization, of course, which means not accurate for all, it's true in large part that those in this world (well, America, at least) who exhibit facility with math and the hard sciences are treated as social outcasts by the fraternity guys and the cheerleaders. And that behavior is deplorable. But no more so than using your superior knowledge -- in THIS ONE FIELD (After all, I have no reason to believe that you are any smarter than I, nor any better versed in any, other subject than math.) -- to denigrate me. And what had I done to deserve it? I asked a question, substantiated with logical rationale, and requested an answer from someone with better training. How horrible of me... Give it a rest you say. And to you I reply: Grow up. And learn some manners. ==== Subject: Re: Online poker RNG... Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >Hey, N. Silver, >Did you really write Give it a rest, Give it a rest was a rather mild comment in response to what the poster considered your over-reaction. And in the context of his helpful and informative message there was no reason to take offence at it. Your post on the other hand was extremely rude. -- Richard ==== Subject: Re: Online poker RNG... On 20 Feb 2006 19:53:14 -0800, Gerry Did you really write Give it a rest, in response to part of my -- >Jeez, I've written this line so many times in the past three days, my >fingers are numb -- simple request for an answer from someone with a >better understanding of math than I have? >WTF is your problem? Seriously. What in the WORLD (I think you can >sense my growing frustration here.) can possess someone to be insulting >toward anyone who comes to a gathering of scientific experts and asks >with honesty and humility for a bit of the insight from their shared >knowledge? >Who raised you to speak to strangers like this? My god, WHAT IS WRONG >WITH MOST OF THE PEOPLE WHO RESPONDED TO MY POST????? Um. I haven't seen anyone being insulting here. Except for you, asking people what's wrong with them. Hint: In general, when it seems like almost everyone has a problem, you should consider the possibility that the problem is actually yours. >Yes, there were a few people who, out of kindness or generosity or the >spirit of shared education, gave a mathematical analysis to my >question. >But YOU, and the others LIKE you, who deign to speak to another human >being that way? You should be ashamed. >And here's the sad part: while it's a generalization, of course, which >means not accurate for all, it's true in large part that those in this >world (well, America, at least) who exhibit facility with math and the >hard sciences are treated as social outcasts by the fraternity guys >and the cheerleaders. And that behavior is deplorable. But no more so >than using your superior knowledge -- in THIS ONE FIELD (After all, I >have no reason to believe that you are any smarter than I, nor any >better versed in any, other subject than math.) -- to denigrate me. And >what had I done to deserve it? I asked a question, substantiated with >logical rationale, and requested an answer from someone with better >training. How horrible of me... >Give it a rest you say. And to you I reply: Grow up. And learn some >manners. David C. Ullrich ==== Subject: Re: Online poker RNG... <5ojhv118jvd2f6frtvhrt94h3op1401ohm@4ax.com> <-v6dnSRKl4YFr2fenZ2dnUVZ_vydnZ2d@comcast.com Just for the sake of argument (I'm unskilled in math, not in logic.), > let's assume hypothetically that it's possible, through (a) creative > coding and (b) distributed processing, a real-time way to crack the > code of the deck. Yes, the author would likely use it to make money. > And...then what? Clearly it is best to keep the process a secret (as long as possible). Given any known unfairness, the poker houses will quickly change the prng to a more secure one, rendering the 'cracker' useless. This may be why pokerrng is expected to be 'sold out' in the near future. Rich ==== Subject: Re: Online poker RNG... <5ojhv118jvd2f6frtvhrt94h3op1401ohm@4ax.com> <-v6dnSRKl4YFr2fenZ2dnUVZ_vydnZ2d@comcast.com> True...Though, Occam's Razor suggests that the better reason may have to do with your earlier post; the author of the lizard with no legs might do well to get while the gettin's good... ==== Subject: LINEAR REGRESSION BY METHOD OF LEAST SQUARES ok....i've got my formula set up to read a set of data in either Corel Quattro or Microsoft Exel where n is the number of readings per time period. For example: if we've got 2 readings taken at time A and the other at time B the value of n will be 2. Similarly if there are readings at time A, time B and time C, then the value of n will be 3. Manually manipulating the formula to read the data at either n=2 or n=3 is fine if it's a one time deal, but how can this formula be composed and put into a macro in Quattro or Exel when n=2 vs n=3 vs n=4 and so on? ==== Subject: Re: The gap, why it's important > Well, you know, if you think you're going to take on the world, you might > want to sober up or get off of whatever it is you're on. Besides, anyone who > does a little research on you can see the type of person you are (i.e. not > rational) and how many times you've been wrong. So, why would someone listen > to you versus someone who is formally trained in Mathematics? If you want to > learn about psychology, for instance, then you'd talk to a psychologist; if > you want to learn about math, you talk to a mathematician. > Dave Ahh, that crucial if... *** Free account sponsored by SecureIX.com *** *** Encrypt your Internet usage with a free VPN account from http://www.SecureIX.com *** ==== Subject: Numbers as abstract concepts I believe that concepts, as ideas that unite several objects under one item of recognition, express themselves through numbers. When we see three blue objects, or three circles, it is as if the concept of blue or circle expresses itself through the number three. It is possible, however, that two concepts which express themselves through two objects each can in their conjunction (through the and function) express themselves through three or less objects, as in the overlapping of concepts, but this does not relate to my question. Can numbers rightfully be called concepts of the exact same type as blue and circle, or is this an error? First of all, numbers have the *unique* property I mentioned above: they are expressed through concepts as the idea that unifies several objects into one group to which a concept may apply universally. While the concept of blue can express itself through the number three, as in three blue objects, it cannot express itself through the concept red or circle, in the way I mentioned above. Second, aggregates can be grouped in different ways. Three objects may be considered two objects through the treatment of two objects as a single, unified object. Numbers, then, seem to be more special or pure than other, more empirical concepts. Do you agree or disagree? ==== Subject: Bivariate gaussian volume Is there a 2D equivalent to the erf function (a simple definition of one)? I need to find the double integral of a bivariate gaussian in a rectangular area defined by points (x0,y0) and (x1,y1). I can integrate it on x and get a function that is the product of a 1D gaussian on y and an erf difference on x0, x1 and y, but then I can't integrate this function on y (I tried integration by parts, but I get into some nasty recursions... because erf(z) integral is a sum of z.erf(z) and another gaussian). Edson Manoel ==== Subject: Re: Bivariate gaussian volume Usually when you have to integrate a normal pdf converting to polar or spherical coordinates usually makes things easier. Since you have a bivariate normal, i'd try spherical coordinates, you should not have to be dealing with error functions, just trig integrals. Eric B ==== Subject: Re: Bivariate gaussian volume > Usually when you have to integrate a normal pdf converting to polar or > spherical coordinates usually makes things easier. Since you have a > bivariate normal, i'd try spherical coordinates, you should not have to > be dealing with error functions, just trig integrals. > Eric B Hold on... when you integrate a (1D) normal pdf over an interval you DO end up with the error function, or something equivalent to it, don't you? There's no way around that is there? I thought the conversion to polar coordinates was just a neat trick relevant only to computing the total integral over -infinity to +infinity. I seem to recall from ages ago that there is no way to express a general rectangular integration of a bivariate pdf in terms of familiar functions (including erf). There are no doubt various approximation methods ranging from brute force numerical integration to something a bit more subtle and efficient. Like I say, though, it's been a while so I could be wrong. ==== Subject: Re:Was: The factorization of odd composites using the triangle numbers! As promised, the number of iterations to find the smallest triangle number to satisfy t(n) - rsa100 = a smaller t(n) The search has to go through this many iterations -- 2.839151482941759638047211429676019512903229638844e+48 Note:Also < than the smallest factor of rsa100 So the range can probably be from the 1st triangle number up to the sqrt(n) of any composite (n) with just 2 prime factors. Or could it be up to the smallest prime factor of the composite? I will check into this further with smaller toy composites. Dan ==== Subject: Re: Was: The factorization of odd composites using the trianglenumbers! > As promised, the number of iterations to find > the smallest triangle number to satisfy > t(n) - rsa100 = a smaller t(n) ... > So the range can probably be from the 1st triangle > number up to the sqrt(n) of any composite (n) with > just 2 prime factors. > Or could it be up to the smallest prime factor of > the composite? > I will check into this further with smaller toy > composites. Since you start looking at t(k) where k ~ sqrt(2n), I don't see what you mean by from the 1st triangle number up to ... sqrt(n). That is, the search starts at about 1.41 * sqrt(n), so is beyond sqrt(n) from the outset. As I mentioned in an earlier thread, if n=p*q and t(k)-t(j)=n, you depend on the relation that p*q = (k*(k+1)-j*(j+1))/2 = (k*k-j*j+k-j)/2 = (k+j+1)*(k-j)/2, hence k-j is a factor of 2*n. Given particular p and q, and knowing that p*q = (k+j+1)*(k-j)/2, it is simple to figure out the number of iterations your method takes. The test case in the earlier thread was n=11111= 41*271; your search begins at t(149) and goes up to t(291), and 11111 = t(291)-t(250). Also, sqrt(11111) ~ 105.4. Note that 291-149 = 142 exceeds the smaller factor and the square root, and exceeds half of the larger factor. I think your method cannot bound the number of iterations in terms of sqrt(n) or the smaller factor p. For example, consider p=7, q=100003, n = pq = 700021. You start looking from near t(sqrt(2n)) = 1183.23, so you try in turn the 48825 numbers t(1183), t(1184), ... t(50008), to find that t(50008)-t(49994) = n. It looks like when q >> p, you will test in the neighborhood of q/2 numbers. -jiw ==== Subject: Re:Was: The factorization of odd composites using the triangle numbers! >> Re:Was: The factorization of odd composites using the triangle numbers: >> Is an equation possible? >> I also found in factoring using the triangle numbers this--- >> Toy composite 4351559, finding both factors . >> A little rehash -- >> 4367490 - 4351559 = 15931 a triangle number >> 4367490 = line 2955 >> 15931 = line 178 >> 2955 - 178 = 2777 a prime factor of 4351559 [...] >Okay, here's the $200,000 question (literally): >How long does it take to factor a number N = p*q? >Certainly there is an algorithm which requires at most >sqrt(N) >divisions. >How many triangle numbers T do you have to search >through to find one >such that T-N is also a triangle number? I suspect it >may be sqrt(N). >--- Christopher Heckman Funny you asked that question because I was just thinking about the sqrt(n) this morning but in a different context. More on the line of the relation of the first triangle numbers (line number)> n in relation to the sqrt(n). I will look into your thoughts on this. I will give you the number of line numbers > than the first line number that rsa100 resides on in my next post That will give you an idea on the size of the integer involved thus the number of iterations. Did you figure out how I found the right triangle number for rsa100 to arrive at its largest factor? I am sure you did, but I am just curious! No way would a brute force method be practical on rsa100 because I would probably be waiting (and my ancestors) for hundreds of years to get the smallest triangle number > rsa100 that would work! Dan ==== Subject: Re: Re:Was: The factorization of odd composites using the triangle numbers! > Re:Was: The factorization of odd composites using the triangle numbers: > Is an equation possible? > I also found in factoring using the triangle numbers this--- > Toy composite 4351559, finding both factors . > A little rehash -- > 4367490 - 4351559 = 15931 a triangle number > 4367490 = line 2955 > 15931 = line 178 > 2955 - 178 = 2777 a prime factor of 4351559 [...] Okay, here's the $200,000 question (literally): How long does it take to factor a number N = p*q? Certainly there is an algorithm which requires at most sqrt(N) divisions. How many triangle numbers T do you have to search through to find one such that T-N is also a triangle number? I suspect it may be sqrt(N). --- Christopher Heckman ==== Subject: another Crossover of AT&T to BCE; Optimal Strategy of VonNeumann Gametheory in playing the stockmarket Portfolio of PAF as of 21Feb06 BCE 14,000 $23.61 $330,540.00 BLS 10 T 9,000 $28.40 $255,600.00 VZ 10 total share-wealth-units last reported 22,800 total share-wealth-units today 23,020 realestate land 3APR03 of 3 lots $19,000. science-art of pictures,porcelain etc starting JAN03 for $17,556. realestate land 30JUL03 another lot $11,500. realestate land Sept05 another lot $75,000. Last Friday of 17FEB, I sold another 1,000 shares of AT&T at 28.25 and with the proceeds plus a trifle of cash extras, put in an order to buy 1,200 BCE at 23.61 and had only a partial fill of 1,000 shares. So today of 21FEB I went back and put in an order to buy 220 shares of BCE at 23.55. And so the portfolio today is back above 23,000 share wealth units for which it had sunk below in Sept of 2005 due to my purchase of a piece of realestate. So one can appreciate how effective this Crossover technique is. I gain in two respects. I gain by the increase of money dollar value and also increase in share-wealth-units. By playing AT&T against BCE if one crosses over significantly to the other I sell some of the gainer to buy more of the other and wait for the other to crossover and repeat the exercise. In the case above I netted 220 free shares of BCE, before taxes. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies ==== Subject: Re: Do mathematicians know their axioms? >|>I'll give a concrete example. There was some discussion on the fom >|>board several years back whether Wiles' proof of FLT went through in >|>ZFC. I had no knowledge of the subject so I couldn't participate, >|>but I *wanted* to say, Why don't you just ask Wiles? There would >|>seem to be two reasons why no one did: no one knew him well enough >|>or was bold enough to send off an email; or anyone who was willing >|>and able to send off a mail didn't think Wiles would know the answer >|>anyway. It's the second possibility that worries (if not >|>scandalizes!) me. >|It would be very surprising indeed if it did not go through. >|Classical analysis and number theory, which are essentially the >|entire substance of the proof, are well authenticated. >what part of the proof did you have in mind as being classical >analysis? Elliptic modular functions and modular forms, and properties of elliptic curves. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 ==== Subject: Re: Do mathematicians know their axioms? turn their informal reasonings into formal ones. > You are right that most could not. I could have replied to any of several messages; this isn't a direct reply to the previous one. But it fits into the thread about here. When research in the foundations of math tries to establish which axioms are required to prove some theorem, there are many methods that do not require any proofs to be formalized, if formalized means written in the style similar to Principia Mathematica. For example, many people have seen Kelley's proof that the Tychonoff theorem (products of compact spaces are compact) for arbitrary spaces is equivalent to AC over ZF, or that Zorn's Lemma and AC are equivalent over ZF, without seeing any formal proofs. The method for equivalence results such as these is: work in the weak system, assume theorem A, prove theorem B in the usual expository style. Then assume theorem B and the weak system and prove theorem A. It does require a good background understanding of exactly what can be done in the weak system, but everything is done in natural language instead of a formalized proof calculus. About half of Simpson's book ``Subsystems of second order arithmetic'' is spent using this method. There is not one formal proof in that book, The proofs are written in a style analogous to any other book in advanced pure mathematics. ==== Subject: Re: Do mathematicians know their axioms? About Grothendieck universes in the proof of Fermats last theorem, and whether they can be eliminated in ZFC, consider http://www.cs.nyu.edu/pipermail/fom/1999-April/003008.html were Simpson says: >It's pretty clear to me that Grothendieck universes can be >straightforwardly eliminated wherever they have been cited in number >theory, and the number theorists who have cited them are well aware of >this. So the only right answer to Shipman's original question about >Grothendieck universes is a straightforward ``no''. The rest of the thread there seems to validate Simpson's point. I think that there is little doubt that Wiles' proof goes through in ZFC. ==== Subject: Re: A convolution integral involving Bessel functions It's the intersection of the two circles of unit radius. If w=0, the centres coincide at (0,0) and the intersection is the entire unit disc. If w=2, the centres are at (-1,0) and (1,0) and the unit radius discs just touch. (The integral is zero.) ==== Subject: Re: A convolution integral involving Bessel functions > I've been trying to evaluate the following integral analytically, and > am now well and truly stuck. Does anybody have any ideas? > The domain of integration is the intersection of two unit circles with > centres at (x,y)= (+-w,0). The integrand is a product of two functions, > which are each Bessel functions of the first kind (J_0) squared. Let a > denote the first zero of J_0(z). > For w a given real number, 0<=w<=2, the double integral is > f(w/2) = Integrate[Integrate[ J_0(a sqrt((x - w/2)^2 + y^2))^2 J_0( > a sqrt((x + w/2)^2 + y^2))^2 , {x, 0, sqrt(1-y^2) - w/2}, {y, 0, > sqrt(1 - (w/2)^2)}]]. I don't know if this will help. But there is a nice change of coordinates that may make the problem more tractable. We can use the so-called bipolar coordinates, where the variable r1 and r2 represent the distances from two select points in the plane. Usually, these are taken to be at (0,+1) and (0,-1). The parameters of your integral can be rescaled to place the centers of the two circles at the latter coordinates and vary their radii instead. The relation between bipolar and cartesian coordinates are: r1^2 = (x-1)^2 + y^2, r2^2 = (x+1)^2 + y^2; x = (r2^2 - r1^2)/4, where s = (r1+r2+2)/2. y = sqrt(s(s-r1)(s-r2)(s-2)), The coordinates (r1,r2) take values in a semi-infinite strip extending diagonally from around the origin in the first quadrant. It is bounded by the lines r1+r2=2, r1-r2=2, and r2-r1=2. The integration region now acquires a simple form. It is the intersection of the range of the coordinates and the half planes r1 <= r and r2 <= r, with r being the radius of the two circles. To write down the new integral we need the Jacobian of the cartesian coordinates with respect to the bipolar ones. It turns out to have the nice form 1 dr1 dr2 dx dy = - --------, where y is as defined above. 2 y(r1,r2) Beside the Jacobian, the integrand becomes J_0(a r1/r)^2 J_0(a r2/r)^2. Unfortunately, the denominator of the Jacobian does not separate cleanly. But this form may be a starting point for some sort of series expansion. An alternate direction is another change of variables: r1 = p1 + p2, so that dr1 dr2 = 2 dp1 dp2, r2 = p1 - p2, and y = sqrt((p1^2-1)(1-p2^2)). The advantage is that the Jacobian of the previous transformation is now factored in the (p1,p2) coordinates. The disadvantage is that the arguments of the Bessel functions become slightly more complicated. There are some addition formulas for Bessle functions [1], but unfortunately they give an infinite number of terms. Hope some of this helps. Igor [1] http://functions.wolfram.com/BesselAiryStruveFunctions/BesselJ/16/02/ ==== Subject: Fermat's last theorem and a counter example FLM states simply that the following equation: x^n + y^n = z^n Doesn't have solution in the whole positive integer number system for (x, y, z, n), where (n>2) Iff, exists a counter example to FLM, then the following conditions must obey: All prime factors of (x+y) must be some prime factors of (z) All prime factors of (z-x) must be some prime factors of (y) All prime factors of (z-y) must be some prime factors of (x) 2*z > x+y > z Do You Think this combinations are possible? Feb,21st,2006 Bassam Karzeddin AL-Hussein Bin Talal University JORDAN ==== Subject: Re: Fermat's last theorem and a counter example > FLM states simply that the following equation: > x^n + y^n = z^n > Doesn't have solution in the whole positive integer number system for (x, y, z, n), where (n>2) > Iff, exists a counter example to FLM, There does not. The theorem has been proven. > then the following conditions must obey: > All prime factors of (x+y) must be some prime factors of (z) > All prime factors of (z-x) must be some prime factors of (y) > All prime factors of (z-y) must be some prime factors of (x) > 2*z > x+y > z > Do You Think this combinations are possible? Certainly. Consider any x+y = z, for instance x=2, y=3, z = 5. Then all your conditions hold. But this is not a FLT counterexample. While it is true that FLT counterexample implies your observations, it is not true that your observations imply FLT counterexample. In the language of mathematics, they are necessary but not sufficient conditions. - ==== Subject: Re: Fermat's last theorem and a counter example But see Hofstadter's Godel, Escher, Bach, wherein he supposes both a proof of FLT AND a counterexample, from which the sounds of Bach playing at his pianoforte may be pulled from the current motion of air molecules in our present atmosphere (see Prelude...Ant Fugue) ==== Subject: The Joy of Numbers by shakunthala Devi Hi folks, I am a computer science student from india.recently i was reading the book entitled The Joy of Numbers by shakunthala Devi. In this book there is a chapter named cubes and cube roots. Here first a table of cubes are given as follows number cubes 1 1 2 8 3 27 4 64 5 125 6 216 7 343 8 512 9 729 10 1000 then the following generalizations are given:- 1) if a cubic no: ends in 1,4,5,6,9,0 its cube root will end in the same no: 2) if a cubic no: ends in 2 the cube root will end in 8 and if it ends in 8 the cube root will end in 2 3) if a cubic no: ends in 3 the cube root will end in 7 and if it ends in 7 the cube root will end in 3 4) if a cube has 4 or 5 or 6 digits the cube root will have 2 digits Then a working example is given find the cube root of 50653 ?? first divide the no: into groups of 3 from right to left so we get the groups as 50 653 now for the 1st group 50 falls between 27 and 64 (2 cubed and 3 cubed) the tens figure of the cube root will be the root of the lower no: that is 3. now to find units digit,the cubic no: ends in 3 its cube root will end in 7. the cube root you are seeking is 37 then the author gives the following explanations to extract cube root of longer no:s with 7,8 or 9 digits you will have to remember a table X Y 1 1 2 7 3 9 4 5 5 3 6 8 7 6 8 2 9 4 10 10 Then a working example is given find the cube root of 92345408 ?? first divide the no: into groups of 3 from right to left so we get the groups as 92 345 408 to find the hundreds unit of a cube root ,we can use table of cubes 92 falls between 4 cubed and 5 cubed so 4 is the hundreds unit of the cube we are extracting the units digit - the terminal figure of the cube root must be 2,for the last digit of the cubic no: is 8 now to find the tens digit we add up all the odd digits of the orginal number from right to left(1st,3rd,5th,7th).subtract from this number the sum of even digits in the no:(2nd,4th,6th,8th) if the total of even figures is greater than or equal to that of odd figures add 11 to the latter before subtracting. here we get sum of odd digits as 18 and that of even digits as 17. now after subtraction we get 1.now we refer back to the table with columns X and Y,we get the Y value of 1 as 1 itself after adding the two digits of the final root which you already have (the hundreds and units digits) ,subtract from the sum the figure you have just obtained from the Y column of the table, to get 6 - 1 = 5 so the complete root is 452 if the sum of digits(the hundreds and units digits) is less than the figure obtained from column Y of the table add 11 to the sum before making the subtraction. well my question is how far this method is true when it comes to extracting cube root of longer no:s with 7,8 or 9 digits ??? as the method seems to fail with certain digits like 365525875 (that is when the difference b/w sum of odd digits and even digits is zero) ==== Subject: Partial Fraction Expansion with complex roots I was reading a book on electrical circuits and it lists ways of working with inverse laplace transforms. Trying to expand a fraction the author reaches: 1 = A(s+2-i)+B(s+2+i) Then he goes on to say that he find A and B by equating the factors (forgive me if this is the incorrect term, I'm Greek) of the equal powers of s which gives him A=i/2 and B=-i/2. He doesn't write all the steps, just the results. How does he do it? What about i? Shouldn't we treat it like s? If I try to solve it I get: 1= s(A+B) + i(B-A) + (2A+2B) How should I equate now? ==== Subject: Re: Partial Fraction Expansion with complex roots > I was reading a book on electrical circuits and it lists ways of > working with inverse laplace transforms. > Trying to expand a fraction the author reaches: > 1 = A(s+2-i)+B(s+2+i) > Then he goes on to say that he find A and B by equating the factors > (forgive me if this is the incorrect term, I'm Greek) of the equal > powers of s which gives him A=i/2 and B=-i/2. See what the equations tell you when s=-2+i and s=-2-i. ==== Subject: Re: Partial Fraction Expansion with complex roots > I was reading a book on electrical circuits and it lists ways of > working with inverse laplace transforms. > Trying to expand a fraction the author reaches: > 1 = A(s+2-i)+B(s+2+i) > Then he goes on to say that he find A and B by equating the factors > (forgive me if this is the incorrect term, I'm Greek) of the equal > powers of s which gives him A=i/2 and B=-i/2. > He doesn't write all the steps, just the results. > How does he do it? What about i? Shouldn't we treat it like s? > If I try to solve it I get: > 1= s(A+B) + i(B-A) + (2A+2B) > How should I equate now? It seems like your confusion is that you want to look at the system of equations: A+B = 0 B-A = 0 2A+2B = 1 However, this isn't quite right as you don't want to treat i like s. In fact, solving this system yields 1 = 0, which clearly doesn't make sense. Since A and B may be complex, your constant on the right hand side of your equation is i(B-A) + (2A+2B). So you should actually be looking at the two equations: (1) A+B = 0 (2) i(B-A) + (2A+2B) = 1 2Bi = 1, so B = -i/2. Since A = -B, we get that A = i/2. Mike ==== Subject: Re: Partial Fraction Expansion with complex roots > However, this isn't quite right as you don't want to treat i > like s. In fact, solving this system yields 1 = 0, which > clearly doesn't make sense. Since A and B may be complex, your > constant on the right hand side of your equation is i(B-A) + > (2A+2B). So you should actually be looking at the two > equations: I understand now. I was treating i like a variable when it's a constant... Looks like I'm not very familiar with complex numbers. ==== Subject: Rational numbers as coordinates The attention given to quaternions and not to four dimensional fields over the rational numbers, or over the integers congruence modulo a prime number, by so many people, suggests that there is some kind of dead end for rational numbers or integers congruence modulo a prime; that is, something you need real numbers for as coordinates. What can be done with exact rational numbers or integers congruence modulo a prime as coordinates and what can't be done? Cliff Nelson Dry your tears, there's more fun for your ears, Forward Into The Past 2 PM to 5 PM, Sundays, California time, at: http://www.kspc.org/ Don't be a square or a blockhead; see: http://bfi.org/node/574 ==== Subject: Group theory question: Suppose that G is a finite group and x is in a (maximal) subgroup M of G. Suppose further that x is in the soluble radical of M, Sol(M) (i.e. the largest subgroup of M that is both normal in M and soluble). Is it true that if x^g is in M then x^g is in Sol(M)? If anyone has any ideas, they would be much appreciated. Graham. ==== Subject: Re: Group theory question: days. My association with the Department is that of an alumnus. >Suppose that G is a finite group and x is in a (maximal) subgroup M of >G. Suppose further that x is in the soluble radical of M, Sol(M) (i.e. >the largest subgroup of M that is both normal in M and soluble). Is it >true that if x^g is in M then x^g is in Sol(M)? >If anyone has any ideas, they would be much appreciated. Certainly not as stated: take x which is in M but not in Sol(M); then x^x=x lies in M, but not in Sol(M). -- ' ==== Subject: Re: Group theory question: >>Suppose that G is a finite group and x is in a (maximal) subgroup M of >>G. Suppose further that x is in the soluble radical of M, Sol(M) (i.e. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ >>the largest subgroup of M that is both normal in M and soluble). Is it >>true that if x^g is in M then x^g is in Sol(M)? >>If anyone has any ideas, they would be much appreciated. >Certainly not as stated: take x which is in M but not in Sol(M); then >x^x=x lies in M, but not in Sol(M). But he said that x is in Sol(M). For finding counterexamples to this kind of problem, the ATLAS of Finite Groups can be useful. Try G = PSL(3,4), M = 2^4:A_5, and note that G has a single conjugacy class of elements of order 2. Derek Holt. ==== Subject: Re: Group theory question: >> Suppose that G is a finite group and x is in a (maximal) subgroup M of >> G. Suppose further that x is in the soluble radical of M, Sol(M) (i.e. >> the largest subgroup of M that is both normal in M and soluble). Is it >> true that if x^g is in M then x^g is in Sol(M)? >> If anyone has any ideas, they would be much appreciated. > Certainly not as stated: take x which is in M but not in Sol(M); then > x^x=x lies in M, but not in Sol(M). Obviously, you meant x^1 here. Jose CArlos Santos ==== Subject: Re: Group theory question: days. My association with the Department is that of an alumnus. > Suppose that G is a finite group and x is in a (maximal) subgroup M of > G. Suppose further that x is in the soluble radical of M, Sol(M) (i.e. > the largest subgroup of M that is both normal in M and soluble). Is it > true that if x^g is in M then x^g is in Sol(M)? > If anyone has any ideas, they would be much appreciated. >> Certainly not as stated: take x which is in M but not in Sol(M); then >> x^x=x lies in M, but not in Sol(M). >Obviously, you meant x^1 here. No; I meant x^x = x^{-1}(x)x=x; but x^1 also works. -- ' ==== Subject: Re: Group theory question: Sorry for the confusion - I want to take x in Sol(M) and some g in G such that x^g is in M. Is it true that x^g is also in Sol(M). Graham. >>Suppose that G is a finite group and x is in a (maximal) subgroup M of >>G. Suppose further that x is in the soluble radical of M, Sol(M) (i.e. >>the largest subgroup of M that is both normal in M and soluble). Is it >>true that if x^g is in M then x^g is in Sol(M)? >>If anyone has any ideas, they would be much appreciated. >Certainly not as stated: take x which is in M but not in Sol(M); then >x^x=x lies in M, but not in Sol(M). >>Obviously, you meant x^1 here. > No; I meant x^x = x^{-1}(x)x=x; but x^1 also works. ==== Subject: Re: Explanation of this hypothesis of SUBSTITUTION RULE requested >My question deals with this following RULE/THEOREM: >The Substitution Rule: If u=g(x) is a differentiable function whose >range is an inteval on which f is continuous, then >integral[ f (g(x)) g' (x) ] = integral[ f (u) du ] >I can't understand why the condition whose range is an interval on >which f is continuous is needed. What are some cases that could >break this, and not allow us to use substitution (yet look on the >surface as a good place to use substit.) ? >>If g is defined for some values for which f is not continuous, then the >>integral does not exist. > May fail to exist. >> Integration requires continuity. > No, it does not require continuity. Continuity guarantees the > existence of the integral, but is not required. A step function, for > example, is easily shown to be integrable even if it is not exontinuous. Spivak's calculus (last edition, pg 365) says that f and g' have to be continuous to use the substitution rule: def. integral from g(a) to g(b) of: f = def. integral from a to b of: (fog) g' For indefinite integration, Spivak (pg 367) poses no such hypothesis/restriction for one to use substition to change the original integral to a form with x's into one with u's (using u=g(x)), then convert it back to an expression with x after antidifferentiation has been done. This suggests that the caveats associated with u=g(x) and it's relationship to f only apply to definite integration. CB4Ever ==== Subject: Re: Pattern Recognition In Statistics This is a variation of my code that plays the Ohio Pick 4 Game. Preliminary tests are very positive. It hit todays Mid-Day Pick 4 twice in 300 trials. It got 2, 24 way boxes, the odds normally are 1 in 417, but my programs odds so far are 1/150. I didn't calculate the actual odds the program has yet, and that is just from real life testing. I've found that the more random of a seed you give the program the more accurate it is. #include #include #include int gen_random() { double r; long int M; double x; int y; int count; M = 268; /* Choose M. Upper bound */ for(count=1; count<=20; ++count) { r = ( (double)rand() / ((double)(RAND_MAX)+(double)(1)) ); x = (r * M); y = (int) x; } return(y); } int main(int argc, char *argv[]) { int N = ((argc == 2) ? atoi(argv[1]) : 1); int result = 0; srand(N); /*initialize random number generator*/ int pick4a[]={8, 9, 4, 8, 2, 3, 5, 6, 4, 9, 6, 4, 9, 6, 3, 6, 7, 3, 4, 1, 3, 1, 5, 9, 7, 3, 0, 3, 3, 2, 0, 3, 9, 7, 6, 1, 0, 2, 7, 1, 4, 2, 0, 7, 0, 6, 7, 8, 2, 5, 1, 6, 0, 9, 0, 1, 9, 5, 1, 2, 9, 6, 4, 7, 6, 4, 3, 4, 4, 1, 4, 4, 7, 5, 7, 4, 5, 4, 6, 4, 6, 0, 6, 3, 4, 7, 2, 0, 7, 5, 6, 0, 1, 4, 8, 7, 5, 5, 5, 2, 8, 3, 9, 3, 3, 8, 7, 0, 5, 1, 9, 2, 9, 5, 7, 8, 3, 9, 5, 9, 6, 9, 2, 6, 4, 1, 3, 7, 0, 7, 6, 9, 4, 7, 8, 9, 8, 1, 2, 9, 0, 5, 1, 7, 1, 4, 3, 5, 3, 0, 1, 5, 5, 6, 7, 2, 3, 7, 3, 7, 4, 4 , 6, 4, 3, 9, 1, 2, 2, 7, 6, 1, 6, 9, 2, 6, 7, 0, 6, 5, 6, 4, 1, 8, 1, 4, 1, 3, 8, 1, 7, 7, 7, 8, 6, 7, 4, 8, 2, 6, 9, 8, 6, 8, 9, 7, 6, 4, 5, 5, 8, 7, 5, 9, 4, 4, 3, 9, 2, 3, 2, 0, 4, 4, 1, 7, 5, 7, 5, 1, 4, 0, 3, 3, 2, 8, 3, 0, 5, 9, 8, 7, 8, 0, 2, 0, 4, 0, 5, 3, 6, 3, 2, 5, 1, 2, 9, 0, 5, 3, 5, 4, 5, 7, 7, 3, 8, 5, 3, 6, 7, 4, 3, 5, 5, 7, 7, 4, 1, 3, 0, 0, 4, 9, 6, 4, 7, 9, 4, 0, 7, 0, 5, 7, 9, 6, 6, 1, 9, 4, 1, 0, 2, 8, 7, 1, 1, 8}; int pick4b[]={4, 2, 8, 6, 3, 7, 3, 2, 3, 8, 5, 7, 7, 5, 3, 9, 8, 2, 5, 1, 3, 9, 0, 8, 1, 3, 3, 2, 4, 2, 5, 0, 3, 2, 6, 5, 7, 0, 1, 0, 1, 0, 5, 3, 2, 6, 9, 4, 7, 2, 5, 4, 9, 5, 5, 2, 0, 1, 0, 4, 5, 3, 0, 4, 3, 5, 2, 4, 1, 1, 7, 5, 5, 4, 2, 0, 9, 1, 0, 5, 4, 0, 6, 9, 0, 7, 0, 2, 3, 8, 2, 7, 8, 2, 0, 5, 5, 7, 9, 5, 0, 8, 6, 3, 5, 2, 4, 3, 9, 9, 5, 5, 8, 1, 6, 5, 9, 3, 4, 9, 2, 0, 1, 1, 3, 5, 7, 8, 3, 9, 9, 3, 9, 7, 3, 4, 8, 4, 7, 6, 4, 3, 7, 1, 6, 3, 7, 7, 0, 2, 7, 8, 0, 5, 4, 9, 1, 6, 5, 8, 8, 9, 4, 6, 4, 7, 3, 6, 6, 5, 6, 3, 0, 5, 4, 1, 9, 5, 4, 3, 3, 4, 4, 5, 5, 8, 0, 4, 7, 2, 3, 1, 1, 2, 7, 8, 0, 9, 3, 0, 2, 2, 6, 2, 6, 3, 4, 5, 7, 1, 0, 9, 7, 1, 8, 5, 5, 2, 5, 0, 7, 2, 8, 6, 6, 2, 0, 6, 9, 3, 7, 0, 8, 1, 5, 8, 4, 1, 8, 1, 2, 6, 9, 1, 5, 0, 0, 0, 9, 6, 9, 1, 2, 5, 5, 1, 1, 5, 0, 8, 9, 8, 7, 0, 5, 0, 9, 5, 4, 1, 2, 6, 2, 8, 0, 0, 3, 5, 4, 1, 3, 2, 8, 6, 2, 4, 6, 8, 0, 7, 0, 8, 6, 6, 4, 9, 9, 3, 6, 2, 0, 9, 5, 5, 9, 0, 1, 8}; int pick4c[]={1, 6, 9, 3, 5, 9, 2, 0, 5, 3, 7, 0, 3, 9, 1, 0, 8, 1, 6, 9, 0, 5, 4, 1, 9, 1, 8, 4, 0, 5, 0, 4, 9, 3, 1, 5, 0, 3, 7, 6, 1, 6, 0, 6, 5, 7, 8, 6, 4, 2, 6, 4, 5, 3, 1, 0, 9, 2, 7, 4, 3, 6, 1, 9, 5, 0, 6, 9, 0, 8, 0, 0, 3, 6, 3, 9, 8, 6, 0, 1, 5, 2, 5, 8, 1, 8, 2, 3, 5, 0, 5, 2, 8, 9, 9, 6, 4, 3, 0, 1, 0, 5, 3, 8, 1, 1, 3, 2, 9, 1, 8, 9, 1, 0, 5, 2, 2, 6, 5, 6, 4, 1, 3, 1, 5, 2, 2, 0, 9, 7, 1, 2, 4, 1, 2, 8, 9, 0, 8, 8, 5, 4, 5, 7, 4, 5, 5, 8, 9, 6, 9, 5, 7, 5, 4, 7, 1, 6, 7, 2, 5, 1 , 9, 1, 2, 5, 2, 5, 8, 4, 1, 1, 6, 4, 0, 1, 9, 0, 3, 1, 3, 8, 3, 5, 9, 6, 1, 1, 0, 3, 5, 5, 8, 0, 7, 5, 4, 4, 9, 4, 1, 5, 8, 4, 1, 1, 9, 1, 8, 0, 8, 7, 0, 6, 3, 5, 7, 6, 0, 4, 0, 8, 4, 4, 5, 8, 0, 4, 2, 9, 1, 8, 4, 6, 6, 1, 7, 4, 5, 6, 2, 3, 8, 7, 7, 0, 4, 1, 0, 7, 3, 5, 5, 4, 1, 8, 2, 2, 1, 8, 9, 6, 4, 7, 1, 9, 1, 6, 1, 0, 2, 2, 7, 4, 4, 4, 5, 4, 7, 6, 0, 1, 7, 3, 6, 6, 5, 6, 2, 8, 7, 9, 5, 4, 6, 8, 3, 4, 9, 7, 7, 6, 5, 1, 4, 0, 4, 3}; int pick4d[]={3, 1, 4, 8, 8, 3, 9, 1, 9, 6, 4, 5, 7, 1, 1, 6, 4, 7, 7, 3, 6, 7, 3, 1, 4, 4, 7, 0, 3, 7, 9, 8, 1, 7, 8, 8, 5, 9, 6, 5, 7, 8, 6, 2, 5, 2, 1, 9, 4, 4, 0, 0, 0, 8, 3, 6, 1, 7, 2, 0, 2, 1, 3, 0, 8, 1, 8, 0, 4, 7, 6, 3, 2, 9, 9, 3, 9, 3, 1, 1, 0, 5, 8, 1, 0, 4, 5, 9, 1, 2, 1, 9, 4, 9, 4, 6, 9, 0, 4, 8, 1, 0, 6, 0, 0, 8, 7, 4, 8, 8, 2, 9, 7, 3, 1, 3, 7, 5, 8, 0, 6, 1, 3, 3, 8, 8, 5, 7, 3, 9, 8, 0, 7, 6, 6, 9, 0, 1, 1, 0, 8, 6, 9, 7, 7, 0, 5, 1, 5, 8, 3, 9, 8, 3, 4, 7, 8, 1, 3, 0, 3, 0 , 6, 8, 3, 2, 1, 5, 3, 7, 0, 3, 9, 2, 0, 7, 6, 6, 2, 1, 8, 5, 3, 6, 9, 2, 8, 9, 2, 3, 8, 7, 0, 3, 5, 8, 3, 9, 6, 2, 6, 3, 7, 8, 5, 4, 8, 6, 2, 2, 8, 6, 9, 8, 5, 4, 2, 7, 8, 0, 7, 4, 1, 6, 8, 5, 1, 4, 2, 6, 2, 1, 8, 7, 0, 3, 0, 1, 8, 6, 9, 5, 0, 0, 2, 4, 4, 4, 9, 6, 7, 0, 2, 7, 5, 2, 9, 7, 5, 0, 2, 8, 0, 4, 1, 9, 0, 3, 1, 8, 5, 5, 7, 2, 0, 3, 5, 8, 5, 0, 6, 8, 6, 2, 7, 1, 0, 7, 6, 9, 2, 3, 2, 9, 1, 9, 2, 1, 8, 7, 2, 7, 4, 8, 8, 9, 6, 1}; int cnt=0; while(cnt<100){ result=gen_random(); printf(%d , pick4a[result]); result=gen_random(); printf(%d , pick4a[result]); result=gen_random(); printf(%d , pick4a[result]); result=gen_random(); printf(%d n, pick4a[result]); cnt++; } return(0); } ==== Subject: Re: Pattern Recognition In Statistics Here is another hypothesis test... I figured out what the odds of getting 4 of tonights numbers correct with my system, and at random. I found there are 5 ways you can win, and with a 2.5% chance of getting any number randomly, the odds are much smaller than what the lottery publishes. With my algorithm my odds went up over 11%. If someone wanted to check my work so I could be certain it would help. 6 14 17 any 25 5.6% 6.7% 6.3% 1.49% total: 0.00035220024% at random: 0.000000390625 6 14 17 25 any 5.6% 6.7% 6.3% 3.37% 0.00079658712% at random: 0.000000390625 6 14 15 17 any 5.6% 6.7% 2.24% 0.749% 0.0000629495552% at random: 0.000000390625 6 14 15 any 25 5.6% 6.7% 2.24% 1.49% 0.000125226752% at random: 0.000000390625 6 14 15 25 any 0.0003331776% 5.6% 6.7% 2.24% 3.37% at random: 0.000000390625 Total: 0.0016701413172% Odds of winning randomly: 0.0001953125% My odds are 11.69% better than at random. The lottery says the odds are 0.295% ==== Subject: Re: Pattern Recognition In Statistics Tonights lotter picks were: 6, 14, 15, 17, 25 My program guesses each of these numbers X percent of the time: 5.6%, 7.1%, 2.25% 0.75% 1.49% The probability of all five numbers being is usually 0.0000009765625% (1 / 575,757), with a 2.5% chance of each number being hit . With my program the average chance of hitting them all is 0.0000009997155% That is 2.3% better than expected. Unfortunately I didn't hit the number in my testing tonight.. But I will keep these tests going for awhile. How is this for hypothesis testing? ==== Subject: Re: JSH: Modified Decker equations, direct solution See comments below - > and substituting out for g_1, I have > g_2^2 - sqrt(3) (7x - 5) g_2 + 17 = 0 > so > g_2 = (sqrt(3)(7x - 5) +/- sqrt(3(7x-5)^2 - 4(17)))/2 > and I have factors of 3 shared with the a's must be factors of > sqrt(7) +/- sqrt(3(7x-5)^2 - 68) But what you want to conclude about the a's is not about factors of 3 - what you need to know is about divisibility of the a's by 3 itself. That is why it is not sufficient to simply drop the term sqrt(3) (7x - 5) from consideration. Note, incidentally, that if x is chosen to be a root of 21 x^2 - 30 x + 1 = 4, [see below] then the numerator of x can be assumed to be coprime to 3. In fact x can be written as x = 1/u, where u is an algebraic integer which is a divisor of 7. In particular, the numerator of x is coprime to 3. > so > sqrt(7) +/- sqrt(3(49x^2 -5 (7) x + 25) - 68) > which is > sqrt(7) +/- sqrt(7) sqrt(21x^2 -15x + 1) No, there is a little error here - it should be sqrt(7) +/- sqrt(7) sqrt(21 x^2 - 30 x + 1). This error doesn't matter much - you can just choose x to be a solution to 21 x^2 - 30 x + 1 = 4, i.e., to 7 x^2 - 10 x - 1 = 0. > and now I can just let > 21x^2 - 15x + 1 = 4 See above. This is wrong, but easily fixed. The main problem is your dropping of sqrt(3) (7 x - 5) as noted above. Marcus. > which gives either g_1 or g_2 coprime to 3, ==== Subject: Re: JSH: Modified Decker equations, direct solution <7boKf.49440$dW3.32262@newssvr21.news.prodigy.com> <-v6dnSVKl4YFr2fenZ2dnUVZ_vydnZ2d@comcast.com [JSH] and substituting back now I have >> 2a = (-3(((g_1 + g_2)/sqrt(3) + 5)/7) - 1) +/- (g_1 - g_2)/sqrt(7) >> multiplying both sides by 7, I have sentence. Don't think so. In the extract I've quoted, multiplying seems to start a new sentence, despite the lack of a capital. So at least two sentences. Brian Chandler http://imaginatorium.org ==== Subject: Re: JSH: Modified Decker equations, direct solution >> In the ring of algebraic integers, modifying the equations given by a >> Rick Decker of Hamilton College, I have >> f Q(x) = f((x^2 + fx)(5^2) + (-1 + fx)(5) + f) >> = f(25 x^2 + 30fx + (f - 5)) >> and >> f Q(x) = (5a_1(x) + f)(5a_2(x) + f) >> where the a's are defined by >> a^2 - (fx - 1)a + f(x^2 + fx) = 0 >> and the direct solution is to pick f=-3 and solve for the a's using the >> quadratic formula, which gives >> a = ((-3x - 1) +/- sqrt((-3x - 1)^2 + 12(x^2 - 3x)))/2 >> and focusing inside the square root I have >> (3x + 1)^2 + 12(x^2 - 3x) = 9x^2 + 6x + 1 + 12x^2 - 36x >> which is >> (3x + 1)^2 + 12(x^2 - 3x) = 21x^2 - 30x + 1 >> and completing the square gives >> (3x + 1)^2 + 12(x^2 - 3x) = 21x^2 - 30x + 75/7 - 75/7+ 1 >> so >> (3x + 1)^2 + 12(x^2 - 3x) = (3(7x - 5)^2 - 68)/7 >> and now let g_1 g_2 = 17, and >> g_1 + g_2 = sqrt(3) (7x - 5) >> then >> x = ((g_1 + g_2)/sqrt(3) + 5)/7 >> and substituting back now I have >> 2a = (-3(((g_1 + g_2)/sqrt(3) + 5)/7) - 1) +/- (g_1 - g_2)/sqrt(7) >> multiplying both sides by 7, I have >> 14a = (-sqrt(3)(g_1 + g_2) + 15 - 7) +/- sqrt(7) (g_1 - g_2) >> so divisibility of the a's by 3 is decided by >> -sqrt(7) +/- (g_1 - g_2) >> but g_1 = - g_2 + sqrt(3)(7x - 5), so I can make that substitution and >> get >> -sqrt(7) +/- (-2g_2 + sqrt(3)(7x - 5)) >> and get that divisibility is a matter of >> -sqrt(7) +/- -2g_2 >> which is >> sqrt(7) +/- 2g_2 >> where again g_1 g_2 = 17, so I can use from before >> g_1 + g_2 = sqrt(3) (7x - 5) >> and substituting out for g_1, I have >> g_2^2 - sqrt(3) (7x - 5) g_2 + 17 = 0 >> so >> g_2 = (sqrt(3)(7x - 5) +/- sqrt(3(7x-5)^2 - 4(17)))/2 >> and I have factors of 3 shared with the a's must be factors of >> sqrt(7) +/- sqrt(3(7x-5)^2 - 68) >> so >> sqrt(7) +/- sqrt(3(49x^2 -5 (7) x + 25) - 68) >> which is >> sqrt(7) +/- sqrt(7) sqrt(21x^2 -15x + 1) >> and now I can just let >> 21x^2 - 15x + 1 = 4 >> which gives either g_1 or g_2 coprime to 3, though you get a >> non-algebraic integer x, which can be substituted into the equation for >> the a's, which gives you a non-monic polynomial with non-rational >> coefficients, but you can get to integer coefficients, and either have >> a reducible over Q polynomial or one that is irreducible, but the >> leading coefficient will have 7 as a factor and not 3, which finally >> shows that hey, I'm right! > Ha. You still don't get it: it doesn't matter what convoluted > arithmetic you do. The proofs that are in place are valid. You > haven't looked at them, so you have no way of knowing that. You > *could* learn enough to understand the basics, but it's so much > fun having delusions of adequacy. >> The solution for x is found from >> 21x^2 - 15x - 3 = 0 >> and the a's given by the x, if my algebra is right, are roots of >> 9604 a^4 + 39788 a^3 + 56644 a^2 + 6559a - 18513 = 0 >> and using a = c/14, I can get the monic polynomial >> c^4 + 58c^3 + 1156c^2 + 1874c - 74052 = 0 >> and its irreducibility over Q completes the direct proof that Galois >> Theory as taught is false, hoping I got the algebra right! > Yet again, huh? > I'll assume your arithmetic is correct (which is presumably all you mean > by 'hoping I got the algebra right', since once you get beyond simple > arithmetic, you are out of your depth). I hate to see you waste your time, Dale, but James did make an error. Where he had 21x^2 - 15x - 3, he should have 21x^2 - 30x - 3. That skunks his quartic in a, which works out to be (7a^2 + 8a - 8)(7a^2 + 36 a + 36) = 0 Two of the a's are divisible by 3, in a sense, but that's meaningless, since none of the four are algebraic integers. Rick ==== Subject: Re: JSH: Modified Decker equations, direct solution <7boKf.49440$dW3.32262@newssvr21.news.prodigy.com> In the ring of algebraic integers, modifying the equations given by a >> Rick Decker of Hamilton College, I have >> f Q(x) = f((x^2 + fx)(5^2) + (-1 + fx)(5) + f) >> = f(25 x^2 + 30fx + (f - 5)) >> and >> f Q(x) = (5a_1(x) + f)(5a_2(x) + f) >> where the a's are defined by >> a^2 - (fx - 1)a + f(x^2 + fx) = 0 >> and the direct solution is to pick f=-3 and solve for the a's using the >> quadratic formula, which gives >> a = ((-3x - 1) +/- sqrt((-3x - 1)^2 + 12(x^2 - 3x)))/2 >> and focusing inside the square root I have >> (3x + 1)^2 + 12(x^2 - 3x) = 9x^2 + 6x + 1 + 12x^2 - 36x >> which is >> (3x + 1)^2 + 12(x^2 - 3x) = 21x^2 - 30x + 1 >> and completing the square gives >> (3x + 1)^2 + 12(x^2 - 3x) = 21x^2 - 30x + 75/7 - 75/7+ 1 >> so >> (3x + 1)^2 + 12(x^2 - 3x) = (3(7x - 5)^2 - 68)/7 >> and now let g_1 g_2 = 17, and >> g_1 + g_2 = sqrt(3) (7x - 5) >> then >> x = ((g_1 + g_2)/sqrt(3) + 5)/7 >> and substituting back now I have >> 2a = (-3(((g_1 + g_2)/sqrt(3) + 5)/7) - 1) +/- (g_1 - g_2)/sqrt(7) >> multiplying both sides by 7, I have >> 14a = (-sqrt(3)(g_1 + g_2) + 15 - 7) +/- sqrt(7) (g_1 - g_2) >> so divisibility of the a's by 3 is decided by >> -sqrt(7) +/- (g_1 - g_2) >> but g_1 = - g_2 + sqrt(3)(7x - 5), so I can make that substitution and >> get >> -sqrt(7) +/- (-2g_2 + sqrt(3)(7x - 5)) >> and get that divisibility is a matter of >> -sqrt(7) +/- -2g_2 >> which is >> sqrt(7) +/- 2g_2 >> where again g_1 g_2 = 17, so I can use from before >> g_1 + g_2 = sqrt(3) (7x - 5) >> and substituting out for g_1, I have >> g_2^2 - sqrt(3) (7x - 5) g_2 + 17 = 0 >> so >> g_2 = (sqrt(3)(7x - 5) +/- sqrt(3(7x-5)^2 - 4(17)))/2 >> and I have factors of 3 shared with the a's must be factors of >> sqrt(7) +/- sqrt(3(7x-5)^2 - 68) >> so >> sqrt(7) +/- sqrt(3(49x^2 -5 (7) x + 25) - 68) >> which is >> sqrt(7) +/- sqrt(7) sqrt(21x^2 -15x + 1) >> and now I can just let >> 21x^2 - 15x + 1 = 4 >> which gives either g_1 or g_2 coprime to 3, though you get a >> non-algebraic integer x, which can be substituted into the equation for >> the a's, which gives you a non-monic polynomial with non-rational >> coefficients, but you can get to integer coefficients, and either have >> a reducible over Q polynomial or one that is irreducible, but the >> leading coefficient will have 7 as a factor and not 3, which finally >> shows that hey, I'm right! > Ha. You still don't get it: it doesn't matter what convoluted > arithmetic you do. The proofs that are in place are valid. You > haven't looked at them, so you have no way of knowing that. You > *could* learn enough to understand the basics, but it's so much > fun having delusions of adequacy. >> The solution for x is found from >> 21x^2 - 15x - 3 = 0 >> and the a's given by the x, if my algebra is right, are roots of >> 9604 a^4 + 39788 a^3 + 56644 a^2 + 6559a - 18513 = 0 >> and using a = c/14, I can get the monic polynomial >> c^4 + 58c^3 + 1156c^2 + 1874c - 74052 = 0 >> and its irreducibility over Q completes the direct proof that Galois >> Theory as taught is false, hoping I got the algebra right! > Yet again, huh? > I'll assume your arithmetic is correct (which is presumably all you mean > by 'hoping I got the algebra right', since once you get beyond simple > arithmetic, you are out of your depth). > I hate to see you waste your time, Dale, but James did make an error. > Where he had 21x^2 - 15x - 3, he should have 21x^2 - 30x - 3. > That skunks his quartic in a, which works out to be > (7a^2 + 8a - 8)(7a^2 + 36 a + 36) = 0 > Two of the a's are divisible by 3, in a sense, but that's meaningless, > since none of the four are algebraic integers. > But your claim that it's meaningless since none of the a's are algebraic integers is meaningless as you can get to algebraic integers using a = c/14 and since 14 is coprime to 3, I have the disproof. Understand? The only thing that can save standard usage of ideal theory is an integer root. So, it's over. I have been right all along. Understand now? James Harris ==== Subject: Re: JSH: Modified Decker equations, direct solution >>In the ring of algebraic integers, modifying the equations given by a >>Rick Decker of Hamilton College, I have >>f Q(x) = f((x^2 + fx)(5^2) + (-1 + fx)(5) + f) >> = f(25 x^2 + 30fx + (f - 5)) >>and >>f Q(x) = (5a_1(x) + f)(5a_2(x) + f) >>where the a's are defined by >>a^2 - (fx - 1)a + f(x^2 + fx) = 0 >>and the direct solution is to pick f=-3 and solve for the a's using the >>quadratic formula, which gives >> The solution for x is found from >>21x^2 - 15x - 3 = 0 >>and the a's given by the x, if my algebra is right, are roots of >>9604 a^4 + 39788 a^3 + 56644 a^2 + 6559a - 18513 = 0 >>and using a = c/14, I can get the monic polynomial >>c^4 + 58c^3 + 1156c^2 + 1874c - 74052 = 0 >>and its irreducibility over Q completes the direct proof that Galois >>Theory as taught is false, hoping I got the algebra right! >Yet again, huh? >I'll assume your arithmetic is correct (which is presumably all you mean >by 'hoping I got the algebra right', since once you get beyond simple >arithmetic, you are out of your depth). >>I hate to see you waste your time, Dale, but James did make an error. >>Where he had 21x^2 - 15x - 3, he should have 21x^2 - 30x - 3. >>That skunks his quartic in a, which works out to be >> (7a^2 + 8a - 8)(7a^2 + 36 a + 36) = 0 >>Two of the a's are divisible by 3, in a sense, but that's meaningless, >>since none of the four are algebraic integers. >> algebraic integers is meaningless as you can get to algebraic integers > using > a = c/14 All that would prove is 14a is an algebraic integer. That's no more useful than claiming that from 7a + 8 = 0 you can get to integers by using a = c/14. True enough, but meaningless. Rick ==== Subject: Re: JSH: Modified Decker equations, direct solution <7boKf.49440$dW3.32262@newssvr21.news.prodigy.com> In the ring of algebraic integers, modifying the equations given by a >>Rick Decker of Hamilton College, I have >>f Q(x) = f((x^2 + fx)(5^2) + (-1 + fx)(5) + f) >> = f(25 x^2 + 30fx + (f - 5)) >>and >>f Q(x) = (5a_1(x) + f)(5a_2(x) + f) >>where the a's are defined by >>a^2 - (fx - 1)a + f(x^2 + fx) = 0 >>and the direct solution is to pick f=-3 and solve for the a's using the >>quadratic formula, which gives > The solution for x is found from >>21x^2 - 15x - 3 = 0 >>and the a's given by the x, if my algebra is right, are roots of >>9604 a^4 + 39788 a^3 + 56644 a^2 + 6559a - 18513 = 0 >>and using a = c/14, I can get the monic polynomial >>c^4 + 58c^3 + 1156c^2 + 1874c - 74052 = 0 >>and its irreducibility over Q completes the direct proof that Galois >>Theory as taught is false, hoping I got the algebra right! Yet again, huh? I'll assume your arithmetic is correct (which is presumably all you mean >by 'hoping I got the algebra right', since once you get beyond simple >arithmetic, you are out of your depth). >>I hate to see you waste your time, Dale, but James did make an error. >>Where he had 21x^2 - 15x - 3, he should have 21x^2 - 30x - 3. >>That skunks his quartic in a, which works out to be >> (7a^2 + 8a - 8)(7a^2 + 36 a + 36) = 0 >>Two of the a's are divisible by 3, in a sense, but that's meaningless, >>since none of the four are algebraic integers. >> algebraic integers is meaningless as you can get to algebraic integers > using > a = c/14 > All that would prove is 14a is an algebraic integer. That's > no more useful than claiming that from 7a + 8 = 0 you can get > to integers by using a = c/14. True enough, but meaningless. > Rick BUT the point of the exercise is to FORCE one of the a's, as a ratio of algebraic integers, to have a numerator that is coprime to 3, which I do, successfully. Therefore, when you get to a monic polynomial with integer coefficients, it's proven that one of its roots is coprime to 3, but it's also possible to prove that none of its roots can be coprime to 3 ***in the ring of algebraic integers***. I suggest you think carefully Professor Decker. You are about to be a rather well-known figure around the world, and you can go that route the front way, or dragged in. I think you'd want it with your head held high, and with some pride, versus fighting to be the last dolt to get it. James Harris ==== Subject: Re: JSH: Modified Decker equations, direct solution <7boKf.49440$dW3.32262@newssvr21.news.prodigy.com> >Rick Decker of Hamilton College, I have >>f Q(x) = f((x^2 + fx)(5^2) + (-1 + fx)(5) + f) >> = f(25 x^2 + 30fx + (f - 5)) >>and >>f Q(x) = (5a_1(x) + f)(5a_2(x) + f) >>where the a's are defined by >>a^2 - (fx - 1)a + f(x^2 + fx) = 0 >>and the direct solution is to pick f=-3 and solve for the a's using the >>quadratic formula, which gives > The solution for x is found from >>21x^2 - 15x - 3 = 0 >>and the a's given by the x, if my algebra is right, are roots of >>9604 a^4 + 39788 a^3 + 56644 a^2 + 6559a - 18513 = 0 >>and using a = c/14, I can get the monic polynomial >>c^4 + 58c^3 + 1156c^2 + 1874c - 74052 = 0 >>and its irreducibility over Q completes the direct proof that Galois >>Theory as taught is false, hoping I got the algebra right! Yet again, huh? I'll assume your arithmetic is correct (which is presumably all you mean >by 'hoping I got the algebra right', since once you get beyond simple >arithmetic, you are out of your depth). >>I hate to see you waste your time, Dale, but James did make an error. >>Where he had 21x^2 - 15x - 3, he should have 21x^2 - 30x - 3. >>That skunks his quartic in a, which works out to be >> (7a^2 + 8a - 8)(7a^2 + 36 a + 36) = 0 >>Two of the a's are divisible by 3, in a sense, but that's meaningless, >>since none of the four are algebraic integers. >> algebraic integers is meaningless as you can get to algebraic integers > using a = c/14 > All that would prove is 14a is an algebraic integer. That's > no more useful than claiming that from 7a + 8 = 0 you can get > to integers by using a = c/14. True enough, but meaningless. Rick > BUT the point of the exercise is to FORCE one of the a's, as a ratio of > algebraic integers, to have a numerator that is coprime to 3, which I > do, successfully. At one point in your argument you have 14a = (-sqrt(3)(g_1 + g2) + 15 - 7) +/- sqrt(7) (g_1 - g_2). You then discard the part that is multiplied by sqrt(3) and focus on the rest, which (mod 3) is -7 +/- sqrt(7)(g_1 - g_2). And eventually you show that this part is essentially congruent to 0 or -1 mod 3. But that does not show that 14a is congruent to 0 or -1 mod 3. Why? Because of the sqrt(3) term that you discarded. What you actually have is that 14a + sqrt(3)(g_1 + g_2) is either coprime to 3 or is divisible by 3. This is actually a statement about the two roots, a_1 and a_2. Assume that a_1 is the root which is such that 14a_1 + sqrt(3)(g_1 + g_2) is divisible by 3. Thus 14a_1 = sqrt(3)*A + 3*B, where A and B are algebraic integers. All you can conclude from this is that 14a_1 is divisible by sqrt(3). You cannot conclude it is divisible by 3. Your error goes back to point where you discarded the term involving sqrt(3). Here is a simple analogy to what you have done: Let K = 21 + sqrt(3). Then K - sqrt(3) is divisible by 3. But K itself is divisible only by sqrt(3) in the algebraic integers. Marcus > Therefore, when you get to a monic polynomial with integer > coefficients, it's proven that one of its roots is coprime to 3, but > it's also possible to prove that none of its roots can be coprime to 3 > ***in the ring of algebraic integers***. > I suggest you think carefully Professor Decker. You are about to be a > rather well-known figure around the world, and you can go that route > the front way, or dragged in. > I think you'd want it with your head held high, and with some pride, > versus fighting to be the last dolt to get it. > James Harris ==== Subject: Re: JSH: Modified Decker equations, direct solution > Therefore, when you get to a monic polynomial with integer > coefficients, it's proven that one of its roots is coprime to 3, but > it's also possible to prove that none of its roots can be coprime to 3 > ***in the ring of algebraic integers***. I see, you admit that there is a common factor in the ring of algebraic integers. However, you have not proven what you think. All you've done is to exhibit a form of argument that is invalid in the ring of algebraic integers. This is not a stroke of brilliance, despite your interest in having that be the case. Here is your likely point of error. You state it's proven that one of its roots is coprime to 3 but fail to provide the direct proof, namely the demonstration of [integers? algebraic integers? what sorts of entities?] u and v for which u A + v 3 = 1, where A is the root in question. Demonstrate these quantities u and v, and then one can discuss whether you've proven anything. I'm inclined to believe that you make a number of unintentional assumptions. If you cannot state the domain in which this coprimeness holds, then you are fooling no one except yourself. > I suggest you think carefully Professor Decker. You are about to be a > rather well-known figure around the world, and you can go that route > the front way, or dragged in. > I think you'd want it with your head held high, and with some pride, > versus fighting to be the last dolt to get it. I see. I read this years ago, in the KJV Bible, Matthew 4:8-10 8. Again, the devil took him to a very high mountain and showed him all the kingdoms of the world and their splendor. 9. All this I will give you, he said, if you will bow down and worship me. 10. Jesus said to him, Away from me, Satan! For it is written: 'Worship the Lord your God, and serve him only.' > James Harris You, therefore, are the devil. Professor Decker is Jesus, and sci.math is the Lord your God. Dale ==== Subject: Re: JSH: Modified Decker equations, direct solution ... stuff deleted ... > I see. I read this years ago, in the KJV Bible, Matthew 4:8-10 > 8. Again, the devil took him to a very high mountain and > showed him all the kingdoms of the world and their splendor. > 9. All this I will give you, he said, if you will bow down > and worship me. > 10. Jesus said to him, Away from me, Satan! For it is written: > 'Worship the Lord your God, and serve him only.' CORRECTION. The above is the New International Version. I meant to copy the KJV version: 8. Again, the devil taketh him up into an exceeding high mountain, and sheweth him all the kingdoms of the world, and the glory of them; 9. And saith unto him, All these things will I give thee, if thou wilt fall down and worship me. 10. Then saith Jesus unto him, Get thee hence, Satan: for it is written, Thou shalt worship the Lord thy God, and him only shalt thou serve. The conclusion still holds. > You, therefore, are the devil. Professor Decker is Jesus, and sci.math > is the Lord your God. Dale ==== Subject: Re: JSH: Modified Decker equations, direct solution >>In the ring of algebraic integers, modifying the equations given by a >>Rick Decker of Hamilton College, I have >>f Q(x) = f((x^2 + fx)(5^2) + (-1 + fx)(5) + f) >> = f(25 x^2 + 30fx + (f - 5)) >>and >>f Q(x) = (5a_1(x) + f)(5a_2(x) + f) >>where the a's are defined by >>a^2 - (fx - 1)a + f(x^2 + fx) = 0 >>and the direct solution is to pick f=-3 and solve for the a's using the >>quadratic formula, which gives >>a = ((-3x - 1) +/- sqrt((-3x - 1)^2 + 12(x^2 - 3x)))/2 >>and focusing inside the square root I have >>(3x + 1)^2 + 12(x^2 - 3x) = 9x^2 + 6x + 1 + 12x^2 - 36x >>which is >>(3x + 1)^2 + 12(x^2 - 3x) = 21x^2 - 30x + 1 >>and completing the square gives >>(3x + 1)^2 + 12(x^2 - 3x) = 21x^2 - 30x + 75/7 - 75/7+ 1 >>so >>(3x + 1)^2 + 12(x^2 - 3x) = (3(7x - 5)^2 - 68)/7 >>and now let g_1 g_2 = 17, and >>g_1 + g_2 = sqrt(3) (7x - 5) >>then >>x = ((g_1 + g_2)/sqrt(3) + 5)/7 >>and substituting back now I have >>2a = (-3(((g_1 + g_2)/sqrt(3) + 5)/7) - 1) +/- (g_1 - g_2)/sqrt(7) >>multiplying both sides by 7, I have >>14a = (-sqrt(3)(g_1 + g_2) + 15 - 7) +/- sqrt(7) (g_1 - g_2) >>so divisibility of the a's by 3 is decided by >>-sqrt(7) +/- (g_1 - g_2) >>but g_1 = - g_2 + sqrt(3)(7x - 5), so I can make that substitution and >>get >>-sqrt(7) +/- (-2g_2 + sqrt(3)(7x - 5)) >>and get that divisibility is a matter of >>-sqrt(7) +/- -2g_2 >>which is >>sqrt(7) +/- 2g_2 >>where again g_1 g_2 = 17, so I can use from before >>g_1 + g_2 = sqrt(3) (7x - 5) >>and substituting out for g_1, I have >>g_2^2 - sqrt(3) (7x - 5) g_2 + 17 = 0 >>so >>g_2 = (sqrt(3)(7x - 5) +/- sqrt(3(7x-5)^2 - 4(17)))/2 >>and I have factors of 3 shared with the a's must be factors of >>sqrt(7) +/- sqrt(3(7x-5)^2 - 68) >>so >>sqrt(7) +/- sqrt(3(49x^2 -5 (7) x + 25) - 68) >>which is >>sqrt(7) +/- sqrt(7) sqrt(21x^2 -15x + 1) >>and now I can just let >>21x^2 - 15x + 1 = 4 >>which gives either g_1 or g_2 coprime to 3, though you get a >>non-algebraic integer x, which can be substituted into the equation for >>the a's, which gives you a non-monic polynomial with non-rational >>coefficients, but you can get to integer coefficients, and either have >>a reducible over Q polynomial or one that is irreducible, but the >>leading coefficient will have 7 as a factor and not 3, which finally >>shows that hey, I'm right! >Ha. You still don't get it: it doesn't matter what convoluted >arithmetic you do. The proofs that are in place are valid. You >haven't looked at them, so you have no way of knowing that. You >*could* learn enough to understand the basics, but it's so much >fun having delusions of adequacy. >>The solution for x is found from >>21x^2 - 15x - 3 = 0 >>and the a's given by the x, if my algebra is right, are roots of >>9604 a^4 + 39788 a^3 + 56644 a^2 + 6559a - 18513 = 0 >>and using a = c/14, I can get the monic polynomial >>c^4 + 58c^3 + 1156c^2 + 1874c - 74052 = 0 >>and its irreducibility over Q completes the direct proof that Galois >>Theory as taught is false, hoping I got the algebra right! >Yet again, huh? >I'll assume your arithmetic is correct (which is presumably all you mean >by 'hoping I got the algebra right', since once you get beyond simple >arithmetic, you are out of your depth). >>I hate to see you waste your time, Dale, but James did make an error. >>Where he had 21x^2 - 15x - 3, he should have 21x^2 - 30x - 3. >>That skunks his quartic in a, which works out to be >> (7a^2 + 8a - 8)(7a^2 + 36 a + 36) = 0 >>Two of the a's are divisible by 3, in a sense, but that's meaningless, >>since none of the four are algebraic integers. >> algebraic integers is meaningless as you can get to algebraic integers > using > a = c/14 > and since 14 is coprime to 3, I have the disproof. > Understand? > The only thing that can save standard usage of ideal theory is an > integer root. What do you mean? The fact that a quartic is reducible over Q does not imply that it has a rational root. > So, it's over. I have been right all along. Understand now? That's what you think; it's what you've always thought, even though your arguments, *all of them*, have been wrong. No, you have never been right. Understand yet? > James Harris Dale ==== Subject: Re: JSH: Modified Decker equations, direct solution > In the ring of algebraic integers, modifying the equations given by a > Rick Decker of Hamilton College, I have > f Q(x) = f((x^2 + fx)(5^2) + (-1 + fx)(5) + f) > = f(25 x^2 + 30fx + (f - 5)) > and > f Q(x) = (5a_1(x) + f)(5a_2(x) + f) > where the a's are defined by > a^2 - (fx - 1)a + f(x^2 + fx) = 0 > and the direct solution is to pick f=-3 and solve for the a's using the > quadratic formula, which gives > a = ((-3x - 1) +/- sqrt((-3x - 1)^2 + 12(x^2 - 3x)))/2 > and focusing inside the square root I have > (3x + 1)^2 + 12(x^2 - 3x) = 9x^2 + 6x + 1 + 12x^2 - 36x > which is > (3x + 1)^2 + 12(x^2 - 3x) = 21x^2 - 30x + 1 > and completing the square gives > (3x + 1)^2 + 12(x^2 - 3x) = 21x^2 - 30x + 75/7 - 75/7+ 1 > so > (3x + 1)^2 + 12(x^2 - 3x) = (3(7x - 5)^2 - 68)/7 > and now let g_1 g_2 = 17, and > g_1 + g_2 = sqrt(3) (7x - 5) > then > x = ((g_1 + g_2)/sqrt(3) + 5)/7 > and substituting back now I have > 2a = (-3(((g_1 + g_2)/sqrt(3) + 5)/7) - 1) +/- (g_1 - g_2)/sqrt(7) > multiplying both sides by 7, I have > 14a = (-sqrt(3)(g_1 + g_2) + 15 - 7) +/- sqrt(7) (g_1 - g_2) > so divisibility of the a's by 3 is decided by > -sqrt(7) +/- (g_1 - g_2) > but g_1 = - g_2 + sqrt(3)(7x - 5), so I can make that substitution and > get > -sqrt(7) +/- (-2g_2 + sqrt(3)(7x - 5)) > and get that divisibility is a matter of > -sqrt(7) +/- -2g_2 > which is > sqrt(7) +/- 2g_2 > where again g_1 g_2 = 17, so I can use from before > g_1 + g_2 = sqrt(3) (7x - 5) > and substituting out for g_1, I have > g_2^2 - sqrt(3) (7x - 5) g_2 + 17 = 0 > so > g_2 = (sqrt(3)(7x - 5) +/- sqrt(3(7x-5)^2 - 4(17)))/2 > and I have factors of 3 shared with the a's must be factors of > sqrt(7) +/- sqrt(3(7x-5)^2 - 68) > so > sqrt(7) +/- sqrt(3(49x^2 -5 (7) x + 25) - 68) Nope. Check your results before you overthrow all mathematics. > which is > sqrt(7) +/- sqrt(7) sqrt(21x^2 -15x + 1) Nope. > and now I can just let > 21x^2 - 15x + 1 = 4 Nope. > which gives either g_1 or g_2 coprime to 3, though you get a > non-algebraic integer x, which can be substituted into the equation for > the a's, which gives you a non-monic polynomial with non-rational > coefficients, but you can get to integer coefficients, and either have > a reducible over Q polynomial or one that is irreducible, but the > leading coefficient will have 7 as a factor and not 3, which finally > shows that hey, I'm right! > The solution for x is found from > 21x^2 - 15x - 3 = 0 Nope. > and the a's given by the x, if my algebra is right, are roots of > 9604 a^4 + 39788 a^3 + 56644 a^2 + 6559a - 18513 = 0 Nope. > and using a = c/14, I can get the monic polynomial > c^4 + 58c^3 + 1156c^2 + 1874c - 74052 = 0 Certainly not. > and its irreducibility over Q completes the direct proof that Galois > Theory as taught is false, hoping I got the algebra right! The correct polynomial is neither monic nor irreducible. Rick ==== Subject: Re: JSH: Modified Decker equations, direct solution > In the ring of algebraic integers, modifying the equations given by a > Rick Decker of Hamilton College, I have > f Q(x) = f((x^2 + fx)(5^2) + (-1 + fx)(5) + f) > = f(25 x^2 + 30fx + (f - 5)) > and > f Q(x) = (5a_1(x) + f)(5a_2(x) + f) > where the a's are defined by > a^2 - (fx - 1)a + f(x^2 + fx) = 0 > and the direct solution is to pick f=-3 and solve for the a's using the > quadratic formula, which gives > a = ((-3x - 1) +/- sqrt((-3x - 1)^2 + 12(x^2 - 3x)))/2 > and focusing inside the square root I have > (3x + 1)^2 + 12(x^2 - 3x) = 9x^2 + 6x + 1 + 12x^2 - 36x > which is > (3x + 1)^2 + 12(x^2 - 3x) = 21x^2 - 30x + 1 > and completing the square gives > (3x + 1)^2 + 12(x^2 - 3x) = 21x^2 - 30x + 75/7 - 75/7+ 1 > so > (3x + 1)^2 + 12(x^2 - 3x) = (3(7x - 5)^2 - 68)/7 > and now let g_1 g_2 = 17, and > g_1 + g_2 = sqrt(3) (7x - 5) > then > x = ((g_1 + g_2)/sqrt(3) + 5)/7 > and substituting back now I have > 2a = (-3(((g_1 + g_2)/sqrt(3) + 5)/7) - 1) +/- (g_1 - g_2)/sqrt(7) > multiplying both sides by 7, I have > 14a = (-sqrt(3)(g_1 + g_2) + 15 - 7) +/- sqrt(7) (g_1 - g_2) > so divisibility of the a's by 3 is decided by > -sqrt(7) +/- (g_1 - g_2) > I don't get this step. It appears to me that in the equation > above for 14a, the term of interest was > sqrt(7) (g_1 - g_2). > Now it has somehow been transformed into > sqrt(7) +/- (g_1 - g_2). > Please explain. Further, how does the 15 - 7 play > into this ? > Marcus. I focus on terms NOT multiplied by factors of 3, so I drop 15 as it is. Looking over the argument I can see how I can say that explicitly using mod so I am going to start a new thread, which should clear up all areas of confusion. Note, for those not clear, I have found the direct refutation of common usage of Galois Theory. By breaking open the radical by solving the square root I can force one of the a's to be coprime to 3 in the ring of algebraic integers, which gives me a non-algebraic integer a's, BUT importantly, this time, they are roots of a non-monic polynomial with a leading coefficient coprime to 3. So the technicality creates the block, preventing them from being algebraic integers, but they can be roots of a non-monic with a leading coefficient coprime to 3, which allows me to get a monic polynomial where it's proven that one of its roots is coprime to 3. It is a brilliant proof. James Harris ==== Subject: Re: JSH: Modified Decker equations, direct solution >In the ring of algebraic integers, modifying the equations given by a >Rick Decker of Hamilton College, I have >f Q(x) = f((x^2 + fx)(5^2) + (-1 + fx)(5) + f) > = f(25 x^2 + 30fx + (f - 5)) >and >f Q(x) = (5a_1(x) + f)(5a_2(x) + f) >where the a's are defined by >a^2 - (fx - 1)a + f(x^2 + fx) = 0 >and the direct solution is to pick f=-3 and solve for the a's using the >quadratic formula, which gives >a = ((-3x - 1) +/- sqrt((-3x - 1)^2 + 12(x^2 - 3x)))/2 usage of Galois Theory. > By breaking open the radical by solving the square root I can force one > of the a's to be coprime to 3 in the ring of algebraic integers, which > gives me a non-algebraic integer a's, BUT importantly, this time, they > are roots of a non-monic polynomial with a leading coefficient coprime > to 3. > So the technicality creates the block, preventing them from being > algebraic integers, but they can be roots of a non-monic with a leading > coefficient coprime to 3, which allows me to get a monic polynomial > where it's proven that one of its roots is coprime to 3. > It is a brilliant proof. Save for the tiny problem that it's wrong. Check your arithmetic. Rick > James Harris ==== Subject: a letter to richard j fateman rjf> This issue actually illustrates Vladimir's basic problem. rjf> he does not go the epsilon step to find the source of this rjf> behavior. 1) this message shows me that being a born researcher and indefatigable promoter of computer algebra (bravo! i applaud you, i take off my hat to you!), you have nothing to do with business. okay, a single person cannot have all possible gifts; and you are endowed much already, to my white envy ;) now if you like to work for free, it's your internal choice, pray continue. however both our team and me personally are in business. we already have done much for public for free, and will do even more for free. 2) if you want us to do even more than we have already done for free, and are going to do for free, why not considering paying us out of your pocket or convincing someone to pay us? 3) this message also shows that you seem to ignore or do not take into account my messages i had sent you in answer to your requests. okay, this is your viewpoint; in my opinion, you have already done more than enough for the world to stick to your own vision and habits. at the same time, actually, at present, i am wondering is there much sense in continuing further explanations to you; all the same you do not hear me, so why bother? 4) in your private messages, this year, last year, you repeatedly told me not polite things. if you are going to continue this mode of behavior i warn you explicitly that i publish your letters to me so the folks here would see more about your, let put it mildly, not polite behavior. maybe this would cool you down a bit. 5) as i have explained you in private massages, right away i am immersed deeply into really time consuming stuff, and 24 hours a day is not too much time; so i reserve my right to answer selectively, and maybe not so often as i would wish to. 6) as you write, Carey = PITA, VB = PITA... who are you, dick? vvb ==== Subject: Re: JSH: Important milestone Mail-To-News-Contact: abuse@dizum.com >> Remember, some of you are in the United States taking federal funds. >> Read the fine print on those fundings. >So what does the fine print say? (For those of us who don't >take federal funds.) This note is legal tender for all debts, public and private. HTH, HAND -- Michael F. Stemper #include You can lead a horse to water, but you can't make him talk like Mr. Ed by rubbing peanut butter on his gums. ==== Subject: attractors_arch hi, i.m an m.arch student who is trying to understand what actually an attractor is and find possible ways of generating those beautiful forms of them _for example, lorenz attractor_ by using mathematica/rhino/maya... The problem is, i.m not a mathematician and it has been only 6 days since i started to search attractors.:) To understand the logic, i.ve read many papers on web and i have a serious of images. what i.m trying to do is, since they show the status that a dynamic system eventually settles down to and are a collection of points such that each point is a function of another point, to use their logic: a small change in one point effects all the system and changes it dramatically. They are interesting because within a specific initial point and dynamical system, they are following the rule which repeatedly defines a motion through the state space_ a trajectory. In the long run, they symbolize the observable states of a natural system which is dynamic. it.s very exciting for me to simulate their behaviours and generate new forms which are working in the same way for my architectural project: it is about learning environments and we think information should be seen as a dynamical system as well. so, may be, architecture is the attractor itself in this case. I want to start with trying to draw them three dimensionally which is a very literal start but, there is no other way for me to understand their logic completely. And i don't know how to do because i want to use their codes/formulas to generate their forms. So, for those of you who already used mathematica to visualize any of the attractors 2d or 3d, i would appreciate if you help me with it...i.m in big trouble with them.. i hope to hear from you very soon. *sorry for this long message... marien k. ==== Subject: Re: Control and Usenet > The most effective tool on Usenet for controlling other posters is > insults. To the contrary, that just burns up bandwidth and clutters our screens with flame wars. The most effective tool for Usenet is to have a newsgroup moderator. The next most effective tool is for the newsreader to have an effective kill file mechanism so that the reader does not have to wade through the flood of postings like the myriad jstevh-generated threads that keep saying essentially the same few things, over and over. > The battles now are playing out because my research is such a kick in > the groin for modern number theorists in all their cherished areas. You're deluding yourself. The battles are mainly threads initiated by you, with a few frustrated respondents who are tired of these threads, and occasionally a respondent who points out what is wrong with your technical proposals. As to the feasibility of your surrogate factoring methodology, so far it seems to have always boiled down to there being a missing step that you gloss over, find a T such that ..., where there is no known efficient way to find such a T. If you had a method that really works for practical problems, it would be of great theoretical and practical interest; the resistance you encounter is not because nobody is interested in a genuine breakthrough, but because you haven't shown us that you really have one. Of course, this has all been pointed out to you before. ==== Subject: Re: linear constant coefficient nonhomogenous difference equation but, is there some another method than using generating functions for these equations? Garen. ==== Subject: Re: linear constant coefficient nonhomogenous difference equation >but, is there some another method than using generating functions for >these equations? Sure. Consider e.g. the equation sum_{j=0}^m a_j y(n+j) = f(n) Suppose f is of the form f(n) = r^n. If r is not a root of the polynomial P(x) = sum_{j=0}^m a_j x^j, then y(n) = r^n/P(r) is a solution. If r is a root of the polynomial with multiplicity k, then y(n) = n!/(n-k)! r^(n-k)/P^(k)(r) is a solution. More generally, you might try to write f as a linear combination of sequences of this form, or an integral of them. Thus if f is square-summable, we can write f(n) = (2 pi)^(-1) int_0^{2 pi} g(t) exp(-int) dt where g(t) = sum_{n=-infty}^infty f(n) exp(int). And then, at least if P has no roots on the unit circle, a solution of the difference equation is y(n) = (2 pi)^(-1) int_0^{2 pi} g(t) exp(-int)/P(exp(-it)) dt Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada ==== Subject: Re: Question of P=NP, Anatoly Plotnikov > His paper if correct proves P=NP by solving an NP hard problem. Either you don't understand his paper, or it is on the wrong track. Solving a single NP-hard problem doesn't prove P=NP. Finding a algorithm that solves all instances of an NP-hard problem and that also has running time bounded by some polynomial function of the problem size would prove P=NP. > So why hasn't the Clay Institute rolled out the red carpet and given > him a check for a $1 million? Why not ask them? Presumably, experienced workers in the field are not convinced that he has found a solid proof. Perhaps it is like your surrogate factoring techniques, which always seem to leave out some crucial detail without which it just cannot work efficiently. > ... Is this Plotnikov yet another crackpot mathematician ...? I don't know, but even reputable mathematicians have been known to publish results which contain errors. Recall that the 4-color theorem was considered proved for quite some time before a subtle but important error was found in Kempe's proof, and it was not a simple matter to patch up the flawed attempt. > Think about it, over 10 years possibly with a known and published, peer > reviewed proof that P=NP, with very little commentary on it, and the > entire world just going on about its business, building up huge > infrastructure around cryptology techniques that depend on factoring > being a hard problem. Factoring in the *general* case *is* a hard problem; at least, the best methods that have been proven and demonstrated aren't feasible for arbitrary numbers with many thousands of digits. Of course there are subsets that are readily factored, and those are typically carefully avoided when one picks RSA keys. Most public-key cryptographic infrastructures support a variety of algorithms, so if RSA is found to be irremediably broken an alternative (not based on the difficulty of factoring) can be used instead. > I say because there is a consistent pattern of using denial and > ignoring research results on the part of the math community as a tool. If you were actually part of the math research community you would know better than to make such a claim. > The same pattern can be seen in the astronomical field where a Dr. > Halton Arp has compelling research challenging the claims of other > astronomers about how far quasars and other objects in the sky ... When I was studying the subject it was mainly the Burbages who were suggesting such things. Their key point was that there are multiple possible causes for observed red-shift, not merely recessional Doppler effect. Indeed we went through something similar with Cepheid variables, where the distance scale had to be radically revised when it turned out that variable stars seen in distant galaxies were of a different kind than the ones seen nearby. Any reputable astrophysicist should be aware of such things and open to reasonable arguments. > But he [Arp] is also considered a crackpot in his field. If so, perhaps his arguments haven't been reasonable? ==== Subject: Re: Question of P=NP, Anatoly Plotnikov off with Let there be a graph G = (X, Gamma), where X is the set of > graph vertices, and Gamma is a map X into X. I don't see how this > matches with the definition of a graph (vertices and edges) that the > rest of the world uses. >>If map X into X is simply a relation from X to X, or in other >>words a subset of X x X, then that matches the vertices and edges >>definition. > Well, it had better be a special map X into X, ie one which is symmetric. > Ie > if x' is in Gamma(x) then x had better be in Gamma(x') for Gamma to the > equivalent of the edges. (The set X is clearly the vertices). > Not if it is a directed graph. The alleged P vs NP solver > is certainly using non-standard and/or sloppy terminology, but it > is not too hard to figure out what is meant by > Let there be a graph G = (X,Gamma), where X is the set > of graph vertices, and Gamma is a map X into X. > , assuming of course that it is meant to correspond with > what the rest of the world means by a 'graph'. Of course > the persistent use of non-standard and/or sloppy terminology > will likely make his alleged proofs impossible to understand. Or at least confusing, requiring one to keep a dictionary of common terms while going through the paper. Plotnikov is Russian, so maybe this is a literal translation into English of the Russian axiom for the standard definition of a graph. I had equated map with function, but I could see that it could also mean relation. --- Christopher Heckman ==== Subject: Re: Probability question <20060218.101618@whim.org> <20060218.160322@whim.org> <20060218.202735@whim.org a is uniformly random in [0,1) > f(a) is defined as: > f(a) = a, 0 <= a < 1/2 > f(a) = a - 1/2, 1/2 <= a < 3/4 > f(a) = 2a - 5/4, 3/4 <= a < 1 > what is Pr(a = 1/4 | f(a) = 1/4) ? > See my other post. Formally, the probability is not defined and not of > practical interest. Probabilities of this nature actually did arise in an admittedly idealised game strategy problem that I was looking at (see However, for practical implementations of the game the precision with which the variables can be expressed is limited, so in effect we are taking probabilities over small ranges, as you say. > Using the precision argument, let X be the > uniform random variable, since I prefer the usual convention of using > capital letters for random variables. You may want > P(|X - 1/4| < e | X in (1/4 - e, 1/4 + e) U (3/4 - e, 3/4 + e)), > which is 1/2. This corresponds to the informal > g(1/4) dx / [g(1/4) dx + g(3/4) dx], > where g is the density of X. > Or you may want > P(|X - 1/4| < e | |f(X) - 1/4| < e), > which is 4/7 (modulo my calculation errors). This corresponds to > g(1/4) dx / [h(1/4) dx], > where h is the density of Y. > In other words, for a practical answer, you need to specify the > precision of your measuring instrument. > -- > Stephen J. Herschkorn > Math Tutor on the Internet and in Central New Jersey and Manhattan ==== Subject: =?UTF-8?Q?Re:_$BAu$$$b=3F7$=3F$KKF|%*!<%W%s=1B(B?= Perhaps this URL ... http://www.okdaily.com/go/svc/ecdict.html Fernando. ==== Subject: Frobenius method for system of equations I am trying to solve the following system for u and w in terms of s: s^2 u'' + s u' - mu^2 u + mu^2 Cot[alpha] (nu s w' - w) = 0 eta^2 (s^2 w'' - s w') - Cot[alpha] (nu s^3 u' + u + Cot[alpha] s^2 w) = 0 where all greek variables are known positive constants and ' denotes differentiation. Application of the Frobenius method, i.e. assuming u(s) = Summation( U_k s^(sigma+k), k=0,infinity) w(s) = Summation( W_k s^(sigma+k), k=0,infinity) leads to an indicial equation with roots 0, 2 and +/- mu. Note also that mu may equal integral values 1,2... depending on physical properties of the system, while nu is bounded to be between 0 and 1/2. For non-integral values of mu, I can use the Frobenius method to find the recurrence relations for the expansion coefficients. This involves straightforward solutions for sigma = +/- mu and sigma = 2, with a slighter more complicated solution for sigma = 0 since it differs by an integer from the sigma = 2 solution (involving taking the limit as sigma approaches zero, similar to Bessel function derivations for integral roots to generate a second independent solution). My question is, what happens when mu is an integer (it is often equal to 1, but could possibily be any positive odd or even integer). I am mu, but not sure which solutions have precedence or if that even matters. For example, if mu = 1, then I have four roots of 2, 1, 0, -1, so do I need to start with the highest root and then generate successsive independent solutions for each root in decreasing order? Will this alter the (previously independent) solution for sigma=0? Alternatively, if mu = 2 (roots 2 [twice], 0, -2) or mu > 2 (but integral), must I begin with the highest root and now alter my previous solutions for sigma=2 and 0? My predilection is to use the sigma = 2 and sigma = 0 solutions as two independent solutions, and then alter the +/-mu solutions as they approach any integral value, but I want to make sure it will always furnish a suitable solution. I know I can always check for independence for each case after I generate the four solutions, but after all that work I do not want to find out that they are NOT linearly independent and then not know how to proceed. Any insights or references to point to? I have not found a good reference for SYSTEMS that require Frobenius method solutions. bt ==== Subject: A Formula for Generating Irrational Normal Numbers? This may be a very stupid question, but could there possibly be a formula that could spit out an endless (infinite) sequence of finite irrational, normal numbers? ==== Subject: Re: Prime Counting: O(.8+e) time, O(e) space. Can this be improved? <11965067.1140476551018.JavaMail.jakarta@nitrogen.mathforum.org> =0 and sum_i(Zi} = 1}. Under what condition for matirx A, can we say for any x, y belongs to set K, we can have x'Ax >= x'Ay? Fan ==== Subject: Meaning of f(dx) Does anybody know how to interpret (and integrate, if possible) quantity of the form f(dx) the associated integral being int( f(dx) ) I'm encountering this type of quantity frequently in martingales