mm-84 === Be unhappy all you want, I don't care for your nitpicking Maple bugreports. Why don't you submit some code to solve these bugs, eh?If you search the archives of Basti's postings from a year ago, you will?d he made statements like : I am the leader of modern mathematicsNobody but a crank makes statements like these - I rest my case.Caesar Garcia> Dear Caesar Garcia,While I have not read these papers yet, and, thus, have none > idea on account of if Dr Basti is right or wrong there, I must> admit that I am feeling fairly unhappy with the way you were> expressing your opinion this time.I'd say, I am hearing strong malevolence in your voice.> AFAIK, torrent of emotion tends to break human reasoning.Publishing or not publishing a paper not necessarily means> the paper is wrong; for example, it simply can be written in> a dif?ult to read fashion etc. As you most probably know,> the ultimate example of publishers' blunder is Galois' papers:> nobody published them over decades.http://www.galois-group.net/I was wondering have you read all Dr Basti's papers yourself?> If not which seems to be the case, then - no comment.CG> I wonder if your new employers would have been so keen > CG> to hire you if they knew you were such a nutcase.I am certain that if I would ever hire a person like you I > would fall into a serious error as such a person is obviously > not a team player and would make constant attempts to break> communication within the company. Speaking so I do not mean that you are a bad or good > person, IMHO, such statements are of no use; I also have> none idea what talents you are endowed by. I just mean you> looks like an individual player; there are and certainly> there will be always individual players of superlative power,> for example, a Nobel prize winner Frederick Soddy who once> worked with Ernest Rutherfordhttp://www.nobel.se/chemistry/laureates/1921/ soddy-bio.html> > Could I kindly ask you to report us rather more facts about > Dr Basti's papers? I am very interested in hearing from you> on this and other points.Best wishes,Vladimir BondarenkoCo-founder, CEO, Mathematical Director> Cyber Tester, LLChttp://www.cybertester.com/. === >> >> So the strategy I am using is to implement a user friendly syntax for> >> metamathematics plus axiomatic set theory. But I doubt it will hold for> >> any serious proof veri?ation system. So on top of that, I may add a> >> system for user de?ed syntax. Then this may in fact be asimpli?ation> >> for further implementations.> > >> >A very ambitious project. Good luck!>> It is not as ambitious as it may sound, as I decided to not work with> optimizations. But I have implemented things like uni?ation branching in> order to achieve an proof-engine that is as general as possible. This way> I do not need Gentzen sequents, it seems.>> One idea that comes to my mind is that you might look at using the> semantic trees to write an interface on top of a program like Qu-Prolog.> This might saving you some time. :-)>Save me some time? As the subject heading indicates, I have alreadyimplemented my system. If you haven't already done so, please have a look atit and let me know what you think. (System requirements: Windows 95 orlater, and IE4 or later.) Try out the tutorial (click the Help button) toget a general idea of the system's capabilities. Examples in the tutorialinclude Russell's Paradox and what I called the Paradox of the UniversalSet.DanVisit DC Proof Online at http://www.dcproof.com -- FREE download === I posted this in another group. But I wanted to get some feedbacksfrom you guys as well.I was wondering. what are some criteria mathematicians use in order to?ure out if there is (an) anlytical solution(s) to a given ode orode system?if there are many, then what books are available that shows suchmethods?what can you suggest? any thoughts are welcome. thanks in advancejohn === See for example the list of references here:http://lie.uwaterloo.ca/odetools/references.htmlI posted this in another group. But I wanted to get some feedbacks> from you guys as well.I was wondering. what are some criteria mathematicians use in order to> ?ure out if there is (an) anlytical solution(s) to a given ode or> ode system?if there are many, then what books are available that shows such> methods?what can you suggest? any thoughts are welcome. thanks in advancejohn === 1. Given two arbitary symbolic expressions involving only real algebraicnumbers of constant value (e.g. 1/6, Sqrt(37/6) are simple examples), is itpossible to determine computationally if one expression is less than orequal to the other? (Presumably this is at least as hard as the zeroequivalence problem?).2. If it is possible, is it well-understood and computationally tractible?3. If 2., then is it also the sort of thing that one might reasonably expecta computer algebra package to handle?TIA,James. === Download beta 1 - www.master-graph.com/mgraph20b1.exe (only 477KB).All who will help me to improve this program will get a freeregistration key.You can do the following:-?d bugs-feedback about you impression by the program-advice me to change or add some new features-translate it to the other language (seewww.master-graph.com/instructions/interface.html)Feel free to contact with me - roman@master-graph.comSincerely,Roman. === Hoops 1-4 developed Hoop division algebras, collapsed from Moufangloops with r-fold symmetry and having additive elimination asgeneralized subtraction. Sizes are functions that are conserved onhoop multiplication; the sizes of a product are the products of thecorresponding sizes of the multiplicands. I now explore theconsequences of sizes becoming zero.(5a) Sizes can become zero. New possibilities exist in non-degenerate algebras that conservemore than one size. Sizes of vecs (generalized vectors) can becomezero without the vec becoming a null vec. C3 conserves a+b+c and((a-b)^2+ (b-c)^2+ (c-a)^2)/2 and so the zeroes occur when eithera+b=-c or a=b=c. D3 conserves a+b+c+d+e+f, a-b+c-d+e-f, & ((a-b)^2+(b-c)^2+ (c-a)^2 -(d-e)^2- (e-f)^2- (f-d)^2)/2, and so has 3non-trivial zeroes. (5b) Zero sizes constrain calculations to a sub-algebra. A multiplicand with a zero size gives a product with the same sizeequal to zero. The result is projected onto a sub-algebra (to whichthe multiplicand belongs) in which this size is constrained to zero.The sub-algebra has lost a symmetry. (5c) Operations with a zero size renormalize. Division by a vector in the constrained algebra propagates the sameconstraint. This is effected by the genInverse function; partialfractions are ignored if the denominator is less than some prescribedh (i.e. if the size approximates to zero). Conic sections provide ananalogy. The bi-cone is highly 3D symmetrical. Restrictingconsideration to a 2D plane gives various less symmetrical 2D ?ures.The system is constrained to have zero distance from the plane. This corresponds to renormalization, the meta-mathematical trickused by mathematical physicists to dispose of un-physical in?ities.It also corresponds to sub- & super-symmetry and the use of gradedalgebras. Roger Beresford.Now you see it. Now you don't! (Magician's patter.) === In the program Mathematica, one can factor an equation Modulus a Primenumber. However, I am having a hard time trying to understand how tointerpret the output. Could someone offer a simple explanation ofFor example, the following normally cannot be factored any further.Factor[5 + x^2]5 + x^2However, if we factor this using the option Modulus a Prime number, we geta different output.Factor[5 + x^2, Modulus -> 3](1 + x)*(2 + x)..or2 + 3*x + x^2I have been having a hard time understanding just what the output is tryingto tell me.I understand how Mod[7, 3] returns 1, but I do not see the relationship-- Dana === > In the program Mathematica, one can factor an equation Modulus a Prime> number. However, I am having a hard time trying to understand how to> interpret the output. Could someone offer a simple explanation of> For example, the following normally cannot be factored any further.> Factor[5 + x^2]5 + x^2However, if we factor this using the option Modulus a Prime number, we get> a different output.> Factor[5 + x^2, Modulus -> 3](1 + x)*(2 + x)> ..or> 2 + 3*x + x^2I have been having a hard time understanding just what the output is trying> to tell me.> I understand how Mod[7, 3] returns 1, but I do not see the relationshipCalculating mod 3, Mathematica maps, -6,-3,0,3,6,9, . to 0, -5,-2,1,4,7,10, to 1, -4,-1,2,5,8,11, to 2so you asked for the factors of x^2+2, when reduced mod 3.It gave you the factors 1+x and 2+x.multiplied together gives 2+3x+x^2, but note that 3 is the same as 0.so the result is 2+x^2. That's what you asked to factor.> X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i09NO2k28322; === I am trying to solve PDEs in the following manner: setup the PDEsin Maple (v.9), generate a C code of the corresponding set of ODEs,mex them to make a .dll ?e and use MATLABs ODE45() to solve thesystem of ODEs. I have tested an example problem successfully consisting of 3ODEs. Here is the Maple code for that.//// Start Code /////> restart;> with(CodeGeneration):Warning, the protected name Matlab has been rede?ed and unprotected> with(DEtools,convertsys):> Lorenz := [ diff(x(t),t) = 1.0*(y(t)-x(t)),> diff(y(t),t) = 1.0*x(t)-y(t)-x(t)*z(t),> diff(z(t),t) = x(t)*y(t)-1.0*z(t) ]:> init := x(0)=1.0, y(0)=1.0, z(0)=1.0; init := x(0) = 1.0, y(0) = 1.0, z(0) = 1.0> S := convertsys( Lorenz, [init], [x(t),y(t),z(t)], t, y, yp );S := [[yp[1] = 1.0 y[2] - 1.0 y[1], yp[2] = 1.0 y[1] - y[2] - y[1]y[3], yp[3] = y[1] y[2] - 1.0 y[3]], [y[1] = x(t), y[2] = y(t), y[3] =z(t)], 0, [1.0, 1.0, 1.0]]> yp := array(map(rhs,S[1]));yp := [1.0 y[2] - 1.0 y[1], 1.0 y[1] - y[2] - y[1] y[3], y[1] y[2] -1.0 y[3]]> example_3ode := codegen[makeproc](yp, parameters=[yp,t,y]);example_3ode := proc(yp::array(1 .. 3), t, y) yp[1] := 1.0*y[2] - 1.0*y[1]; yp[2] := 1.0*y[1] - y[2] - y[1]*y[3];yp[3] := y[1]*y[2] - 1.0*y[3]; return ;end proc;> C(example_3ode,output=example_3ode.c);//// End Code ////I could mex the output ?e example_3ode.c and make it a .dll ?eusing a mexFunction and solve the ODEs using ODE45() of MATLAB.I have also considered a sample PDE problem, x(r,t). I converted thePDE into a set of ODEs. But I have not been able to convert the systemof ODEs as done in the above case. I am providing the correspondingcode below. Can somebody please take a look at it and let me know howto go about the problem.//// Start Code ////> restart;> read`D:/madhu/FuelCells/PEMFCModeling/ModelingUsingMaple9/pde /fdpack.mpl`:> read`D:/madhu/FuelCells/PEMFCModeling/ModelingUsingMaple9/ utils/utils.mpl`:> read`D:/madhu/FuelCells/PEMFCModeling/ModelingUsingMaple9/ plots/plots.mpl`:> read `D:/madhu/FuelCells/PEMFCModeling/ModelingUsingMaple9/BESIRK` :> with(DEtools,convertsys):> PDE1 := diff(x(r,t),t) = nu*diff(x(r,t),r): PDE1;> IC1 := t=0, x(r,t) = 0.0: IC1; t = 0, x(r, t) = 0.> BC1 := r=0, x(r,t) = 1.0: BC1; r = 0, x(r, t) = 1.0> BC2 := r=1.0, x(r,t) = 10.0: BC2; r = 1.0, x(r, t) = 10.0> n_x := 11; n_x := 11> rspec := h_r = 1.0/(n_x-1): rspec; h_r = 0.1000000000> PDE1a :=convert(PDE1,fddiff,order=[1,0],forward=[1,0],indexletters= [j,none]):PDE1a; d nu (x[j + 1] - x[j]) -- x[j] = -------------------- dt h[r] > PDE1b := subs(r[j]=(j-1)*h[r],h[r]=h_r,PDE1a): PDE1b; d nu (x[j + 1] - x[j]) -- x[j] = -------------------- dt h_r > BC1a :=convert(BC1[2],fddiff,order=[2,0],forward=[2,0], indexletters=[1,none]):BC1a := subs(h[r]=h_r,BC1a): BC1a; x[1] = 1.0> BC2a :=convert(BC2[2],fddiff,order=[2,0],forward=[2,0], indexletters=[n_x,none]):BC2a := subs(h[r]=h_r,BC2a): BC2a; x[11] = 10.0> params := {nu=0.05}: params; {nu = 0.05}> xeqns := [seq(PDE1b,j=2..n_x-1)];> whattype(xeqns); list> S := convertsys(xeqns, [], [seq(x[j],j=2..n_x-1)], t, x, xp ):Error, (in DEtools/convertsys) invalid system of differentialequations//// End Code ////Not sure how do I go about the above error message.madhu === > 87,160 formulas and 10,828 graphics about mathematical functions> But is there 5 times as much as in, say, Abramowitz> and Stegun? No way. 5 times as much data, perhaps, but data> is not information.Well, i just compared a single charpter of HMF: 23. Bernoulli and Euler Polynomials - Riemann Zeta FunctionI downloaded the three notebooks ?BernoulliB', ?EulerE'and ?Zeta'. They sum up to very well 5 times the information given in the HMF on these subjects.But I did not like to use the information online.I spend most of the time going up and down the'information-tree', thus loading tons of ?HTML'just to ?d a little formula.WR claims: with new searching capabilities.Actually the search facility did not know ?zeta','euler polynomials', ?bernoulli polynomials'and was totally useless in these cases.I think the site is an important knowledge basefor mathematical functions, but best used inconnection with notebooks and MathReader (or Mathematica). === we are planing to start with a project refering imaging software forpattern recognition. In a ?st step we are looking for some goodbooks in this area e.g. math-background, standard algorithms.You aren't speci? about what it is that you want the computer torecognize, but here are a few good references:Applied Pattern Recognition, by Paulus and Hornegger (includesmaterial speci? to images)Both of these title are heavier on the image processing side ofthings:Algorithms for Image Processing and Computer Vision, by ParkerDigital Image Processing, by Gonzalez and WoodsYou might consider machine learning, which can be used to developpattern recognizers. If so, consider:Computer Systems That Learn, by Weiss and KulikowskiIt might help if you were to provide more speci? information aboutyour problem.good luck,Will Dwinnellhttp://will.dwinnell.com <8cd70836.0312271457.1da6690b@posting.google.com> <8cd70836.0312272043.6e4441e@posting.google.com> <8cd70836.0312280217.55e9ad52@posting.google.com> === On Mon, 12 Jan 2004 01:17:07 +0100> The point is, functional programming is inherently a layer of> complication on top of the imperative machine architecture.> > It's a step further away, but it's quite far from being a> complication! Functional languages have a far, far simpler semantics> than their imperative counterparts. (That's one of the reasons why> optimizing them is easier.)Just to add in the usual response: For wider de?itions of machine,i.e. parallel, distributed, global computing, functional languagesappear to be closer to the machine architecture. The typicalobligatory reference in this case is Sisal. However, it is dormant ifnot dead, but someone else also pointed out Single Assignment Crecently (www.sac-home.org).X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0C1kJo05152; Sun, 11 Jan 2004 20:46:19 -0500 === An integer is a set of all whole #'s and their inverse. An Integer isin?ate and it can have and in?ate amount of digits. === >> Have you remembered to halve the t=0 input datum?> I`m sorry i don`t understand. why is this necessary? > > Think of the ?ite fft as an approximation to the integral fourier > transform. If you make a trapezoidal approximation to the integral, > the ?st and last points must have weight 0.5 relative to the > internal points. For a signal that decays to 0, the last point doesn't > matter, but the ?st one does.I don't think so; the periodic boundary conditions of the DFT nullifyyour argument about the trapezoidal rule, since they remove anyspecialness about the ?st and last sample points (except for thearbitrary choice of phase of the Fourier kernel, of course, whichdepends on what you call t=0). === > Have you remembered to halve the t=0 input datum?>> I`m sorry i don`t understand. why is this necessary? >> >> Think of the ?ite fft as an approximation to the integral fourier >> transform. If you make a trapezoidal approximation to the integral, >> the ?st and last points must have weight 0.5 relative to the >> internal points. For a signal that decays to 0, the last point doesn't >> matter, but the ?st one does.> > I don't think so; the periodic boundary conditions of the DFT nullify> your argument about the trapezoidal rule, since they remove any> specialness about the ?st and last sample points (except for the> arbitrary choice of phase of the Fourier kernel, of course, which> depends on what you call t=0).I don't think the OP is interested in a ?strict' dft - he wants a dft thatapproximates the integral ft - since he is expecting an exponential totransform to a lorentzian. This doesn't work right unless you halve theweight of the t=0 point.For a discussion in the context of 2d NMR, which may be the OP's realproblem, see Wutrich et al. J. Mag. Res. 66, 187 (1986). === Is there a relationship known for the distance between the largest root ofan orthogonal polynomial and the endpoint of the orthogonality interval? Iwould like to know how fast this distance shrinks as a function of thedegree of the polynomialgert === >> Is there a relationship known for the distance between the largest root of> an orthogonal polynomial and the endpoint of the orthogonality interval? I> would like to know how fast this distance shrinks as a function of the> degree of the polynomial>I'm not an expert on this but glanced through Abramowitz & Stegun. There isno general formula for all orthogonal polynomials (well, unless if such hasbeen found during last 30 years; you might want to search mathscinethttp://www.ams.org/mathscinet . I can't do it from home), but let me picktwo examples. Here x_n is the biggest root of the polynomial of order n:1) Chebyshev 1st kind, orthogonal on [-1,1]: x_n = cos((2n-1)pi/2n)2) Legendre, orthogonal on [-1,1]: x_n = 1-(j_n)^2/2 * 1/n^2 + O(1/n^3).Here j_n is the nth biggest positive zero of Bessel function J_0(x).All the best,Teijo === Does anyone have any algorithms for calculating an optimum blackjack betting strategy that will make me rich?I've had great success lowering my bet after winning. My thinking is that if you have a base bet of say $8 and you win, then betting only $5 the next hand, basically means that the house has to beat you 2 hands to get your original bet back. What I'm wondering is this:I understand that the odds of ?g a coin and getting heads is still 50:50 even if you got heads 10 times in a row. However, in theory, if you ? the coin an in?ite amount of times, then wouldn't it be safe to say that you could expect to get heads 50% of the time? So, if the house has 50:50 odds, because you are wisely counting cards, then won't my betting strategy work if I play for several hours on end? === here is pointer where you ?d some comparisons: E.F.F. Botta, K. Dekker, Y. Notay, A. van der Ploeg, C. Vuik, F.W. Wubs, P.M. de Zeeuw, How fast the Laplace equation was solved in 1995, J. Applied Numerical Mathematics 24 (1997) 439--455.It is available with me as: http://homepages.cwi.nl/~pauldz/journals/laplace9597. pdfAnother pointer that might be relevant, because of the comparisonof multigrid and Bi-CGSTAB, reads: Chapter 5, Application of Bi-CGSTAB to discretized coupled PDEs, in: Acceleration of Iterative Methods by Coarse Grid Corrections, (Ph.D. Thesis, University of Amsterdam, 1997).It is available with me as: http://ftp.cwi.nl/pauldz/Thesis_PDF/Chapter5.pdf Paul de Zeeuw http://homepages.cwi.nl/~pauldz/ === > > My ?ted values, and approx. std. errors using a Fortran program are:> a = 0.000250(0.000070)> b = 6.95 (0.20)> c = 25.69 (0.49)> I see that someone else agrees with these values.Many thanks to those who have responded to my request.What sort of algorithm are you using in your Fortran program?Levenburg-Marquardt was mentioned by another poster, and I have codefor that in _Numerical Recipes_, though I was hoping for a simplermethod. I'm not sure how to port Fortran code into Labview (theprogramming is being contracted out), though I imagine there is a wayto do it.-- Allen Windhorn (507) 345-2782 FAX (507) 345-2805Kato Engineering (Though I do not speak for Kato)P.O. Box 8447, N. Mankato, MN 56002Allen.Windhorn@LSUSA.com === > Is Nature a Logical System?Error, missing operator or `;`> Is; Nature; a; Logical; System; Is Nature a Logical System === Jon Briggs,> Is Nature a Logical System?> The answer is obviously that we don't know and that we can't know.True, we cannot know with perfect certainty, but we can still observethat it does happen to have the properties of a logical system.At this point, we can ask the question: Is Nature a Logical System?While we cannot perform an experiment that will tell us, it still maybe useful to discuss what the consequences of an answer would mean. If the answer is no, the discussion probably ends right there. But ifthe answer is yes, we are now allowed to make more statements aboutit. Such as (from the original post):Unless nature is presumed to be an exception to theIncompletenessTheorem, to contain all the statements of nature completely andconsistently the system of all observations would require additionalstatements beyond what is observed. Because these new statements liebeyond the set of all observations these statements cannot be madeusing the scienti? method, and therefore, physics is concluded to beunable to provide a ?al theory of nature.> You don't need to invoke an incompleteness theorem to know that> we have no complete and certain knowledge about the universe. That's> a physical fact of life from the start.If what you say is true, this fact must be demonstrated in amathematical model of the universe.And if what I say is true then we can create a model of the universethat produces this fact.Mike Helland === Is Nature a Logical System?I think it is. The following explains my position and makes > several insightful conclusions beginning with this premise.Any challenges to my reasoning [or] con?mations of its > validity are enthusiastically anticipated.[]> The above is section II from the following paper:> http://www.techmocracy.net/science/time.htmIs Nature Logical?I think a better question would be, Is Logic Natural?I would answer that Logic seems to be Natural, so far.If it ever ceases to be, I suppose it will be up tothe survivors to decide how it will need to be reformed.Jim Burns === >> Is Nature a Logical System?>> >> I think it is. The following explains my position and makes >> several insightful conclusions beginning with this premise.>> >> Any challenges to my reasoning [or] con?mations of its >> validity are enthusiastically anticipated.>[]>> The above is section II from the following paper:>> http://www.techmocracy.net/science/time.htm>>Is Nature Logical?>I think a better question would be, Is Logic Natural?>quoting from http://www.druidic.org/bennish/essays/status.htmfor the thesis that logic is empirical. By this he means that the lawsof logic, like any other scienti? laws, can be tested empiricallyand modi?d if they do not ? with our experience.(later in the essay)A rather more serious objection was raised by Crispin Wright in hispaper Inventing Logical Necessity (1986). Wright argues that Quinestheory of knowledge cannot treat logical beliefs simply as another setof beliefs clustered near the centre of our array. Logic is needed toform the links between our beliefs, and as such they have an entirelydifferent role to speci? beliefs about states of affairs (the catsits on the mat) or scienti? theories (cats tend to fall frommantelpieces to mats with a constant acceleration of 9/8ms-2).(the essay concludes:)Quines picture falls apart if we have no logic to provide connectionsbetween the beliefs in our array. Nevertheless, that is only anargument for the need for some sort of logic for us to be able tothink coherently it does not prove that the speci? logical beliefsthat we hold at any one time could not be changed.Nevertheless, the thought is persuasive that logical statements are inprinciple open to such veri?ation within the sort of structure thatQuine has proposed. John Bailey === >> Is Is Nature a Logical System? a mathematical question?>> Is Is Is Nature a Logical System? a mathematical question?> a logical question?>> Is Is Is Is Nature a Logical System? a mathematical question?>> a logical question? a grammatically-correct question?>>Is Is Is Is Is Nature a Logical System? a mathematical question?>a logical question? a grammatically-correct question? a genuine>question?Yes. Oh, sorry, was I supposed to say something different? === >> Is Is Nature a Logical System? a mathematical question?>> Is Is Is Nature a Logical System? a mathematical question?>> a logical question?>> Is Is Is Is Nature a Logical System? a mathematical question?> a logical question? a grammatically-correct question?>>Is Is Is Is Is Nature a Logical System? a mathematical question?>>a logical question? a grammatically-correct question? a genuine>>question?>>Yes. >>Oh, sorry, was I supposed to say something different?Well, in the interest of grammatical correctness you were supposedto spell it grammatically correct instead ofgrammatically-correct.>************************David C. Ullrich === > Is Is Nature a Logical System? a mathematical question?>> Is Is Is Nature a Logical System? a mathematical question?> a logical question?>> Is Is Is Is Nature a Logical System? a mathematical question?>> a logical question? a grammatically-correct question?>>Is Is Is Is Is Nature a Logical System? a mathematical question?>a logical question? a grammatically-correct question? a genuine>question?>> Yes.>> Oh, sorry, was I supposed to say something different?Ah, you spoiled it :-(Dirk Vdm === >> Is Is Nature a Logical System? a mathematical question?>>Is Is Is Nature a Logical System? a mathematical question?>>a logical question?>>Is Is Is Is Nature a Logical System? a mathematical question?>a logical question? a grammatically-correct question?Is Is ?[]' a grammatically-correct question a _grammatically correct_ question?************************David C. Ullrich === In sci.math, David C Ullrich< qkcrpvkbftffmheq0s36mbi4pvmri9kdni@4ax.com>:>> Is Is Nature a Logical System? a mathematical question?>>Is Is Is Nature a Logical System? a mathematical question?>a logical question?>>Is Is Is Is Nature a Logical System? a mathematical question?>>a logical question? a grammatically-correct question?Is Is ?[]' a grammatically-correct question a > _grammatically correct_ question?************************David C. UllrichIs all this leading anywhere? My brain is beginning to hurt:-)-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === > Is Nature a Logical System?It seems pretty illogical to me.> Any challenges to my reasoning are con?mations of its validity are> enthusiastically anticipated.One of the most common ?r p seudoscience.> what is meant by the de?ition. If nature will be de?ed as the set> of all observed phenomena, what is all observed phenomena supposed to> mean?Observed by who or what? This is ill-de?ed. How do you decide who or what counts as an observer, what counts as an observation, and why is this a reasonable decision?> Statement b is a little tricker. More initial conditions will be> required beyond the ladybug simply being in the picture. I need to> de?e what an inch is and then produce a 12-inch ruler with 1/8> marks along it. Once I have these, I then follow the simple procedure> of placing the ladybug along the edge of the ruler, counting the> number of marks covered by the ladybug, and multiplying that number by> 1/8. When follow those steps I observed that the ladybug covered 2> marks (similarly to the way I observed the lady bug to produce> statement a) and the measurement resulted in 1/4. Which was the> statement I needed to produce.The result of a measurement is depends on many factors. For example, the measurement method (in this case, the ruler, which only a 1/8 resolution), the observer's judgement (the end of the ladybug will almost certainly fall between two marks on the ruler and it is a matter of judgement which mark to choose), and other unpredictable factors.> If the generic rules underlying statement b produce a system of> statements, is statement a also contained in a system of statements?What the hell is a system of statements?> 1. The logical rules can produce many statementsThe system of rules used to measure the lady bug was also used to> measure my laptop. The collection of all the statements produced by> these rules can be considered all the true facts of that logical> system.This system not nearly all of nature.> According to G?del's Incompleteness Theorem this logical system can> only contain all the true facts consistently with each other by> creating a larger encompassing system.This is not true. Goedoel's incompleteness theorem dealt with axiomatic systems. Your system does not seem to be axiomatic. It is also not clear that your system contains natural number arithmetic in the usual sense.> 4. At basic levels, all systems are governed by the laws of physicsThis is not true. Even physicists admit that the laws of physics do not cover all situations, or even all situations that are obviously physical. For example, none of the laws of physics say anything about what happens inside the event horizon of a black hole.> can be produced from the initial conditions of the universe and the> rules of the universe which we have attempted to describe with our> laws of physics.Are you trying to refute quantum mechanics? Not that it can't be done but you'll need to do better than that.> But metaphysics and philosophy are not limited to making statements> about what can be observed and might be capable of providing a> consistent theory of nature. This consistent theory of nature can alsoNor is physics. It is important in physics to eliminate statements about non-observable things as much as possible, but no statement that claims to be a universal law of nature can avoid this completelyHave a tolerable existence. Eli === > Robin Chapman,> Is Is Nature a Logical System? a mathematical question?>> If mathematics is:>> The study of the measurement, properties, and relationships of> quantities and sets, using numbers and symbols.>> And if my question is:>> can nature (de?ed as the collection of statements we make about the> universe using measurement and general observation) be considered a> logical system that is governed by rules of mathematics such a the> incompleteness theorem.>> Then I'm thinking that my question can most likely be answered by a> mathematician.>> The answer is obviously that we don't know and that we can't know.>> If Nature is a logical system, we can't know. Because even if we> stumble upon the correct formulation for it, we cannot> know if the next experiment we perform will produce a result that> is inconsistent with that formulation.>> If Nature is not a logical system, we can't know. Because no> matter what ?ite set of measurements we perform, there will always> be a ?ite logical system (perhaps requiring extremely large numbers> of ad hoc parameters and hidden variables) consistent with those> measurements.>> You don't need to invoke an incompleteness theorem to know that> we have no complete and certain knowledge about the universe. That's> a physical fact of life from the start.But no one, sob sob, bothered to reply in sci.physics, sob, sob. ;-)Dirk Vdm === Dirk Van de moortel> But no one, sob sob, bothered to reply in sci.physics, sob, sob.You're absolutely correct that I posted this here after no one repliedin the physics group. I have something I'd like to discuss with peopleand I tried the topic here because the heavy use of mathematicalmodels and logical systems.It seems to have generated some discussion (with some noise even) andthats all I'm looking for.Mike === It seems to have generated some discussion (with some noise even) and> thats all I'm looking for.The discussion, or the noise?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === >>There are 3000 lucky draw tickets priced at $2 each. The>grand prize is a $500 color TV and there are ?e $200 black-and-white>TV's as consolation prizes. What is the expected value of a $2 ticket>in this draw?The answer is: ?ty three gazillion dollars.Go write that on your homework also, this problem is verydated there's not a black and white TV in the world that'sworth 200 dollars.adam === > Logic seems so central to our thinking, I want to say it's biological.> But if that's true it is also subjective. Which I'm not sure I agree> with.Where do you stand? A Chomskian theory appeals. What's the consensus of the mathematical> community?I don't know, I'm not a mathematician. Maybe there isn't a consensus.I sense this objective vs. subjective nature of logic is still a hottopic for debate.I have the feeling that Modus Ponens (or A => B, A then B) emergesfrom the causality of our temporal perceptions.I think the Pavlovian Response is empircal evidence for this:Ring a bell, Feed the Dog. Ring a bell, Feed the Dog. Ring a bell, Dog salivatesA, BA, B A, B (I've begun to assume A implies B)A, Where's B?Do you think this has merit? Are there other elements of logicexhibited in very basic behavioral patterns? === Logic seems so central to our thinking, I want to say it's biological.> But if that's true it is also subjective. Which I'm not sure I agree> with.>> Where do you stand?>> A Chomskian theory appeals.>> What's the consensus of the mathematical> community?>> I don't know, I'm not a mathematician. Maybe there isn't a consensus.I sense this objective vs. subjective nature of logic is still a hot> topic for debate.I have the feeling that Modus Ponens (or A => B, A then B) emerges> from the causality of our temporal perceptions.I think the Pavlovian Response is empircal evidence for this:Ring a bell, Feed the Dog.> Ring a bell, Feed the Dog.> Ring a bell, Dog salivatesA, B> A, B> A, B (I've begun to assume A implies B)> A, Where's B?Do you think this has merit? Are there other elements of logic> exhibited in very basic behavioral patterns?Many years ago when I was a student of philosophy I read Popper writingagainst this Pavlovian idea. The observation (made by the dog) of A isfollowed by B does _not_ give rise to the hypothesis A causes B. Can't remember why.I would warn you against confusing A causes B with the logician's ifA then B. For the logician, if A then B is de?ed by the tableA B if A then Bt t tt f ff t tf f twhere t means true and f means false. The ancient Greeks discussedthis and it was taken up again in the late nineteenth century.-- G.C. === > Many years ago when I was a student of philosophy I read Popper writing> against this Pavlovian idea. The observation (made by the dog) of A is> followed by B does _not_ give rise to the hypothesis A causes B. > Can't remember why.I would warn you against confusing A causes B with the logician's if> A then B. For the logician, if A then B is de?ed by the tableA B if A then B> t t t> t f f> f t t> f f twhere t means true and f means false. The ancient Greeks discussed> this and it was taken up again in the late nineteenth century.I agree that the Pavlovian response is an inference, whereas theLogician's assertion that A implies B, is necessarily true byde?ition.But isn't this how we learn and develop logical models of ourenvironment? I don't think we could live without making inferencessuch as these from experience.Furthermore, isn't all reasoning basically based upon this idea o?ference?SomeEquation => SomeOtherEquationSomeTruth => SomeOtherTruthOther than that, it appears variable subtitution is all that isnecessary.Of course I may be terribly confused and misinformed, but this seemsto me to be the way things work.I have a deep respect for the basic ideas of Karl Popper, Thomas Kuhnand the Logical Positivists camp in general. I like to learn moreabout his argument against Pavlovian inferences.-Steve === >May I ask, then, in your opinion, is Newton's calculus more effective than>Leibniz's? I realize he has won the debate over the years, as we are taught>in school that he is the founder of the calculus. However, I am>interested to know how your personally feel about it, being a physicist.>>thanks for your time,>>Kavon>P.S. I agree, dy/dx instantly has me thinking of a fraction. However, my>math skills are surely limited compared to yours.Like Thomas said in his post, there is no difference betweenNewton's calculus and Leibniz's calculus.If someone is just starting to learn calculus then I think thatthe notation of Leibniz is better to use. It's de?itly moreintuitive because you can think fraction which is notalways bad in fact, if you have to think about what a dirivative is, then you might a well think of it as a fraction. Just don't take the analogy too far.As for how I personally feel about the Newton/Leibniz debateI don't really have an opinion one way or the other. MaybeNewton came up with it ?st, maybe Leibniz did, but no onewill ever know for sure. The best thing is probably just to give them both equal credit.adam === Adam,Kavon>>May I ask, then, in your opinion, is Newton's calculus more effectivethan>Leibniz's? I realize he has won the debate over the years, as we aretaught>in school that he is the founder of the calculus. However, I am>interested to know how your personally feel about it, being a physicist.>>thanks for your time,>>Kavon>P.S. I agree, dy/dx instantly has me thinking of a fraction. However, my>math skills are surely limited compared to yours.>> Like Thomas said in his post, there is no difference between> Newton's calculus and Leibniz's calculus.>> If someone is just starting to learn calculus then I think that> the notation of Leibniz is better to use. It's de?itly more> intuitive because you can think fraction which is not> always bad in fact, if you have to think about what a> dirivative is, then you might a well think of it as a> fraction. Just don't take the analogy too far.>> As for how I personally feel about the Newton/Leibniz debate> I don't really have an opinion one way or the other. Maybe> Newton came up with it ?st, maybe Leibniz did, but no one> will ever know for sure. The best thing is probably just to> give them both equal credit.>> adam> === See comment on Tony Smith's conformal gravitons below.Message 1Let me take a moment away from the TV, where we've been watching the progress of the horrendous local ?es and hoping they don't come any closer to Pasadena, to comment on Robert Sheaffer's recent comment that It is exceedingly unlikely that UFOs are real entities, they are mostly due to ?random errors' occurring in the human perceptual and social-belief formation system. There is noise in every information channel, and our society has agreed to describe a certain kind of that noise as UFOs. While we may argue about their origin and purpose, it has long since become abundantly clear that what we call UFOs are not simply noise. Unless the thousands of internally consistent reports by credible witnesses, including abductees, are some form of mass hallucination, these phenomena clearly have an objective existence. Moreover, as a student of comparative mythology and folklore, I can ?mly attest that the patterns reported by contemporary witnesses and experiencers jibe remarkably well with pre-modern accounts of ?shields, abductions to fairy-land by short creatures with big eyes and pointed, el? chins, missing-time episodes, and rides in ?wheels a la Ezekiel's famous trip.To ignore all this evidence, soft though it most of it may be (although I strongly suspect that despite a half-century of stonewalling, some members of the U.S. and other intelligence communities-the CIA, MI6, the old KGB, etc.-have in their possession an abundance of hard evidence garnered from Roswell, Kecksburg, and other UFO crash sites), is to wallow in a state of denial so all-embracing that it beggars the imagination. Or do debunkers like Sheaffer, Oberg, Klass, and the rest have another, more devious agenda, that is, to further the Government's long-standing policy of keeping the lid on by systematically ridiculing those of us who suspect the truth about this phenomenon, all the while being privy to above-top-secret knowledge that would prove us right? One > wonders.All best wishes &ScottC. SCOTT LITTLETONPresident, Phi Beta Kappa Alumni in Southern CaliforniaProfessor of Anthropology, EmeritusOccidental CollegeLos Angeles, CA 90041TEL (323) 255-5477FAX (323) 982-0264http://www.oxy.edu/~yokatta/home.htmAny suf?iently advanced technology isindistinguishable from magic --Sir Arthur C. ClarkeI think we're property. . . --Charles FortAgain, the basic exotic physics of Chapter 9 of Sir Martin Rees's WMD book Our Final Hour as well as how UFOs, if they are real, would have to ? metrically engineered weightless warp drive (local absence of g-force) in my opinion is summarized in http://qedcorp.com/APS/StarGate1.mov However, again I emphasize that the physics there in no way depends on whether or not UFOs are real. If they are not real now, if my physics is on the right path, they will be!Message 2minimization of collateral damage to civilian population, maximizing at the same time ef?iency of in?ng of decision-making of national leaders of target countries.Dr. Sergeyev told me that similar data were in possession of our Western colleagues; - with whom Russian military elite routinely shared most p. 94-95 of ?97 edition of his book), so as not to break mutual trust which during Cold War years had been the principal basis of Global Stability. (Of course, it wasn't necessary to carry across borders carloads of technical papers - for intelligent enough people, comprising intellectual elite, it was suf?ient few words. Here can be drawn parallels with ?st years of Atomic Era: sources of Soviet atomic spies in USA - including such top scientists as Einstein, Oppenheimer, Bohr & others - needed just to tell few words to such Soviet scienti? agents as Ya. Terletsky etc.)See my comments on this in Pavel Sudoplatov's Special Tasks paperback ed. with foreword by Robert Conquest of the Hoover Inst. Stanford.Majority of competent Soviet (and post-Soviet) experts in this area feel it to be morally permissible to use in cases of extreme necessity surgical psychotronic strikes against selected human targets instead of destructive & inef?ient massive interventions like Vietnam, Afghanistan & Iraq; - which bring much needless suffferings to innocent civilians, lots of political problems, terrible ?ancial expenses - and almost no positive results. Regretfully, collapse of f. USSR stopped development of selective varieties of psychotronic genome weapons which might be precisely aimed at genome patterns of some chosen human target - this was long before lists of death genes became freely obtainable from Internet. As I've mentioned, in 1989 in Kiev - then an important center of Soviet psychotronic research - there were performed tests of non-selective torsionic weapons, capable to interfere with some of biochemical reactions inside the cell; - whose in? could not be shielded by any material screens. However, these - compact enough to be installed on a satellite & requiring very little electric power - devices couldn't function as an ef?ient weapon, as their effects took too much time (perhaps, years) to manifest themselves (because of this such weapons weren't used in Afghanistan & Chechnya). Soviet experts guessed that in order to create really ef?ient swift weapon there must be undertaken more thorough research, involving use of supercomputers for precise calculation of resonance frequencies of vulnerable genes (in fact, they believe that this is the main task of recently built in USA for genetic research Blue Gene supercomputer with tera?ce ssing speed).My comments on above. I cannot comment on the particulars of the above allegations. We did have Gennady Shipov here from Moscow for extended visits in 1999-2000 as part of the ISSO effort to try to ?ure out the torsion physics. I did not make much progress until perhaps only a few days ago because of a remark from Professor Thomas Angelides in London about the structure of the Penrose Masseless Twistor Conformal Group SU(2,2) or SU(4) depending on topological signature in 4 complex dimensions.Professor Angelides pointed out that 4 of the 5 extra elements of the Lie Algebra of the Conformal Group are a kind of mirrored version of the translation subgroup of the Poincare group. The Fifth Element is the dilation group implicit in in? where in the FRW metricR(t) ~ e^HtH is the Hubble not the Hamiltonian. ;-)Now in my toy model, the Dirac vacuum for the electron in globally ?ace-time is unstable to the BCS formation of a ODLRO BEC of virtual electron-positron pairs due to virtual photon exchange near the edge of the -mc^2 Fermi surface.This results in the microscopic derivation of the Vacuum Coherence Field, which in the large scale is the In? FieldSince the Vacuum Coherence Field is a complex scalar ?ld the ONLY topological defects in 3D space must be 1D strings sweeping out a world sheet as further explained in this thread by Saul-Paul Sirag. This is a nice after thought.This is a toy model. The rest of the lepto-quarks will have to be considered later.Note that Haisch, Puthoff, Rueda, Marshall, Cole et-al zero point energy gravity program fails to ask the right questions because vacuum coherence is entirely lacking in their model.Utiyama in the 60's showed that local gauging of the classical Poincare group of special relativity gave Einstein's curved space-time geometrodynamic symmetric tensor ?ld from the 5 energy-momentum generators of the Lie algebra of the translation subgroup. Locally gauging the 6 space-time rotations of the Lorentz subgroup of the invariant light cones of null geodesics ds = 0 gave the torsion antisymmetric tensor ?ld years before Shipov rederived it in Moscow probably without knowing of Utiyama's work in Japan.It is common knowledge that the rest mass of elementary lepto-quarks and the weak radioactive gauge force bosons required breaking down the 15 parameter Conformal Group to the 10 parameter Poincare group.In my theoryEinstein's Diff(4) symmetric geometrodynamic ?ld comes from the modulation or rippling of the coherent Goldstone phase of the Vacuum Coherence Field.Indeed, the 4 generators of the translation subgroup of the Poincare group of special relativity are Pu ~ h/id/dx^uThe Sakharov metric elasticity ?ld is simply (without holographic t'Hooft-Sussking renormalization at ?st) (Lp^2/ih)Pu(Goldstone Phase).Lp^2 = hG(Newton)/c^3 ~ 10^-66 cm^2I need a second set of generators P'u from the Conformal Group operating vacuum ?ldwhere ifLp* = Lp^2/3(c/H*)^1/3where H is the Hubble , which clearly needs a more a-temporal meaning in terms of the constant total mass M of the Universe inside the expanding Hubble horizon.That isGM/c^2 = c/H* independent of cosmic time t and the FRW scaling R(t) of the dimensionless comoving r coordinate of the Hubble ?= M(on mass shell real matter + radiation) + M(near EM ?lds) + M(exotic vacuum ?lds)M is a contingent WAP initial condition of the Level 1 parallel universe M(exotic vacuum) = M(dark energy) + M(dark matter)M(dark energy) ~ 0.73MM(dark matter) ~ 0.23MM(on mass shell real matter + radiation) + M(near EM ?lds) ~ 0.04MNot to be confused with the LARGE-SCALE ONLY cosmic time-dependentH(t) = R(t)^-1(dR(t)/dt)Note that the time dependent densities scale asR(t)^-3 for real matter with w ~ 0R(t)^-4 for real EM radiation like the CMB with w = +1/3and independent of R(t) for exotic vacuum with w = -1 from Einstein's equivalence principle + Heisenberg's uncertainty principleWhere the equation of state of ALL STUFF REAL OR VIRTUAL isw = pressure/(energy density)The electron rest mass ism ~ (e/c)^2/zpf*^1/2from the strongly attractive dark matter vortex ring core of positive vortex string of a super?hat is, in the sense of a Taylor series expansion about the normal vacuumfrom pouring macro-quantum coherent oil on the random chaotic micro-quantum zero point ?tions of ALL quantum ?lds. This stillness was the calm before the storm of the Big Bang reheating post-in? in this decoding of The Cipher of Genesis (Carlo Suares). ;-)that can be positive or negative giving Dark Energy or Dark Matter respectively.The Pu' are from the mirror translation subgroup of the Conformal Group.Let PSI = Vacuum Coherence Field= Translation Gradient Ripple + Mirrored Conformal Translation Gradient Ripple= Einstein Gravity Ripple + Exotic Vacuum (Dark Energy/Matter) RippleThe quanta normal ?oise of these ripples in the emergent More is different (PW Anderson) vacuum super?ignal areEinstein's gravitons and Tony Smith's conformal gravitons. === > >> --snip> The above formula also implies, that your expression for e(r) will> not be bounded for r->1. However, differentiating phi wrt r does> not give [2] (btw, do you really call the potential differentiated> wrt r the electric ?ld? For me, the electric ?ld is the gradient> of the potential, so the electric ?ld in this case should be> the potential differentiated wrt r times the gradient of r.)> What I get when [1] is differentiated wrt r is>> 1/(pi*r^2) ((a^2+2*r^2)*E[z^2]-(2*a^2+r^2)*K[z^2]).>> However, this expression is still unbounded as r->a.> yes sorry. i forgot the gradient of r. however it is still unbounded> which>> is very disturbing. it should be bounded!>> This may be a stupid question, but my physics knowledge is somewhat>> limited. Why must the electric ?ld be bounded?>>well to me - it would seem that anything where the force was becomming> singular would imply that we were approaching a point like source generating> the electric ?ld. but why would this make sense when we are dealing with> a *constant* distribution of charge. It seems reasonable that if you have a> ?ite amount of charge spread evently over the disc, then forces for test> charges in that domain should be ?ite as well. shouldnt they? check my> calculations:1. start with poissons formula in axially symmetric cylindrical coords -> with a source term ofLaplacian(phi) = -sigma(r)*delta(z)*U(a-r)where U = heaviside step function. delta = dirac delta. sigma charge> density. Solve using hankel transforms and then integrate the dual bessel> functions to get the elliptic integral - integrate again to get the result.i should note that above you say my differentiation is incorrect. i fear> that this comment was probably due to bad notation on my part. remember> that z=a/r in the arguments of the elliptic funcitons. i probably should> have made it s = a/r instead or something similarso we havephi = 2/(pi*r) * [ r^2*E(s^2) + (a^2-r^2)*K(s^2) ]> I had not too much time, but yesterday I did a quickback-of-the-envelope calculation and obtained a resultwhich looks quite different. The main difference was,that the arguments of the elliptic functions lookdifferent. In fact, for a=1 I got expressions of theformE(-4 r/(1+r^2+z^2)) and K(-4 r/(1+r^2+z^2)).Furthermore, I am now quite sure that the electric ?ldwill get unbounded as you approach the edge of the disc.I am not sure at all, that I did not miscalculate, butI recommend that you go over your computations again.HTH,Michael.-- &&&&&&&&&&&&&&&&#@#&&&&&&&&&&&&&&&&Dr. Michael UlmFB Mathematik, Universitaet Rostockmichael.ulm@mathematik.uni-rostock.de === JacoIf I have an element g(x) in R[x], where R[x] is a polynomial ring> with coef?ients from a ?ite ?ld GF(2^m), how do I de?e the> multiplicative inverse of g(x)? (The multiplicative identity would be> e, the multiplicative identity of GF(2^m).)What I do not understand is how two polynomials with positive degree> greater or equal to one can be multiplied to give e (which is a> polynomial of degree 0 in R[x])?Your time, effort and suggestions will be greatly appreciated> Jaco === > If I have an element g(x) in R[x], where R[x] is a polynomial ring> with coef?ients from a ?ite ?ld GF(2^m), how do I de?e the> multiplicative inverse of g(x)? (The multiplicative identity would be> e, the multiplicative identity of GF(2^m).)What I do not understand is how two polynomials with positive degree> greater or equal to one can be multiplied to give e (which is a> polynomial of degree 0 in R[x])?Is it possible that you mean to be working, not in R[x], where R = GF(2^m), but in R[x] / < f(x) >, where < f(x) > is the ideal generated by some polynomial in R[x]? If so, then after you multiply your two polynomials you get to reduce the product modulo f, making it more likely you'll get a polynomial of degree zero.-- === > What I do not understand is how two polynomials with positive degree> greater or equal to one can be multiplied to give e (which is a> polynomial of degree 0 in R[x])?If the ring of coef?ients has zero divisors, products of polynomials can be very odd.Let the ring be Z/9Z, and consider the product of 3 x^2 + 3 x + 3 with itself, for example. === >What is the ?st number that can be divided byt he frist three prime>numbers?>Go through all the numbers from one to twenty and askIs this number divisable by 2 AND 3 AND 4?If you ?d one that is, let us know.adam === >What is the ?st number that can be divided byt he frist three prime>numbers?>> Go through all the numbers from one to twenty and ask> Is this number divisable by 2 AND 3 AND 4?If you ?d one that is, let us know.Seems like a little bit of attitude from one who thinks that 4 is prime!adam === >>What is the ?st number that can be divided byt he frist three prime>>numbers?>Go through all the numbers from one to twenty and ask>Is this number divisable by 2 AND 3 AND 4?>>If you ?d one that is, let us know.>>adam>Whoops, I meant to say:Go through all the numbers from one to thirty and askIs this number divisable by 2 AND 3 AND 5?..and of course you could say zero since0/2 = 0/3 = 0/5 = 0adam === Suppose I have points X(x_0,,x_n) and Y(y_0,,y_n) with y_i and x_iin R^3. I want to ?d an approximate nonlinear warping based onlyon translation, rotation, and shearing that maps X->Y, whatis the best way of doing this? Can I produce a warping whichis optimal int he least squares sense? === >Suppose I have points X(x_0,,x_n) and Y(y_0,,y_n) with y_i and x_i>in R^3. I want to ?d an approximate nonlinear warping based only>on translation, rotation, and shearing that maps X->Y, what>is the best way of doing this? Can I produce a warping which>is optimal int he least squares sense?I suppose you want a map F: R^3 -> R^3 such that F(x_i) = y_i fori=0n. But I'm not sure what you mean by based only ontranslation, rotation, and shearing, as those are all linear maps.I think what you want to do is write a general formula for your mapwith various unknown parameters, and then the constraints F(x_i) = y_irepresent a set of equations in those parameters. The best situationis where the map is linear in the parameters (even though it's nonlinearin the x coordinates), in which case you have a good chance of beingable to solve for the parameters, if there are enough of them. Or you can do the standard linear least-squares if you'd be satis?d with an approximation rather than a perfect ?.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === > A long time ago, I vaguely recall an elementary school teacher telling> us some trick for quickly multiplying two numbers (2 digits, I believe)> with one or both having 5 as the second digit. Anyone recall the trick?> Numbers like 35*45 or maybe even 20*35.>The trick for multipling by 5 is multiply by 10 and divide by two.That is add 0 to the end of the number and then divide by two. === Dear Ladies and Gentlemen,Im searching for a function of either velocity or accelaration ( d(t) = f(v)or/and d(t) = f(a)), which expresses the damped oscillation d(t) caused byvarying velocity or accelaration.I found this formula, which express a mass m in [kg] hanging in a spring kin [N/m],d(t) = C*[e^(-t/tau)]*cos(o*t - phi)where:d(t) = distance in [m] to the time t in [s],C = distance different from neutral position in [m],o = SQRT(kg/m), resonance frequency in [rad/s]tau = time constant for damping oscillationphi = initial phasebut it does only express oscillation beginning in t=0. I need a formulawhich express d(t) to any time t, for the purpose of simulation in acomputer.I'm amatuer and not very well in the english language, but I hope myexplanation is understandable.Torben W. HansenDenmark === > Dear Ladies and Gentlemen,>> Im searching for a function of either velocity or accelaration ( d(t) =f(v)> or/and d(t) = f(a)), which expresses the damped oscillation d(t) causedby> varying velocity or accelaration.>> I found this formula, which express a mass m in [kg] hanging in a spring k> in [N/m],>> d(t) = C*[e^(-t/tau)]*cos(o*t - phi)>> where:> d(t) = distance in [m] to the time t in [s],> C = distance different from neutral position in [m],> o = SQRT(kg/m), resonance frequency in [rad/s]> tau = time constant for damping oscillation> phi = initial phase>> but it does only express oscillation beginning in t=0. I need a formula> which express d(t) to any time t, for the purpose of simulation in a> computer.>> I'm amatuer and not very well in the english language, but I hope my> explanation is understandable.>> Torben W. Hansen> Denmark>Simply replace t with t-t0 ied(t)=C*[e^(-(t-t0)/tau)]*cos(o*(t-t0) - phi), t>=t0 and t0 is starting timeI hope this helpsGoran === Torben W. Hansen schrieb:Dear Ladies and Gentlemen,Im searching for a function of either velocity or accelaration ( d(t) = f(v)> or/and d(t) = f(a)), which expresses the damped oscillation d(t) caused by> varying velocity or accelaration.I found this formula, which express a mass m in [kg] hanging in a spring k> in [N/m],d(t) = C*[e^(-t/tau)]*cos(o*t - phi)to start in t0 you should take d(t) = C*[e^(-(t-t0)/tau)]*cos(o*(t-t0) - phi)hth (and I hope you wanted this simple solution)klauswhere:> d(t) = distance in [m] to the time t in [s],> C = distance different from neutral position in [m],> o = SQRT(kg/m), resonance frequency in [rad/s]> tau = time constant for damping oscillation> phi = initial phasebut it does only express oscillation beginning in t=0. I need a formula> which express d(t) to any time t, for the purpose of simulation in a> computer.I'm amatuer and not very well in the english language, but I hope my> explanation is understandable.Torben W. Hansen> Denmark === Martin>> First, I am not a mathematician :-) I will thank you for any simple> explanation you may want to give.>> I have a list of edges in a graph:>> EdgeID: nodeA -> nodeB.>> I have a given start node and want to get a list of all edges that are> connected directly or indirectly to this node. I _don't_ need thetravelling> salesman to ?d the shortest possible route, it's just that there maybe> separate closed graphs in this edge cloud and I just need all theedges in> the special sub-graph the given node is in.>> Hope, I could make myself understandable.>> I will also appreciate any hint on literature understandable for a> non-mathematician.> Martin>> Have a look at:>> http://planetmath.org/encyclopedia/ConnectedGraph.html> http://www.cs.sunysb.edu/~algorith/?es/dfs-bfs.shtml>> Or any standard implementation of Depth First Search or Breadth First> Search.>> Jack> === > quantify Baseball to other sports; Optimal Strategy for> Baseball> Archimedes Plutonium > NOdtgEMAIL> whole entire Universe is just one big atom where dots of> the electron-dot-cloud are galaxies> sci.logic, soc.history, sci.math> A friend asked me to watch this years World Series to render my> opinion.> I sketchedly watched and here is my opinion, as I usually do not have> time for such recreation.I watched only parts of game 2 & 3 where the Florida Marlins were> being> overpowered, and missed game 1. And from game 3 I decided it was a> waste> of time to watch games 4 or 5 in that the New York Yanks would> probably> overpower the Marlins and so only watched a few innings of pitching. I> watched nearly the full game of 6.I had seen some clips of Baseball sluggers such as B. Bonds and S.> Sosa> (excuse me if name is incorrect spelling) from the sports section of> the> local TV news when getting the weather report so I have some awareness> of the best hitters of Baseball.I am not interested in sports only to the extent in which my> analytical> mind can> recast the sport as to Optimal Strategies.Baseball Optimal Strategy: after watching this last game of World> Series> which includes these threads.(1) The pitcher in Baseball, unlike many other sports, is the dominate> feature of baseball when you consider that of 9 players, the pitcher> has> a say in every offensive action.(2) We can math quantify the dominance of Baseball pitcher with other> sports such as Football. Where the quarterback has a big control of> the> offense but not as much of control of the overall game as the Baseball> pitcher for defense. The quarterback if he ran every play would have> as> much control as the Baseball pitcher. But he does not and has a> handoff> to a running back or a receiver. So we can say mathwise that the> Baseball pitcher has 100% control of defense Say-of-Action whereas in> Football the quarterback at most has 25% to 33% Say-of-Action concept.> Because in each offensive play in Football, the quarterback either> runs> himself or hands to a runningback or throws to a receiver. But in> Baseball, every offensive play has a Say by the pitcher with his> pitch.(3) So, unlike every other sport, the pitcher in Baseball is the> dominate feature because the Say-of-Action is 100% the pitcher.So, the OS of Baseball in order to assemble a team that will win the> World Series for that year, is key in having as many pitchers that are> ace pitchers such as Beckett and Pavano of Marlins. Beckett pitching> was> superior to that of Pettite and Pavano to that of Clemens.For 2004, if a team had 2 pitchers of the quality of Beckett, they> would> have the highest chances of reaching the World Series.By the way, I did not see the pitchers batting, so I guess the game of> Baseball has made some progressive rule change where it is optional> for> the pitchers to go to bat; and in the old days I remember the pitcher> was usually the 9th spot hitter and usually an easy out. I guess the> new> rule is that when the option is picked that the team rotates the 8> hitters and leaves out the pitchers as hitters.Getting back to this idea that pitching is the key to baseball> winning.> What does it matter for a team to have great hitting such as a Bonds> or> Sosa if they choke on pitching in that the other team with average> hitters but with great pitching. By the way, did Beckett pitch to> BondsNow suppose the Yankees had both Bonds and Sosa in their lineup.> Facing> had 3 pitchers of the quality of Beckett then I would guess that the> Marlins would have still won the game.I did see clips of the Boston game versus the Yankees and the Boston> ace> pitcher of Martinez (forgive the spelling). So I am guessing that> teams> that have at least one or two good pitchers make it to the playoffs.> Pitching is number one key.So, these concepts would then ask for a Baseball historian to look at> the winning teams and to see whether every World Series champ had 2So, the above should guide all club owners who aspire to win the World> Series that if your team has at least 2 excellent pitchers, is the> basic> prerequisite. And to concentrate the effort more in getting great> pitching than in getting great batting.> The batting of the Marlins was often frustrating and it seemed as> though> homeruns were a rarity for the Marlins so it goes to show that teams> that focus on hitting are not really the Optimal Strategy thread.And ?ally a discussion of Pitching itself. I would hate to be a> pitcher personally because throwing a ball at 95 mph for many hours> has> a toll on the arm and is a job that does not last too long. I think> there is a Optimal-Strategy-Subset for pitching itself in baseball.> The> key is to get the batter out in fast quick time. That means it is no> good to strike out a batter rather than to get him to ?. If a> pitcher is so good at devising a pitch that is popped up for a ?t,> then that is better than going for 3 strikes because if you pitch a> ball> with a great spin on it such that the probability when hit popps up> then> conceivably 3 pitches for the inning can retire the side whereas> strikeouts require at least 9 pitches. By the way is there any> statistic> where a pitcher retires a complete side with only 3 pitches in all???So, the pitcher that can devise a pitch that is prone to pop up is> superior to the pitcher that relies on fastballs and strikes. I am> guessing that some pitch has such a spin on it that the batter is> likely> to pop up or ground out.The perfect pitched Baseball game would have 9 innings and only 9 X 3> => 27> pitches where all batters swung and hit the ?st pitch and popped out> or grounded out. Not the game where the pitcher made 9 X 9 = 81> pitches> and all strikes and all strikeouts.whole entire Universe is just one big atom where dots> of the electron-dot-cloud are galaxiesNow physics tells me that a 95mph pitch is added onto the hit vectorof a homerun hit. So, I wonder if all major league pitchers were tobegin slow pitching such as an underarm lob, whether batters such asBonds or Sosa are unable to hit a slow pitch as a homerun??If so, then why in the world does not a club owner or manager everadvise his pitchers to begin lobbing the ball at the plate as insoftball and here the aim is to force the batter to ? or groundout.I think the entire game of baseball ought to be reexamined as towhether pitching at high speeds is actually the best way to pitch.I suspect no physicist or scientist has researched the Maximumgreatest pitch for our modern stadium distance to homeruns.I would not be at all surprized if a slow underarm pitch with a spinon the ballrenders the most popups.hit pitches end up as foul balls. Going from that fact, then devise apitch that has the greatest likelihood of fouling and where a ?lderor the catcher or the in?ld can get to the ball and make the out.I suspect that if a physicist closely and carefully examines thephysics of speed of pitch and a spin on the ball will render a ?ost often. And that this speed is not a fast pitch but rather insteada slow pitch.Question: I just wonder if a Barry Bonds (excuse any misspelling) werepitched the slowest strikes whether he can hit those as a homerun. Itis certain that he can hit fastballs as homeruns for the past seasonsproves that fact. If it is impossible for Bonds to hit a slow pitchwith a spin on it as a homerun, then it is obviously clear what thefuture trend of baseball will be--- pitchers adopting slow pitchingwith some wicked spins on the ball. And a gradual phase out offastball pitching.Another physics to use on pitching is that when a ball is pitched highand lofty such that it comes and falls down onto the homeplate andwhen such a ball is hitit is usually grounded and not airbound ? I believe the physicsexplanation for that is that the hitter's bat general hits the ballnot in the middle of the ball but the upper hemisphere of the ball isstruck by the bat causing the ball to head for the ground direction.If that is the case, then a pitch that is slow and lofty and comesfalling down onto the homeplate as a strike pitch and if hit by thebatter would generally be a ground ball to an in?lder and a easy outat ?st or even a double play.I do not know if the rules of Baseball make it such that a pitchermust have a certain speed on the ball or whether the rules allow for asoftball pitcher to come into the sport and pitch the slowest pitchespossible.I feel that no physicist has really involved himself thoroughly in thesport of baseball and beginning to answer some of these questions thatI have raised. And if someone serious would get involved, I expectmany surprizes to issue forth from such a research. I believe the ideaof fast ball pitching is more of a faith than a science in that almosteveryone believes that fast ball pitching is superior to slow speedpitching. But no-one has proven that belief. I am testing that beliefand saying it is faith and not science.Prove: prove that fastball pitching is superior in getting hitters outmore than slow speed pitching. Additionally, spin on slow speedpitching must be includedin the research.As I said in my original post that a team that wants to win the WorldSeries must have 2 ace pitchers such as Beckett of the Florida Marlinswinning the Series because if the ?st ace pitched game 1, and thesecond ace pitched game 2, then the ?st ace would come back andpitch game 3 and the second ace game 4 and thus a done deal.And I suspect anyone can go back in baseball history since so much o? is ?med, that one can see that a team that won the series had 2pitchers that were higher aces than the defeated team.Now if Florida had used Beckett in game 1, Pavano in game 2 andBeckett in game 34 games. But since ace pitchers are not perfect and since the hittersof the Marlins are not perfect that it took 6 games to defeat theYankees. But the Yankees were totally inferior in pitching to theMarlins and due to that inferiority it was prone to defeat.The World Series in baseball is a tournament to uncover what team inbaseball has the 2 most best ace pitchers. It is not about hitting.And that is the sick thing about movies on baseball for rarely domovies spend any time on the pitcher. But all movies on baseballshould be concentrated on the pitcher.So, if anyone right now wants to ?ure out what team will win theWorld Series in 2004, it is easier than you think. All you have to dois scout and look at the teams for which team has the Two Best Acepitchers. And if they do not get injured during the season, then those2 ace pitchers should nail down the World Series for you.Archimedes Plutoniumwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxies === Whether a pitcher is hittable or not depends solely on whether the battercan predict when the ball will cross the plate and where the ball will be atthat moment.Fastballs are dif?ult to hit because their inherent spread of velocities(typically 80-100 MPH) make it dif?ult for the batter to know when toswing -- the change up is a slow ball delivered with exactly the samemotion as a fastball and its velocity (typically 60-80 MPH) tempts thebatter to swing long before the ball reaches the plate.Curve balls are dif?ult to hit because it's very dif?ult for the batterto know how much the ball will curve -- i.e. how much below the point of thetrajectory of a fastball of the same speed where it will cross the plate.The most unhittable pitchers are (a) knuckleball pitchers and (b) theso-called junk-ball pitchers. Knuckleball pitchers succeed because eventhey don't know where the ball will cross the plate; the batter is even moreperplexed. Junk-ball pitchers simply throw such a variety of curving andswerving pitchers that the batter also can't predict what the next pitchwill be.In general, it really doesn't matter what kind or speed of pitch is thrownif the batter can predict when to swing to hit it. Norm === > quantify Baseball to other sports; Optimal Strategy for> Baseball> I think the entire game of baseball ought to be reexamined as to> whether pitching at high speeds is actually the best way to pitch.>> I suspect no physicist or scientist has researched the Maximum> greatest pitch for our modern stadium distance to homeruns.>> I would not be at all surprized if a slow underarm pitch with a spin> on the ball> renders the most popups.Archimedes might have a good point!Ewell Blackwell was a blooper ball pitcherwho had a low earned run average,and he appeared in several All Star games.Ted Williams was the only person to hita home run off of one of his blooper balls.--Tom Potter http://tompotter.us === I have a 2nd order DDEy''(x) + ay'(x) + b(y(x) - y(x-a) = 0with 2 bc'sy(0) = 1 and y(x) = 0 when x->InfI know y(x) for 0Inf , and I am looking for a analytic solution so I was wondering if there is any shortcuts for getting the solution at x=Large without going through all xI have a 2nd order DDE>y''(x) + ay'(x) + b(y(x) - y(x-a) = 0I assume this should be y''(x) + ay'(x) + b(y(x) - y(x-a)) = 0but do you really want the two a's to be the same?>with 2 bc's>>y(0) = 1 and y(x) = 0 when x->Inf>>I know y(x) for 0so I can easily solve for each interval na>My problem is that to enforce the second bc I need the solution as >x->Inf , and I am looking for a analytic solution so I was wondering if >there is any shortcuts for getting the solution at x=Large without going >through all x 0 as x -> in?ity, you'll want to use a linearcombination of the solutions for roots with negative real part.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === so what happened to Zeno Mr Gerl?Herc === What is the most POPULAR C++ library for mathematical calculations(ordinary differential equations, matrices, mathematical statistics,partial differential equations, dynamic optimization)? Please indicateone free library and one commercial library. I need to include thisinformation in my lecture. === Consider a power series sum(i=1 to oo,a_i*x^i) mod p.I think this can only make sense if for all i biggerthan a ?ed value j, p|a_i and the tail vanishesidentically. a_i=i! comes to mind.Now I was wondering if such a power series can bede?ed and converges mod p AND as a normal series.i! evidently is far too big, but I wonder if thebinomial coef?ient (i*k over i) would work.(If I thought correctly, 1. it grows like (k^k)^iand the series can converge at least on an intervaland b) i|(ik k) for all k. Correct? If not - do youhave an alternative example?)-- Hauke Reddmann <:-EX8 For our chemistry workgroup,remove math from the addressFor spamming, remove anything else === >Consider a power series sum(i=1 to oo,a_i*x^i) mod p.>I think this can only make sense if for all i bigger>than a ?ed value j, p|a_i and the tail vanishes>identically. a_i=i! comes to mind.>Now I was wondering if such a power series can be>de?ed and converges mod p AND as a normal series.>i! evidently is far too big, but I wonder if the>binomial coef?ient (i*k over i) would work.>(If I thought correctly, 1. it grows like (k^k)^i>and the series can converge at least on an interval>and b) i|(ik k) for all k. Correct? If not - do you>have an alternative example?) Try a_i = ?n(i))!.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israelUniversity of British ColumbiaVancouver, BC, Canada V6T 1Z2 === In sci.math, David Moran:> >> I'm an independent researcher, which means that I use my *own*>> funding, and my *own* direction to go out and see what knowledge I can>> obtain. Some of my research has been in the area of mathematics.>> Getting important research ?dings is one thing, and getting them>> noticed, is another.[JSH stuff snipped]>> What's in it for you?>> James HarrisJames, Again, the problem is not with the de?ition; It is with the subsets of> numbers. If I create a ring with all even integers [2,4,6] and want to> introduce 9 into the ring, I can't because 9 is odd. If the numbers you are> describing do not belong to the ring, then they must be in another ring. I> think you need to write an argument without all this crap about how> mathematicians are evil and are hiding an error. That is beside the point> and I don't think you're educated enough in mathematics to make that call.Pedant point: you can easily construct the set gen[2] union {9}(I write gen[2] to indicate the set of all even numbers, for brevity).This is of course not a closed set under addition, so it'sde?itely not a ring.If one attempts to construct (gen[2])[9], to deal withthis issue, one gets J. [*]I should also note that Z[1/2], according to James, is also toosmall, if I read his logic correctly (such as it is). Z[1/2],according to James, should equal the set of all reals.(Yes, this is preposterous. 1/3 is not in this set, letalone an arbitrary real number such as pi or e orRe(r_i) where r_i are the roots of the equation 3x^5+2x^3-12x+17=0,whatever they are. :-) )David Moran[*] there's probably a more standard notation for this somewhere. :-)-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === In sci.math, James Harris<3c65f87.0310260707.3326c555@ posting.google.com>:> I'm an independent researcher, which means that I use my *own*> funding, and my *own* direction to go out and see what knowledge I can> obtain. Some of my research has been in the area of mathematics.Getting important research ?dings is one thing, and getting them> noticed, is another.> > At least here on Usenet I can talk freely to people around the world.But will anyone listen? :-)What I'd like to explain is my disturbing and to me fascinating> ?ding of a problem with a math de?ition that's over a hundred> years old. In looking over various replies to my previous posts on> this subject, I've seen assertions that de?itions can't cause> problems, which is something that I can address quickly at the start.Over a hundred years ago, the great German mathematicians Karl Gauss> played with numbers of the form a+bi, where ?a' and ?b' are integers. > In his honor they were later called gaussian integers, though a number> like 1+2i is not an integer. The gaussian up-front is important. > Later mathematicians came up with other numbers they called algebraic> integers, which include gaussian integers.They thought they'd found THE set, or superset you might call it,> which includes all numbers with certain special properties of> integrality.The most important property to point out is the ability to have> primeness between numbers.For instance, with integers, 2 and 3 are coprime, that is, they don't> share non-unit factors, that is, factors other than 1 or -1, with each> other.Just be clear here, factors of 1, are called units or unit factors.But notice that with rationals, you have 2(3/2) = 3, so 2 and 3 *do*> share a factor and are not coprime in that ring, which is typically> called a ?ld *because* every element except 0, has a multiplicative> inverse.What Gauss had started considering, which other mathematicians> extended, was the idea of sets of numbers where you kept interesting> properties of the set of integers, like being able to say two numbers> were coprime.What I've found is a problem with their set of algebraic integers, as> unfortunately, despite what many mathematicians think, it's too small.That's it. The de?ition they use is too small to do what they think> it does, which is include all these interesting numbers with special> properties.But because they *think* it's big enough, mathematicians have an error> in their discipline based on their false assumption, as they've come> up with more arguments based on that assumption, which then aren't> actually proven.It's like when the Greeks with their word atom thought they had the> smallest thing, and later our civilization used it, and broke atoms> apart, though part of the de?ition is that they are *indivisible*,> as people can de?e things, and later *re?e* their de?itions.Now my research ?ding isn't hard to show quickly in broad strokes.On of my important analysis tools is a simple technique to factor> polynomials into non-polynomial factors.For instance, with the polynomial P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078The roots of this polynomial are, approximately:r_1 = -0.02320596700418064467907344992r_2 = 0.04221522840004950601300611271 - 0.03710343776564665666915421379*Ir_3 = 0.04221522840004950601300611271 + 0.03710343776564665666915421379*Icourtesy of GP/PARI.This equation is divisible by 7^2, yieldingP(x) = 7^2 * (300125 * x^3 - 18375 * x^2 + 360 * x + 22).that technique gives youP(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3For some reason I cannot verify this in GP/PARI:7^2*(2401*x^3-147*x^2+3*x)*(5^3)-3*(-1+49*x)*(5)*(7^2 )+7^3 -(14706125*x^3-900375*x^2+17640*x+1078)= -35280 * x.Were your claim correct I should get identically 0. I suspect a sign errorduring equation transcription.Fix.so I can factor to getP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7).where the a's are the roots of the cubica^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).Ermwhat exactly are you doing? Did you mean that you couldfactor P(x) intoP(x) = (5 * a_1 * x + 7) * (5 * a_2 * x + 7) * (5 * a_3 * x + 7) ?We know we can factor P(x) = 14706125 * (x - r_1) * (x - r_2) * (x - r_3)(as I've done above). It turns out 14706125 = 5^3 * 7^6, andtherefore we can writeP(x) = 5^3 * 7^3 * (5 * a_1 * x + 7) * (5 * a_2 * x + 7) * (5 * a_3 * x + 7)wherea_i = -7 / (5 * r_i)ora_1 = 60.32931098056739402658477747a_2 = -18.71011003573824246783784328 - 16.44452558021931061668416939*Ia_3 = -18.71011003573824246783784328 + 16.44452558021931061668416939*IIt turns out these a's empirically satisfy the equationE(a) = 22*a^3 - 504*a^2 - 36015*a - 823543 = 0and the gcd of the coef?ients of this equation is 1.I'd have to do some work to prove this precisely, though, as GP/PARIin this case is doing the equivalent of a toy calculator'sdivision of 1/3 = 0.33333333 ; it's not giving me a nicerepresentation here but simply approximating it to the best o?s ability.Unfortunately, it turns out the roots of this equation are in factb_1 = 60.32931098056739402658477746b_2 = -18.71011003573824246783784328 + 16.44452558021931061668416939*Ib_3 = -18.71011003573824246783784328 - 16.44452558021931061668416939*Iand b_1 - a_1 = -1.21169034 E-26, which is a very tinyerror, but not identically 0. So I can't construe thisas a mathematically iron-clad proof. (Basically, the lastdigit is a 6 instead of a 7.) Even if b_1 = a_1 to 53 bitsof precision I couldn't construe it as such.But I can check the equation by deriving it using adifferent method. It's worth noting that, if we assumer_i are the roots of P(x) = 0, (we have already computedapproximations to r_i above, but let's pretend that wehaven't) whereP(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078then 1/r_i are the roots of Q(x) = 0, whereQ(x) = x^3 * P(1/x) = 1078 x^3 + 17640 x^2 - 900375 x + 14706125and we can manufacture an equation with roots a_i = -7/(5*r_i), withthis technique; if we set R(x) = 0, whereR(x) = x^3 * P(-7/(5*x)) = 1078*x^3 - 24696*x^2 - 1764735*x - 40353607Changing independent variables and dividing by 49, we getR(a)/49 = 22 * a^3 - 504 * a^2 - 36015 * a - 823543which turns out to be equal to E(a) above. So GP/PARI wasn'ttoo far off after all.Now despite the complexity, you can rely on *simple* ideas still, by> noticing that setting x=0, pulls out constant terms, asP(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)as the cubic de?ing the a's with x=0 isa^3 - 3a^2, which has roots, 0, 0 and 3.Well, now you've completely lost *me*, for starters; are you solvingP(x) = 0 or deriving a cubic in a using x as a parameter insteadof a variable to be solved for?Ow, my brain hurts.However, you are correct in that P(0) = 7^2 * 22. These factorsmust be accounted for in some fashion -- although since theroots of P(x) = 0 aren't algebraic integers it's a bit likeasking for the factors of 6 (2 and 3) to be accounted for in theequation35 y^2 + 11y - 6 = 0In short, such a statement doesn't appear all that meaningful.You may not realize it, but what you just saw is revolutionary, both> in the special techniques, and most importantly with the consequences> that quickly follow.That's because P(x) has another special feature as P(x) = 49(300125 x^3 - 18375 x^2 + 360 x + 22)where that 49 is just begging to be divided off, which gives, of> course,P(x)/49 = 300125 x^3 - 18375 x^2 + 360 x + 22.This equation is correct, fortunately for you. ;-)But remember, my three factors with the a's from before had *constant*> terms of 7, 7 and 22, so dividing by 49 must give constant terms of 1,> 1, and 22, which is the result that is so earth shattering.The roots of the equation E(a) = 0 aren't algebraic integers anyway.Here the principle is like if you have S(x) = 7x^2 + 14x + 7 = (7x+7)(x+1) in that setting x=0 gives you *constant* terms within the expression,> which you can conveniently, also look at to see how it works.S(0) = (7(0) + 7)(0 + 1) = 7(1).The point is that the 7 is constant, so x's value means nothing to it.So from before withP(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)I know that dividing through by 49, it must go likeP(0)/49 = (5(0)/7 + 1)(5(0)/7 + 1)(5(3) + 7) = 1(1)(22)and as the constant terms are *independent* of the value of x, it MUST> be that in generalP(x)/49 = (5a_1/7 + 1)(5a_2/7 + 1)(5a_3 + 7).Your logic appears ? as is easily shown by example. Considerthe equationF(x) = x^3 - 49 = 0.This equation must have 3 algebraic integer roots. One can factor itF(x) = 49 * (7 * d_1 * x - 1) * (7 * d_2 * x - 1) * (d_3 * x - 1)where I've arbitrarily selected two d's to be divided by 7 inthe ?al representation, and used -1 for simplicity; as onecan easily see,d_1 = 1/(7 * c_1)d_2 = 1/(7 * c_2)d_3 = 1/c_3for some permutation of the c's.What can I conclude? Not much.It turns out c_1 = 7^(2/3) so clearly it's not divisible by 7.c_2 = 7^(2/3) * exp(2 * pi * i / 3) andc_3 = 7^(2/3) * exp(4 * pi * i / 3).None of these are divisible by 7, in the sense that c_i / 7yields an algebraic integer. (exp(i * theta) is a unitin the ring of algebraic integers, for any theta = 2 *pi * m / n, m, n > 0.) Therefore, at most one of the d'sis an algebraic integer, and since d_3 = 7^(-2/3) * exp(2 * pi *i * n /3)where n is 0, 1, or 2 (ah yes, this is a hand-picked equation!),d_3 is not an algebraic integer (it satis?s the equation49 y^3 - 1). The other two aren't, either.The problem now though is that conclusion can be used to show that> unequivocally beyond any reasonable doubt the de?ition of algebraic> numbers is TOO SMALL, as at times 5a_1/7 and 5a_2/7 are not included.You see, they get left out, which is a problem because from the> *assumption* of mathematicians, they should be included, if the ring> of algebraic integers is the ring that mathematicians thought it was.And what is the set of algebraic integers, then?It's clearly not a ?ld:7 is in the set (x - 7 = 0)2 is in the set (x - 2 = 0)7/2 is not in the set (2x - 7 = 0 is not monic in its highest non-zero x power).Some of you may ?d yourselves fearful of using your own mathematical> understanding, if you realize I'm right, and then realize that> mathematicians are disputing the result, especially if you see posters> tossing out far more complicated math in reply to my post, but> remember, math isn't magic.Logic rules mathematics, so look for what makes sense. And remember> that you can't assume that posters are on your side. I don't want you> to assume that I'm on your side either.You see, I don't need you to assume anything, as what I need you to do> is check.While some mathematicians may erroneously believe now that it's in> their interest to hide the problem I've revealed, that mistake in> thinking does not help the rest of the world. After all, what good> does it do everyone else for mathematicians to hide their de?ition> problem?What's in it for you?Understanding. So far, you've not extended mine. :-) You'll needto ? a number of glaring logic errors, to start.> James Harris-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === he either can't, or won't, or doesn't get the question, orthe de?itions behind the question, or all or partof the trivium (logic, grammar, rhetoric well,he seems to have some elements of rhetoric, down-cold .-) > Can you write down a1, a2, a3 explicitly for your polynomial> as others have done (not another polynomial, YOUR polynomial)> and show they have the behavior you want them to have?--les ducs d'Enron! === have you considered a career in acting? I mean,about half of the time you fail to follow-upon questions to your own postings, immediately, butthat applies to virtually all of your originations,eventually, in the ten-year programmeto prove some thing -- what is it that you're trying to prove? incidentally,the book taht I quoted on rings made at least one mistake,in saying that Fermat proved the case of FLT for n=3,wehn he actually used in?ite descent for n=4,which apparently did not comply to his more-general proof. so,the question is, do all cases fall to your peculiar method, ordid you ?d a counterexample in Object Ring Theory? that's the 64-centivo question for me, and i?'s answered, maybe something will be in itfor me, two!> What I'd like to explain is a disturbing and fascinating ?ding of a> problem with a math de?ition that's over a hundred years old. > But notice that with rationals, you have 2(3/2) = 3, so 2 and 3 do> share a non-unit factor and are not coprime in that ring, which is> typically called a ?ld because every element except 0, has a> multiplicative inverse. > That's it. The de?ition they use is too small to do what they think> it does, which is include all these interesting numbers with special> properties. > What's in it for you?--les ducs d'Enron! === another error that it made was:said that Cusa made an erroneous proofthat the circle was incomeasurable with teh tetragon(skware). > the book taht I quoted on rings made at least one mistake,> in saying that Fermat proved the case of FLT for n=3,> wehn he actually used in?ite descent for n=4,> which apparently did not comply to his more-general proof. so,> the question is, do all cases fall to your peculiar method, or> did you ?d a counterexample in Object Ring Theory?--les ducs d'Enron! === > I'm an independent researcher, which means that I use my *own*> funding, and my *own* direction to go out and see what knowledge I can> obtain. Some of my research has been in the area of mathematics.> [mostly non-math SNIPPED]On of my important analysis tools is a simple technique to factor> polynomials into non-polynomial factors.For instance, with the polynomial P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078that technique gives you[1] P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3> I really think this was clearer when you were writing it as[2] P(m) = f^2*((m^3*f^4 - 3*m^2*f^2 + 3*m)*x^3 - 3*(-1 + m*f^2)*x*u^2 + u^3*f). Your expression [1] is obtained from [2] by the following substitions: u = 1 f = 7 x = 5 m = x It is clear that in your factorization of [1], you want to treat5 not as a number, but as a polynomial variable. Therefore it was clearer when you had x, as in [2], rather than 5.> so I can factor to getP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7).> That is, P(x) = (a1*y + 7)*(a2*y + 7)*(a3*y + 7),where y is the polynomial variable for which youhave substituted 5.> where the a's are the roots of the cubica^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).> It is worth noting that the roots are all algebraicintegers since this polynomial is monic.> Now despite the complexity, you can rely on *simple* ideas still, by> noticing that setting x=0, pulls out constant terms, asP(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)as the cubic de?ing the a's with x=0 isa^3 - 3a^2, which has roots, 0, 0 and 3.> OK so far.> You may not realize it, but what you just saw is revolutionary, both> in the special techniques, and most importantly with the consequences> that quickly follow.That's because P(x) has another special feature as P(x) = 49(300125 x^3 - 18375 x^2 + 360 x + 22)where that 49 is just begging to be divided off, which gives, of> course,P(x)/49 = 300125 x^3 - 18375 x^2 + 360 x + 22.> Also OK.> But remember, my three factors with the a's from before had *constant*> terms of 7, 7 and 22, so dividing by 49 must give constant terms of 1,> 1, and 22, which is the result that is so earth shattering.> All you are saying is that P(0)/49 = 1*1*22. Why is that earth-shattering ?> Here the principle is like if you have S(x) = 7x^2 + 14x + 7 = (7x+7)(x+1) in that setting x=0 gives you *constant* terms within the expression,> which you can conveniently, also look at to see how it works.S(0) = (7(0) + 7)(0 + 1) = 7(1).The point is that the 7 is constant, so x's value means nothing to it.> What you have done in this little example is not parallel towhat you have done with P(x). You are not factoring P(x)into linear terms involving x: that is, your factors forP(x) are terms g1, g2, and g3, where, for example, g1 = 5*a1 + 7where it is a1 which is a function of x. Moreover, a1 isnot a *polynomial* function of x. You have made this point yourself many, many times before:you have often said very proudly that you were factoring P(m) (as you were calling it previously) into NONPOLYNOMIAL factors. You were right, in the sense that the coef?ients ai were functions of m, but not polynomial functions. This is important because when you try to draw the parallel with S(x), you ARE factoring it into linearfactors involving x. It looks like the old shell game. It looks like youare pulling the old switcheroo on the variable of factorization.To maintain the parallel between S(x) and P(x), you shouldnot be using x: you should be using 5 ! > So from before withP(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)I know that dividing through by 49, it must go likeP(0)/49 = (5(0)/7 + 1)(5(0)/7 + 1)(5(3) + 7) = 1(1)(22)and as the constant terms are *independent* of the value of x, it MUST> be that in generalP(x)/49 = (5a_1/7 + 1)(5a_2/7 + 1)(5a_3 + 7).> No - ?st of all, a_1, a_2, and a_3 are dependenton x. Secondly, in order for P(x) to factor in theform you have given, the number 49 is split up inways that ALSO depend on x. In symbols, you would have 49 = f1(x)*f2(x)*f3(x),where a1(x)/f1(x), a2(x)/f2(x), a3(x)/f3(x) and7/f1(x), 7/f2(x), and 7/f3(x) are ALL algebraic integers. When x = 0, f1(0) = f2(0) = 7, and f3(0) = 1.This works just ?e, because a1(0) = a2(0) = 0. But for other values of x, there is no reason tothink that f1(x) = f2(x) = 7 and f3(x) = 1. Theyare just three algebraic integer divisors of 49.Their values are dependent on x, and they do infact have the values you expect when x = 0. === ========================================== === ** The following section is the important stuff in this post.* === ======================================= === === Now: you may say, yes, but if I consider[3] g1(x) = 5*a1 + 7, the constant term is 7. Moreover, the constantterm of g1(x)/7 is 1. This is true no matterwhat x is. I always get the same constant term.It has the value 7 INDEPENDENT of x. Of course that is true. But so what? If f1is an algebraic integer factor of both 7 and a1,then g1(x)/f1(x) = 5*a1(x)/f1(x) + 7/f1(x). To which you say, AHA! NOW you are saying that the constantterm is 7/f1(x) !!! Right? To which I say: not at all. *By your own de?ition*,the constant term of g1(x)/f1(x) is g1(0)/f1(0).Recalling that f1(0) = 7, I have g1(0)/f1(0) = 5*a1(0)/f1(0) + 7/f1(0) = 5*a1(0)/7 + 7/7 = 5*0 + 1 = 1.of g1 is: g1(0) = f1(0)*(5*a1(0)/7 + 7/7) = 7,just as in [3] above. No problem at all! Here is *my* conjecture about where your thinkinggoes wrong. You are using constant term in twodifferent ways. It goes back to the expression h1(x) = g1(x)/f1(x) = 5*a1(x)/f1(x) + 7/f1(x).You are accustomed to thinking that the terms thereon the end comprise the constant term, and has the same value for all x. THAT IS NOT TRUE. The constant term of h1(x) is nothing but 7/f1(0). Realizing that f1(x) can be a nonconstant function of x whose values are algebraic integer divisorsof 7 for all values of x does NOT lead to any contradiction or problem. Put it another way: 7^2 is factored differentlyinto 3 pieces depending on the value of x. Thosethree pieces are algebraic integers. You appear to be assuming that that part on the end, 7/f1(x), is the constant term of the function h1(x). IT ISN'T. The constant term asyou have consistently and correctly used it is h1(0). Since, however, because of its placement, you are thinkingof 7/f1(x) as the constant term, you think it must be *constant*. That is, it must have the same value for all x. THAT IS NOT TRUE. The constant term of h1(x) is nothing but 7/f1(0). Realizing that f1(x) can be a nonconstant functionof x whose values are algebraic integer divisorsof 7 for all values of x does NOT lead to any contradiction or problem. Put it another way: 7^2 is factored differentlyinto 3 pieces depending on the value of x. Thosethree pieces are algebraic integers. Assuming thatthey are not always equal to the values that theyhave when x = 0 does not lead to any contradictionof the independence of the constant term. Put yet another way: the constant term in g1is not 7/f1(x) *unless* you assume f1(x) is itselfconstant, and thus always equal to f1(0). But that is assuming what you want to prove! The constant term in g1(x) is 7/f1(0). That is *all* it is, and there is no way to show that it is7/f1(x) for other values of x. Whether this is where your thinking has gone off the track, I don't know for sure,though it seems consistent with many things you have said.Arturo has a related theory, possibly even the same theory, about what you are thinking. But it's all academic. The various counterexamples that we have produced show that yourthinking goes wrong SOMEWHERE. You are whistling pastgraveyard when you persist in your belief that there issome error in core mathematics, or that there is a problem with the de?ition of algebraic integers.The far simpler (and correct) explanation is that your argument iswrong. We know exactly *where* it is wrong, but only you can reveal what you are actually thinking. Andso far you have not completely done that. Until you do,we will not be able to resolve to your own satisfaction whatyour mistake is. But we know enough, I would say, to satisfyeveryone else.> The problem now though is that conclusion can be used to show that> unequivocally beyond any reasonable doubt the de?ition of algebraic> numbers is TOO SMALL, as at times 5a_1/7 and 5a_2/7 are not included.> Face up to what you are saying. You are claiming above tohave an airtight argument that a1 has 7 as a factor, which ofcourse means a1/7 is an algebraic integer. Now you say you can conclude that 5*a1/7 is not an algebraic integer. That is not a de?ition problem. That is an out-and-out CONTRADICTION. Face up to it! Bite the bullet! And a contradiction like this canonly mean one of two things: 1. Mathematics is inconsistent, or 2. You have an error in your reasoning. We have pointed this out to you over and over again. Wehave pointed out exactly where your error is: it is insaying that because a1(0) is divisible by 7, that thesame must be true for a1(x) when x is not zero. You refuseto listen. You just keep repeating your mantra about the constant terms of P(x) and its factors being *independent* of x, as if that leads to any conclusion. Get this through your incredibly thick head: the factthat the constant terms, when x = 0, satisfy certain divisibility criteria tells you NOTHING about what happens when x is nonzero. Nor is factorization ofoverly simplistic REDUCIBLE quadratic polynomials a reliable guideto what happens with IRREDUCIBLE polynomials. Algebraicintegers, as you yourself have pointed out, are really different from ordinary integers: they cannot be factoredinto little indivisible atoms called ?primes'; they are*in?itely divisible* and they can factor in ways that are quite strange, compared to factorization of integers. Nora B.> You see, they get left out, which is a problem because from the> *assumption* of mathematicians, they should be included, if the ring> of algebraic integers is the ring that mathematicians thought it was.Some of you may ?d yourselves fearful of using your own mathematical> understanding, if you realize I'm right, and then realize that> mathematicians are disputing the result, especially if you see posters> tossing out far more complicated math in reply to my post, but> remember, math isn't magic.Logic rules mathematics, so look for what makes sense. And remember> that you can't assume that posters are on your side. I don't want you> to assume that I'm on your side either.You see, I don't need you to assume anything, as what I need you to do> is check.While some mathematicians may erroneously believe now that it's in> their interest to hide the problem I've revealed, that mistake in> thinking does not help the rest of the world. After all, what good> does it do everyone else for mathematicians to hide their de?ition> problem?What's in it for you?> James Harris === > I'm an independent researcher, which means that I use my *own*> funding, and my *own* direction to go out and see what knowledge I can> obtain. Some of my research has been in the area of mathematics.> [mostly non-math SNIPPED]> On of my important analysis tools is a simple technique to factor> polynomials into non-polynomial factors.For instance, with the polynomial P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078that technique gives you[1] P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3> I really think this was clearer when you were writing it as[2] P(m) = f^2*((m^3*f^4 - 3*m^2*f^2 + 3*m)*x^3 - 3*(-1 + m*f^2)*x*u^2 +> u^3*f).I want to point out posters who are busiy working to ADD confusion,like Nora Baron as notice this poster has decided to go back to amore complex expression.Now I used the more complex expression for MONTHS and Nora Baronalong with others had a ?ld day confusing people, and lying aboutthe expression as I explained things more than once in posts replyingto this poster and others.Basically they'd focus people on the x above and claim IT was thekey variable, and refuse to back down, so I took it away. And now youcan see the poster wants it back.For those on sci.physics and sci.skeptic who still don't understandwhat I'm dealing with, consider the attempts to ADD complexity andconfusion, while I emphsize simplicity.James Harris === > For instance, with the polynomial > > P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078Why would your technique work with your polynomial and fail withQ(x) = 6125 x^3 + 6125 x^2 - 6370 x + 1078With this polynomial and your technique you get 1/7 in your ring.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === I parafrase you and ahve a question at the end: > On of my important analysis tools is a simple technique to factor > polynomials into non-polynomial factors. > > For instance, with the polynomial > > Q(x) = 6125 x^3 + 6125 x^2 - 6370 x + 1078 > > that technique gives you > > Q(x)= 7^2(x^3 + x^2 - x)(5^3) - 3(-1 + x)(5)(7^2) + 7^3 > > so I can factor to get > > Q(x) = (5 b_1 + 7)(5 b_2 + 7)(5 b_3 + 7). > > where the b's are the roots of the cubic > > b^3 + 3(-1 + x)b^2 - 49(x^3 + x^2 - x)). > > Now despite the complexity, you can rely on simple ideas still, by > noticing that setting x=0, pulls out constant terms, as > > Q(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) > > as the cubic de?ing the b's with x=0 is > > b^3 - 3b^2, which has roots, 0, 0 and 3. > > You may not realize it, but what you just saw is revolutionary, both > in the special techniques, and most importantly with the consequences > that quickly follow. > > That's because Q(x) has another special feature as > > Q(x) = 49(125 x^3 + 125 x^2 - 130 x + 22) > > where that 49 is just begging to be divided off, which gives, of > course, > > Q(x)/49 = 125 x^3 + 125 x^2 - 130 x + 22. > > But remember, my three factors with the b's from before had constant > terms of 7, 7 and 22, so dividing by 49 must give constant terms of 1, > 1, and 22, which is the result that is so earth shattering. > > Of course, the distributive property is important here. > > Distributive Property: > > a(b+c) = ab + ac > > The point is that the 7 is constant, so x's value means nothing to it. > > So from before with > > Q(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) > > I know that dividing through by 49, it must go like > > Q(0)/49 = (5(0)/7 + 1)(5(0)/7 + 1)(5(3) + 7) = 1(1)(22) > > and as the constant terms are independent of the value of x, it MUST > be that in general > > Q(x)/49 = (5b_1/7 + 1)(5b_2/7 + 1)(5b_3 + 7). > > The problem now though is that conclusion can be used to show that > unequivocally beyond any reasonable doubt the de?ition of algebraic > numbers is TOO SMALL, as at times 5b_1/7 and 5b_2/7 are not included. > > You see, they get left out, which is a problem because from the > assumptions of mathematicians, they should be included, if the ring of > algebraic integers is the ring that mathematicians thought it was.Now, James, is the above argument correct or not. If it is not correct,which step is incorrect, and why would that step be correct with yourpolynomial and incorrect with mine?If it is correct, set x = 1 above and ?d that when you include inyour ring *any one* of b_1/7, b_2/7 or b_3/7, you get them all three,and as an added bonus you also get 1/7 in your ring.(In my opinion, it is the last paragraph but 2 where you talk aboutMUST. You assume that it MUST be true in the algebraic integers,that is wrong.)-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === > I parafrase you and ahve a question at the end:You don't have my permission to parafrase and I don't ?d mockingposts of interest.Now I've looked over some of your posts and can rather easily explainwhat you seem to ?d signi?ant.Actually, part of the reason I haven't been terribly worried aboutanswering you in detail is I've wondered how many people may begin todoubt algebra and think that maybe math is inconsistent if they can't?ure out what's happening in your examples.I will say that it's quite simple.After all, constants *are* constants, and math is consistent, so therehas to be a rational explanation.James Harris === > I parafrase you and ahve a question at the end: > > You don't have my permission to parafrase and I don't ?d mocking > posts of interest.I was not aware that I needed permission to parafrase. > Now I've looked over some of your posts and can rather easily explain > what you seem to ?d signi?ant.Ah, that's good. > Actually, part of the reason I haven't been terribly worried about > answering you in detail is I've wondered how many people may begin to > doubt algebra and think that maybe math is inconsistent if they can't > ?ure out what's happening in your examples. > > I will say that it's quite simple. > > After all, constants *are* constants, and math is consistent, so there > has to be a rational explanation.But actually you do *not* explain, why your reasoning should work foryour polynomial but not for mine.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === > > I parafrase you and ahve a question at the end:> > You don't have my permission to parafrase and I don't ?d mocking> > posts of interest.I was not aware that I needed permission to parafrase.Dik: FYI, this is one of those English words of Greekorigin, where we use ph for the f sound, viz.paraphrase.It's only in rational languages like Spanish (and apparentlyyours) where phrase is spelled the way it sounds. - Randy === > > I parafrase you and ahve a question at the end:> > You don't have my permission to parafrase and I don't ?d mocking> > posts of interest.I was not aware that I needed permission to parafrase.It's mocking. Now if you want to be insulting from the start, that'syour business. But it's just another show of bad character.You can write out the steps in your own words to try and make yourpoint, without attempting to insult me from the start.It's your decision.James Harris === > > I parafrase you and ahve a question at the end: > > > > You don't have my permission to parafrase and I don't ?d mocking > > posts of interest. > > I was not aware that I needed permission to parafrase. > > It's mocking. Now if you want to be insulting from the start, that's > your business. But it's just another show of bad character. > > You can write out the steps in your own words to try and make your > point, without attempting to insult me from the start.I can't write them in my own words, because I wish to show that *your*reasoning leads to nonsense with my polynomial. Attempting to do yourreasoning in my own words may lead to something different than what *you*mean.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === > > I parafrase you and ahve a question at the end:> >> > You don't have my permission to parafrase and I don't ?d mocking> > posts of interest.>> I was not aware that I needed permission to parafrase.>> It's mocking. Now if you want to be insulting from the start, that's> your business. But it's just another show of bad character.>> You can write out the steps in your own words to try and make your> point, without attempting to insult me from the start.>> It's your decision.> James HarrisTake a bit of your own advice and cut the insults. I've yet to see a postfrom you insult free. === > I parafrase you and ahve a question at the end:You don't have my permission to parafrase and I don't ?d mocking> posts of interest.>>I was not aware that I needed permission to parafrase.>> Now I've looked over some of your posts and can rather easily explain> what you seem to ?d signi?ant.>>Ah, that's good.>> Actually, part of the reason I haven't been terribly worried about> answering you in detail is I've wondered how many people may begin to> doubt algebra and think that maybe math is inconsistent if they can't> ?ure out what's happening in your examples.I will say that it's quite simple.After all, constants *are* constants, and math is consistent, so there> has to be a rational explanation.>>But actually you do *not* explain, why your reasoning should work for>your polynomial but not for mine.Well no he doesn't, but he assures you that he could if he felt like it. Isn't that enough? I mean it's not like he says things that arenot so - if he says he can explain that's enough for me.************************David C. Ullrich === > Actually, part of the reason I haven't been terribly worried about > answering you in detail is I've wondered how many people may begin to > doubt algebra and think that maybe math is inconsistent if they can't > ?ure out what's happening in your examples. > > I will say that it's quite simple. > > After all, constants *are* constants, and math is consistent, so there > has to be a rational explanation. > >But actually you do *not* explain, why your reasoning should work for >your polynomial but not for mine. > > Well no he doesn't, but he assures you that he could if he felt like > it. Isn't that enough? I mean it's not like he says things that are > not so - if he says he can explain that's enough for me.I think this is kind of scary. Just a few days ago I read the storyabout somebody who had found a way to speed up transmission of videothrough the internet thousandfold. People were interested, but hedied before he could reveal what he had done. Now that invention islost for eternity Suppose now that James dies before he could givehis simple explanation?-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === >> I parafrase you and ahve a question at the end:>>You don't have my permission to parafrase you're using, except in a slightly different context where one can seethat the conclusion is false. Hence either there's something wrongwith the reasoning (hint: this is the correct explanation) or there issome difference between the case he considered and the caseyou consider.If you don't want to look like you really can't explain your proof youneed to say what the difference is, not state that he doesn't havepermission to show that you're wrong.>and I don't ?d mocking>posts of interest.Wasn't anything mocking about his post.>Now I've looked over some of your posts and can rather easily explain>what you seem to ?d signi?ant.Really? Then why _don't_ you?>Actually, part of the reason I haven't been terribly worried about>answering you in detail is I've wondered how many people may begin to>doubt algebra and think that maybe math is inconsistent if they can't>?ure out what's happening in your examples.>>I will say that it's quite simple.>>After all, constants *are* constants, and math is consistent, so there>has to be a rational explanation.Of course there's a rational explanation. The rational explanation that's clear to everyone but you is that the things you say MUSTbe so are simply not so.>James Harris************************David C. Ullrich === > I parafrase you and ahve a question at the end:You don't have my permission to parafrase and I don't ?d mocking> posts of interest.Now I've looked over some of your posts and can rather easily explain> what you seem to ?d signi?ant.Actually, part of the reason I haven't been terribly worried about> answering you in detail is I've wondered how many people may begin to> doubt algebra and think that maybe math is inconsistent if they can't> ?ure out what's happening in your examples.I will say that it's quite simple.After all, constants *are* constants, and math is consistent, so there> has to be a rational explanation.In the past, you have experienced subjective feelings of absolute certainty. You have then realized that you were wrong. This has happened many times. Is it possible that you are wrong now? Many people have looked at your argument and have made a sincere effort to understand what you are saying. Those that have understood in detail what you're trying to say, have pointed out your error. Is it possible that now, just as in the past, they are right and you are wrong? If so, it's an opportunity to learn something. === > I parafrase you and ahve a question at the end:>> You don't have my permission to parafrase and I don't ?d mocking> posts of interest.Permission? > Now I've looked over some of your posts and can rather easily explain> what you seem to ?d signi?ant.>> Actually, part of the reason I haven't been terribly worried about> answering you in detail is I've wondered how many people may begin to> doubt algebra and think that maybe math is inconsistent if they can't> ?ure out what's happening in your examples.>> I will say that it's quite simple.>> After all, constants *are* constants, and math is consistent, so there> has to be a rational explanation.I thought that the whole premise of your discovery is that some parts ofmath (as currently de?ed) are NOT consistent. === > >>I parafrase you and ahve a question at the end:> You don't have my permission to parafrase and I don't ?d mocking> posts of interest.Now I've looked over some of your posts and can rather easily explain> what you seem to ?d signi?ant.Can, perhaps. Will not, certainly.Gib === > So we have:> nonconstant term constant term> P(m)/49 = [ (5a_1(m)/w_1(m) + (7/w_1(m)) -1) + 1 ]> * [ (5a_2(m)/w_2(m) + (7/w_2(m)) -1) + 1 ]> * [ ( ( (h_3(m)+22)/w_3(m) ) -22 ) + 22 ] Readers should note that Arturo Magidin added and subtracted 1 in twocases and 22 in another.Well adding and subtracting the same number gives you 0, of course, sohe didn't do anything, but he acts like he did.Also notice that he *switched* variable names, as if it matterswhether or not it's x or m, though I used x in my original post inthis thread. And, yes, I switched recently from using m, as itoccurred to me that posters like Arturo Magidin might be oddly ?atedon a variable name, when it's algebra, so they shouldn't be.The fact is that setting x=0, gives the constant terms, which areindependent of x, and for the three factors of P(x) that I focus on,you get 7, 7 and 22 respectively. Dividing 49 from P(x), which hasthat factor for all x, gives you factors with 1, 1, and 22,respectively, for the constants. > As you can see, the constant terms are 1, 1, and 22. But the> nonconstant terms are NOT what you claim they are. What did I claim them to be Arturo Magidin?>Note that even> though there are those -1 and -22 in the nonconstant term,> that's not a problem. Just as we had above, when we noticed that the> nonconstant term of g_3(m) had a -15 in it; not every summand has> to be mulitplied by something that changes with m for something to be> the nonconstant term. Remember:IT IS NOT TRUE THAT THE CONSTANT TERM NECESSARILY CONSISTS OF EVERY> SUMMAND WHICH DOES NOT CHANGE AS m CHANGES. IT IS NOT TRUE THAT IN THE> NON-CONSTANT TERM EVERY SUMMAND MUST BE MULTIPLIED BY SOMETHING THAT> CHANGES AS m CHANGES.Setting x=0 gives you an expression which has lost x's footprint. That is mathematicially, x is just gone. Arturo Magidin hasapparently gone bye-bye, but for those of you who wonder, if you havex+2, and set x=0, you just have 2, which is not associated with x.SURE, you may REMEMBER that you had x+2, but mathematically, it's just2, which is why setting a variable to 0 is such a powerful technique.After all, if 2 had a permanent association, don't you think thosewould add up? I mean, it'd never just be 2, now would it?James Harris === >> So we have:>> nonconstant term constant term>> P(m)/49 = [ (5a_1(m)/w_1(m) + (7/w_1(m)) -1) + 1 ]>> * [ (5a_2(m)/w_2(m) + (7/w_2(m)) -1) + 1 ]>> * [ ( ( (h_3(m)+22)/w_3(m) ) -22 ) + 22 ] >>Readers should note that Arturo Magidin added and subtracted 1 in two>cases and 22 in another.What I did was calculate explicitly the constant and nonconstantterms, ACCORDING TO YOUR DEFINITION.Your de?ition of constant term is evaluate at 0. That's what Idid.If you'll remember from your little Advance Polynomial Factorizationthing, what you had was that if you have an arbitrary function h(x),then the constant term is h(0), and the non-constant term ish(x)-h(0).So, if h_1(m) = g_1(m)/w_1(m) = (5a_1(m))/w_1(m) + (7/w_1(m)), anda_1(0)=0, and w_1(0)=7, thenh_1(0) = 5(a_1)0/7 + 7/7 = 1and the nonconstant part of h_1(m) ish_1(m) - h_1(0) = (5a_1(m)/w_1(m)) + (7/w_1(m)) - h_1(0) = (5a_1(m)/w_1(m)) + (7/w_1(m)) - 1.Analogous with h_2, and with h_3.Your error is claiming that 7/w_1(m) is constant. That's theCONCLUSION you want, so you cannot claim it yet. We do NOT know that7/w_1(m)=1; that's what you are trying to conclude.>Well adding and subtracting the same number gives you 0, of course, so>he didn't do anything, but he acts like he did.I followed your de?itions TO THE LETTER.And that's also what YOU DO. You argue about constant term andnonconstant term, but all you do is say thatf(x) = (f(x)-f(0)) + f(0).So all you do is add and subtract the same number, and act like youdid a big deal.Remember: that's what YOU do.>Also notice that he *switched* variable names, as if it matters>whether or not it's x or m, though I used x in my original post in>this thread.I followed your usual de?ition, which has m as the variable you setto 0. If you changed to x's, then surely you can ?ure out what needsto be done.> And, yes, I switched recently from using m, as it>occurred to me that posters like Arturo Magidin might be oddly ?ated>on a variable name, when it's algebra, so they shouldn't be.Oh, big deal. Here it is with x, you simpleton.So, if h_1(x) = g_1(x)/w_1(x) = (5a_1(x))/w_1(x) + (7/w_1(x)), anda_1(0)=0, and w_1(0)=7, thenh_1(0) = 5(a_1)0/7 + 7/7 = 1and the nonconstant part of h_1(x) ish_1(x) - h_1(0) = (5a_1(x)/w_1(x)) + (7/w_1(x)) - h_1(0) = (5a_1(x)/w_1(x)) + (7/w_1(x)) - 1.So we have: nonconstant term constant term P(x)/49 = [ (5a_1(x)/w_1(x) + (7/w_1(x)) -1) + 1 ] * [ (5a_2(x)/w_2(x) + (7/w_2(x)) -1) + 1 ] * [ ( ( (h_3(x)+22)/w_3(x) ) -22 ) + 22 ] Do you deny that these calculations are accurate?>The fact is that setting x=0, gives the constant terms, which are>independent of x, and for the three factors of P(x) that I focus on,>you get 7, 7 and 22 respectively. Dividing 49 from P(x), which has>that factor for all x, gives you factors with 1, 1, and 22,>respectively, for the constants.Which is exactly what I have above.>> As you can see, the constant terms are 1, 1, and 22. But the>> nonconstant terms are NOT what you claim they are. >>What did I claim them to be Arturo Magidin?You claim that they are also equal to 7/w_1(x), 7/w_2(x), and22/w_3(x). This follows because you CLAIM that the nonconstant term is5a_1(x)/w_1(x), and since the function is (nonconstant)+(constant),and the function is equal to5a_1(x)/w_1(x) + 7/w_1(x),then it follows that you claim that the constant term is 7/w_1(x);likewise for the second and third ones.They are not equal to that. They are 1, 1, and 22, but they are NOTalso equal to 7/w_1(x), 7/w_2(x), and 22/w_3(x).>> Note that even>> though there are those -1 and -22 in the nonconstant term,>> that's not a problem. Just as we had above, when we noticed that the>> nonconstant term of g_3(m) had a -15 in it; not every summand has>> to be mulitplied by something that changes with m for something to be>> the nonconstant term. Remember:>> >> IT IS NOT TRUE THAT THE CONSTANT TERM NECESSARILY CONSISTS OF EVERY>> SUMMAND WHICH DOES NOT CHANGE AS m CHANGES. IT IS NOT TRUE THAT IN THE>> NON-CONSTANT TERM EVERY SUMMAND MUST BE MULTIPLIED BY SOMETHING THAT>> CHANGES AS m CHANGES.>>Setting x=0 gives you an expression which has lost x's footprint. This is a non-sequitur. While it is true that every summand in theconstant term does not change as x changes, it is NOT true that everysummand that does not change as x changes has to be in the constantterm. You had that in g_3(x) = 5a_3(x)+7. Since a_3(0)=3, what you getis that g_3(0) = 22, sog_3(x) = (5a_3(x) - 15) + 22 and the -15, which does not change as x changes, is NOT part of theconstant term.That's why you can have a -1 and a -22 in the nonconstant termsabove. >That is mathematicially, x is just gone. Arturo Magidin has>apparently gone bye-bye, but for those of you who wonder, if you have>x+2, and set x=0, you just have 2, which is not associated with x.LOL! Please re read what you are replying to, James. I did not saythat the independent terms vary with x; I said that not everythignwhich does not vary with x is in the independent term.Again, the correct expression is: nonconstant term constant term P(x)/49 = [ (5a_1(x)/w_1(x) + (7/w_1(x)) -1) + 1 ] * [ (5a_2(x)/w_2(x) + (7/w_2(x)) -1) + 1 ] * [ ( ( (h_3(x)+22)/w_3(x) ) -22 ) + 22 ] Taken from YOUR de?ition, that for a function h(x), the constantterm is h(0) and the nonconstant term is h(x)-h(0).You are the one who added and subtracted 1 and claimed it was a bigdeal when you noted simply that5a_1(x)/w_1(x) + 7/w_1(x) = (5a_1(x)/w_1(x) + 7/w_1(x) - 1) + 1. === ========================================== === ======Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of ?ures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the inde?ite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan === ======================================= === =========Arturo Magidinmagidin@math.berkeley.edu===The math I can use is VERY simple, thankfully, as mathematicians are> ?hting this result, so I use very basic math to give them less room> to obfuscate.Here what I'm using is the distributive property. Remember that one?> Um, no, what you're using is an intuitive believe that> your non polynomial factors should have certain properties> in common with polynomial factors. And they don't. You> keep saying they MUST, but a number of people have actually> worked out explicit expressions for those terms which> makes it abundantly clear that they DON'T.Constant terms are constant. So the 7 that appears in two factors is*independent* of x, or m, of whatever variable name I use.That's just a fact.So there's no intuitive belief on my part and in fact, the posterRandy Poe can't point to the actual math argument to validate hisclaims, which may be why he deleted out all the math.Mathematicians are *desperate* here and the truth is not as importantto them as keeping up their fantasy world, which is why they have tokeep replying to my posts.You see, they MUST keep replying to me because otherwise they'reafraid people will actually trace out the argument and see I'm right.They *have* to keep replying to in? you. > Can you write down a1, a2, a3 explicitly for your polynomial> as others have done (not another polynomial, YOUR polynomial)> and show they have the behavior you want them to have? - RandyHuh? These posters aren't even making sense. They're not *rational*as rather than trying to ?ure it all out, they've made up theirminds to preserve their status quo by denying reason.If you look at my posts explaining the problem, you'll notice ratherbasic algebra.Mathematicians live in a fantasy world, where math was just their toy,something they owned.I've burst their bubble, and now they're showing they can't handle thetruth.James Harris === > Constant terms are constant. So the 7 that appears in two factors is> *independent* of x, or m, of whatever variable name I use.That's just a fact.> Several people have written down the expressions fora1, a2, a3. They don't behave the way you say they do.That's just a fact.> So there's no intuitive belief on my part and in fact, the poster> Randy Poe can't point to the actual math argument to validate his> claims, which may be why he deleted out all the math.Yeah, I can. See the work by A. Magidin and C. Bond forexample. They've shown you expressions for the a's. Theydon't behave the way you want. Those are actual math arguments.They contain actual equations, actual expressions. Thoseactual expressions can be substituted into your actualpolynomial to show with actual math arguments that they area correct factorization of your actual mathematicalpolynomial.So in what sense is that not an actual math argument?> Huh? These posters aren't even making sense.They're not? Does that mean the expressions given for thea1, a2, and a3 are incorrect? I haven't seen you say yeaor nay to those expressions.If they are the correct expressions, then in what sensedo the expressions not make sense? - Randy === > >>[deletia]>Here the principle is like if you have>>S(x) = 7x^2 + 14x + 7 = (7x+7)(x+1)>>in that setting x=0 gives you *constant* terms within the expression,>which you can conveniently, also look at to see how it works.>>S(0) = (7(0) + 7)(0 + 1) = 7(1).>>The point is that the 7 is constant, so x's value means nothing to it.>>I don't understand what you mean by this. Your usages of 7 is confusing>>for me. Could you please restate this with the example >>S(x) = 7x^2 + 19x + 10 = (7x+5)(x+2)>>S(0) = (7(0)+5)(0+2) = 5(2)>>And tell me what is constant and thus independent of what? I just wish to>>understand what all this fuss is about.[deletia]> Here what I'm using is the distributive property. Remember that one?Distributive Property:a(b+c) = ab + acAnd believe me if you have to break it down to the distributive> property then you have some SERIOUS denial on the part of> mathematicians.Can you believe that people? Having to break it all the way down to> the freaking *distributive property*?> James HarrisPlease don't confuse a lack of clarity on your part with inability to comprehend on the part of others. You've been told how to make your writing clearer and consistently refuse to.-- Will Twentyman === > What you just presented and others may have seen was completely wrong!Your original polynomial:P(x) = 14706125x^3 - 900375x^2 + 157640x + 1708can, indeed be factored asP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7).But the next logical step, which you did not take, is to determine a_1,> a_2, and a_3 so that the equality holds for *all* ?x'. Otherwise,> calling both expressions, P(x) is false and misleading.Since you did not post the values of a_1, a_2, and a_3 which satisfy the> stated equality, I have taken the liberty of posting them for you. They> are:a_1 = -(7/5)+(1/5)(1078+175640x-900375x^2+14706125x^3)^(1/3)a_2 => -(7/5)-(1/10)(1-3^(1/2)I)(1078+175640x-900375x^2+14706125x^3) ^(1/3)a_3 => -(7/5)-(1/10)(1+3^(1/2)I)(1078+175640x-900375x^2+14706125x^3) ^(1/3)For these values of a_1, a_2 and a_3 (and only these values), does P(x)> = 14706125x^3 - 900375x^2 + 175640x + 1078 = (5 a_1 + 7)(5 a_2 + 7)(5> a_3 + 7).In fact, the following speci? cases can be easily con?med:x=0, P(0) = 1078> x=1, P(1) = 13982468> x=2, P(2) = 114399858which holds for either the original representation of P(x) or the other> representation.Since you did not provide the expressions for a_1, a_2, and a_3, but> merely asserted *incorrectly* that two of them go to zero when x = 0,> there is no way to determine the equivalence of the two representations> for P(x) for any other values of x from your post.Note, however, that *none* of the correct a_1, a_2 or a_3 go to zero> when x = 0! You have arbitrarily picked numbers that produce the> constant term you are looking for and then claim that this results from> setting x = 0. This is false.In fact, for the case x= 0:a_1 => (1/5)(-7+7^(2/3) 22^(1/3) = 0.650704>> a_2 => (1/10)(-14+I7^(2/3)> 22^(1/3)(I+3*(1/2))) = -2.42535+1.77596I>> a_3 => (1/10)(-14-7^(2/3)> 22^(1/3)(1+3^(1/2)I)) = -2.42535-1.77596INone of these values is zero at x=0, but the value of P(0) is correctly> given as 1078.You also failed to state that the constant term in the original> expression for P(x) is also ?pulled out' by setting x = (105 +> (129487)^(1/2) I)/3430 or setting x = (105 - (129487)^(1/2) I)/3430.> These values work for both of the representations for P(x) as long as> the values I have given above for a_1, a_2, and a_3 are used in the> second representational form.[snip]The rest of your post is based on the serious gap which occurred when> you jumped to the conclusion, without any proof, that exactly two of the> ?a's go to zero when x = 0. They do not. Before you repost, please> correct the problem by using the correct values of a_1, a_2 and a_3,> instead of just waving your hands and declaring that two of the ?a's go> to 0 when x = 0.Wacky, isn't it. But, hey it's just basic math. Yup, yup, yup!--> A man with integrity identi?s, acknowledges and corrects his errors.> One with it ignores, denies or defends them. -- Democracy: The triumph> of popularity over principle. -- http://www.crbond.comOK, for the record I'm not a mathematician, neither have I been followingall these JSH threads that much. But I was wondering.If we were to de?ebond_1(x) := -(7/5)+(1/5)(1078+175640x-900375x^2+14706125x^3)^(1/3) bond_2(x) :=(7/5)-(1/10)(1-3^(1/2)I)(1078+175640x-900375x^2+14706125x^3 )^(1/3)bond_3(x) :=(7/5)-(1/10)(1+3^(1/2)I)(1078+175640x-900375x^2+14706125x^3 )^(1/3)Then de?ea1(x):= 0, when x=0 := bond_1(x), otherwisea2(x) := 0, when x=0 := bond_2(x), otherwisea3(x) := 3, when x=0 := bond_3(x), otherwise(where a1 and a2 and a3 are understood to be functions with domain C)Wouldn't it still be true thatP(x)=(5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) for all x in C? Not thatit will make any difference to anything. I'm probably not followingproperly. === [snip]> OK, for the record I'm not a mathematician, neither have I been following> all these JSH threads that much. But I was wondering.>> If we were to de?e>> bond_1(x) := -(7/5)+(1/5)(1078+175640x-900375x^2+14706125x^3)^(1/3)>> bond_2(x) :=> (7/5)-(1/10)(1-3^(1/2)I)(1078+175640x-900375x^2+14706125x^3)^ (1/3)>> bond_3(x) :=> (7/5)-(1/10)(1+3^(1/2)I)(1078+175640x-900375x^2+14706125x^3)^ (1/3)>> Then de?e> a1(x):= 0, when x=0> := bond_1(x), otherwise>> a2(x) := 0, when x=0> := bond_2(x), otherwise>> a3(x) := 3, when x=0> := bond_3(x), otherwise>> (where a1 and a2 and a3 are understood to be functions with domain C)>> Wouldn't it still be true that> P(x)=(5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) for all x in C? Not that> it will make any difference to anything. I'm probably not following> properly.I can't imagine what you are getting at. I simply derived the expressions fora_1, a_2, and a_3 from the requirement that:P(x) = 14706125x^3 - 900375x^2 + 157640x + 1708 = (5 a_1 + 7)(5 a_2 + 7)(5a_3 + 7)as stated in James' post. If you de?e different values for a_1 at x=0 thanthose values required by the expressions which represent them and which aregiven by the polynomial in ?x', the equality doesn't hold.--There are two things you must never attempt to prove: the unprovable -- andthe obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === I can't imagine what you are getting at. I simply derived the expressions for> a_1, a_2, and a_3 from the requirement that:P(x) = 14706125x^3 - 900375x^2 + 157640x + 1708 = (5 a_1 + 7)(5 a_2 + 7)(5> a_3 + 7)as stated in James' post. If you de?e different values for a_1 at x=0 than> those values required by the expressions which represent them and which are> given by the polynomial in ?x', the equality doesn't hold.> Alright, help me to understand then.If I de?eP(x) := 14706125x^3 - 900375x^2 + 157640x + 1708bond_1(x) := -(7/5)+(1/5)(1078+175640x-900375x^2+14706125x^3)^(1/3)bond_2( x) :=(7/5)-(1/10)(1-3^(1/2)I)(1078+175640x-900375x^2+14706125x^3 )^(1/3)bond_3(x) :=(7/5)-(1/10)(1+3^(1/2)I)(1078+175640x-900375x^2+14706125x^3 )^(1/3)Then de?e a1(x):= 0, when x=0 := bond_1(x), otherwise a2(x) := 0, when x=0 := bond_2(x), otherwise a3(x) := 3, when x=0 := bond_3(x), otherwiseQ(x) := (5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7)Can you give me a c where P(c) != Q(c)? (If P andQ are different functions there must be one such c) === > I can't imagine what you are getting at. I simply derived the expressions for> a_1, a_2, and a_3 from the requirement that:>> P(x) = 14706125x^3 - 900375x^2 + 157640x + 1708 = (5 a_1 + 7)(5 a_2 + 7)(5> a_3 + 7)>> as stated in James' post. If you de?e different values for a_1 at x=0 than> those values required by the expressions which represent them and which are> given by the polynomial in ?x', the equality doesn't hold.>> Alright, help me to understand then.> If I de?e>> P(x) := 14706125x^3 - 900375x^2 + 157640x + 1708>> bond_1(x) := -(7/5)+(1/5)(1078+175640x-900375x^2+14706125x^3)^(1/3)>> bond_2(x) :=> (7/5)-(1/10)(1-3^(1/2)I)(1078+175640x-900375x^2+14706125x^3)^ (1/3)>> bond_3(x) :=> (7/5)-(1/10)(1+3^(1/2)I)(1078+175640x-900375x^2+14706125x^3)^ (1/3)>> Then de?e> a1(x):= 0, when x=0> := bond_1(x), otherwise>> a2(x) := 0, when x=0> := bond_2(x), otherwise>> a3(x) := 3, when x=0> := bond_3(x), otherwise>> Q(x) := (5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7)>> Can you give me a c where P(c) != Q(c)? (If P and> Q are different functions there must be one such c)Pythagoras (if that really is your name), I didn't mean to imply that you could*not* do as you suggested -- de?ing speci? values of a variable at speci?values of its argument -- I simply meant that I was responding to James' post.There is no evidence that he intended to de?e the values of a_1, a_2 and a_3 atspeci? values of ?x'. Instead, he proposed and posted two representations of P(x)which he claimed were equivalent, and from which he obtained the result thatexactly two of the ?a's go to zero when x=0. But that result does not follow fromthe given equations. If you want to make a case that at x=0 we can discard thevalues of a_1, a_2 and a_3 and replace them by de?ing those values to be 0, 0 and3, by all means go ahead. You can rest assured that P(x) will equal Q(x) asrequired. However, note that for James results, the values of P(x) and Q(x) are*only* equal at x=0. For other values of ?x' they disagree.If, on the other hand, you are interested in what the values of a_1, a_2 and a_3are at x=0 from the posted equations, you might be better off just to go ahead andevaluate them at x=0.--There are two things you must never attempt to prove: the unprovable -- and theobvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === [snip technobabble]> On of my important analysis tools is a simple technique to factor> polynomials into non-polynomial factors.>> For instance, with the polynomial>> P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078>> that technique gives you>> P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3>> so I can factor to get>> P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7).>> where the a's are the roots of the cubic>> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).>> Now despite the complexity, you can rely on simple ideas still, by> noticing that setting x=0, pulls out constant terms, as>> P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)>> as the cubic de?ing the a's with x=0 is>> a^3 - 3a^2, which has roots, 0, 0 and 3.>> You may not realize it, but what you just saw is revolutionary, both> in the special techniques, and most importantly with the consequences> that quickly follow.You may not realize it, but what you just posted and others may have seenwas dead wrong! As I spelled out clearly in a previous post, theappropriate next step in your argument, after establishing two forms forthe same function, is to determine the expressions giving a_1, a_2 anda_3 and then substituting x=0 in them to determine their behavior. Yousimply *assert* that two of them go to zero when x=0, and this is provablyfalse. In fact, I posted the values for a_1, a_2 and a_3 which meet yourrequirement thatP(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078 =(5a_1+7)(5a_2+7)(5a_3+7)for all ?x' and the values are:a_1 = -(7/5)+(1/5)(1078+175640x-900375x^2+14706125x^3)^(1/3)a_2 = -(7/5)-(1/10)(1-3^(1/2)I)(1078+175640x-900375x^2+14706125x^3) ^(1/3)a_3 = -(7/5)-(1/10)(1+3^(1/2)I)(1078+175640x-900375x^2+14706125x^3) ^(1/3)none of which go to zero when x=0. It is easy to verify by hand, or withMathematica, Macsyma, Maple, Reduce or another symbolic algebra packagethe the above values for a_1, a_2 and a_3 make the equation:P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078 =(5a_1+7)(5a_2+7)(5a_3+7)true for all ?x'.At x=0, in fact, a_1 = (1/5)(-7+7^(2/3) 22^(1/3), a_2 =(1/10)(-14+I7^(2/3) 22^(1/3)(I+3*(1/2))), and a_3 = (1/10)(-14-7^(2/3)22^(1/3)(1+3^(1/2)I)).The rest of your post is based on a gap in your argument which hasresulted from your leaping triumphantly to a false conclusion with noproof in evidence. Please correct your errors before revising theargument.Wacky, isn't it. But, hey, it's just basic math. Yup, yup, yup!--A man of integrity identi?s, acknowledges and corrects his errors. Onewithout it ignores, denies or defends them.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === [snip technobabble]>>On of my important analysis tools is a simple technique to factor>>polynomials into non-polynomial factors.>>For instance, with the polynomial>>P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078>>that technique gives you>>P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3>>so I can factor to get>>P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7).>>where the a's are the roots of the cubic>>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).>>Now despite the complexity, you can rely on simple ideas still, by>>noticing that setting x=0, pulls out constant terms, as>>P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)>>as the cubic de?ing the a's with x=0 is>>a^3 - 3a^2, which has roots, 0, 0 and 3.>>You may not realize it, but what you just saw is revolutionary, both>>in the special techniques, and most importantly with the consequences>>that quickly follow.>>You may not realize it, but what you just posted and others may have seen> was dead wrong! As I spelled out clearly in a previous post, the> appropriate next step in your argument, after establishing two forms for> the same function, is to determine the expressions giving a_1, a_2 and> a_3 and then substituting x=0 in them to determine their behavior. You> simply *assert* that two of them go to zero when x=0, and this is provably> false. In fact, I posted the values for a_1, a_2 and a_3 which meet your> requirement thatP(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078 => (5a_1+7)(5a_2+7)(5a_3+7)for all ?x' and the values are:a_1 = -(7/5)+(1/5)(1078+175640x-900375x^2+14706125x^3)^(1/3)a_2 = -(7/5)-(1/10)(1-3^(1/2)I)(1078+175640x-900375x^2+14706125x^3) ^(1/3)a_3 = -(7/5)-(1/10)(1+3^(1/2)I)(1078+175640x-900375x^2+14706125x^3) ^(1/3)none of which go to zero when x=0. It is easy to verify by hand, or with> Mathematica, Macsyma, Maple, Reduce or another symbolic algebra package> the the above values for a_1, a_2 and a_3 make the equation:P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078 => (5a_1+7)(5a_2+7)(5a_3+7)true for all ?x'.At x=0, in fact, a_1 = (1/5)(-7+7^(2/3) 22^(1/3), a_2 => (1/10)(-14+I7^(2/3) 22^(1/3)(I+3*(1/2))), and a_3 = (1/10)(-14-7^(2/3)> 22^(1/3)(1+3^(1/2)I)).The rest of your post is based on a gap in your argument which has> resulted from your leaping triumphantly to a false conclusion with no> proof in evidence. Please correct your errors before revising the> argument.Wacky, isn't it. But, hey, it's just basic math. Yup, yup, yup!> Well, sort of. Your choices for the a's are correct as far as theygo, but this assumes that the three cube roots of 1078++14706125x^3are distributed among the three terms (5a_i+7) in a way that allocatesone cube root for each of the three terms. Of course, this need not bethe case.In fact, you're making somewhat the same mistake that James does when hesays that P(0)/49 MUST [sic] take the form (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7). He mistakenly assumes that two of the a's are divisibleby 7 when in fact *all three* of the a's are divisible by 7^{2/3}in the ring of algebraic integers (that is, a_i / 7^{2/3} is analgebraic integer, as can easily be veri?d. In fact, IIRC, Noraalready showed this.) The correct factorization over the algebraicintegers, then, isP(0)/49 = (5 a_1/7^{2/3} + 7^{1/3})(5 a_2/7^{2/3} + 7^{1/3}) (5 a_3/7^{2/3} + 7^{1/3})What James perceives as a problem with the de?ition of the algebraicintegers is really a re?n of the fact that he needs a 2-termdivisor for his FLT argument, has convinced himself that it MUST [sic]be true, and interprets the numerous counterexamples as evidence thatthere's something wrong with the algebraic integers, rather than therebeing something wrong with the result he needs. If past events areany indication, he'll eventually understand it, but only after monthsof effort on our part. Oh well, it's a way to pass the time.Rick (trimming those followups) === > [snip technobabble][snip]At x=0, in fact, a_1 = (1/5)(-7+7^(2/3) 22^(1/3), a_2 => (1/10)(-14+I7^(2/3) 22^(1/3)(I+3*(1/2))), and a_3 = (1/10)(-14-7^(2/3)> 22^(1/3)(1+3^(1/2)I)).The rest of your post is based on a gap in your argument which has> resulted from your leaping triumphantly to a false conclusion with no> proof in evidence. Please correct your errors before revising the> argument.Wacky, isn't it. But, hey, it's just basic math. Yup, yup, yup!Well, sort of. Your choices for the a's are correct as far as theygo, but this assumes that the three cube roots of 1078++14706125x^3> are distributed among the three terms (5a_i+7) in a way that allocates> one cube root for each of the three terms. Of course, this need not be> the case.In fact, you're making somewhat the same mistake that James does when he> says that P(0)/49 MUST [sic] take the form (5 a_1/7 + 1)(5 a_2/7 + 1)> (5 a_3 + 7). Actually I think he is right about that, at least up to permutations.> He mistakenly assumes when x <> 0> that two of the a's are divisible> by 7 when in fact *all three* of the a's are divisible by 7^{2/3}> in the ring of algebraic integers (that is, a_i / 7^{2/3} is an> algebraic integer, as can easily be veri?d. In fact, IIRC, Nora> already showed this.) I didn't show it. I thought it might be true at one time,but I was wrong. In general there are three factors w1, w2, andw3 of 49 such that a1/w1, a2/w2, a3/w3, 7/w1, 7/w2, and 7/w3 areall algebraic integers which are functions of x. But the expressions for w1, w2, and w3 are not simple for x <> 0.> The correct factorization over the algebraic> integers, then, isP(0)/49 = (5 a_1/7^{2/3} + 7^{1/3})(5 a_2/7^{2/3} + 7^{1/3})> (5 a_3/7^{2/3} + 7^{1/3}) No - when x = 0, P(0)/49 = 3*5 + 7 = 22. JSH is basicallyright about this part. > What James perceives as a problem with the de?ition of the algebraicintegers is really a re?n of the fact that he needs a 2-term> divisor for his FLT argument, has convinced himself that it MUST [sic]> be true, and interprets the numerous counterexamples as evidence that> there's something wrong with the algebraic integers, rather than there> being something wrong with the result he needs. If past events are> any indication, he'll eventually understand it, but only after months> of effort on our part. Oh well, it's a way to pass the time.> It is now really on a knife edge, whether he will understand it or not.The emotional investment this time is truly enormous. He has gonevery far down this road; starting over would be almost unthinkable. Nora B.> Rick (trimming those followups) === > [snip technobabble]>>On of my important analysis tools is a simple technique to factor>>polynomials into non-polynomial factors.>>For instance, with the polynomial>>P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078>>that technique gives you>>P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3>>so I can factor to get>>P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7).>>where the a's are the roots of the cubic>>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).>>Now despite the complexity, you can rely on simple ideas still, by>>noticing that setting x=0, pulls out constant terms, as>>P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)>>as the cubic de?ing the a's with x=0 is>>a^3 - 3a^2, which has roots, 0, 0 and 3.>>You may not realize it, but what you just saw is revolutionary, both>>in the special techniques, and most importantly with the consequences>>that quickly follow.>> You may not realize it, but what you just posted and others may have seen> was dead wrong! As I spelled out clearly in a previous post, the> appropriate next step in your argument, after establishing two forms for> the same function, is to determine the expressions giving a_1, a_2 and> a_3 and then substituting x=0 in them to determine their behavior. You> simply *assert* that two of them go to zero when x=0, and this is provably> false. In fact, I posted the values for a_1, a_2 and a_3 which meet your> requirement that>> P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078 => (5a_1+7)(5a_2+7)(5a_3+7)>> for all ?x' and the values are:>> a_1 = -(7/5)+(1/5)(1078+175640x-900375x^2+14706125x^3)^(1/3)>> a_2 = -(7/5)-(1/10)(1-3^(1/2)I)(1078+175640x-900375x^2+14706125x^3) ^(1/3)>> a_3 = -(7/5)-(1/10)(1+3^(1/2)I)(1078+175640x-900375x^2+14706125x^3) ^(1/3)>> none of which go to zero when x=0. It is easy to verify by hand, or with> Mathematica, Macsyma, Maple, Reduce or another symbolic algebra package> the the above values for a_1, a_2 and a_3 make the equation:>> P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078 => (5a_1+7)(5a_2+7)(5a_3+7)>> true for all ?x'.>> At x=0, in fact, a_1 = (1/5)(-7+7^(2/3) 22^(1/3), a_2 => (1/10)(-14+I7^(2/3) 22^(1/3)(I+3*(1/2))), and a_3 = (1/10)(-14-7^(2/3)> 22^(1/3)(1+3^(1/2)I)).>> The rest of your post is based on a gap in your argument which has> resulted from your leaping triumphantly to a false conclusion with no> proof in evidence. Please correct your errors before revising the> argument.>> Wacky, isn't it. But, hey, it's just basic math. Yup, yup, yup!>> Well, sort of. Your choices for the a's are correct as far as they>> go, but this assumes that the three cube roots of 1078++14706125x^3> are distributed among the three terms (5a_i+7) in a way that allocates> one cube root for each of the three terms. Of course, this need not be> the case.I don't recall making that assumption. I took the two equations for P(x) asgiven, and submitted them to a symbolic algebra package to determine theexpressions for a_1, a_2 and a_3. Since it (Mathematica) produced those valueswithout complaint, and I con?med the equivalence to my satisfaction, I postedthem.> In fact, you're making somewhat the same mistake that James does when he> says that P(0)/49 MUST [sic] take the form (5 a_1/7 + 1)(5 a_2/7 + 1)> (5 a_3 + 7).Again, I took his posted factored form as a given. I'm not sure why think I havecertain beliefs about these forms or that these forms are necessary. Jamesposted them. I simply solved for the values of a_1, a_2 and a_3.> He mistakenly assumes that two of the a's are divisible> by 7 when in fact *all three* of the a's are divisible by 7^{2/3}> in the ring of algebraic integers (that is, a_i / 7^{2/3} is an> algebraic integer, as can easily be veri?d. In fact, IIRC, Nora> already showed this.) The correct factorization over the algebraic> integers, then, is>> P(0)/49 = (5 a_1/7^{2/3} + 7^{1/3})(5 a_2/7^{2/3} + 7^{1/3})> (5 a_3/7^{2/3} + 7^{1/3})>> What James perceives as a problem with the de?ition of the algebraic>> integers is really a re?n of the fact that he needs a 2-term> divisor for his FLT argument, has convinced himself that it MUST [sic]> be true, and interprets the numerous counterexamples as evidence that> there's something wrong with the algebraic integers, rather than there> being something wrong with the result he needs. If past events are> any indication, he'll eventually understand it, but only after months> of effort on our part. Oh well, it's a way to pass the time.> Rick (trimming those followups)--There are two things you must never attempt to prove: the unprovable -- and theobvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === >>P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 >>P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7). >>where the a's are the roots of the cubic >>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > You may not realize it, but what you just posted and others may have seen > was dead wrong! > P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078 = > (5a_1+7)(5a_2+7)(5a_3+7) > > for all ?x' and the values are: > a_1 = -(7/5)+(1/5)(1078+175640x-900375x^2+14706125x^3)^(1/3) > a_2 = -(7/5)-(1/10)(1-3^(1/2)I)(1078+175640x-900375x^2+14706125x^3) ^(1/3) > a_3 = -(7/5)-(1/10)(1+3^(1/2)I)(1078+175640x-900375x^2+14706125x^3) ^(1/3)That is one solution, note however that there are multiple solutions!Working out James' factorisation gives us:P(x) = 125 a1 a2 a3 + 7.25 (a1 a2 + a1 a3 + a2 a3) + 49.5 (a1 + a2 + a3) + 7^3Equating (remember the original P(X): >>P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3) as follows: a1 a2 a3 = 7^2(2401 x^3 - 147 x^2 + 3x) a1 a2 + a1 a3 + a2 a3 = 0 a1 + a2 + a3 = -3(-1 + 49x)gives an identity and James cubic for the values of the a's. So what Jamesdid post was not dead wrong upto this point. You can use differentequalities of course and come at a different cubic. I do not know whichcubic your a's satisfy (and am not really interested), except that they are*not* algebraic integers in general, I think. > He mistakenly assumes that two of the a's are divisible > by 7 when in fact *all three* of the a's are divisible by 7^{2/3} > in the ring of algebraic integers (that is, a_i / 7^{2/3} is an > algebraic integer, as can easily be veri?d. In fact, IIRC, Nora > already showed this.)No, Nora did not show this, and actually she was shown wrong. That iseasy to see. When m=0, in James' polynomial a3 is coprime to 7. *Ingeneral* (as shown by Arturo), 7 can be factored as p.q.r where allthree are mutually coprime and a1 is divisible by p.q, a2 by p.r anda3 by p.s. With James' polynomial this is in all cases were thepolynomial in a is irreducible.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === [.snip.]>No, Nora did not show this, and actually she was shown wrong. That is>easy to see. When m=0, in James' polynomial a3 is coprime to 7. *In>general* (as shown by Arturo), 7 can be factored as p.q.r where all>three are mutually coprime and a1 is divisible by p.q, a2 by p.r and>a3 by p.s. With James' polynomial this is in all cases were the>polynomial in a is irreducible.Typo alert:You mean 7 is factored as p.q.r, with p,q,r pairwise coprime, a1divisible by p.q, a2 divisible by p.r, and a3 divisible by q.r (orsome other combination, of course).IN addition, one can show that a1 will be coprime to r, a2 coprime toq, and a3 coprime to p. In the irreducible case, one can further showthat none of p,q,r are units; in the reducible case, however, one canshow that at least one, and possibly two of them will be units, butthe factorization and (I think also the) properties still remain(though for other reasons). === ==================================== === ===========Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of ?ures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the inde?ite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan === ======================================= === =========Arturo Magidinmagidin@math.berkeley.edu=== [.snip.]> by 7 when in fact *all three* of the a's are divisible by 7^{2/3}> in the ring of algebraic integers (that is, a_i / 7^{2/3} is an> algebraic integer, as can easily be veri?d. In fact, IIRC, Nora> already showed this.)>>No, Nora did not show this, and actually she was shown wrong. That is>easy to see. When m=0, in James' polynomial a3 is coprime to 7. *In>general* (as shown by Arturo), 7 can be factored as p.q.r where all>three are mutually coprime and a1 is divisible by p.q, a2 by p.r and>a3 by p.s. With James' polynomial this is in all cases were the>polynomial in a is irreducible.And also when it is reducible. The difference is that when it isirreducible, none of p, q, or r are units; when it is reducible, it ispossible for one or two of them to be units (depending on how thepolynomial splits). === ===================================== === ==========Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of ?ures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the inde?ite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan === ======================================= === =========Arturo Magidinmagidin@math.berkeley.edu === > gives an identity and James cubic for the values of the a's. So what James> did post was not dead wrong upto this point. You can use different> equalities of course and come at a different cubic. I do not know which> cubic your a's satisfy (and am not really interested), except that they are> *not* algebraic integers in general, I think.Correct, they aren't in general.> > He mistakenly assumes that two of the a's are divisible> > by 7 when in fact *all three* of the a's are divisible by 7^{2/3}> > in the ring of algebraic integers (that is, a_i / 7^{2/3} is an> > algebraic integer, as can easily be veri?d. In fact, IIRC, Nora> > already showed this.)>Recast my ?st sentence as He mistakenly assumes that two of the a's[must be] divisible by 7 when in fact *all three* of the a's [are inafter that:What James perceives as a problem with the de?ition of the algebraicintegers is really a re?n of the fact that he needs a 2-termdivisor for his FLT argument, has convinced himself that it MUST [sic]be true, and interprets the numerous counterexamples as evidence thatthere's something wrong with the algebraic integers, rather than therebeing something wrong with the result he needs.Rick === gives an identity and James cubic for the values of the a's. So what James> did post was not dead wrong upto this point. You can use different> equalities of course and come at a different cubic. I do not know which> cubic your a's satisfy (and am not really interested), except that they are> *not* algebraic integers in general, I think.> Correct, they aren't in general.I found the posts here interesting as C. Bond made a rather basicmistake, which I thought Rick Decker caught which is that the posterassumes that a_1, a_2 and a_3 are equal, when, in general, they arenot.> > He mistakenly assumes that two of the a's are divisible> > by 7 when in fact *all three* of the a's are divisible by 7^{2/3}> > in the ring of algebraic integers (that is, a_i / 7^{2/3} is an> > algebraic integer, as can easily be veri?d. In fact, IIRC, Nora> > already showed this.)>Recast my ?st sentence as He mistakenly assumes that two of the a's> [must be] divisible by 7 when in fact *all three* of the a's [are in> after that:What James perceives as a problem with the de?ition of the algebraic> integers is really a re?n of the fact that he needs a 2-term> divisor for his FLT argument, has convinced himself that it MUST [sic]> be true, and interprets the numerous counterexamples as evidence that> there's something wrong with the algebraic integers, rather than there> being something wrong with the result he needs.And that comment re?delusion on the part of Rick Decker.Readers can look at the argument, and see what actually is in it.Notice how I'll be strongly emphasizing constant terms all the waydown.P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078which has a constant term that is 1078.Well P(x) can also be written out asP(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3so I can factor to getP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)where the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).Notice it *appears* that the constant terms for the three factors areall 7, which can't be right, as the constant term of P(x) is 1078, sosetting x=0, revealsP(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)as the cubic de?ing the a's with x=0 isa^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1 and a_2for 0, so that leaves a_3 with a value of 3 when x=0.So let a_3 = b_3 + 3, where I keep indices matched. Then I haveP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 5(3) + 7)P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 22)and now my constant terms work out correctly.But P(x) has 49 as a factor as every term in P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078has 49 as a factor, so I can divide by 49, and dividing 1078 by 49gives me 22, as the new constant term.Well that means thatP(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22)is the only way that the constant terms keep matching.James Harris === Dear Group,I want to compare arrays of values to determine thedifference. This is currently done using a correlationcoef?ient in a 2 dimensional array with the originalvalues transformed to z scores. Each of the values hasa different degree of explanation within the model -both globally and geographically. Beta coef?ientshave been used to adjust the values. The sum of the zscore is then used to pull out similar records. Thisworks to a point.What I need is a better way of comparing the valuesand pulling out similar records. Any ideas or suggestionswould be very welcome?Ian.X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oft === at 05:06 PM, Steven Margolin said:>What is the most succinct axiomization of math.Well, ?st of all Mathematics covers a multitude of sins. Thecustomary way is to start with a logical system and axioms of settheory, then de?e various structures in terms of sets, with theproperties in the de?itions serving as axioms for those systems.Keep in mind that there are a lot of topics that have nothing to dowith Geometry or numbers.>Do the Peano Axioms do it?That depends on what is is. Peano's postulates in a proper logic allowyou to model the theorems of Mathematics, so in a sense they areadequate, but probaly not what you have in mind.>How about Peano and Euclid together? Geomtry and are inappropriate for lots of other Geometries.If all you want is Euclidean Geometry, you don't even need Peano; thegeometric axioms are enough by themselves. They're also enough tomodel the real numbers. at 05:30 PM, Steven Margolin said:>How could you have both Euclidean and Hyberbolic Geometry. Because the axioms wouldn't refer to the same things. One common modelof Hyperbolic Plane Geometry uses speci? arcs in the Euclidean Planeto represent lines. The points and permissible arcs satisfy all theaxioms for points and lines in the Hyperbolic Plane.>Wouldn't you>want to be able to say that there, say, 5pi/4 radians in a triangleThe measurements in the model are not the standard Euclideanmeasurements. They are de?ed in such a way as to make theirproperties Hyperbolic rather than Euclidean. at 05:50 PM, Steven Margolin said:>If Choice can be true or false, how can ZFC be complete? ZFC can't be complete if Peano is consistent.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === > >> >> []>> >> Out of curiosity, do you think that Dr. Ullrich's statement was>> complimentary?>complimentary. Classic misrepresentation. Locate the ?st instance you can of me> referring to you as a dumb fuck and I'll show you the previous posts> in which you called me the same and/or worse.Dishonest fuck as well as dumb - for _you_ to complain about> people being less than complimentary is very funny.> >He underestimates my intelligence and refuses to>wish to admit it.> ************************David C. UllrichDavid Ullrich:(Ok, although it seems a little tacky: I'm a professor hereat OSU. I got a PhD at Wisconsin (with Walter Rudin). Then I got ajob here, a few years later I got tenure, and a few years ago I gotpromoted to full professor. The committee suggested I apply for thepromotion - I hadn't meant to. _They_ must think I have _some_ ideawhat math is about or they wouldn't pay me.Mitchell Jones:They may have thought that in the past. Perhaps they bought into yourBS. Or perhaps Oklahoma State is a fourth rate, backwater school of thesort that attracts academic fakers. I couldn't say. I can say this,though: it is not in their interest, or in any academic institution'sinterests, to employ professors who verbally abuse students. Askingquestions and stating arguments is an essential part of the learningprocess, and professors who habitually respond to questions and argumentswith the kind of sneering condescension you have exhibited in thisnewsgroup are the academic kiss of death for the institution that employsthem. Undergraduate students hate them and try to avoid their courses, andthe best graduates ? other institutions to do their postgraduatework, or else become so disgusted with academic life that they turn tocareers in private industry. The result is a brain drain that destroysacademic life at the institution.Mitchell Jones is perceptive indeed.John Correy === 2 Eldon says that is not true.> This suggests that you are using a private de?ition of> compl[i/e]ment. This is not helpful.>> When HH was ?, at least one is a head was true and when TT was> ?, at least one is a tail was true. Those are complements.> When HT, or TH was ?. Either statement was true, and they are> complementary.I don't believe this is the conventional usage of complement in> probability. The complement of tossing a coin and getting a H is T,> because that is the total set of alternatives. The _complement_ of HH> is therefore { HT, TH, TT}; you are using the same word to refer to> the opposite (not a generally well-de?ed notion). This means you> are using a private language, which is a Bad Start if you hope to> persuade anyone of anything.>I'm not arguing with math language, nor trying to change it. Where Ihave private language, I will gladly alter the language. We can argueabout what to call the relationship, but the relationship cannot bedenied.Situation a: A coin was ? and it landed H. (We didn't see the? re ad about it in the statement. We would not expect a demurhad it landed T)Situation b: A coin was ? and it landed T. (We would not expecta demur had it landed H)Situation c: Two coins were ? and they landed HH. (We should notexpect a demur at TT)Situation d: Two coins were ?, they landed HH, the statement wasgenerated, Two coins were ? and at least one is a head. What isthe probability that both are heads? (We should not expect a demur atTT)Situation e: Two coins were ?, they landed TT, the statement wasgenerated, Two coins were ? and at least one is a tail. What isthe probability that both are tails? (This question should have thesame answer as the one in Situation d)As a is related to b, so is b related to c. Maybe it isn'tcomplementary. When a coin is ? two things can happen. a and bare them. When two coins are ?, four things can happen, c and dare two of them. There are two peas in the ?st pod, four in thesecond and we've seen two of them. The ?peas in a pod' are related,maybe not complementary, but related.> Consider the three problem statements (different problems):>> No, you don't. Consider that two coins were tossed. We can say Two> coins were tossed and at least one is a ________. We can't ?l in> that blank, prior to some kind of inspection, or ?look'.The short answer, which I'm sure you've been given many times, is> simply Sorry, Chum, no, we don't mean the statement with a blank was> made, and then the blank was unprejudicially ?led. If we _did_ mean> that, of course your value of 50% is correct. But we hereby clarify> that that is _not_ what we mean. End of discussion.>We need to clarify who _we_ is. When a question is stated, should theanswerer of the question answer the question as stated, or what _we_meant? If it's a test question and the student knows what the profmeant, and how it will be graded, then that's another discussion. Acorrect answer to a question is not according to what we meant, butaccording to what we said. > But why does your penchant for antiprejudice only extend to head/tail?>It doesn't. > Consider that two coins were tossed. We can say Two coins were tossed> and at _____ _____ ____ a ________. We have several blanks, (though> the third one is only for grammatical consistency, since it's> automatically ?led by is or are) and we can't ?l in these> blanks, prior to some kind of inspection, or ?look'.>i)Two coins were ? and at least one is a head. That statementwas made. It can all be said, prior to the look, except _head_. i')Two coins were ? and at least one is a tail.Prior to the look, we can always say that one, maybe both of thosestatements is true. We can make neither, prior to a look, and theycouldn't either.When one is made, the question is, what is the probability that theother statement is also true? Prior to the look the chances for bothto be true was 1/2. Now that there has been a look and there isevidence of a head, does the chance for there also being a tail jumpto 2/3? Think about it.> The ?st might by most, and might be least. The second might be> one, and might be two, or it might be zero. It seems clear that> most and least might be regarded as opposites (or ?complements'> in the sense in which you're using it). It's not clear at all how one> might go about assigning proportions to the various possible> statements that might be made. I am fairly sure that by having> slightly more complicated situations (such as a three-colour die) it's> impossible to make any such meaningful assignment. It's not reasonable> to ask you to solve this wider problem, but you could indicate whether> you think that your claims can ever be extended beyond this one> statement of this one problem?> Whatever we are doing, we must be consistent. We can re? manytimes as we wish, prior to the look, without altering the odds. Afterthe look, we have to do whatever it was that we were doing. If we wishto change at least one to at most one, we must make up our mindsprior to the look. Think about that a while before you argue with it.> Snipping, because this is dreadfully repetitive; but you make> irritatingly spurious responses to simple statements. here's a> couple of examples.(1)> What about the second question? We ?might' presumably have said any of> four things - the ?st/second is a head/tail, except that the> ?st/tail combination makes a contradiction.No contradiction if you make the statement, the ?st coin is a tail,> what are the chances for two tails?(2)> But is the last plausible? Suppose we are using this system to> generate real questions, does it make sense to ask: Two coins were> tossed, and the second is a tail. What is the probability that both> are heads? > No, but it makes sense if you aske, what is the probability that both> are tails?It would help if you studied a little the way people make statements> in mathematical descriptions. When I say Question P does not make> sense I mean that Question P does not make sense. Saying Ah, but if> you change it to something else it does make sense is just silly.> Two coins were ? and at least one is a tail. What are thechances for two heads? I thought this was the question you said didn'tmake sense, and I agree, but,Two coins were ? and at least one is a tail. What are thechances for two tails? makes sense to me.Two coins were ? and at least one is a head. What are thechances for two heads? and Two coins were ? and at least one isa tail. What are the chances for two tails? These two questions getthe same argument, all over the world. They are the same question, orsomewhat related, I'm not sure what we call the relationship. Theensuing argument is the same.Here, it will take some thought to realize what the complements are.> You said, and at least two. Did you mean until at least two?> That's the mistake the conventioneers make on the other question. They> see and at least one; they take it to mean until at least one.No. Here you show again that you simply do not understand what maths> is about. It's different from hermeneutics [again! hope I got the> spelling right] - we are not interested in nit-picking over texts by> dead authors. Ask a conventioneer to clarify and he* will explain> that yes, we mean the interpretation until at least one (in your> terms anyway).>Yes, and I agree that when the ?until' is in there, the answer is 1/3.When the ?until' is left out, it changes the question, and the answer.We can say, well, we mean for it to mean the same thing but, then werun into the real question at least one is and can't answer it correctly.We're on the naive side of a counter-intuitive question. That'scomfortable, so long as we are in the majority. Then we run into dumbfuck Moritz, who won't shut up. The problem is, he's correct. If themajority ever agrees with him, then we have to change, or we becomethe dumb fucks.When we change from at least one to until at least one we have toalter the coin ?quence. That's why we can't make a workingmodel, and Moritz can. He's ?g two coins four ways. We're?g two coins three ways. It's harder, and takes a little moreexplaining. We can't ?o coins, then look, then change it to oneof three. We have to pick a null, prior to the look. That's what Icall prejudice, or prior prejudice. Two coins were ?, theylanded HT and the heads statement was made. Heads were chosen, for theanswer to be 1/3, they must have been pre chosen, chosen prior to thelook. > * Has to be male. The others are conventioneeresses.>I like that, I agree, and they both make the same arguments.Eldon Moritz Brian Chandler> ----------------> geo://Sano.Japan.Planet_3> Jigsaw puzzles from Japan at:> http://imaginatorium.org/shop/ === > I'm not arguing with math language, nor trying to change it. Where I> have private language, I will gladly alter the language. We can argue> about what to call the relationship, but the relationship cannot be> denied.Why not call it opposite? Call it anything you like, as long as itisn't easily confused with the usual meaning of ?complement'.> [me]> Consider the three problem statements (different problems):> [EM]> No, you don't. Consider that two coins were tossed. We can say Two> coins were tossed and at least one is a ________. We can't ?l in> that blank, prior to some kind of inspection, or ?look'.The short answer, which I'm sure you've been given many times, is> simply Sorry, Chum, no, we don't mean the statement with a blank was> made, and then the blank was unprejudicially ?led. If we _did_ mean> that, of course your value of 50% is correct. But we hereby clarify> that that is _not_ what we mean. End of discussion.>> We need to clarify who _we_ is.I hope it's not too presumptuous of me to inform you that by we Imean, er, everyone involved in the discussion except yourself.When a question is stated, should the> answerer of the question answer the question as stated, or what _we_> meant?This is a silly thing to say - obviously the answerer can only answerthe question as understood. Sometimes, inevitably, there ismisunderstanding. Clari?ation helps to eliminate this. In thisparticular case, you argue that more or less everyone except yourselfhas failed to understand the words, even though we have in factunderstood exactly what the questioner meant. So your quest is not amathematical one, it is (as I've said several times) one ofhermeneutics, a branch of human endeavour which I think particulardevoid of value.The other problem with your quest is that as far as one can see it isfor ever going to be restricted to the providing of the Correct Answerto the One Question. You seem to have no notion that in any sense it'sgeneralisable. Suppose you succeed? Students learn that the CorrectAnswer to the Eldon question is 50%, because part of the question_might_ have been different. They obviously ask: what of playingcards?Two decks of cards are cut and* at least one is black; what is theprobability both are?Two decks of cards are cut and* at least one is a spade; what is theprobability both are?Two decks of cards are cut and* at least one is a queen; what is theprobability both are?Two decks of cards are cut and* at least one is a court card; what isthe probability both are?How do you answer these?(* I hope and is OK to trigger the Eldon answer.)> But why does your penchant for antiprejudice only extend to head/tail?>> It doesn't. > Consider that two coins were tossed. We can say Two coins were tossed> and at _____ _____ ____ a ________. We have several blanks, (though> the third one is only for grammatical consistency, since it's> automatically ?led by is or are) and we can't ?l in these> blanks, prior to some kind of inspection, or ?look'.>> i)Two coins were ? and at least one is a head. That statement> was made. It can all be said, prior to the look, except _head_.No. Prior to the look, it can all be said except *ANY* of least,one, or head. Why single out head?> Whatever we are doing, we must be consistent. We can re? many> times as we wish, prior to the look, without altering the odds. After> the look, we have to do whatever it was that we were doing. If we wish> to change at least one to at most one, we must make up our minds> prior to the look. Think about that a while before you argue with it.Why? Just because you say so? I toss two coins. I decide I am going tomake either the statement At least one is a head, what are thechances both are? OR At most one is a head, what are the chancesneither is? I then look. I then make the only statement that is true,or if both are true, I will make the ?st statement if the day of themonth is divisible by three, the second if it's not.The point is that I then make this Statement to which you claim tohave a funny answer. I say the same words in the same sequence.Somehow I suppose you intend to claim that although I say these words,it doesn't count as your Statement, because your argument is based onspecious claims about what's going on in other people's minds.I repeat that I do think there is a valid point lurking here:mathematics _cannot_ answer the question - You are on a tv show in astrange country, and the statement is made that two coins were tossed,and coins were tossed and at least one is a head; what is theprobability both are? Who knows? Perhaps the custom in this countryis that the answers to all questions are Yes. That's also why theMarilyn von Goat question has to be very carefully phrased if you wantthe Marilyn answer to be clearly correct.Brian Chandler----------------geo://Sano.Japan.Planet_3Jigsaw puzzles from Japan at:http://imaginatorium.org/shop/ === > >> >> []>> >> Out of curiosity, do you think that Dr. Ullrich's statement was>> complimentary?>complimentary. Classic misrepresentation. Locate the ?st instance you can of me> referring to you as a dumb fuck and I'll show you the previous posts> in which you called me the same and/or worse.Dishonest fuck as well as dumb - for _you_ to complain about> people being less than complimentary is very funny.> >He underestimates my intelligence and refuses to>wish to admit it.> ************************David C. UllrichI told him you were not complimentary to me. I rest my case.Eldon Moritz === >[]Out of curiosity, do you think that Dr. Ullrich's statement was> complimentary?>complimentary. >> >> Classic misrepresentation. Locate the ?st instance you can of me>> referring to you as a dumb fuck and I'll show you the previous posts>> in which you called me the same and/or worse.>> >> Dishonest fuck as well as dumb - for _you_ to complain about>> people being less than complimentary is very funny.>> >>He underestimates my intelligence and refuses to>>wish to admit it.>> >> >> ************************>> >> David C. Ullrich>>I told him you were not complimentary to me. I rest my case.Tee-hee. I suspect that the many people who've been complainingthat you don't seem to read more than a few key words in messagesbefore replying to them are resting _their_ cases at this point aswell>Eldon Moritz************************David C. Ullrich === I've been somewhat wary of giving fodder to people who like to lieabout mathematics, so I've just talked about resolving an issue thatI'd noticed brought up by Nora Baron, Dik Winter, and possiblyothers.But I decided to talk about it now, so here's quick explanation, as Imake sure that I'm right.I have an expression that is P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3where the key term for this discussion is- 3(-1 + 49 x )(5)(7^2)because if x=1/49, it goes to 0.Some people noticed I guess that you can just take the 49 away, andhave7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + x )(5)(7^2) + 7^3so now, x=1, will work to 0 out that term.But then you have7^2(2401 - 147 + 3) (5^3) + 7^3which factors easily enough.The answer is that what may seem like a simple maneuver gives a resultthat is also given with a different expression by something likex=1/49 with the ?st.That is, there exists an expression for which x=1/49 will give you7^2(2401 - 147 + 3) (5^3) + 7^3so mathematically, that possibility has precedence because forconsistency, you can have 7 a unit in the ring, and cover bothpossibilities.That is, given7^2(2401 - 147 + 3) (5^3) + 7^3there's no way to tell mathematically whether or not it came from oneexpression or the other, so the math defaults to the more generalcondition, which is that 7 is a unit.James Harris === > I've been somewhat wary of giving fodder to people who like to lie> about mathematics, so I've just talked about resolving an issue that> I'd noticed brought up by Nora Baron, Dik Winter, and possibly> others.>> But I decided to talk about it now, so here's quick explanation, as I> make sure that I'm right.>> I have an expression that is>> P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3Wrong! Did you ever actually expand that version of P(x) to see if itmatched your original equation? Evidently not. Your original equation was:P(x) = 14706125x^3 - 900375x^2 + 17640x + 1078Your factored representation expands to:P(x) = 14706125x^3 -900375x^2 - 17640x + 1078Notice anything funny? The very 1st factorization in your argument has asign error -- and one that you have neither acknowledged or corrected fromthe beginning. Furthermore, it invalidates all your subsequentcalculations. (Do you suffer from some sort of cognitive disorder?)Wacky isn't it? But, hey, it's just basic math. Yup, yup, yup!--What a maroon! -- Bugs Bunny.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === > I've been somewhat wary of giving fodder to people who like to lie > about mathematics, so I've just talked about resolving an issue that > I'd noticed brought up by Nora Baron, Dik Winter, and possibly > others.Ah, here comes the simple explanation. > But I decided to talk about it now, so here's quick explanation, as I > make sure that I'm right. > > I have an expression that is > > P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 > > where the key term for this discussion is > > - 3(-1 + 49 x )(5)(7^2) > > because if x=1/49, it goes to 0. > > Some people noticed I guess that you can just take the 49 away, and > have > > 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + x )(5)(7^2) + 7^3 > > so now, x=1, will work to 0 out that term. > > But then you have > > 7^2(2401 - 147 + 3) (5^3) + 7^3 > > which factors easily enough. > > The answer is that what may seem like a simple maneuver gives a result > that is also given with a different expression by something like > x=1/49 with the ?st. > > That is, there exists an expression for which x=1/49 will give you > > 7^2(2401 - 147 + 3) (5^3) + 7^3 > > so mathematically, that possibility has precedence because for > consistency, you can have 7 a unit in the ring, and cover both > possibilities. > > That is, given > > 7^2(2401 - 147 + 3) (5^3) + 7^3 > > there's no way to tell mathematically whether or not it came from one > expression or the other, so the math defaults to the more general > condition, which is that 7 is a unit.Alas, I do understand nothing of the explanation. Do you mean that 7is a unit in the object ring?-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === that's an interesting question. perhaps James Harris could rephrase itfor a mass Usenet reader audience, a.k.a. this local ringof the googolplex. > > so mathematically, that possibility has precedence because for> > consistency, you can have 7 a unit in the ring, and cover both> > possibilities.> > That is, given> > 7^2(2401 - 147 + 3) (5^3) + 7^3> > there's no way to tell mathematically whether or not it came from one> > expression or the other, so the math defaults to the more general> > condition, which is that 7 is a unit.Alas, I do understand nothing of the explanation. Do you mean that 7> is a unit in the object ring?--les ducs d'Enron! === > > I've been somewhat wary of giving fodder to people who like to lie> > about mathematics, so I've just talked about resolving an issue that> > I'd noticed brought up by Nora Baron, Dik Winter, and possibly> > others.Ah, here comes the simple explanation. > But I decided to talk about it now, so here's quick explanation, as I> > make sure that I'm right.> > I have an expression that is > > P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3And I didn't just make that expression up out of the blue or by usingtrial and error, as there's a trail to it all, but for now I have tohandle mathematicians trying to run from the conclusion to which itleads, before going back to the interesting details like how I ?uredit out.> > where the key term for this discussion is> > - 3(-1 + 49 x )(5)(7^2)> > because if x=1/49, it goes to 0.I've been thinking of what to call it and I kind of like calling it abalanced factorization. Or maybe I should call it a perfectfactorization.The point is that there are *many* ways to factor a particularexpression, and my way keeps you in the ring of objects.Others have deliberately broken out of it by using a slightlydifferent factorization.> > Some people noticed I guess that you can just take the 49 away, and> > have> > 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + x )(5)(7^2) + 7^3> > so now, x=1, will work to 0 out that term.> > But then you have> > 7^2(2401 - 147 + 3) (5^3) + 7^3> > which factors easily enough.> > The answer is that what may seem like a simple maneuver gives a result> > that is also given with a different expression by something like> > x=1/49 with the ?st.> > That is, there exists an expression for which x=1/49 will give you> > 7^2(2401 - 147 + 3) (5^3) + 7^3> > so mathematically, that possibility has precedence because for> > consistency, you can have 7 a unit in the ring, and cover both> > possibilities.That is, once you get to that expression, you don't have a footprintback from whence it came. Since it can come from a perfectfactorization using something like 1/49 or some other number thatwould force other integers besides 1 or -1 to be units, the mathdefaults to that situation.That is, you are in a ?ld.> > That is, given> > 7^2(2401 - 147 + 3) (5^3) + 7^3> > there's no way to tell mathematically whether or not it came from one> > expression or the other, so the math defaults to the more general> > condition, which is that 7 is a unit.Alas, I do understand nothing of the explanation. Do you mean that 7> is a unit in the object ring?No. There are several ways to factor a polynomial into non-polynomialfactors and my way is balanced.In trying to test the idea, you and others have looked at variantfactorizations.Now it turns out that I didn't just pullP(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3out of a hat.So the mathematics is based on *other* mathematics, and it's not likeyou can just play with it, and make changes that don't reverbate back.Now, unfortunately, I have to give information out at the tail end soto speak, as rather than get intrigued by my research mathematiciansare ?hting to hide it from view.James Harris === > > I've been somewhat wary of giving fodder to people who like to lie > > about mathematics, so I've just talked about resolving an issue that > > I'd noticed brought up by Nora Baron, Dik Winter, and possibly > > others. > > Ah, here comes the simple explanation. > > > But I decided to talk about it now, so here's quick explanation, as I > > make sure that I'm right. > > > > I have an expression that is > > > > P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 > > And I didn't just make that expression up out of the blue or by using > trial and error, as there's a trail to it all, but for now I have to > handle mathematicians trying to run from the conclusion to which it > leads, before going back to the interesting details like how I ?ured > it out. > > > > > where the key term for this discussion is > > > > - 3(-1 + 49 x )(5)(7^2) > > > > because if x=1/49, it goes to 0. > > I've been thinking of what to call it and I kind of like calling it a > balanced factorization. Or maybe I should call it a perfect > factorization. > > The point is that there are *many* ways to factor a particular > expression, and my way keeps you in the ring of objects. > > Others have deliberately broken out of it by using a slightly > different factorization.*Why* do you break out of it or keep in of it? I thought your factorisationwas such that a1, a2 and a3 were algebraic integers. In *my* polynomial Iuse also a factorisation that keeps you in the algebraic integers. So*where* is the difference? (Note that also in my case the polynomial forthe a's is monic with integer cof?ients.) You get your monic theway you do by equating (a1 + a2 + a3), (a1 a2 + a1 a3 + a2 a3) and(a1 a2 a3) to factors of indivual terms of the original polynomial. Ido the same. So *what* is the difference. (And what part of yourargument fails with my polynomial.) > > Some people noticed I guess that you can just take the 49 away, and > > have > > > > 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + x )(5)(7^2) + 7^3 > > > > so now, x=1, will work to 0 out that term. > > > > But then you have > > > > 7^2(2401 - 147 + 3) (5^3) + 7^3 > > > > which factors easily enough. > > > > The answer is that what may seem like a simple maneuver gives a result > > that is also given with a different expression by something like > > x=1/49 with the ?st. > > > > That is, there exists an expression for which x=1/49 will give you > > > > 7^2(2401 - 147 + 3) (5^3) + 7^3 > > > > so mathematically, that possibility has precedence because for > > consistency, you can have 7 a unit in the ring, and cover both > > possibilities. > > That is, once you get to that expression, you don't have a footprint > back from whence it came. Since it can come from a perfect > factorization using something like 1/49 or some other number that > would force other integers besides 1 or -1 to be units, the math > defaults to that situation. > > That is, you are in a ?ld.But I do not get at an expression, I get at a polynomial in a: a^3 + 3(-1 + m).a^2 - 7^2.2(.m^2 - 3.m)which for m = 0 equals: a^3 - 3.a^2and so for m = 0, the roots are 0, 0 and 3. Just like you do.Why can *you* extrapolate the behaviour from that for m = 1,while I can do not such an extrapolation? > > That is, given > > > > 7^2(2401 - 147 + 3) (5^3) + 7^3 > > > > there's no way to tell mathematically whether or not it came from one > > expression or the other, so the math defaults to the more general > > condition, which is that 7 is a unit. > > Alas, I do understand nothing of the explanation. Do you mean that 7 > is a unit in the object ring? > > No. There are several ways to factor a polynomial into non-polynomial > factors and my way is balanced. > > In trying to test the idea, you and others have looked at variant > factorizations.I have looked at a factorisation of other polynomials, not at variantfactorisations of your polynomial. And I did that factorisation in away that is similar to what you did. > Now it turns out that I didn't just pull > > P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 > > out of a hat. > > So the mathematics is based on *other* mathematics, and it's not like > you can just play with it, and make changes that don't reverbate back. > > Now, unfortunately, I have to give information out at the tail end so > to speak, as rather than get intrigued by my research mathematicians > are ?hting to hide it from view.So you pull a factorisation out of your hat, claims various things aboutit, without ever specifying why what you claim is correct for yourfactorisation and not for factorisations of other polynomials. And *you*think we have to accept that on the face of it?-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === > And I didn't just make that expression up out of the blue or by using> trial and error, as there's a trail to it all, but for now I have to> handle mathematicians trying to run from the conclusion to which it> leads, before going back to the interesting details like how I ?ured> it out.This exhibits a striking lack of awareness of what is mathematicallyinteresting.How someone happened to ?ure something out is perhapspsychologically interesting, socially interesting, and pedagogicallyinteresting. But it is not mathematically interesting.What is mathematically interesting is the theorems, the proofs, thede?itions, and also the ways that those things open the ground formore mathematics. How a person happens to discover a particularly beautiful theorem or anifty proof is not, actually, mathematically interesting. Thomas === I'm taeching the beginning epsilon-delta course. Lookingahead in the book (Ross, Elementary Analysis: the Theoryof Calculus) I see the following de?ition:Suppose that f : A -> R. Then lim_{x->a} f(x) = L if forevery sequence (x_n) in A such that x -> a we havef(x_n) -> L.At ?st I thought this must be a typo. But it turns outhe means it - later when he shows that this de?itionis equivalent to the one in terms of epsilon and deltathe condition is |x - a| < delta, not 0 < |x-a| < delta.I'm shocked. Is this version of the de?ition actuallystandard in some circles? My impression is that thede?ition is inconsistent with what the students arealmost certainly going to see in later courses - is this correct? (I hate to cause confusion by using ade?ition different from what's in the book unlessI have a very good reason, but if this de?ition isas rare as it seems to me it is that could be a goodenough reason - hence the question whether itreally is extremely uncommon.)************************David C. Ullrich === > I'm taeching the beginning epsilon-delta course. Looking> ahead in the book (Ross, Elementary Analysis: the Theory> of Calculus) I see the following de?ition:Suppose that f : A -> R. Then lim_{x->a} f(x) = L if for> every sequence (x_n) in A such that x -> a we have> f(x_n) -> L.At ?st I thought this must be a typo. But it turns out> he means it - later when he shows that this de?ition> is equivalent to the one in terms of epsilon and delta> the condition is |x - a| < delta, not 0 < |x-a| < delta.What is the space A? does A include or exclude the point a? If Aexcludes the point a, then you could have your 0 < |x - a| byde?ition of A.-- Johan KULLSTAM sysengr === > I'm taeching the beginning epsilon-delta course. Looking> ahead in the book (Ross, Elementary Analysis: the Theory> of Calculus) I see the following de?ition:>> Suppose that f : A -> R. Then lim_{x->a} f(x) = L if for> every sequence (x_n) in A such that x -> a we have> f(x_n) -> L.This is a good de?ition to use for proving thata limit doesn't exist (for example, sin(1/x) as x->0+,and the sequences 1/(2*n*pi) and 1/((2*n+1/2)*pi) )Obviously it depends on ?st de?ing the limit of a sequence.>> At ?st I thought this must be a typo. But it turns out> he means it - later when he shows that this de?ition> is equivalent to the one in terms of epsilon and delta> the condition is |x - a| < delta, not 0 < |x-a| < delta.This is a good de?ition to use for proving that a limit does exist.>> I'm shocked. Is this version of the de?ition actually> standard in some circles? My impression is that the> de?ition is inconsistent with what the students are> almost certainly going to see in later courses - is this> correct? (I hate to cause confusion by using a> de?ition different from what's in the book unless> I have a very good reason, but if this de?ition is> as rare as it seems to me it is that could be a good> enough reason - hence the question whether it> really is extremely uncommon.)Have you considered using a different text?Or even giving both de?itions and proving their equivalence?-- P.A.C. Smith'If the Apocalypse comes, beep me.' <*> http://www.srcf.ucam.org/~pas51 === > I'm taeching the beginning epsilon-delta course. Looking>> ahead in the book (Ross, Elementary Analysis: the Theory>> of Calculus) I see the following de?ition:>> Suppose that f : A -> R. Then lim_{x->a} f(x) = L if for>> every sequence (x_n) in A such that x -> a we have>> f(x_n) -> L.>>This is a good de?ition to use for proving that>a limit doesn't exist (for example, sin(1/x) as x->0+,>and the sequences 1/(2*n*pi) and 1/((2*n+1/2)*pi) )>>Obviously it depends on ?st de?ing the limit of a sequence.> At ?st I thought this must be a typo. But it turns out>> he means it - later when he shows that this de?ition>> is equivalent to the one in terms of epsilon and delta>> the condition is |x - a| < delta, not 0 < |x-a| < delta.>>This is a good de?ition to use for proving that a limit does exist.> I'm shocked. Is this version of the de?ition actually>> standard in some circles? My impression is that the>> de?ition is inconsistent with what the students are>> almost certainly going to see in later courses - is this>> correct? (I hate to cause confusion by using a>> de?ition different from what's in the book unless>> I have a very good reason, but if this de?ition is>> as rare as it seems to me it is that could be a good>> enough reason - hence the question whether it>> really is extremely uncommon.)>>Have you considered using a different text?>>Or even giving both de?itions and proving their equivalence?Edgar has pointed out that I didn't read the section carefullyenough - the right de?ition comes later.The two de?itions I was talking about are _not_ equivalent.The problem is not sequences versus epsilons, the problemis whether f(a) should be relevant:Say f(x) = 0 for x <> 0, f(0) = 1. Then lim_{x->0} f(x) shouldbe 0, but by one of the de?itions above the limit doesnot exist.(At least that's what I thought the problem was - in factI missed a tiny bit of notation, the wrong de?itionabove is a de?ition of something other thanlim_{x->a} f(x).)************************David C. Ullrich === > Edgar has pointed out that I didn't read the section carefully> enough - the right de?ition comes later.>Eheheh remember my question some months ago about the same topic ;) ? Okbasically those are the same de?itions; over here when we want to refer toyour standard de?ition we write: lim (x->a, x<>a) f(x) instead of lim(x->a) f(x).--Julien Santini,France === > I'm taeching the beginning epsilon-delta course. Looking> ahead in the book (Ross, Elementary Analysis: the Theory> of Calculus) I see the following de?ition:Suppose that f : A -> R. Then lim_{x->a} f(x) = L if for> every sequence (x_n) in A such that x -> a we have> f(x_n) -> L.My copy doesn't have this. Instead is a De?ition 20.1with a little S in there. Followed by De?ition 20.3where he uses it to get 4 of the usual de?itions.But mine is only the 1980 edition. Has it been loused up since then?At ?st I thought this must be a typo. But it turns out> he means it - later when he shows that this de?ition> is equivalent to the one in terms of epsilon and delta> the condition is |x - a| < delta, not 0 < |x-a| < delta.In my copy, the one with |x-a| < delta has the S in it,and the one with 0 < |x-a| < delta is the usual limit(without the S).I'm shocked. Is this version of the de?ition actually> standard in some circles? My impression is that the> de?ition is inconsistent with what the students are> almost certainly going to see in later courses - is this > correct? (I hate to cause confusion by using a> de?ition different from what's in the book unless> I have a very good reason, but if this de?ition is> as rare as it seems to me it is that could be a good> enough reason - hence the question whether it> really is extremely uncommon.)************************David C. Ullrich-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === > I'm taeching the beginning epsilon-delta course. Looking>> ahead in the book (Ross, Elementary Analysis: the Theory>> of Calculus) I see the following de?ition:>> >> Suppose that f : A -> R. Then lim_{x->a} f(x) = L if for>> every sequence (x_n) in A such that x -> a we have>> f(x_n) -> L.>>My copy doesn't have this. Instead is a De?ition 20.1>with a little S in there. Followed by De?ition 20.3>where he uses it to get 4 of the usual de?itions.>But mine is only the 1980 edition. Has it been loused up since then?what was coming up (never got that far the last time I taughtthe course) - missed the little S. Sure enough in20.3(a) he speci?s S = J{a} and everything's ?e.>> At ?st I thought this must be a typo. But it turns out>> he means it - later when he shows that this de?ition>> is equivalent to the one in terms of epsilon and delta>> the condition is |x - a| < delta, not 0 < |x-a| < delta.>>In my copy, the one with |x-a| < delta has the S in it,>and the one with 0 < |x-a| < delta is the usual limit>(without the S).Yes - I saw 20.6, didn't notice the little S, and missed 20.7.That's a relief, thanks. Remind me next time to actuallylook at the whole section before jumoing to conclusions.************************David C. Ullrich