Be unhappy all you want, I don't care for your nitpicking Maple bug reports. 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 find he made statements like : I am the leader of modern mathematics... Nobody 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 difficult 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 Rutherford http://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 Bondarenko Co-founder, CEO, Mathematical Director > Cyber Tester, LLC http://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 verification system. So on top of that, I may add a >> system for user defined syntax. Then this may in fact be a simplification >> 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 unification 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 already implemented my system. If you haven't already done so, please have a look at it and let me know what you think. (System requirements: Windows 95 or later, and IE4 or later.) Try out the tutorial (click the Help button) to get a general idea of the system's capabilities. Examples in the tutorial include Russell's Paradox and what I called the Paradox of the Universal Set. Dan Visit DC Proof Online at http://www.dcproof.com -- FREE download ==== I 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 figure 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 advance john ==== See for example the list of references here: http://lie.uwaterloo.ca/odetools/references.html I 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 > figure 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 advance john ==== 1. Given two arbitary symbolic expressions involving only real algebraic numbers of constant value (e.g. 1/6, Sqrt(37/6) are simple examples), is it possible to determine computationally if one expression is less than or equal to the other? (Presumably this is at least as hard as the zero equivalence 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 expect a 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 free registration key. You can do the following: -find bugs -feedback about you impression by the program -advice me to change or add some new features -translate it to the other language (see www.master-graph.com/instructions/interface.html) Feel free to contact with me - roman@master-graph.com Sincerely, Roman. ==== Hoops 1-4 developed Hoop division algebras, collapsed from Moufang loops with r-fold symmetry and having additive elimination as generalized subtraction. Sizes are functions that are conserved on hoop multiplication; the sizes of a product are the products of the corresponding sizes of the multiplicands. I now explore the consequences of sizes becoming zero. (5a) Sizes can become zero. New possibilities exist in non-degenerate algebras that conserve more than one size. Sizes of vecs (generalized vectors) can become zero 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 either a+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 3 non-trivial zeroes. (5b) Zero sizes constrain calculations to a sub-algebra. A multiplicand with a zero size gives a product with the same size equal to zero. The result is projected onto a sub-algebra (to which the 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 same constraint. This is effected by the genInverse function; partial fractions are ignored if the denominator is less than some prescribed h (i.e. if the size approximates to zero). Conic sections provide an analogy. The bi-cone is highly 3D symmetrical. Restricting consideration to a 2D plane gives various less symmetrical 2D figures. The system is constrained to have zero distance from the plane. This corresponds to renormalization, the meta-mathematical trick used by mathematical physicists to dispose of un-physical infinities. It also corresponds to sub- & super-symmetry and the use of graded algebras. Roger Beresford. Now you see it. Now you don't! (Magician's patter.) ==== 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^2 However, 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^2 I 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 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^2 However, 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^2 I 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 relationship Calculating 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 2 so 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 PDEs in Maple (v.9), generate a C code of the corresponding set of ODEs, mex them to make a .dll file and use MATLABs ODE45() to solve the system of ODEs. I have tested an example problem successfully consisting of 3 ODEs. Here is the Maple code for that. //// Start Code ///// > restart; > with(CodeGeneration): Warning, the protected name Matlab has been redefined 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 file example_3ode.c and make it a .dll file using a mexFunction and solve the ODEs using ODE45() of MATLAB. I have also considered a sample PDE problem, x(r,t). I converted the PDE into a set of ODEs. But I have not been able to convert the system of ODEs as done in the above case. I am providing the corresponding code below. Can somebody please take a look at it and let me know how to 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 differential equations //// 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 Function I 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 find 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 base for mathematical functions, but best used in connection with notebooks and MathReader (or Mathematica). ==== we are planing to start with a project refering imaging software for pattern recognition. In a first step we are looking for some good books in this area e.g. math-background, standard algorithms. You aren't specific about what it is that you want the computer to recognize, but here are a few good references: Applied Pattern Recognition, by Paulus and Hornegger (includes material specific to images) Both of these title are heavier on the image processing side of things: Algorithms for Image Processing and Computer Vision, by Parker Digital Image Processing, by Gonzalez and Woods You might consider machine learning, which can be used to develop pattern recognizers. If so, consider: Computer Systems That Learn, by Weiss and Kulikowski It might help if you were to provide more specific information about your problem. good luck, Will Dwinnell http://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 definitions of machine, i.e. parallel, distributed, global computing, functional languages appear to be closer to the machine architecture. The typical obligatory reference in this case is Sisal. However, it is dormant if not dead, but someone else also pointed out Single Assignment C recently (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 is infinate and it can have and infinate 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 finite fft as an approximation to the integral fourier > transform. If you make a trapezoidal approximation to the integral, > the first 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 first 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 first 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). ==== > Have you remembered to halve the t=0 input datum? >> I`m sorry i don`t understand. why is this necessary? >> >> Think of the finite fft as an approximation to the integral fourier >> transform. If you make a trapezoidal approximation to the integral, >> the first 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 first 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 first 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 that approximates the integral ft - since he is expecting an exponential to transform to a lorentzian. This doesn't work right unless you halve the weight of the t=0 point. For a discussion in the context of 2d NMR, which may be the OP's real problem, see Wutrich et al. J. Mag. Res. 66, 187 (1986). ==== 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... gert ==== 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 is no general formula for all orthogonal polynomials (well, unless if such has been found during last 30 years; you might want to search mathscinet http://www.ams.org/mathscinet . I can't do it from home), but let me pick two 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 flipping a coin and getting heads is still 50:50 even if you got heads 10 times in a row. However, in theory, if you flipped the coin an infinite 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 find 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.pdf Another pointer that might be relevant, because of the comparison of 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 fitted 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 code for that in _Numerical Recipes_, though I was hoping for a simpler method. I'm not sure how to port Fortran code into Labview (the programming is being contracted out), though I imagine there is a way to do it. -- Allen Windhorn (507) 345-2782 FAX (507) 345-2805 Kato Engineering (Though I do not speak for Kato) P.O. Box 8447, N. Mankato, MN 56002 Allen.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 observe that 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 may be useful to discuss what the consequences of an answer would mean. If the answer is no, the discussion probably ends right there. But if the answer is yes, we are now allowed to make more statements about it. Such as (from the original post): Unless nature is presumed to be an exception to the Incompleteness Theorem, to contain all the statements of nature completely and consistently the system of all observations would require additional statements beyond what is observed. Because these new statements lie beyond the set of all observations these statements cannot be made using the scientific method, and therefore, physics is concluded to be unable to provide a final theory of nature. 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 a mathematical model of the universe. And if what I say is true then we can create a model of the universe that 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] confirmations 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? I would answer that Logic seems to be Natural, so far. If it ever ceases to be, I suppose it will be up to the 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] confirmations 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.htm for the thesis that logic is empirical. By this he means that the laws of logic, like any other scientific laws, can be tested empirically and modified if they do not fit with our experience. (later in the essay) A rather more serious objection was raised by Crispin Wright in his paper Inventing Logical Necessity (1986). Wright argues that Quineās theory of knowledge cannot treat logical beliefs simply as another set of beliefs clustered near the centre of our array. Logic is needed to form the links between our beliefs, and as such they have an entirely different role to specific beliefs about states of affairs (Īthe cat sits on the matā) or scientific theories (Īcats tend to fall from mantelpieces to mats with a constant acceleration of 9/8ms-2ā). (the essay concludes:) Quineās picture falls apart if we have no logic to provide connections between the beliefs in our array. Nevertheless, that is only an argument for the need for some sort of logic for us to be able to think coherently ö it does not prove that the specific logical beliefs that we hold at any one time could not be changed. Nevertheless, the thought is persuasive that logical statements are in principle open to such verification within the sort of structure that Quine 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 supposed to spell it grammatically correct instead of grammatically-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 : >> 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 Is all this leading anywhere? My brain is beginning to hurt... :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== > Is Nature a Logical System? It seems pretty illogical to me. > Any challenges to my reasoning are confirmations of its validity are > enthusiastically anticipated. One of the most common flags for pseudoscience. > what is meant by the definition. If nature will be defined as the set > of all observed phenomena, what is all observed phenomena supposed to > mean? Observed by who or what? This is ill-defined. 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 > define 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 statements The 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 physics This 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 also Nor 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 completely Have 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 (defined 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 finite set of measurements we perform, there will always > be a finite 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 replied in the physics group. I have something I'd like to discuss with people and I tried the topic here because the heavy use of mathematical models and logical systems. It seems to have generated some discussion (with some noise even) and thats 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.html Needless 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 five $200 black-and-white >TV's as consolation prizes. What is the expected value of a $2 ticket >in this draw? The answer is: fifty three gazillion dollars. Go write that on your homework... also, this problem is very dated... there's not a black and white TV in the world that's worth 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 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 salivates A, 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? ==== 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 salivates A, 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 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 defined by the table A B if A then B t t t t f f f t t f f t where t means true and f means false. The ancient Greeks discussed this 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 defined by the table A B if A then B > t t t > t f f > f t t > f f t where 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 the Logician's assertion that A implies B, is necessarily true by definition. But isn't this how we learn and develop logical models of our environment? I don't think we could live without making inferences such as these from experience. Furthermore, isn't all reasoning basically based upon this idea of inference? SomeEquation => SomeOtherEquation SomeTruth => SomeOtherTruth Other than that, it appears variable subtitution is all that is necessary. Of course I may be terribly confused and misinformed, but this seems to me to be the way things work. I have a deep respect for the basic ideas of Karl Popper, Thomas Kuhn and the Logical Positivists camp in general. I like to learn more about 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 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 definitly 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 first, maybe Leibniz did, but no one will 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 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 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 definitly 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 first, 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 1 Let me take a moment away from the TV, where we've been watching the progress of the horrendous local fires 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 firmly attest that the patterns reported by contemporary witnesses and experiencers jibe remarkably well with pre-modern accounts of flying shields, abductions to fairy-land by short creatures with big eyes and pointed, elfin chins, missing-time episodes, and rides in flying 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 & Scott C. SCOTT LITTLETON President, Phi Beta Kappa Alumni in Southern California Professor of Anthropology, Emeritus Occidental College Los Angeles, CA 90041 TEL (323) 255-5477 FAX (323) 982-0264 http://www.oxy.edu/~yokatta/home.htm Any sufficiently advanced technology is indistinguishable from magic --Sir Arthur C. Clarke I think we're property. . . --Charles Fort Again, 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 fly via 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 2 minimization of collateral damage to civilian population, maximizing at the same time efficiency of influencing 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 sufficient few words. Here can be drawn parallels with first 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 scientific 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 & inefficient massive interventions like Vietnam, Afghanistan & Iraq; - which bring much needless suffferings to innocent civilians, lots of political problems, terrible financial 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 influence 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 efficient 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 efficient 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 teraflop processing 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 figure 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 inflation where in the FRW metric R(t) ~ e^Ht H is the Hubble not the Hamiltonian. ;-) Now in my toy model, the Dirac vacuum for the electron in globally flat space-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 Inflation Field Since the Vacuum Coherence Field is a complex scalar field 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 field 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 field 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 theory Einstein's Diff(4) symmetric geometrodynamic field 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^u The Sakharov metric elasticity field is simply (without holographic t'Hooft-Sussking renormalization at first) (Lp^2/ih)Pu(Goldstone Phase). Lp^2 = hG(Newton)/c^3 ~ 10^-66 cm^2 I need a second set of generators P'u from the Conformal Group operating vacuum field where if Lp* = Lp^2/3(c/H*)^1/3 where 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 is GM/c^2 = c/H* independent of cosmic time t and the FRW scaling R(t) of the dimensionless comoving r coordinate of the Hubble flow. M = M(on mass shell real matter + radiation) + M(near EM fields) + M(exotic vacuum fields) 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) ~ 0.73M M(dark matter) ~ 0.23M M(on mass shell real matter + radiation) + M(near EM fields) ~ 0.04M Not to be confused with the LARGE-SCALE ONLY cosmic time-dependent H(t) = R(t)^-1(dR(t)/dt) Note that the time dependent densities scale as R(t)^-3 for real matter with w ~ 0 R(t)^-4 for real EM radiation like the CMB with w = +1/3 and independent of R(t) for exotic vacuum with w = -1 from Einstein's equivalence principle + Heisenberg's uncertainty principle Where the equation of state of ALL STUFF REAL OR VIRTUAL is w = pressure/(energy density) The electron rest mass is m ~ (e/c)^2/zpf*^1/2 from the strongly attractive dark matter vortex ring core of positive vortex string of a superfluid. That is, in the sense of a Taylor series expansion about the normal vacuum from pouring macro-quantum coherent oil on the random chaotic micro-quantum zero point fluctuations of ALL quantum fields. This stillness was the calm before the storm of the Big Bang reheating post-inflation 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) Ripple The quanta normal fluid noise of these ripples in the emergent More is different (PW Anderson) vacuum superfluid signal are Einstein'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 field? For me, the electric field is the gradient > of the potential, so the electric field 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 field 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 field. but why would this make sense when we are dealing with > a *constant* distribution of charge. It seems reasonable that if you have a > finite amount of charge spread evently over the disc, then forces for test > charges in that domain should be finite as well. shouldnt they? check my > calculations: 1. start with poissons formula in axially symmetric cylindrical coords - > with a source term of Laplacian(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 similar so we have phi = 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 quick back-of-the-envelope calculation and obtained a result which looks quite different. The main difference was, that the arguments of the elliptic functions look different. In fact, for a=1 I got expressions of the form E(-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 field will get unbounded as you approach the edge of the disc. I am not sure at all, that I did not miscalculate, but I recommend that you go over your computations again. HTH, Michael. -- &&&&&&&&&&&&&&&&#@#&&&&&&&&&&&&&&&& Dr. Michael Ulm FB Mathematik, Universitaet Rostock michael.ulm@mathematik.uni-rostock.de ==== Jaco If I have an element g(x) in R[x], where R[x] is a polynomial ring > with coefficients from a finite field GF(2^m), how do I define 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 coefficients from a finite field GF(2^m), how do I define 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 coefficients 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 first 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 find one that is, let us know. adam ==== >What is the first 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 find 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 first 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 find one that is, let us know. adam > Whoops, I meant to say: Go through all the numbers from one to thirty and ask Is this number divisable by 2 AND 3 AND 5? ..and of course you could say zero since 0/2 = 0/3 = 0/5 = 0 adam ==== 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 find 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? ==== >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 find 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 for i=0...n. But I'm not sure what you mean by based only on translation, rotation, and shearing, as those are all linear maps. I think what you want to do is write a general formula for your map with various unknown parameters, and then the constraints F(x_i) = y_i represent a set of equations in those parameters. The best situation is where the map is linear in the parameters (even though it's nonlinear in the x coordinates), in which case you have a good chance of being able to solve for the parameters, if there are enough of them. Or you can do the standard linear least-squares if you'd be satisfied with an approximation rather than a perfect fit. Robert Israel israel@math.ubc.ca Department 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 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) 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 ==== > 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) 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 ie d(t)=C*[e^(-(t-t0)/tau)]*cos(o*(t-t0) - phi), t>=t0 and t0 is starting time I hope this helps Goran ==== 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) klaus 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 ==== 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 the travelling > salesman to find the shortest possible route, it's just that there may be > separate closed graphs in this edge cloud and I just need all the edges 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/files/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 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 > Bonds Now 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 2 So, 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 finally 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 fly out. If a > pitcher is so good at devising a pitch that is popped up for a fly > out, > 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 first 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 galaxies Now physics tells me that a 95mph pitch is added onto the hit vector of a homerun hit. So, I wonder if all major league pitchers were to begin slow pitching such as an underarm lob, whether batters such as Bonds 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 ever advise his pitchers to begin lobbing the ball at the plate as in softball and here the aim is to force the batter to fly out or ground out. 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. hit pitches end up as foul balls. Going from that fact, then devise a pitch that has the greatest likelihood of fouling and where a fielder or the catcher or the infield can get to the ball and make the out. I suspect that if a physicist closely and carefully examines the physics of speed of pitch and a spin on the ball will render a flyout most often. And that this speed is not a fast pitch but rather instead a slow pitch. Question: I just wonder if a Barry Bonds (excuse any misspelling) were pitched the slowest strikes whether he can hit those as a homerun. It is certain that he can hit fastballs as homeruns for the past seasons proves that fact. If it is impossible for Bonds to hit a slow pitch with a spin on it as a homerun, then it is obviously clear what the future trend of baseball will be--- pitchers adopting slow pitching with some wicked spins on the ball. And a gradual phase out of fastball pitching. Another physics to use on pitching is that when a ball is pitched high and lofty such that it comes and falls down onto the homeplate and when such a ball is hit it is usually grounded and not airbound flight. I believe the physics explanation for that is that the hitter's bat general hits the ball not in the middle of the ball but the upper hemisphere of the ball is struck 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 comes falling down onto the homeplate as a strike pitch and if hit by the batter would generally be a ground ball to an infielder and a easy out at first or even a double play. I do not know if the rules of Baseball make it such that a pitcher must have a certain speed on the ball or whether the rules allow for a softball pitcher to come into the sport and pitch the slowest pitches possible. I feel that no physicist has really involved himself thoroughly in the sport of baseball and beginning to answer some of these questions that I have raised. And if someone serious would get involved, I expect many surprizes to issue forth from such a research. I believe the idea of fast ball pitching is more of a faith than a science in that almost everyone believes that fast ball pitching is superior to slow speed pitching. But no-one has proven that belief. I am testing that belief and saying it is faith and not science. Prove: prove that fastball pitching is superior in getting hitters out more than slow speed pitching. Additionally, spin on slow speed pitching must be included in the research. As I said in my original post that a team that wants to win the World Series must have 2 ace pitchers such as Beckett of the Florida Marlins winning the Series because if the first ace pitched game 1, and the second ace pitched game 2, then the first ace would come back and pitch 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 of it is filmed, that one can see that a team that won the series had 2 pitchers that were higher aces than the defeated team. Now if Florida had used Beckett in game 1, Pavano in game 2 and Beckett in game 3 4 games. But since ace pitchers are not perfect and since the hitters of the Marlins are not perfect that it took 6 games to defeat the Yankees. But the Yankees were totally inferior in pitching to the Marlins and due to that inferiority it was prone to defeat. The World Series in baseball is a tournament to uncover what team in baseball has the 2 most best ace pitchers. It is not about hitting. And that is the sick thing about movies on baseball for rarely do movies spend any time on the pitcher. But all movies on baseball should be concentrated on the pitcher. So, if anyone right now wants to figure out what team will win the World Series in 2004, it is easier than you think. All you have to do is scout and look at the teams for which team has the Two Best Ace pitchers. And if they do not get injured during the season, then those 2 ace pitchers should nail down the World Series for you. Archimedes Plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies ==== Whether a pitcher is hittable or not depends solely on whether the batter can predict when the ball will cross the plate and where the ball will be at that moment. Fastballs are difficult to hit because their inherent spread of velocities (typically 80-100 MPH) make it difficult for the batter to know when to swing -- the change up is a slow ball delivered with exactly the same motion as a fastball and its velocity (typically 60-80 MPH) tempts the batter to swing long before the ball reaches the plate. Curve balls are difficult to hit because it's very difficult for the batter to know how much the ball will curve -- i.e. how much below the point of the trajectory of a fastball of the same speed where it will cross the plate. The most unhittable pitchers are (a) knuckleball pitchers and (b) the so-called junk-ball pitchers. Knuckleball pitchers succeed because even they don't know where the ball will cross the plate; the batter is even more perplexed. Junk-ball pitchers simply throw such a variety of curving and swerving pitchers that the batter also can't predict what the next pitch will be. In general, it really doesn't matter what kind or speed of pitch is thrown if 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 pitcher who had a low earned run average, and he appeared in several All Star games. Ted Williams was the only person to hit a home run off of one of his blooper balls. -- Tom Potter http://tompotter.us ==== I have a 2nd order DDE y''(x) + ay'(x) + b(y(x) - y(x-a) = 0 with 2 bc's y(0) = 1 and y(x) = 0 when x->Inf I 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) = 0 I assume this should be y''(x) + ay'(x) + b(y(x) - y(x-a)) = 0 but 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 nax->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 -> infinity, you'll want to use a linear combination of the solutions for roots with negative real part. Robert Israel israel@math.ubc.ca Department 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 indicate one free library and one commercial library. I need to include this information 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 bigger than a fixed 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 defined and converges mod p AND as a normal series. i! evidently is far too big, but I wonder if the binomial coefficient (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?) -- Hauke Reddmann <:-EX8 For our chemistry workgroup,remove math from the address For 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 fixed 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 >defined and converges mod p AND as a normal series. >i! evidently is far too big, but I wonder if the >binomial coefficient (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 = floor(ln(i))!. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, 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 findings is one thing, and getting them >> noticed, is another. [JSH stuff snipped] >> What's in it for you? >> James Harris James, Again, the problem is not with the definition; 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's definitely not a ring. If one attempts to construct (gen[2])[9], to deal with this issue, one gets J. [*] I should also note that Z[1/2], according to James, is also too small, 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, let alone an arbitrary real number such as pi or e or Re(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.net It'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 findings 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 > finding of a problem with a math definition that's over a hundred > years old. In looking over various replies to my previous posts on > this subject, I've seen assertions that definitions 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 field *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 definition 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 definition is that they are *indivisible*, > as people can define things, and later *refine* their definitions. Now my research finding 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 + 1078 The roots of this polynomial are, approximately: r_1 = -0.02320596700418064467907344992 r_2 = 0.04221522840004950601300611271 - 0.03710343776564665666915421379*I r_3 = 0.04221522840004950601300611271 + 0.03710343776564665666915421379*I courtesy of GP/PARI. This equation is divisible by 7^2, yielding P(x) = 7^2 * (300125 * x^3 - 18375 * x^2 + 360 * x + 22) . 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 For 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 error during equation transcription. Fix. 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). Erm...what exactly are you doing? Did you mean that you could factor P(x) into P(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, and therefore we can write P(x) = 5^3 * 7^3 * (5 * a_1 * x + 7) * (5 * a_2 * x + 7) * (5 * a_3 * x + 7) where a_i = -7 / (5 * r_i) or a_1 = 60.32931098056739402658477747 a_2 = -18.71011003573824246783784328 - 16.44452558021931061668416939*I a_3 = -18.71011003573824246783784328 + 16.44452558021931061668416939*I It turns out these a's empirically satisfy the equation E(a) = 22*a^3 - 504*a^2 - 36015*a - 823543 = 0 and the gcd of the coefficients of this equation is 1. I'd have to do some work to prove this precisely, though, as GP/PARI in this case is doing the equivalent of a toy calculator's division of 1/3 = 0.33333333 ; it's not giving me a nice representation here but simply approximating it to the best of its ability. Unfortunately, it turns out the roots of this equation are in fact b_1 = 60.32931098056739402658477746 b_2 = -18.71011003573824246783784328 + 16.44452558021931061668416939*I b_3 = -18.71011003573824246783784328 - 16.44452558021931061668416939*I and b_1 - a_1 = -1.21169034 E-26, which is a very tiny error, but not identically 0. So I can't construe this as a mathematically iron-clad proof. (Basically, the last digit is a 6 instead of a 7.) Even if b_1 = a_1 to 53 bits of precision I couldn't construe it as such. But I can check the equation by deriving it using a different method. It's worth noting that, if we assume r_i are the roots of P(x) = 0, (we have already computed approximations to r_i above, but let's pretend that we haven't) where P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078 then 1/r_i are the roots of Q(x) = 0, where Q(x) = x^3 * P(1/x) = 1078 x^3 + 17640 x^2 - 900375 x + 14706125 and we can manufacture an equation with roots a_i = -7/(5*r_i), with this technique; if we set R(x) = 0, where R(x) = x^3 * P(-7/(5*x)) = 1078*x^3 - 24696*x^2 - 1764735*x - 40353607 Changing independent variables and dividing by 49, we get R(a)/49 = 22 * a^3 - 504 * a^2 - 36015 * a - 823543 which turns out to be equal to E(a) above. So GP/PARI wasn't too 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, as P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) as the cubic defining the a's with x=0 is a^3 - 3a^2, which has roots, 0, 0 and 3. Well, now you've completely lost *me*, for starters; are you solving P(x) = 0 or deriving a cubic in a using x as a parameter instead of a variable to be solved for? Ow, my brain hurts. However, you are correct in that P(0) = 7^2 * 22. These factors must be accounted for in some fashion -- although since the roots of P(x) = 0 aren't algebraic integers it's a bit like asking for the factors of 6 (2 and 3) to be accounted for in the equation 35 y^2 + 11y - 6 = 0 In 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 with P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) I know that dividing through by 49, it must go like P(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 P(x)/49 = (5a_1/7 + 1)(5a_2/7 + 1)(5a_3 + 7). Your logic appears flawed, as is easily shown by example. Consider the equation F(x) = x^3 - 49 = 0. This equation must have 3 algebraic integer roots. One can factor it F(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 in the final representation, and used -1 for simplicity; as one can easily see, d_1 = 1/(7 * c_1) d_2 = 1/(7 * c_2) d_3 = 1/c_3 for 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) and c_3 = 7^(2/3) * exp(4 * pi * i / 3). None of these are divisible by 7, in the sense that c_i / 7 yields an algebraic integer. (exp(i * theta) is a unit in the ring of algebraic integers, for any theta = 2 * pi * m / n, m, n > 0.) Therefore, at most one of the d's is 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 satisfies the equation 49 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 definition 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 field: 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 find 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 definition > problem? What's in it for you? Understanding. So far, you've not extended mine. :-) You'll need to fix a number of glaring logic errors, to start. > James Harris -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== he either can't, or won't, or doesn't get the question, or the definitions behind the question, or all or part of 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-up on questions to your own postings, immediately, but that applies to virtually all of your originations, eventually, in the ten-year programme to 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 infinite 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 find a counterexample in Object Ring Theory? that's the 64-centivo question for me, and if it's answered, maybe something will be in it for me, two! > What I'd like to explain is a disturbing and fascinating finding of a > problem with a math definition 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 field because every element except 0, has a > multiplicative inverse. > That's it. The definition 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 proof that 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 infinite 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 find 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 + 1078 that 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 treat 5 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 get P(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 you have substituted 5. > 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). > It is worth noting that the roots are all algebraic integers 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, as P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) as the cubic defining the a's with x=0 is a^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 to what you have done with P(x). You are not factoring P(x) into linear terms involving x: that is, your factors for P(x) are terms g1, g2, and g3, where, for example, g1 = 5*a1 + 7 where it is a1 which is a function of x. Moreover, a1 is not 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 coefficients 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 linear factors involving x. It looks like the old shell game. It looks like you are pulling the old switcheroo on the variable of factorization. To maintain the parallel between S(x) and P(x), you should not be using x: you should be using 5 ! > So from before with P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) I know that dividing through by 49, it must go like P(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 P(x)/49 = (5a_1/7 + 1)(5a_2/7 + 1)(5a_3 + 7). > No - first of all, a_1, a_2, and a_3 are dependent on x. Secondly, in order for P(x) to factor in the form you have given, the number 49 is split up in ways 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) and 7/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 fine, because a1(0) = a2(0) = 0. But for other values of x, there is no reason to think that f1(x) = f2(x) = 7 and f3(x) = 1. They are just three algebraic integer divisors of 49. Their values are dependent on x, and they do in fact 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 constant term of g1(x)/7 is 1. This is true no matter what 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 f1 is 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 constant term is 7/f1(x) !!! Right? To which I say: not at all. *By your own definition*, 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 thinking goes wrong. You are using constant term in two different 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 there on 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 divisors of 7 for all values of x does NOT lead to any contradiction or problem. Put it another way: 7^2 is factored differently into 3 pieces depending on the value of x. Those three 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 as you have consistently and correctly used it is h1(0). Since, however, because of its placement, you are thinking of 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 function of x whose values are algebraic integer divisors of 7 for all values of x does NOT lead to any contradiction or problem. Put it another way: 7^2 is factored differently into 3 pieces depending on the value of x. Those three pieces are algebraic integers. Assuming that they are not always equal to the values that they have when x = 0 does not lead to any contradiction of the independence of the constant term. Put yet another way: the constant term in g1 is not 7/f1(x) *unless* you assume f1(x) is itself constant, 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 is 7/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 your thinking goes wrong SOMEWHERE. You are whistling past graveyard when you persist in your belief that there is some error in core mathematics, or that there is a problem with the definition of algebraic integers. The far simpler (and correct) explanation is that your argument is wrong. We know exactly *where* it is wrong, but only you can reveal what you are actually thinking. And so far you have not completely done that. Until you do, we will not be able to resolve to your own satisfaction what your mistake is. But we know enough, I would say, to satisfy everyone else. > The problem now though is that conclusion can be used to show that > unequivocally beyond any reasonable doubt the definition 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 to have an airtight argument that a1 has 7 as a factor, which of course 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 definition problem. That is an out-and-out CONTRADICTION. Face up to it! Bite the bullet! And a contradiction like this can only 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. We have pointed out exactly where your error is: it is in saying that because a1(0) is divisible by 7, that the same must be true for a1(x) when x is not zero. You refuse to 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 fact that the constant terms, when x = 0, satisfy certain divisibility criteria tells you NOTHING about what happens when x is nonzero. Nor is factorization of overly simplistic REDUCIBLE quadratic polynomials a reliable guide to what happens with IRREDUCIBLE polynomials. Algebraic integers, as you yourself have pointed out, are really different from ordinary integers: they cannot be factored into little indivisible atoms called 'primes'; they are *infinitely 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 find 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 definition > 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 + 1078 that 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 a more complex expression. Now I used the more complex expression for MONTHS and Nora Baron along with others had a field day confusing people, and lying about the expression as I explained things more than once in posts replying to this poster and others. Basically they'd focus people on the x above and claim IT was the key variable, and refuse to back down, so I took it away. And now you can see the poster wants it back. For those on sci.physics and sci.skeptic who still don't understand what I'm dealing with, consider the attempts to ADD complexity and confusion, while I emphsize simplicity. James Harris ==== ... > For instance, with the polynomial > > P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078 Why would your technique work with your polynomial and fail with Q(x) = 6125 x^3 + 6125 x^2 - 6370 x + 1078 With this polynomial and your technique you get 1/7 in your ring. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: 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 defining 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 definition 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 your polynomial and incorrect with mine? If it is correct, set x = 1 above and find that when you include in your 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 about MUST. You assume that it MUST be true in the algebraic integers, that is wrong.) -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: 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 find mocking posts of interest. Now I've looked over some of your posts and can rather easily explain what you seem to find significant. 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 figure 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. James Harris ==== > I parafrase you and ahve a question at the end: > > You don't have my permission to parafrase and I don't find 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 find significant. 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 > figure 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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: 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 find mocking > > posts of interest. I was not aware that I needed permission to parafrase. Dik: FYI, this is one of those English words of Greek origin, where we use ph for the f sound, viz. paraphrase. It's only in rational languages like Spanish (and apparently yours) 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 find 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 Harris ==== > > I parafrase you and ahve a question at the end: > > > > You don't have my permission to parafrase and I don't find 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 your reasoning in my own words may lead to something different than what *you* mean. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: 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 find 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 Harris Take a bit of your own advice and cut the insults. I've yet to see a post from 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 find 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 find significant. 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 > figure 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. ************************ 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 > figure 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 story about somebody who had found a way to speed up transmission of video through the internet thousandfold. People were interested, but he died before he could reveal what he had done. Now that invention is lost for eternity... Suppose now that James dies before he could give his simple explanation? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: 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 see that the conclusion is false. Hence either there's something wrong with the reasoning (hint: this is the correct explanation) or there is some difference between the case he considered and the case you consider. If you don't want to look like you really can't explain your proof you need to say what the difference is, not state that he doesn't have permission to show that you're wrong. >and I don't find 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 find significant. Really? Then why _don't_ you? 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 >figure 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 MUST be 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 find mocking > posts of interest. Now I've looked over some of your posts and can rather easily explain > what you seem to find significant. 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 > figure 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 find mocking > posts of interest. Permission? > Now I've looked over some of your posts and can rather easily explain > what you seem to find significant. 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 > figure 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 of math (as currently defined) 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 find mocking > posts of interest. Now I've looked over some of your posts and can rather easily explain > what you seem to find significant. Can, perhaps. Will not, certainly. Gib ==== 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. Well adding and subtracting the same number gives you 0, of course, so he didn't do anything, but he acts like he did. 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. And, yes, I switched recently from using m, as it occurred to me that posters like Arturo Magidin might be oddly fixated on a variable name, when it's algebra, so they shouldn't be. 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. > 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 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. SURE, you may REMEMBER that you had x+2, but mathematically, it's just 2, 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 those would 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 nonconstant terms, ACCORDING TO YOUR DEFINITION. Your definition of constant term is evaluate at 0. That's what I did. If you'll remember from your little Advance Polynomial Factorization thing, 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 is h(x)-h(0). So, if h_1(m) = g_1(m)/w_1(m) = (5a_1(m))/w_1(m) + (7/w_1(m)), and a_1(0)=0, and w_1(0)=7, then h_1(0) = 5(a_1)0/7 + 7/7 = 1 and the nonconstant part of h_1(m) is h_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 the CONCLUSION you want, so you cannot claim it yet. We do NOT know that 7/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 definitions TO THE LETTER. And that's also what YOU DO. You argue about constant term and nonconstant term, but all you do is say that f(x) = (f(x)-f(0)) + f(0). So all you do is add and subtract the same number, and act like you did 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 definition, which has m as the variable you set to 0. If you changed to x's, then surely you can figure out what needs to be done. > And, yes, I switched recently from using m, as it >occurred to me that posters like Arturo Magidin might be oddly fixated >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)), and a_1(0)=0, and w_1(0)=7, then h_1(0) = 5(a_1)0/7 + 7/7 = 1 and the nonconstant part of h_1(x) is h_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), and 22/w_3(x). This follows because you CLAIM that the nonconstant term is 5a_1(x)/w_1(x), and since the function is (nonconstant)+(constant), and the function is equal to 5a_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 NOT also 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 the constant term does not change as x changes, it is NOT true that every summand that does not change as x changes has to be in the constant term. You had that in g_3(x) = 5a_3(x)+7. Since a_3(0)=3, what you get is that g_3(0) = 22, so g_3(x) = (5a_3(x) - 15) + 22 and the -15, which does not change as x changes, is NOT part of the constant term. That's why you can have a -1 and a -22 in the nonconstant terms above. >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 say that the independent terms vary with x; I said that not everythign which 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 definition, that for a function h(x), the constant term 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 big deal when you noted simply that 5a_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 figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite 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 Magidin magidin@math.berkeley.edu ==== The math I can use is VERY simple, thankfully, as mathematicians are > fighting 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 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. Mathematicians are *desperate* here and the truth is not as important to them as keeping up their fantasy world, which is why they have to keep replying to my posts. You see, they MUST keep replying to me because otherwise they're afraid people will actually trace out the argument and see I'm right. They *have* to keep replying to influence 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? - Randy Huh? These posters aren't even making sense. They're not *rational* as rather than trying to figure it all out, they've made up their minds to preserve their status quo by denying reason. If you look at my posts explaining the problem, you'll notice rather basic 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 the truth. 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 for a1, 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 for example. They've shown you expressions for the a's. They don't behave the way you want. Those are actual math arguments. They contain actual equations, actual expressions. Those actual expressions can be substituted into your actual polynomial to show with actual math arguments that they are a correct factorization of your actual mathematical polynomial. 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 the a1, a2, and a3 are incorrect? I haven't seen you say yea or nay to those expressions. If they are the correct expressions, then in what sense do 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 + ac And 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 Harris Please 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 + 1708 can, indeed be factored as P(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 specific cases can be easily confirmed: x=0, P(0) = 1078 > x=1, P(1) = 13982468 > x=2, P(2) = 114399858 which 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...+1.77596...I a_3 = > (1/10)(-14-7^(2/3) > 22^(1/3)(1+3^(1/2)I)) = -2.42535...-1.77596...I None 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 identifies, acknowledges and corrects his errors. > One with it ignores, denies or defends them. -- Democracy: The triumph > of popularity over principle. -- http://www.crbond.com 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 define 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 define 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. ==== [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 define 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 define > 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 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 define 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. -- There are two things you must never attempt to prove: the unprovable -- and the 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 define 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 define 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 define 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) ==== > 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 define 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 define 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 define > 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 -- defining specific values of a variable at specific values of its argument -- I simply meant that I was responding to James' post. There is no evidence that he intended to define the values of a_1, a_2 and a_3 at specific 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 that exactly two of the 'a's go to zero when x=0. But that result does not follow from the given equations. If you want to make a case that at x=0 we can discard the values of a_1, a_2 and a_3 and replace them by defining those values to be 0, 0 and 3, by all means go ahead. You can rest assured that P(x) will equal Q(x) as required. 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_3 are at x=0 from the posted equations, you might be better off just to go ahead and evaluate them at x=0. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- 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 defining 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! -- A man of integrity identifies, acknowledges and corrects his errors. One without 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 defining 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. 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). 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 verified. 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 definition of the algebraic integers is really a reflection 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) ==== > [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 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. 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 verified. 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, and w3 of 49 such that a1/w1, a2/w2, a3/w3, 7/w1, 7/w2, and 7/w3 are all 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, 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}) No - when x = 0, P(0)/49 = 3*5 + 7 = 22. JSH is basically right about this part. > What James perceives as a problem with the definition of the algebraic integers is really a reflection 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 gone very 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 defining 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) as given, and submitted them to a symbolic algebra package to determine the expressions for a_1, a_2 and a_3. Since it (Mathematica) produced those values without complaint, and I confirmed the equivalence to my satisfaction, I posted them. > 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 have certain beliefs about these forms or that these forms are necessary. James posted 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 verified. 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 definition of the algebraic integers is really a reflection 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 the obvious. -- 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) + 7^3 Equating (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 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. > 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 verified. 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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: 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, a1 divisible by p.q, a2 divisible by p.r, and a3 divisible by q.r (or some other combination, of course). IN addition, one can show that a1 will be coprime to r, a2 coprime to q, and a3 coprime to p. In the irreducible case, one can further show that none of p,q,r are units; in the reducible case, however, one can show that at least one, and possibly two of them will be units, but the 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 figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite 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 Magidin magidin@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 verified. 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 is irreducible, none of p, q, or r are units; when it is reducible, it is possible for one or two of them to be units (depending on how the polynomial 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 figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite 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 Magidin magidin@math.berkeley.edu ==== 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 verified. In fact, IIRC, Nora > > already showed this.) > Recast my first 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 definition of the algebraic integers is really a reflection 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. 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 basic mistake, which I thought Rick Decker caught which is that the poster assumes that a_1, a_2 and a_3 are equal, when, in general, they are not. > > 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 verified. In fact, IIRC, Nora > > already showed this.) > Recast my first 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 definition of the algebraic > integers is really a reflection 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 reflects 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 way down. P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078 which has a constant term that is 1078. Well P(x) can also be written out as 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 roots of a^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 are all 7, which can't be right, as the constant term of P(x) is 1078, so setting x=0, reveals P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) as the cubic defining the a's with x=0 is a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1 and a_2 for 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 have P(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 + 1078 has 49 as a factor, so I can divide by 49, and dividing 1078 by 49 gives me 22, as the new constant term. Well that means that P(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 the difference. This is currently done using a correlation coefficient in a 2 dimensional array with the original values transformed to z scores. Each of the values has a different degree of explanation within the model - both globally and geographically. Beta coefficients have been used to adjust the values. The sum of the z score is then used to pull out similar records. This works to a point. What I need is a better way of comparing the values and pulling out similar records. Any ideas or suggestions would be very welcome? Ian. X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft ==== at 05:06 PM, Steven Margolin said: >What is the most succinct axiomization of math. Well, first of all Mathematics covers a multitude of sins. The customary way is to start with a logical system and axioms of set theory, then define various structures in terms of sets, with the properties in the definitions serving as axioms for those systems. Keep in mind that there are a lot of topics that have nothing to do with Geometry or numbers. >Do the Peano Axioms do it? That depends on what is is. Peano's postulates in a proper logic allow you to model the theorems of Mathematics, so in a sense they are adequate, 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; the geometric axioms are enough by themselves. They're also enough to model 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 model of Hyperbolic Plane Geometry uses specific arcs in the Euclidean Plane to represent lines. The points and permissible arcs satisfy all the axioms 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 triangle The measurements in the model are not the standard Euclidean measurements. They are defined in such a way as to make their properties 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 JOAT not reply to spamtrap@library.lspace.org ==== > >> >> [...] >> >> Out of curiosity, do you think that Dr. Ullrich's statement was >> complimentary? >complimentary. Classic misrepresentation. Locate the first 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 David Ullrich: (Ok, although it seems a little tacky: I'm a professor here at OSU. I got a PhD at Wisconsin (with Walter Rudin). Then I got a job here, a few years later I got tenure, and a few years ago I got promoted to full professor. The committee suggested I apply for the promotion - I hadn't meant to. _They_ must think I have _some_ idea what math is about or they wouldn't pay me. Mitchell Jones: They may have thought that in the past. Perhaps they bought into your BS. Or perhaps Oklahoma State is a fourth rate, backwater school of the sort that attracts academic fakers. I couldn't say. I can say this, though: it is not in their interest, or in any academic institution's interests, to employ professors who verbally abuse students. Asking questions and stating arguments is an essential part of the learning process, and professors who habitually respond to questions and arguments with the kind of sneering condescension you have exhibited in this newsgroup are the academic kiss of death for the institution that employs them. Undergraduate students hate them and try to avoid their courses, and the best graduates flee to other institutions to do their postgraduate work, or else become so disgusted with academic life that they turn to careers in private industry. The result is a brain drain that destroys academic 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 definition of > compl[i/e]ment. This is not helpful. When HH was flipped, at least one is a head was true and when TT was > flipped, at least one is a tail was true. Those are complements. > When HT, or TH was flipped. 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-defined 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 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. Situation a: A coin was flipped and it landed H. (We didn't see the flip, we read about it in the statement. We would not expect a demur had it landed T) Situation b: A coin was flipped and it landed T. (We would not expect a demur had it landed H) Situation c: Two coins were flipped and they landed HH. (We should not expect a demur at TT) Situation d: Two coins were flipped, they landed HH, the statement was generated, Two coins were flipped and at least one is a head. What is the probability that both are heads? (We should not expect a demur at TT) Situation e: Two coins were flipped, they landed TT, the statement was generated, Two coins were flipped and at least one is a tail. What is the probability that both are tails? (This question should have the same answer as the one in Situation d) As a is related to b, so is b related to c. Maybe it isn't complementary. When a coin is flipped two things can happen. a and b are them. When two coins are flipped, four things can happen, c and d are two of them. There are two peas in the first pod, four in the second 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 fill 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 filled. 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 the answerer of the question answer the question as stated, or what _we_ meant? If it's a test question and the student knows what the prof meant, and how it will be graded, then that's another discussion. A correct answer to a question is not according to what we meant, but according 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 filled by is or are) and we can't fill in these > blanks, prior to some kind of inspection, or 'look'. > i)Two coins were flipped and at least one is a head. That statement was made. It can all be said, prior to the look, except _head_. i')Two coins were flipped and at least one is a tail. Prior to the look, we can always say that one, maybe both of those statements is true. We can make neither, prior to a look, and they couldn't either. When one is made, the question is, what is the probability that the other statement is also true? Prior to the look the chances for both to be true was 1/2. Now that there has been a look and there is evidence of a head, does the chance for there also being a tail jump to 2/3? Think about it. > The first 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 reflip as 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. > 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 first/second is a head/tail, except that the > first/tail combination makes a contradiction. No contradiction if you make the statement, the first 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 flipped and at least one is a tail. What are the chances for two heads? I thought this was the question you said didn't make sense, and I agree, but, Two coins were flipped and at least one is a tail. What are the chances for two tails? makes sense to me. Two coins were flipped and at least one is a head. What are the chances for two heads? and Two coins were flipped and at least one is a tail. What are the chances for two tails? These two questions get the same argument, all over the world. They are the same question, or somewhat related, I'm not sure what we call the relationship. The ensuing 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 we run 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's comfortable, so long as we are in the majority. Then we run into dumb fuck Moritz, who won't shut up. The problem is, he's correct. If the majority ever agrees with him, then we have to change, or we become the dumb fucks. When we change from at least one to until at least one we have to alter the coin flip sequence. That's why we can't make a working model, and Moritz can. He's flipping two coins four ways. We're flipping two coins three ways. It's harder, and takes a little more explaining. We can't flip two coins, then look, then change it to one of three. We have to pick a null, prior to the look. That's what I call prejudice, or prior prejudice. Two coins were flipped, they landed HT and the heads statement was made. Heads were chosen, for the answer to be 1/3, they must have been pre chosen, chosen prior to the look. > * 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/ ==== 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 it isn't easily confused with the usual meaning of 'complement'. 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 fill 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 filled. 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 I mean, 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 answer the question as understood. Sometimes, inevitably, there is misunderstanding. Clarification helps to eliminate this. In this particular case, you argue that more or less everyone except yourself has failed to understand the words, even though we have in fact understood exactly what the questioner meant. So your quest is not a mathematical one, it is (as I've said several times) one of hermeneutics, a branch of human endeavour which I think particular devoid of value. The other problem with your quest is that as far as one can see it is for ever going to be restricted to the providing of the Correct Answer to the One Question. You seem to have no notion that in any sense it's generalisable. Suppose you succeed? Students learn that the Correct Answer to the Eldon question is 50%, because part of the question _might_ have been different. They obviously ask: what of playing cards? Two decks of cards are cut and* at least one is black; what is the probability both are? Two decks of cards are cut and* at least one is a spade; what is the probability both are? Two decks of cards are cut and* at least one is a queen; what is the probability both are? Two decks of cards are cut and* at least one is a court card; what is the 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 filled by is or are) and we can't fill in these > blanks, prior to some kind of inspection, or 'look'. i)Two coins were flipped 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? 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 to make either the statement At least one is a head, what are the chances both are? OR At most one is a head, what are the chances neither is? I then look. I then make the only statement that is true, or if both are true, I will make the first statement if the day of the month is divisible by three, the second if it's not. The point is that I then make this Statement to which you claim to have 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 on specious 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 a strange country, and the statement is made that two coins were tossed, and coins were tossed and at least one is a head; what is the probability both are? Who knows? Perhaps the custom in this country is that the answers to all questions are Yes. That's also why the Marilyn von Goat question has to be very carefully phrased if you want the Marilyn answer to be clearly correct. Brian Chandler ---------------- geo://Sano.Japan.Planet_3 Jigsaw 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 first 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. Eldon Moritz ==== > [...] Out of curiosity, do you think that Dr. Ullrich's statement was > complimentary? >complimentary. >> >> Classic misrepresentation. Locate the first 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 complaining that you don't seem to read more than a few key words in messages before replying to them are resting _their_ cases at this point as well... >Eldon Moritz ************************ David C. Ullrich ==== 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) + 7^3 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 first. 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. 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) + 7^3 Wrong! Did you ever actually expand that version of P(x) to see if it matched your original equation? Evidently not. Your original equation was: P(x) = 14706125x^3 - 900375x^2 + 17640x + 1078 Your factored representation expands to: P(x) = 14706125x^3 -900375x^2 - 17640x + 1078 Notice anything funny? The very 1st factorization in your argument has a sign error -- and one that you have neither acknowledged or corrected from the beginning. Furthermore, it invalidates all your subsequent calculations. (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 first. > > 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 7 is a unit in the object ring? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== that's an interesting question. perhaps James Harris could rephrase it for a mass Usenet reader audience, a.k.a. this local ring of 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) + 7^3 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 figured 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. > > 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 first. > > 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 field. > > 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. 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 fighting 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 figured > 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 factorisation was such that a1, a2 and a3 were algebraic integers. In *my* polynomial I use also a factorisation that keeps you in the algebraic integers. So *where* is the difference? (Note that also in my case the polynomial for the a's is monic with integer cofficients.) You get your monic the way 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. I do the same. So *what* is the difference. (And what part of your argument 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 first. > > > > 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 field. 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^2 and 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 variant factorisations of your polynomial. And I did that factorisation in a way 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 fighting to hide it from view. So you pull a factorisation out of your hat, claims various things about it, without ever specifying why what you claim is correct for your factorisation 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, +31205924131 home: 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 figured > it out. This exhibits a striking lack of awareness of what is mathematically interesting. How someone happened to figure something out is perhaps psychologically interesting, socially interesting, and pedagogically interesting. But it is not mathematically interesting. What is mathematically interesting is the theorems, the proofs, the definitions, and also the ways that those things open the ground for more mathematics. How a person happens to discover a particularly beautiful theorem or a nifty proof is not, actually, mathematically interesting. Thomas ==== I'm taeching the beginning epsilon-delta course. Looking ahead in the book (Ross, Elementary Analysis: the Theory of Calculus) I see the following definition: 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 first I thought this must be a typo. But it turns out he means it - later when he shows that this definition is equivalent to the one in terms of epsilon and delta the condition is |x - a| < delta, not 0 < |x-a| < delta. I'm shocked. Is this version of the definition actually standard in some circles? My impression is that the definition 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 definition different from what's in the book unless I have a very good reason, but if this definition 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 ==== > I'm taeching the beginning epsilon-delta course. Looking > ahead in the book (Ross, Elementary Analysis: the Theory > of Calculus) I see the following definition: 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 first I thought this must be a typo. But it turns out > he means it - later when he shows that this definition > 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 A excludes the point a, then you could have your 0 < |x - a| by definition 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 definition: 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 definition 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 first defining the limit of a sequence. At first I thought this must be a typo. But it turns out > he means it - later when he shows that this definition > 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 definition to use for proving that a limit does exist. I'm shocked. Is this version of the definition actually > standard in some circles? My impression is that the > definition 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 > definition different from what's in the book unless > I have a very good reason, but if this definition 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 definitions 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 definition: >> 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 definition 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 first defining the limit of a sequence. > At first I thought this must be a typo. But it turns out >> he means it - later when he shows that this definition >> 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 definition to use for proving that a limit does exist. > I'm shocked. Is this version of the definition actually >> standard in some circles? My impression is that the >> definition 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 >> definition different from what's in the book unless >> I have a very good reason, but if this definition 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 definitions and proving their equivalence? Edgar has pointed out that I didn't read the section carefully enough - the right definition comes later. The two definitions I was talking about are _not_ equivalent. The problem is not sequences versus epsilons, the problem is whether f(a) should be relevant: Say f(x) = 0 for x <> 0, f(0) = 1. Then lim_{x->0} f(x) should be 0, but by one of the definitions above the limit does not exist. (At least that's what I thought the problem was - in fact I missed a tiny bit of notation, the wrong definition above is a definition of something other than lim_{x->a} f(x).) ************************ David C. Ullrich ==== > Edgar has pointed out that I didn't read the section carefully > enough - the right definition comes later. > Eheheh... remember my question some months ago about the same topic ;) ? Ok basically those are the same definitions; over here when we want to refer to your standard definition 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 definition: 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 Definition 20.1 with a little S in there. Followed by Definition 20.3 where he uses it to get 4 of the usual definitions. But mine is only the 1980 edition. Has it been loused up since then? At first I thought this must be a typo. But it turns out > he means it - later when he shows that this definition > 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 definition actually > standard in some circles? My impression is that the > definition 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 > definition different from what's in the book unless > I have a very good reason, but if this definition 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 definition: >> >> 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 Definition 20.1 >with a little S in there. Followed by Definition 20.3 >where he uses it to get 4 of the usual definitions. >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 taught the course...) - missed the little S. Sure enough in 20.3(a) he specifies S = J{a} and everything's fine. >> At first I thought this must be a typo. But it turns out >> he means it - later when he shows that this definition >> 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 actually look at the whole section before jumoing to conclusions. ************************ David C. Ullrich