mm-3699 === Subject: Re: Antidiagonal, Infinity About Euclid's postulates there, they define Euclidean geometry. Disagreeing with the parallel postulate, number five, allows non-Euclidean geometry. About the compass and rule to trisect an angle, for example to form an angle of pi/3 radians from lines intesecting a point on a straight line, an angle of pi radians, it's easy to bisect an angle with the compass, and 1/3 is the sum of 1 / 22x for each positive integer x, eg .010101(01).... A theoretical rule and compass bijection that takes infinitesimal time to bisect an angle may well trisect an angle in some finite time. When you cross the room, you get all the way there. You can begin to sample a random real number from the unit interval by flipping a coin a given number of times. What you get is a discretized sample. About y/x, say that all that is known is that y is greater than x. Is y thus a dependent variable of x? All that is known is that y > x. The value of y/x as x goes to infinity is indefinite. It might be a finite value, it might diverge. Are x and y thus interdependent? About the sets of the hyperreals and the reals, somebody else says that each set contains the same elements, that is x E R <-> x E *R. The square root of two: the length of the diagonal of the unit square, and the length of the side of the square with diagonal of length two. Two: the only integer whose sum with itself is its product with itself. Then again multiplication is just the sum of a mutiplicand with itself as many times as the other. About the itty bitty line segments, infinitesimal line segments might be comprised of a pair of adjacent points. Yet, that's not acceptible. About the representation of a continuous function as a signal, Fourier might have much to tell us about it. The real number is a point! Identify a real number. Point to it on the real number line. Where are the negative integers in the ordinals? 2 - 4 = -2. Chapman, if you're going to go outside put some pants on. I don't get that function that's continuous at all irrationals. It doesn't have an inverse, it's partial. Ross === Subject: Re: Antidiagonal, Infinity >About Euclid's postulates there, they define Euclidean geometry. Actually, they don't. >About the compass and rule to trisect an angle, for example to form >an angle of pi/3 radians The issue isn't whether there is an angle that can be trisected; the issue is whether you can trisect an arbitrary angle. >1/3 is the sum of 1 / 22x for each positive integer x, There is no such sum. There is a limit of sums, but you can't take a limit with compass and straightedge. >A theoretical rule and compass bijection that >takes infinitesimal time A meaningless term. Mathematics is not about time. Nor is it about Zeno. >You can begin to sample a random real number You still haven't defined what you mean by a random real number. >About y/x, say that all that is known is that y is greater than x. What sorts of things are x and y? >Is y thus a dependent variable of x? What does that question mean? >The value of y/x as x goes to infinity is indefinite. What does that sentence mean? >About the sets of the hyperreals and the reals, somebody else says >that each set contains the same elements, You said that, and it's false. >Then again multiplication is just the sum of a >mutiplicand with itself as many times as the other. Pi*Pi >infinitesimal line segments Please define. >might be comprised of a pair of adjacent points. There are non. And line segments in Euclidean spaces contain more than two points. >About the representation of a continuous function as a signal, This is sci.math, not sci.ee; we don't do signals. Besides, even an EE would consider that to be nonsense. A signal is not the same thing as a sine wave. >Fourier might have much to tell us about it. Yes, he would tell you about approximating continuous functions as sums of trigonometric functions. >Point to it on the real number line. Pointing may be a valuable skill in kindergarten; it has nothing to do with Mathematics. >Where are the negative integers in the ordinals? Where are the orange groves in the Bessemer converter? There are none. >I don't get that function that's continuous at all irrationals. >It doesn't have an inverse, Why is it relevant whether it has an inverse? >it's partial. Where is it undefined? -- spamtrap@library.lspace.org === Subject: Re: Antidiagonal, Infinity > About Euclid's postulates there, they define Euclidean geometry. > Disagreeing with the parallel postulate, number five, allows > non-Euclidean geometry. Actually, Euclid's axioms have been shown insufficient. > About the compass and rule to trisect an angle, for example to form an > angle of pi/3 radians from lines intesecting a point on a straight > line, an angle of pi radians, it's easy to bisect an angle with the > compass, and 1/3 is the sum of 1 / 22x for each positive integer x, > eg .010101(01).... A theoretical rule and compass bijection that > takes infinitesimal time to bisect an angle may well trisect an angle > in some finite time. When you cross the room, you get all the way > there. The trisection problem specifies certain rules to be followed in any attempt. Trying to do it by breaking those rules is disallowed. The remainder of Ross' post was to incoherent to comment on. === Subject: Re: Antidiagonal, Infinity <3f82e66f$1$fuzhry+tra$mr2ice@news.patriot.net> <3f85c273$18$fuzhry+tra$mr2ice@news.patriot.net> <3f89f684$43$fuzhry+tra$mr2ice@news.patriot.net> linux) >> About Euclid's postulates there, they define Euclidean geometry. >> Disagreeing with the parallel postulate, number five, allows >> non-Euclidean geometry. > Actually, Euclid's axioms have been shown insufficient. In what sense? -- Sale or rental of this disc is ILLEGAL. If you have rented or purchased this disc, please call the MPAA at 1-800-NO-COPYS. -- The MPAA begins a new anti-piracy program, found on a DVD purchased in China === Subject: Re: Antidiagonal, Infinity @news.patriot.net> <87oewk6kn5.fsf@phiwumbda.org> at 02:58 PM, jesse@phiwumbda.org (Jesse F. Hughes) said: >In what sense? In the sense that some of the statements Euclid purported to prove from the listed axioms and postulates don't follow from them. See, e.g., Foundations of Geometry[1] for an explanation. Google for, e.g;., Axiom of Archimedes. [1] You may see me jokingly refer to it as groundlaager, but it was a seminal book by one of the major Mathematicians of the[2] century, and still quite relevant today. [2] Whether you count him as 19th or 20th. -- spamtrap@library.lspace.org === Subject: Re: Antidiagonal, Infinity >> About Euclid's postulates there, they define Euclidean geometry. >> Disagreeing with the parallel postulate, number five, allows >> non-Euclidean geometry. > Actually, Euclid's axioms have been shown insufficient. > In what sense? See Hilbert's work on the subject. === Subject: Re: Antidiagonal, Infinity >> About Euclid's postulates there, they define Euclidean geometry. >> Disagreeing with the parallel postulate, number five, allows >> non-Euclidean geometry. > Actually, Euclid's axioms have been shown insufficient. > In what sense? See http://www.beva.org/math323/asgn7/dec12.htm or do a Google search for Euclid and Hilbert at === Subject: Re: Antidiagonal, Infinity <3f82e66f$1$fuzhry+tra$mr2ice@news.patriot.net> <3f85c273$18$fuzhry+tra$mr2ice@news.patriot.net> <3f89f684$43$fuzhry+tra$mr2ice@news.patriot.net> <87oewk6kn5.fsf@phiwumbda.org> linux) >> Actually, Euclid's axioms have been shown insufficient. >> In what sense? > See > http://www.beva.org/math323/asgn7/dec12.htm > or do a Google search for Euclid and Hilbert at -- But remember, as long as one human being follows the rules of mathematics, then mathematics as a human discipline survives. Right now I'm that one human being, so mathematics survives. -- James S. Harris === Subject: Re: Antidiagonal, Infinity >> Actually, Euclid's axioms have been shown insufficient. >In what sense? Presumably, in the sense that they don't model what they are (pre-mathematically) intended to model: without some sort of axiom(s) of completeness in the style of Hilbert, various points that Euclid purportedly constructs just may not *be* there! (Or so I've been told.) Lee Rudolph === Subject: Re: Antidiagonal, Infinity > Actually, Euclid's axioms have been shown insufficient. >>In what sense? > Presumably, in the sense that they don't model what they are > (pre-mathematically) intended to model: without some sort of > axiom(s) of completeness in the style of Hilbert, various > points that Euclid purportedly constructs just may not *be* > there! (Or so I've been told.) I have only a passing interest, not a deep one, here, but I'd still like to have a reference. I think I understand what you're talking about, but, nonetheless, a concise and well-presented explanation would be wonderful. -- s Harris, render of worlds === Subject: Re: Antidiagonal, Infinity >> Actually, Euclid's axioms have been shown insufficient. >In what sense? >> Presumably, in the sense that they don't model what they are >> (pre-mathematically) intended to model: without some sort of >> axiom(s) of completeness in the style of Hilbert, various >> points that Euclid purportedly constructs just may not *be* >> there! (Or so I've been told.) >I have only a passing interest, not a deep one, here, but I'd still >like to have a reference. >I think I understand what you're talking about, but, nonetheless, a >concise and well-presented explanation would be wonderful. I don't have a reference either, but it's a well-known fact that Euclid's axioms were not complete. === Subject: Star Gate Topology of the Universe Jack, You say, In particular I do not as yet see any physical necessity for Saul-Paul's Ansatz that physical 3D expanding accelerating cosmic space is actually SU(2)/ID rather than the SU(2) shown in Fig 3... The idea that 3D cosmological space SU(2)/ID rather than SU(2) is the view of Luminet et al. in their *Nature* (9 October) paper. They say (p. 593-594): The Poincare dodecahedral space is a dodecahedral block of space with opposite faces abstractly glued together, so objects passing out of the dodecahedron across any face return from the opposite face. Light travels across the faces in the same way, so if we sit inside the dodecahedron and look outward across a face, our line of sight re-enters the dodecahedron from the opposite face. We have the ILLUSION [emphasis added] of looking into an adjacent copy of the dodecahedron. Part of the problem here is what is meant by illusion physically? On the one hand we have very pretty topological formal algorithms in which, for example, a cylinder is topologically equivalent to a flat rectangle with periodic boundary conditions on one pair of opposite edges. One can generalize this to multiply-connected 2D surfaces with wormhole handles or toroidal surfaces and take it to 3D etc. However, my point is that physically the Universe could be either, in the simplest toy model, in 2D like the curved cylinder or like the flat rectangle with a pair of Star Gate edges of the world in which material objects are instantly teleported across space. Formally, the two situations are equivalent or indistinguishable from the POV of the topology, but the local metrical physics is more than the pre-metrical global topology and the actual situation is empirical and cannot be decided apriori on purely esthetic grounds. Note that a simply-connected ruled 2D surface like the cylinder has zero intrinsic Gaussian curvature as I recall. What about the doubly-connected 2D torus? A Flatlander insect crawling on either the 2D cylinder or torus without boundary is equivalent to the insect being instantly teleported from one edge to the opposite edge across a distance c/H(t) when we make the conceptual cuts to a flat rectangle. So the issue is whether there is any physical way to distinguish the two ideas or are they absolutely degenerate or indistinguishable by any imaginable physical measurements such as the presence or absence of a local tuv(zpf) =/= 0 exotic vacuum tensor field? Now as far as the WMAP data is concerned this distinction between rubber expanding accelerated versions of 1. we are literally trapped like Flatlanders inside a closed multiply connected 3D space without boundary but non-trivial Betti number with no sharply detectable 2D Star Gate walls vs 2. we are literally trapped like Flatlanders inside a 3D spherical dodecahedron with 12 pentagonal edged Star Gate walls or faces on the expanding scale c/H(t) with essentially instant teleportation to the opposite face through a traversable wormhole that is short compared to c/H(t). 2 requires the unified exotic macro-quantum vacuum dark energy/matter local zero point stress-energy density Diff(4) tensor field tuv(vac) = (c4/8piG*)/zpfguv not to vanish at the faces in such a way as to support and maintain this huge network of traversable cosmic scale wormholes that implement the inferred global topology from the ~ 2.7 deg K CMB temperature fluctuation spherical harmonic multipole resolved statistics of WMAP. However, if we are limited only to remote-sensing of this global topology as in the WMAP space probe the above distinction is probably moot or degenerate. We will only be able to decide this perhaps by sending out warp drive space probes to see if the world has an edge or not. The local /zpf zero point dark energy/matter exotic vacuum c-number ODLRO geometrodynamic kernel of tuv(vac) also permits the implementation of globally faster-than-light Alcubierre warp drives or, equivalently, Bondi-Terletskii negative matter propulsion that is a locally self-generated weightless free float slower-than-light timelike geodesic with small curvature tidal forces inside the ship, but which appears as spacelike world line trajectory to the outside Earth bound observer using remote sensors on the craft. Since Omega(dark energy) ~ 0.666 (:-)) to 0.73 depending on which Pundit you believe, this is not such a nutty idea as it might first appear to the more faint-hearted conservative reader. :-) *Note the above kind of hypothetical warp drive technology is reported by military intelligence connected UFO investigators that IMHO pass the crackpot filter. Whether or not those reports are true is not the gedankenexperiment what is significant is that we can conceive of such a technology within mainstream physics. Kip Thorne used this method in his 1986 paper on traversable wormholes. One way to decide between 1 and 2 is to look for a time-lag in what John emphasis) but in different regions of the sky p. 567 and what Luminet et-al describe as temperature correlations in matching circles on the sky. Is there a physical distinction between a curved multiply-connected 3D space with traversable wormholes and no boundary and a flat space with Star Gate boundaries that are the 12 pentagonal faces of the spherical dodecahedron http://mathworld.wolfram.com/Dodecahedron.html ? In principle yes, since spatial curvature is a local observable and the proper distance through the alternate worm hole paths connecting opposite faces is a variable physical parameter. When the topologist has the bug or crawling insect exit on the right edge and enter on the left edge, this is pre-metrical with no concept of metric time, hence it is an incomplete physical description leaving out measurable properties of the phenomenon. Since the detailed paper on pp 593 - 595 is not available to many readers on the bcc list, let's look at some excerpts from Dodecahedral space-topology as an explanation for weak wide-angle temperature correlations in the cosmic microwave background: The current 'standard model' of cosmology posits an infinite flat universe forever expanding under the pressure of dark energy. This follows from Einstein's classical principle of equivalence combined with Heisenberg's quantum principle of uncertainty to deduce w = -1 for the equation of state for the zero point vacuum fluctuations of ANY local micro-quantum field of any spin where w = pressure/energy density. Dark energy is w = -1 exotic vacuum with a negative zero point pressure because, the effective gravitation from a region of any stuff, real or exotic vacuum, in the weak field limit obeys a Poisson equation Laplacian of the stuff ~ (G/c2)(energy density + 3 pressure) = (G/c2)(energy density)(1 + 3w) This universally anti-gravitates causing the expansion of 3D co-moving 3D space in the FRW metric to accelerate when the net residual micro-quantum zero point energy density from all spins is positive. The gravitating dark matter exotic vacuum has positive zero point pressure with negative zero point energy density. IMHO dark matter is not made net residual zero point energy density of the physical vacuum is its local macro-quantum coherent order parameter PSI(x,L) = |Higgs field(x,L)|eiGoldstone phase(x,L) at scale L, coarse-grained space-time event x in the sense of a wavelet transform generalized version of the Wigner phase space density with ODLRO in the virtual electron-positron pair reduced micro-quantum density matrix. /zpf(x,L) = Lp*-2[1 - Lp*3|Higgs field(x,L)|2] Lp*2 = Lp4/3(c/H(t))2/3 H(now) ~ 1028 cm Lp2 = hG(Newton)/c3 Lp*(now) ~ 1 fermi, i.e. 1 Gev energy scale The ordinary non-gravitating vacuum has /zpf(x,L) = 0. Einstein's gravity is from the modulation of the Goldstone phase guv(x,L) = Minkowski metric + (1/2)(Sakharov's metric elasticity tensor of Hagen Kleinert's world crystal lattice) = nuv(Minkowski) + (1/2)[du(x,L),v + dv(x,L),u] ,u is partial derivative with respect to xu du(x,L) is the world crystal local distortion field at scale L du(x,L) = Lp*2(Spin 1 gauge-force invariant Bohm-Aharonov Goldstone phase),u Where also [DuDu + V(|Higgs field(x,L)|)]|Higgs field(x,L)|eGoldstone Phase(x,L) = 0 Is the local nonlinear Diff(4)+ spin 1 gauge invariant Landau-Ginzburg equation for the macro-quantum coherent vacuum More is different local order parameter that damps down the random zero point energy density contribution to the Einstein cosmological constant / in the large-scale limit. Einstein's Riemann 4th rank tensor curvature is from disclination defect density of string topological defects of the Goldstone phase in the Kleinert world crystal lattice with unit cells of scale Lp* using the t'Hooft-Susskind world hologram conjecture. This can be generalized to include torsion fields from Suv(x,L) = (1/2)[du(x,L),v - dv(x,L),u] =/= 0 corresponding to dislocation defects in the world crystal lattice as shown by Hagen Kleinert (Free University of Berlin). Note that Lp* depends on the cosmological FRW H(t) = R(t),t/R(t). Therefore, it starts out as 10-33 cm at the Big Bang micro -> MACRO quantum vacuum phase transition, but gets larger making the energy scale for quantum gravity lower as the universe 3D co-moving space in the FRW limit expands. This has falsifiable consequences. On the other hand, my basic theory does not require the additional world hologram conjecture of t'Hooft-Susskind Lp* = Lp2/3L2/3 where I take L = c/H(t) as a kind of Mach principle. Returning to the Nature paper: The WMAP data confirms the k = 0 flat space model on small scales but it is alleged that the model breaks down at large scales where Temperature correlations across the microwave sky match expectations on angular scales narrower than 60 degrees but, contrary to predictions, vanish on scales wider than 60 degrees ... The observed lack of temperature correlations on scales beyond 60 degrees means that the broadest waves are missing ... This is a bit like a cut-off in a wave guide and like the Casimir effect between two conducting plates. ... perhaps because space itself is not big enough to support them. They predict FRW Omega zero ~ 1.013 and temperature correlations in matching circles on the sky on the basis of a global topological model of a finite 3D space with no boundary that is multiply-connected with the pattern of holes, like in a more complicated 3D version of a 2D torus, called Poincare dodecahedral space. These holes IMHO can be thought of as huge traversable wormholes supported by tuv(zpf) = tuv(exotic vacuum) =/= 0 local Diff(4) tensor fields. The ordinary non-gravitating vacuum has tuv(zpf) = 0. Omega(Dark Matter) = 0.28. The claim is made that the L =2, 3, 4 multipole terms in the temperature fluctuation power spectrum (expanded in 2D spherical polar coordinated harmonics of latitude and longitude over the sky) fits their dodecahedral space k = 1 model better than the k = 0 inflation flat space model -- especially for the L = 2 quadrupole where the data is calibrated to fit their model exactly at L = 4. WMAP found a quadrupole only about one-seventh as strong as what would be expected in infinite flat space ... for large values of L, ranging up to L ~ 900 corresponding to small scale temperature fluctuations, the spectrum tracks the infinite universe [k = 0, Omega zero = 1]predictions exceedingly well. The CMB temperature fluctuations arise primarily (but not exclusively) from density fluctuations in the early universe. photons traveling from denser regions do a little extra work against gravity and therefore arrive cooler, while photons from less dense regions arrive warmer We need to be more careful about density here. The usual meaning is density. The repulsive dark energy phase /zpf > 0 of exotic vacuum acts like less dense because of the anti-gravity blue shift causing real far field photons from such an exotic vacuum region to arrive warmer with more energy per quantum. Just the opposite for the gravitating dark matter /zpf < 0 region of exotic vacuum. The density fluctuations across space split into a sum of three-dimensional harmonics ... just as temperature fluctuations split into a sum of two-dimensional spherical harmonics ... The low quadrupole implies a cut-off on the wavelengths of the three-dimensional harmonics. Such a cut-off presents an awkward problem in infinite flat space, because it defines a preferred length scale in an otherwise scale-invariant space. A more natural explanation invokes a finite universe ... Whereas most potential spatial topologies fail to fit the WMAP results, the Poincare dodecahedral space fits them very well. The Poincare dodecahedral space is a dodecahedral block of space with opposite faces abstractly glued together, so objects passing out of the dodecahedron across any space return from the opposite face. *This abstract topological idea seems to be, upon further reflection, absolutely physically equivalent to a closed 3D space without boundary that is multiply-connected by giant traversable wormholes or Star Gates that require tuv(exotic vacuum) to support them. The Poincare dodecahedral block of space therefore requires a network of 6 giant star gates of scale c/H(t), analogous to a 2D spherical surface with 6 wormhole handles, or a DeRham integral domain chain 3D Betti number of 6, I would imagine, for the present case. Light travels across the faces in the same way, so if we sit inside the dodecahedron and look outward across a face, our line of sight re-enters the dodecahedron from the opposite face. We have the illusion of looking into an adjacent copy of the dodecahedron. This illusion is, I suppose, physically indistinguishable from looking into the 2D face of a giant cosmic scale traversable very short Star Gate wormhole to what is on the other side. If we take the original dodecahedral block of space not as a euclidean dodecahedron (with edge angles ~ 117 degrees) but as a spherical dodecahedron (with edge angles exactly 120 degrees), then adjacent images of the dodecahedron fit together snugly to tile the hypersphere (Fig 3b), analogously to the way adjacent images of spherical pentagons (with perfect 120 degree angles) fit snugly to tile an ordinary sphere (Fig 3a). p. 594 OK my question here is whether or not there is any physics to this new formal construction such as 120 parallel finite universes next door that we can never access, or maybe we can? This led to the issue what is physical space? Is it SU(2)/ID? Note that: 12 spherical pentagons tile the surface of an ordinary sphere. They fit together snugly because their corner angles are exactly 120 degrees. Note that each spherical pentagon is just a pentagonal piece of a sphere. Note also that we cannot identify opposite edges of a 5-sided 2D pentagon to make a multiply-connected 2D surface. The 3D dodecahedron has an even number of 2D faces 12 allowing 6 traversable wormhole handles of opposite faces. Each face is a 2D star gate portal. Note also that a 2D sphere has no boundary, but that its 12 spherical pentagonal pieces do have boundaries. 120 spherical dodecahedra tile the surface of a hypersphere. The hypersphere is simply-connected with no boundary and no holes. Therefore you only have to slice it once to get it to split into two disjoint pieces. This is unlike the single 3D spherical dodecahedron that is 6-fold multiply-connected also without boundary. You have to slice it seven times to get it to split into 2 disjoint 3D pieces. Note that a 2D torus has only one hole and you have to slice it twice to make it break apart into two disjoint pieces. The problem here now is whether or not there is any physical measurable consequence of the number 120 here? Does the simply-connected 3D hypersphere with no boundary play a physical role or is it simply an esthetic formal nicety? This is one of the key questions that motivated this original thread. to be continued Poincare dodecahedral space as they describe it is exactly the same as SU(2)/ID. These authors are quite aware of this equivalence as they show in the paper Cosmic microwave background constraints on multi-connected spherical spaces, which you can download from: http://xxx.lanl.gov/abs/astro-ph/0303580 On page 2 (of this 5 page paper), they say: The finite subgroups of S3 are the cyclic groups Zn, the binary dihedral groups D*m, the binary tetrahedral, octahedral and icosahedral groups, respectively of order n, 4m, 24, 48 and 120. Now since you are looking for a physically significant difference between SU(2)/ID and SU(2) tiled with 120 spherical dodecahedra, I should point out the fundamental significance of the volume of the 3D space, which affects both the light travel times as well as the density of the space. As these authors say: In all cases the volume of the space S3/G is the volume of the 3-sphere S3 divided by the order |G| of the holonomy group. structures S3/G, and that has to do with the A-D-E classification of these spaces. This is because A-D-E Coxeter graphs classify at least 20 physically relevant mathematical objects. As I have mentioned previously, these include Coxeter (relfection) groups, Lie algebras, gravitational instanton spaces, catastrophe bundles, 2D conformal field theories, Heisenberg algebras, and much more. I have started writing this up in a Word file, and will send it to you as soon as I finish it. Saul-Paul === Subject: Simple curve question... If I have the following three data sets: A B -------------- 1500 300 4000 500 6000 300 What is the best way to calculate the corresponding B for any given A? I cannot assume any specific curve shape, meaning B could be 300 100 500, but A will always be increasing. B (BTW: this is -not- homework, it is related to mass spec analysis and my math sucks) === Subject: Re: Simple curve question... Well, I would try as below: 1) assume that B reach its maximum, say M (>500), at A=3750 (mid-point of 1500-6000), 2) fit a parabolic curve open downwards to these data. Michael Leung BCC .b9.a6.b9g©.97.a6l«.97.87séD :u2Hib.6$AX1.1990766@newssvr21.news.prodigy.com... > If I have the following three data sets: > A B > -------------- > 1500 300 > 4000 500 > 6000 300 > What is the best way to calculate the corresponding B for any given A? I > cannot assume any specific curve shape, meaning B could be 300 100 500, but > A will always be increasing. > (BTW: this is -not- homework, it is related to mass spec analysis and my > math sucks) === Subject: Re: Simple curve question... > BCC .b9.a6.b9g©.97.a6l«.97.87séD > :u2Hib.6$AX1.1990766@newssvr21.news.prodigy.com... > If I have the following three data sets: > A B > -------------- > 1500 300 > 4000 500 > 6000 300 > What is the best way to calculate the corresponding B for any given A? I > cannot assume any specific curve shape, meaning B could be 300 100 500, > but > A will always be increasing. > B > (BTW: this is -not- homework, it is related to mass spec analysis and my > math sucks) > Well, I would try as below: > 1) assume that B reach its maximum, say M (>500), > at A=3750 (mid-point of 1500-6000), > 2) fit a parabolic curve open downwards to these data. Fitting not needed. Three points specifies a unique interpolating parabola (see Lagrange or Newton interpolation). The Lagrange quadratic for these points is: y = 300*(x-4000)(x-6000)/(1500-4000)(1500-6000) + 500*(x-1500)(x-6000)/(4000-1500)(4000-6000) + 300*(x-1500)(x-4000)/(6000-1500)(6000-4000) = (x-4000)(x-6000)/37500 - (x-1500)(x-6000)/10000 + (x-1500)(x-4000)/30000 This parabola does reach a peak of 502.5 at x=3750. However, it's terribly dangerous to draw any conclusions from such a model if all you know are three points, especially if you're planning on extrapolating the model outside the range 1500-6000. Any additional info you can incorporate (such as knowledge of the location of the peak) would be helpful. - Randy === Subject: Re: (matrix analysis) Is this true: A*X*B=X => A=I and B=I? >What else results about A and B can I obtain from A*X*B=X? >(this is not a HW problem!) Is X some particular matrix, or is the equation A X B=X supposed to be true for all matrices X of a certain size? Suppose A = (a_{ij}) is an m x m matrix, B = (b_{ij}) is n x n, and A X B = X for all m x n matrices X. Then taking X = e_{kl} (the matrix with 1 in position (k,l) and 0 elsewhere) we get a_{ik} b_{lj} = 1 for i=k, j=l and 0 otherwise, which implies A = cI and B = c(-1)I for some c <> 0. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: (matrix analysis) Is this true: A*X*B=X => A=I and B=I? >What else results about A and B can I obtain from A*X*B=X? >(this is not a HW problem!) > Is X some particular matrix, or is the equation A X B=X supposed to be > true for all matrices X of a certain size? > Suppose A = (a_{ij}) is an m x m matrix, B = (b_{ij}) is n x n, and > A X B = X for all m x n matrices X. Then taking X = e_{kl} (the matrix > with 1 in position (k,l) and 0 elsewhere) we get a_{ik} b_{lj} = 1 for > i=k, j=l and 0 otherwise, which implies A = cI and B = c(-1)I for some > c <> 0. I really appreciate your help! For this problem, the specification is that A, B, X are all square with size of NxN... The condition that X=A*X*B is imposed to all inputs X. Basically, this is a linear 2-D separable transform, Y=A*X*B, but now I want the output to be the input, X=A*X*B. Under this condition, what should be A, and what should be B, or they have other hidden relations? complete, there are no other A and B that can satisfy the requirement... am I right? -Walala === Subject: Re: (matrix analysis) Is this true: A*X*B=X => A=I and B=I? >For this problem, the specification is that A, B, X are all square with size >of NxN... The condition that X=A*X*B is imposed to all inputs X. >complete, there are no other A and B that can satisfy the requirement... am >I right? Yes, that's what I said. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: how about A1*X*B1+A2*X*B2=X? >For this problem, the specification is that A, B, X are all square with size >of NxN... The condition that X=A*X*B is imposed to all inputs X. >complete, there are no other A and B that can satisfy the requirement... am >I right? > Yes, that's what I said. Can you say something about A1*X*B1+A2*X*B2=X? Basically, I want to approximate the input X with a bunch of transforms of the input X, try to find optimal approximation in the MSE sense.... For the above problem, can I say it equals to (A1+A2)*X*(B1+B2)=X... and then A1+A2=cI, B1+B2 = c(-1)I... Is this true? Or what else relation you can say about A1, A2, B1, B2? -Walala === Subject: Re: Slow Factoring Method The method I described works in any base. It doesn't appear to be any more efficient in base 10. 1023 base 10 1023 mod 10 = 3 1*3 = 3 7*9 = 63 1023 mod 100 = 23 01*23 = 23 11*93 = 1023 *** 21*63 = 1323 31*33 = 1023 *** 41*03 = 123 51*73 = 3723 61*43 = 2623 71*13 = 923 81*83 = 6723 91*53 = 4823 07*89 = 623 17*19 = 323 27*49 = 423 37*79 = 2823 47*09 = 423 57*39 = 2223 67*69 = 4623 77*99 = 7623 87*29 = 2523 97*59 = 5723 Russell - 2 many 2 count === Subject: stable marriage problem (matching) with constraints Suppose we have B boys, G girls. B=G. The boys b1 and b3 are neighbors of b2, b2 and b4 are neighbors of b3, etc. Same for the girls: g1 and g3 are neighbors of g2, g2 and g4 are neighbors of g3, etc. We want to find a matching such that it is stable, AND if the boy m(i) is matched to the girl g(j), then the neighbors of m(i) (that is, m(i-1) and m(i+1) ) must match to the neighbors of g(j) (that is, g(j-1) or g(j+1) ). Anyone can think of a good algo? === Subject: Re: stable marriage problem (matching) with constraints Sorry, my problem was ill defined, please ignore it. > Suppose we have B boys, G girls. B=G. > The boys b1 and b3 are neighbors of b2, b2 and b4 are neighbors of b3, > etc. > Same for the girls: g1 and g3 are neighbors of g2, g2 and g4 are > neighbors of g3, etc. > We want to find a matching such that it is stable, AND if the boy m(i) > is matched to the girl g(j), > then the neighbors of m(i) (that is, m(i-1) and m(i+1) ) must match to > the neighbors of g(j) (that is, g(j-1) or g(j+1) ). > Anyone can think of a good algo? === Subject: Re: Q wrt a number-theoretic generating function > I'm reading some introductory material about Dirichlet's generating > functions of arithmetical functions, but could not find anything about > the obvious generating function > Sum_p p{-x}, > where the sum is extended over all primes. Has it been studied? Is it > an independent function or can it be expressed in terms Riemann's > zeta? Anything else interesting about it? References? > Michele By the way, I might as well add this: exp(sum{p=primes} 1/px) = sum{k=1 to oo} A(k)/kx, where A(k) = product{p=primes} 1/(a(p,k))! , and where each a(p,k) is a nonnegative integer such that pa(p,k) is the highest power of the prime p which divides k. (I think...) Leroy Quet === Subject: Re: FUNctions/Continued-Fraction Puzzle UGGG! I used the wrong word (twice). Reposting with 2 solutions replaced with proofs. Sorry. --- >... > But, anyway, my solution is below the replied-to message. > (This might be actually trivial. But it does not seem to be with the > little thought I have given it. In any case, perhaps I should not have > cross-posted this to rec.puzzles {if I should have even posted it to > sci.math}; but what the...) > For all real x > 1, and for some function of x, y(x); > where each y is a real, y =y(x), based on x: it is so that: f([x; x2, x3, x4,...,xm]) = [y; y2, y3, y4,...,ym], for EVERY positive integer m; where: [x; x2, x3, x4,...,xm] is the continued-fraction 1 > x + ------------------ ; > 1 > x2 + -------------- > 1 > x3 + --------- > .... > + 1/xm and [y; y2, y3, y4,...,ym] is also a continued-fraction (obviously); and [x; x2, x3, x4,...,xm] converges to X; and f(w) is a real -> real function, such that f'(X) exists and is finite nonzero. So, what are the possible f(w)'s, given all of the conditions above?? > First, by the way, f'(X) is the (1st) derivative of f(w) at w = X, in > case this is not obvious. I should mention that f can equate to an infinite number of functions > if it need not be analytic. If it need by analytic, however, there are > a finite number of possible functions that can equal f(w). > (So, find the set of analytic f(w)'s.) This puzzle seems to be more difficult than I first assumed. > I will wait until Friday, at least, to post the answer if no one else > posts the solution before that. .... > ...the solution: > I get that the only possible analytic f is: > f(w) = w. > Proof: > limit{m -> oo} (x/y)(2m-1) = 1. > So, x must = y. And, consequently, f(w) must = w. > * earlier result at: I highly suspect my PROOF is far from the simplest. Is there a PROOF which is any simpler, even trivial? (Perhaps my result itself, that f(w) = w is the ONLY analytic function, is wrong.) Leroy Quet === Subject: Re: FUNctions/Continued-Fraction Puzzle >... > But, anyway, my solution is below the replied-to message. > (This might be actually trivial. But it does not seem to be with the > little thought I have given it. In any case, perhaps I should not have > cross-posted this to rec.puzzles {if I should have even posted it to > sci.math}; but what the...) > For all real x > 1, and for some function of x, y(x); > where each y is a real, y =y(x), based on x: it is so that: f([x; x2, x3, x4,...,xm]) = [y; y2, y3, y4,...,ym], for EVERY positive integer m; where: [x; x2, x3, x4,...,xm] is the continued-fraction 1 > x + ------------------ ; > 1 > x2 + -------------- > 1 > x3 + --------- > .... > + 1/xm and [y; y2, y3, y4,...,ym] is also a continued-fraction (obviously); and [x; x2, x3, x4,...,xm] converges to X; and f(w) is a real -> real function, such that f'(X) exists and is finite nonzero. So, what are the possible f(w)'s, given all of the conditions above?? > First, by the way, f'(X) is the (1st) derivative of f(w) at w = X, in > case this is not obvious. I should mention that f can equate to an infinite number of functions > if it need not be analytic. If it need by analytic, however, there are > a finite number of possible functions that can equal f(w). > (So, find the set of analytic f(w)'s.) This puzzle seems to be more difficult than I first assumed. > I will wait until Friday, at least, to post the answer if no one else > posts the solution before that. .... > ...the solution: > I get that the only possible analytic f is: > f(w) = w. > Proof: > limit{m -> oo} (x/y)(2m-1) = 1. > So, x must = y. And, consequently, f(w) must = w. > * earlier result at: I highly suspect my solution is far from the simplest. Is there a solution which is any simpler, even trivial? (Perhaps my result itself, that f(w) = w is the ONLY analytic function, is wrong.) Leroy Quet === Subject: Re: Euler books > where can I find english or german translations of some of euler's > works in the internet? > I can't afford to buy those. I know for instance that Euler's opera > omnia is very expensive. This would be a _really_ good place to start: http://math.dartmouth.edu/~euler/welcome.html This Google search will help you find others: HTH xanthian. -- === Subject: Re: Euler books > where can I find english or german translations of some of euler's > works in the internet? Oh, Internet. Try the Gutenberg Project, which has online full text of many classic books. One problem with many English translations of Euler is that they are so recent (published by Springer Verlag, in some cases I know) that they are still under copyright for the English text. The original Latin text of Euler's works is, of course, long since in the public domain. Hope this helps! -- Karl M. Bunday Christ has set us free. Galatians 5:1 Learn in Freedom (TM) http://learninfreedom.org/ kmbunday AT earthlink DOT net (preferred email address) === Subject: Re: Euler books > they are still under copyright for the English text. The original Latin text > of Euler's works is, of course, long since in the public domain. Euler was among the last scholars to write his papers in Latin. During the 19-th century and later most papers were published in national languages. Bob Kolker === even see it) === Subject: f(x+1) = f(x) + 1/f(x) What is the family of real -> real analytic functions f(x), such that: f(x+1) = f(x) + 1/f(x) for all real x? Leroy Quet === Subject: Re: f(x+1) = f(x) + 1/f(x) >What is the family of real -> real analytic functions f(x), such that: >f(x+1) = f(x) + 1/f(x) >for all real x? There cannot be any such function on all reals. If we let g=f2, the equation yields g(x+1) = g(x) + 2 + 1/g(x), so g(x+1) > g(x) + 2. As g is non-negative, one cannot go far in the negative direction. If we look at the equation for g and ignore its positivity, it cannot be continuous for the same reason; it must be 0 somewhere. If g starts at some x_0, the asymptotics of g are g(x) ~ 2x + C + ln(x+C/2)/2 + O(ln(x)/x), where C can be a periodic function of the fractional part of x. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: f(x+1) = f(x) + 1/f(x) > What is the family of real -> real analytic functions f(x), such that: > f(x+1) = f(x) + 1/f(x) > for all real x? > Leroy Quet It can't have any roots since then the relation is undefined. It also has to diverge to the right since lim f(x) = lim f(x+1) = lim f(x) + lim 1/f(x) ==> lim 1/f(x) = 0 ==> lim f(x) = +/-infinity where the limit is taken as x-->+infinity That's all I got right now. Have a tolerable existence. Eli -- === Subject: Re: f(x+1) = f(x) + 1/f(x) What is the family of real -> real analytic functions f(x), such that: >> f(x+1) = f(x) + 1/f(x) >> for all real x? >> Leroy Quet > It can't have any roots since then the relation is undefined. It also has > to diverge to the right since > lim f(x) = lim f(x+1) = lim f(x) + lim 1/f(x) > ==> lim 1/f(x) = 0 ==> lim f(x) = +/-infinity > where the limit is taken as x-->+infinity > That's all I got right now. > Have a tolerable existence. Eli > -- Both your statements are also true for all derivatives of f? .... Quaternion === Subject: Re: Factorial/Exponential Identity, Infinity Besides two, there actually is another integer where ths sum of it and itself is equal to the product of it and itself: zero. A binary number that is normal to base 2 has equal probability of a given element being a zero or a one, right? That is to say, statistically any contiguous finite subsequence is expected to have equal numbers of ones and zeros. Is that the same thing as its sequence having zero-density of one half? In any even length contiguous subsequence of (01)..., half the values are one and half zero. The current state of opinion appears to be that almost all real numbers are absolutely normal, that is, normal to each integer base greater than one. Then, there are uncountably many abnormal numbers. In the unit interval, then, what if normal numbers are almost half of them, with the rational numbers with zero-density of one half rounding the proportion to exactly one half? Bizarre, no? Then again, the only normal numbers thus far contrived are not shown to appear in the wild, I hear Sierpinski designed a construction of one. Yet, there is evidence that pi, for example, is not abnormal to base ten, eg, in the first million decimal digits each digit appears around 1/10 of the time. MathPages' Is e normal?: http://www.mathpages.com/home/kmath519.htm . Many (most?) sequences with equal densities of zeros and ones are not normal to base two. They don't contain every other expected sequence, ie being normal in base four, eight, etcetera. A number normal to base two is normal to base 2x for positive integer x. I read this quote of Euler off the Internet the other day and thought it was pretty good, Euler's rather caustic: Notable enough, however, are the controversies over the series 1 - 1 + 1 - 1 + 1 - ... whose sum was given by Leibniz as 1/2, although others disagree. ... Understanding of this question is to be sought in the word sum; this idea, if thus conceived -- namely, the sum of a series is said to be that quantity to which it is brought closer as more terms of the series are taken -- has relevance only for convergent series, and we should in general give up the idea of sum for divergent series. - L. Euler I think the sum is zero. Notice that Euler has the idea of sums of divergent series. Anyways, I got to thinking about the factorizations of ((sum n)x - sum(nx)) / s(n+1, n-x+1). I noticed the patterns of some of the factors, for example how for five values of x in a row the possible rational function, or ratio of polynomials, if it exists, that goes to x!, has the given factor. Then again, I'm still trying to determine an efficient method to determine a type of additive partitioning of a number, for use in enumeration. As well, I'd like to read more about the consideration of hypermatrices. I started this thread because I had the notion that half the sequences had densities of one half. Yet we have seen that n!/(n/2)!2 2n evaluates to a different asymptotic expression than that, yet now I read that multitudinous numbers are normal. Ross === Subject: Re: Factorial/Exponential Identity, Infinity > Besides two, there actually is another integer where ths sum of it and > itself is equal to the product of it and itself: zero. x*x = x + x being a quadratic equation, how many solutions did you expect to find? > The current state of opinion appears to be that almost all real > numbers are absolutely normal, that is, normal to each integer base > greater than one. Then, there are uncountably many abnormal > numbers. I do not see that your conclusion follows from your premise. > In the unit interval, then, what if normal numbers are almost half of > them, with the rational numbers with zero-density of one half rounding > the proportion to exactly one half? Since almost all and almost half are incompossible, why ask? > Bizarre, no? Your thought processes are indeed bizarre. > Then again, the only normal numbers thus far contrived are not shown > to appear in the wild, I hear Sierpinski designed a construction of > one. Yet, there is evidence that pi, for example, is not abnormal to > base ten, eg, in the first million decimal digits each digit appears > around 1/10 of the time. > MathPages' Is e normal?: http://www.mathpages.com/home/kmath519.htm > Many (most?) sequences with equal densities of zeros and ones are not > normal to base two. They don't contain every other expected sequence, > ie being normal in base four, eight, etcetera. A number normal to > base two is normal to base 2x for positive integer x. > I read this quote of Euler off the Internet the other day and thought > it was pretty good, Euler's rather caustic: > Notable enough, however, are the controversies over the series 1 - 1 > + 1 - 1 + 1 - ... whose sum was given by Leibniz as 1/2, although > others disagree. ... Understanding of this question is to be sought in > the word sum; this idea, if thus conceived -- namely, the sum of a > series is said to be that quantity to which it is brought closer as > more terms of the series are taken -- has relevance only for > convergent series, and we should in general give up the idea of sum > for divergent series. - L. Euler > I think the sum is zero. Notice that Euler has the idea of sums of > divergent series. Even Homer nods. === Subject: Integer-Alteration Game Each player takes turns coming up with algorithmic steps, one step at a time in sequence, each step giving a rule to transform an integer into another. (such as: multiply m by greatest prime p, where p divides m and is <= sqrt(|m|). If no such prime exists, leave m unchanged.) (or such as: m = floor(|m|/d(|m|)), where d(m) is number of positive divisors of m.) (or rules may involve meta-transformations, such as sending players to previous rules, depending somehow upon the value of the integer when reaching the step.) Anyway, players are encouraged to be creative when inventing rules! Each step is capable of transforming any integer, always giving an integer as output. After a predetermined number of steps have been created (the same number for each player), a random integer is generated somehow. Players then try to guess what the output intger will be. The algorithm is run using the random start-integer. The winner of the game is the player who comes closest to guessing the final output integer. (needs work....) Leroy Quet === Subject: Re: absolute value graph > Sketch the graph of f(x)=2abs(x) > ------ > 1+x2 > How would I start this question? > This function is even, i.e. f(-x) = f(x). Then sketch the graph of f(x) = > 2x/(1 + x2) for x >= 0. and reflect it on the OY axis. Not that the above is a bad answer, but here's another perspective. Near the zero of f[x] (@ x=0 obviously), the graph looks like an absolute value function - v-shaped. When x is big - in either direction, f[x] is positive and near zero - like a reciprocal function, in fact. Apparently then, for some x in between 0 and really big, your function f must have had a maximum value... hth, cdj === Subject: {Field Theory} Please help me understand this special case (was This could get confusing fast...) In my other thread I pointed out that in a field F of integers, the notation ab of two elements of that field is ambiguous, namely because it has the 3 possible and not necessarily equal interpretations: 1. ab = a * b where * is the multiplication of F 2. ab = (a + a + ... + a) (b times) 3. ab = (b + b + ... + b) (a times) Some people were replying to the effect that the 3 are in fact equal. And they are, when F is well behaved. But since it was confusing me so much, I sought a counterexample, and lo and behold I found one with little difficulty. This, I exhibit thus: let F as a set (not a field yet) simply equal Z, ie, all integers (do not confuse with Z_n) Now let p:N->Q be the mapping of N to Q which is used in Cantor's famous proof that the rationals are countable. For example, p(1)=1, p(2)=1/2, p(3)=2, p(4)=3, p(5)=1/3, etc. Now enhance p into a better function q defined by: for x > 0, q(x) = p(x) for x = 0, q(x) = q(0) = 0 for x < 0, q(x) = -p(-x) Finally let q' be the inverse of q. (I'll leave the existence proof details to you since it is all quite easy) Now associate with F the following addition and multiplication: a + b = q'[q(a)+q(b)] a * b = q'[q(a)q(b)] It is not hard to prove that F is thus turned into a field (with 1 as its unit and 0 as its zero) What, then, can we make of 5*3? Let us check the 3 possible interpretations: 1. 5*3 = f'[f(5)*f(3)] = f'[(1/3) * 2] = f'(2/3) = 7 2. 5*3 = 5 + 5 + 5 = f'[f(5)*3] = f'[1] = 1 3. 5*3 = 3 + 3 + 3 + 3 + 3 = f'[f(3)*5] = f'[10]. I didn't bother to write out Cantor's table all the way needed to calculate f'[10] but it is obviously way larger than 7. I have thus exhibited that the notation in question is indeed ambiguous. I am greatly muddled on the matter and am seeking clarification... Some have responded that there is no such thing as a field of integers... but this seems to make no sense to me. Yes, I can see how you can argue that Z_p is really a field of sets, not of integers, although I see nothing forbiding us from using {0,1,...,p-1} as the set and simply modifying the addition and multiplication so the modular arithmetic takes place there, and thus obtain a finite field of integers. Is it the case that the definition of a field is really some arcane, ethereal thing taught in Ph.D. level math and that Herstein's definition is just a handwaving gesture similar to an algebra 101 definition of the rationals? === Subject: Re: {Field Theory} Please help me understand this special case (was This could get confusing fast...) > In my other thread I pointed out that in a field F of integers, the One can indeed label the congruence classes of the finite field Z/(p) by integers, e.g. the congruence class representatives 0,1,2,...,p-1. The proper terminology here is a residue field of the ring of integers, or image field..., not field of integers. > notation ab of two elements of that field is ambiguous, namely > because it has the 3 possible and not necessarily equal > interpretations: > 1. ab = a * b where * is the multiplication of F > 2. ab = (a + a + ... + a) (b times) > 3. ab = (b + b + ... + b) (a times) > Some people were replying to the effect that the 3 are in fact equal. > And they are, when F is well behaved. But since it was confusing me Well-behaved field? That is not a mathematical term. In any case, the example I give below should help to alleviate any confusion here. > so much, I sought a counterexample, and lo and behold I found one with > little difficulty. This, I exhibit thus: > let F as a set (not a field yet) simply equal Z, ie, all integers (do > not confuse with Z_n) > Now let p:N->Q be the mapping of N to Q which is used in Cantor's > famous proof that the rationals are countable. For example, > p(1)=1, p(2)=1/2, p(3)=2, p(4)=3, p(5)=1/3, etc. > Now enhance p into a better function q defined by: > for x > 0, q(x) = p(x) > for x = 0, q(x) = q(0) = 0 > for x < 0, q(x) = -p(-x) > Finally let q' be the inverse of q. (I'll leave the existence proof > details to you since it is all quite easy) > Now associate with F the following addition and multiplication: > a + b = q'[q(a)+q(b)] > a * b = q'[q(a)q(b)] > It is not hard to prove that F is thus turned into a field (with 1 as > its unit and 0 as its zero) > What, then, can we make of 5*3? Let us check the 3 possible > interpretations: > 1. 5*3 = f'[f(5)*f(3)] = f'[(1/3) * 2] = f'(2/3) = 7 > 2. 5*3 = 5 + 5 + 5 = f'[f(5)*3] = f'[1] = 1 > 3. 5*3 = 3 + 3 + 3 + 3 + 3 = f'[f(3)*5] = f'[10]. You are somehow confused. For an enlightening example let's consider the isomorphism between the ring of Roman numerals and the ring of integers: V times III = q'(q(V) * q(III)) = q'(5*3) = q'(15) = XV V plus V plus V = q'(q(V) + q(V) + q(V)) = q'(5+5+5) = q'(15) = XV III plus...plus III = q'(q(III)+...+q(III)) = q'(3+3+3+3+3) = q'(15) = XV The roman numerals are simply new names for the integers, and the operations on them may be performed by translating the numerals to their standard names, performing the standard operations, then mapping the standard result to its Roman numeral. For further discussion of such isomorphisms see my prior post -Bill Dubuque > I didn't bother to write out Cantor's table all the way needed to > calculate f'[10] but it is obviously way larger than 7. > I have thus exhibited that the notation in question is indeed > ambiguous. I am greatly muddled on the matter and am seeking > clarification... > Some have responded that there is no such thing as a field of > integers... but this seems to make no sense to me. Yes, I can see > how you can argue that Z_p is really a field of sets, not of integers, > although I see nothing forbiding us from using {0,1,...,p-1} as the > set and simply modifying the addition and multiplication so the > modular arithmetic takes place there, and thus obtain a finite field > of integers. Is it the case that the definition of a field is really > some arcane, ethereal thing taught in Ph.D. level math and that > Herstein's definition is just a handwaving gesture similar to an > algebra 101 definition of the rationals? === Subject: Re: {Field Theory} Please help me understand this special case (was This could get confusing fast...) at 07:00 PM, snizpilbor@yahoo.com (Sniz Pilbor) said: >In my other thread I pointed Claimed. >field F of integers, The integers don't form a field. >and not necessarily equal interpretations: Incorrect. >Some people were replying to the effect that the 3 are in fact >equal. And they are, when F is well behaved. What do you mean by well behaved? If F is a filed then there are a well defined mappings *_1: ZxF -> F, *_2: FxZ -> F and I: Z -> F such that all three are equivalent once you make the functions explicit. >I found one Nope. It's not a counter example. You just got confused about what operator to apply when. >Now associate with F the following addition and multiplication: >a + b = q'[q(a)+q(b)] >a * b = q'[q(a)q(b)] You'll need to make everything explicit to see your error. Use distinct symbols for the operations in Z and in F. >It is not hard to prove that F is thus turned into a field (with 1 >as its unit and 0 as its zero) Only if p(1)=1. >What, then, can we make of 5*3? Nothing, without context. It might refer to any of your interpretations 1-3, all of which give the same result, or it might refer to a 4th interpretation, which gives a different result. >Some have responded that there is no such thing as a field of >integers... Let me rephrase that: it is not a standard Mathematical term, and you have not defined it. >I see nothing forbiding us from using {0,1,...,p-1} as the set or {0,1,3,4,...p}. Would that satisfy your definition of filed of integers? >simply modifying the addition and multiplication so the modular >arithmetic takes place there, If you chose different addition and multiplication operators, would you still call it a field of integers? >Is it the case that the definition of a field is really some arcane, >ethereal thing taught in Ph.D. level math No: what is the case is that when you introduce new nomenclature you are obligated to define it. Until you have a subject well and truly under you belt, introducing new nomenclature in a question is more likely to confuse the issue than to help. Introducing new nomenclature without a definition is guarantied to confuse the issue, since the readers are not likely to interpret it the same way that you do. -- spamtrap@library.lspace.org === Subject: Re: {Field Theory} Please help me understand this special case (was This could get confusing fast...) > In my other thread I pointed out that in a field F of integers, the > notation ab of two elements of that field is ambiguous, namely > because it has the 3 possible and not necessarily equal > interpretations: > 1. ab = a * b where * is the multiplication of F > 2. ab = (a + a + ... + a) (b times) > 3. ab = (b + b + ... + b) (a times) > Some people were replying to the effect that the 3 are in fact equal. > And they are, when F is well behaved. This has nothing to do with F being well beheved or not. It is a consequence of the way in which integers are defined within a ring R (with identity) in general: one uses the unique ringhomomorphism from Z to R. More generally (anticipating your example below) one can define the integers within a monoid M (= semigroup with identity) by using the unique monoid-homomorphism from N to M. > But since it was confusing me > so much, I sought a counterexample, and lo and behold I found one with > little difficulty. This, I exhibit thus: > let F as a set (not a field yet) simply equal Z, ie, all integers (do > not confuse with Z_n) > Now let p:N->Q be the mapping of N to Q which is used in Cantor's > famous proof that the rationals are countable. For example, > p(1)=1, p(2)=1/2, p(3)=2, p(4)=3, p(5)=1/3, etc. This is interesting but beside the point. Your p is not a homomorphism. Marc === Subject: Re: {Field Theory} Please help me understand this special case (was This could get confusing fast...) Some people were replying to the effect that the 3 are in fact equal. > And they are, when F is well behaved. > This has nothing to do with F being well beheved or not. > It is a consequence of the way in which integers are defined > within a ring R (with identity) in general: one uses the unique > ringhomomorphism from Z to R. > More generally (anticipating your example below) one can define the > integers within a monoid M (= semigroup with identity) by using > the unique monoid-homomorphism from N to M. replying? By integers I was meaning the normal integers, that is, 1, 2, 3, -1, -2, -3, 0, etc., encountered in basic high school algebra, not generalized integers of a ring. I apologize if this caused additional confusion. As for the ambiguity, in the post to which you replied I gave a specific example where the 3 interpretations of 3*5 all ended up being completely different. Your reply seems to be saying that no, no, they are all equal, whether or not F is well-behaved. Where, then, was my mistake? Was the F which I exhibited in my post not in fact a field, and if so why not, which axiom(s) of the field did it fail to uphold? The previous response to my post, whose writer seems to have actually read it through before replying, seemed to clear things up, but now the waters are rendered as muddy as ever. I sincerely want to understand what you are saying, but you must not begrudge me to remain skeptical when a counterexample is right in front of us which you have not explained. > This is interesting but beside the point. Your p is not a homomorphism. I never claimed that it was (why would it matter?). I was using it as a means of defining an addition and multiplication which would make F a field. I get the impression that you read just that far and then stopped. The original confusion of the ambiguity seems to be even greater when learned field theoreticians rise in abundance to say it doesn't exist, *blatantly ignoring the counterexample right there*. The original post in this thread (with some minor corrections): > In my other thread I pointed out that in a field F of integers, the > notation ab of two elements of that field is ambiguous, namely > because it has the 3 possible and not necessarily equal > interpretations: > 1. ab = a * b where * is the multiplication of F > 2. ab = (a + a + ... + a) (b times) > 3. ab = (b + b + ... + b) (a times) > Some people were replying to the effect that the 3 are in fact equal. > And they are, when F is well behaved. But since it was confusing me > so much, I sought a counterexample, and lo and behold I found one with > little difficulty. This, I exhibit thus: > let F as a set (not a field yet) simply equal Z, ie, all integers (do > not confuse with Z_n) > Now let p:N->Q+ be the mapping of N to Q+ (the positive rationals) > which is used in Cantor's > famous proof that the rationals are countable. For example, > p(1)=1, p(2)=1/2, p(3)=2, p(4)=3, p(5)=1/3, etc. > Now enhance p into a better function q defined by: > for x > 0, q(x) = p(x) > for x = 0, q(x) = q(0) = 0 > for x < 0, q(x) = -p(-x) > Finally let q' be the inverse of q. (I'll leave the existence proof > details to you since it is all quite easy) > Now associate with F the following addition and multiplication: > a + b = q'[q(a)+q(b)] > a * b = q'[q(a)q(b)] > It is not hard to prove that F is thus turned into a field (with 1 as > its unit and 0 as its zero) > What, then, can we make of 5*3? Let us check the 3 possible > interpretations: > 1. 5*3 = q'[q(5)*q(3)] = q'[(1/3) * 2] = q'(2/3) = 7 > 2. 5*3 = 5 + 5 + 5 = q'[q(5)*3] = q'[1] = 1 > 3. 5*3 = 3 + 3 + 3 + 3 + 3 = q'[q(3)*5] = q'[10]. > I didn't bother to write out Cantor's table all the way needed to > calculate q'[10] but it is obviously way larger than 7. > I have thus exhibited that the notation in question is indeed > ambiguous. I am greatly muddled on the matter and am seeking > clarification... > Some have responded that there is no such thing as a field of > integers... but this seems to make no sense to me. Yes, I can see > how you can argue that Z_p is really a field of sets, not of integers, > although I see nothing forbiding us from using {0,1,...,p-1} as the > set and simply modifying the addition and multiplication so the > modular arithmetic takes place there, and thus obtain a finite field > of integers. Is it the case that the definition of a field is really > some arcane, ethereal thing taught in Ph.D. level math and that > Herstein's definition is just a handwaving gesture similar to an > algebra 101 definition of the rationals? === Subject: Re: {Field Theory} Please help me understand this special case (was This could get confusing fast...) > In my other thread I pointed out that in a field F of integers, the > notation ab of two elements of that field is ambiguous, namely > because it has the 3 possible and not necessarily equal > interpretations: > 1. ab = a * b where * is the multiplication of F > 2. ab = (a + a + ... + a) (b times) > 3. ab = (b + b + ... + b) (a times) > Some people were replying to the effect that the 3 are in fact equal. > And they are, when F is well behaved. But since it was confusing me > so much, I sought a counterexample, and lo and behold I found one with > little difficulty. This, I exhibit thus: > let F as a set (not a field yet) simply equal Z, ie, all integers (do > not confuse with Z_n) > Now let p:N->Q be the mapping of N to Q which is used in Cantor's > famous proof that the rationals are countable. For example, > p(1)=1, p(2)=1/2, p(3)=2, p(4)=3, p(5)=1/3, etc. > Now enhance p into a better function q defined by: > for x > 0, q(x) = p(x) > for x = 0, q(x) = q(0) = 0 > for x < 0, q(x) = -p(-x) For this to give a bijection you probably meant that p is a bijection between N and the POSITIVE rationals. > Finally let q' be the inverse of q. (I'll leave the existence proof > details to you since it is all quite easy) > Now associate with F the following addition and multiplication: > a + b = q'[q(a)+q(b)] > a * b = q'[q(a)q(b)] > It is not hard to prove that F is thus turned into a field (with 1 as > its unit and 0 as its zero) This is correct, since the addition and multiplication is now that of the rationals. We simply decided to give them funny and unnatural names via q and q'. > What, then, can we make of 5*3? Let us check the 3 possible > interpretations: > 1. 5*3 = f'[f(5)*f(3)] = f'[(1/3) * 2] = f'(2/3) = 7 > 2. 5*3 = 5 + 5 + 5 = f'[f(5)*3] = f'[1] = 1 > 3. 5*3 = 3 + 3 + 3 + 3 + 3 = f'[f(3)*5] = f'[10]. Since in your field 1+1+1 is not equal 3 nor is 1+1+1+1+1 equal to 5, the computations 2. and 3. are incorrect, unless interpreted as below (when there is no reason to expect them to be equal to computation 1.). Nothing exciting about it. > I didn't bother to write out Cantor's table all the way needed to > calculate f'[10] but it is obviously way larger than 7. > I have thus exhibited that the notation in question is indeed > ambiguous. I am greatly muddled on the matter and am seeking > clarification... You seem to suffer from the following confusion: When the notation m*r, where m is an integer and r is an element of ring, is defined, it is done using ONLY the operations of the ring. Remember, only the ring addition and ring multiplication exist (at that moment). Let me add tags _R and _Z to denote that the element or operation is that of the ring R or of the integers Z. Observe that we actually have a third multiplication taking place between an element of a ring and an element of the ring, denote this by *_M (M stands for module, you will learn about these in your next algebra course). Multiplication by adding together copies of an element of the ring necessitates that there is an integer participating to the proceedings, i.e. this is the module multiplication. So, we define module multiplication as follows: 1) Start with defining multiplication by the integers 0 and 1: 0_Z *_M r_R = 0_R, 1_Z *_M r_R = r_R 2) Extend the definition (by induction) to larger positive integers so as to satisfy one of the Z-module axioms: (n_Z +_Z m_Z)*_M r_R= (n_Z *_M r_R) +_R (m_Z *_M r_R) 3) Extend the definition to negative integers by the usual trick of using the additive inverse (within the ring) Your algebra textbook then hopefully shows a list of natural looking consequencies of this definition of module multiplication, such as (n_Z *_Z m_Z) *_M r_R = n_Z *_M (m_Z *_M r_R). As a consequence of this definition we then end up with e.g. 3_Z *_M r_R = (1_Z +_Z 1_Z +_Z 1_Z)*_M r_R = (1_Z *_M r_R) +_R (1_Z *_M r_R) +_R (1_Z *_M r_R) = r_R +_R r_R +_R r_R as expected (here again the module multiplication axiom was used). Observe that of the various multiplications present: 1) *_Z applies, when both sides are integers 2) *_R applies, when both sides are ring elements 3) *_M applies, when one is integer and the other from the ring. Your example consisted of the following 3 different multiplications 1. 5_R *_R 3_R 2. 5_R *_M 3_Z 3. 5_Z *_M 3_R There is no reason, why these should be the same, when *_R is very unnatural as was the case here. You muddied the waters by making ring multiplication by 3 (really by 3_R) to be different from module multiplication by 3 (really by 3_Z). Algebra textbooks often denote all of these simply by *. You are correct in that, if you cook up something really weird (such as your example), then a confusion may result. When a confusion is a possibility, it is IMHO the responsibility of the cook to clearly indicate what is meant (e.g. by adding the subscripts to the operations as above). You may feel that it is a disservice to the students not to emphasize this distinction. I DO notify my freshman algebra students of these distinctions, but I won't unduly emphasize it and stick to the standard notation in class room usage. The reason to this is that it is part of their education to: A) be aware of these distinctions, B) be able to interpret a possibly ambiguous formula correctly from the context (this is necessary for them to be able to study more algebra on their own), C) be aware of the fact that usually the distinctions don't result in any changes (which is why the same notation is used). Good luck in your studies, Jyrki Lahtonen, Turku, Finland === Subject: 1/m +1/(m+1) +...+1/n close to x Let m be a fixed positive integer. Let x be a fixed positive real. Let n(m,x) be the highest positive integer such that 1/m + 1/(m+1) + 1/(m+2) + ...+ 1/n(m,x) <= x. A lot of questions can be asked about this. But what I am wondering now is, what is: E(m,x) = x - (1/m + 1/(m+1) + 1/(m+2) + ...+ 1/n(m,x)) asymptotical to? All this is highly related to the topic at: === Subject: Re: Two-planes in Four-Space > Let G(2,4) be the Grassmanian of 2-dimensional subspaces in R4. > I map G(2,4) -> RP2 as follows. Given a plane g, choose an > orthogonal basis e1, e2. Identifying R4 with the quaternions in > the obvious way, form the quaternion u = e1 * e2(-1). Then > u is a square root of -1, well defined up to a sign. The square > roots of -1 are naturally identified with the unit ball S2 sitting > in R3 sitting in R4 via (a,b,c) |-> (0,a,b,c), so u gives a > well-defined element of S2/(plus or minus 1) = RP2. One might add that the fiber F_u over the point in RP2 represented by the quaternion u has a natural group structure: Given two planes g1 and g2 in F_u, it follows that the set g1 g2 = {p q |p in g1 and q in g2} is also a plane in F_u. This fiber is naturally isomorphic to H*/(R+Ru)* . Another characterization of F_u is that it consists of all planes that are fixed by the 90 degree rotation represented by u. Now I repeat my two questions: 1) We now have a fibration G(2,4) -> RP2 in which the fibers are naturally isomorphic to the various groups H*/(R+Ru)*, where u ranges over square roots of -1 in the quaternions. Is there a standard name for this fibration? 2) Two planes g1 and g2 are in the same fiber F_u if and only if there exists a quaternion p such that p g1 = g2 (or equivalently, if and only if there exists a 90 degree rotation that fixes both g1 and g2). Is there a standard name for this equivalence relation? === Subject: Re: help me...my problem?? > sequence {An} > A1 = 1 > A2 = 8 > An = root (An-1 * An-2) (n=3,4,5....) > find lim An (n->infinite) > --------------------------------- > i want your warm advice. > help me...please === Subject: Re: help me...my problem?? Original Format Nice question ! Denote D=(0,infty). Suppose that f:(0,infty)--->f(D) is continuous and strictly monotonic. Let F: f(D) --->(0,infty) be its inverse function. Instead of geometric mean let us consider the ,,f-quasi mean M_f(a,b) of two positive numbers a,b, that is (*) M_f(a,b)=F((f(a)+f(b))/2) . For instance The last mean can be generalized as / ((ar+br)/2){1/r} for real r , r=/=0 , and MEAN_r(a,b):=/ lim_{r-->0}A_r(a,b)=G(a,b):= sqrt(ab) when r=0 . Let 0=1} defined by (1) A(n)=M_f(A(n-1),A(n-2)) , n=3,4,... , with A(1)=p , A(2)= q . By summing equalities 2*Y(k)=Y(k-1)+Y(k-2) , (k=3,4,...,n) we obtain 2*SUM_{k=3 to k=n}Y(k)=SUM_{k=2 to k=n-1}Y(k) + SUM_{k=1 to k=n-2}Y(k) that is (2) 2*Y(n) + Y(n-1) = Y(1) +2*Y(2) . Because min{a,b} =< M_f(a,b) =< max{a,b} for a,b >0, it's easy to prove that (X(n))_{n>=1}, and therefore also (Y(n))_{n>=1}, are convergent. Let LIM:=lim_{n-->infty}A(n), L:= lim_{n-->infty} Y(n)= f(LIM), or LIM=F(L) . and thus L = lim_{n-->infty}Y(n)=(Y(1)+2*Y(2))/3 . In conclusion LIM:= lim_{n--->infty}A(n)=F((Y(1)+2*Y(2))/3)= F( (f(A(1)+2*f(A(2))/3) . For instance , when r is a real number f(x)=xr (when r=/=0) , and f(x)= ln(x) when r=0 , we find for (A(n))_{n>=1} generated by A(1)=p , A(2)=q , 0infty} A(n)= Sqrt[3](64) =4 . Note: Let D=(0,infty), f:D--->f(D)) be a continuous strictly monotonic function and F: f(D)-->D its inverse function. An interesting question may be : Suppose as fixed the following numbers : k = a positive integer , k>=2 , p_1,p_2,...,p_k , (p_j >0 , j=1,2,...,k) , w_1,w_2,...,w_k are positiv numbers such that w_1+w_2+...+w_k = 1 . Furher let A_f(a_1,a_2,...,a_k):=F(w_1*f(a_1)+w_2*f(a_2)+...+w_k*f(a_k)) , a_j > 0 . Define the sequence (S(n))_{n>=1} by S(1)=p_1 , S(2)=p_2 ,...,S(k)=p_k , and S(n)= A_f(S(n-1),S(n-2),S(n-3),...,S(n-k)) , n in {k+1,k+2,...} . NEW QUESTION: what about the convergence of sequence (S(n))_{n>=1} ? === Subject: Re: JSH: About time > Now that I've revealed the odd and you could say esoteric error in > core mathematics with such a short, and rather simple argument, the > issue now is how long until mathematicians decide that they'd rather > have correct mathematics versus the *belief* that they had been > perfect in keeping error out of the collected body of work that is > called mathematics. > Really, I can not understand your work, not even the core error problem. > I apologize for replying in a public forum, but I've chosen, or been > guided, to work under certain constraints. The danger of e-mail is the > temptation to say different things to different people. It would be > better for me, professionally, to hide my support for you, but I refuse > to do it. Don't worry about it, as I find I'm liking your posts. Actually, I kind of got a kick out of some of your previous posts back when you were insulting me. My thinking is that you appreciate quality work when you see it. Now for those who wonder, being rather pissed at my current predicament, with *some* mathematicians--I have to remember not to characterize all of them--lying about my work or running away from it, I sent Jim Ferry and Keith Ramsay a testy email, where I informed them I considered them to be runners as well, and I would chase them down with the rest, by getting a computerized proof check of my work. And I also told Ferry to look for a job outside of mathematics. Now, I'm not so angry, and I think that maybe I shouldn't have sent that email, but then again, it all revolves around the question of whether or not Jim Ferry does understand the math. > My work is out there and rather easy to go over as can be seen at the > Hong Konk math site: See http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782 And I send people there because their allowal of the use of LaTeX > makes for a *much* better presentation, and given the *social* issues > I'm facing, I need all the help I can get. > The social issues. Indeed. What help could I be to you with the > mathematical issues? Even herd mathematicians like Magidin are better > at math than I. The fact is, you don't *need* any help with the > mathematical issues. Ah yes, rather prescient of you to emphasize need. It seems to me that things are about to get interesting. > I'm still working on your Plan. I've very busy right now, but I'm > finding ways to make the time for it. It will take several weeks > at least. As I began to conceive/receive it, I realized that it > would do no good if you had a Bush-like outlook on life. You seemed > annoyed by my detour into politics, but this is what I mean: you > being concerned with *mathematical* issues is like Bush trying to > build bigger bombs to fight terrorism. America is already powerful > enough, and doesn't need to become all-mighty so that all nations of > the world cower at the thought of incurring its wrath. According to > Bush, terrorists and the people who cheer for them are insane and evil. > According to Bush, a vast segment of the Islamic world is insane and evil. Bush is having issues with his father, so he reversed his father on everything. Bush Sr. said he wouldn't raise taxes and did. So Bush the younger lowers them repeatedly, even when it's cuckoo. Bush Sr. didn't invade Iraq beyond the UN resolutions, so Bush the younger flouts the UN and invades the country. Bush Sr. refuted Saddam Hussein for invading a country and Bush the younger emulated him by invading against international law. It's not as complicated as you put it Jim Ferry, as it's just a rather dysfunctional family playing out its issues on the world stage. It's happened before, I'm sure. > But George Bush is simpleminded, so perhaps we should forgive him his > simplistic viewpoint. Hatred for America is engendered by . . . America's > better teeth. Now here's what I want you to do . . . Only a suggestion, > mind you: I'm all for freedom and human rights and democracy (my ideas > are also better than yours), so do whatever you want. What? What are you > doing? Now I have to beat the crap out of you! Oh, boo hoo, you want > your natural resources back? Too late, I ate them. Ha ha. Hey! Don't > hit me there! Evil terrorist! I have a date with Brittany Spears tonight! Bush isn't simpleminded and in fact he's quite intelligent, but lazy. I think part of what he does is out of anger at his father, and the rest is just following along with Cheney, who appears to be, um, not quite sane. > I'm not saying that you've installed any puppet dictatorships, James, but > there are aspects to America's foreign policy arrogance that remind me of > you. Sure, it's a natural reaction to being better than everyone else. > Better at building weapons, better at doing math, easy to think better in > every way. Easy to wonder, I'm so great, but they all hate me . . . they > must be wrong. They must be evil. Well Bush and I seem to have similarities, which I've noticed to my chagrin. I've decided that it comes from my being brought up as a fundamentalist Christian, so I understand that rigid thinking. > This is why I doubt that working out your Plan is even worth the effort. > No one likes being told they have to change their behavior to achieve their > goals. My guide (or muse, or the Holy Spirit, or whatever it is that seems > to be sending this Plan to me) argues that because you so fervently want to > achieve your goal and because you realize that what you're doing isn't > working, that you are receptive to new ideas. But I think your attitude is > more like, Let Ferry write his little plan, and if I don't like it, I'll > just laugh at it. So, yes, I'm resisting having to do this pointless work, > but still the Holy Spirit (or guide, or muse, or whatever) is prodding me on. Uh oh. Maybe I need to help you out a bit, in case you're losing it. Human beings are in the object ring. You're just a pack of ideas, in one sense. As ideas, you can be infected by ideas, like in the movie The Matrix. It turns out that very complex ideas, like human beings, exist in other forms, and they can infect a human mind. Thing is, such infections are rather hard to fix, as it's not so easy like in that movie. I'll consider your future replies Jim Ferry, and I may cut you loose. My suggestion, see someone, like a priest (not a Catholic priest), or even better, a Zen master. > So please, just tell me, Oh, you want me to roll over and play Mr. Nice > Guy! No dice! and then I can forget this whole hassle. At first I was so > honored that the Holy Spirit was choosing to work through me . . . oh, it's > probably a load of rubbish anyway. Me wanting to be important and imagining > that I was getting divine inspiration. The whole thing just seems ridiculous. > Never mind. I understand the feeling, and I take you seriously. You need to talk to someone, and not on Usenet. > Some of you are now facing the reality of the human brain versus any > fantasy you might have had about being completely rational. Human > beings are NOT rational creatures but necessarily rely on social > forces to determine what they believe. You are creatures of society. You may have believed that your mathematical knowledge was based > completely on logic and rationality, but human beings don't work that > way; it's built-in to your wiring NOT to work that way. Some of you must learn to be more than human. > You know, I don't think genetics come into it. I'm in no position to > know, but here's what the little voice in my head says: > The chief difference between James Harris's and the Establishment's > mathematical systems lies not in the validity of each: rather it is > simply that the world has chosen to accept the Establishment viewpoint. > Reality is created by consciousness more than you know. James Harris > is attempting to create a new Reality to displace the old, but his > arguments simply *do not pertain* within the current Reality. The > current Reality is flawed, of course, by Goedelian indeterminancies, > and is ripe for replacement by something superior . . . Yop, you're infected Jim Ferry. > Yeah. Sounds like a load of crap. Like Langan's CTMU consciousness > creates the world crap. Forget it. > Mathematics is an absolute truth. Cbeck. That's Mathematics as opposed to mathematics. What most people call mathematics is a body of discoveries by people like me--discoverers. It can be flawed. But Mathematics is perfect. > The physical world is what it is. Check. The physical world is a finite bit of Mathematics. And like I said above, human beings are in the object ring. Yup, people, you are mathematical objects, just highly complicated ones. > I apologize for all my inconsistencies. I started writing this in one > frame of mind, and ended up in another. A little voice in my head > telling me how James Harris can conquer the (mathermatical) world? > People are going to start telling *me* to take my meds . . . Go to a Zen master. A good one can handle that mental virus you have. > You must learn to be truly rational, for the first times in your > lives. So it's about time, as I wait, and wonder, how many of you can handle > the truth. And how many of you prefer the fantasy which was the world you > believed in, which actually never existed, except in your > imaginations; your wishes for a nicer world, where your wishes matter. > James Harris > My imagination. Sigh. I'm sure that's all it was. Sorry for wasting > your time. Now that was interesting. James Harris === Subject: Re: JSH: About time > Now that I've revealed the odd and you could say esoteric error in > core mathematics with such a short, and rather simple argument, the > issue now is how long until mathematicians decide that they'd rather > have correct mathematics versus the *belief* that they had been > perfect in keeping error out of the collected body of work that is > called mathematics. > Really, I can not understand your work, not even the core error problem. > I apologize for replying in a public forum, but I've chosen, or been > guided, to work under certain constraints. The danger of e-mail is the > temptation to say different things to different people. It would be > better for me, professionally, to hide my support for you, but I refuse > to do it. > Don't worry about it, as I find I'm liking your posts. Actually, I > kind of got a kick out of some of your previous posts back when you > were insulting me. > My thinking is that you appreciate quality work when you see it. > Now for those who wonder, being rather pissed at my current > predicament, with *some* mathematicians--I have to remember not to > characterize all of them--lying about my work or running away from it, > I sent Jim Ferry and Keith Ramsay a testy email, where I informed them > I considered them to be runners as well, and I would chase them down > with the rest, by getting a computerized proof check of my work. > And I also told Ferry to look for a job outside of mathematics. > Now, I'm not so angry, and I think that maybe I shouldn't have sent > that email, but then again, it all revolves around the question of > whether or not Jim Ferry does understand the math. > My work is out there and rather easy to go over as can be seen at the > Hong Konk math site: See http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782 And I send people there because their allowal of the use of LaTeX > makes for a *much* better presentation, and given the *social* issues > I'm facing, I need all the help I can get. > The social issues. Indeed. What help could I be to you with the > mathematical issues? Even herd mathematicians like Magidin are better > at math than I. The fact is, you don't *need* any help with the > mathematical issues. > Ah yes, rather prescient of you to emphasize need. > It seems to me that things are about to get interesting. > I'm still working on your Plan. I've very busy right now, but I'm > finding ways to make the time for it. It will take several weeks > at least. As I began to conceive/receive it, I realized that it > would do no good if you had a Bush-like outlook on life. You seemed > annoyed by my detour into politics, but this is what I mean: you > being concerned with *mathematical* issues is like Bush trying to > build bigger bombs to fight terrorism. America is already powerful > enough, and doesn't need to become all-mighty so that all nations of > the world cower at the thought of incurring its wrath. According to > Bush, terrorists and the people who cheer for them are insane and evil. > According to Bush, a vast segment of the Islamic world is insane and evil. > Bush is having issues with his father, so he reversed his father on > everything. > Bush Sr. said he wouldn't raise taxes and did. So Bush the younger > lowers them repeatedly, even when it's cuckoo. > Bush Sr. didn't invade Iraq beyond the UN resolutions, so Bush the > younger flouts the UN and invades the country. > Bush Sr. refuted Saddam Hussein for invading a country and Bush the > younger emulated him by invading against international law. > It's not as complicated as you put it Jim Ferry, as it's just a rather > dysfunctional family playing out its issues on the world stage. > It's happened before, I'm sure. > But George Bush is simpleminded, so perhaps we should forgive him his > simplistic viewpoint. Hatred for America is engendered by . . . America's have > better teeth. Now here's what I want you to do . . . Only a suggestion, > mind you: I'm all for freedom and human rights and democracy (my ideas > are also better than yours), so do whatever you want. What? What are you > doing? Now I have to beat the crap out of you! Oh, boo hoo, you want > your natural resources back? Too late, I ate them. Ha ha. Hey! Don't > hit me there! Evil terrorist! I have a date with Brittany Spears tonight! > Bush isn't simpleminded and in fact he's quite intelligent, but lazy. > I think part of what he does is out of anger at his father, and the > rest is just following along with Cheney, who appears to be, um, not > quite sane. > I'm not saying that you've installed any puppet dictatorships, James, but > there are aspects to America's foreign policy arrogance that remind me of > you. Sure, it's a natural reaction to being better than everyone else. > Better at building weapons, better at doing math, easy to think better in > every way. Easy to wonder, I'm so great, but they all hate me . . . they > must be wrong. They must be evil. > Well Bush and I seem to have similarities, which I've noticed to my > chagrin. > I've decided that it comes from my being brought up as a > fundamentalist Christian, so I understand that rigid thinking. > This is why I doubt that working out your Plan is even worth the effort. > No one likes being told they have to change their behavior to achieve their > goals. My guide (or muse, or the Holy Spirit, or whatever it is that seems > to be sending this Plan to me) argues that because you so fervently want to > achieve your goal and because you realize that what you're doing isn't > working, that you are receptive to new ideas. But I think your attitude is > more like, Let Ferry write his little plan, and if I don't like it, I'll > just laugh at it. So, yes, I'm resisting having to do this pointless work, > but still the Holy Spirit (or guide, or muse, or whatever) is prodding me on. > Uh oh. Maybe I need to help you out a bit, in case you're losing it. > Human beings are in the object ring. > You're just a pack of ideas, in one sense. > As ideas, you can be infected by ideas, like in the movie The > Matrix. > It turns out that very complex ideas, like human beings, exist in > other forms, and they can infect a human mind. > Thing is, such infections are rather hard to fix, as it's not so easy > like in that movie. > I'll consider your future replies Jim Ferry, and I may cut you loose. > My suggestion, see someone, like a priest (not a Catholic priest), or > even better, a Zen master. > So please, just tell me, Oh, you want me to roll over and play Mr. Nice > Guy! No dice! and then I can forget this whole hassle. At first I was so > honored that the Holy Spirit was choosing to work through me . . . oh, it's > probably a load of rubbish anyway. Me wanting to be important and imagining > that I was getting divine inspiration. The whole thing just seems ridiculous. > Never mind. > I understand the feeling, and I take you seriously. You need to talk > to someone, and not on Usenet. > Some of you are now facing the reality of the human brain versus any > fantasy you might have had about being completely rational. Human > beings are NOT rational creatures but necessarily rely on social > forces to determine what they believe. You are creatures of society. You may have believed that your mathematical knowledge was based > completely on logic and rationality, but human beings don't work that > way; it's built-in to your wiring NOT to work that way. Some of you must learn to be more than human. > You know, I don't think genetics come into it. I'm in no position to > know, but here's what the little voice in my head says: > The chief difference between James Harris's and the Establishment's > mathematical systems lies not in the validity of each: rather it is > simply that the world has chosen to accept the Establishment viewpoint. > Reality is created by consciousness more than you know. James Harris > is attempting to create a new Reality to displace the old, but his > arguments simply *do not pertain* within the current Reality. The > current Reality is flawed, of course, by Goedelian indeterminancies, > and is ripe for replacement by something superior . . . > Yop, you're infected Jim Ferry. > Yeah. Sounds like a load of crap. Like Langan's CTMU consciousness > creates the world crap. Forget it. > Mathematics is an absolute truth. Cbeck. > That's Mathematics as opposed to mathematics. > What most people call mathematics is a body of discoveries by people > like me--discoverers. > It can be flawed. But Mathematics is perfect. > The physical world is what it is. Check. > The physical world is a finite bit of Mathematics. > And like I said above, human beings are in the object ring. > Yup, people, you are mathematical objects, just highly complicated > ones. > I apologize for all my inconsistencies. I started writing this in one > frame of mind, and ended up in another. A little voice in my head > telling me how James Harris can conquer the (mathermatical) world? > People are going to start telling *me* to take my meds . . . > Go to a Zen master. A good one can handle that mental virus you have. > You must learn to be truly rational, for the first times in your > lives. So it's about time, as I wait, and wonder, how many of you can handle > the truth. And how many of you prefer the fantasy which was the world you > believed in, which actually never existed, except in your > imaginations; your wishes for a nicer world, where your wishes matter. > James Harris > My imagination. Sigh. I'm sure that's all it was. Sorry for wasting > your time. > Now that was interesting. > James Harris James, What do you expect to accomplish with all these insults you're dishing out? Not like you'll answer me anyway. These insults are childish, grow up. David Moran === Subject: Re: JSH: About time >>Now that was interesting. >>James Harris > James, What do you expect to accomplish with all these insults you're > dishing out? Not like you'll answer me anyway. These insults are childish, > grow up. > David Moran ??? Were you and I reading the same post? James reply was insightful and fascinating. Sure, he and I disagree on a few points, but I don't understand how you could categorize it as insulting. It was actually quite helpful to me in my distress. Maybe James's and my upbringings are the root of some of our differences of opinion. James was brought up in a fundamentalist Christian household; my household was a jumble of various and/or no beliefs. My struggle is to find attunement with God in an empty, profane world. James probably has to work to shut out that oppressive, thundering, Old Testament Jehovah. Thus, whereas I welcome an inspiration of the Holy Spirit as fresh air, James seems to see such things as infections, mental diseases. Now that I see where you're coming from, James, let me say that I'm not oppressed by thunder, nor looking to be cured. Rather, I'm straining to hear something like far off music, like elven-song, and am amazed that I can still hear it if I try. I fear that it will abandon me if I fail to honor it, and that it would be a terrible shame to waste it. Though it merely moves through me toward you, when it moves through me, I am purified. I see now that it comes to me because you could never hear it, perhaps because of Jehovah's thunder, but more simply because of the hammer and tongs of your mathworks. === Subject: Re: JSH: About time >[...] >James, What do you expect to accomplish with all these insults you're >dishing out? It's an interesting question. Sometimes I think that he's actually going to get people to agree he's right this way: You're wrong. Idiot. Well, you're still wrong. Liar. No, agreeing you were right would be lying. Satan's going to kill you. Oh my, I just can't tolerate these insults! Ok, you're right. I mean it's hard to believe anyone could think that it's going to work out that way, but he does believe lots of things that it's hard to believe anyone would believe... >Not like you'll answer me anyway. These insults are childish, >grow up. >David Moran === Subject: Re: JSH: About time >>[...] >>James, What do you expect to accomplish with all these insults you're >>dishing out? >It's an interesting question. Sometimes I think that he's actually >going to get people to agree he's right this way: >You're wrong. >Idiot. >Well, you're still wrong. >Liar. >No, agreeing you were right would be lying. >Satan's going to kill you. >Oh my, I just can't tolerate these insults! Ok, you're right. >I mean it's hard to believe anyone could think that it's going >to work out that way, but he does believe lots of things that >it's hard to believe anyone would believe... But that's [bullying until somebody cried Uncle] is a successful tactic among children. MostW[sigh!]some people figure out this doesn't work to one's advantage when they attain adulthood. It is bothersome that elementary schooling encourages this kind of persistent behaviour; teachers aren't supposed to correct mistakes because that may bruise the little egos and not build their confidence. /BAH Subtract a hundred and four for e-mail. === Subject: Re: JSH: About time >[...] >James, What do you expect to accomplish with all these insults you're >dishing out? > It's an interesting question. Sometimes I think that he's actually > going to get people to agree he's right this way: > You're wrong. > Idiot. > Well, you're still wrong. > Liar. > No, agreeing you were right would be lying. > Satan's going to kill you. > Oh my, I just can't tolerate these insults! Ok, you're right. > I mean it's hard to believe anyone could think that it's going > to work out that way, but he does believe lots of things that > it's hard to believe anyone would believe... >Not like you'll answer me anyway. These insults are childish, >grow up. >David Moran > ************************ > David C. Ullrich Sometimes I wonder what his mathematical background is. He strikes me as someone who has had very little. If he actually has a physics degree, you would think that there'd be enough math there to set him straight; I'm a physics major myself (Meteorology). David Moran === Subject: Re: JSH: About time > Sometimes I wonder what his mathematical background is. He strikes me as > someone who has had very little. If he actually has a physics degree, you > would think that there'd be enough math there to set him straight; I'm a > physics major myself (Meteorology). You can see that he doesn't get it at all if you mention anything that isn't simple arithmetic or that he hasn't defined himself. I'd say his mathematical background is extremely limited. The fact that he is completely unwilling (incapable?) of learning makes things worse. === Subject: Re: JSH: About time >>[...] >>James, What do you expect to accomplish with all these insults you're >>dishing out? >> It's an interesting question. Sometimes I think that he's actually >> going to get people to agree he's right this way: >> You're wrong. >> Idiot. >> Well, you're still wrong. >> Liar. >> No, agreeing you were right would be lying. >> Satan's going to kill you. >> Oh my, I just can't tolerate these insults! Ok, you're right. >> I mean it's hard to believe anyone could think that it's going >> to work out that way, but he does believe lots of things that >> it's hard to believe anyone would believe... >>Not like you'll answer me anyway. These insults are childish, >>grow up. >>David Moran >> ************************ >> David C. Ullrich >Sometimes I wonder what his mathematical background is. None even though he might have received As in el-hi. I have a nephew who has been told all his young life that he knows a lot about computers. He knows nothing except how to point and click very fast. He has grown up thinking he knows it all just because he knew how to maniputate a mouse quickly. > .. He strikes me as >someone who has had very little. If he actually has a physics degree, you >would think that there'd be enough math there to set him straight; I'm a >physics major myself (Meteorology). My father claims he majored in history. His highest education level is 12th grade. When I went to high school, they also used the major word. It has nothing to do with the discipline associated with university terms. I suspect that some people may equate a high school major with a college-level degree. Did he really say physics degree or did you extrapolate that from him saying physics major? /BAH Subtract a hundred and four for e-mail. === Subject: Re: JSH: About time > My father claims he majored in history. His highest education level > is 12th grade. When I went to high school, they also used the > major word. It has nothing to do with the discipline associated > with university terms. I suspect that some people may equate > a high school major with a college-level degree. Did he really > say physics degree or did you extrapolate that from him saying > physics major? He claims to have a BS in Physics from Vanderbilt. I give him the benefit of the doubt on that. I can assure you that you can get a BS in physics without ever seeing math at a very abstract level. He also was apparently in gifted-talented programs as a youngster and had a lot of people telling him how smart he was. He seems unable to comprehend that a lot of the people in the newsgroups he frequents had the same life experiences, but managed to make the transition from big fish in little pond to little fish in big pond which, from all evidence, he did not. - Randy === Subject: Re: JSH: About time > He also was apparently in gifted-talented programs as a > youngster and had a lot of people telling him how smart > he was. He seems unable to comprehend that a lot of the > people in the newsgroups he frequents had the same life > experiences, but managed to make the transition from big fish > in little pond to little fish in big pond which, from > all evidence, he did not. Maybe he made a different transition. Out of the water . . . === Subject: Re: JSH: About time [...] James, What do you expect to accomplish with all these insults you're >dishing out? > It's an interesting question. Sometimes I think that he's actually > going to get people to agree he's right this way: > You're wrong. > Idiot. > Well, you're still wrong. > Liar. > No, agreeing you were right would be lying. > Satan's going to kill you. > Oh my, I just can't tolerate these insults! Ok, you're right. > I mean it's hard to believe anyone could think that it's going > to work out that way, but he does believe lots of things that > it's hard to believe anyone would believe... >Not like you'll answer me anyway. These insults are childish, >grow up. David Moran ************************ > David C. Ullrich >>Sometimes I wonder what his mathematical background is. >None even though he might have received As in el-hi. I have >a nephew who has been told all his young life that he >knows a lot about computers. He knows nothing except how >to point and click very fast. He has grown up thinking he >knows it all just because he knew how to maniputate a mouse >quickly. Presumably he doesn't know how to find usenet... >> .. He strikes me as >>someone who has had very little. If he actually has a physics degree, you >>would think that there'd be enough math there to set him straight; I'm a >>physics major myself (Meteorology). >My father claims he majored in history. His highest education level >is 12th grade. When I went to high school, they also used the >major word. It has nothing to do with the discipline associated >with university terms. I suspect that some people may equate >a high school major with a college-level degree. Did he really >say physics degree or did you extrapolate that from him saying >physics major? He definitely has a degree in physics (or at least that's the story). Every once in a while it's explained that this is why we should believe he's right. >/BAH >Subtract a hundred and four for e-mail. === Subject: Re: JSH: About time >>[...] >>James, What do you expect to accomplish with all these insults you're >>dishing out? >> It's an interesting question. Sometimes I think that he's actually >> going to get people to agree he's right this way: >> You're wrong. >> Idiot. >> Well, you're still wrong. >> Liar. >> No, agreeing you were right would be lying. >> Satan's going to kill you. >> Oh my, I just can't tolerate these insults! Ok, you're right. >> I mean it's hard to believe anyone could think that it's going >> to work out that way, but he does believe lots of things that >> it's hard to believe anyone would believe... >>Not like you'll answer me anyway. These insults are childish, >>grow up. >>David Moran >> ************************ >> David C. Ullrich >Sometimes I wonder what his mathematical background is. > None even though he might have received As in el-hi. I have > a nephew who has been told all his young life that he > knows a lot about computers. He knows nothing except how > to point and click very fast. He has grown up thinking he > knows it all just because he knew how to maniputate a mouse > quickly. > .. He strikes me as >someone who has had very little. If he actually has a physics degree, you >would think that there'd be enough math there to set him straight; I'm a >physics major myself (Meteorology). > My father claims he majored in history. His highest education level > is 12th grade. When I went to high school, they also used the > major word. It has nothing to do with the discipline associated > with university terms. I suspect that some people may equate > a high school major with a college-level degree. Did he really > say physics degree or did you extrapolate that from him saying > physics major? > /BAH > Subtract a hundred and four for e-mail. I remember him saying he had a physics degree, however, I could claim I have a math degree and you wouldn't know; that's the internet for you. By the way I don't have a math degree. David Moran === Subject: qual vs quan research boundary=----=_NextPart_000_0034_01C391E6.5E7AD640 --------------------------------------------------------------------- #1 - Is environmental dumping more likely to occur in impoverished neighborhoods compared to middle neighborhoods? (I think thit is quantative question) #2 - Do children raised in single parent families have greater academic difficulties than children raised in two parent families? (I think thit is quantative question) #3 - How does motivation lead to success in the workforce? (I think thit is qualatative question) === Subject: Re: qual vs quan research > following questions and have to state whether they are qualatative or > quantative. HEre are the questions and what I think they are. Can anyone > tell me if I'm correct? THANKS! > #1 - Is environmental dumping more likely to occur in impoverished > #neighborhoods compared to middle neighborhoods? (I think thit is > #quantative question) > #2 - Do children raised in single parent families have greater academic > #difficulties than children raised in two parent families? (I think thit > #is quantative question) > #3 - How does motivation lead to success in the workforce? (I think thit > #is qualatative question) -- I'm not speaking as an expert here but this is what I think. Third is definately qualitative. The first two could be considered quantitative because the contain comparative words like more but I consider them qualitative because they are really questions of cause. Nobody really cares whether more or less of something happens here or there by accident. Any satisfactory answer to these questions would have to suggest a reason why more or less of something happens here or there. Besides, for a quantitative question, I would expect something more specific than more or less. I would expect an equation or something. Have a tolerable existence. Eli === Subject: Re: Minimal Graph, Four Color Theorem I'd hesitate to ask you to go there, but it's Harris Steven James' 10-year program to prove the last theorem of Fermat. of course, it was actually his first one! he's wroking on the definition of divisibility, now. > I hesitate to ask, but what is HSJ? --les ducs de Enron! http://larouchepub.com === Subject: Re: Quadratic System Help from slick_shoes@punkass.com: >18=a(5)2+b(5)+c >69=a(12)2+b(12)+c >74=a(14)2+b(14)+c > I'm not sure how to go about solving this. In class, we've only solved >systems like this when on of the equations had x=0, so it rather easy to >solve for c. Can anybody point me in the right direction on this one? >slick_shoes My earlier response never entered the newgroup, so I try again: You have three equations and three unknown values, now a, b, and c. If you understand a little about matrices, then you already can figure out what..... but you are asking for help, so you probably know at most, parts of intermediate algebra. If this is so, then either use an equation to find expression of one variable in terms of the others and back substitute; or use elimination by adding or subtracting multiples of one equation to another. G C === Subject: Re: Quadratic System Help > from slick_shoes@punkass.com: >18=a(5)2+b(5)+c >69=a(12)2+b(12)+c >74=a(14)2+b(14)+c > I'm not sure how to go about solving this. One way is as a system of linear equations in a,b,and c as unknowns. 25*a + 5*b + c = 18 144*a + 12*b + c = 69 196*a + 14*b + c = 74 Another is by row reduction of the augmented matrix [[ 25 5 1 18 ] [ 144 12 1 69 ] [ 196 14 1 74 ]] === Subject: harmonic functions How do you prove: Suppose D is a connected domain and {f_n} is a sequence of non-negative harmonic functions on D. Then we have one the following: (1) The sequence {f_n} contains a subsequence diverging to infinitely pointwise on D, (2) The sequence {f_n} contains a subsequence converging uniformly on compact subsets of D. My intuition is that all that is needed is Harnack's inequality (in one form or another) and the Arzela-Ascoli theorem, but I can't seem to combine them in just the right way... === Subject: Re: harmonic functions >How do you prove: >Suppose D is a connected domain and {f_n} is a sequence of non-negative >harmonic functions on D. Then we have one the following: >(1) The sequence {f_n} contains a subsequence diverging to infinitely >pointwise on D, >(2) The sequence {f_n} contains a subsequence converging uniformly on >compact subsets of D. >My intuition is that all that is needed is Harnack's inequality (in one form >or another) and the Arzela-Ascoli theorem, but I can't seem to combine them >in just the right way... The fact that you seem to know what's needed to prove this makes it smell like homework - you can't quite figure out how to use the hint. So I'll give a few more hints: First, a relevant form of Harnack's inequality would be this: For any compact set K in D and any point p in D there is a constant c such that u(x) <= c u(p) for all x in K. Second, either the family is unbounded at some point or it isn't. Third, if K1, K2 are compact and K1 is contained in the interior of K2 then there exists c such that the sup of grad(u) over K1 is less than c times the sup of u over K2. (Note that bounds on grad(u) imply equicontinuity...) === Subject: inversion via lagrange theory hello, is it possible to invert something of a form say... f(t) = G(f(t)) + H(f(t),t) (G, H arbitrary functions, where as shown, H is explicitly dependant on t.) via say the lagrange inversion theorem to solve for f(t) ? (im not too familiar with the theorem, which is why i am asking...however on inspection of the theorem i would assume it is not possible....is there a theorem that allows for its solution?) cheers moth === Subject: Re: Value of PI using trigonometry and calculus yess, thats pure calculus way to prove !! in my earlier post same has been proved using geometry too. atul > pi = lim n*sin(180/n) > n->inf > lim(n->oo) n*sin 180/n = 180 lim(n->oo) (sin 180/n)/(180/n) > = 180 lim(x->0) (sin x)/x > = 180 provided x is in radians > I believe this thing is known and exists. > Basic calculus > lim(x->0) (sin x)/x = 1 === Subject: Re: Value of PI using trigonometry and calculus William, What I was saying, was correct. Finally I found it in one of the answers on Dr. Math website. It reads as follows : === Subject: Re: Pi and Polygons This method of finding successive approximations to Pi is one of the oldest methods known, because it is one of the easiest to understand, and you can draw nice pictures to explain it. If a regular polygon has n sides, then we can draw lines from its center to all the vertices, and these lines will divide the pie-shaped picture into n wedges. Each wedge's central angle will have a measure of 360/n degrees. So, we can draw this picture of one wedge, where C is at the center of the polygon: A /| / | / | / | / | C /__________| D | | | | | | B Segment AB here is one side of the original regular polygon. Since angle ACB is 360/n, angle ACD is 180/n. Therefore, if the length of AC is 1/2, the length of AD is sin(180/n)/2. Therefore, the length of AB is sin(180/n), and the perimeter of the entire polygon is n*sin(180/n). So, you are correct: when you let n go toward infinity, sin(180/n) will tend towards zero. Since we know that a circle whose radius is 1/2 has a circumference of Pi, the 0 and infinity balance each other in the limit. Of course, there is a practical problem with all of this. In order to calculate the sines, you need to know a thing or two about Pi. It is true that for some special angles like 30, 45, and 60 degrees (and their sums and differences, etc.) you can write down an explicit elementary expression for their sines, this is not true of some other angles like 180/11. So, how do we calculate the perimeters without knowing Pi already? We need to find some trick, or we need to find some other method entirely of approximating Pi. And that is where a great part of the glorious history of mathematics starts. For more about Pi, its role in history, and the various attempts to know it better, check out the excellent book _A History of Pi_ by Petr Beckmann. People have done some pretty clever things to get to know Pi. - Doctor Ken, The Math Forum http://mathforum.org/dr.math/ cheers, Atul > pi = lim n*sin(180/n) > n->inf > lim(n->oo) n*sin 180/n = 180 lim(n->oo) (sin 180/n)/(180/n) > = 180 lim(x->0) (sin x)/x > = 180 provided x is in radians > I believe this thing is known and exists. > Basic calculus > lim(x->0) (sin x)/x = 1 === Subject: JVC Anyone who knows the difference between the Jonker Volgenant (JV) and the Jonker Volgenant Castanion (JVC) Algorithm? === Subject: Re: Groups with 16 elements > I found an interesting program for doing things > with groups of order < 32 at http//math.ucsd.edu/~jwavrick . > program (or using telnet); [...] > ORDERS for Groups Number 35 and 38 > Group number 35 of Order 16 > 1 elements of order 1: A > 3 elements of order 2: C E G > 12 elements of order 4: B D F H I J K L M N O P > 0 elements of order 8: > 0 elements of order 16: > Both 35 and 38 have the same center; Z = {A E C G}. > This is the same distribution of orders as in Z_2 x Q, so > 35 or 38 is Z_2 x Q. > For 35 and 38, there are 6 = 12/2 elements of order 4, You mean 6 *cyclic subgroups* of order 4, not 6 elements. > and if we call these x_i; i=1,6, then let > x_12 = x_22, x_32 = x_42, x_52 = x_62, > which gives 3 elements of order 2, all of which are in the center. No. All elements of order 4 in Z_2 x Q square to the same element, so 2 of the 3 involutions in the center are not the square of an element of order 4. Likewise, only 2 of the 3 involutions in the center of the other non-abelian group with this distribution of element orders are the squares of elements of order 4. This group is the semidirect product Z_4 x| Z_4 with presentation . > I would like to look first at the 5 groups 34 thru 38 groups more > closely. (Note that none of these 5 have an element of order 8.) > Also, #36 and 37 have the same distribution of orders. > ORDERS for Group Number 36 > Group number 36 of Order 16 > 1 elements of order 1: A > 7 elements of order 2: C E G I K N P > 8 elements of order 4: B D F H J L M O > 0 elements of order 8: > 0 elements of order 16: > ORDERS for Group Number 37 > Group number 37 of Order 16 > 1 elements of order 1: A > 7 elements of order 2: C E G I K M O > 8 elements of order 4: B D F H J L N P > 0 elements of order 8: > 0 elements of order 16: > CENTER of Group Number 36 { A B C D } (cyclic--Z_4) > CENTER of Group Number 37 { A C E G } (Z_2 x Z_2) [...] > This shows the structure of 36, but I haven't figured out how to > write it concisely, as one does for D_8, e.g. This is the central product Z_4 * D_4 =~ Z_4 * Q that Derek mentioned. It has rank 3, and one easy presentation is . Group 37 has rank 2 with, for example, presentation . > I plan to look at the rest of the groups. I don't yet see > how one decides how many groups there are with each distribution of > orders. It didn't turn out quite like I thought--there are more > groups with no element of order 8 than I would have guessed. I know of no way to decide except to actually construct all of the groups of order 16 and count the elements of each order in each group. > There must be a systematic way of doing this--I know there are > papers, and I think some books, on groups of order 2n, but > I am not at a University, so I don't have the easy access > that I used to have. As I posted earlier in the thread, the systematic way is to look for all possible ways that 2 subgroups of order 8 can intersect, necessarily in a subgroup of order 4, such that one normalizes the other. You correctly started on this path when you looked at all the ways to have a cyclic subgroup of order 8. -- Jim Heckman === Subject: Re: JSH: $100,000 US offer, Abel Prize > Tee-hee. Nothing makes a pig happier than a roll in the wallow, as this tittering porker knows! http://www.shop4egifts.com/target.asp?item=/jpg/25112.jpg&width=314&height=4 00# > Heh-heh. Nothing makes a donkey happier than a rut in the barnyard, as this hee-hawing hoofer knows. http://www.rubylane.com/ni/shops/viperswife/iteml/GW-684 > David C. Ullrich === Subject: abstract math I was wondering if I could get some help on this problem. We know that multiplication of integers is commutative; i.e. ab=ba for all pairs of integers a and b. Prove that, for every natural number n, the product of n integers is independant of the order of the factors. === Subject: Re: abstract math >I was wondering if I could get some help on this problem. >We know that multiplication of integers is commutative; i.e. ab=ba for >all pairs of integers a and b. Prove that, for every natural number n, >the product of n integers is independant of the order of the factors. This requires associative as well. If f(a,b) = (a+b)2, this operation is commutative but not associative, and it is false, as f(1, f(1, 3)) = (1 + 10)2, and f(f(1,1), 3) = (4 + 3)2. For commutative and associative operations, it is an exercise in induction. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: abstract math > I was wondering if I could get some help on this problem. > We know that multiplication of integers is commutative; i.e. ab=ba for > all pairs of integers a and b. Prove that, for every natural number n, > the product of n integers is independant of the order of the factors. Hint: Given any permutation of ABC... normalize it by first moving A to the first place via transpositions then inductively doing the same on the tail after A, moving B to the 2nd place, C to the third place, etc. e.g. DCBA DCAB DACB ADCB now A is in normal place, work on rest DBC BDC now B is in normal place, work on rest CD now C is in normal place, so too is D ---- ABCD If you've studied permutation groups you'll recognize the relationship to the representation of a permutation as a product of transpositions. -Bill Dubuque === Subject: Re: abstract math >We know that multiplication of integers is commutative; i.e. ab=ba for >all pairs of integers a and b. Prove that, for every natural number n, >the product of n integers is independant of the order of the factors. I hate this type of homework: it is more difficult to neatly write down what it is that you have to prove [*] than to actually prove it. [*] For all natural numbers n, for all integers x_1,...,x_n, for all bijections f : {1,...,n} -> {1,...,n} x_1 * x_2 * ... * x_n = x_f(1) * x_f(2) * ... * x_f(n) and even then I should probably make clear where the parethesis are supposed to be in the products. Peter van Rossum === Subject: Re: abstract math > I was wondering if I could get some help on this problem. > We know that multiplication of integers is commutative; i.e. ab=ba for > all pairs of integers a and b. Prove that, for every natural number n, > the product of n integers is independant of the order of the factors. Mathematical Induction and the UFT. Bob Pease === Subject: Re: abstract math >> I was wondering if I could get some help on this problem. >> We know that multiplication of integers is commutative; i.e. ab=ba for >> all pairs of integers a and b. Prove that, for every natural number n, >> the product of n integers is independant of the order of the factors. > Mathematical Induction and the UFT. You don't need unique factorization. The property holds in any commutative ring, not just in a UFD. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: abstract math Amin schrieb im Newsbeitrag > I was wondering if I could get some help on this problem. > We know that multiplication of integers is commutative; i.e. ab=ba for > all pairs of integers a and b. Prove that, for every natural number n, > the product of n integers is independant of the order of the factors. Sorry for asking: But why would you want to prove this? I mean - I believe in it, that's enough for me. Don't want to offend you - I was just currious. -Gernot === Subject: Re: abstract math > Amin schrieb im Newsbeitrag > I was wondering if I could get some help on this problem. > We know that multiplication of integers is commutative; i.e. ab=ba > for > all pairs of integers a and b. Prove that, for every natural number > n, > the product of n integers is independant of the order of the > factors. > Sorry for asking: But why would you want to prove this? I mean - I > believe in it, that's enough for me. > Don't want to offend you - I was just currious. > -Gernot It is not so much offensive as indicative of potentially dangerous Snerdism. Would you post a message like this to a newsgroup like alt.religion.Christian? Sorry for asking: But why would you want to believe this? I mean - I believe in God, that's enough for me. Don't want to offend you - I was just currious. The CENTRAL reason for studying Math is to DO proofs. Whazzamatta you??? RJ Pease === Subject: Re: abstract math > The CENTRAL reason for studying Math is to DO proofs. I would just like to quote Laurent Schwartz : The reason for studying math is to discover some interesting properties, and to prove them just to be sure. Well, I would consider proofs as mere tools... Yann === Subject: Re: abstract math > Amin schrieb im Newsbeitrag > I was wondering if I could get some help on this problem. > We know that multiplication of integers is commutative; i.e. ab=ba > for > all pairs of integers a and b. Prove that, for every natural number > n, > the product of n integers is independant of the order of the > factors. > Sorry for asking: But why would you want to prove this? I mean - I > believe in it, that's enough for me. > Don't want to offend you - I was just currious. It's a standard approach in mathematics: Start from as few axioms as possible, and build things up. In this case, it shows you don't need to take equivalence of arbitrary permutation as an axiom, you just need to take commutation of pairs. And, I believe, associativity. To the OP: Try induction on n. Consider this at the induction step: (product of n-1 factors)*x Now, from your assumptions, you can do two thing without changing the product: - interchange x and the last element in (product of n-1 factors) - rearrange (product of n-1 factors) in any way you like. Can you convince yourself (and your teacher) that these two operations are sufficient to generate all permutations of n factors? Try it for 3 or 4 factors. If that's not sufficient, note that you can also do this operation: - regroup as (first factor)*(n-2 factors * x) and this: - regroup as (first factor)*(any permutation of n-2 factors and x) - Randy === Subject: Re: abstract math >Amin schrieb im Newsbeitrag >> I was wondering if I could get some help on this problem. >> We know that multiplication of integers is commutative; i.e. ab=ba for >> all pairs of integers a and b. Prove that, for every natural number n, >> the product of n integers is independant of the order of the factors. >Sorry for asking: But why would you want to prove this? I mean - I >believe in it, that's enough for me. >Don't want to offend you - I was just currious. This is only a guess. Beginning of school year = beginning of home work in class writing proofs. Can't see the answer immediately = ask the net for the answer. More advanced math work = believes only what can be proven. === Subject: Re: abstract math > I was wondering if I could get some help on this problem. > We know that multiplication of integers is commutative; i.e. ab=ba for > all pairs of integers a and b. Prove that, for every natural number n, > the product of n integers is independant of the order of the factors. Would an induction proof work? Bill === Subject: Intersection area of 2 circles assume 2 circles in 2D with given centers cx1, cy1 and cx2, cy2 and radiuses r1, r2. Now if dist(c1, c2) < r1+r2, the circles intersect. What I need is the intersection's center of mass (if 2d can have mass), and the area of the intersection. -- -Gernot In order to reply, revert my forename from: tonreG.Frisch.at.Dream-D-Sign.de@invalid.com ________________________________________ Looking for a good game? Do it yourself! GLBasic - you can do www.GLBasic.com === Subject: Re: Intersection area of 2 circles > assume 2 circles in 2D with given centers cx1, cy1 and cx2, cy2 and > radiuses r1, r2. > Now if dist(c1, c2) < r1+r2, the circles intersect. What I need is the > intersection's center of mass (if 2d can have mass), and the area of > the intersection. Draw the common chord and all the radii to its endpoints. The area of the intersection is a sector minus an isosceles triangle on each side of the chord. Add 'em up. I can't think of a much slicker way to do the CoM than the same construction. It's not very hard to work out the CoM of a sector[1] and the CoM of a triangle is its centroid, then use linearity. [1] if the sector has radius r and angle 2t then I make the CoM a distance (2 sin t)/(3t) from the centre. But that's just a quick guess. You are welcome :) Dave -- Remove the opinion on spam to reply. === Subject: Re: Intersection area of 2 circles >> assume 2 circles in 2D with given centers cx1, cy1 and cx2, cy2 and >> radiuses r1, r2. >> Now if dist(c1, c2) < r1+r2, the circles intersect. What I need is the >> intersection's center of mass (if 2d can have mass), and the area of >> the intersection. > Draw the common chord and all the radii to its endpoints. The area of the > intersection is a sector minus an isosceles triangle on each side of the > chord. Add 'em up. > I can't think of a much slicker way to do the CoM than the same > construction. It's not very hard to work out the CoM of a sector[1] and > the CoM of a triangle is its centroid, then use linearity. > [1] if the sector has radius r and angle 2t then I make the CoM a distance > (2 sin t)/(3t) from the centre. But that's just a quick guess. > You are welcome :) > Dave I'm not certain, but I would think that the center of mass would be the average of the centers of the two circles weighted by their areas, that is, ( (x1,y1)*pi*r12 + (x2,y2)*pi*r22 ) / ( pi*r12 + pi*r22 ) Have a tolerable existence. Eli -- === Subject: Re: Intersection area of 2 circles >> assume 2 circles in 2D with given centers cx1, cy1 and cx2, cy2 and >> radiuses r1, r2. >> Now if dist(c1, c2) < r1+r2, the circles intersect. What I need is the >> intersection's center of mass (if 2d can have mass), and the area of >> the intersection. > Draw the common chord and all the radii to its endpoints. The area of > the intersection is a sector minus an isosceles triangle on each side > of the chord. Add 'em up. > I can't think of a much slicker way to do the CoM than the same > construction. It's not very hard to work out the CoM of a sector[1] and > the CoM of a triangle is its centroid, then use linearity. > [1] if the sector has radius r and angle 2t then I make the CoM a > distance (2 sin t)/(3t) from the centre. But that's just a quick guess. > I'm not certain, but I would think that the center of mass would be the > average of the centers of the two circles weighted by their areas, that > is, ( (x1,y1)*pi*r12 + (x2,y2)*pi*r22 ) / ( pi*r12 + pi*r22 ) The center of mass of the intersection cannot be located as you thought. Consider, for example, a degenerate case when one disk is completely contained within another disk of larger radius. Then the center of mass of the intersection is always just the center of the smaller disk; it is independent of the center and radius of the larger disk. David Cantrell === Subject: Re: Intersection area of 2 circles assume 2 circles in 2D with given centers cx1, cy1 and cx2, cy2 and > radiuses r1, r2. > Now if dist(c1, c2) < r1+r2, the circles intersect. What I need is the > intersection's center of mass (if 2d can have mass), and the area of > the intersection. >> Draw the common chord and all the radii to its endpoints. The area of >> the intersection is a sector minus an isosceles triangle on each side >> of the chord. Add 'em up. >> I can't think of a much slicker way to do the CoM than the same >> construction. It's not very hard to work out the CoM of a sector[1] and >> the CoM of a triangle is its centroid, then use linearity. >> [1] if the sector has radius r and angle 2t then I make the CoM a >> distance (2 sin t)/(3t) from the centre. But that's just a quick guess. >> I'm not certain, but I would think that the center of mass would be the >> average of the centers of the two circles weighted by their areas, that >> is, ( (x1,y1)*pi*r12 + (x2,y2)*pi*r22 ) / ( pi*r12 + pi*r22 ) > The center of mass of the intersection cannot be located as you thought. > Consider, for example, a degenerate case when one disk is completely > contained within another disk of larger radius. Then the center of mass > of the intersection is always just the center of the smaller disk; it is > independent of the center and radius of the larger disk. > David Cantrell You're right. Oh, well. Have a tolerable existence. Eli -- === Subject: Re: Intersection area of 2 circles > assume 2 circles in 2D with given centers cx1, cy1 and cx2, cy2 and > radiuses r1, r2. > Now if dist(c1, c2) < r1+r2, the circles intersect. What I need is the > intersection's center of mass (if 2d can have mass), and the area of > the intersection. > Draw the common chord and all the radii to its endpoints. The area of the > intersection is a sector minus an isosceles triangle on each side of the > chord. Add 'em up. Or see formula (11) at . > I can't think of a much slicker way to do the CoM than the same > construction. It's not very hard to work out the CoM of a sector[1] and > the CoM of a triangle is its centroid, then use linearity. > [1] if the sector has radius r and angle 2t then I make the CoM a > distance (2 sin t)/(3t) from the centre. But that's just a quick guess. But of course that distance can't be independent of r. I suppose you intended to write (2 r sin t)/(3t); if so, your quick guess was correct. David Cantrell === Subject: Re: should Gauss's Law of Magnetism be: Integral B dot dA = q/p + q/p_t > So the new Gauss Law of Magnetism looks somewhat like this: > Integral B dot dA = q/p + q/p_t where t is temperature related. > This new law restores complete symmetry to the Maxwell Equations. It > dismisses the monopole idea completely. > It just says that we were just not advanced enough since the 1800s > when > the Gauss laws were built to realize that temperature and (probably > pressure as well) needs to be fitted into the Maxwell Equation > structure. > As you can see, the q/p + q/p_t is algebraically equivalent to u i + u > i_d > and thereby we have *complete symmetry* within the Maxwell Equations. I suspect this new Gauss Law of Magnetism would need a negative term and that the q/p_t term should be negative. Makes sense in that the Meissner Effect is a exclusion of a magnetic field inside the entire volume. And this issue would suggest that Graham Lee is wrong when he says the Maxwell Eq. conform to the Meissner Effect because the Gauss Law of Magnetism really does not conform to the Meissner Effect in that there is nothing in the Maxwell Equations to give a exclusion of magnetism from all volume. So the Maxwell Equations do indeed have a large gap in that there is nothing in the Maxwell Equations to account for a exclusion of magnetism from volume. If the above new Gauss Law of Magnetism is correct, whether it has a negative term or not, then, it should predict other phenomenon. And although I do not have temperature (and perhaps pressure also) included, I should be able to cast some sort of prediction just on form alone. So, let us say this new Gauss law of Magnetism, either Integral B dot dA = q/p + q/p_t or Integral B dot dA = q/p - q/p_t was correct. Then what sort of new prediction would that law give? Well, it should give some prediction concerning normal conductivity in that the magnetism inside a normal conductor such as copper at room temperature. That the internal magnetism of the term ( - q/p_t ) does not exclude an outside magnet from its volume and thus that term represents what physically? I suspect it represents resistance to the current flow. So that perhaps resistance to current flow is all contained within a revised Maxwell theory. Which makes sense, because if the true Maxwell Equations have a temperature and pressure parameters, then that translates into resistance. Archimedes Plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: predictions of a newly revised Gauss Law of Magnetism: Integral B dot dA = q/p - q/p_t Another prediction of this newly revised Gauss Law of Magnetism would be to give the numbers of the Quantized Hall Effect. And that makes sense on another level of logic. If I remember correctly, the algebraic form of spectral lines series such as Balmer or Rydberg follows a form such as this: 1/x - 1/y And this new Gauss law is not much different with its q/p - q/p_t and that makes alot of sense also in that spectral lines are not much different from Quantum Hall Effect, in fact, one can argue that the Quantum Hall Effect is spectral lines. So, if this above equation for Gauss Law of Magnetism is correct and true would predict that the math numbers that arise in Quantized Hall Effect should follow from this new equation. For we all know that superconductivity and Meissner Effect and Quantum Hall Effect are all linked. And going in the reverse direction of starting with Spectral line formulas, one should be able to derive a generalized Maxwell Equations, without ever knowing what they were at the outset. Archimedes Plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: The Passing of a Mathematician >In which particular way does the NSA keep alive the value of democracy? It helps protect the United States of America, a democratic nation, from being invaded and conquered by foreign non-democratic nations. John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: The Passing of a Mathematician > In which particular way does the NSA keep alive the > value of democracy? By protecting the security of the United States, a major democratic power, and its citizens. -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: polysigned numbers > Using my notation of (-,+,*,#), there are two possible reductions for > (a,b,c,d). I presume that in this notation (*1)2 = -1, (*1)(-1) = #1 and (*1)(#1) = +1 etc. > The tetrahedral interpretation uses: (a,b,c,d) = (a-m,b-m,c-m,d-m) where > m = min(a,b,c,d). > The tetrahedral maps to R3, with no obvious interpretation of > multiplication (to me), and addition corresponds to vector addition of > (i,j,k) points. I didn't work out the mapping in detail since you > posted it elsewhere. If my assumption is correct these 4-signed numbers form a ring isomorphic to R x C: *1 will correspond to (-1, i) etc. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: polysigned numbers >>Using my notation of (-,+,*,#), there are two possible reductions for >>(a,b,c,d). > I presume that in this notation (*1)2 = -1, (*1)(-1) = #1 and (*1)(#1) = +1 > etc. As Timothy has defined it, no. -1(a,b,c,d) = (d,a,b,c) +1(a,b,c,d) = (c,d,a,b) *1(a,b,c,d) = (b,c,d,a) #1(a,b,c,d) = (a,b,c,d) >>The tetrahedral interpretation uses: (a,b,c,d) = (a-m,b-m,c-m,d-m) where >>m = min(a,b,c,d). >>The tetrahedral maps to R3, with no obvious interpretation of >>multiplication (to me), and addition corresponds to vector addition of >>(i,j,k) points. I didn't work out the mapping in detail since you >>posted it elsewhere. > If my assumption is correct these 4-signed numbers form a ring > isomorphic to R x C: *1 will correspond to (-1, i) etc. No. You'll have to apply trig functions to get the exact values. You are going from norm 1 to norm sqrt(2). The norm is preserved under the isomorphism. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: polysigned numbers >Using my notation of (-,+,*,#), there are two possible reductions for >(a,b,c,d). >> I presume that in this notation (*1)2 = -1, (*1)(-1) = #1 and (*1)(#1) = >> +1 etc. >> > As Timothy has defined it, no. > -1(a,b,c,d) = (d,a,b,c) > +1(a,b,c,d) = (c,d,a,b) > *1(a,b,c,d) = (b,c,d,a) > #1(a,b,c,d) = (a,b,c,d) Bonkers: so #1 is the multiplicative identity not +1 :-( >The tetrahedral interpretation uses: (a,b,c,d) = (a-m,b-m,c-m,d-m) where >m = min(a,b,c,d). >The tetrahedral maps to R3, with no obvious interpretation of >multiplication (to me), and addition corresponds to vector addition of >(i,j,k) points. I didn't work out the mapping in detail since you >posted it elsewhere. >> If my assumption is correct these 4-signed numbers form a ring >> isomorphic to R x C: *1 will correspond to (-1, i) etc. > No. Yes: Still isomorphic to R x C. > You'll have to apply trig functions to get the exact values. You > are going from norm 1 to norm sqrt(2). The norm is preserved under the > isomorphism. Sorry, that makes no sense? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: polysigned numbers >>Using my notation of (-,+,*,#), there are two possible reductions for >>(a,b,c,d). >I presume that in this notation (*1)2 = -1, (*1)(-1) = #1 and (*1)(#1) = >+1 etc. >As Timothy has defined it, no. >>-1(a,b,c,d) = (d,a,b,c) >>+1(a,b,c,d) = (c,d,a,b) >>*1(a,b,c,d) = (b,c,d,a) >>#1(a,b,c,d) = (a,b,c,d) > Bonkers: so #1 is the multiplicative identity not +1 :-( Correct. >>The tetrahedral interpretation uses: (a,b,c,d) = (a-m,b-m,c-m,d-m) where >>m = min(a,b,c,d). >>The tetrahedral maps to R3, with no obvious interpretation of >>multiplication (to me), and addition corresponds to vector addition of >>(i,j,k) points. I didn't work out the mapping in detail since you >>posted it elsewhere. >If my assumption is correct these 4-signed numbers form a ring >isomorphic to R x C: *1 will correspond to (-1, i) etc. >>No. > Yes: Still isomorphic to R x C. >> You'll have to apply trig functions to get the exact values. You >>are going from norm 1 to norm sqrt(2). The norm is preserved under the >>isomorphism. > Sorry, that makes no sense? You're right, I had forgotten how multiplication in RxC works. So #1 = (1,1) What would -1,+1,*1 be? -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: JSH your ship has come in!!!! > But, I expect you remember that I did ask you a question that I was unable > to answer. > I was using the notation [a,b,c] to represent an ordered triple of complex > numbers. And I was wondering if the ordered triple [1,2,8] was an element of > the Object Ring. > You replied: > Quit being lazy!!! You have the definition, figure it out for yourself!!! > What amazes me is how often people are willing to ask someone else to do > their work for them. > If you're smart enough, answer your own question. > I'm curious to see if you can. > I've given the definition for the object ring, so no excuses. > Well, I am ashamed to say I still cannot figure it out. > Sounds like a ploy. James... I am having genuine problems with my question. I don't really see how it could be called a ploy. If you know the answer you could tell me, and put me out of my misery. If you don't know the answer, fair enough, you can't be expected to know everything. -- Clive Tooth http://www.clivetooth.dk === Subject: Re: JSH your ship has come in!!!! interpreted to mean that you are just humoring me. > Backtracked? After the initial jolt I went a little overboard, calling you the > messiah, etc. I've certainly backed off that stance. Please consider how > dizzying it is to be shaken out of the conventional mindset. You did? When? I thought you just had some wild and scary dream with me in a castle and stuff. Cool dream. Any more? > Well then, how about the core error problem? Can't you understand how independent terms are independent? > I just don't get it! How many times do I have to tell you! Your > mathematical writings make no sense to me! Are you going to be calling me a > retard next? You are quite intelligent, and in fact are one of the most intelligent people, by your IQ, on the planet. So no, I will not call you a retard Jim Ferry, as you are not one. > Oh, excuuuuse me for not being as smart as you, James! What happens to me in > the New World Order? Do you send me to the slaughterhouse with all the other > meat animals? Huh? Look around you Jim Ferry. There's more than enough slaughter going on now to distress the cognizant, and I assure you that I'm not doing any of it. My efforts to try and prevent some of it, like my emails about Iraq's supposed weapons of mass destruction stockpiles, failed. Other efforts of mine have failed as well, and I've learned not to carry the world on my shoulders, not even in fantasy. Sigh. I feel a bit guilty now. And I sent you that nasty email talking about putting you out of mathematics as well. Maybe there IS a problem with how I've gone about things. > Sure, I've got some trinkets of intellect: M.I.T. degree, some Putnam > medals, Ph.D. from Brown, joined Mega once, mathematician, rocket > scientist, etc. But that's a hell of long way from the serious hardware > that you're interested in: > Abel Prize, Fields Medal, Nobel Prize, Clay Prize, etc. And even the people > who have such hardware aren't necessarily able to follow in your bold > footsteps. > So why are you on *my* case? Because you *backtracked*!!! It would have been better if you'd said nothing at all!!! Now I have to toss you into the pile with Barry Mazur, Andrew Granville, Ralph McKenzie, David Ullrich, and Arturo Magidin. > I never claimed to understand your work. Maybe I had a fleeting flash of > intuition that it was correct, but that was all. Are you being honest Jim Ferry? Please explain to me why you ever claimed my work was correct *in detail*. I think you owe me that as I'm worried it was some kind of cruel prank. > I was offering a different kind of help. Something to complement and > complete the intense laser of your rational thought. Something to help > bring your theories into the human realm. You don't believe in the efficacy > of such things, I guess, or you just want to sear your way through with that > laser of yours, so fine. Go get 'em, tiger. You don't need my help. Actually I don't need your help, but I *would* appreciate it greatly. But how do I know if you're sincere, or just playing some elaborate joke? As for understanding my work, math proofs begin with a truth and proceed by logical steps to a conclusion which then must be true. I give you the steps, you follow, and then I'd think you should understand. James Harris === Subject: Re: JSH your ship has come in!!!! >[...] >> I just don't get it! How many times do I have to tell you! Your >> mathematical writings make no sense to me! Are you going to be calling me a > retard next? >You are quite intelligent, and in fact are one of the most intelligent >people, by your IQ, on the planet. Just curious: How do you know anything about his IQ? >So no, I will not call you a retard Jim Ferry, as you are not one. >> Oh, excuuuuse me for not being as smart as you, James! What happens to me in >> the New World Order? Do you send me to the slaughterhouse with all the other >> meat animals? >Huh? Look around you Jim Ferry. There's more than enough slaughter >going on now to distress the cognizant, and I assure you that I'm not >doing any of it. >My efforts to try and prevent some of it, like my emails about Iraq's >supposed weapons of mass destruction stockpiles, failed. >Other efforts of mine have failed as well, and I've learned not to >carry the world on my shoulders, not even in fantasy. Same hint as always: if you don't like it when... oh, never mind. >Sigh. I feel a bit guilty now. And I sent you that nasty email >talking about putting you out of mathematics as well. >Maybe there IS a problem with how I've gone about things. Giggle. Perhaps. >> Sure, I've got some trinkets of intellect: M.I.T. degree, some Putnam > medals, Ph.D. from Brown, joined Mega once, mathematician, rocket >> scientist, etc. But that's a hell of long way from the serious hardware >> that you're interested in: >> Abel Prize, Fields Medal, Nobel Prize, Clay Prize, etc. And even the people >> who have such hardware aren't necessarily able to follow in your bold >> footsteps. >> So why are you on *my* case? >Because you *backtracked*!!! >It would have been better if you'd said nothing at all!!! >Now I have to toss you into the pile with Barry Mazur, Andrew >Granville, Ralph McKenzie, David Ullrich, and Arturo Magidin. >> I never claimed to understand your work. Maybe I had a fleeting flash of >> intuition that it was correct, but that was all. >Are you being honest Jim Ferry? >Please explain to me why you ever claimed my work was correct *in >detail*. >I think you owe me that as I'm worried it was some kind of cruel >prank. For heaven's sake. Yes, when he says he doesn't understand your work but thinks it's correct he's joking. Regarding whether it's cruel: I would have never imagined in my wildest dreams that even _you_ could possibly be so blinded by megalomania and so desparate for approval that you'd think someone was being sincere when he says he doesn't understand your work but nonetheless thinks it's right. I mean that's such an _utterly_ ridiculous thing to say in a mathematical context that it's hard to see how anyone could possibly not take it as a joke. So I'm certain he assumed that you'd realize all along he was making fun of you. Making fun of you is perhaps a little nasty, but it's certainly no nastier than you deserve given your obnoxious nehavior for years, so it doesn't bother me. But lately when you start talking like you think he's being sincere it becomes a whole different story. Actually encouraging a lunatic's delusions is not a nice thing to do. On the other hand the fact that you've thought he was sincere is _the_ most hilarious thing I've ever seen - we need to consider the greatest good of the greatest number and all that, being a little nasty to you versus people all over the planet laughing their asses off, I think it's a wash. Honest. I point this out just in case you might decide to try not to be _quite_ so hilarious: People have been laughing at Jim's parodies for years, but lately when you seem to believe he's being sincere, or _might_ be sincere, in this and related threads, it really does add a whole new dimension - goes from hilarious to absolutely world-class hilarious. Now you know. >> I was offering a different kind of help. Something to complement and >> complete the intense laser of your rational thought. Something to help >> bring your theories into the human realm. You don't believe in the efficacy >> of such things, I guess, or you just want to sear your way through with that >> laser of yours, so fine. Go get 'em, tiger. You don't need my help. >Actually I don't need your help, but I *would* appreciate it greatly. >But how do I know if you're sincere, or just playing some elaborate >joke? >As for understanding my work, math proofs begin with a truth and >proceed by logical steps to a conclusion which then must be true. >I give you the steps, you follow, and then I'd think you should >understand. >James Harris === Subject: Re: JSH your ship has come in!!!! >>[...] >I just don't get it! How many times do I have to tell you! Your >mathematical writings make no sense to me! Are you going to be calling me a >retard next? >>You are quite intelligent, and in fact are one of the most intelligent >>people, by your IQ, on the planet. > Just curious: How do you know anything about his IQ? James is referring to the fact that I was able to join Mega, which purports to be a society for people whose IQ is at the one-in-a-million level or higher. In point of fact, Mega is a society for people who achieved high enough scores on certain tests -- in my case, the Mega Test. The idea that these tests actually test IQ (whatever that is), and that they have any validity to discern those denizens of such an aethereal realm as one-in-a-million seems laughable to me, as does the behavior of some of the people in Mega. It makes me a little embarrassed to be associated with them, but the Mega Test looked fun (half of it is math), so I took it, got a pretty high score (46 out of 48), and thought I'd check out their lofty Society. I don't claim to be all that smart, however. I know plenty of people smarter than I am. And James Harris . . . well, I can't hold a candle to him. >>So no, I will not call you a retard Jim Ferry, as you are not one. is so intimidating that I forget there are other ways in which I, perhaps, surpass you, rather than vice versa. It would be difficult to have an equable relationship with someone who exceeds me in every way. >Oh, excuuuuse me for not being as smart as you, James! What happens to me in >the New World Order? Do you send me to the slaughterhouse with all the other >meat animals? >>Huh? Look around you Jim Ferry. There's more than enough slaughter >>going on now to distress the cognizant, and I assure you that I'm not >>doing any of it. >>My efforts to try and prevent some of it, like my emails about Iraq's >>supposed weapons of mass destruction stockpiles, failed. Yes, that's too bad. >>Other efforts of mine have failed as well, and I've learned not to >>carry the world on my shoulders, not even in fantasy. As they say, Think globally, act locally. I hope that you don't give up on the world just because you can't save it in one fell swoop. And whatever you do, don't start reading Ayn Rand. Yikes. > Same hint as always: if you don't like it when... > oh, never mind. >>Sigh. I feel a bit guilty now. And I sent you that nasty email >>talking about putting you out of mathematics as well. >>Maybe there IS a problem with how I've gone about things. Yes. Too confrontational. Too alpha male. The hammer (that you use to shape the mathematics forged in your mind) is not the tool to wield in the human realm. > Giggle. Perhaps. Oh tee hee hee yourself. What's your problem, Ullrich? Oh, never mind. >Sure, I've got some trinkets of intellect: M.I.T. degree, some Putnam >medals, Ph.D. from Brown, joined Mega once, mathematician, rocket >scientist, etc. But that's a hell of long way from the serious hardware >that you're interested in: >Abel Prize, Fields Medal, Nobel Prize, Clay Prize, etc. And even the people >who have such hardware aren't necessarily able to follow in your bold footsteps. >So why are you on *my* case? >>Because you *backtracked*!!! Okay, yes! I backtracked! I received a sudden insight, and it just blew me away! I said a lot of crazy things. Now I realize that I'm just not capable of following your intricate logic. So, again, does that make me worthless? I wonder about your moral value system, James Harris, where the value of a life is equated solely with its intelligence. A little convenient, no?, that *you* end up valued most this way. >>It would have been better if you'd said nothing at all!!! >>Now I have to toss you into the pile with Barry Mazur, Andrew >>Granville, Ralph McKenzie, David Ullrich, and Arturo Magidin. >I never claimed to understand your work. Maybe I had a fleeting flash of >intuition that it was correct, but that was all. >>Are you being honest Jim Ferry? >>Please explain to me why you ever claimed my work was correct *in >>detail*. >>I think you owe me that as I'm worried it was some kind of cruel >>prank. Well, I do like a good prank, even if it is cruel, but I don't see how this could be mistaken for one. I mean, I'm the one acting like a fool, right? If someone walks up to you on the street going, flibber jibber jibber and slapping his head while bouncing in circles on one foot, he's not really playing a prank on you. Maybe it's funny (or maybe not), but he's not doing anything *to you*. He's making an ass . . . of himself. I simply had a flash of insight that your work was correct. This jolted me out of the ossified mentality of my herdmates. I came to realize that the reasons I *assumed* it was incorrect had to do with cultural assumptions, the same way that I *assumed* Wiles's proof was valid. I started to ponder the implications of your work being correct, and it was frightening. I'd thought I was standing on firm ground. The ground turned to sand, the sand to dust, and the dust scattered in the wind. The world was gone, and there stood James Harris to create a new and better world. So yes, I kind of freaked out there. I deified you. But I've calmed down since then. I realized that if I could not turn my flash of insight into a logical chain of reasoning, then it was worthless. I couldn't base anything on it. Your proof is left on equal status with Wiles's: something I can't understand. I felt a little silly. And useless. Then something occurred to me: I was not left in the same place as before. Sure, my mind was not sharpened to the point where I could understand your work, but I *was* at least free of the herd! Free from the outlook of the herd! Free from their incessant mooing! Oh wait, I still read sci.math, so I'm not really free from their incessant mooing. I now have a theory about the meaning behind that initial flash of insight. I believe that I have a purpose: a purpose to help you, James. I can't help you mathematically (and, again, there's no need of that), but I can help you in other ways. With a Plan. In a perfect world, you'd just receive the Plan from the Holy Spirit yourself. But, because of your upbringing perhaps, you can't. So it's up to me. I have been chosen (for some bizarre reason) to receive the Plan. Now I can't receive it perfectly either (I've already posted some pretty foolish things as the Spirit prepared me), nor can you readily accept such a ridiculous thing as the Holy Spirit communicating a plan to someone. Well, maybe you can -- that would make things easier -- but I sort of doubt it. I probably shouldn't have brought up the whole Divine Guidance aspect of this in the first place. Hmmm. > For heaven's sake. Yes, when he says he doesn't understand your > work but thinks it's correct he's joking. For heaven's sake? Interesting choice of words. > Regarding whether it's cruel: I would have never imagined in my > wildest dreams that even _you_ could possibly be so blinded by > megalomania and so desparate for approval that you'd think someone > was being sincere when he says he doesn't understand your > work but nonetheless thinks it's right. I mean that's such an > _utterly_ ridiculous thing to say in a mathematical context that > it's hard to see how anyone could possibly not take it as a joke. > So I'm certain he assumed that you'd realize all along he was > making fun of you. Making fun of you is perhaps a little nasty, > but it's certainly no nastier than you deserve given your obnoxious > nehavior for years, so it doesn't bother me. > But lately when you start talking like you think he's being sincere > it becomes a whole different story. Actually encouraging a > lunatic's delusions is not a nice thing to do. On the other > hand the fact that you've thought he was sincere is _the_ > most hilarious thing I've ever seen - we need to consider > the greatest good of the greatest number and all that, > being a little nasty to you versus people all over the > planet laughing their asses off, I think it's a wash. > Honest. I point this out just in case you might decide > to try not to be _quite_ so hilarious: People have been > laughing at Jim's parodies for years, but lately when > you seem to believe he's being sincere, or _might_ > be sincere, in this and related threads, it really does > add a whole new dimension - goes from hilarious to > absolutely world-class hilarious. Now you know. What baffles me is that I've been one of the particularly sarcastic Harris haters over the years. Why would the Holy Spirit come to me? I'm overwhelmed by the sense of my own unworthiness. But throughout the Bible, especially the New Testament, God is seen to relish working through the biggest sinners. Mary Magdalene. Paul. I'm sorry that the beauty of this was destroyed by your parents. It explains a lot. Interesting that Ullrich is laughing so hard. When one laughs that hard, it's often because he is hiding from a deeper pain. Even were he correct about everything, just why would it be funny? I think maybe David sees something of you in himself, James, which sets up a great deal of conflict. I'd like to tell him to love his own inner Harris but that's just too touchy-feely for my taste. >I was offering a different kind of help. Something to complement and >complete the intense laser of your rational thought. Something to help >bring your theories into the human realm. You don't believe in the efficacy >of such things, I guess, or you just want to sear your way through with that >laser of yours, so fine. Go get 'em, tiger. You don't need my help. >>Actually I don't need your help, but I *would* appreciate it greatly. >>But how do I know if you're sincere, or just playing some elaborate >>joke? I don't know. The herd seems to think it's a joke too. I guess you just need to evaluate what I say. If I were telling you to go out and kill prostitutes to cleanse the Earth of sin, well, that would be a bad message. But what I'm saying should resonate with a certain part of you yearning for all this bickering to be over. >>As for understanding my work, math proofs begin with a truth and >>proceed by logical steps to a conclusion which then must be true. >>I give you the steps, you follow, and then I'd think you should >>understand. Right. In a perfect world. Have you looked at Wiles's proof? Think how much training that would take to understand. Your requirements are no less demanding, but it's difficult for you to see that. For you, it's easy as pie. -- | Jim Ferry | Center for Simulation | +------------------------------------+ of Advanced Rockets | | http://www.uiuc.edu/ph/www/jferry/ +------------------------+ | jferry@[delete_this]uiuc.edu | University of Illinois | === Subject: Re: JSH your ship has come in!!!! >But enough silliness. I don't take orders from James, nor does James give them >at all, as far as I know. At first I didn't realize what my role in all this >was to be, but it's becoming clear to me now. James needs a plan, and for >reasons not clear to me, I have been receiving a plan. It keeps me awake at >night. It distracts me from my work. It is vast and beautiful His plan is not completely vast. About half, I'd say. >Here's to the revolution! I believe in this context, MY role is to be part of the ancien regime. All right then, Ferry, is this to be a call to arms? En garde! If you're not with us, you're against us. I should have suspected you'd side with the upstart, rather than keeping him on the outside of the establishment. And here I thought we could count on you, a compatriot from the Land of Lincoln. Ya just can't trust anyone any more. dave === Subject: Talk.Origin banned Subject: Does Mathamitcs prove a Universal designer? A professor of mathematics from the University of Cambridge, P. Dirac, said, in the magazine Scientific American: One could perhaps describe the situation by saying that God is a mathematician of a very high order, and He used very advanced mathematics in constructing the universe. === Subject: Re: Talk.Origin banned Subject: Does Mathamitcs prove a Universal designer? > A professor of mathematics from the University of Cambridge, P. Dirac, said, > in the magazine Scientific American: One could perhaps describe the > situation by saying that God is a mathematician of a very high order, and He > used very advanced mathematics in constructing the universe. Generally speaking I find myself in disagreement with Prof. Dirac. It does, I suppose, depend upon exactly what one means by the term advanced mathematics. If one makes the connection that God = laws of nature (physics) the statement that God is a mathematician is basically saying that physics is mathematically based. This makes some sense. However, generally speaking, the more one understands the true basis of these laws the simpler becomes the mathematics. The complexity viewed by man being a result of his improper viewpoint. God as creator, on the other hand, having the central and correct viewpoint defines nature in the most simple of terms and hence used not very advanced mathematics meaning very complex and esoteric, but rather the most simple of mathematics appled to the proper viewpoints and and expanded fractal-like being self-similar at all levels. Just because man has not yet dicovered the proper viewpoint doesn't make God some brain. If I might use the example of the advanced mathematics of planetary motions using epicycles and the greater non-earth centered (simpler) math of sun-centered calculations. In essence the simpler description is in fact more advanced simply because it is simpler, though both may be in a sense correct. === Subject: Re: Talk.Origin banned Subject: Does Mathamitcs prove a Universal designer? > A professor of mathematics from the University of Cambridge, P. Dirac, said, > in the magazine Scientific American: One could perhaps describe the > situation by saying that God is a mathematician of a very high order, and He > used very advanced mathematics in constructing the universe. I would have no trouble with that. But what all mathematicians know, and most non-maths people do not know, is that the output of a function will vary with its input. Nor are we dealing with a simple function in a single variable, but a recursive function in almost infinite variables in which you do not get a single output, but a matrix output. Simply put, this accounts for order and randomness. Evolution is easily accommodated within a recursive function of high order. And if random processes are part of the inputs, it is easily seen that the universe is not deterministic. Dirac, btw, was a devout atheist. He may have said what you quoted (above) as a general statement, but he did not himself believe in God. In fact, Dirac was so evangelistic about his nonbelief that one of his contemporaries said Dirac has his own religion: 'There is no God, and Dirac is his prophet.' Dirac also said this: The steady progress of physics requires for its theoretical formulation a mathematics which get continually more advanced. This is only natural and to be expected. What however was not expected by the scientific workers of the last century was the particular form that the line of advancement of mathematics would take, namely it was expected that mathematics would get more and more complicated, but would rest on a permanent basis of axioms and definitions, while actually the modern physical developments have required a mathematics that continually shifts its foundation and gets more abstract. Non-euclidean geometry and noncommutative algebra, which were at one time were considered to be purely fictions of the mind and pastimes of logical thinkers, have now been found to be very necessary for the description of general facts of the physical world. It seems likely that this process of increasing abstraction will continue in the future and the advance in physics is to be associated with continual modification and generalisation of the axioms at the base of mathematics rather than with a logical development of any one mathematical scheme on a fixed foundation. Paper on Magnetic Monopoles (1931) (http://www-gap.dcs.st-and.ac.uk/~history/Quotations/Dirac.html) === Subject: Re: Talk.Origin banned Subject: Does Mathamitcs prove a Universal designer? > A professor of mathematics from the University of Cambridge, P. Dirac, said, > in the magazine Scientific American: One could perhaps describe the > situation by saying that God is a mathematician of a very high order, and He > used very advanced mathematics in constructing the universe. Since Scotty likes to name drop and claims to have met many people, does anyone want to take odds that Scotty has already recently spoken to God about this very point and put God straigth on the matter!. === Subject: Re: Talk.Origin banned Subject: Does Mathamitcs prove a Universal designer? > A professor of mathematics from the University of Cambridge, P. Dirac, said, > in the magazine Scientific American: One could perhaps describe the > situation by saying that God is a mathematician of a very high order, and He > used very advanced mathematics in constructing the universe. No, the existence of mathematics does not prove the existence of a Universal designer - in the sense I think you mean it anyway. For example, one possibility is that at the big bang an infinite (or a whole lot) of universes were created and only those that were internally self consistent survived, in some sense. Or another way of looking at it is that things like arithmetic just must be. 10 apples and 10 oranges must both be dividable into two rows of 5. However, the reverse could well be true. If you believe in God, in some sense he was/is a mathematician. Bill === Subject: Another optimization question hi everyone. given a real valued function f that is: 1. defined on a convex set C of $mathbb{R}n$ (but not defined outside C, we may assume that f is $+infty$ outside of C) and positive (>0) everywhere on C. 2. non-convex in C. 3. continuous and has directional derivatives of all orders for any feasible direction from a point x in C. 4. at each x in C, f(x) can only be computed numerically (has no known closed form). which numerical search algorithms or nonlinear programming techniques would be the best/most effective for finding the local minima and, if possible, a global minimum for this function? J. === Subject: Re: Another optimization question no numerical procedure can reasonably deal with infinity. hence you must use a barrier-approach (adding terms like -log(g_i(x)) for each function g_i which describes a boundary part of C as g_i(x)=0 , with g_i(x)>0 in interior(C), provided such interior exists. otherwise you first must reduce the problem to a subspace where this is the case) . next you could use unconstrained optimization using only function values (e.g. uobyqa from powell or its newer successor, described in a recently published paper in but you must modify this code in order to maintain feasiblitiy, that means you must restrict the possible moves such that feasibility is maintained (via checking the g_i individually). also a grid-search a la torczon comes into mind, again with the moves restricted to the feasible part of the grid. hth peter === Subject: Re: Quadratic Sieve > Does anyone have or know of a Quadratic sieve implementation that could be > distributed? For Windows, there's an implementation of PPMPQS included with Yuji Kida's UBASIC. I've used it to do composite numbers up to 107 digits. 100-digit composites take about a week with a 2GHz machine. http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- === Subject: Re: Quadratic Sieve > For Windows, there's an implementation of PPMPQS included with Yuji Kida's > UBASIC. > I've used it to do composite numbers up to 107 digits. > 100-digit composites take about a week with a 2GHz machine. I'm interested to read that, as it seemed to me that with the standard quadratic sieve (using Montgomery quadratics) there was a limitation on the size of the number one could factorise. I should say that I am not an expert, but gave a course on factorisation a couple of years ago, where I looked at the quadratic sieve as an introduction to the number field sieve. As I explained it, one used quadratics Q(x) = ax2 + 2bx + c with ac - b2 = n so that aQ(x) = (ax + b)2 - n, and one sieved for smooth numbers Q(j) on a sieve centred at the positive root of Q(x). With this method, one has to harvest an even number of smooth numbers from each quadratic (because the factor ar comes in if there are r factors, and one needs an exact square). But with a sieve size of say 2 million, and a number with 100 digits, one would only get an occasional smooth number from each quadratic and as far as I could see would hardly ever if ever get two. Obviously the algorithm must be modified in some way for numbers this large. Do you know how it was modified? [I haven't looked very carefully through the literature on this -- I'm interested because a student is doing an Msc trying to use the quadratic sieve with parallel processing.] -- Timothy Murphy e-mail: tim /at/ birdsnest.maths.tcd.ie (all email over 80k dispatched to /dev/null) tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: How much longer must physics put up with F=ma? > The total force [F] used to move an object against friction or any other > restraint, is the total force exerted on it minus the frictional - or other > restraining force; so that the _net force_ is what does the actual work, and > is [f = F-uw]; which leaves F = wa/g + uw; or f = wa/g! > The coefficient of restraint [u] against moving something on a level surface > is usually a fraction of the object's weight [w], so that the product [uw] > is the measure of the restraining frictional force: On slopes the > coefficient of restraint is proportional to the sin of the angle of the > slope: For a 90Á vertical slope the coefficient of restraint against driving > something pointed directly into it will depend on the hardness of the > material: As is the case for driving stakes and such into the ground. Yes, the total force needed to move an object would seem to include the force necessary to overcome friction, and any other impeding resistance, as well as the inertia of the object that results from mass. But in physics, is force really defined to include the total effort, or just that part that overcomes inertia to give acceleration? When pushing a rock across the ground, frictional force usually seems to arise quickly to prevent the rock moving beyond a certain speed without applying additional force. When force is lessened, the rock slows immediately. It is easy to see how Aristotle came to think that force was needed to keep an object moving at a constant speed. But perhaps this is not what modern physics considers force. If you are pushing an object along the ground, there is also air resistance, which is another form of friction. If you pound a stake into the ground, you get ground resistance, which is really the forces of atomic electric fields that you are running up against. If I push against a brick wall, nothing will move. Am I really exerting a force, or is this only tension, or stress? But when we consider an object in a frictionless state, there is another question that arises. Newton said in his third law of motion: To every action there is always opposed an equal reaction; or the mutual actions of two bodies upon each other are always equal, and directed to contrary parts. Now when a force is applied to a mass in a frictionless state, the inertial force opposes the motive force, which limits the acceleration according to a=F/m. But if the inertial force is an equal and opposite force to the motive force, as Newton's 3rd law seems to say, then why do we have acceleration at all? === Subject: Re: a puzzle related to artinian group > # > #> # > #> # Suppose n people sit around a table and n-1 cards are dealt to them. > #> # There is no asumption on the number of cards a player receive. In > #> # each round, all players with 2 or more cards pass one card to the > #> # left and one card to the right. Prove that eventually, all players > #> # but one have exactly one card. > # #> Your puzzle is a special case of the so-called chip-firing games > #> in > # #> A Bjoerner, L Lovasz, PW Shor > #> Chip-firing games on graphs > #> European Journal of Combinatorics 12, 1991, pp 283-291 > # #> Bjoerner, Lovasz, and Shor prove that for every chip-firing game, > #> the initial configuration determines whether or not the game will > #> continue forever. If the game eventually terminates, then the final > #> configuration is fully determined by the initial configuration. > # #> For your game, it is easy to prove that there always exists > #> SOME sequence of moves for which the game terminates. > #> [BLS] then yields that your game ALWAYS terminates. > # # > # I think you are taking the rules to be that in each round the players > # with two or more cards make their moves one after the other (in some > # unspecified order). > Yes, I do. > The analysis of [BLS] applies to games, where the players move one by one. > In one move, some player with at least two cards passes one card to the > left and one card to the right. My statements above all refer to this > situation. > The puzzle of Yannick is a very special case of the [BLS] games: > Yannick compresses many moves into a single super-move, and he makes > all players with at least two cards move SIMULTANEOUSLY. However, > this does not change the combinatorics of the game. If you decompose > a super-move into its one-player moves, you will have the same effect > on the distribution of cards. > ... super-move down into a sequence of one-player moves. I also now see from chip-firing games. So that gives Yannick his proof. The bit you say is easy really is fairly easy: by induction on k with 2 <= k <= n-1, you can show that the game with k-cards terminates (in the inductive step, you pick any card, C say, and then play as if it wasn't there: this play will terminate by induction in a state where the player holding card C has at most 2 cards and all the other players have at most one card. If the player holding card C has 1 card you've finished, if 2, then at least two players hold no cards and you have an arrangement like 0, 1, ..., 1, 2, 1, ..., 1, 0 which you can check terminates, no matter how you play.) Now by [BLS], as you say, the original game with super-moves must terminate with the same final position. So there is the proof Yannick asked for. Perhaps, in return, Yannick could now explain why this puzzle is related to artinian groups. Rob. === Subject: Re: a puzzle related to artinian group # # So that gives Yannick his proof. The bit you say is easy really is fairly # easy: by induction on k with 2 <= k <= n-1, you can show that the game with Actually, the paper of Bjoerner, Lovasz & Shor also contains a proof for the termination in this special case: They show that if the number of chips is smaller than (twice the number of edges minus the number of vertices), then the chip firing game always terminates. In our case, we have a ring with n vertices and n edges, and n-1 chips. --Gerhard # k-cards terminates (in the inductive step, you pick any card, C say, and # then play as if it wasn't there: this play will terminate by induction in a # state where the player holding card C has at most 2 cards and all the other # players have at most one card. If the player holding card C has 1 card # you've finished, if 2, then at least two players hold no cards and you have # an arrangement like 0, 1, ..., 1, 2, 1, ..., 1, 0 which you can check # terminates, no matter how you play.) Now by [BLS], as you say, the original # game with super-moves must terminate with the same final position. # # So there is the proof Yannick asked for. Perhaps, in return, Yannick could # now explain why this puzzle is related to artinian groups. # # # Rob. # === Subject: Re: a puzzle related to artinian group PS: There is 1.5 page paper by Michael Thorup with a delighfully neat proof of (a generalisation of) the necessary result on card-firing games on a finite undirected graph at: http://citeseer.nj.nec.com/thorup96firing.html Rob. === Subject: Re: Feller's Direct Proof of Stirling's Formula > Stirling's Formula_. His proof hinges on showing that > phi(t) = sin(pi*t)/pi, > where phi(t) is in product form. That is, > phi(t) = t * Product(k = 1, 2, 3, ...; (1 - t*t/(k*k))). > The last step in his proof of this stymies me. Feller first shows that: > phi(2x) = 2phi(x) * phi(x + 1/2) / phi(1/2) > In other words, he proves the double-angle formula for phi. So far, so > good. > He then sets > f(x) = log(pi * phi(t) / sin(pi*t)), > asserts that > f(2x) = (f(x) + f(x + 1/2)) / 2, > and loses me. Can anyone explain? He says that it follows from the > double-angle formula for sin(pi*x)/pi and the above formula for phi(2x). > (I know there are other proofs that phi(t) = sin(pi*t)/pi. My question > is about Fellers'.) Feller published a correction in the Monthly the year following the http://www.jstor.org/browse?config=jstor#Mathematics > not keep track of my steps. Now I can't find it again. Does anyone > have a URL? Please note, I'm not talking about Feller's book, just the -- A. === Subject: Re: Feller's Direct Proof of Stirling's Formula > [...] >> f(2x) = (f(x) + f(x + 1/2)) / 2, >> and loses me. Can anyone explain? He says that it follows from the >> double-angle formula for sin(pi*x)/pi and the above formula for phi(2x).... > Presumably your definition of f should have t instead of x. Correct. > I can't get Feller's formula either, but instead come up with > f(2x) = f(x) + f(x + 1/2) - f(1/2). > Would the rest of his proof follow from that, or is there a more serious > difficulty? (*) f(2t) = (f(t) + f(t + 1/2)) / 2 he proceeds as follows: Let 2x be the point at which f is maximized. Then repeatedly applying (*) shows that f(x) = f(2x) f(x/2) = f(2x) f(x/4) = f(2x) f(x/8) = f(2x) ... and so on Since f is continuous (which he proves), we have f(0) = f(2x). In a is 0. Thus f's maximum is 0. Similarly, f's minimum is 0. Thus, f is identically 0 and the desired result follows. I got the same result you did but, as you see, his argument rests on his recurrence. -- Opinions expressed above are not necessarily my employer's. James M. Stern stern@itgssi.com (213) 270-7955 ITG Software Solutions, Inc. Culver City, CA 90230 === Subject: Re: Million Dollar Prize for Solving Kepler's Equation > Does anybody else agree with me that a mathematical solution to Kepler's > Equation, in closed fom, is a sufficiently important unsolved mathematical > problem to merit inclusion in the Clay Foundation's list of million dollar > prizes for mathematics? Let |e| < 1. Then define a new function kep as follows: key(e,y) is the unique number z such that z - e sin z = y. The solution to Kepler's equation E - e sin E = M in closed form is E = kep(e, M). Gimme my money. === Subject: Re: Contractible Spaces Originator: grubb@lola >If a topological space S is contractible to some point p in S, >is S contractible to every point in S? >If a topological space S is strongly contractible to some point >p in S, is S strongly contractible to every point in S? >What's an example of a space that is >contractible but not strongly contractible? >S is contractible to a when there's some continuous h:Sx[0,1] -> S with > for all x in S, h(x,0) = x, h(x,1) = a, >and strongly contractible when in addition > for all t, h(a,t) = a The answer to the first question is yes. to contract to b given a contraction to a, note that h(b, t) is a path from b to a. Now contract to a, then move backwards along the path to b. The answer to the second question is no. The comb space is a counterexample. Let X=([0,1]x{0}) union (({1/n: n natural} union {0})x[0,1]). Then X is strongly contractible to (0,0) but not to (0,1). The difficulty is that if (0,1) is fixed, all (1/n,1) for large n have to stay close to (0,1) and hence can't move down and around and up to (0,1). This example is in Spanier's book. --Dan Grubb === Subject: Re: Contractible Spaces === Subject: Re: Contractible Spaces >> If a topological space S is strongly contractible to some point >> p in S, is S strongly contractible to every point in S? >I don't think so. Consider the comb space in R2: >X = [0,1] x {0} union {0} x [0,1] union {1/n} x [0,1], n >= 1. >Then X is strongly contractible to the origin (collapse everything >down to [0,1]x{0}, and then left to (0,0)), but not to the point >(0,1): during the course of any possible homotopy, the points >(1/n,1) on the teeth of the comb would have to go down to the x-axis >and so would eventually get far away from the point (0,1), so there >is no continuous homotopy fixing (0,1). Don't understand. The distance within the comb from any point to any other is at most 3. As before, collapse all of the comb, except for {0}x[0,1] down to [0,1]x{0} and left along the bottom to (0,0) and thence from there upto (0,1). >(See Sieradski's book _An__Introduction to Topology and Homotopy_, >examples 6-7 on page 308 and example 4 on page 318.) Don't have access. >> What's an example of a space that is >> contractible but not strongly contractible? >Have you looked in the book _Counterexamples in Topology_? Yes, Arthur Steen's book doesn't touch homotopy. > S is contractible to a when there's some continuous h:Sx[0,1] -> S with > for all x in S, h(x,0) = x, h(x,1) = a, > and strongly contractible when in addition > for all t, h(a,t) = a ---- === Subject: Re: Contractible Spaces Originator: grubb@lola >Then X is strongly contractible to the origin (collapse everything >down to [0,1]x{0}, and then left to (0,0)), but not to the point >(0,1): during the course of any possible homotopy, the points >(1/n,1) on the teeth of the comb would have to go down to the x-axis >and so would eventually get far away from the point (0,1), so there >is no continuous homotopy fixing (0,1). >Don't understand. The distance within the comb from any point to any >other is at most 3. As before, collapse all of the comb, except for >{0}x[0,1] down to [0,1]x{0} and left along the bottom to (0,0) and thence >from there upto (0,1). You can't do that contraction to [0,1]x{0} continuously and keep {0}x[0,1] fixed. Those top points have to stay close to (0,1) by continuity. --Dan Grubb === Subject: Please help on this Combinatorial Optimization problem I have 2n vertices, n for group A, n for group B. The vertices are fully connected by edges, each edge has a different weight. What I want to solve is this: Go trough all vertices, without repetition, and minimize the sum of weights, under this rule: ... -A-A-B-B-A-A-B-B-A-A-B-B-... where A represents some vertex of A. This is different than Bipartite TSP, which is ...-A-B-A-B-A-... Can anyone tell me if there are algorithms to solve my problem? === Subject: Re: Please help on this Combinatorial Optimization problem > I have 2n vertices, n for group A, n for group B. > ... -A-A-B-B-A-A-B-B-A-A-B-B-... No solution help, just a comment, you really need _4n_ vertices for that to work, since both A and B must each have an even number of elements. xanthian. -- === Subject: Group action ... If Z is the set of integers and SL(2,Z) is the group of unimodular matrices with integer coefficients, we can define naturally an action of SL(2,Z) on Z x Z by (a,b;c,d).(m,n)=(am+bn, cm+dn). Is it possible to describe, simply, the orbits of this action ? and the same question for the stabilizer of an element in Z x Z ? -- Ce message a ete poste via la plateforme Web club-Internet.fr This message has been posted by the Web platform club-Internet.fr http://forums.club-internet.fr/ === Subject: Re: Group action ... > If Z is the set of integers and SL(2,Z) is the group of unimodular > matrices with integer coefficients, we > can define naturally an action of SL(2,Z) on Z x Z by > (a,b;c,d).(m,n)=(am+bn, cm+dn). > Is it possible to describe, simply, the orbits of this action ? Yes. The m and n with a given greatest common divisor form an orbit. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Is it mass; or is it weight > A slug [...] [sic] doesn't even have an official definition. BUSTED! By official definition: 1 slug = 1 pound second2/foot 1 pound = 453.59237 grams * (32.1740 feet/second2) 1 foot = 0.3048 centimeters 1 centimeter = 1/100 meter = 1/29979245800 light-seconds 1 gram (I believe) = the mass of 1 milliliters of water at the triple point. 1 milliliter = 1 centimeter3 Other units: pound-mass = 1/32.1740 slugs poundal = 1/32.1740 pounds = 1 pound-mass feet/second2 Newton = 1 kilogram meter/second2 dyne = 1 gram centimeter/second2 kilogram = 1000 grams meter = 1/299792458 light-seconds The inch is 2.54 centimeters. There is also the survey inch, defined as 1/0.3937 centimeters, and survey foot, defined as 1200/3937 meters. Aside from avoirdupois pounds (defined above), there are also troy pounds (troy ounces, etc.). 1 troy pound = 144/175 pound = 343.2417216 grams * (32.1740 feet/second2) 1 grain = 1/7000 pounds (= 1/5760 troy pounds) 1 pennyweight = 24 grains = 1/20 troy ounce 1 dram (avoiddupois) = 27.34375 grains 1 troy ounce = 1/12 troy pound === Subject: Re: Is it mass; or is it weight >> A slug [...] [sic] doesn't even have an official definition. >BUSTED! >By official definition: >1 slug = 1 pound second2/foot Agreed. But you need a standard for one of them, either the pound force or the slug, or it is a meaningless tautology. I'll grant you that seconds and feet are well defined for the purpose of this argument, so you don't need to worry about them. >1 pound = 453.59237 grams * (32.1740 feet/second2) No. 1 pound = 453.59237 grams. Period. http://www.ngs.noaa.gov/PUBS_LIB/FedRegister/FRdoc59-5442.pdf http://gssp.wva.net/html.common/refine.pdf 1 pound force = 453.59237 grams times whatever value you choose to use as your standard acceleration of gravity. The pound force is such a recent bastardization, that it is uniquely identified by that name. Of all the hundreds of different pounds used at various times and places throughout history, only one has spun off a force unit of the same name that has seen any significant use. Now all you need to show me is that somebody, somewhere, has OFFICIALLY adopted that value of 32.1740 ft/sî for the purpose of defining a pound force, and then I'll gladly agree that you have shown both the pound force and the slug to have an official definition--at least for somebody somewhere, within whatever scope your defining agency has the authority to set this standard. Can you do so? Tell me who set the standard, and when--a cite to some printed source or one on the web would be preferred. Show me this in the laws or regulations of some country. Or in some standard by an international standards organization. Or in some officially adopted standard of some national standards laboratory. Or in some officially adopted standard of some professional organization. Put up or shut up. >1 foot = 0.3048 centimeters >1 centimeter = 1/100 meter = 1/29979245800 light-seconds >1 gram (I believe) = the mass of 1 milliliters of water at the triple point. Wrong. Grams haven't been defined in terms of milliliters of water since 1799 at least. That's when the French first defined the kilogram by a particular platinum artifact they called the Kilogram of the Archives. When the 44 or so standard kilograms were constructed after the Treaty of the Meter, and one of them chosen as the International Prototype Kilogram, the target in their construction wasn't some mass of water but rather the mass of the previous standard which had been maintained by the French government. did have liters defined as the volume occupied by a kilogram of water. However, it wasn't at the triple point of water (0.01 ÁC or 273.16 K), but rather at the temperature of maximum density (about 277.13 K or 3.98 ÁC). During that period, of course, a liter wasn't exactly equal to 1000 cubic centimeters. It was about 1000.028 cubic centimeters. >1 milliliter = 1 centimeter3 >Other units: >pound-mass = 1/32.1740 slugs That's would be fine if slugs had an official definition. When you show me that somebody has in fact officially adopted that 32.1740 ft/sî you mentioned above, then you will be able to say that a slug is defined as 32.1740 pounds. >poundal = 1/32.1740 pounds = 1 pound-mass feet/second2 There you go, stupid. This system with poundals is a different system from the one with slugs. It is a much older system, in fact. Both of these systems, the absolute fps system and the gravitational fps systems, as well as some others such as a gravitational inch-pound-second system, are a very limited subset of the English units, a subset which like SI forms a coherent system of units. That means, for example, that there are no pints or gallons in any of those subsystems. Now fill in the blank: The base unit of mass in this oldest fps system of mechanical units is the ______________. >Newton = 1 kilogram meter/second2 >dyne = 1 gram centimeter/second2 >kilogram = 1000 grams >meter = 1/299792458 light-seconds >The inch is 2.54 centimeters. There is also the survey inch, defined >as 1/0.3937 centimeters, and survey foot, defined as 1200/3937 meters. >Aside from avoirdupois pounds (defined above), there are also >troy pounds (troy ounces, etc.). >1 troy pound = 144/175 pound = 343.2417216 grams * (32.1740 feet/second2) Completely and absolutely false. http://w0rli.home.att.net/youare.swf 1 troy pound = 144/175 pound = 343.2417216 grams Unlike their avoirdupois cousins, and unlike grams and kilograms, the troy units of weight have never spawned units of force of the same name. There is no troy pound force and no troy ounce force, and there never have been. Gene Nygaard http://ourworld.compuserve.com/homepages/Gene_Nygaard/ === Subject: Does this require assumptions?: I conclude that 1+1=2 opposite kind of virtue--economy. Economy, in humour and art, does not mean mechanical brevity but implicit hints instead of explicit statements--the oblique allusion in lieu of the frontal attack. Does this require assumptions?: I conclude that 1+1=2. === Subject: Re: Does this require assumptions?: I conclude that 1+1=2 > Does this require assumptions?: I conclude that 1+1=2. No, because 2 is simply a name for the successor of 1. It requires no assumptions beyond the basic axioms and the definitions of '+' and 'successor' to prove that 1+1 is the successor of 1, or 1+1 = s(1). -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Does this require assumptions?: I conclude that 1+1=2 Bill Bonde ( the oblique allusion in lieu of the frontal attack ) > Does this require assumptions?: I conclude that 1+1=2. Nope. It's the usual DEFINITION of 2. you must. of course assume the definition of + , your universe of discourse and principal existential axioms. Bob Pease === Subject: Re: How long must physics put up w/f=ma? In sci.physics, Gene Nygaard : >>In sci.physics, tadchem >>: > The answer to the question How long must physics put up w/f=ma? is: >> As long a F = ma describes reality And just what does that have to do with *w/f=ma*? >>I think the 'w/' in this case is an abbreviation for 'with', >>not part of the formula proper. Of course writing it that >>way isn't exactly helpful. > Of course it isn't helpful. > But if you haven't figured out yet, Shead is an idiot. He'll never > change. Yet we other idiots keep replying to a thread whose subject > asks about w/f=ma? > It might be understandable that after all these years, Shead still > hasn't the foggiest idea what a factorial is, or the symbol for this > operation, a symbol he keeps throwing into his formulas. But he damn > sure out to be held to a standard of knowing that a slash / is a > symbol for a mathematical operation. Don't be blaming this confusion > on anyone other than the culprit, Dense Donny Shead. > Donald Shead is also a crackpot who keeps using ft as a symbol for > the multiplication of force by time--and is too stupid to figure out > why people don't understand what he is saying. I was under the impression that he was using 'ft' as an abbreviation for 'foot' (since he highly prefers English units, for some bizarre reason). As for being an idiot -- I prefer not to make judgements on the person, but certainly Shead's postings are consistent with being an intractable quack, unable to see that SI, for all of its warts (one of them being the requirement of an existent of the mass-artifact of 1 kg; another being the liter/litre controversy) doesn't have the conversion problems and/or other difficulties of contemporary English/Imperial units: 1 km = 1,000 m = 1,000,000 mm and that's more or less that, all throughout SI, regardless of the item being measured. (Some artifacts, however, still exist: 1 cal = 4.180 J, for instance. I'm not sure where 'cal' came from apart from its relationship to water (1 cal heats 1 gm water 1 degree C) and it may not be, strictly speaking, part of SI. There are also some issues on the electromagnetic side of the issues, which I'm not that familiar with. Of course this is a far cry from the grains/oz/lbs [pick a flavor!], the mil/in/ft/rod/mile/furlough, or the dry vs wet quart issues perambulating all over Imperial measurements. Even pints are different.) I base this primarily on his non-response to the elementary observation that a mass artifact (the aforementioned 1 kg item) is far easier to handle than an apparatus that can be proven to generate a 1 Newton (or, for those so inclined, 1 lb-force) force. (Arguably, the simplest method to reliably produce this force would rely on the ideal gas law, and require precise temperature measurements.) Nor is he treading with the crowd on his notions regarding inertia, mass, and force. I don't see why it makes any difference when writing F = ma a = F/m m = F/a assuming scalar quantities. (In vector quantities, the third must be tossed, for obvious reasons; F and a are inherently vectors.) Use whichever one is appropriate for the problem solution. Side issue: I'll admit it is a pity we're not willing to spend money on the gigantic conversion efforts necessary to bring our roadway and speedometer systems consistent with Europe, though. :-/ But oh well. :-) > Gene Nygaard > http://ourworld.compuserve.com/homepages/Gene_Nygaard/ -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Naive User Stories Followups. In sci.physics, Richard Henry <0zPib.33608$La.24164@fed1read02>: >> There are subtler effects, however. In princple you can get kT ln 2 >> joules of useful energy by randomizing a bit that starts out in a >> known state. And of course energy has mass. So perhaps a blank >> diskette, or one that contains the digits of pi or some other >> determinate pattern, weighs slightly more than one that contains >> random data. >> How, pray tell, does one distinguish the digits of pi from random data? > In what radix? 10, 16, 2, pi, etc? > In what representation? Binary, ascii, BCD, etc? Does it matter? Given any finite digit sequence, pi embeds it. Somewhere. :-) (At least, such is my understanding.) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Naive User Stories > Followups. > In sci.physics, Richard Henry > : >> There are subtler effects, however. In princple you can get kT ln 2 >> joules of useful energy by randomizing a bit that starts out in a >> known state. And of course energy has mass. So perhaps a blank >> diskette, or one that contains the digits of pi or some other >> determinate pattern, weighs slightly more than one that contains >> random data. >> How, pray tell, does one distinguish the digits of pi from random data? > In what radix? 10, 16, 2, pi, etc? > In what representation? Binary, ascii, BCD, etc? > Does it matter? Given any finite digit sequence, pi embeds it. > Somewhere. :-) > (At least, such is my understanding.) I'm afraid I don't understand your understanding. === Subject: Re: Jordan measure Any takers??? -- === Subject: Re: How to calculate the total coverage area of a few circles? > late. The method is very skillful. I believe it has a excellent efficiency. > Thaks again. _ It's true that Arcs (the list of arcs between x=xleft and x=xright) can be arranged so that each even/odd pair of entries in Arcs gives bounds of a contiguous area, bounded below by the first arc, above by the second arc, and left and right by xleft and xright. But the step Sort Arcs [e] so both ends are in non-decreasing y-order (which is always possible because of the Event list construction). In one pass, flag arcs that are inside other circles, and in another pass, remove the flagged arcs is wrong and (for example) fails where an overlap of two circles fills between two other circles; for example, with circles of radius 13 at (13,31), (19,13), (25,47), and (33,29). The only way around this I have thought of is O(n2), and leads to a worst case complexity of O(n4) and an average complexity of O(n3). I will post a program later illustrating it. -jiw === Subject: Re: How to calculate the total coverage area of a few circles? 3QLpj-NoP*NzsIC,boYU]bQ]H'y<#4ga3$21: late. The method is very skillful. I believe it has a excellent efficiency. > Thaks again. _ > It's true that Arcs (the list of arcs between x=xleft and x=xright) > can be arranged so that each even/odd pair of entries in Arcs gives > bounds of a contiguous area, bounded below by the first arc, above by > the second arc, and left and right by xleft and xright. But the step > Sort Arcs [e] so both ends are in non-decreasing y-order (which is > always possible because of the Event list construction). In one pass, > flag arcs that are inside other circles, and in another pass, remove > the flagged arcs is wrong and (for example) fails where an overlap of > two circles fills between two other circles; for example, with circles > of radius 13 at (13,31), (19,13), (25,47), and (33,29). The only way > around this I have thought of is O(n2), and leads to a worst case > complexity of O(n4) and an average complexity of O(n3). I will > post a program later illustrating it. The method of Edelsbrunner that I referred to earlier gives you complexity O(n log n). Basically you get one term in the sum for every feature of the power diagram of the circles (Voronoi diagram of their centers in the case that all the circles have the same radius). -- David Eppstein http://www.ics.uci.edu/~eppstein/ Univ. of California, Irvine, School of Information & Computer Science === Subject: Re: Fun and Mental; Raisin for High; Achievements of Shoes >You mean, you still don't realize that the above is a fake and that >>there is no such institution as The Lovenstein Institute of Scranton >>Pennsylvania. Even the Guardian which, at the time, jumped >>enthusiastically on this crap, had to retract and admit that they were >>taken for a ride. I suggest you contact them. >I wonder if this is the reason for liberals to work on dumbing >down public schools? Might be, might be:-) > Even people who appear to have an ability >to think use this rumor of Bush dumbness to get a Democrat >elected next year. With a stress on appear. Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: Fun and Mental; Raisin for High; Achievements of Shoes there is no such institution as The Lovenstein Institute of Scranton >Pennsylvania. Even the Guardian which, at the time, jumped >enthusiastically on this crap, had to retract and admit that they were >taken for a ride. I suggest you contact them. I wonder if this is the reason for liberals to work on dumbing down public schools? Even people who appear to have an ability to think use this rumor of Bush dumbness to get a Democrat elected next year. /BAH Subtract a hundred and four for e-mail. === Subject: Re: Fun and Mental; Raisin for High; Achievements of Shoes >You mean, you still don't realize that the above is a fake and that >>there is no such institution as The Lovenstein Institute of Scranton >>Pennsylvania. Even the Guardian which, at the time, jumped >>enthusiastically on this crap, had to retract and admit that they were >>taken for a ride. I suggest you contact them. >I wonder if this is the reason for liberals to work on dumbing >down public schools? 'Scuse me? I happen to be a knee-jerk liberal, and I have no interest at all in dumbing down schools. > Even people who appear to have an ability >to think use this rumor of Bush dumbness to get a Democrat >elected next year. Every time I've ever seen a picture of him next to a senior advisor, I get the impression of a 4-year-old on career day, visiting the grown ups. - Randy === Subject: Re: Proving the Four Color Theorem > Given what you know about graphs and vertex coloring, which do you > think would be the easiest task?. > 1. Prove that a 5-chroma planar graph cannot exist. > 2. Prove that all planar graphs are 4-colorable. > 3. Prove that a 5-chroma graph cannot be planar. This is as meaningful a question as asking which of the following is the easiest task? 1. Prove 4 + 5 = 9. 2. Prove 9 - 5 = 4. 3. Prove 9 - 4 = 5. J === Subject: Re: Proving the Four Color Theorem >> Given what you know about graphs and vertex coloring, which do you >> think would be the easiest task?. >> 1. Prove that a 5-chroma planar graph cannot exist. >> 2. Prove that all planar graphs are 4-colorable. >> 3. Prove that a 5-chroma graph cannot be planar. > This is as meaningful a question as asking which of the following is > the easiest task? > 1. Prove 4 + 5 = 9. > 2. Prove 9 - 5 = 4. > 3. Prove 9 - 4 = 5. Which is it? :-) -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Proving the Four Color Theorem > Given what you know about graphs and vertex coloring, which do you > think would be the easiest task?. > 1. Prove that a 5-chroma planar graph cannot exist. > 2. Prove that all planar graphs are 4-colorable. > 3. Prove that a 5-chroma graph cannot be planar. They are all the same problem. Therefore, they are equally hard/easy. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Naive Q: Set theory, logic - which comes first? <1xUcb.14254$O85.6040@pd7tw1no> <3f79e264$7$fuzhry+tra$mr2ice@news.patriot.net> <3f82ebb8$3$fuzhry+tra$mr2ice@news.patriot.net> that its truth has been established beyond all doubt. No. That is not either a Mathematical concept or an achievable goal. >The integers have more properties than can be captured by any axiom >system. And what would those be? How does one distinguish an axiom system in which those proerties are true from one in which they are not? >You said that there are statements about natural numbers that can >neither be proven nor disproven, The term Natural Numbers usually refers to the axiom system known as the Peano Postulates. >which may or may not be true, but >does not follow from what G.9adel showed. It certainly does, unless you are using the term natural numbers as some vague metaphysical concept having nothing to do with any axiom system. What do *you* mean by natural numbers? >G.9adel did show that in a given >formal system for natural numbers there are statements about natural >numbers that can neither be proven nor disproven. Which means that if you stick to Mathematics my statement is true. It doesn't matter which axiom system you pick, as long as it includes Peanos' Postulates among its consequences, there will be statements that can neither be proven nor disproven. -- spamtrap@library.lspace.org === Subject: Re: Homeomorphism and boundaries <3f85d5c4$21$fuzhry+tra$mr2ice@news.patriot.net> <3f89fd84$48$fuzhry+tra$mr2ice@news.patriot.net> at 01:53 PM, magidin@math.berkeley.edu (Arturo Magidin) said: >Then you mapped (0,2pi) to C by sending x to exp(ix), and that map is >also not an open map. Of course, if your domain changed to [0,2pi], >compact, then it would be a closed continuous mapping, but it would >not be injective. It's the range I should have changed, not the domain. Specifically, I should have stated that the map was an injection into the unit circle whose image was missing one point. Or, equivalently, given the map as the natural injection of the open interval into its one-point compactification. -- spamtrap@library.lspace.org === Subject: Re: Fundamental Reason for High Achievements of Jews at 11:59 PM, Noah Roberts said: >Hell, during WW2 my country commited similar horrific acts right >here on our own soil - Japanese concentration camps. What we did to the Nissei was shameful, but it wasn't remotelcy close to what Germany or even Japan did. -- spamtrap@library.lspace.org === Subject: Re: Fundamental Reason for High Achievements of Jews > What do you think will happen when all these > bitter Black, Latino and Redneck trained warriors, > get back to America, after wasting many months > in a foreign land, and they begin to take stock??? > They might make the Nazi look like choir boys, > because these guys are not as sophisticated > as the Germans were/are. > Tom Potter > OOOOOOahahaha, Tom Potter the (mostly self) righteous > dude, who sees himself as the true epidemy of MORALS, > nonchalantly describes Blacks, Latinos and Rednecks as > not as sophisticated as Germans.......AHAhahahhaa..... > AHAHAHAHAh.........ahahahahah........ahahahahah.. > Potter, ....spoken like a true Jew........welcome into the fold. > ahahahahaha.........ahahahahanson Sophisticated: Having or appealing to those having worldly knowledge and refinement and savoir faire Maybe Harry Hyena hanson is right, and that when the Americans return from the Middle East, they will have more worldly knowledge, but I dare say that they will not be as refined. But the question is: how will they react to the fact that their buddies have been killed and maimed, and that they have been deprived of the joys of being with their friends and families, etc. when they begin to digest all this worldly knowledge, and see that others instigated the events that lead to their misfortune, and profited from it? -- Tom Potter http://tompotter.us === Subject: Re: Fundamental Reason for High Achievements of Jews >>Actually, the masses have been brainwashed >>to give the German people a bad rap, >>for trying to defend themselves from the Bolsheviks. Mr. Potter's attempt to trivialize Nazi's is rather sickening, > but I must also say that giving German people a bad rap is > guilty of the same mentality. it wasn't just the Germans who > murdered Jews during WW2....... > ....Several groups trying to completely whipe out each other > in ongoing wars and mass genocides [today]. > Hitler's manipulations could be just as effective today, > and in fact I have feared such a thing happening in my country > ever since 9/11. Hell, during WW2 my country commited similar > horrific acts right here on our soil - Japanese concentration camps. > When you mix a fearful there is no bottom to how bad it can get. > The human capacity for great evil is very great, and it lies within > each and it lies within each of us no matter what our culture > or belief systems stipulate. Mass murder and wanton cruelty > are unfortunately very human traits. > This is none of the fundamental aspects that Tom does not see, does > not comprehend and much less accept. But, Noah, it is even worse than > what you describe, re wanton cruelty. Take all the wanton cruelty > arising from connections with politics/culture/religion magically away > and you will still see a substantial permanent residue of people who > demand and PAY to receive cruel treatment and men and women who > get PAID and enjoy torturing others. As you say, Noah, wanton cruelty > is unfortunately a very human trait ....and it is are here to stay..... > and cruelty hurts just the same whether it comes from hormonal or > institutional aberrations. > ......And now comes Tom's response to Noah, > Noah who did NOT suggest that ALL Jewish people instigate..... > It is interesting to see that Noah Roberts > seems to be suggesting that all Jewish people > instigate conflict and war, for power and riches. > Now, that Tom believes that he has placed the turd successfully > onto Noah quickly washes his own dirty hands with strident > backpedaling (for the moment): > All Jews are not instigators of conflict and war for profit, > any more than all Muslims are terrorists, > or all Italians are Mafia members, > or all Gypsies are fortune tellers. > And of course, all instigators of conflict and war > for power and riches are not Jews, as can be > seen by the examples of Bush, Blair and Rumfeld. > Now, him feeling clean and righteous, Tom is ratcheting > up the bar again from NOT ALL Jews to MOST were Jews > .....ahahahaha...ahahahaha...... > The problem is, that most of the instigators > of the class wars of the 1900's were Jews, > and most of the instigators of the religious wars > of the 2000's are Jews, and it seems to me, > that the good, moral Jews, who believe > that all folks are entitled to life, liberty, > and the pursuit of happiness, should take > an aggressive stand against the war instigators, > as it will bite all Jews in the ass, just as it did > in the class wars. > The Bolsheviks, who were mostly Jews , > caused a lot of problems for Germany and the > world during the 1900's, and over one million > of them migrated to Israel after the native Russians > regained control of their government, and as can be seen, > they are desperately trying to get a world religious war > going, so they can get back in the chips. > Tom, now back in business with MOST of the Jews as the culprit > instigators, the time has come for him to explain who and > why one ought to react harshly against ALL Jews.......... > ahahahaha......ahahahaha......and that one should not be > surprised (which, all fun aside, is a possible real scenario): > It is not surprising that many Germans reacted harshly > against all Jews, good and bad, and it will not be > surprising if many Americans react the same way, > if America is pulled into a war that costs many > Blacks, Latinos, and Rednecks their lives, limbs, > liberties and fortunes. > Most people are intimidated by the character assassination blitz > put out by a small immoral Jewish element when anyone suggests > that the motives and actions of some Jews are less than noble, > but the truth needs to be told to prevent an enormous amount of > suffering by Jews and the peoples of the entire world. > In the 1900's, the German leaders and people > were also intimidated by the Bolsheviks, > because they were not only engaging in character assassination, > the Bolsheviks did anything to silence opposition, including > beatings and real assassinations. This is why some Germans formed > strong arm groups to protect themselves and their leaders > for being physically attacked by the Bolsheviks. > AHAHAHahhahah........ahahahaha.....IAROTFLMAO....... > Tom must have seen Hitler's propaganda ca. 1940's films > Der ewige Jude being a Symphonie des Ekels und des Grauens, > and in Jud S.9fss in which Hitler depicted the Jews to be like > rats, but Potter makes another, novel leap and pits worms against > rats........ahahahah.....What Potter does not see is that in his world > in a clash between worms and rodents, the rats will win hands > down. .....Unless of course the worms infect the rats, and > cause rat extermination by internal systems collapse....ahahahaha... > Even intimidated and brainwashed worms turn > when the hits the fan, and there are a couple of billion > intimidated and brainwashed worms, and the > is beginning to hit the fan. > What do you think will happen when all these > bitter Black, Latino and Redneck trained warriors, > get back to America, after wasting many months > in a foreign land, and they begin to take stock??? > They might make the Nazi look like choir boys, > because these guys are not as sophisticated > as the Germans were/are. > Tom Potter > OOOOOOahahaha, Tom Potter the (mostly self) righteous > dude, who sees himself as the true epidemy of MORALS, > nonchalantly describes Blacks, Latinos and Rednecks as > not as sophisticated as Germans.......AHAhahahhaa..... > AHAHAHAHAh.........ahahahahah........ahahahahah.. > Potter, ....spoken like a true Jew........welcome into the fold. > ahahahahaha.........ahahahahanson === Subject: Re: Fundamental Reason for High Achievements of Jews > Harry Hyena hanson has lost the ability > to make a rational, intelligent, MORAL > response to my messages. > The only MORAL response to your messages is to > condemn you [Tom Potter] for the ignoramabus and bigot > that you are. You don't like Jews. > Bob Kolker > Easy, Bob, easy. It is not a crime for anyone not to like Jews. > It is anyone's fundamental right to like or dislike whatever they wish. > There is neither a law nor obligation which says that you should > or must be liked. The luxury of being liked must be earned. > As a matter of fact this maybe one of the core-problems that Jews > suffer from: Their bloated and loudly advertised need to be liked. > May that be as is, what makes me ROTFL about > Peeping Tom Crackpotter is that he has the chutzpah to use the > word MORAL, when in fact he is simple quite angry at me because > I refused to join him in: > utters terrorist threats against the President of the United States, > & calls and recruits for the violent overthrow of the government: 888888888888 - in Potter's own words - 888888888888 > [Tom Potter:] > Hopefully, hanson will join with me,.......... > ... and begin to slaughter the instigators of conflict and war. > ... most Americans will turn against the Bush's.... > ... and join in the blood fest. In fact,..... > ... Bush and his friends, associates, and family and property > ... will be pissed on and destroyed, like Saddam's statues. > 8888888888888888888888888888888888888888888888 [Tom Potter] > Frankly, I'd like to see ..... the American government to investigate > and try to make a case against me...., Tom Potter. ...... > ************************* > Ahahahahaha........ahahahahaha..... > Peeping Tom Crackpotter, the MORAL man, ahahahaha.....hahahaha > ahahahaha.....hahahanson As can be seen, by Harry Hyena hanson's post, the bull machine is on. Like to see bull in motion? Just expose the people who defend and support the instigators of conflict and war, and lo!!!! The bull machine turns on. -- Tom Potter http://tompotter.us === Subject: Re: Fundamental Reason for High Achievements of Jews > As can be seen, Bush is willing partner > in the Bolshevik game. -- Tom The Bolsheviks are dead or hiding in the closet > or under your bed. --- Bob It is interesting to see that Bob Kolker and > Harry Hyena hanson see anti-Bolsheviks under > their beds, and they it is driving them mad. > I wonder why???? > Tom Potter Ahahahaha.......ahahaha.......so, it is interesting to/for you > to check under other peoples beds, Tom. Hmm, that then > is a new revelation about your character, you PEEPING TOM. > But your behavior here is not very surprising as it belongs > into the same level of your apparently very wide ranging > prurient interests, like when you bragged a while ago that a > two bit whore hit you over the head with her earnings,...... > a bag full of pennies. ..... man, you are hanging out in very > strange places, Tom. It looks more like it's YOU who is driving > mad from peeping and cavorting with hookers at the age of 71. > AHhahahahhaha.......ahahahanson > Yep! > Harry Hyena hanson has lost it. (That is, if he ever had it.) > Harry's problem is that he gets all bent out of shape, > when I point out that his beloved Bolsheviks > migrated to Israel and New York,from where > they are instigating the religious wars. -- Tom > Ahahahaha....AHAHAHAH......They are beloved to you, alright. > But I never said that I liked them. So, to make you happy for > the week, Tom, here's a battle song from the Landsers of WWII > on their march against the Bolsheviks. Sang your German soldier: > --- Hundertzehn Patronen in der Tasche, > --- and der Seite das Gewehr, > --- in der Hand die Handgranate, > --- Bolscheviki komm daher! > Isn't this balsam for your scrotal soul, Tom? > Tom, if the Bolsheviks, the object & subjects of your fetish, were > not there, then people like you will find or invent other groups for > the goal of their visions. There are/will always be people with > agendas that instigate, just like yours. All have the same goal, > power and riches, they just cloak it differently, like i.e. you do. > Or have you lost the ability to discriminate and see that you are > no different then the ones you attack, save the odor and tint? > Say, where are you from hanson, > and what nations are you a citizen of????? > Tom Potter > See, ....you are at it again, .....you are PEEPING again, > like a senile Peeping Tom who can't see the obvious any > longer. Consider an apology to your 2 bit whore and then > have her cleanse your plumbing. It may re-normalize you. > But don't be a cheap , Potter, pay her more that 2 bits. > or she may hit you over the head again........ahahahaha > AHAHAHAH.....hahahahaha.........ahahahahanson I am pleased to see that Harry Hyena hanson is beginning to understand the parable about the whore who hit the person who called her a two bit whore, over the head with a bag of quarters. As intelligent, rational, MORAL folks know, and as can be seen in the news, on the street, at work, at play, and in the news groups, call some people's hyping of conflict and war bull, and they dump all kinds of bull on you. Exposing the war for profit game is like turning on the tap of a high pressure bull faucet. Watch out for the quarters! Watch out for the bull! -- Tom Potter http://tompotter.us === Subject: Re: Fundamental Reason for High Achievements of Jews > I prefer to give any interested parties the benefit > of my experiences to see if it has any bearing > on their life. liberty and happiness. > Tom Potter This is too much....you wanna dole out benefits now , Tom? > I doubt that you'll get many takers, the way you turned out. > especially if you need to see if it has any bearing on their life > admitting here and now that you are a ing PEEPING TOM ...... > Ahahahaha........ahahahanson > I think that Harry Hyena hanson has lost it. > (That is, if he ever had it.) > Note his desperate efforts to negate my messages: > the way you turned out. > PEEPING TOM > This is too much > ahhahaha........ahahahahahaha > Tom-Tom, Tom, this is not a negation, it is an EVALUATION > of your message. Your complaint looks like a bitter reaction > to the fact that you haven't **found** it yet, > weven after 71 years...... > ahahahahaha......ahahahaha.......ahahahahah If you don't think I've **found** it take a look at my web site, and take a look at my recent pictures that show me doing my natural thing. (Enjoying life.) Why don't you post a lot of pictures of you engaged in your regular activities?? A picture is worth a thousand words. (Or really about 100,000 bytes in JPG.) I'd be more than happy to compare my life, accomplishments, and morality to the people who support and defend the people who instigate conflict and war, for power and riches. The point being, that morality, and commitment to life, liberty, and happiness for ALL people, rather than selfishness and greed, brings happiness. Spread the word. -- Tom Potter http://tompotter.us === Subject: Re: Fundamental Reason for High Achievements of Jews >> You are an utter ass, Tom Potter. Blame the victim is a totally played >> out strategy. You know this so you attempt to surround it with large >> doses of unprovable accusations to make the victim appear to be an >> incredibly effective plotter. >> Aside from all that, the Bolsheviks are dead and gone. Communism died >> (no tears shall we shed). Red fascism mostly died with Stalin and the >> Soviet Union is gone, and let us rejoice over that. So Bolshevism at >> this stage is a boogey man and a boojum. >> Now there are collectivists, socialists and people who want to >> redistribute what is not theirs, but this is not particularly Jewish. >> Politically savvy folk have been playing Lord Bountiful with other >> people's wealth and property since God invented dirt. >The Bolsheviks are dead and gone, >they migrated to Israel and New York, yes I hear they all live in a cold water flat in Harlem. >from where they are instigating the religious wars of the 2000's, >as the loot from their instigation of the class wars of the 1900's >is almost gone. cold water flats in NYC are expensive so the loot must be gone by now. >No doubt Robert J. Kolker knows that Communism, >like Socialism, Republics, Democracies, Monarchies, etc. >are just forms of governments, and they can be good or bad, >and all have their strong and weak points. quite true and the idea is to get rid of the ones that don't work isn't it. >Forms of government don't instigate conflict and war for >power and riches, the Bolsheviks do, and they have >had a long history of doing so. Look out! I just saw one run under your bed!!!! >As can be seen, I don't Blame the victims, >I blame the people who instigate conflict and war >for power and riches. Like Bush? >The Blacks, Latino's, Rednecks, Muslims, etc. >who sacrifice their lives, limbs, liberties and money, >to enrich these immoral instigators, like Bush? (hey, this is alt.politics.bush... gotta keep on topic) THOM >are the victims >Tom Potter http://tompotter.us === Subject: Re: Fundamental Reason for High Achievements of Jews >(hey, this is alt.politics.bush... gotta keep on topic) No. This crap is cross-posted to: - soc.college.admissions - sci.math - sci.physics - sci.chem - comp.lang.c - alt.politics.bush And nothing of your drivel is topical in (at very least) one of these groups. Are you able to spot? And now stop polluting c.l.c. -- Irrwahn -- Irrwahn (irrwahn33@freenet.de) === Subject: Re: Fundamental Reason for High Achievements of Jews >> You are an utter ass, Tom Potter. Blame the victim is a totally played >> out strategy. You know this so you attempt to surround it with large >> doses of unprovable accusations to make the victim appear to be an >> incredibly effective plotter. >Aside from all that, the Bolsheviks are dead and gone. Communism died >(no tears shall we shed). Red fascism mostly died with Stalin and the >Soviet Union is gone, and let us rejoice over that. So Bolshevism at >this stage is a boogey man and a boojum. >Now there are collectivists, socialists and people who want to >redistribute what is not theirs, sorry but socialists aren't into that. Modern socialism is a balance of public and private enterprise with an adiquate social safety net. THOM >but this is not particularly Jewish. >Politically savvy folk have been playing Lord Bountiful with other >people's wealth and property since God invented dirt. >Bob Kolker === Subject: Re: Fundamental Reason for High Achievements of Jews > sorry but socialists aren't into that. Modern socialism is a balance > of public and private enterprise with an adiquate social safety net. That means stealing from one and giving to other via taxes. Taxation is Theft (because it is enforced at gun point and is not voluntary like charity). The adequate social safety is hardley ever adequate, is almost never safe and is very unsocial to those who are taxed to pay for it. Bob Kolker === Subject: Re: Fundamental Reason for High Achievements of Jews >> sorry but socialists aren't into that. Modern socialism is a balance >> of public and private enterprise with an adiquate social safety net. >That means stealing from one and giving to other via taxes. Taxation is >Theft Stupidity alert! >(because it is enforced at gun point and is not voluntary like >charity). The adequate social safety is hardley ever adequate, is almost >never safe and is very unsocial to those who are taxed to pay for it. >Bob Kolker === Subject: Re: Fundamental Reason for High Achievements of Jews >> The German people did not advocate >> Do unto others, before they do unto you. >> as the poster suggests. >> What they did was respond as best they could >> to a greedy, immoral attempt to take over, >> and rape their nation. >> Ah, that explains the gas chambers. It was all self defense! >> Dude, you are a moron. >> If FDR had joined with the German people, >> rather than selling out to the Bolsheviks, >> WWII would have lasted two months, >> and there would have been no nuclear weapons, >> no Cold War, no Korean War, no Vietnam, and no >> instigation of religious wars. >> You are pretty silly too. >> bob >> - who would have been put on a train if he'd been living in Germany >> during the round-ups (one grandparent is all it took). >First of all, >the gas chambers have been vastly hyped, Why? isn't 6 million enough for you? Shall we throw in the SS Einzotgruppe that roamed around Russia just shooting Jews rather than gasing them? >and have become the a key part of an over all strategy, >as can be seen by looking at the facts, >and by taking an intelligent, rational, unbiased look at the facts. >Secondly, there is no doubt that people >use tit for tat as a social strategy, and no doubt >some Germans took tit for tat to extreme levels, >just as Israel is taking tit for tat to extreme levels today. yes they have been suckered into that haven't they by the Palistianians. >Most Germans recognized that their plight, >and the destruction of their nation, >had been brought about because many Jews >in Germany and around the world, had supported >the class war instigations of the Bolsheviks, Wrong again, the Bolsheviks were long gone by the 20's and replaced by Stalinists. >and had waged a propaganda war against >Germany and the nations who took action >to control the Bolshevik terrorists, >and they resented this. wrong again. >It is interesting to note that and that after the death of FDR, Truman, Hoover, and others, >who recognized the Bolshevik threat did a 180, >and began to counter them, just as the German people had >tried to do. ZZZZZZZZZZZZZZzzzzzzzzzzzzzz... I'm sorry did you say something? Hoover, Hoover... wasn't that the dude that threw us into the Great Depression? >As can be seen, Bush is a willing partner of the Bolsheviks, >who are instigating the religious wars of the 2000's, >to get back into the chips, as the loot from their >class wars is rapidly running out. >If Bush had not gotten into bed with Sharon >and the Bolsheviks, that bed must be in a grave yard since the Bolsheviks have been gone an awful long time. >there would have been no 911, >no war with Afghanistan and Iraq, Why, right wing wackos aren't the result of Bolshevikism or communism, they are the result of right wing wackoism. >no movement of America toward a police and military state, >no weakening of America's relationships with most nations on Earth, >no weakling of the United nations, NATO, the G8, etc. >no dollar flight, no meltdown of the economy, etc. no jelly in my peanut butter and jelly sandwich no rust or decay no All in the Family or Married with Children on TV no Gypsie fortune tellers no.... _________________ (please insert your favorite predjudice) >Also note, that thanatos@coldmail.nu >uses the standard tactic of people who support, >or have been brainwashed by Bolshevik propaganda, >of attacking the messenger. >They are masters of character and real assassination. >These are their major weapons. yah, that Alexander Korinski is a real terrorist. I'll bet he'se hiding out with Bin Laden or Saddam. THOM >Tom Potter http://tompotter.us === Subject: Re: Fundamental Reason for High Achievements of Jews >> The German people did not advocate >> Do unto others, before they do unto you. >> as the poster suggests. >> What they did was respond as best they could >> to a greedy, immoral attempt to take over, >> and rape their nation. >> Ah, that explains the gas chambers. It was all self defense! >> Dude, you are a moron. >> If FDR had joined with the German people, >> rather than selling out to the Bolsheviks, >> WWII would have lasted two months, >> and there would have been no nuclear weapons, >> no Cold War, no Korean War, no Vietnam, and no >> instigation of religious wars. >> You are pretty silly too. >> bob >> - who would have been put on a train if he'd been living in Germany >> during the round-ups (one grandparent is all it took). >First of all, >the gas chambers have been vastly hyped, > Why? isn't 6 million enough for you? Shall we throw in the SS > Einzotgruppe that roamed around Russia just shooting Jews rather than > gasing them? >and have become the a key part of an over all strategy, >as can be seen by looking at the facts, >and by taking an intelligent, rational, unbiased look at the facts. >Secondly, there is no doubt that people >use tit for tat as a social strategy, and no doubt >some Germans took tit for tat to extreme levels, >just as Israel is taking tit for tat to extreme levels today. > yes they have been suckered into that haven't they by the > Palistianians. Thom makes an interesting point when he suggests that the Germans were suckered into tit for tat responses against all Jews because of the class war instigations of a few Jews, that were so harmful to the German people, and the entire world. No doubt the religious war instigations of the Bolsheviks will be far more disastrous to mankind, and the tit for tat responses will be much greater against innocent people. If FDR had joined with Germany, Spain, Japan, and other nations, in going after the Bolshevik terrorists, rather than backing them, WWII would has lasted two months, and there would have been no millions of deaths, no Cold War, no Korea, no Vietnam , no nuclear weapons, and no instigation of global religious war. And if Bush had gone after the Bolshevik instigators, rather than selling out to them and Sharon, there would have been no 9/11, no movement of America toward a police and military state, no loss of freedoms to Americans, no danger to American's at home and abroad, no collapse of the economy, no shattering of the national budget, no war against the Iraqi people, with its' attendant cost, depletion of non-renewable resources, environmental damage, etc, no loss of trust by most nations of the world, no hatred of America by hundreds of millions of Muslims, no raid on Social Security, no freeze of Civil Service employees salaries, no bankruptcy of major airlines, no dollar flight, no meltdown of the dollar, no meltdown of America, no weakening of the United Nations, no weakening of NATO, the G8, no independent European GPS system, no loss of trillion dollar markets by Boeing, G.E. and other American firms, etc. -- Tom Potter http://tompotter.us === Subject: Re: Fundamental Reason for High Achievements of Jews > Hoover, Hoover... wasn't that the dude that threw us into the Great > Depression? Actually not. A lot of things were happening beyond the control of Hoover or anyone else. The world was on a binge during the 20's and the bubble went pop. So no matter who was in charge there would have been a recession. Puting the Smoot-Hawley tarrif in place in 1931 did not help. Many people think the stock market crash caused the depression. Not so. The recession was underway before that. Stock market crashes do not cause depressions. We had a real sharp crash in 1986 and no depression resulted. By the way, T. Potter sees Bolshies under the bed, behind the drapes, in the woodshed and inhabiting the shadows. He equates Bolshiveks with Jews. Never mind the the Bolshies that did the most damage were the gentiles, Lenin and Stalin. Poor Trotsky, aka Lev Bronshteyn (who was not very nice, btw) did only a fraction of the damage done by the other two. It was that ex seminary student Joe Stalin, a Groozysian thug, who forced farm collectivization on his country and caused at least 7.5 millions to starve. Stalin also created the Gulag. But Potter will blame that on the Jews too. Look at all those Jewish-Bolshivek villains. Beria, Kruschev who spilled the blood of their countrymen. Potter will blame all the bad weather on the Jews too. Bob Kolker === Subject: Re: Fundamental Reason for High Achievements of Jews > Hoover, Hoover... wasn't that the dude that threw us into the Great > Depression? > Actually not. A lot of things were happening beyond the control of > Hoover or anyone else. The world was on a binge during the 20's and the > bubble went pop. So no matter who was in charge there would have been a > recession. Puting the Smoot-Hawley tarrif in place in 1931 did not help. > Many people think the stock market crash caused the depression. Not so. > The recession was underway before that. Stock market crashes do not > cause depressions. We had a real sharp crash in 1986 and no depression > resulted. > By the way, T. Potter sees Bolshies under the bed, behind the drapes, in > the woodshed and inhabiting the shadows. He equates Bolshiveks with > Jews. Never mind the the Bolshies that did the most damage were the > gentiles, Lenin and Stalin. Poor Trotsky, aka Lev Bronshteyn (who was > not very nice, btw) did only a fraction of the damage done by the other > two. It was that ex seminary student Joe Stalin, a Groozysian thug, who > forced farm collectivization on his country and caused at least 7.5 > millions to starve. Stalin also created the Gulag. But Potter will blame > that on the Jews too. Look at all those Jewish-Bolshivek villains. > Beria, Kruschev who spilled the blood of their countrymen. > Potter will blame all the bad weather on the Jews too. It is interesting to see that Robert J. Kolker continues to suggest that ALL Jews are instigators of conflict and war for profit. I assert that all Jews are not instigators of conflict, any more than all Muslims are terrorists, or all Gypsies are fortune tellers. And as can be seen in the cases of Bush, Blair and Rumfeld, all instigators of conflict and war for power and riches are not Jewish. Robert J. Kolker does bring up a good point when he points out that it was a long up hill struggle for the native Russians to regain control of their country after the Bolshevics assassinated the native Russian leaders, and the Royal family, and co-opted the Russian government. Although Stalin, who had a Jewish wife, was a front man for the Bolshevics, he was motivated by self-survival, and had enough power to eliminate any one who was a threat to him. Robert J. Kolker brings up another good point when he suggest that FDR, who inherited a recession, and turned it into a long depression, by meddling with the economy in order to cater to the people who were instigating class warfare in America, and all over the world. Note that there was a more drastic stock market crash when Reagan was president, and as he let market forces make the necessary corrections, a year or two later, there was prosperity, rather than depression. And Robert J. Kolker makes another good point when he points out that Stalin, like FDR and many other international leaders, were forced by the class war instigations, to make drastic changes to his nations market system, with disastrous results. Of course, the historical data shows that the Bolshevics were responsible for co-opting the food from the area of Russia where millions starved, as punishment for anti-Semitic activities in that area. It amazes me, how so much spin has been put on history, and even current events, right in front of people's eyes, and that anyone buys the spin. It goes to show you that Pavlov was right. -- Tom Potter http://tompotter.us === Subject: Re: Fundamental Reason for High Achievements of Jews >> If thine enemy smite thee on thy cheek, tear his head off and down >> his neck. [...] A real man >> kills his enemies, takes his land and cattler, rapes his women, enslaves >> his children and contemplates how good life can be. >> In this evil world, if you >> are not on top, some other part is and is humping you. The price of >> liberty is efficiently killing the enemies of liberty. The rule is very >> simple. Support and help your friends. Kill your enemies. That is the >> only way to survive. >> Do unto others, before they do unto you. >> Fascinating, even if it's off topic. >> Robert Kolker, who claims to be Jewish, advocates a philosophy >> of life which sounds a lot like what we normally would associate >> with Nazis and supremacists. >> If someone claimed the Jews believe what Kolker claims to >> believe, that person would surely be labelled an anti-Semite. >> And yet, Kolker seems to be serious, and has received some support >> from others in this thread. >> What gives? >Actually, the masses have been brainwashed >to give the German people a bad rap, >for trying to defend themselves from >the Bolsheviks.who were instigating >conflict and war, for power and riches in the 1900's. Please give examples. Stalin was no angel but except for his defeat in Finland never tried what the right wing Hitler did. All his crimes were internial Stalin did not invade Germany in June 1941, Hitler Invaded Russia, Stalin did however cooperate with Hitler in the occupation of Poland. >As the Bolsheviks.were assassinating leaders and breaking >up the meetings of political rivals, Please give examples including dates. For the most part it was the Nazi's that that broke up and attacked rivals. >a few nations, >such as Germany, Italy, Spain, Japan, and Turkey >turned to strong leaders to counteract the Bolshevik menace, >which sought to co-opt their governments, >as they had the Russian government. Russia didn't have a government, it had a dictatorship run by the Tzars. Next you need to learn the difference between the Trioika Communists and the Bolsheviks. The Bolshevik Government under Alexander Korenski was over thown by the Troika (and keep in mind that Lenin was sent to Russia by the Kiaser to take power and knock Russia out of the war so why not Blame Willheilm for the left wing mess. >As can be seen by their contributions to society, >and their relationships with folks all over the world, >the German people are probably the most intelligent, >most moral, and most productive people on Earth. Yes they produced 6 million bodies in the gas chambers and in some cases produced consumer goods from the bodies like wigs and lamp shades made from human skin, plus they recycled all the gold in teeth and the like into the Reich's bank. >The Bolsheviks, who have a history of instigating >conflict and war for profit, excuse me that you have that wrong. Its the Nazi's and republicans that do stuff like for money and profit. >have a vested interest >in tagging their critics and adversaries, with negative boilerplate >as this keeps their scam alive. >Note that the Bolsheviks.instigated the class wars of the 1900's, >and after the native Russians regained control of their nation, >they migrated to Israel and New York, >from where they are instigating the religious wars of the 2000's, >to get back into the chips. are you drunk? >The German people did not advocate >Do unto others, before they do unto you. >as the poster suggests. >What they did was respond as best they could >to a greedy, immoral attempt to take over, >and rape their nation. By Hitler, the Nazi's and the rich industrialists that backed them. >If FDR had joined with the German people, >rather than selling out to the Bolsheviks, sorry but the Bolsheviks were long gone by 1940. >WWII would have lasted two months, >and there would have been no nuclear weapons, Except German ones >no Cold War, no Korean War, no Vietnam, OK. lets look at VN. Ho Chi Minn was a nationalists who even in the 30's was tryiung to get the French out of his country. When the Japs invaded the OSS gave him aid and he sided with us but when we broke our promises to him and gave SEA back to France to bribe them to join the UN, Ho turned to the only people left, the commies. Now given that, why do you think Ho Chi Minn and other Vietnamese Nationalists would not have fought for independence from the French??? > and no >instigation of religious wars. Yah right the Koran and bible would have gone puff and disappeared. >As can be seen, >Bush has also sold out to the Bolsheviks hard to do since the Bolsheviks haven't existed for 80+ years and the USSR for 13+ years. Bush sold out to the New World Order. THOM >Tom Potter http://tompotter.us === Subject: Re: Fundamental Reason for High Achievements of Jews >> If thine enemy smite thee on thy cheek, tear his head off and down >> his neck. [...] A real man >> kills his enemies, takes his land and cattler, rapes his women, enslaves >> his children and contemplates how good life can be. >> In this evil world, if you >> are not on top, some other part is and is humping you. The price of >> liberty is efficiently killing the enemies of liberty. The rule is very >> simple. Support and help your friends. Kill your enemies. That is the >> only way to survive. >> Do unto others, before they do unto you. >> Fascinating, even if it's off topic. >> Robert Kolker, who claims to be Jewish, advocates a philosophy >> of life which sounds a lot like what we normally would associate >> with Nazis and supremacists. >> If someone claimed the Jews believe what Kolker claims to >> believe, that person would surely be labelled an anti-Semite. >> And yet, Kolker seems to be serious, and has received some support >> from others in this thread. >> What gives? >Actually, the masses have been brainwashed >to give the German people a bad rap, >for trying to defend themselves from >the Bolsheviks.who were instigating >conflict and war, for power and riches in the 1900's. > Please give examples. Stalin was no angel but except for his defeat > in Finland never tried what the right wing Hitler did. All his crimes > were internial > Stalin did not invade Germany in June 1941, Hitler Invaded Russia, > Stalin did however cooperate with Hitler in the occupation of Poland. >As the Bolsheviks.were assassinating leaders and breaking >up the meetings of political rivals, > Please give examples including dates. For the most part it was the > Nazi's that that broke up and attacked rivals. It is interesting to see that Thom does not comprehend, that Germany, and many nations, tried to control the class war instigations of the Bolsheviks by many peaceful means, and by making treaties to oppose the instigation of conflict and war, and when all measures failed, they went after the Bolsheviks in the nations they were using as bases of operations, just as America went after real terrorists, and imagined terrorists in Iraq. Note that the Germans were welcomed in most nations as their local leaders had been bought or intimidated by the Bolsheviks and were unable to maintain the social, financial and cultural integrity of their nations. Even France had little opposition to the Germans. If FDR had joined with Germany, Spain, Japan, and other nations, in going after the class war instigators, rather than backing them, WWII would has lasted two months, and there would have been no millions of deaths, no Cold War, no Korea, no Vietnam , no nuclear weapons, and no instigation of global religious war. Bush has also sold out to the religious war instigators, and brought much harm to America and the world. And it is interesting to see that Thom does not know that the Bolsheviks were assassinating leaders, and assaulting people, and breaking up meetings, and that the Black Shirts and Nazi were reactions to this. I suggest that he read some factual, concurrent history of those times, rather than allow himself to be brainwashed by present day spin motivated more by agenda, than by facts or a respect for mankind. -- Tom Potter http://tompotter.us === Subject: Re: Fundamental Reason for High Achievements of Jews > Yes they produced 6 million bodies in the gas chambers and in some > cases produced consumer goods from the bodies like wigs and lamp > shades made from human skin, plus they recycled all the gold in teeth > and the like into the Reich's bank. Twelve million. The Nazis killed untermenschen other than Jews. Stalin killed 7.5 million Ukranians and only God knows how many others he worked to death in the Gulags. Stalin was a Monster of the same magnitude as Hitler. The Chinese Communists win the prize however. They liquidated 40 to 60 million class enemies to acheive their revolution. Nazi-ism was racial insanity but Lenninist Communistm was sanitized by Western intellectuals and still was death and destruction. Pinko-stinko fellow travelers and useful idiots put a good face on Stalin's doings. Bob Kolker === Subject: Re: Fundamental Reason for High Achievements of Jews ... > Stalin killed 7.5 million Ukranians and only God knows how many others > he worked to death in the Gulags. Stalin was a Monster of the same > magnitude as Hitler. > The Chinese Communists win the prize however. They liquidated 40 to 60 > million class enemies to acheive their revolution. Are you objecting to this? I think it was you who said that it is OK to kill all his enemies. I wonder how a brainless individual like you manages to write. And now, *please* stop posting to clc. Jirka === Subject: Re: Fundamental Reason for High Achievements of Jews >... >> Stalin killed 7.5 million Ukranians and only God knows how many others >> he worked to death in the Gulags. Stalin was a Monster of the same >> magnitude as Hitler. >> The Chinese Communists win the prize however. They liquidated 40 to 60 >> million class enemies to acheive their revolution. >Are you objecting to this? I think it was you who said that it is OK to >kill all his enemies. I wonder how a brainless individual like you manages >to write. It's all a matter of probability: let a bunch of bonobos play around with computer terminals for a suitable amount of time, and sooner or later something like what you are replying to is likely to find its way to usenet. >And now, *please* stop posting to clc. Seconded. -- Irrwahn === Subject: Re: Fundamental Reason for High Achievements of Jews : It's all a matter of probability: let a bunch of bonobos play around : with computer terminals for a suitable amount of time, and sooner or : later something like what you are replying to is likely to find its : way to usenet. And just what makes you think that this does not already occur on a daily basis? ----- Richard Schultz schultr@mail.biu.ac.il Department of Chemistry, Bar-Ilan University, Ramat-Gan, Israel Opinions expressed are mine alone, and not those of Bar-Ilan University ----- an optimist is a guy/ that has never had/ much experience === Subject: Re: Fundamental Reason for High Achievements of Jews >: It's all a matter of probability: let a bunch of bonobos play around >: with computer terminals for a suitable amount of time, and sooner or >: later something like what you are replying to is likely to find its >: way to usenet. >And just what makes you think that this does not already occur on a >daily basis? Nothing, I don't think so, and I didn't say so. In fact, it merely seems to occur even more frequently. :) Just as I said: it's a matter of probability... > an optimist is a guy/ that has never had/ much experience True enough. Irrwahn -- The generation of random numbers is too important to be left to chance. === Subject: Re: Fundamental Reason for High Achievements of Jews > Yes they produced 6 million bodies in the gas chambers and in some > cases produced consumer goods from the bodies like wigs and lamp > shades made from human skin, plus they recycled all the gold in teeth > and the like into the Reich's bank. > Twelve million. The Nazis killed untermenschen other than Jews. > Stalin killed 7.5 million Ukranians and only God knows how many others > he worked to death in the Gulags. Stalin was a Monster of the same > magnitude as Hitler. > The Chinese Communists win the prize however. They liquidated 40 to 60 > million class enemies to acheive their revolution. > Nazi-ism was racial insanity but Lenninist Communistm was sanitized by > Western intellectuals and still was death and destruction. Pinko-stinko > fellow travelers and useful idiots put a good face on Stalin's doings. It is interesting to see how many people have been conditioned to blame Communism for events orchestrated by a group of people who instigate conflict and war, for power and riches. Communism, like Socialism, Monarchies, Democracies, etc. are just forms of government, and they can be good or bad, and ALL forms of government have their good and bad points. Of course, as can be seen by historical examples, the major weakness of a Democracy is that it can be bought by special interest groups, or manipulated by demogogues. As can be seen by the historial evidence, the Bolshevics were the major instigators of the class wars of the 1900's, the colonial wars of the 1800's, and they are right in there instigating the religious wars of the 2000's, and naturally when wars are instigated, bad things happen. The instigation of conflict and war, for power and riches is the stock in trade of Bolsheviks, much as fortune telling is the stock in trade of Gypsies. They netted trillions of dollars from their class wars, and they expect to net far more from their religious wars. The shame of it all is, that most of the Bolsheviks, and after these conflicts rage for a while, the combatants eventually turn against the good Jews. Can you imagine what will happen in America, when hundreds of thousands of bitter Blacks, Latino's, and Rednecks return from the religious wars, and who they will blame for their wounds, the deaths of their buddies, their loss of liberty, separation from their children and spouses, and their financial and social misfortunes, and the degeneration of their nation???? -- Tom Potter http://tompotter.us === Subject: Re: Fundamental Reason for High Achievements of Jews >> AHhahahhaha......Bob, are you having a bout of genocidal >> upwellings again? As a Jew, you are supposed to be better >> and more lovable then the Muslims...and 5.5 times smarter. >> You said so yourself!. >> Now, you've become like them, a Jewish Homicide bomber, >> a Wailing-Wall-Martyr....a Sui-Holocauster........ >> ahahahhha............ahahahahanson >There is nothing wrong with killing people who have harmed you, are will >harm you eventually. We send our soldiers to die regularly. It is >business as usual. I just want a high body count. what kind of stupid are you ? Bruce ----------------------------------------------------------------------- It was so much easier to blame it on Them. It was bleakly depressing to think that They were Us. If it was Them, then nothing was anyone's fault. If it was Us, what did that make Me ? After all, I'm one of Us. I must be. I've certainly never thought of myself as one of Them. No-one ever thinks of themselves as one of Them. We're always one of Us. It's Them that do the bad things. <=> Terry Pratchett. Jingo. === Subject: Limit point compacness help? Is a limit point compact subspace of a Hausdorff space necessarily closed? Well since I can't seem to prove this and since if this is true, the proof would look nothing like the proof that a compact subspace of a Hausdorff space is closed, I am trying to come up with a counterexample. There is an example of a limit point compact space that isn't compact in my Munkres book which is : Let Y consist of two points and give Y the topology consisting of Y and the empty set. Then X = Z_+ x Y is limit point compact but not compact. (Z_+ is the positive integers). Can I use this to show a counterexample? I guess the question is does X sit inside of a Hausdorff space making X open? I don't think it does because X itself is not Hausdorff since the open sets consist of unions of {n} x Y and thus if Y = {a,b}, then 1 x a and 1 x b do not have disjoint neighborhoods. Thoughts? Steve === === === Subject: Re: Is ...9999.9999... = 0 ? This is a much more intelligent question than most of the responses indicate. The simplest answer is that ...999.999... is not a real number, because the series sum[i=0..inf]{9*10i} does not converge over the real numbers. However, there are situations in which number representations like ...999.999... do make sense. For example, in the binary two's-complement notation used by all digital computers, the number representation ...1111 corresponds to -1. The reason for that is that, as with your example, adding 1 to ...1111 causes all digits to be wiped out by carries. This can be similarly extended to ten's complement arithmetic, where ...999 equals -1 and, equivalantly, ...999.999... equals 0. So the question you've asked shows some great intuition along these lines. To be precise, constructs like two's complement and ten's complement are number representations (as are decimal numbers and Roman numerals), rather than number systems (as are the real numbers and the integers). So when you talk about ...999.999... you are talking about a scheme for encoding real numbers into written symbols, and not about a fundamental property of real numbers themselves. However, there exists a number system in which concepts like ...999 are more fundemantal: the p-adic numbers (for any prime number p, for example for p=7 there exist the 7-adic numbers). There, instead of repeating decimals like 0.272727.., you can have numbers like ..27272727.0. p-adic numbers are of theoretical interest and have some counterintuitive properties, but they aren't very applicable to everyday use. === Subject: Re: Is ...9999.9999... = 0 ? Brian Quincy Hutchings scribbled the following: > there's no opposite of a projection, but > two or more of them is a perspectivity, > viz painting (Brunelleschi et al). > the real meet (sik) of this is just in ten's complimentation, Spelling nitpick: (sic), not (sik). > in the base of ten. or nine's complimentation. that is, > when you subtract a larger number from a smaller, > this is whta you get, til you do the complimentation > to get the negative number. -- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ To err is human. To really louse things up takes a computer. - Anon === Subject: Re: Is ...9999.9999... = 0 ? Chapman (I guess) had me correctly: I use sik when I make a mistake on purpose, and sic when others make a mistake, on prupose or not. > viz painting (Brunelleschi et al). > the real meet (sik) of this is just in ten's complimentation, > Spelling nitpick: (sic), not (sik). --Dec.2000 'WAND' Chairman Paul O'Neill, reelected to Board. Newsish? http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac === Subject: Re: Is ...9999.9999... = 0 ? > Brian Quincy Hutchings scribbled the following: >> there's no opposite of a projection, but >> two or more of them is a perspectivity, >> viz painting (Brunelleschi et al). >> the real meet (sik) of this is just in ten's complimentation, > Spelling nitpick: (sic), not (sik). >> in the base of ten. or nine's complimentation. that is, >> when you subtract a larger number from a smaller, >> this is whta you get, til you do the complimentation >> to get the negative number. Annd another spelling nitpick: complementation not complimentation. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Is ...9999.9999... = 0 ? Robin Chapman scribbled the following: >> Brian Quincy Hutchings scribbled the following: > there's no opposite of a projection, but > two or more of them is a perspectivity, > viz painting (Brunelleschi et al). > the real meet (sik) of this is just in ten's complimentation, >> Spelling nitpick: (sic), not (sik). > in the base of ten. or nine's complimentation. that is, > when you subtract a larger number from a smaller, > this is whta you get, til you do the complimentation > to get the negative number. > Annd another spelling nitpick: complementation not complimentation. Ignoring the nitpick on the first word of your reply, who says it wasn't intentional? Perhaps you have to say something like: Please, O most illustrious holy one, the beacon of wisdom, emperor for eternity, may I get the 'negative' number?. -- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ The truth is out there, man! Way out there! - Professor Ashfield === Subject: Mathematical Olympics I've always wondered about Mathematical Olympics. Don't the judges have to be smarter than any of the contestants, to be able to judge them accordingly? In normal Olympics, the judges can be any skinny little wimps off the street, as they can merely watch the athletes without competing themselves. But mathematics is different. Unlike athletics, it's not plain to see who is better than who. You have to be a mathematician yourself. There's *one* thing that suggest that the judges don't have to be the smartest ones after all - in certain theoretical calculations, checking whether something has been done is far easier than actually doing it. For example consider the simple array sort. Actually sorting the array can't be done faster than O(nlog n) time, this has been proven. But checking if the array has been sorted can be done in O(n) time with a simple algorithm any child could come up with. Maybe this can be expanded to more complicated calculations? -- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ Roses are red, violets are blue, I'm a schitzophrenic and so am I. - Bob Wiley === Subject: OLD HP CALUCLATORS I am looking for some old HP calculators like HP 41CV, HP 41CX, HP 71B, HP 15C, HP 16C, HP 67 and any others in the 1980's era...If you have one you would like to sell please email me at === Subject: Re: OLD HP CALUCLATORS > I am looking for some old HP calculators like HP 41CV, HP 41CX, HP > 71B, HP 15C, HP 16C, HP 67 and any others in the 1980's era...If you > have one you would like to sell please email me at Here's a good source of older HP calculators: http://www.hpmuseum.org/cgi-sys/cgiwrap/hpmuseum/adforum.cgi -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: OLD HP CALUCLATORS I've got an old 97 but it's not a wreck so I'm not selling. Sorry > I am looking for some old HP calculators like HP 41CV, HP 41CX, HP > 71B, HP 15C, HP 16C, HP 67 and any others in the 1980's era...If you > have one you would like to sell please email me at === Subject: Re: OLD HP CALUCLATORS > I am looking for some old HP calculators like HP 41CV, HP 41CX, HP > 71B, HP 15C, HP 16C, HP 67 and any others in the 1980's era...If you > have one you would like to sell please email me at I have an HP55 and no, I'm not selling. Bought it in 1975 and it still works :-) http://www.dotpoint.com/xnumber/hp55.htm Dirk Vdm === Subject: differential equation Hi! I have to solve this differential equation: r'' = k/(|r|2) where r is a 2-d vector. Marco === Subject: Re: differential equation >Hi! I have to solve this differential equation: >r'' = k/(|r|2) >where r is a 2-d vector. What is k? (My guess would be that k was a constant, but if so this can't be exactly the equation you need to solve, because the left side is a vector and the right side is a scalar...) > Marco === Subject: Re: how to prove that this is a measurable function >>Show that >>1) (x,y) |-> x+y is measurable; >>2) if f: X -> Y, g: Y -> Z are measurable, then so is g op f >>(composition); and >>3) if f, g: X -> R are measurable, so is fg (product). >> >2) is false. The composition of Lebesgue measurable (LM) functions need not >be LM . However, compositions work for Borel measurable functions, so one >could work with that. Another approach: X_B(y) is clearly LM on R2 and >X_A(x+y) is the characteristic function of T(R x A), where T is a linear >transformation on R2. > > How could it be false? I am talking of abstract measurable spaces > here. Given a set A in the sigma field on Z, isn't (g o f)-1 (A) > = f-1 (g-1 (A)) in the sigma field on X? Is the definition of LM > only that the inverse maps Borel sets to Lebesgue sets? The problem with saying a function f is measurable with this definition is that continuous functions need not be measurable when both sigma-algebras are the collection of Lebesgue measurable sets. In fact, let g be the 'Cantor staircase' or 'devil's staircase' and let f(x)=x+g(x). Then f is strictly increasing, continuous, and takes a set of measure 0, C, to a set of positive measure f(C). Now let A be any non-measurable subset of f(C). Then f(A) is measurable, but f-1 (f(A)) is not. The point is that, while your definition is perfectly good when dealing with Borel sets, or generic sigma-algebras, it has unpleasant effects when dealing with Lebesgue measurable sets. In particular, the composition of a measurable function and a continuous one need not be measurable. However, this is the situation the OP wanted. What is needed is that inverse images of sets of measure 0 via the continuous function have to again be of measure 0. This does happen for the OP. --Dan Grubb === Subject: Re: how to prove that this is a measurable function There's a problem in a real analysis book I'm reading that asks to show that (x,y) |-> f(xy) is measurable in R2 if f(x) is measurable in R. For that, I want to show that k{-1}(E) is lebesgue measurable if E is measurable in R. I seem to be unable to show this... Any ideas? Any help appreciated, cheers, nojb. How can I prove that the function g(x,y) = X_A(x+y) X_b(y) is lebesgue > measurable in R2 where A, Bsubset R are measurable sets and X_A, X_B > are the charactersitic functions? > If A and B are Borel measurable subsets of R, then the proof you request > is easy: both (x,y) --> x+y and (x,y) --> y are continuous maps of R2 > into R, hence they are both Borel measurable. Therefore so are the > compositions (x,y) --> X_A(x+y) and (x,y) --> X_B(y) Borel measurable, > hence so is their product. > If A and B are Lebesgue measurable, then so is your function. This is > because both of the maps (x,y) --> x+y and (x,y) --> y have the property > that the inverse image (with respect to either map) of a Lebesgue > measurable set is Lebesgue measurable. The same composition argument as > before finishes the matter. === Subject: Re: how to prove that this is a measurable function > There's a problem in a real analysis book I'm reading that asks to > show that (x,y) |-> f(xy) is measurable in R2 if f(x) is measurable > in R. If E is a Borel set in R of measure 0, use Fubini's theorem to show that {(x,y) : xy lies in E} has measure 0 in R2. This implies f(xy) is a.e. equal to a Borel measurable function on R2, hence is Lebesgue measurable on R2. === Subject: Re: how to prove that this is a measurable function >2) is false. The composition of Lebesgue measurable (LM) functions need >not >be LM . However, compositions work for Borel measurable functions, so one >could work with that. Another approach: X_B(y) is clearly LM on R2 and >X_A(x+y) is the characteristic function of T(R x A), where T is a linear >transformation on R2. > >How could it be false? I am talking of abstract measurable spaces >>here. Given a set A in the sigma field on Z, isn't (g o f)-1 (A) >>= f-1 (g-1 (A)) in the sigma field on X? Is the definition of LM >>only that the inverse maps Borel sets to Lebesgue sets? >> >If f : R -> R, then a standard definition is that f is LM iff >f(-1)[(a,oo)] is LM for all a in R. > If so, the answer to my latter question is affirmative. gave. > In general > measurable spaces, a funciton is measurable iff its inverse map sets > from one sigma field into the other. Contrary to your declaration, my > original claim, > 2) if f: X -> Y, g: Y -> Z are measurable, then so is g op f > (composition) > is true. In that context, yes you are of course right. The OP was asking about LM functions, and compositions of those can give headaches. Yes, we can use your 2) in this context if we're careful to specify that the map T(x,y) = x + y is more than just LM; T(-1)(E) should be LM for every LM E contained in R. === === Subject: Re: why is Hamel dimension well-defined? > [[ This message was both posted and mailed: see > the To, Cc, and Newsgroups headers for details. ]] > Would someone please point me to a proof of this theorem? Any two Hamel > bases of a vector space have the space have the same cardinality. This is > driving me crazy. Have a tolerable existence. Eli There is a proof in Real and Abstract Analysis by Hewitt & Stromberg. My favorite one is in Jacobson's Basic Algebra II. It uses the idea of an abstract dependence relation. The nice thing about it is that it can be generalized to 'countable dependence relation' and obtain that the dimension of a Hilbert space is well-defined also. Jacobson uses it to obtain results about completely reducible modules. --Dan Grubb === Subject: Re: Rotate Quaternion numbers? > Is it possible, using quaternion numbers, to rotate 3 dimensions > around the fourth. A bit like rotating x and y around the z axis. Or > am i being stupid? > I have an object that exists in 4D Quaternion space. I can view it in > 3D (b using a simple ray-tracing technique) and i can rotate in 3D > around either of the other axis. Any ideas? The (continuous) automorphisms of the quaternions (these are the maps that preserve the structures, addition and multiplication as well as the topology) correspond to the 3D rotations, i.e. the group SO(3). Given an arbitrary nonzero quaternion u, we can construct an automorphism in defining the map q |--> u*q*(u-1) where u-1 is the inverse of u. Because any real number commutes with all quaternions, the the u deletes out when q is real; so the whole real axis is mapped onto itself. And the other three (imaginary) dimensions are rotated around the axis given by u (because u itself is also mapped to itself, so it is a fixed point) by an angle which is given by 2*arctan(Im(u)/Re(u)) (sketch of proof: calculate the trace of the above mapping as an endomorphism of a real vector space and compare to the trace 1+2*cos(phi) of a rotation of angle phi in 3D). Given two nonzero quaternions u1 and u2, they induce the same rotation if and only if their quotient is a real number. === Subject: Re: Rotate Quaternion numbers? I'm still not 100% fully understanding though. Perhaps it's the Math syntax: q' is the rotated Quaternion number yes? q is the original? That leaves a and b. Are these the 2 angles you mentioned? I need to think about this a lot more! Mat > Hello everybody, Is it possible, using quaternion numbers, to rotate 3 dimensions > around the fourth. A bit like rotating x and y around the z axis. Or > am i being stupid? > Not at all, but as you probably reaslise it's a bit more complicated in > 4D. In 4D rotations take place about planes. the simplest rotation in 4D > fixes a plane (e.g. the x-y plane) and rotates in the other two > dimensions, so a 90 degree rotation takes z -> w and w -> -z. > A more general 4D rotation taks place about two orthogonal planes and > rotates about both of them simultaneously through two different angles. To > completely specify the rotation you need to give one plane (the second is > uniquely defined by being orthogonal to the first) and both angles, + > senses of rotation if it is not obvious from the way the plane(s) are > defined. > This has a number of consequences. Unlike in 3D (or 5D or 7D) a general 4D > rotation has no fixed point except the origin. Also above 3D rotations are > not simple, meaning that for rotation R it is not generally true that Rt > == I for some t. > You can use quaternions to rotate in 4D. The formula is > q' = aqb > The pair (a, b) of unit quaternions specifies the rotation, multiplication > is just quaternion multipilication, and the pairing is unique up to > replacing (a, b) with (-a, -b). You can of course use special orthogonal > matrices, but they are in my opinion the least user friendly way to get > things done in 4D. > John === Subject: Re: Rotate Quaternion numbers? > I'm still not 100% fully understanding though. > Perhaps it's the Math syntax: > q' is the rotated Quaternion number yes? > q is the original? > That leaves a and b. Are these the 2 angles you mentioned? No. a and b are unit quaternions. The formula q' = a * q * b gives a 4D rotation, from q to q'. '*' is quaternion multiplication. It is obvious that this is a linear transformation, and the properties of quaternion multiplication and the fact that a and b are unit quaternions mean it preseves lengths and so is orthogonal. It's less obvious that all 4D rotations can be written this way, and that for each rotation the pair of rotations that generates it is unique, up to multiplying both quaternions by -1. I'm not sure how useful this method of generating 4D rotations is: it's mathematical compactness and symmetery suggests there's more to it than meets the eye, but I've not come across anything more about it than given above. I mentioned it as your post metioned 'using quaternion numbers' to do rotations in 4D, and this one way to do so. John === Subject: Re: Core error, FEAR is a natural response >: Here's a demonstration of how you can get a problem, using xy=2, where >: x and y are algebraic integers. Consider x=2a, and y = b, so b is an >: algebraic integer, but 'a' is not, so x does not technically have 2 as >: a factor in the ring of algebraic integers, as that requires *both* >: factors be algebraic integers, while 'a' is not one. So b *should* be >: a unit, but because ab=1, it is NOT a unit, because the unit >: definition would require that both 'a' and b be algebraic integers. >I don't understand how you are deciding what should and should >not be in the algebraic integers. >>Here's a simpler example, consider 2 and 6 in the ring of evens, but 2 >>is NOT a factor of 6 because 3 is not even, understand? > I'm going to join with the others in asking what point > you meant to make here regarding what should be in > the ring of algebraic integers. > You have the situation that 2a = 6. Yet a is not in the > ring of evens. Clearly 2 is a factor of 6, but 2 is not > a factor of 6 in the ring of evens. Are you saying > that a should be in the ring of evens, that there's > an error in the definition of the ring, that the ring > is incomplete? > What is the should test? The fundamental problem is that should is not a word with a generally accepted meaning in mathematics. JSH follows his own inscrutable rules, not those of mathematics. Gib === Subject: Re: Core error, FEAR is a natural response >I think many of you *should* be terrified beyond the capacity for >rational thought, which is what I'm seeing. Is this the same should as in your proof, or do the asterisks change the logical meaning? >[...]*belief*[...]*selectively*[...]*has* I see a definite pattern of usage here. Could you clarify the asterisk operator for us? === Subject: Re: Core error, FEAR is a natural response linux) > I see a definite pattern of usage here. Could you clarify the asterisk > operator for us? The asterisk operator is a standard means of emphasis (boldface) on Usenet. Some newsreaders (like Gnus) even render the boldface and omit the asterisks. Maybe you knew that, but maybe not. -- When I am grown to man's estate I shall be very proud and great, and tell the other girls and boys not to meddle with my toys. --Robert Louis Stevenson === Subject: Re: Core error, FEAR is a natural response is this the Real JSH, at MSN? anyway, I saw the problem with it, the first time you'd stated it: it's nonsensical, because you've simply exchanged the even numbers for the units; that is, it's just a matter of labelling: 2 is your unit, and 4 is your first even mumber, 6 is odd. or are the evens really a ring, since they add, subtract and multiply? there's a very thin line between trembling in fear, and shaking with laughter, dood. as with I could, but I don't think it's worth the effort, and it wouldn't really show you anything. maybe you should move on to greener pastures; I know, I said, I would! I mean, I'd spend a lot less time in sci.math, if i didn't indulge in this ridiculous pasttime. > Here's a simpler example, consider 2 and 6 in the ring of evens, but 2 > is NOT a factor of 6 because 3 is not even, understand? > Similarly, because 'a' above is not an algebraic integer, then though > b should be a unit, it's not because the *definition* for unit > requires that both 'a' and b be in the ring of algebraic integers, so > b is not a unit, which means it is a non-unit factor of 2, which gives > a false implication, since x = 2a, like with 6, 2(3) = 6. > I don't understand this definition. Could you specify a numerical > value for a, to make the example more concrete? > Well, I could, but I don't think it's worth the effort, and it > wouldn't really show you anything. > As for the definition it suffices to note that it doesn't allow 'a' to > be 1/2, as that would mean that 2 is a unit, since 2(1/2) = 1. > Given that xy=2, with x=2a, y=b, with b an algebraic integer, x an > algebraic integer, while 'a' is not, it *should* be that y does not > share non-unit factors with 2, since x *should* have 2 as a factor, > but in the ring of algebraic integers, because the definition > arbitrarily excludes 'a', neither of those is the case. --les ducs de Buffet; vote NONE OF THE BELOW on Trickier Dick Cheney's California Recall & e-Dereg! http://larouchepub.com === Subject: Re: Core error, FEAR is a natural response >What I'm facing is a situation where the math--rather basic >algebra--says I'm right. > Of course, you believe it's right. And if other people believed it > was right, then your explanation of their actions might be plausible. > But it seems to me that they *don't* believe you're right; rather they > believe you are mistaken, confused, deluded or however they think of > it. Then isn't there a rather more obvious explanation for their > actions? It's MATH. It's not about belief. A math proof begins with a truth and proceeds by logical steps to a conclusion which then MUST be true! That's what makes this situation obscene as I can step through the proof from begining to end, and mathematicians simply decide to avoid or deny. They run away. Remember besides posting on newsgroups I've spoken to a mathematician in-person and explained it step-by-step, and he was quite skeptical and challenging as to the math, though otherwise the conversation was congenial. I handled every challenge, explained every step, and showed how it flowed logically from the previous. The proof is in fact that--a proof. There is no mathematical basis for challenging it at this point, and as it is a short proof, there is no RATIONAL way that mathematicians can avoid its conclusion. It's mathematicians who are acting on belief, not me. After all, who needs belief, when you have algebra? James Harris === Subject: Re: Core error, FEAR is a natural response What I'm facing is a situation where the math--rather basic >algebra--says I'm right. Of course, you believe it's right. And if other people believed it > was right, then your explanation of their actions might be plausible. > But it seems to me that they *don't* believe you're right; rather they > believe you are mistaken, confused, deluded or however they think of > it. Then isn't there a rather more obvious explanation for their > actions? > It's MATH. It's not about belief. A math proof begins with a truth > and proceeds by logical steps to a conclusion which then MUST be true! The Harris kind of proof proceeds by steps whose meaning can only be inferred by guessing, has the odd error in it, plus some f***ing big gap right in the middle where things get interesting, and ends with a conclusion that only you could belief. === Subject: Re: Core error, FEAR is a natural response > The Harris kind of proof proceeds by steps whose meaning can only be > inferred by guessing, has the odd error in it, plus some f***ing big gap > right in the middle where things get interesting, and ends with a > conclusion that only you could belief. A concise description of Harristotelian logic... Gib === Subject: Re: Core error, FEAR is a natural response > It's MATH. It's not about belief. A math proof begins with a truth > and proceeds by logical steps to a conclusion which then MUST be true! Show me the logical step that produces the word *should*. I know how to prove formal statements A is B and A is not B with formal logic. But how to prove A should be B with logic? > I handled every challenge, explained every step, and showed how it > flowed logically from the previous. The proof is in fact that--a > proof. Explained every step? Really? And how exactly A should be B followed from the previous? === Subject: Re: Core error, FEAR is a natural response > It's MATH. It's not about belief. A math proof begins with a truth > and proceeds by logical steps to a conclusion which then MUST be true! > Show me the logical step that produces the word *should*. > I know how to prove formal statements A is B and A is not B with formal logic. > But how to prove A should be B with logic? Readers should note that I have given the proof step-by-step and it doesn't have should in it at all. In fact, besides writing a paper, which I received email confirmation yesterday is still under review at a mainstream math journal, I've given repeated explanations, in detail. Rather than deal with the actual math argument, posters pick and choose from various posts I've made explaining various aspects of the definition problem currently sitting at the heart of mathematics in an area called algebraic number theory. It has been there for over a hundred years, so necessarily it has its subtleties. > I handled every challenge, explained every step, and showed how it > flowed logically from the previous. The proof is in fact that--a > proof. > Explained every step? Really? > And how exactly A should be B followed from the previous? And again, I point out that posters pick and choose from discussions where I explain, rather than going to the posts that actually do step through the argument carefully, and in detail. It'd be one thing if it were just a Usenet problem, but my experience with Professor McKenzie of Vanderbilt University, where I explained everything in-person and handled objections as he brought them up, yet he still sought to run away by claiming it was out of his area, highlights that it's a systemic problem. Mathematicians *can* decide to ignore a correct math argument, and keep doing so, despite realizing that the negative consequences if they are caught are huge. You should think about that reality. Basically, clearly certain mathematicians know they can lie to you about mathematics. James Harris === Subject: Re: Core error, FEAR is a natural response > It's MATH. It's not about belief. A math proof begins with a truth > and proceeds by logical steps to a conclusion which then MUST be true! > Show me the logical step that produces the word *should*. > I know how to prove formal statements A is B and A is not B with formal logic. > But how to prove A should be B with logic? > Readers should note that I have given the proof step-by-step and it > doesn't have should in it at all. In fact, besides writing a paper, > which I received email confirmation yesterday is still under review at > a mainstream math journal, I've given repeated explanations, in > detail. All your short explanations of the error in core are using extremely imprecise and consufing word should. You even use asterisks, like that - *should*. So can you give the short explanation without that non-math word? > Rather than deal with the actual math argument, posters pick and > choose from various posts I've made explaining various aspects of the > definition problem currently sitting at the heart of mathematics in an > area called algebraic number theory. > It has been there for over a hundred years, so necessarily it has its > subtleties. > I handled every challenge, explained every step, and showed how it > flowed logically from the previous. The proof is in fact that--a > proof. > Explained every step? Really? > And how exactly A should be B followed from the previous? > And again, I point out that posters pick and choose from discussions > where I explain, rather than going to the posts that actually do step > through the argument carefully, and in detail. > It'd be one thing if it were just a Usenet problem, but my experience > with Professor McKenzie of Vanderbilt University, where I explained > everything in-person and handled objections as he brought them up, yet > he still sought to run away by claiming it was out of his area, > highlights that it's a systemic problem. > Mathematicians *can* decide to ignore a correct math argument, and > keep doing so, despite realizing that the negative consequences if > they are caught are huge. > You should think about that reality. > Basically, clearly certain mathematicians know they can lie to you > about mathematics. > James Harris === Subject: Re: {Field Theory} This notation could get very confusing very fast... > I am reading Herstein's Abstract Algebra and teaching myself field > theory. I came to the following and it gave me a bit of trouble (I am > paraphrasing it for people who don't own that book): > If F is a finite field with q elements, then viewing F simply as an > abelian group under its addition, +, we have from group theory that > qx=0 for all x in F. > Now after a bit of consideration I came to the conclusion that, yes, > this is true, if we go by the notation (which Herstein did not > explicitly define) that qx, where q is a counting number and x is an > abstract object, means x+x+x+...+x, q times. Yes, Herstein could have made this more explicit in the discussion on groups. The point is that xn for a multiplicative group is written n*x for an additive group. > In terms of fields > this takes a little bit of chewing before swallowing since now we have > not one but two very different kinds of multiplication, each of which > uses the same symbol. This happens often. For example, when a field K contains a subfield F, you can view K as a vector space with F as scalars. Then, f*k could either be field multiplication in K or scalar multiplication in vector space K. > This is still unambiguous when the elements of our field are abstract > things, or at least anything BUT counting numbers. But if our field > is in fact counting numbers, it brings up a very gaping problem, at > least as I see things (although, true, I am very much a newbie!) [cut] > I am probably just overlooking something obvious, and am sure I will > be corrected! I don't think you are overlooking anything. Yes, the above notation can lead to ambiguity. If that happens, then you have to use a more precise, explicit notation to distinguish the field multiplication from the group repetition operation. Such sloppiness is very common in mathematics. -- Bill Hale === Subject: Re: {Field Theory} This notation could get very confusing very fast... at 09:48 PM, snizpilbor@yahoo.com (Sniz Pilbor) said: >In terms of fields >this takes a little bit of chewing before swallowing since now we >have not one but two very different kinds of multiplication, each of >which uses the same symbol. Correct; it is quite common in Mathematics to use the same symbol for related notions and to epect the reader to identify by context which is meant. >This is still unambiguous when the elements of our field are >abstract things, It's unambiguous across the board. >But if our field is in fact counting numbers, What do you mean by counting numbers? Integers? They don't form a field. >To be specific, suppose F is a field of counting numbers, There is no such thing. What is the multiplicative inverse of 2? Now, perhaps you meant the ring of integers, but there is still no ambiguity. >What, then, are we to make of ab? It is ambiguous No it isn't. >has the three possible meanings: No. If you apply the distributive law you will see that all three expressions have the same value. >1. a mulitplied by b by the multiplication operation of F (1+1+...1 (a times))*(1+1+...1 (b times)) >2. a+a+a+...+a (b times), (1+1+...1 (a times))*(1+1+...1 (b times)) >3. b+b+b+...+b (a times) (1+1+...1 (a times))*(1+1+...1 (b times)) >Was Herstein's choice of symbols merely a bad fluke? No. It was normal and consistent. >else it seems that the study of fields of integers There are none. Perhaps you mean rings. >would soon be rendered hopelessly ambiguous No. >and context-dependant. Yes, but it is a convenient and harmless context dependence, and can easily be removed if your willing to use more cumbersome expressions. >I am probably just overlooking something obvious, See above. > When you say 2 = 1+1, by your + are you talking about >normal addition, or the field's addition? It doesn't matter, because you get the same result either way. There is only one nondegenerate homomorphism of Z into a ring or field, so it is convenient to use the same symbol for the integer 1 and the identity of the ring or field. >In a very well behaved field, What is a well behaved field? The relevant porperties apply to *ALL* fields. >but in a field of integers What is that? >whose addition and multiplication have nothing to do with normal >addition and multiplication, It doesn't matter, as long as they satisfy the requirements for a field. >it becomes ambiguous. Take your field F. It has a unique additive identity, all it O, and a unique multiplicative Identity, call it I. There is a unique nondegenerate homomorphism E: Z->F, since E(0) must be O, E(1) must be I and all other values are determined by the fact that it is a homomorphism. Now if q is a positive integer and b is in F, b+b+...b (q times) is b*I+b*I+...b*I (q times). Now apply the distributive law and you get b*(I+I+...I (q times)). But (I+I+...I (q times)) is G(q). So there is no ambiguity in overloading the symbols. -- spamtrap@library.lspace.org === Subject: Re: {Field Theory} This notation could get very confusing very fast... > As F is a field, it has an multiplicative identity usually notated 1. > Thus 1x = x is no problem. If you consider 2 = 1+1, etc., > then '2'x = (1+1)x = 1x + 1x = x+x = 2x, etc. > Thus you've your situation. > When you say 2 = 1+1, by your + are you talking about > normal addition, or the field's addition? In the above all was within the field except for = 2x where I use the notation (2)x = x+x where (2) is the integer 2 and '2' = 1+1 in field. By convention nx = x+x+..+x unto n x's but xn isn't used where n is integer and x is group element. === Subject: Re: JSH: Harris's Big Fat Blunder > James Harris posted a new version of his proof of a core > error on October 12. He made a simple algebra mistake > which he has since acknowledged. > The interesting thing about this is that, when the algebra > mistake is corrected, it leads immediately to a proof that > him main conclusions are wrong. ... JSH will ignore your post. Gib === Subject: Re: JSH: Harris's Big Fat Blunder Nora Badboobda Baron, you stole that line from Captain Arnie! oh, wait; that was Peter Townsend. nevermind. PS: just as I've been saying (for years, in hte larger aspect), Captain Arnie has immediately proposed complete dereg of energy, along with green supplies, no offshore drilling, and not happy with the President on the Clean Air Act. at least, he didn't come out against nuclear power. Where's Warren? > the ball game is over: because in that case, b1 and > b2 CANNOT be algebraic integers. --les ducs de Buffet; vote NONE OF THE BELOW on Trickier Dick Cheney's California Recall & e-Dereg! http://larouchepub.com === Subject: Simple combinations/Permutations question I'm not much of a mathmatician, but was reviewing the definitions of combinations and permutations today. permutations is that with permutations, order matters. The question I have is if thisis the case, why do we talk about a safe(e.g. in a bank) as having a combination ? Shouldn't it be a permutation ? To open a bank safe, order matters, so why don't we say the permutation to the safe ? Is the phrase combination to the safe mathmatically incorrect ? Any help in enlightenting me in this would most likely ensure I sleep soundly tonight. -NH === Subject: Re: Simple combinations/Permutations question Visiting Assistant Professor at the University of Montana. >I'm not much of a mathmatician, but was reviewing the definitions of >combinations and permutations today. >permutations is that with permutations, order matters. >The question I have is if thisis the case, why do we talk about a >safe(e.g. in a bank) as having a combination ? Because you are talking about different things. When we say a safe has a combination, we are not talking about mathematics, and we are not talking about the ->technical<- meaning of combinations and permutations. Just as when you talk about adding someone to the group you do not mean that you will put a big + sign between that person and the group, and then come up with a number... > Shouldn't it be a >permutation ? To open a bank safe, order matters, so why don't we say >the permutation to the safe ? Is the phrase combination to the safe >mathmatically incorrect ? The phrase has no mathematical meaning in the first place, so it is neither correct nor incorrect. Combination to the safe is an English sentence, not a mathematical statement. It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Variation on an impossible problem >Allowing only the use of the + operator, is it possible to use each of >the digits 0 1 2 3 4 5 6 7 8 9 once only to make a total of 1000? >One can combine digits to create sums like 210 + 345 + 6 + 78 + 9 = >648 etc. >>No. sum(i,i=0..9) = 0 mod 9, whereas 1000 = 1 mod 9. > I could possibly form must be divisible be nine? I can see that if all > such sums must be divisible by nine then 1000 cannot be a possible > sum. > Moreover, I can see that sum(i,i=0..9) = 0 mod 9 but how does this > generalise to all possible sums formed? > Mitch. Note that every number N with digits d(n)n, d(n-1), ..., d(1), d(0): N = d(n) d(n-1) d(n-2) ... d(1) d(0) can be written as follows: N = sum(10(k) d(k)), where the index runs from 0 to n. Since 10 is congruent to 1 modulo 9, one also has 10k congruent to 1 modulo 9, and thus N is congruent to sum(d(k)) modulo 9. Any sum of numbers thus obtained (from the digits 0, ... ,9) is congruent to the sum of the individual numbers, and each individual is congruent to the sum of its digits. Thus, the sum of all the numbers is congruent to the sum of the full collection of digits used. Dale. === Subject: help me....my teacher....my problem is~~ let |z-10i | =6 let x : argument of z find max * min of 6cos(x) + 8sin(x) -------------------------- i wait your hot-advice...please... thank in advance. === Subject: Re: help me....my teacher....my problem is~~ > let |z-10i | =6 A circle radius 6 centred at 10i > let x : argument of z > find max * min of 6cos(x) + 8sin(x) 6 and 8, eh? sin and cos (at right angles) eh? Combine those to get 10sin(x+a) for some a. Then you just want to find the max and the min. If (x+a) stays within (-Pi/2, Pi/2), then the the min and max occur at extremal values of x, as 10sin(x+a) is monotone increasing in that range. If however, (x+a) exceeds either or both those bounds then the max and/or min will be +10 or -10 respectively. Phil -- Unpatched IE vulnerability: window.open search injection Description: cross-domain scripting, cookie/data/identity theft, command execution Reference: http://safecenter.net/liudieyu/WsFakeSrc/WsFakeSrc-Content.HTM Exploit: http://safecenter.net/liudieyu/WsFakeSrc/WsFakeSrc-MyPage.htm === Subject: Re: help me....my teacher....my problem is~~ > If (x+a) stays within (-Pi/2, Pi/2), then the the min > and max occur at extremal values of x, as 10sin(x+a) > is monotone increasing in that range. > If however, (x+a) exceeds either or both those bounds then > the max and/or min will be +10 or -10 respectively. ------------------ i am not understand. i think that interval of x is less than pi. addition explanation ...please === Subject: signed graph = directed graph ? A signed graph is just an ordinary graph with each of its edges labeled with either a + or a -. ... Those of you who are really quick may have noticed that this labeling is basically just labeling each edge with an element of GF(2). What is the difference between signed graph and directed graph? === Subject: Re: Finishing argument, core error proven > For me there have been two perspectives as I work to figure out how to > explain the definition problem in mathematics with LOTS of opposition, > and I wonder about mathematicians so dedicated to attacking an > argument that is clearly correct. I remind of that as I present what should finish their ability to > distract, as I've seen a strange and dedicated effort to ignore the > actual math, and simply toss up just about anything rather than face > the truth. All variables are in the ring of algebraic integers unless otherwise > stated. Let P(m) = f2((m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f) and let R(m) = (m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f so P(m) = f2 R(m). Now consider P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where the a's are given by the following cubic: a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). Then it must be true that the following factorization exists R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where the b's are given by the following cubic: b3 + 3(-1+mf2)b2 - (m3 f4 - 3m2 f2 + 3m), > should be > b3+...+ 3(-1+mf2)b2+...- (m3 f4 - 3m2 f2 + 3m) > where a_3 = b_3, and at m=0, b_3 = 3. > It's not true in general that a_3 = b_3. Um, here's one spot where I got it wrong, when I was right the first time, as pointed out in this thread on sci.math by the poster Arturo Magidin. And in fact, in general, a_3 = b_3. The cubic I gave was wrong though, again, as the correct cubic is b3+ (...)b2 + (...)b+...- (m3 f4 - 3m2 f2 + 3m) which may seem strange, but consider m=1, f=sqrt(2), gives b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 where you can see what's hidden in general. James Harris === Subject: Re: Finishing argument, core error proven > For me there have been two perspectives as I work to figure out how to > explain the definition problem in mathematics with LOTS of opposition, > and I wonder about mathematicians so dedicated to attacking an > argument that is clearly correct. I remind of that as I present what should finish their ability to > distract, as I've seen a strange and dedicated effort to ignore the > actual math, and simply toss up just about anything rather than face > the truth. All variables are in the ring of algebraic integers unless otherwise > stated. Let P(m) = f2((m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f) and let R(m) = (m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f so P(m) = f2 R(m). Now consider P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where the a's are given by the following cubic: a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). Then it must be true that the following factorization exists R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where the b's are given by the following cubic: b3 + 3(-1+mf2)b2 - (m3 f4 - 3m2 f2 + 3m), > should be > b3+...+ 3(-1+mf2)b2+...- (m3 f4 - 3m2 f2 + 3m) > where a_3 = b_3, and at m=0, b_3 = 3. > It's not true in general that a_3 = b_3. Um, here's one spot where I got it wrong, when I was right the first time, as pointed out in this thread on sci.math by the poster Arturo Magidin. And in fact, in general, a_3 = b_3. The cubic I gave was wrong though, again, as the correct cubic is b3+ (...)b2 + (...)b+...- (m3 f4 - 3m2 f2 + 3m) which may seem strange, but consider m=1, f=sqrt(2), gives b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 where you can see what's hidden in general. James Harris === Subject: Re: Finishing argument, core error proven > [.snip.] >> This is not correct. Note that a1/f = b1, or a1 = f*b1. >> Putting this into the equation above that the a's >> satisfy and simplifying, one obtains >> f*b13 + 3*(-1 + m*f2)*b12 - (m3*f4 - 3*m2*f2 + 3*m). >> Clearly b2 is a root of the same polynomial. Since >> b3 = a3, b3 satisfies a different equation: >> b33 + 3*(-1+mf2)b32 - f2(m3 f4 - 3m2 f2 + 3m) = 0. >Um, Nora Baron do you accept that there exists a SINGLE cubic which >has the b's as it's roots? >That is, ONE cubic with the roots b_1, b_2, and b_3? > Um, James Harris, since b3=a3, but b2 is not equal to a2 and b1 is > not equal to a1, they cannot be roots of the same cubic UNLESS the > cubic that defines the a's is reducible. It's not true that a_3 = b_3 in general, which I noted in a follow-up post. What is true is that at m=0, a_3 = b_3 = 3. Now then, there MUST exist a *single* cubic that has b_1, b_2 and b_3 as its roots, and in fact, at m=0, I can give that single cubic as b3 - 3b2, whose roots correctly give b_1 = b_2 = 0, and b_3 = 3. No other primitive cubic exists with the same roots. The SINGLE cubic is important as it can't be non-monic. If it were non-monic, then how would it give a monic at m=0? Well you may try to find some function of m that behaves that way, but then you have to reconcile that with R(m) = (m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where clearly, there's no way that a non-monic will work. For readers confused here, consider that a non-monic is something like P(x) = 2x + 1, where the leading coefficient is not 1 or -1. My point is that the SINGLE cubic defining the b's MUST be monic. > If the cubic that defines the a's is irreducible, then any cubic with > integer coefficients that has a3 as a root must have a1 and a2 as > roots, and that's all the roots, and none of them are equal to b1 or > to b2. So in order to be able to have a cubic that has a3 as a root, > and also b1 and b2, then the original cubic defining a1, a2, and a3 > must have a linear factor corresponding to a3, which means that a3 > must be an integer. It is true that the SINGLE cubic defining the b's can't have all algebraic integers coefficients for all m, though at m=0, it is b3 - 3b2, so it CAN for certain values of m. In general the defining cubic is b3+...+ 3(-1+mf2)b2 +...- (m3 f4 - 3m2 f2 + 3m) where in general certain terms are inexpressible and they are not, in general, algebraic integers. James Harris === Subject: Re: Finishing argument, core error proven Visiting Assistant Professor at the University of Montana. Cc: >> [.snip.] > This is not correct. Note that a1/f = b1, or a1 = f*b1. > Putting this into the equation above that the a's > satisfy and simplifying, one obtains f*b13 + 3*(-1 + m*f2)*b12 - (m3*f4 - 3*m2*f2 + 3*m). Clearly b2 is a root of the same polynomial. Since > b3 = a3, b3 satisfies a different equation: b33 + 3*(-1+mf2)b32 - f2(m3 f4 - 3m2 f2 + 3m) = 0. >>Um, Nora Baron do you accept that there exists a SINGLE cubic which >>has the b's as it's roots? >>That is, ONE cubic with the roots b_1, b_2, and b_3? >> Um, James Harris, since b3=a3, but b2 is not equal to a2 and b1 is >> not equal to a1, they cannot be roots of the same cubic UNLESS the >> cubic that defines the a's is reducible. >It's not true that a_3 = b_3 in general, which I noted in a follow-up >post. And unless you are in a specific family of cases, you must have a_3=b_3 in every other case. >What is true is that at m=0, a_3 = b_3 = 3. >Now then, there MUST exist a *single* cubic that has b_1, b_2 and b_3 >as its roots, and in fact, at m=0, I can give that single cubic as Sigh. Given any three COMPLEX numbers, there exists a cubic that has those three complex numbers as roots. Namely, if you have the complex numbers a, b, and c, then the single cubic that has those three complex numbers as roots is (x-a)(x-b)(x-c). That's basic precalculus stuff. The point, however, is that any old cubic is not good enough. If you want to conclude stuff about the integrality of b_1, b_2, and b_3 (e.g., whether or not they are algebraic integers) then you need the cubic to be monic and to have algebraic integer (or better yet, integer) coefficients. In any case when b_3=a_3 (which is every case in which a_1x is not equal to -uf and a_2x is not equal to -uf), you have b_3=a_3 but b_2 different from a_1 and a_2, and b_1 different from a_1 and a_2 (assuming f is not a unit, which you do). IF the polynomial that defines the a's is irreducible over Q, then ANY polynomial with integer coefficients that has a_3 as a root must be a multiple of the polynomial that defines the a's. And therefore, it is either of degree strictly larger than 3, or else it is a scalar multiple of the polynomial that defines the a's and has exactly a_1, a_2, and a_3 as roots. Therefore, in all such cases, it is impossible for there to be a single cubic WITH INTEGER COEFFICIENTS which has b_1, b_2, and b_3 as roots. This is simple, basic, precalculus algebra of polynomials. >The SINGLE cubic is important as it can't be non-monic. >If it were non-monic, then how would it give a monic at m=0? Let f(x) = (2m+1)x3 - 3x2 +1. It's a single cubic. It gives a monic cubic when m=0. Are you claiming that it is ALWAYS monic? [.rest deleted.] 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 critize. A great many people are staggered to this extend, 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 === Subject: Re: Finishing argument, core error proven >For me there have been two perspectives as I work to figure out how to >explain the definition problem in mathematics with LOTS of opposition, >and I wonder about mathematicians so dedicated to attacking an >argument that is clearly correct. I remind of that as I present what should finish their ability to >distract, as I've seen a strange and dedicated effort to ignore the >actual math, and simply toss up just about anything rather than face >the truth. All variables are in the ring of algebraic integers unless otherwise >stated. Let P(m) = f2((m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f) and let R(m) = (m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f so P(m) = f2 R(m). Now consider P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where the a's are given by the following cubic: a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). Then it must be true that the following factorization exists R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where the b's are given by the following cubic: b3 + 3(-1+mf2)b2 - (m3 f4 - 3m2 f2 + 3m), > This is not correct. Note that a1/f = b1, or a1 = f*b1. > Putting this into the equation above that the a's > satisfy and simplifying, one obtains f*b13 + 3*(-1 + m*f2)*b12 - (m3*f4 - 3*m2*f2 + 3*m). Clearly b2 is a root of the same polynomial. Since > b3 = a3, b3 satisfies a different equation: b33 + 3*(-1+mf2)b32 - f2(m3 f4 - 3m2 f2 + 3m) = 0. Um, Nora Baron do you accept that there exists a SINGLE cubic which > has the b's as it's roots? That is, ONE cubic with the roots b_1, b_2, and b_3? If so, then your objection fails as provably that cubic must be > non-monic. Do you understand? No, no, no. You made a simple mistake in algebra. Check it > over again. One clue is that your equation for the b's > is different from your equation for the a's, even though > b3 = a3. > As for b1 and b2: just take your equation for the a's > (which is correct) and make the substitution a = f*b, and > do the algebra. > Another clue: note that YOUR equation for the b's is monic. > If it were correct, you could conclude immediately, without > having to think about R(0) etc., that the b's were all > algebraic integers. Easier than you thought, eh? But the > correct equation for b1 and b2 however is *non-monic* ! That's not possible. Readers should note that the b's vary with the value of m, and at m=0, you have b3 - 3b2 as the SINGLE cubic which correctly gives the value for ALL of the b's, as then you have b_1 = b_2 = 0, and b_3 = 3. Now Nora Baron needs that SINGLE cubic to be non-monic for some values of m, but that leads to a problem, there's no mathematical way for the cubic to be non-monic, and have b3 - 3b2, which is clearly monic, at m=0. Those of you who think m=0 is a special case can have fun trying to find some function of m, such that the cubic will be non-monic, as needed for *some* values. Now then Nora Baron is correct in noting that if the cubic had all algebraic integer coefficients, it would have algebraic integer roots, so guess what? It's coefficients, in general, are not all algebraic integers, which is why you need ... to in any way display the cubic in general. James Harris === Subject: Re: Finishing argument, core error proven ... > Now Nora Baron needs that SINGLE cubic to be non-monic for some > values of m, but that leads to a problem, there's no mathematical way > for the cubic to be non-monic, and have b3 - 3b2, which is clearly > monic, at m=0. f_0(m).b3 + f_1(m).b2 + f_2(m).b + f_3(m), where the f_i(m) are algebraic integer functions of m and f_0(0) = 1, f_1(0) = -3, f_2(0) = 0 and f_3(0) = 0? I think that is the situation you have. -- === Subject: Re: Finishing argument, core error proven > ... > Now Nora Baron needs that SINGLE cubic to be non-monic for some > values of m, but that leads to a problem, there's no mathematical way > for the cubic to be non-monic, and have b3 - 3b2, which is clearly > monic, at m=0. > f_0(m).b3 + f_1(m).b2 + f_2(m).b + f_3(m), > where the f_i(m) are algebraic integer functions of m and f_0(0) = 1, > f_1(0) = -3, f_2(0) = 0 and f_3(0) = 0? Which would require that f_3(m)/f_0(m) = -(m3 f4 - 3m2 f2 + 3m) so then, does anyone else besides this poster accept that possibility? The poster deleted out important information, which I'll replace. R(m) = (m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) > I think that is the situation you have. Well then do you believe that f_3(m)/f_0(m) can equal what it does and ever have f_0(m) NOT be a factor of f_3(m)? Readers, this post is a checkmate post. Oh yeah, I also need to correct my previous posts as I've apparently given the wrong cubic for the b's as verified by a test. The proper cubic is b3 + (...)b2 + (....)b -(m3 f4 - 3m2 + 3m) and I found out that my cubic was wrong by using m=1, f=sqrt(2), which gives the following cubic for the b's: b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 Those of you with sufficient mathematical sophistication will realize this is a checkmate post, and will hopefully not force me to spell it all out in detail before conceding to the math. I don't know why any of you decided to fight the math anyway. James Harris === Subject: Re: Finishing argument, core error proven > ... > Now Nora Baron needs that SINGLE cubic to be non-monic for some > values of m, but that leads to a problem, there's no mathematical way > for the cubic to be non-monic, and have b3 - 3b2, which is clearly > monic, at m=0. > > f_0(m).b3 + f_1(m).b2 + f_2(m).b + f_3(m), > > where the f_i(m) are algebraic integer functions of m and f_0(0) = 1, > f_1(0) = -3, f_2(0) = 0 and f_3(0) = 0? > Which would require that > f_3(m)/f_0(m) = -(m3 f4 - 3m2 f2 + 3m) > so then, does anyone else besides this poster accept that possibility? Oh, yes, for each integer m. But are f_1(m)/f_3(m) and f_2(m)/f_3(m) algebraic integers for all m? Consider the cubic (x-b1)(x-b2)(x-b3), obviously this is a monic cubic with constant term -b1.b2.b3 = -(m3.f4 - 3m2.f2 + 3m). However we do not know whether the coefficients of b2 and b are algebraic integer for all m. But for each m they are algebraic numbers. So choose an arbitrary m: m0. Say the coefficient for b2 is c2 and the coefficient for b is c1. We can write c2 and c1 as the quotient of an algebraic integer and a normal integer. Take r0 the lowest common multiple of the two normal integers. Multiply the polynomial by r0, then r0.(b-b1)(b-b2)(b-b3) has algebraic integer coefficients for m=m0. Define the f_i(m0) accordingly. Do that for each integer m and you get the given functions f_i(m). Note that the r's depend on m! > I think that is the situation you have. > Well then do you believe that f_3(m)/f_0(m) can equal what it does and > ever have f_0(m) NOT be a factor of f_3(m)? O, yes for each integer m, f_0(m) is a factor of f_3(m). So what? > Readers, this post is a checkmate post. O. Why? This is not sufficient to show that the b's are algebraic integers. To show that you must also have that f_0(m) is for each m a factor (in the algebraic integers) of f_1(m) and f_2(m). To show that you must have that the coefficients of (b-b1)(b-b2)(b-b3) are algebraic integers, and to show that you must have that b1, b2 and b3 are algebraic integers. Looks a bit circular to me. -- === Subject: Re: Finishing argument, core error proven > ... > Now Nora Baron needs that SINGLE cubic to be non-monic for some > values of m, but that leads to a problem, there's no mathematical way > for the cubic to be non-monic, and have b3 - 3b2, which is clearly > monic, at m=0. f_0(m).b3 + f_1(m).b2 + f_2(m).b + f_3(m), where the f_i(m) are algebraic integer functions of m and f_0(0) = 1, > f_1(0) = -3, f_2(0) = 0 and f_3(0) = 0? Which would require that > f_3(m)/f_0(m) = -(m3 f4 - 3m2 f2 + 3m) > so then, does anyone else besides this poster accept that possibility? > Oh, yes, for each integer m. But are f_1(m)/f_3(m) and f_2(m)/f_3(m) > algebraic integers for all m? Consider the cubic (x-b1)(x-b2)(x-b3), > obviously this is a monic cubic with constant term -b1.b2.b3 = > -(m3.f4 - 3m2.f2 + 3m). However we do not know whether the > coefficients of b2 and b are algebraic integer for all m. But for each > m they are algebraic numbers. So choose an arbitrary m: m0. Say the > coefficient for b2 is c2 and the coefficient for b is c1. We can write > c2 and c1 as the quotient of an algebraic integer and a normal integer. > Take r0 the lowest common multiple of the two normal integers. Multiply > the polynomial by r0, then > r0.(b-b1)(b-b2)(b-b3) > has algebraic integer coefficients for m=m0. Define the f_i(m0) > accordingly. Do that for each integer m and you get the given functions > f_i(m). Note that the r's depend on m! Well readers, did anyone see what's wrong with this poster's position? He suggests > f_0(m).b3 + f_1(m).b2 + f_2(m).b + f_3(m), where the f_i(m) are algebraic integer functions of m and f_0(0) = 1, > f_1(0) = -3, f_2(0) = 0 and f_3(0) = 0? But the poster fails to point out that his position requires that for certain values of m and another important variable I call f, that his f_0(m) *must* have f as a factor, so it's basically a step function. Is that the right word step function? Here's important information left out by the poster, including where the b's fit into the big picture: P(m) = f2((m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f) P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). R(m) = (m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) Here reducibility over rationals of the cubic defining the a's a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m) has been given as an issue, and the posters position requires that his f_0(m) vary such that it's 1 if that cubic is reducible over rationals, and f, if it's irreducible over rationals. > I think that is the situation you have. Well then do you believe that f_3(m)/f_0(m) can equal what it does and > ever have f_0(m) NOT be a factor of f_3(m)? > O, yes for each integer m, f_0(m) is a factor of f_3(m). So what? I'll direct a question to the poster now. Well, then, what is f_0(m) if a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m) is irreducible over Q? > Readers, this post is a checkmate post. > O. Why? This is not sufficient to show that the b's are algebraic > integers. To show that you must also have that f_0(m) is for each m > a factor (in the algebraic integers) of f_1(m) and f_2(m). To show that > you must have that the coefficients of (b-b1)(b-b2)(b-b3) are > algebraic integers, and to show that you must have that b1, b2 and b3 > are algebraic integers. Looks a bit circular to me. The poster has no room to run and has walked into the checkmate. All that's left is for the poster to concede (or run away) that his position requires that f_0(m) behave as described. Checkmate. James Harris === Subject: Re: Finishing argument, core error proven > ... >b3 + 3(-1+mf2)b2 - (m3 f4 - 3m2 f2 + 3m), This is not correct. Note that a1/f = b1, or a1 = f*b1. > Putting this into the equation above that the a's > satisfy and simplifying, one obtains f*b13 + 3*(-1 + m*f2)*b12 - (m3*f4 - 3*m2*f2 + 3*m). Clearly b2 is a root of the same polynomial. Since > b3 = a3, b3 satisfies a different equation: b33 + 3*(-1+mf2)b32 - f2(m3 f4 - 3m2 f2 + 3m) = 0. Um, Nora Baron do you accept that there exists a SINGLE cubic which > has the b's as it's roots? > Nora Baron *only* state that the single cubic you give is not one which > has the b's as its roots. Um, do *you* accept that there exists a SINGLE cubic which has the b's as its roots? Readers, I'm emphasizing this point because I know that the characteristics of that single cubic must be such as to end the debate. Now, this poster and others would like to throw TWO cubics at you because you can't tell from TWO cubics where for one only two of the roots give you b_1 and b_2, while for the other only one of the roots gives you b_3. So what good are their cubics, when I can give ONE which brings b_1, b_2 and b_3 together? I'm emphasizing the *single* cubic because it must be monic. > If so, then your objection fails as provably that cubic must be > non-monic. > There is such a cubic indeed. Possibly it is non-monic, more likely though, > it does not have integer coefficients. It can't be non-monic as then you wouldn't get b_1=b_2=0, b_3=3 at m=0, which requires that the cubic, at m=0, MUST be b3 -3b2 = 0, as no other primitive cubic exists that will give the appropriate roots. So at a minimum you MUST concede the cubic at m=0. Do you concede it Dik T. Winter? > ... Not correct, as noted above. And as I noted above, this poster seems to be lost on the idea that > given b_1, b_2 and b_3 there must exist a *single* cubic for which > they are the roots, as this poster tried to give TWO, which is rather > interesting nonsense. > I thought it is *your* obsession that there must exist a *single* cubic > for which they are the roots. Good for you, there is such a polynomial, > alas, most likely not one with integer coefficients. It has integer coefficients at m=0, as then it's just b3 - 3b2 and readers note the dancing. I'm emphasizing this single cubic because its very important, as if it's conceded that it's monic in general, then the debate is over. > It doesn't, except when m = 0. For example, if m = 1 > and f = 5 and u = 1, the CORRECT equation for b1 and b2 > becomes 5*b3 + 72*b2 - 553 = 0. a primitive irreducible *non-monic* polynomial with integer > coefficients. This implies that b1, for example, > cannot be an algebraic integer. Since b1 = a1/f = a1/5, > this implies that a1 does not have a factor of f = 5. But must there not exist a SINGLE cubic for which b_1, b_2 and b_3 are > roots? > There is: (x - b1)(x - b2)(x - b3). But, does it have integer coefficients, > or rational coefficients? Not when f = 5 and u = 1, as shown above. But is it monic? > Then why are you the one giving two cubics in one case, when you > didn't try to give two cubics for the a's? > Because the a's satisfy a single cubic *with integer coefficients*. > The b's do *not* so in general. The important question for the SINGLE cubic for which the b's are roots is, IS IT MONIC? > Actually with a *single* cubic, the algebra shows quite clearly that I > can't be wrong, which is probably why you tossed out two cubics, as if > no one would realize that a single cubic will suffice for three > values. > Pray give the single cubic with b1, b2 and b3 as roots. And *show* that > the coefficients are integer. The coefficients aren't integers, and in fact, they're not even algebraic integers, in general, though they may be algebraic integers for specific values of m and f. But you see, that's the point, the ring of algebraic integers isn't big enough. Considering that SINGLE cubic proves it. James Harris === Subject: Re: Finishing argument, core error proven > ... >b3 + 3(-1+mf2)b2 - (m3 f4 - 3m2 f2 + 3m), > > This is not correct. Note that a1/f = b1, or a1 = f*b1. > Putting this into the equation above that the a's > satisfy and simplifying, one obtains > > f*b13 + 3*(-1 + m*f2)*b12 - (m3*f4 - 3*m2*f2 + 3*m). > > Clearly b2 is a root of the same polynomial. Since > b3 = a3, b3 satisfies a different equation: > > b33 + 3*(-1+mf2)b32 - f2(m3 f4 - 3m2 f2 + 3m) = 0. > > Um, Nora Baron do you accept that there exists a SINGLE cubic which > has the b's as it's roots? > > Nora Baron *only* state that the single cubic you give is not one which > has the b's as its roots. > Um, do *you* accept that there exists a SINGLE cubic which has the b's > as its roots? Yes, I do, and I think everyone does. (x-b1)(x-b2)(x-b3) serves just fine. It is a cubic in b, but as the b's depend on m, it is not necessarily a cubic in m. > Readers, I'm emphasizing this point because I know that the > characteristics of that single cubic must be such as to end the > debate. Ah. > So what good are their cubics, when I can give ONE which brings b_1, > b_2 and b_3 together? > I'm emphasizing the *single* cubic because it must be monic. Pray show. (I think you mean a cubic polynomial?) > If so, then your objection fails as provably that cubic must be > non-monic. > > There is such a cubic indeed. Possibly it is non-monic, more likely though, > it does not have integer coefficients. > It can't be non-monic as then you wouldn't get b_1=b_2=0, b_3=3 at > m=0, which requires that the cubic, at m=0, MUST be b3 -3b2 = 0, as > no other primitive cubic exists that will give the appropriate roots. > So at a minimum you MUST concede the cubic at m=0. At m=0 it is b2.(b-3). > Do you concede it Dik T. Winter? So, yup, no problem. > I thought it is *your* obsession that there must exist a *single* cubic > for which they are the roots. Good for you, there is such a polynomial, > alas, most likely not one with integer coefficients. > It has integer coefficients at m=0, as then it's just > b3 - 3b2 Oh, you noted that already. > But must there not exist a SINGLE cubic for which b_1, b_2 and b_3 are > roots? > > There is: (x - b1)(x - b2)(x - b3). But, does it have integer coefficients, > or rational coefficients? Not when f = 5 and u = 1, as shown above. > But is it monic? As written, yes it is monic. No problem. But it does not necessarily have integer coefficients, so what do you conclude from it being monic? > Then why are you the one giving two cubics in one case, when you > didn't try to give two cubics for the a's? > > Because the a's satisfy a single cubic *with integer coefficients*. > The b's do *not* so in general. > The important question for the SINGLE cubic for which the b's are > roots is, IS IT MONIC? No. The important question is: is it monic and does it have integer coefficients? Otherwise you can not conclude anything about the b's being algebraic integers. > Actually with a *single* cubic, the algebra shows quite clearly that I > can't be wrong, which is probably why you tossed out two cubics, as if > no one would realize that a single cubic will suffice for three > values. > > Pray give the single cubic with b1, b2 and b3 as roots. And *show* that > the coefficients are integer. > The coefficients aren't integers, and in fact, they're not even > algebraic integers, in general, though they may be algebraic integers > for specific values of m and f. They can be made algebraic integers for *all* values of m and f. That is not the problem. Let's take (x-b1)(x-b2)(x-b3); all b's are algebraic numbers, so all coefficients are algebraic numbers. Each algebraic number can be written as the quotient of an algebraic integer and a normal integer. Do so. Multiply by the LCM of the denominators and you have a polynomial with algebraic integer coefficients of which the three b's are roots. The question is: is that polynomial monic? The answer is: in general, no. > But you see, that's the point, the ring of algebraic integers isn't > big enough. Big enough for what? For b1, b2 and b3 all three being algebraic integers? Yes, that is right. This simply means that your argument that they are algebraic integers fails. They may be objects in some other ring (the algebraic numbers for instance), but your proof hinge on the concept of co-primeness, and when you use that concept you have to show what ring you are using, and what definition of co-primeness you are using and a whole lot of other things. Your definition of object ring is incomplete as it is impossible to determine whether numbers are in it or not. -- === Subject: Re: Finishing argument, core error proven > ... >b3 + 3(-1+mf2)b2 - (m3 f4 - 3m2 f2 + 3m), This is not correct. Note that a1/f = b1, or a1 = f*b1. > Putting this into the equation above that the a's > satisfy and simplifying, one obtains f*b13 + 3*(-1 + m*f2)*b12 - (m3*f4 - 3*m2*f2 + 3*m). Clearly b2 is a root of the same polynomial. Since > b3 = a3, b3 satisfies a different equation: b33 + 3*(-1+mf2)b32 - f2(m3 f4 - 3m2 f2 + 3m) = 0. Um, Nora Baron do you accept that there exists a SINGLE cubic which > has the b's as it's roots? Nora Baron *only* state that the single cubic you give is not one which > has the b's as its roots. Um, do *you* accept that there exists a SINGLE cubic which has the b's > as its roots? > Yes, I do, and I think everyone does. (x-b1)(x-b2)(x-b3) serves just fine. > It is a cubic in b, but as the b's depend on m, it is not necessarily a > cubic in m. The question is to the solution for the b's, just like earlier in the post a solution for the a's is given as they are roots of the cubic a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). Readers should note the need for me to put *back* in math that is needed to properly evaluate various claims, as this poster left them out. My assessment of that behavior of selectively deleting out key bits of information is that it is done to try and make a convincing argument against the facts. In my experience, posters seem adept at producing posts calculated to convince other readers on the newsgroup that I'm wrong, by referring selectively to information. It's basically posting sleight-of-hand. Here's more pertinent information: P(m) = f2((m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f) P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) R(m) = (m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) > Readers, I'm emphasizing this point because I know that the > characteristics of that single cubic must be such as to end the > debate. > Ah. The reason is that posters like this one have worked to convince readers of mathematically false things, and I can show that their assertions would force the cubic that has the b's for roots to be *selectively* non-monic, where if they are correct, the cubic has to be non-monic if a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). is irreducible over Q, and monic if it's not. That is, you'd need a function of m that would equal 1, when a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m) is reducible over Q, and f, when it is not, and that function would be the leading coefficient of the cubic defining the b's. > So what good are their cubics, when I can give ONE which brings b_1, > b_2 and b_3 together? I'm emphasizing the *single* cubic because it must be monic. > Pray show. (I think you mean a cubic polynomial?) The cubic cannot, in general, be given without using ..., as then it is b3 + (...)b2 + (...)b - (m3 f4 - 3m2 f2 + 3m). However, for certain values of m and f, the terms are algebraic integers, and the cubic is easily displayed. For instance, for m=1, f=sqrt(2), it is b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 and note that here a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m), gives a3 + 3a2 - 2, which IS reducible, but intriguingly, isn't even reducible to linear terms over Q. The answer is that posters arguing against the cubic defining the b's being in general monic, are just wrong, and reducibility over Q just determines whether or not you can easily *see* that they are wrong. Of course, also for m=0, and any f, it is, as I've mentioned before b3 - 3b2, and a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m), gives a3 - 3a2. > If so, then your objection fails as provably that cubic must be > non-monic. There is such a cubic indeed. Possibly it is non-monic, more likely though, > it does not have integer coefficients. It can't be non-monic as then you wouldn't get b_1=b_2=0, b_3=3 at > m=0, which requires that the cubic, at m=0, MUST be b3 -3b2 = 0, as > no other primitive cubic exists that will give the appropriate roots. So at a minimum you MUST concede the cubic at m=0. > At m=0 it is b2.(b-3). > Do you concede it Dik T. Winter? > So, yup, no problem. What's important about that readers is it forces the poster to either accept that the cubic is monic in general, which means accepting the core problem, or claim that there's a *function* of m, which varies in some odd way. > I thought it is *your* obsession that there must exist a *single* cubic > for which they are the roots. Good for you, there is such a polynomial, > alas, most likely not one with integer coefficients. It has integer coefficients at m=0, as then it's just b3 - 3b2 > Oh, you noted that already. > But must there not exist a SINGLE cubic for which b_1, b_2 and b_3 are > roots? There is: (x - b1)(x - b2)(x - b3). But, does it have integer coefficients, > or rational coefficients? Not when f = 5 and u = 1, as shown above. But is it monic? > As written, yes it is monic. No problem. But it does not necessarily > have integer coefficients, so what do you conclude from it being monic? Well, no, it doesn't necessarily have integer coefficients, like I showed before with f=sqrt(2), m=1, where b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 but readers can see just how misleading Nora Baron was in posts showing TWO cubics, where one must exist. Now for certain values, the cubic has algebraic integer coefficients, while for others, it does not because the ring of algebraic integers doesn't include certain numbers that it should, which leads to a problem in core as you can have the appearance of two proofs contradicting each other. > Then why are you the one giving two cubics in one case, when you > didn't try to give two cubics for the a's? Because the a's satisfy a single cubic *with integer coefficients*. > The b's do *not* so in general. The important question for the SINGLE cubic for which the b's are > roots is, IS IT MONIC? > No. The important question is: is it monic and does it have integer > coefficients? Otherwise you can not conclude anything about the b's > being algebraic integers. That's bogus, as I can show with b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1. Now this poste may *want* a function as the leading coefficient of the general b cubic such that it magically equals f when a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m) is irreducible over Q, and switches to 1, when it is reducible, but there's no logical basis for that position. It's simply some person *wishing* something they cannot prove. > Actually with a *single* cubic, the algebra shows quite clearly that I > can't be wrong, which is probably why you tossed out two cubics, as if > no one would realize that a single cubic will suffice for three > values. Pray give the single cubic with b1, b2 and b3 as roots. And *show* that > the coefficients are integer. The coefficients aren't integers, and in fact, they're not even > algebraic integers, in general, though they may be algebraic integers > for specific values of m and f. > They can be made algebraic integers for *all* values of m and f. That is not > the problem. I split up a paragraph to point out the usual tactic used by such posters as this one. Make a statement, which they cannot prove, and then keep talking. It's like a fast talking technique where you say something false, and hope to snow readers by moving on before they catch you. In fact, the coefficients provably cannot be algebraic integers for all values of m and f, like for instance, with m=1, f=sqrt(5), they are not. >Let's take (x-b1)(x-b2)(x-b3); all b's are algebraic numbers, Here's the second sneak, as yes, the b's are in fact algebraic numbers, even when they're not algebraic integers, but the poster has moved to a field. > so all coefficients are algebraic numbers. Each algebraic number can be > written as the quotient of an algebraic integer and a normal integer. Here's the next point of the attempt at snowing the reader, as the poster is pushing the position that the b's can be written like, say, x/f, where x is some algebraic integer that doesn't have f as a factor, but possibly shares non-unit factors with f. > so. Multiply by the LCM of the denominators and you have a polynomial with > algebraic integer coefficients of which the three b's are roots. The question > is: is that polynomial monic? The answer is: in general, no. Now notice, given all of that, you'd think that there'd be some *logical* argument, but instead you have the poster simply asserted the desired conclusion. It's this type of poster that gives me fits as Nora Baron and Arturo Magidin both engage in this behavior. Some readers just jump to their conclusion, and apparently assume they proved it, when in fact, they simply use tactics. Later I'm accused of not handling their objections, when they never give mathematically valid objections, but instead use various techniques to *appear* to give valid objections. > But you see, that's the point, the ring of algebraic integers isn't > big enough. > Big enough for what? For b1, b2 and b3 all three being algebraic integers? > Yes, that is right. This simply means that your argument that they are > algebraic integers fails. They may be objects in some other ring (the > algebraic numbers for instance), but your proof hinge on the concept of > co-primeness, and when you use that concept you have to show what ring > you are using, and what definition of co-primeness you are using and a > whole lot of other things. Your definition of object ring is incomplete > as it is impossible to determine whether numbers are in it or not. So then, some reader comes along--skimming--doesn't bother to see all the places where the poster cheated, and assumes that someone has proven me wrong, when in fact, nothing of the kind has happened. I'll make my reply and the poster just does the same thing. Readers skim and think I'm just refusing to acknowledge the truth. The thread becomes unmanageably huge and I make a new thread, and readers claim I'm running away. Later, the poster comes back with the same techniques, never actually managing to refute a single point of my argument mathematically. James Harris === Subject: Re: Finishing argument, core error proven Visiting Assistant Professor at the University of Montana. [.snip.] >> Um, do *you* accept that there exists a SINGLE cubic which has the b's >> as its roots? >> Yes, I do, and I think everyone does. (x-b1)(x-b2)(x-b3) serves just fine. >> It is a cubic in b, but as the b's depend on m, it is not necessarily a >> cubic in m. >The question is to the solution for the b's, just like earlier in the >post a solution for the a's is given as they are roots of the cubic >a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). >Readers should note the need for me to put *back* in math that is >needed to properly evaluate various claims, as this poster left them >out. Will you quit your whining already? If people don't trim, you whine that their posts are too long; if they trim, you whine that they trimmed. If they make multiple replies to keep them short, you whine that they make multiple replies; if they make a single reply, you whine that they address different issues and are trying to snow people. And if anybody ->dares<- to say that you should stop grandstanding like you do above, you whine that they are not addressing the math. Well, neither are you: you are just whining. If someone left something out, and you think it is important, then bring it back up and mention it was there already. No need to start whining and telling readers what they should or should not note. >My assessment of that behavior of selectively deleting out key bits of >information is that it is done to try and make a convincing argument >against the facts. My assessment of your behavior is that you are far more interested in appearances than you are in substance. That is why you keep inserting these Appeals to the Gallery in the middle of your posts. [.snip.] >Here's more pertinent information: >P(m) = f2((m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f) >P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) >R(m) = (m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f >R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) So: R(m) = P(m)/f2. Assuming that P(m) is not zero, we have: R(m) = P(m)/f2 = (1/f)(a_1x+uf)(1/f)(a_2x+uf)(a_3x+uf) = (b_1x + u)(b_2x+u)(b_3x+uf) and you have defined b_1=a_1/f, b_2=a_2/f. So b_1x+u = (a_1x+uf)/f, and b_2x + u + (a_2x+uf)/f. Since P(m) is not zero, we can cancel terms and get b_3x + uf = a_3x + uf. So b_3x = a_3 x. Since x is not zero (it is coprime to f, which is not a unit), we have that b_3=a_3. So, UNLESS P(m)=0, we must have b_3=a_3. That means that b_3 is a root of the cubic you mention above: a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). There are two cases: CASE 1. The cubic is irreducible over Q; that is, x3 + 3(-1+mf2)x2 - f2(m3f4 - 3m2f2+3m) has no rational roots; that is, none of a1, a2, a3 are rational numbers (and being monic, they would necessarily be integers). and CASE 2. The cubic irreducible over Q; that is, it has at least one, possibly three, rational roots. That means, exactly one of a1, a2, a3 is an integer, or all three are integers. In CASE 2, which is NOT the case m=0, or the case u=1, f=sqrt(2), m=1 that you've used as well, we have the following: b3 is a root of G(x) = x3 + 3(-1+mf2)x2 - f2(m3f4 - 3m2f2+3m). Assume F(x) is ANY polynomial with rational coefficients that has b3 as a root. Then we can divide F(x) by G(x), to get F(x) = G(x)*Q(x) + r(x) where r(x) and Q(x) are polynomials with rational coefficients, and r(x)=0 or else r(x) has degree strictly less than the degree of G. Since F(x) has b3 as a root, then 0=F(b3)=G(b3)Q(b3) + r(b3) = r(b3), so r(x) also has b3 as a root. If r(x) is not zero, then we can divide G(x) by r(x), and we have G(x) = r(x)Q'(x) + s(x), with s(x) equal to 0 or with strictly smaller degree than r(x), and therefore either constant or degree 1. Then 0 = G(b3) = r(b3)Q'(b3) + s(b3) = s(b3), so s(b3)=0; that means either that r(x) divides G(x) (if s is constant), which is impossible because G(x) was assumed to be irreducible; or else s(b3)=0 and s is a polynomial of degree 1, which means that b3 is a rational number, which is also impossible since G(x) was assumed irreducible. The conclusion is that r(x) must be the constant zero, which means the original F(x) must be a multiple of G(x), which means that either F(x) is the zero polynomial, is a constant multiple of G(x), or else it has strictly larger degree than G(x) (i.e., it is of degree greater than 3). Therefore: If F(X) is ANY cubic polynomial with rational coefficients which has b3 as a root, then it must be a constant multiple of G(x), and its roots are a1, a2, and a3=b3. (Assuming the polynomial that defines the a's is irreducible over Q). So, if there is a cubic polynomial with rational coefficients that defines the b's, it must mean that we have {b1,b2} = {a1, a2}. Which means, either b1=a1 and b2=a2, or else b1=a2 and b2=a1. The first is impossible, since b1=a1/f, so a1=a1/f means a1=0, which would mean the original G(x) had a rational root, impossible. The latter is also impossible, for if b1=a2 and b2=a1, then a1/f = a2, so a1 = f(a1/f) = f*b1= f*a2 = f*(f*b2) = f2*b2 = f2*a1, so a1=0 and again we have a contradiction. Therefore, if the polynomial defining the a's is irreducible over Q, then there is NO polynomial with rational coefficients which is (a) cubic; and (b) has b1, b2, and b3 as roots. Therefore, IF there is a single cubic polynomial with rational coefficients that has b1, b2, and b3 as roots, then the original polynomial was reducible over Q. Which, surprise surprise, is exactly the two cases that you manage to handle. >> Readers, I'm emphasizing this point because I know that the >> characteristics of that single cubic must be such as to end the >> debate. >> Ah. >The reason is that posters like this one have worked to convince >readers of mathematically false things, and I can show that their >assertions would force the cubic that has the b's for roots to be >*selectively* non-monic, where if they are correct, the cubic has to >be non-monic if >a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). >is irreducible over Q, and monic if it's not. ->YOU<- can show? So you're not just repeating what they proved already? I always knew you were not above claiming works of others as your own. >That is, you'd need a function of m that would equal 1, when >a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m) >is reducible over Q, and f, when it is not, and that function would be >the leading coefficient of the cubic defining the b's. Yes, there is a cubic which has exactly b1, b2, and b3 as roots. However, in the case where P(m) is not zero and G(x) is irreducible over Q, this polynomial (which is a scalar multiple of (x-b1)(x-b2)(x-b3)) does NOT have integer coefficients. It does not even have RATIONAL coefficients; to get rational coefficients you need to go to a polynomial of larger degree. But, setting that aside, (1) What do you mean, reducible over Q, and f? What does it mean to be reducible over f? (2) What's wrong with a function which is 1 for some values of m but not 1 for others? Surely you've seen plenty of those? >> So what good are their cubics, when I can give ONE which brings b_1, >> b_2 and b_3 together? >> I'm emphasizing the *single* cubic because it must be monic. >> Pray show. (I think you mean a cubic polynomial?) >The cubic cannot, in general, be given without using ..., as then it >b3 + (...)b2 + (...)b - (m3 f4 - 3m2 f2 + 3m). It cannot? If we cannot see the cubic, then how do you know the cubic has integer coefficients? >However, for certain values of m and f, the terms are algebraic >integers, and the cubic is easily displayed. For instance, for m=1, >f=sqrt(2), it is >b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 >and note that here >a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m), gives >a3 + 3a2 - 2, >which IS reducible, but intriguingly, isn't even reducible to linear >terms over Q. x3+3x2 - 2 = (x+1)(x2+2x-2) Why is that intriguing? In this case, your cubic factors into a linear times an irreducible quadratic; you get lucky that your linear term has constant term coprime to f, so you get the other two factors sharing f2 in equal portions. You did not get so lucky when u=1, f=2, m=1, where the linear term did not avoid the value of 2; then you had a3 + 3(-1+4)a2 - 4(16 - 3(4) + 3) a3 + 9a2 - 28 = (a+2)(a2+7a-14) in that case, the factors of 2 did NOT factor out as you hoped they would, because both the linear term and the irreducible quadratic contributed to the total factor of 22 = 4. >The answer is that posters arguing against the cubic defining the b's >being in general monic, are just wrong, and reducibility over Q just >determines whether or not you can easily *see* that they are wrong. Nonsense. In the case u=1, f=2, m=1, the value of a3=b3=-2, and the other two values are b1 = a1/2 = (-7-sqrt(105))/4 b2 = a2/2 = (-7+sqrt(105))/4 Neither is an algebraic integer: they are roots of (x-b1)(x-b2) = x2 - (b1+b2)x + b1*b2 = x2 - (-14/4)x + (49-105)/16 = x2 + (7/2)x + (56/16) = x2 + (7/2)x + (7/2) so they are roots of the irreducible non-monic 2x2 + 7x + 7. Hence they are not algebraic integers. They cannot be the roots of any monic polynomial with algebraic integer coefficients, let alone a cubic. In fact, any cubic that has them as a roots would be a scalar multiple of (x-b1)(x-b2)(x-b3) = (x2 + (7/2)x+(7/2))(x+2) = x3 + (7/2)x2 + (7/2)x + 2x2 + 7x + 7 = x3 + (11/2)x2 + (21/2)x + 7 and if we clear denominators, we have 2x3 + 11x2 + 21x + 14, so no monic cubic with integer coefficients has all three as roots (it would have to be a scalar multiple of the above; since 2 and 11 are coprime, there is no algebraic integer other than units that divides all coefficients, so you cannot get rid of the 2 in the leading term). In case you argue that I just assigned the b's wrong, here are the two other possibilities (up to exchanging b1 and b2): CASE 2: b1 = -2/2, b2 = (-7-sqrt(105))/4, b3 = (-7+sqrt(105))/2 Then any cubic having all three as roots is a scalar multiple of (x-b1)(x-b2)(x-b3) = (x+1)(x+(7+sqrt(105))/4)(x+(7-sqrt(105))/2) = (x+1)(x2 + ((21-sqrt(105))/4)x -(56/8)) = x3 + ((21-sqrt(105))/4)x2 - 7x + x2 + ((21-sqrt(105))/4)x - 7 = x3 + [(25-sqrt(105))/4]x2 - [(7+sqrt(105))/4]x - 7 and clearing denominators, we have 4x3 + (25-sqrt(105))x2 - (7+sqrt(105))x - 28. Nonmonic with algebraic integer coefficients. It is not hard to see that this is irreducible over Q[sqrt(105)], and so none of its roots are algebraic integers. And of course, there is no cubic with INTEGER coefficients that will work, either. Finally, CASE 3: b1 = -2/2, b2 = (-7+sqrt(105))/4, b3 = (-7-sqrt(105))/2 Then any cubic having all three as roots is a scalar multiple of (x-b1)(x-b2)(x-b3) = (x+1)(x+(7-sqrt(105))/4)(x+(7+sqrt(105))/2) = (x+1)(x2 + ((21+sqrt(105))/4)x -(56/8)) = x3 + ((21+sqrt(105))/4)x2 - 7x + x2 + ((21+sqrt(105))/4)x - 7 = x3 + [(25+sqrt(105))/4]x2 - [(7-sqrt(105))/4]x - 7 and clearing denominators we have 4x3 + (25-sqrt(105))x2 - (7-sqrt(105))x - 28 again nonmonic, irreducible over Q[sqrt(105)], and none of its roots are algebraic integers. No cubic with INTEGER coefficients will work. So we can SEE that you are wrong. >Of course, also for m=0, and any f, it is, as I've mentioned before >b3 - 3b2, >and >a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m), gives >a3 - 3a2. Which is reducible into three linear terms, and reducibility of the polynomial is NECESSARY before there can be a cubic with integer coefficients that defines the b's (though not sufficient, as seen just above). [.snip.] >What's important about that readers is it forces the poster to either >accept that the cubic is monic in general, which means accepting the >core problem, or claim that there's a *function* of m, which varies >in some odd way. Yeah, some really odd way. Like, for example, (2m+1) which has value 1 with m=0 but not when m is not zero. Or a function which is 1 at m=0 and at m=1, but not anywhere else? Something REALLY odd like (2m-1)2? [.snip.] >> There is: (x - b1)(x - b2)(x - b3). But, does it have integer coefficients, >> or rational coefficients? Not when f = 5 and u = 1, as shown above. >> But is it monic? >> As written, yes it is monic. No problem. But it does not necessarily >> have integer coefficients, so what do you conclude from it being monic? >Well, no, it doesn't necessarily have integer coefficients, like I >showed before with f=sqrt(2), m=1, where >b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 You are using values of f that are not integers. But here, you get algebraic integers. However, as seen above, the MONIC version does not necessarily have ALGEBRAIC integer coefficients either. You are being misleading here, since you do not clarify just what it is you are claiming. Which of the following is your claim? POSSIBLE CLAIM 1: There is a cubic polynomial, whose coefficients are functions of u, f, m, with the property that for every algebraic integer value of u, f, and m, the polynomial is monic, has algebraic integer coefficients, and has b1, b2, and b3 as roots. POSSIBLE CLAIM 2: For every algebraic integer value of u, f, and m, there is a cubic monic polynomial with algebraic integer coefficients that has b1, b2, and b3 as roots. POSSIBLE CLAIM 3: For every algebraic integer value of u, f, and m, there is a monic cubic polynomial which has b1, b2, and b3 as roots, but we make no claim about the coefficients of that polynomial ? POSSIBLE CLAIM 3 gives you nothing; only from possible claims 1 or 2 could you conclude that b1, b2, and b3 are algebraic integers. And the situation you are in is 3. >but readers can see just how misleading Nora Baron was in posts >showing TWO cubics, where one must exist. You're wrong. For any value of m for which P(m) is not zero, and the polynomial a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). is irreducible over Q, you cannot have a single cubic with rational coefficients that has b1, b2, and b3 as roots. (NOTE THAT in order to be irreducible over Q, the coefficients MUST BE rational numbers; your example with f=sqrt(2) does not satisfy these conditions, since the polynomial DOES NOT have rational coefficients; the corresponding claim would have to be that the resulting polynomial is irreducible over Q[sqrt(2)] ). [.snip.] >Now for certain values, the cubic has algebraic integer coefficients, >while for others, it does not because the ring of algebraic integers >doesn't include certain numbers that it should, which leads to a >problem in core as you can have the appearance of two proofs >contradicting each other. Ah, yes. So your cubic doesn't have the right numbers, so once again, things that ->should<- be there aren't there. But why should they be there? Because you ->want<- them to be, not for any other reason. Well, tough. [.snip.] >> The important question for the SINGLE cubic for which the b's are >> roots is, IS IT MONIC? >> No. The important question is: is it monic and does it have integer >> coefficients? Otherwise you can not conclude anything about the b's >> being algebraic integers. >That's bogus, as I can show with >b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1. No, it's not bogus. Here, your original polynomial defining the a's DOES NOT have rational coefficients. Here, the question must become wether it is monic and irreducible over Q[sqrt(2)]. And this is all one BIG red herring. You are always assuming f is a prime integer, except when it is convenient for you to distract with other values. [.snip.] >It's simply some person *wishing* something they cannot prove. You mean, like there should be a cubic, even though you cannot prove there is one? [.snip.] >>Let's take (x-b1)(x-b2)(x-b3); all b's are algebraic numbers, >Here's the second sneak, as yes, the b's are in fact algebraic >numbers, even when they're not algebraic integers, but the poster has >moved to a field. There is no second sneak. This is basic precalculus, which even you must have gone over in your gifted program. [.snip.] >> so all coefficients are algebraic numbers. Each algebraic number can be >> written as the quotient of an algebraic integer and a normal integer. >Here's the next point of the attempt at snowing the reader, as the >poster is pushing the position that the b's can be written like, say, >x/f, where x is some algebraic integer that doesn't have f as a >factor, but possibly shares non-unit factors with f. That's a lie, James. Nowhere is he saying or implying that the denominator will be ->f<-; but EVERY algebraic number can be written as a quotient of an algebraic integer and a regular integer, though this expression NEED NOT BE IN LEAST TERMS. Nowhere is it assumed that it is in least terms, it is unnecessary for the argument. So you are erecting strawmen and lying. [.snip.] >It's this type of poster that gives me fits as Nora Baron and Arturo >Magidin both engage in this behavior. Ad hominem. >Some readers just jump to their conclusion, and apparently assume they >proved it, when in fact, they simply use tactics. Yeah, that's what you do. Should be multiples. That's jumping to a conclusion. When asked to justify, instead of explaining the should, you insult, you whine, and eventually explain SOME OTHER THING NOBODY COMPLAINED ABOUT. And then, having committed a strawman and a red herring, you claim to have handled the objection, because you addressed ->something<-... just not what people were asking and/or complaining about. [.snip.] 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 critize. A great many people are staggered to this extend, 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 === Subject: Re: Finishing argument, core error proven there's got to be some sort of *job* that could employ such skills in whining. seriously, it seems like about half of JSH's output is grandstanding, but it could be argued that you're doing the same. I mean, how many times have I seen the math reduced to a simple problemma in the definition of (say) divisibility -- often the first chapter of elementary books on numbertheory -- -- and you've also deomnstrated such, as I recall -- but you make these long & tedious corrections? I can't follow them, thus the quotes, nor do I want to bother with most of JSH's explanations, since I'm quite satisfied with the shorter, unanswered refutations. I'm really glad, though, that everyone takes him on his word, and his apparent pursuit of aggrandizement, since the idea that he *is* being paid to do this is just too much (at least *one*, two, much) to bear. [massive deletives impleted] > And if anybody ->dares<- to say that you should stop grandstanding > like you do above, you whine that they are not addressing the math. --Dec.2000 'WAND' Chairman Paul O'Neill, reelected to Board. Newsish? http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac === Subject: Re: Finishing argument, core error proven > [.snip.] >> Um, do *you* accept that there exists a SINGLE cubic which has the b's >> as its roots? >> Yes, I do, and I think everyone does. (x-b1)(x-b2)(x-b3) serves just fine. >> It is a cubic in b, but as the b's depend on m, it is not necessarily a >> cubic in m. >The question is to the solution for the b's, just like earlier in the >post a solution for the a's is given as they are roots of the cubic >a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). >Readers should note the need for me to put *back* in math that is >needed to properly evaluate various claims, as this poster left them >out. > Will you quit your whining already? If people don't trim, you whine > that their posts are too long; if they trim, you whine that they > trimmed. If they make multiple replies to keep them short, you whine > that they make multiple replies; if they make a single reply, you > whine that they address different issues and are trying to snow > people. And here readers should notice that Arturo Magidin starts right at the beginning, not with mathematics, but with a protest. Beyond that consider the abhorrent *length* of this person's post. I'll leave it all in this time, but I want readers to pay attention to this particular poster's tactics. > And if anybody ->dares<- to say that you should stop grandstanding > like you do above, you whine that they are not addressing the math. > Well, neither are you: you are just whining. If someone left something > out, and you think it is important, then bring it back up and mention > it was there already. No need to start whining and telling readers > what they should or should not note. The previous poster left out information pertinent to his own comments, without which, readers were left guessing. Given that poster's assumption that he'd get the benefit of the doubt, since SO many posters keep saying I'm wrong, though they're incapable of proving it, it's not surprising that the poster would leave out pertinent mathematics, pertinent to his own post, without fear of getting caught. Which is why I'm taking extra effort to highlight the tactics of these posters. >My assessment of that behavior of selectively deleting out key bits of >information is that it is done to try and make a convincing argument >against the facts. > My assessment of your behavior is that you are far more interested in > appearances than you are in substance. That is why you keep inserting > these Appeals to the Gallery in the middle of your posts. > [.snip.] I'm simply noting the behavior, especially given that certain posters--who basically just say I'm wrong--are given the benefit of the doubt. As I've watched what happens when I refute these posters mathematically, I can see that it doesn't pay for me to just hope that readers are catching them. >Here's more pertinent information: >P(m) = f2((m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f) >P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) >R(m) = (m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f >R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) > So: R(m) = P(m)/f2. Assuming that P(m) is not zero, we have: > R(m) = P(m)/f2 = (1/f)(a_1x+uf)(1/f)(a_2x+uf)(a_3x+uf) > = (b_1x + u)(b_2x+u)(b_3x+uf) > and you have defined b_1=a_1/f, b_2=a_2/f. So b_1x+u = (a_1x+uf)/f, > and b_2x + u + (a_2x+uf)/f. Since P(m) is not zero, we can cancel > terms and get > b_3x + uf = a_3x + uf. > So b_3x = a_3 x. > Since x is not zero (it is coprime to f, which is not a unit), we have > that b_3=a_3. Hey, that's correct, which means I had it right the first time, but later second guessed myself about a_3 and b_3. > So, UNLESS P(m)=0, we must have b_3=a_3. > That means that b_3 is a root of the cubic you mention above: > a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). > There are two cases: > CASE 1. The cubic is irreducible over Q; that is, > x3 + 3(-1+mf2)x2 - f2(m3f4 - 3m2f2+3m) > has no rational roots; that is, none of a1, a2, a3 are > rational numbers (and being monic, they would necessarily be > integers). > and > CASE 2. The cubic irreducible over Q; that is, it has at least one, > possibly three, rational roots. That means, exactly one of > a1, a2, a3 is an integer, or all three are integers. > In CASE 2, which is NOT the case m=0, or the case u=1, f=sqrt(2), m=1 > that you've used as well, we have the following: > b3 is a root of G(x) = x3 + 3(-1+mf2)x2 - f2(m3f4 - 3m2f2+3m). > Assume F(x) is ANY polynomial with rational coefficients that has b3 as > a root. Then we can divide F(x) by G(x), to get > F(x) = G(x)*Q(x) + r(x) > where r(x) and Q(x) are polynomials with rational coefficients, and > r(x)=0 or else r(x) has degree strictly less than the degree of G. > Since F(x) has b3 as a root, then 0=F(b3)=G(b3)Q(b3) + r(b3) = r(b3), > so r(x) also has b3 as a root. If r(x) is not zero, then we can > divide G(x) by r(x), and we have > G(x) = r(x)Q'(x) + s(x), with s(x) equal to 0 or with strictly smaller > degree than r(x), and therefore either constant or degree 1. Then > 0 = G(b3) = r(b3)Q'(b3) + s(b3) = s(b3), so s(b3)=0; that means either > that r(x) divides G(x) (if s is constant), which is impossible because > G(x) was assumed to be irreducible; or else s(b3)=0 and s is a > polynomial of degree 1, which means that b3 is a rational number, > which is also impossible since G(x) was assumed irreducible. > The conclusion is that r(x) must be the constant zero, which means the > original F(x) must be a multiple of G(x), which means that either F(x) > is the zero polynomial, is a constant multiple of G(x), or else it has > strictly larger degree than G(x) (i.e., it is of degree greater than > 3). > Therefore: If F(X) is ANY cubic polynomial with rational coefficients > which has b3 as a root, then it must be a constant multiple of G(x), > and its roots are a1, a2, and a3=b3. (Assuming the polynomial that > defines the a's is irreducible over Q). > So, if there is a cubic polynomial with rational coefficients that > defines the b's, it must mean that we have {b1,b2} = {a1, a2}. Which > means, either b1=a1 and b2=a2, or else b1=a2 and b2=a1. > The first is impossible, since b1=a1/f, so a1=a1/f means a1=0, which > would mean the original G(x) had a rational root, impossible. The > latter is also impossible, for if b1=a2 and b2=a1, then a1/f = a2, so > a1 = f(a1/f) = f*b1= f*a2 = f*(f*b2) = f2*b2 = f2*a1, so a1=0 and > again we have a contradiction. > Therefore, if the polynomial defining the a's is irreducible over Q, > then there is NO polynomial with rational coefficients which is (a) > cubic; and (b) has b1, b2, and b3 as roots. That can be the case even when it is reducible over Q, as I've pointed out by giving the defining cubic when m=1, f+sqrt(2), which is b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 as readers can actually see that only two of the coefficients, the leading and last, are rational. So far Arturo Magidin hasn't done anything but take up space. > Therefore, IF there is a single cubic polynomial with rational > coefficients that has b1, b2, and b3 as roots, then the original > polynomial was reducible over Q. > Which, surprise surprise, is exactly the two cases that you manage to > handle. Now the surprise surprise here really doesn't fit in context. Readers can guess what the purpose of the poster was in putting that phrase there. >> Readers, I'm emphasizing this point because I know that the >> characteristics of that single cubic must be such as to end the >> debate. >> Ah. >The reason is that posters like this one have worked to convince >readers of mathematically false things, and I can show that their >assertions would force the cubic that has the b's for roots to be >*selectively* non-monic, where if they are correct, the cubic has to >be non-monic if >a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). >is irreducible over Q, and monic if it's not. > ->YOU<- can show? So you're not just repeating what they proved > already? > I always knew you were not above claiming works of others as your > own. Now readers should consider how long this poster has gone on without actually saying anything. My assessment of this particular poster is that he has learned that most sci.math readers don't read in detail, but skim, and they are *especially* likely to skim a long post. Skimming along, they may feel that Arturo Magidin has said a lot, when he hasn't said anything as of yet, as I've pointed out. Now then, mathematically his position would require that the cubic defining the b's vary such that if the cubic defining the a's is reducible over Q, then that cubic is monic, but illogically, if the cubic defininng the a's is irreducible over Q, then his position requires a leading coefficient of f. Now if this poster were honest, then he'd simply acquiesce to the truth. However, instead, look at the LENGTH of his post!!! >That is, you'd need a function of m that would equal 1, when >a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m) >is reducible over Q, and f, when it is not, and that function would be >the leading coefficient of the cubic defining the b's. > Yes, there is a cubic which has exactly b1, b2, and b3 as > roots. However, in the case where P(m) is not zero and G(x) is > irreducible over Q, this polynomial (which is a scalar multiple of > (x-b1)(x-b2)(x-b3)) does NOT have integer coefficients. It does not > even have RATIONAL coefficients; to get rational coefficients you need > to go to a polynomial of larger degree. Which still isn't saying anything. Again I give the cubic for the b's for m=1, f=sqrt(2), as it is b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 and readers can see that, yeah, it has some irrational coefficients. However, it is monic. > But, setting that aside, > (1) What do you mean, reducible over Q, and f? What does it mean to > be reducible over f? This poster also has a habit of trying to find little ways to distract. Like here he broke up a paragraph where I noted that if the cubic defining the a's is NOT irreducible over Q then somehow supposedly the cubic defining the b's has a leading coefficient with a factor of f, so I wasn't talking about reducibility over f. The posters tactics are rather transparent, but readers typically come in believing that I'm wrong, so this poster can get away with distractions. Just remember readers, the correct math should be *more* concise, and direct. But notice how certain posters, like this one and the poster that goes by Nora Baron have posts that get longer, and longer, and longer. I find it rather annoying, as they keep playing tricks. > (2) What's wrong with a function which is 1 for some values of m but > not 1 for others? Surely you've seen plenty of those? Now you see the delivery from the previous setup, as this poster carefully hides the full position that when the cubic defining the a's is reducible over Q the equation would be monic, but if its irreducible, by his claims, it'd have a leading coefficient with a factor that is f. So you get the question, seemingly rational, but its purpose is to distract rather than be mathematically precise. And notice how LONG this poster has gone on here already and the length of this post!!! >> So what good are their cubics, when I can give ONE which brings b_1, >> b_2 and b_3 together? >> I'm emphasizing the *single* cubic because it must be monic. >> Pray show. (I think you mean a cubic polynomial?) >The cubic cannot, in general, be given without using ..., as then it >is >b3 + (...)b2 + (...)b - (m3 f4 - 3m2 f2 + 3m). > It cannot? If we cannot see the cubic, then how do you know the cubic > has integer coefficients? I didn't say it did. An example, again is b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 which results from m=1, f=sqrt(2). >However, for certain values of m and f, the terms are algebraic >integers, and the cubic is easily displayed. For instance, for m=1, >f=sqrt(2), it is >b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 >and note that here >a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m), gives >a3 + 3a2 - 2, >which IS reducible, but intriguingly, isn't even reducible to linear >terms over Q. > x3+3x2 - 2 = (x+1)(x2+2x-2) > Why is that intriguing? In this case, your cubic factors into a > linear times an irreducible quadratic; you get lucky that your linear > term has constant term coprime to f, so you get the other two factors > sharing f2 in equal portions. You did not get so lucky when u=1, f=2, > m=1, where the linear term did not avoid the value of 2; then you had > a3 + 3(-1+4)a2 - 4(16 - 3(4) + 3) > a3 + 9a2 - 28 = (a+2)(a2+7a-14) > in that case, the factors of 2 did NOT factor out as you hoped they > would, because both the linear term and the irreducible quadratic > contributed to the total factor of 22 = 4. It's intriguing because it shows that the cubic doesn't need to be fully reducible over Q, which makes it even more obvious how quirky the assertions of the poster are. Now then, as to the poster's assertion about factors, I ask that the poster give the cubic defining the b's, the SINGLE cubic. The answer is that the cubic defining the b's in that case do not have all algebraic integer coefficients. It's not complicated, but it's not the answer this poster will want to admit, without possibly trying to move to the field of algebraic numbers, as this poster needs a cubic with algebraic integer coefficients, and a leading coefficient with a factor of f. >The answer is that posters arguing against the cubic defining the b's >being in general monic, are just wrong, and reducibility over Q just >determines whether or not you can easily *see* that they are wrong. > Nonsense. In the case u=1, f=2, m=1, the value of a3=b3=-2, and the > other two values are > b1 = a1/2 = (-7-sqrt(105))/4 > b2 = a2/2 = (-7+sqrt(105))/4 > Neither is an algebraic integer: they are roots of > (x-b1)(x-b2) = x2 - (b1+b2)x + b1*b2 > = x2 - (-14/4)x + (49-105)/16 > = x2 + (7/2)x + (56/16) > = x2 + (7/2)x + (7/2) > so they are roots of the irreducible non-monic > 2x2 + 7x + 7. Hence they are not algebraic integers. They cannot be > the roots of any monic polynomial with algebraic integer coefficients, > let alone a cubic. > In fact, any cubic that has them as a roots would be a scalar multiple > of (x-b1)(x-b2)(x-b3) = (x2 + (7/2)x+(7/2))(x+2) > = x3 + (7/2)x2 + (7/2)x + 2x2 + 7x + 7 > = x3 + (11/2)x2 + (21/2)x + 7 > and if we clear denominators, we have > 2x3 + 11x2 + 21x + 14, > so no monic cubic with integer coefficients has all three as roots (it > would have to be a scalar multiple of the above; since 2 and 11 are > coprime, there is no algebraic integer other than units that > divides all coefficients, so you cannot get rid of the 2 in the > leading term). Reader should note that I've never claimed a cubic with integer coefficients for all values of m and f. This poster is adept at making up requirements, not given, and presenting them as if they are important. Here the poster is presenting an issue around having integer coefficients, when my point is that for values of m and f, coefficients are not even algebraic integers, let alone, integers. Let's get to the end as this post is long enough anyway. Read along and I'll give final comments there. > In case you argue that I just assigned the b's wrong, here are the two > other possibilities (up to exchanging b1 and b2): > CASE 2: b1 = -2/2, b2 = (-7-sqrt(105))/4, b3 = (-7+sqrt(105))/2 > Then any cubic having all three as roots is a scalar multiple of > (x-b1)(x-b2)(x-b3) = (x+1)(x+(7+sqrt(105))/4)(x+(7-sqrt(105))/2) > = (x+1)(x2 + ((21-sqrt(105))/4)x -(56/8)) > = x3 + ((21-sqrt(105))/4)x2 - 7x > + x2 + ((21-sqrt(105))/4)x - 7 > = x3 + [(25-sqrt(105))/4]x2 - [(7+sqrt(105))/4]x - 7 > and clearing denominators, we have > 4x3 + (25-sqrt(105))x2 - (7+sqrt(105))x - 28. Nonmonic with > algebraic integer coefficients. It is not hard to see that this is > irreducible over Q[sqrt(105)], and so none of its roots are algebraic > integers. And of course, there is no cubic with INTEGER coefficients > that will work, either. > Finally, > CASE 3: b1 = -2/2, b2 = (-7+sqrt(105))/4, b3 = (-7-sqrt(105))/2 > Then any cubic having all three as roots is a scalar multiple of > (x-b1)(x-b2)(x-b3) = (x+1)(x+(7-sqrt(105))/4)(x+(7+sqrt(105))/2) > = (x+1)(x2 + ((21+sqrt(105))/4)x -(56/8)) > = x3 + ((21+sqrt(105))/4)x2 - 7x > + x2 + ((21+sqrt(105))/4)x - 7 > = x3 + [(25+sqrt(105))/4]x2 - [(7-sqrt(105))/4]x - 7 > and clearing denominators we have > 4x3 + (25-sqrt(105))x2 - (7-sqrt(105))x - 28 > again nonmonic, irreducible over Q[sqrt(105)], and none of its roots > are algebraic integers. No cubic with INTEGER coefficients will > work. So we can SEE that you are wrong. >Of course, also for m=0, and any f, it is, as I've mentioned before >b3 - 3b2, >and >a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m), gives >a3 - 3a2. > Which is reducible into three linear terms, and reducibility of the > polynomial is NECESSARY before there can be a cubic with integer > coefficients that defines the b's (though not sufficient, as seen just > above). > [.snip.] >What's important about that readers is it forces the poster to either >accept that the cubic is monic in general, which means accepting the >core problem, or claim that there's a *function* of m, which varies >in some odd way. > Yeah, some really odd way. Like, for example, (2m+1) which has value 1 > with m=0 but not when m is not zero. > Or a function which is 1 at m=0 and at m=1, but not anywhere else? > Something REALLY odd like (2m-1)2? > [.snip.] >> There is: (x - b1)(x - b2)(x - b3). But, does it have integer coefficients, >> or rational coefficients? Not when f = 5 and u = 1, as shown above. >> But is it monic? >> As written, yes it is monic. No problem. But it does not necessarily >> have integer coefficients, so what do you conclude from it being monic? >Well, no, it doesn't necessarily have integer coefficients, like I >showed before with f=sqrt(2), m=1, where >b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 > You are using values of f that are not integers. But here, you get > algebraic integers. However, as seen above, the MONIC version does not > necessarily have ALGEBRAIC integer coefficients either. > You are being misleading here, since you do not clarify just what it > is you are claiming. > Which of the following is your claim? > POSSIBLE CLAIM 1: There is a cubic polynomial, whose coefficients are > functions of u, f, m, with the property that for every algebraic > integer value of u, f, and m, the polynomial is monic, has algebraic > integer coefficients, and has b1, b2, and b3 as roots. > POSSIBLE CLAIM 2: For every algebraic integer value of u, f, and m, > there is a cubic monic polynomial with algebraic integer coefficients > that has b1, b2, and b3 as roots. > POSSIBLE CLAIM 3: For every algebraic integer value of u, f, and m, > there is a monic cubic polynomial which has b1, b2, and b3 as roots, > but we make no claim about the coefficients of that polynomial > POSSIBLE CLAIM 3 gives you nothing; only from possible claims 1 or 2 > could you conclude that b1, b2, and b3 are algebraic integers. > And the situation you are in is 3. >but readers can see just how misleading Nora Baron was in posts >showing TWO cubics, where one must exist. > You're wrong. For any value of m for which P(m) is not zero, and the > polynomial > a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). > is irreducible over Q, you cannot have a single cubic with rational > coefficients that has b1, b2, and b3 as roots. > (NOTE THAT in order to be irreducible over Q, the coefficients MUST > BE rational numbers; your example with f=sqrt(2) does not satisfy > these conditions, since the polynomial DOES NOT have rational > coefficients; the corresponding claim would have to be that the > resulting polynomial is irreducible over Q[sqrt(2)] ). > [.snip.] >Now for certain values, the cubic has algebraic integer coefficients, >while for others, it does not because the ring of algebraic integers >doesn't include certain numbers that it should, which leads to a >problem in core as you can have the appearance of two proofs >contradicting each other. > Ah, yes. So your cubic doesn't have the right numbers, so once > again, things that ->should<- be there aren't there. But why should > they be there? Because you ->want<- them to be, not for any other > reason. > Well, tough. > [.snip.] >> The important question for the SINGLE cubic for which the b's are >> roots is, IS IT MONIC? >> No. The important question is: is it monic and does it have integer >> coefficients? Otherwise you can not conclude anything about the b's >> being algebraic integers. >That's bogus, as I can show with >b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1. > No, it's not bogus. Here, your original polynomial defining the a's > DOES NOT have rational coefficients. Here, the question must become > wether it is monic and irreducible over Q[sqrt(2)]. > And this is all one BIG red herring. You are always assuming f is a > prime integer, except when it is convenient for you to distract with > other values. > [.snip.] >It's simply some person *wishing* something they cannot prove. > You mean, like there should be a cubic, even though you cannot prove > there is one? > [.snip.] >>Let's take (x-b1)(x-b2)(x-b3); all b's are algebraic numbers, >Here's the second sneak, as yes, the b's are in fact algebraic >numbers, even when they're not algebraic integers, but the poster has >moved to a field. > There is no second sneak. This is basic precalculus, which even you > must have gone over in your gifted program. > [.snip.] >> so all coefficients are algebraic numbers. Each algebraic number can be >> written as the quotient of an algebraic integer and a normal integer. >Here's the next point of the attempt at snowing the reader, as the >poster is pushing the position that the b's can be written like, say, >x/f, where x is some algebraic integer that doesn't have f as a >factor, but possibly shares non-unit factors with f. > That's a lie, James. Nowhere is he saying or implying that the > denominator will be ->f<-; but EVERY algebraic number can be written > as a quotient of an algebraic integer and a regular integer, though > this expression NEED NOT BE IN LEAST TERMS. Nowhere is it assumed that > it is in least terms, it is unnecessary for the argument. > So you are erecting strawmen and lying. > [.snip.] >It's this type of poster that gives me fits as Nora Baron and Arturo >Magidin both engage in this behavior. > Ad hominem. >Some readers just jump to their conclusion, and apparently assume they >proved it, when in fact, they simply use tactics. > Yeah, that's what you do. Should be multiples. That's jumping to a > conclusion. When asked to justify, instead of explaining the should, > you insult, you whine, and eventually explain SOME OTHER THING NOBODY > COMPLAINED ABOUT. And then, having committed a strawman and a red > herring, you claim to have handled the objection, because you > addressed ->something<-... just not what people were asking and/or > complaining about. > [.snip.] Now then, if you read all the way to here then you're probably a rare reader unless you just jumped to the end. For most readers taking the time to carefully go through such a long post is a major investment, and I think Arturo Magidin knows it is. So he makes overlong posts where he goes on and on and on, basically relying on you finally accepting that simply his act of disagreement means that I'm wrong. James Harris === Subject: Re: Finishing argument, core error proven Visiting Assistant Professor at the University of Montana. [.snip.] >>Here's more pertinent information: >>P(m) = f2((m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f) >>P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) >>R(m) = (m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f >>R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) >> So: R(m) = P(m)/f2. Assuming that P(m) is not zero, we have: >> R(m) = P(m)/f2 = (1/f)(a_1x+uf)(1/f)(a_2x+uf)(a_3x+uf) >> = (b_1x + u)(b_2x+u)(b_3x+uf) >> and you have defined b_1=a_1/f, b_2=a_2/f. So b_1x+u = (a_1x+uf)/f, >> and b_2x + u + (a_2x+uf)/f. Since P(m) is not zero, we can cancel >> terms and get >> b_3x + uf = a_3x + uf. >> So b_3x = a_3 x. >> Since x is not zero (it is coprime to f, which is not a unit), we have >> that b_3=a_3. >Hey, that's correct, which means I had it right the first time, but >later second guessed myself about a_3 and b_3. IF P(m) is not zero, you can conclude this. >> So, UNLESS P(m)=0, we must have b_3=a_3. >> That means that b_3 is a root of the cubic you mention above: >> a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). >> There are two cases: >> CASE 1. The cubic is irreducible over Q; that is, >> x3 + 3(-1+mf2)x2 - f2(m3f4 - 3m2f2+3m) >> has no rational roots; that is, none of a1, a2, a3 are >> rational numbers (and being monic, they would necessarily be >> integers). >> and >> CASE 2. The cubic irreducible over Q; that is, it has at least one, >> possibly three, rational roots. That means, exactly one of >> a1, a2, a3 is an integer, or all three are integers. Drat. This should read The cubic is reducible over Q. >> In CASE 2, which is NOT the case m=0, or the case u=1, f=sqrt(2), m=1 >> that you've used as well, we have the following: And this should be In CASE 1. >> b3 is a root of G(x) = x3 + 3(-1+mf2)x2 - f2(m3f4 - 3m2f2+3m). >> Assume F(x) is ANY polynomial with rational coefficients that has b3 as >> a root. Then we can divide F(x) by G(x), to get >> F(x) = G(x)*Q(x) + r(x) >> where r(x) and Q(x) are polynomials with rational coefficients, and >> r(x)=0 or else r(x) has degree strictly less than the degree of G. >> Since F(x) has b3 as a root, then 0=F(b3)=G(b3)Q(b3) + r(b3) = r(b3), >> so r(x) also has b3 as a root. If r(x) is not zero, then we can >> divide G(x) by r(x), and we have >> G(x) = r(x)Q'(x) + s(x), with s(x) equal to 0 or with strictly smaller >> degree than r(x), and therefore either constant or degree 1. Then >> 0 = G(b3) = r(b3)Q'(b3) + s(b3) = s(b3), so s(b3)=0; that means either >> that r(x) divides G(x) (if s is constant), which is impossible because >> G(x) was assumed to be irreducible; or else s(b3)=0 and s is a >> polynomial of degree 1, which means that b3 is a rational number, >> which is also impossible since G(x) was assumed irreducible. >> The conclusion is that r(x) must be the constant zero, which means the >> original F(x) must be a multiple of G(x), which means that either F(x) >> is the zero polynomial, is a constant multiple of G(x), or else it has >> strictly larger degree than G(x) (i.e., it is of degree greater than >> 3). >> Therefore: If F(X) is ANY cubic polynomial with rational coefficients >> which has b3 as a root, then it must be a constant multiple of G(x), >> and its roots are a1, a2, and a3=b3. (Assuming the polynomial that >> defines the a's is irreducible over Q). >> So, if there is a cubic polynomial with rational coefficients that >> defines the b's, it must mean that we have {b1,b2} = {a1, a2}. Which >> means, either b1=a1 and b2=a2, or else b1=a2 and b2=a1. >> The first is impossible, since b1=a1/f, so a1=a1/f means a1=0, which >> would mean the original G(x) had a rational root, impossible. The >> latter is also impossible, for if b1=a2 and b2=a1, then a1/f = a2, so >> a1 = f(a1/f) = f*b1= f*a2 = f*(f*b2) = f2*b2 = f2*a1, so a1=0 and >> again we have a contradiction. >> Therefore, if the polynomial defining the a's is irreducible over Q, >> then there is NO polynomial with rational coefficients which is (a) >> cubic; and (b) has b1, b2, and b3 as roots. >That can be the case even when it is reducible over Q, as I've pointed >out by giving the defining cubic when m=1, f+sqrt(2), which is >b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 THIS POLYNOMIAL IS NOT REDUCIBLE OVER Q. For a polynomial to be reducible over Q, the very FIRST thing it must satisfy is that its coefficients must be RATIONAL NUMBERS. This polynomial does NOT have rational numbers, and therefore this polynomial is NOT reducible over Q. In addition, you will note that your reply is a complete waste of space. I explained exactly why, when the polynomial defining the a's is IRREDUCIBLE, then there can be no cubic. And then you reply that That can be the case even when it is reducible. So, you address cases that I was not talking about, as if they were relevant. Not to mention that your example is nonsense and inapplicable. >as readers can actually see that only two of the coefficients, the >leading and last, are rational. Which means it is not reducible over Q. It is not even DEFINED over Q. >So far Arturo Magidin hasn't done anything but take up space. No, I have PROVEN that your implicit claim that there is a cubic with rational coefficients defining the b's is false whenever the cubic defining the a's is irreducible over Q. Your example is not applicable, since the polynomial defining the a's is not defined over Q. case), it does not mean that I have simply 'wasted space. >> Therefore, IF there is a single cubic polynomial with rational >> coefficients that has b1, b2, and b3 as roots, then the original >> polynomial was reducible over Q. >> Which, surprise surprise, is exactly the two cases that you manage to >> handle. >Now the surprise surprise here really doesn't fit in context. Yes, it does. The two cases you manage to handle by providing explicitly a cubic for b1, b2, and b3 which is defined over Q are cases where the polynomial defining the a's is reducible. Just as I predicted. [.snip.] >Now readers should consider how long this poster has gone on without >actually saying anything. That's a lie. I just haven't said anything you ->understood<-. >My assessment of this particular poster is that he has learned that >most sci.math readers don't read in detail, but skim, and they are >*especially* likely to skim a long post. No, YOU skim a lot, and that's why you say so much nonsense. If you writing irreducible when I clearly meant reducible, and you might have seen the proof that you cannot have a cubic with integer coefficients defining b1, b2, and b3 when the polynomial defining the a's is irreducible over Q (which requires it to have rational coefficients). [.snip.] >Skimming along, they may feel that Arturo Magidin has said a lot, when >he hasn't said anything as of yet, as I've pointed out. No, you haven't pointed out, you have simply ->lied<- about. That's how you handle objections. By lying about them. [.snip.] >Now if this poster were honest, then he'd simply acquiesce to the >truth. The truth is you don't know what you are talking about, as you amply demonstrate in this post. I've acquiesced to that truth already. >However, instead, look at the LENGTH of his post!!! Quit your whining, sonny. [.snip.] >>That is, you'd need a function of m that would equal 1, when >>a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m) >>is reducible over Q, and f, when it is not, and that function would be >>the leading coefficient of the cubic defining the b's. >> Yes, there is a cubic which has exactly b1, b2, and b3 as >> roots. However, in the case where P(m) is not zero and G(x) is >> irreducible over Q, this polynomial (which is a scalar multiple of >> (x-b1)(x-b2)(x-b3)) does NOT have integer coefficients. It does not >> even have RATIONAL coefficients; to get rational coefficients you need >> to go to a polynomial of larger degree. >Which still isn't saying anything. Aren't you reading? IF P(m) is not zero, AND G(x) is irreducible over Q, then no polynomial with rational coefficients and degree 3 can have all of b1, b2, and b3 as roots. >Again I give the cubic for the b's for m=1, f=sqrt(2), as it is >b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 >and readers can see that, yeah, it has some irrational coefficients. Again, I note that your example is a big fat red herring. This polynomial is not even defined over Q; I stated clearly: THE POLYNOMIAL DOES NOT EVEN HAVE RATIONAL COEFFICIENTS when P(m) is not zero and G(x) is irreducible over Q. The reason why your example is just a waste of space is: this polynomial does not have rational coefficients, so it would SUPPORT my statement above (that the polynomial WILL NOT have rational coefficients), not show it is false as you imply. When m=1 and b=sqrt(2), and u=1, the polynomial defining the a's is a3 + 3(-1+mf2)a2-f2(m3 f4 - 3m2 f2 + 3m) = a3 + 3(-1+2)a2 - 2(4 - 6 + 3) = a3 +3a2 - 2 So there are two reasons why your example SUPPORTS my points, you ninny: (a) I talked about no cubic with rational coefficients existing. So what do you do? You give a cubic that does not have rational coefficients. (b) I specify that I am talking about the case where the polynomial defining the a's is irreducible. IN this case, the polynomial is REDUCIBLE: a3+3a2-2 = (a+1)(a2 + 2a - 2), so your example cannot In short, it was a big fat red herring and a big whopping lie on your part. >Like here he broke up a paragraph where I noted that if the cubic >defining the a's is NOT irreducible over Q then somehow supposedly the >cubic defining the b's has a leading coefficient with a factor of f, >so I wasn't talking about reducibility over f. What does reducibility over f mean? There is NO SINGLE CUBIC that defines the b's. There is an infinite number of cubics, all constant multiples of (x-b1)(x-b2)(x-b3), which is the UNIQUE MONIC POLYNOMIAL WITH COMPLEX COEFFICIENTS that is cubic and has b1, b2, and b3 as roots. When the original polynomial is irreducible over Q (which requires it to be DEFINED over Q), then NO multiple of this polynomial has rational coefficients, let alone integer ones. Your statement above is also a lie. Nobody is saying that if the polynomial defining the a's is reducible over Q then the cubic defining the b's has a leading coefficient with a factor of f. What has been said is exactly the opposite: IF THE CUBIC DEFINING THE a's IS IRREDUCIBLE OVER Q, then there is NO cubic polynomial with rational coefficients that defines the b's. NONE AT ALL. And since any scalar multiple of a polynomial has the same roots as the polynomial, it makes no sense to talk about the leading coefficient of a polynomial with given roots. So you can stop your lies and your distractions, and you appeals to the gallery. Stick to what is actually written, not what you invent to try to make your point. [.snip.] >> (2) What's wrong with a function which is 1 for some values of m but >> not 1 for others? Surely you've seen plenty of those? >Now you see the delivery from the previous setup, as this poster >carefully hides the full position that when the cubic defining the a's >is reducible over Q the equation would be monic, but if its >irreducible, by his claims, it'd have a leading coefficient with a >factor that is f. You are lying. No such position has been given. The ONLY thing I stated here is that if the polynomial defining the a's is IRREDUCIBLE over and P(m) is not zero, then there is NO CUBIC WITH RATIONAL COEFFICIENTS that can have b1, b2, and b3 as roots. I never said anything about an equation being monic or not monic, and I certainly never made a claim as stupid as the leading coefficient has a factor that is f when you have not specified which the infinite number of scalar multiples of a polynomial you might refer to. You are either skimming or you are lying. In either case, your (already non-existent) credibility takes a dip. [.snip.] >And notice how LONG this poster has gone on here already and the >length of this post!!! Quit yer whining, sonny. [.snip.] >b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 >which results from m=1, f=sqrt(2). >>However, for certain values of m and f, the terms are algebraic >>integers, and the cubic is easily displayed. For instance, for m=1, >>f=sqrt(2), it is >>b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 >>and note that here >>a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m), gives >>a3 + 3a2 - 2, >>which IS reducible, but intriguingly, isn't even reducible to linear >>terms over Q. >> x3+3x2 - 2 = (x+1)(x2+2x-2) >> Why is that intriguing? In this case, your cubic factors into a >> linear times an irreducible quadratic; you get lucky that your linear >> term has constant term coprime to f, so you get the other two factors >> sharing f2 in equal portions. You did not get so lucky when u=1, f=2, >> m=1, where the linear term did not avoid the value of 2; then you had >> a3 + 3(-1+4)a2 - 4(16 - 3(4) + 3) >> a3 + 9a2 - 28 = (a+2)(a2+7a-14) >> in that case, the factors of 2 did NOT factor out as you hoped they >> would, because both the linear term and the irreducible quadratic >> contributed to the total factor of 22 = 4. >It's intriguing because it shows that the cubic doesn't need to be >fully reducible over Q, which makes it even more obvious how quirky >the assertions of the poster are. No, it makes nothing obvious about it. It is nothing but a red herring. >Now then, as to the poster's assertion about factors, I ask that the >poster give the cubic defining the b's, the SINGLE cubic. Sigh. ANY cubic defining the b's will be a constant multiple of (x-b1)(x-b2)(x-b3). That's BASIC FREAKING PRECALCULUS, so even YOU must have read about it at some point. But this polynomial is useless for determining whether b1, b2, and b3 are algebraic integers, because you do not know that it has algebraic integer coefficients UNLESS You already know that b1, b2, and b3 are algebraic integers. However, whenever the polynomial defining the a's is IRREDUCIBLE OVER Q, there is NO cubic polynomial that has the b's as roots AND has rational coefficients. ALL the constant multiples of (x-b1)(x-b2)(x-b3) will have NON rational coefficinets. The monic, and all the other ones. >The answer is that the cubic defining the b's in that case do not have >all algebraic integer coefficients. It's not complicated, but it's >not the answer this poster will want to admit, without possibly trying >to move to the field of algebraic numbers, as this poster needs a >cubic with algebraic integer coefficients, and a leading coefficient >with a factor of f. Why would I not want to admit it? IF THE POLYNOMIAL DEFINING THE b'S DOES NOT HAVE ALL ALGEBRAIC INTEGER COEFFICIENTS, AND IS MONIC, THEN THE b'S ARE NOT ALGEBRAIC INTEGERS. It would ->prove my point<-, you silly goose, it would ->prove you are wrong<-. You are so confused that you don't even know what you are arguing for any more. [.snip.] >>The answer is that posters arguing against the cubic defining the b's >>being in general monic, are just wrong, and reducibility over Q just >>determines whether or not you can easily *see* that they are wrong. >> Nonsense. In the case u=1, f=2, m=1, the value of a3=b3=-2, and the >> other two values are >> b1 = a1/2 = (-7-sqrt(105))/4 >> b2 = a2/2 = (-7+sqrt(105))/4 >> Neither is an algebraic integer: they are roots of >> (x-b1)(x-b2) = x2 - (b1+b2)x + b1*b2 >> = x2 - (-14/4)x + (49-105)/16 >> = x2 + (7/2)x + (56/16) >> = x2 + (7/2)x + (7/2) >> so they are roots of the irreducible non-monic >> 2x2 + 7x + 7. Hence they are not algebraic integers. They cannot be >> the roots of any monic polynomial with algebraic integer coefficients, >> let alone a cubic. >> In fact, any cubic that has them as a roots would be a scalar multiple >> of (x-b1)(x-b2)(x-b3) = (x2 + (7/2)x+(7/2))(x+2) >> = x3 + (7/2)x2 + (7/2)x + 2x2 + 7x + 7 >> = x3 + (11/2)x2 + (21/2)x + 7 >> and if we clear denominators, we have >> 2x3 + 11x2 + 21x + 14, >> so no monic cubic with integer coefficients has all three as roots (it >> would have to be a scalar multiple of the above; since 2 and 11 are >> coprime, there is no algebraic integer other than units that >> divides all coefficients, so you cannot get rid of the 2 in the >> leading term). >Reader should note that I've never claimed a cubic with integer >coefficients for all values of m and f. James Harris should note that his claim is empty unless he can say something about the cubic. GIVEN ANY 3 COMPLEX NUMBERS, there is a cubic that has those three numbers as roots and no other complex number as root. Saying that there is a cubic that has b1, b2, and b3 as roots is as empty as saying that b1, b2, and b3 are numbers. It's useless. [.Ad hominem personal attacks and no math deleted.] >Now then, if you read all the way to here then you're probably a rare >reader unless you just jumped to the end. I love the way you accuse everyone of your own shortcomings. You lie, so you accuse everyone of lying. You don't read what people write, so you accuse people of not reading what others write. You misunderstand, so you accuse people of misunderstanding. And here, you jumped to the end without comprehending, so you lie and appeal to the gallery. 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 critize. A great many people are staggered to this extend, 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 === Subject: Re: Finishing argument, core error proven Visiting Assistant Professor at the University of Montana. >> [.snip.] > Um, do *you* accept that there exists a SINGLE cubic which has the b's > as its roots? Yes, I do, and I think everyone does. (x-b1)(x-b2)(x-b3) serves just fine. > It is a cubic in b, but as the b's depend on m, it is not necessarily a > cubic in m. >>The question is to the solution for the b's, just like earlier in the >>post a solution for the a's is given as they are roots of the cubic >>a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). >>Readers should note the need for me to put *back* in math that is >>needed to properly evaluate various claims, as this poster left them >>out. >> Will you quit your whining already? If people don't trim, you whine >> that their posts are too long; if they trim, you whine that they >> trimmed. If they make multiple replies to keep them short, you whine >> that they make multiple replies; if they make a single reply, you >> whine that they address different issues and are trying to snow >> people. >And here readers should notice that Arturo Magidin starts right at the >beginning, not with mathematics, but with a protest. Clearly, the answer is No, I will not quit my whining. And, here, readers, should note that James Harris starts right at the beginning, not with mathematics, but with a protest. === Subject: Re: Finishing argument, core error proven > For me there have been two perspectives as I work to figure out how to > explain the definition problem in mathematics with LOTS of opposition, > and I wonder about mathematicians so dedicated to attacking an > argument that is clearly correct. I remind of that as I present what should finish their ability to > distract, as I've seen a strange and dedicated effort to ignore the > actual math, and simply toss up just about anything rather than face > the truth. All variables are in the ring of algebraic integers unless otherwise > stated. Let P(m) = f2((m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f) and let R(m) = (m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f so P(m) = f2 R(m). Now consider P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where the a's are given by the following cubic: a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). Then it must be true that the following factorization exists R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where the b's are given by the following cubic: b3 + 3(-1+mf2)b2 - (m3 f4 - 3m2 f2 + 3m), > should be > b3+...+ 3(-1+mf2)b2+...- (m3 f4 - 3m2 f2 + 3m) > So, James Harris, how about apologizing to Nora Baron. After all, James > Harris, YOU made a mistake, not Nora Baron. That's an interesting question. > And in general you would come off a lot less pitiful if you would stop > attempting to discredit those who present clear cogent math with personal > attacks. It makes you appear as a pompous ass, which I suppose if we were > to meet you in person you are not. Really, it would serve your cause to > just address the math with math. Most of the time it's clear that you > cannot. So you resort to (a) insults, (b) pontificating, (c) simply > repeating the original discredited post, or (d) abandoning the thread and > starting over. I like arguing. And I like going over my math results, but what I find with certain posters like Nora Baron in particular is that they cheat. Now I like objections that I've handled to not come back up. I like it when basic algebra isn't challenged. I like it when other people play fair. > I suspect after an insulting parting shot we'll see (d) next. > KeithK You sound quite emotional. Math isn't about emotion, but cold, hard logic. Now if you think that Nora Baron is being treated badly by me, then, hey, you have the right to that opinion. But I think the poster can take care of her or himself, as since the poster uses a pseudonym, you can't be sure Nora Baron is female. Remember, this is Usenet and the Internet. People are often not who you think they are. James Harris === Subject: Re: Finishing argument, core error proven > I like it when basic algebra isn't challenged. Your track record proves otherwise. You delight in challenging basic algebra. > I like it when other people play fair. You reserve the right to play unfairly for yourself. Furthermore, you like it best when someone accepts your erroneous arguments. > Now if you think that Nora Baron is being treated badly by me, then, > hey, you have the right to that opinion. > But I think the poster can take care of her or himself, as since the > poster uses a pseudonym, you can't be sure Nora Baron is female. > Remember, this is Usenet and the Internet. People are often not who > you think they are. > James Harris We still don't know if James Harris is your real name, or an anagram for Mr. J. Harie Ass. Wacky, isn't it? But, hey, that's just basic math. Yup, yup, yup! -- 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 === Subject: Re: Big Bertha Thing blogs Big Bertha Thing progress Cosmic Ray Series Possible Real World System Constructs http://web.onetel.com/~tonylance/progress.html Access page to 6K Web page Astrophysics net ring access site Newsgroup Reviews including uk.transport The Progress of Discontent by T. Warton (Written at Oxford in 1746) Half-hours With the Best Authors The Chandos Classics Edited by Charles Knight Published by Frederick Warne and Co. 1890 Big Bertha Thing besides Some people think that Big Bertha is less suitable for the Have they heard the long name for CP Conf? Perspective. They think that is a better place to put Big Bertha. They must be besides themselves, which proves the thing. (C) Copyright Tony Lance 1998 To comply with my copyright, please distribute complete and free of charge. Tony Lance pobox47@big-bertha-thing.com Big Bertha Thing testament New Testament Rosary 1. The Angel Tells of the Virgin Birth. 2. The Blessing of John the Baptist in the Womb by the Unborn Christ.(13 weeks) 3. The Birth of Our Lord. 4. The Purification of Our Lady and the Presentation of Our Lord. 5. The Finding in the Temple. 6. The Baptism of Our Lord in the Jordan. (Luke 3) 7. The Wedding at Cana. (John 2) 9. The Transfiguration of Christ with Moses and Elias. (Matthew 17) 10. The Last Supper. (Matthew 26) 11. The Agony in the Garden. 12. The Scourging at the Pilar. 13. The Crowning With Thorns. 14. The Carrying of the Cross. 15. The Crucifixion. 16. The Resurrection. 17. The Ascension into Heaven. 18. The Coming Down of the Holy Ghost. 19. The Assumption into Heaven. 20. The Crowning of Our Lady Queen of Heaven. (1st five Joyfull, 2nd five Luminous, 3rd five Sorrowful, 4th five Glorious.) Old Testament Rosary by Tony Lance 1. The Promise of Hannah. (I Kings 1) Birth of Samuel. 2. The Three Angels Visit Abraham. (Genesis 18) Birth of Isaac. 3. The Betrothal of Ruth. (Ruth 3) Stranger and great-grandmother of David. 4. The Offering up of Isaac by Abraham. (Genesis 22) 5. The Finding of Joseph in Egypt by Jacob. (Genesis 46) 7. The Feast in the Temple for the Rechabites. (Jeremias 35) 8. The Covenant of Moses on Mount Sinai. (Exodus 19) 9. The Forty Day Walk of Elias. (III Kings 19) 10. The Writing on the Wall. (Daniel 5) The Fall of Empires. 11. The Water Dungeon of Jeremias. (Jeremias 38) The Babylonian Transmigration for 70 years. 12. The Deriding of David by the Kinsman of Saul. (II Kings 16) The Fall of Jerusalem. 14. The Ark is Carried for Forty Years. (Numbers 14) The Entry into the Promised Land. 15. The Execution of Jonah. (Jonah 1) The City of Nineveh Wears Sackcloth and Ashes. 16. The Raising to Life by the Tomb of Eleusius. (IV Kings 13) 17. The Taking up of Elias in the Fiery Chariot. (IV Kings 2) 18. The Spirit of Moses Comes Down on the Judges. (Numbers 11) The Law of Precident Cases. 19. The Slaying of Holofernes by Judith. (Judith 13) The Treasure to the 12 tribes. 20. The Crowning of Esther Queen of Susan. (Esther 2) The Decree of Pogrom is reversed. Rosary Intentions for each decade of St. Louis Marie de Montfort Luminous Mysteries by Tony Lance Lord through thy Holy Mother that we may:- 1. know true humility. 2. have charity one for another. 3. love poverty. 4. have purity of body and soul. 5. obtain Divine Wisdom. 6. be free from Original Sin.(Wet head saying 'I baptise you in the name of the Father and of the Son and of the Holy Ghost. Amen') 7. give grace one to another. 8. enlighten one another. 9. obtain Union with Christ.(3 Nights of Soul) St.John of the Cross Battle vices and deny 5 physical senses. (Pictures, icons and Rock and Roll.) Battle ghosts of old vices and deny spiritual senses. (Old sins plague you for second time.) Visions, Locutions and tears. All else gone except the love of God. (Last of all the memory is perfected. It gets worse first.) 10. break bread together. 11. have perfect contrition. 12. know mortification. 13. have contempt for the world. 14. have patience in crosses. 15. obtain repentance for sinners, perserverance in grace for the just and reief for the Holy Souls in Purgatory. 16. love thee and be fervent in they service. 17. love Heaven our true home. Annual Preparation. 12 days seek Contempt for World, 6 days each know self, Mary and Christ. 19. know the Holy Ghost. 20. obtain perserverance in grace for the just and the crown of Heavenly Glory. NB. Douai-Rheims cross reference links. The Sermon on the Mount. Mark 1,14:15 Matthew 4,12:17 Matthew 5,3 Luke 6,20 NB. Parallels between old and new testament rosary event readings. They had no wine, yet drank it. Others had wine and drank not. One was crowned with thorns and crucified. Another robed as a king and decreed to hang. One was scourged at the pillar. Another derided at the head of a whole column of soldiers. One was found with the wise men. Another was found, as the wisest man in Egypt. Elizabeth and Rachel were barren and both bore child. One was crucified. and descended into Hell (Limbo), for three days. Another walked the plank and was swallowed by a whale, for three days. One was born in a stable. Another betrothed in a barn. One was crowned Queen of Heaven. Another crowned Queen of an empire. One rose up into Heaven. Another was taken up in a fiery chariot. One was baptised in the Jordan. Another was saved from the bullrushes. That is 10 out of 20 direct parallels. The other 10 are more nuanced, but who can tell? NB. The Original Divine Office of the Church was Simply to Read Book Of Psalms Once a Week. (Divisions into days and hours can be the 15 minute segments as follows.) 3 by 7 Each psalm is followed by the prayer. Glory by to the Father and to the Son and to the Holy Ghost, as it was in the beggining, is now and ever shall be, world without end. Amen. 1-9 10-17 18-24 (25-32 33-37 38-44) 45-53 54-60 61-67 (68-72 73-77 78-84) 85-89 90-98 99-104 (105-109 110-117 118-118) 119-131 132-139 140-150 NB. On meditation and contemplation. You have to dig the field before you can watch the flowers grow. Both are problem solving. Dread Visions Six Harms By Tony Lance (Based on the writings of St.John of the Cross; the Mystic Doctor of the Church.) Faith fails, replaced by a vision sensible. What we see is more real, than God invisible. Visions turn the spirit from its flight Heavenward. Had Mary Magdallen touched the feet of the Lord. Then grounded in faith, would she never be. How can we hold this vision dearer than Thee. In nakedness of spirit, hold ourselves renouncible. Grace fails, imperceptible forsaken for perceptible. The gift becomes a right, which denies the gift. Decide the vision not from God, with Satan left. Seeking after visions is a sinfull occasion. Turning into diabolical, that divine vision. With the Devil disguised, as an angel of light, And man too ignorant, to even take fright. Grace cannot be rejected; into the soul infused. The taint of evil can be rejected, by vision refused. === Subject: math and selfstudie Hi there. I've got a question, it's fairly newbe, so bare with me. I want to do math as a selfstudie , for fun and soothing a little fascination/obsession.(well , everybody needs a hobby , eh?) What are actually good books or websites on calculus and algebra one can recommend? (Difficulty level is not important, I have a notion that most things can be learned given time, motivation and persperation). I want to studie the proofs of calculus and algebra theorems and I'm searching for a textbook/site where those proofs are as exact and precise as can be, as close to the original proof as the person who thougt of it. If some one knows such books, I'd be a happy lad. Bye. === Subject: Re: math and selfstudie > Hi there. > I've got a question, it's fairly newbe, so bare with me. > I want to do math as a selfstudie , for fun and soothing a little > fascination/obsession.(well , everybody needs a hobby , eh?) > What are actually good books or websites on calculus and algebra one > can recommend? (Difficulty level is not important, I have a notion > that most things can be learned given time, motivation and > persperation). > I want to studie the proofs of calculus and algebra theorems and I'm > searching for a textbook/site where those proofs are as exact and > precise as can be, as close to the original proof as the person who > thougt of it. > If some one knows such books, I'd be a happy lad. > Bye. if I may recommend a strategy, look at the book: How to Solve It by Polya. I think it gives a good perspective on the art of proof. There are other very good books on the subject too, but I'd start there. It is not so much the topic (math), but the classes of methods used in doing proofs. Get an wonderful orientation in those and I think that will go a very long way in helping you in this new hobby. HTH, Flip mm-3699 === Subject: Re: Antidiagonal, Infinity About Euclid's postulates there, they define Euclidean geometry. Disagreeing with the parallel postulate, number five, allows non-Euclidean geometry. About the compass and rule to trisect an angle, for example to form an angle of pi/3 radians from lines intesecting a point on a straight line, an angle of pi radians, it's easy to bisect an angle with the compass, and 1/3 is the sum of 1 / 22x for each positive integer x, eg .010101(01).... A theoretical rule and compass bijection that takes infinitesimal time to bisect an angle may well trisect an angle in some finite time. When you cross the room, you get all the way there. You can begin to sample a random real number from the unit interval by flipping a coin a given number of times. What you get is a discretized sample. About y/x, say that all that is known is that y is greater than x. Is y thus a dependent variable of x? All that is known is that y > x. The value of y/x as x goes to infinity is indefinite. It might be a finite value, it might diverge. Are x and y thus interdependent? About the sets of the hyperreals and the reals, somebody else says that each set contains the same elements, that is x E R <-> x E *R. The square root of two: the length of the diagonal of the unit square, and the length of the side of the square with diagonal of length two. Two: the only integer whose sum with itself is its product with itself. Then again multiplication is just the sum of a mutiplicand with itself as many times as the other. About the itty bitty line segments, infinitesimal line segments might be comprised of a pair of adjacent points. Yet, that's not acceptible. About the representation of a continuous function as a signal, Fourier might have much to tell us about it. The real number is a point! Identify a real number. Point to it on the real number line. Where are the negative integers in the ordinals? 2 - 4 = -2. Chapman, if you're going to go outside put some pants on. I don't get that function that's continuous at all irrationals. It doesn't have an inverse, it's partial. Ross === Subject: Re: Antidiagonal, Infinity >About Euclid's postulates there, they define Euclidean geometry. Actually, they don't. >About the compass and rule to trisect an angle, for example to form >an angle of pi/3 radians The issue isn't whether there is an angle that can be trisected; the issue is whether you can trisect an arbitrary angle. >1/3 is the sum of 1 / 22x for each positive integer x, There is no such sum. There is a limit of sums, but you can't take a limit with compass and straightedge. >A theoretical rule and compass bijection that >takes infinitesimal time A meaningless term. Mathematics is not about time. Nor is it about Zeno. >You can begin to sample a random real number You still haven't defined what you mean by a random real number. >About y/x, say that all that is known is that y is greater than x. What sorts of things are x and y? >Is y thus a dependent variable of x? What does that question mean? >The value of y/x as x goes to infinity is indefinite. What does that sentence mean? >About the sets of the hyperreals and the reals, somebody else says >that each set contains the same elements, You said that, and it's false. >Then again multiplication is just the sum of a >mutiplicand with itself as many times as the other. Pi*Pi >infinitesimal line segments Please define. >might be comprised of a pair of adjacent points. There are non. And line segments in Euclidean spaces contain more than two points. >About the representation of a continuous function as a signal, This is sci.math, not sci.ee; we don't do signals. Besides, even an EE would consider that to be nonsense. A signal is not the same thing as a sine wave. >Fourier might have much to tell us about it. Yes, he would tell you about approximating continuous functions as sums of trigonometric functions. >Point to it on the real number line. Pointing may be a valuable skill in kindergarten; it has nothing to do with Mathematics. >Where are the negative integers in the ordinals? Where are the orange groves in the Bessemer converter? There are none. >I don't get that function that's continuous at all irrationals. >It doesn't have an inverse, Why is it relevant whether it has an inverse? >it's partial. Where is it undefined? -- spamtrap@library.lspace.org === Subject: Re: Antidiagonal, Infinity > About Euclid's postulates there, they define Euclidean geometry. > Disagreeing with the parallel postulate, number five, allows > non-Euclidean geometry. Actually, Euclid's axioms have been shown insufficient. > About the compass and rule to trisect an angle, for example to form an > angle of pi/3 radians from lines intesecting a point on a straight > line, an angle of pi radians, it's easy to bisect an angle with the > compass, and 1/3 is the sum of 1 / 22x for each positive integer x, > eg .010101(01).... A theoretical rule and compass bijection that > takes infinitesimal time to bisect an angle may well trisect an angle > in some finite time. When you cross the room, you get all the way > there. The trisection problem specifies certain rules to be followed in any attempt. Trying to do it by breaking those rules is disallowed. The remainder of Ross' post was to incoherent to comment on. === Subject: Re: Antidiagonal, Infinity <3f82e66f$1$fuzhry+tra$mr2ice@news.patriot.net> <3f85c273$18$fuzhry+tra$mr2ice@news.patriot.net> <3f89f684$43$fuzhry+tra$mr2ice@news.patriot.net> linux) >> About Euclid's postulates there, they define Euclidean geometry. >> Disagreeing with the parallel postulate, number five, allows >> non-Euclidean geometry. > Actually, Euclid's axioms have been shown insufficient. In what sense? -- Sale or rental of this disc is ILLEGAL. If you have rented or purchased this disc, please call the MPAA at 1-800-NO-COPYS. -- The MPAA begins a new anti-piracy program, found on a DVD purchased in China === Subject: Re: Antidiagonal, Infinity @news.patriot.net> <87oewk6kn5.fsf@phiwumbda.org> at 02:58 PM, jesse@phiwumbda.org (Jesse F. Hughes) said: >In what sense? In the sense that some of the statements Euclid purported to prove from the listed axioms and postulates don't follow from them. See, e.g., Foundations of Geometry[1] for an explanation. Google for, e.g;., Axiom of Archimedes. [1] You may see me jokingly refer to it as groundlaager, but it was a seminal book by one of the major Mathematicians of the[2] century, and still quite relevant today. [2] Whether you count him as 19th or 20th. -- spamtrap@library.lspace.org === Subject: Re: Antidiagonal, Infinity >> About Euclid's postulates there, they define Euclidean geometry. >> Disagreeing with the parallel postulate, number five, allows >> non-Euclidean geometry. > Actually, Euclid's axioms have been shown insufficient. > In what sense? See Hilbert's work on the subject. === Subject: Re: Antidiagonal, Infinity >> About Euclid's postulates there, they define Euclidean geometry. >> Disagreeing with the parallel postulate, number five, allows >> non-Euclidean geometry. > Actually, Euclid's axioms have been shown insufficient. > In what sense? See http://www.beva.org/math323/asgn7/dec12.htm or do a Google search for Euclid and Hilbert at === Subject: Re: Antidiagonal, Infinity <3f82e66f$1$fuzhry+tra$mr2ice@news.patriot.net> <3f85c273$18$fuzhry+tra$mr2ice@news.patriot.net> <3f89f684$43$fuzhry+tra$mr2ice@news.patriot.net> <87oewk6kn5.fsf@phiwumbda.org> linux) >> Actually, Euclid's axioms have been shown insufficient. >> In what sense? > See > http://www.beva.org/math323/asgn7/dec12.htm > or do a Google search for Euclid and Hilbert at -- But remember, as long as one human being follows the rules of mathematics, then mathematics as a human discipline survives. Right now I'm that one human being, so mathematics survives. -- James S. Harris === Subject: Re: Antidiagonal, Infinity >> Actually, Euclid's axioms have been shown insufficient. >In what sense? Presumably, in the sense that they don't model what they are (pre-mathematically) intended to model: without some sort of axiom(s) of completeness in the style of Hilbert, various points that Euclid purportedly constructs just may not *be* there! (Or so I've been told.) Lee Rudolph === Subject: Re: Antidiagonal, Infinity > Actually, Euclid's axioms have been shown insufficient. >>In what sense? > Presumably, in the sense that they don't model what they are > (pre-mathematically) intended to model: without some sort of > axiom(s) of completeness in the style of Hilbert, various > points that Euclid purportedly constructs just may not *be* > there! (Or so I've been told.) I have only a passing interest, not a deep one, here, but I'd still like to have a reference. I think I understand what you're talking about, but, nonetheless, a concise and well-presented explanation would be wonderful. -- s Harris, render of worlds === Subject: Re: Antidiagonal, Infinity >> Actually, Euclid's axioms have been shown insufficient. >In what sense? >> Presumably, in the sense that they don't model what they are >> (pre-mathematically) intended to model: without some sort of >> axiom(s) of completeness in the style of Hilbert, various >> points that Euclid purportedly constructs just may not *be* >> there! (Or so I've been told.) >I have only a passing interest, not a deep one, here, but I'd still >like to have a reference. >I think I understand what you're talking about, but, nonetheless, a >concise and well-presented explanation would be wonderful. I don't have a reference either, but it's a well-known fact that Euclid's axioms were not complete. === Subject: Star Gate Topology of the Universe Jack, You say, In particular I do not as yet see any physical necessity for Saul-Paul's Ansatz that physical 3D expanding accelerating cosmic space is actually SU(2)/ID rather than the SU(2) shown in Fig 3... The idea that 3D cosmological space SU(2)/ID rather than SU(2) is the view of Luminet et al. in their *Nature* (9 October) paper. They say (p. 593-594): The Poincare dodecahedral space is a dodecahedral block of space with opposite faces abstractly glued together, so objects passing out of the dodecahedron across any face return from the opposite face. Light travels across the faces in the same way, so if we sit inside the dodecahedron and look outward across a face, our line of sight re-enters the dodecahedron from the opposite face. We have the ILLUSION [emphasis added] of looking into an adjacent copy of the dodecahedron. Part of the problem here is what is meant by illusion physically? On the one hand we have very pretty topological formal algorithms in which, for example, a cylinder is topologically equivalent to a flat rectangle with periodic boundary conditions on one pair of opposite edges. One can generalize this to multiply-connected 2D surfaces with wormhole handles or toroidal surfaces and take it to 3D etc. However, my point is that physically the Universe could be either, in the simplest toy model, in 2D like the curved cylinder or like the flat rectangle with a pair of Star Gate edges of the world in which material objects are instantly teleported across space. Formally, the two situations are equivalent or indistinguishable from the POV of the topology, but the local metrical physics is more than the pre-metrical global topology and the actual situation is empirical and cannot be decided apriori on purely esthetic grounds. Note that a simply-connected ruled 2D surface like the cylinder has zero intrinsic Gaussian curvature as I recall. What about the doubly-connected 2D torus? A Flatlander insect crawling on either the 2D cylinder or torus without boundary is equivalent to the insect being instantly teleported from one edge to the opposite edge across a distance c/H(t) when we make the conceptual cuts to a flat rectangle. So the issue is whether there is any physical way to distinguish the two ideas or are they absolutely degenerate or indistinguishable by any imaginable physical measurements such as the presence or absence of a local tuv(zpf) =/= 0 exotic vacuum tensor field? Now as far as the WMAP data is concerned this distinction between rubber expanding accelerated versions of 1. we are literally trapped like Flatlanders inside a closed multiply connected 3D space without boundary but non-trivial Betti number with no sharply detectable 2D Star Gate walls vs 2. we are literally trapped like Flatlanders inside a 3D spherical dodecahedron with 12 pentagonal edged Star Gate walls or faces on the expanding scale c/H(t) with essentially instant teleportation to the opposite face through a traversable wormhole that is short compared to c/H(t). 2 requires the unified exotic macro-quantum vacuum dark energy/matter local zero point stress-energy density Diff(4) tensor field tuv(vac) = (c4/8piG*)/zpfguv not to vanish at the faces in such a way as to support and maintain this huge network of traversable cosmic scale wormholes that implement the inferred global topology from the ~ 2.7 deg K CMB temperature fluctuation spherical harmonic multipole resolved statistics of WMAP. However, if we are limited only to remote-sensing of this global topology as in the WMAP space probe the above distinction is probably moot or degenerate. We will only be able to decide this perhaps by sending out warp drive space probes to see if the world has an edge or not. The local /zpf zero point dark energy/matter exotic vacuum c-number ODLRO geometrodynamic kernel of tuv(vac) also permits the implementation of globally faster-than-light Alcubierre warp drives or, equivalently, Bondi-Terletskii negative matter propulsion that is a locally self-generated weightless free float slower-than-light timelike geodesic with small curvature tidal forces inside the ship, but which appears as spacelike world line trajectory to the outside Earth bound observer using remote sensors on the craft. Since Omega(dark energy) ~ 0.666 (:-)) to 0.73 depending on which Pundit you believe, this is not such a nutty idea as it might first appear to the more faint-hearted conservative reader. :-) *Note the above kind of hypothetical warp drive technology is reported by military intelligence connected UFO investigators that IMHO pass the crackpot filter. Whether or not those reports are true is not the gedankenexperiment what is significant is that we can conceive of such a technology within mainstream physics. Kip Thorne used this method in his 1986 paper on traversable wormholes. One way to decide between 1 and 2 is to look for a time-lag in what John emphasis) but in different regions of the sky p. 567 and what Luminet et-al describe as temperature correlations in matching circles on the sky. Is there a physical distinction between a curved multiply-connected 3D space with traversable wormholes and no boundary and a flat space with Star Gate boundaries that are the 12 pentagonal faces of the spherical dodecahedron http://mathworld.wolfram.com/Dodecahedron.html ? In principle yes, since spatial curvature is a local observable and the proper distance through the alternate worm hole paths connecting opposite faces is a variable physical parameter. When the topologist has the bug or crawling insect exit on the right edge and enter on the left edge, this is pre-metrical with no concept of metric time, hence it is an incomplete physical description leaving out measurable properties of the phenomenon. Since the detailed paper on pp 593 - 595 is not available to many readers on the bcc list, let's look at some excerpts from Dodecahedral space-topology as an explanation for weak wide-angle temperature correlations in the cosmic microwave background: The current 'standard model' of cosmology posits an infinite flat universe forever expanding under the pressure of dark energy. This follows from Einstein's classical principle of equivalence combined with Heisenberg's quantum principle of uncertainty to deduce w = -1 for the equation of state for the zero point vacuum fluctuations of ANY local micro-quantum field of any spin where w = pressure/energy density. Dark energy is w = -1 exotic vacuum with a negative zero point pressure because, the effective gravitation from a region of any stuff, real or exotic vacuum, in the weak field limit obeys a Poisson equation Laplacian of the stuff ~ (G/c2)(energy density + 3 pressure) = (G/c2)(energy density)(1 + 3w) This universally anti-gravitates causing the expansion of 3D co-moving 3D space in the FRW metric to accelerate when the net residual micro-quantum zero point energy density from all spins is positive. The gravitating dark matter exotic vacuum has positive zero point pressure with negative zero point energy density. IMHO dark matter is not made net residual zero point energy density of the physical vacuum is its local macro-quantum coherent order parameter PSI(x,L) = |Higgs field(x,L)|eiGoldstone phase(x,L) at scale L, coarse-grained space-time event x in the sense of a wavelet transform generalized version of the Wigner phase space density with ODLRO in the virtual electron-positron pair reduced micro-quantum density matrix. /zpf(x,L) = Lp*-2[1 - Lp*3|Higgs field(x,L)|2] Lp*2 = Lp4/3(c/H(t))2/3 H(now) ~ 1028 cm Lp2 = hG(Newton)/c3 Lp*(now) ~ 1 fermi, i.e. 1 Gev energy scale The ordinary non-gravitating vacuum has /zpf(x,L) = 0. Einstein's gravity is from the modulation of the Goldstone phase guv(x,L) = Minkowski metric + (1/2)(Sakharov's metric elasticity tensor of Hagen Kleinert's world crystal lattice) = nuv(Minkowski) + (1/2)[du(x,L),v + dv(x,L),u] ,u is partial derivative with respect to xu du(x,L) is the world crystal local distortion field at scale L du(x,L) = Lp*2(Spin 1 gauge-force invariant Bohm-Aharonov Goldstone phase),u Where also [DuDu + V(|Higgs field(x,L)|)]|Higgs field(x,L)|eGoldstone Phase(x,L) = 0 Is the local nonlinear Diff(4)+ spin 1 gauge invariant Landau-Ginzburg equation for the macro-quantum coherent vacuum More is different local order parameter that damps down the random zero point energy density contribution to the Einstein cosmological constant / in the large-scale limit. Einstein's Riemann 4th rank tensor curvature is from disclination defect density of string topological defects of the Goldstone phase in the Kleinert world crystal lattice with unit cells of scale Lp* using the t'Hooft-Susskind world hologram conjecture. This can be generalized to include torsion fields from Suv(x,L) = (1/2)[du(x,L),v - dv(x,L),u] =/= 0 corresponding to dislocation defects in the world crystal lattice as shown by Hagen Kleinert (Free University of Berlin). Note that Lp* depends on the cosmological FRW H(t) = R(t),t/R(t). Therefore, it starts out as 10-33 cm at the Big Bang micro -> MACRO quantum vacuum phase transition, but gets larger making the energy scale for quantum gravity lower as the universe 3D co-moving space in the FRW limit expands. This has falsifiable consequences. On the other hand, my basic theory does not require the additional world hologram conjecture of t'Hooft-Susskind Lp* = Lp2/3L2/3 where I take L = c/H(t) as a kind of Mach principle. Returning to the Nature paper: The WMAP data confirms the k = 0 flat space model on small scales but it is alleged that the model breaks down at large scales where Temperature correlations across the microwave sky match expectations on angular scales narrower than 60 degrees but, contrary to predictions, vanish on scales wider than 60 degrees ... The observed lack of temperature correlations on scales beyond 60 degrees means that the broadest waves are missing ... This is a bit like a cut-off in a wave guide and like the Casimir effect between two conducting plates. ... perhaps because space itself is not big enough to support them. They predict FRW Omega zero ~ 1.013 and temperature correlations in matching circles on the sky on the basis of a global topological model of a finite 3D space with no boundary that is multiply-connected with the pattern of holes, like in a more complicated 3D version of a 2D torus, called Poincare dodecahedral space. These holes IMHO can be thought of as huge traversable wormholes supported by tuv(zpf) = tuv(exotic vacuum) =/= 0 local Diff(4) tensor fields. The ordinary non-gravitating vacuum has tuv(zpf) = 0. Omega(Dark Matter) = 0.28. The claim is made that the L =2, 3, 4 multipole terms in the temperature fluctuation power spectrum (expanded in 2D spherical polar coordinated harmonics of latitude and longitude over the sky) fits their dodecahedral space k = 1 model better than the k = 0 inflation flat space model -- especially for the L = 2 quadrupole where the data is calibrated to fit their model exactly at L = 4. WMAP found a quadrupole only about one-seventh as strong as what would be expected in infinite flat space ... for large values of L, ranging up to L ~ 900 corresponding to small scale temperature fluctuations, the spectrum tracks the infinite universe [k = 0, Omega zero = 1]predictions exceedingly well. The CMB temperature fluctuations arise primarily (but not exclusively) from density fluctuations in the early universe. photons traveling from denser regions do a little extra work against gravity and therefore arrive cooler, while photons from less dense regions arrive warmer We need to be more careful about density here. The usual meaning is density. The repulsive dark energy phase /zpf > 0 of exotic vacuum acts like less dense because of the anti-gravity blue shift causing real far field photons from such an exotic vacuum region to arrive warmer with more energy per quantum. Just the opposite for the gravitating dark matter /zpf < 0 region of exotic vacuum. The density fluctuations across space split into a sum of three-dimensional harmonics ... just as temperature fluctuations split into a sum of two-dimensional spherical harmonics ... The low quadrupole implies a cut-off on the wavelengths of the three-dimensional harmonics. Such a cut-off presents an awkward problem in infinite flat space, because it defines a preferred length scale in an otherwise scale-invariant space. A more natural explanation invokes a finite universe ... Whereas most potential spatial topologies fail to fit the WMAP results, the Poincare dodecahedral space fits them very well. The Poincare dodecahedral space is a dodecahedral block of space with opposite faces abstractly glued together, so objects passing out of the dodecahedron across any space return from the opposite face. *This abstract topological idea seems to be, upon further reflection, absolutely physically equivalent to a closed 3D space without boundary that is multiply-connected by giant traversable wormholes or Star Gates that require tuv(exotic vacuum) to support them. The Poincare dodecahedral block of space therefore requires a network of 6 giant star gates of scale c/H(t), analogous to a 2D spherical surface with 6 wormhole handles, or a DeRham integral domain chain 3D Betti number of 6, I would imagine, for the present case. Light travels across the faces in the same way, so if we sit inside the dodecahedron and look outward across a face, our line of sight re-enters the dodecahedron from the opposite face. We have the illusion of looking into an adjacent copy of the dodecahedron. This illusion is, I suppose, physically indistinguishable from looking into the 2D face of a giant cosmic scale traversable very short Star Gate wormhole to what is on the other side. If we take the original dodecahedral block of space not as a euclidean dodecahedron (with edge angles ~ 117 degrees) but as a spherical dodecahedron (with edge angles exactly 120 degrees), then adjacent images of the dodecahedron fit together snugly to tile the hypersphere (Fig 3b), analogously to the way adjacent images of spherical pentagons (with perfect 120 degree angles) fit snugly to tile an ordinary sphere (Fig 3a). p. 594 OK my question here is whether or not there is any physics to this new formal construction such as 120 parallel finite universes next door that we can never access, or maybe we can? This led to the issue what is physical space? Is it SU(2)/ID? Note that: 12 spherical pentagons tile the surface of an ordinary sphere. They fit together snugly because their corner angles are exactly 120 degrees. Note that each spherical pentagon is just a pentagonal piece of a sphere. Note also that we cannot identify opposite edges of a 5-sided 2D pentagon to make a multiply-connected 2D surface. The 3D dodecahedron has an even number of 2D faces 12 allowing 6 traversable wormhole handles of opposite faces. Each face is a 2D star gate portal. Note also that a 2D sphere has no boundary, but that its 12 spherical pentagonal pieces do have boundaries. 120 spherical dodecahedra tile the surface of a hypersphere. The hypersphere is simply-connected with no boundary and no holes. Therefore you only have to slice it once to get it to split into two disjoint pieces. This is unlike the single 3D spherical dodecahedron that is 6-fold multiply-connected also without boundary. You have to slice it seven times to get it to split into 2 disjoint 3D pieces. Note that a 2D torus has only one hole and you have to slice it twice to make it break apart into two disjoint pieces. The problem here now is whether or not there is any physical measurable consequence of the number 120 here? Does the simply-connected 3D hypersphere with no boundary play a physical role or is it simply an esthetic formal nicety? This is one of the key questions that motivated this original thread. to be continued Poincare dodecahedral space as they describe it is exactly the same as SU(2)/ID. These authors are quite aware of this equivalence as they show in the paper Cosmic microwave background constraints on multi-connected spherical spaces, which you can download from: http://xxx.lanl.gov/abs/astro-ph/0303580 On page 2 (of this 5 page paper), they say: The finite subgroups of S3 are the cyclic groups Zn, the binary dihedral groups D*m, the binary tetrahedral, octahedral and icosahedral groups, respectively of order n, 4m, 24, 48 and 120. Now since you are looking for a physically significant difference between SU(2)/ID and SU(2) tiled with 120 spherical dodecahedra, I should point out the fundamental significance of the volume of the 3D space, which affects both the light travel times as well as the density of the space. As these authors say: In all cases the volume of the space S3/G is the volume of the 3-sphere S3 divided by the order |G| of the holonomy group. structures S3/G, and that has to do with the A-D-E classification of these spaces. This is because A-D-E Coxeter graphs classify at least 20 physically relevant mathematical objects. As I have mentioned previously, these include Coxeter (relfection) groups, Lie algebras, gravitational instanton spaces, catastrophe bundles, 2D conformal field theories, Heisenberg algebras, and much more. I have started writing this up in a Word file, and will send it to you as soon as I finish it. Saul-Paul === Subject: Simple curve question... If I have the following three data sets: A B -------------- 1500 300 4000 500 6000 300 What is the best way to calculate the corresponding B for any given A? I cannot assume any specific curve shape, meaning B could be 300 100 500, but A will always be increasing. B (BTW: this is -not- homework, it is related to mass spec analysis and my math sucks) === Subject: Re: Simple curve question... Well, I would try as below: 1) assume that B reach its maximum, say M (>500), at A=3750 (mid-point of 1500-6000), 2) fit a parabolic curve open downwards to these data. Michael Leung BCC .b9.a6.b9g©.97.a6l«.97.87séD :u2Hib.6$AX1.1990766@newssvr21.news.prodigy.com... > If I have the following three data sets: > A B > -------------- > 1500 300 > 4000 500 > 6000 300 > What is the best way to calculate the corresponding B for any given A? I > cannot assume any specific curve shape, meaning B could be 300 100 500, but > A will always be increasing. > (BTW: this is -not- homework, it is related to mass spec analysis and my > math sucks) === Subject: Re: Simple curve question... > BCC .b9.a6.b9g©.97.a6l«.97.87séD > :u2Hib.6$AX1.1990766@newssvr21.news.prodigy.com... > If I have the following three data sets: > A B > -------------- > 1500 300 > 4000 500 > 6000 300 > What is the best way to calculate the corresponding B for any given A? I > cannot assume any specific curve shape, meaning B could be 300 100 500, > but > A will always be increasing. > B > (BTW: this is -not- homework, it is related to mass spec analysis and my > math sucks) > Well, I would try as below: > 1) assume that B reach its maximum, say M (>500), > at A=3750 (mid-point of 1500-6000), > 2) fit a parabolic curve open downwards to these data. Fitting not needed. Three points specifies a unique interpolating parabola (see Lagrange or Newton interpolation). The Lagrange quadratic for these points is: y = 300*(x-4000)(x-6000)/(1500-4000)(1500-6000) + 500*(x-1500)(x-6000)/(4000-1500)(4000-6000) + 300*(x-1500)(x-4000)/(6000-1500)(6000-4000) = (x-4000)(x-6000)/37500 - (x-1500)(x-6000)/10000 + (x-1500)(x-4000)/30000 This parabola does reach a peak of 502.5 at x=3750. However, it's terribly dangerous to draw any conclusions from such a model if all you know are three points, especially if you're planning on extrapolating the model outside the range 1500-6000. Any additional info you can incorporate (such as knowledge of the location of the peak) would be helpful. - Randy === Subject: Re: (matrix analysis) Is this true: A*X*B=X => A=I and B=I? >What else results about A and B can I obtain from A*X*B=X? >(this is not a HW problem!) Is X some particular matrix, or is the equation A X B=X supposed to be true for all matrices X of a certain size? Suppose A = (a_{ij}) is an m x m matrix, B = (b_{ij}) is n x n, and A X B = X for all m x n matrices X. Then taking X = e_{kl} (the matrix with 1 in position (k,l) and 0 elsewhere) we get a_{ik} b_{lj} = 1 for i=k, j=l and 0 otherwise, which implies A = cI and B = c(-1)I for some c <> 0. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: (matrix analysis) Is this true: A*X*B=X => A=I and B=I? >What else results about A and B can I obtain from A*X*B=X? >(this is not a HW problem!) > Is X some particular matrix, or is the equation A X B=X supposed to be > true for all matrices X of a certain size? > Suppose A = (a_{ij}) is an m x m matrix, B = (b_{ij}) is n x n, and > A X B = X for all m x n matrices X. Then taking X = e_{kl} (the matrix > with 1 in position (k,l) and 0 elsewhere) we get a_{ik} b_{lj} = 1 for > i=k, j=l and 0 otherwise, which implies A = cI and B = c(-1)I for some > c <> 0. I really appreciate your help! For this problem, the specification is that A, B, X are all square with size of NxN... The condition that X=A*X*B is imposed to all inputs X. Basically, this is a linear 2-D separable transform, Y=A*X*B, but now I want the output to be the input, X=A*X*B. Under this condition, what should be A, and what should be B, or they have other hidden relations? complete, there are no other A and B that can satisfy the requirement... am I right? -Walala === Subject: Re: (matrix analysis) Is this true: A*X*B=X => A=I and B=I? >For this problem, the specification is that A, B, X are all square with size >of NxN... The condition that X=A*X*B is imposed to all inputs X. >complete, there are no other A and B that can satisfy the requirement... am >I right? Yes, that's what I said. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: how about A1*X*B1+A2*X*B2=X? >For this problem, the specification is that A, B, X are all square with size >of NxN... The condition that X=A*X*B is imposed to all inputs X. >complete, there are no other A and B that can satisfy the requirement... am >I right? > Yes, that's what I said. Can you say something about A1*X*B1+A2*X*B2=X? Basically, I want to approximate the input X with a bunch of transforms of the input X, try to find optimal approximation in the MSE sense.... For the above problem, can I say it equals to (A1+A2)*X*(B1+B2)=X... and then A1+A2=cI, B1+B2 = c(-1)I... Is this true? Or what else relation you can say about A1, A2, B1, B2? -Walala === Subject: Re: Slow Factoring Method The method I described works in any base. It doesn't appear to be any more efficient in base 10. 1023 base 10 1023 mod 10 = 3 1*3 = 3 7*9 = 63 1023 mod 100 = 23 01*23 = 23 11*93 = 1023 *** 21*63 = 1323 31*33 = 1023 *** 41*03 = 123 51*73 = 3723 61*43 = 2623 71*13 = 923 81*83 = 6723 91*53 = 4823 07*89 = 623 17*19 = 323 27*49 = 423 37*79 = 2823 47*09 = 423 57*39 = 2223 67*69 = 4623 77*99 = 7623 87*29 = 2523 97*59 = 5723 Russell - 2 many 2 count === Subject: stable marriage problem (matching) with constraints Suppose we have B boys, G girls. B=G. The boys b1 and b3 are neighbors of b2, b2 and b4 are neighbors of b3, etc. Same for the girls: g1 and g3 are neighbors of g2, g2 and g4 are neighbors of g3, etc. We want to find a matching such that it is stable, AND if the boy m(i) is matched to the girl g(j), then the neighbors of m(i) (that is, m(i-1) and m(i+1) ) must match to the neighbors of g(j) (that is, g(j-1) or g(j+1) ). Anyone can think of a good algo? === Subject: Re: stable marriage problem (matching) with constraints Sorry, my problem was ill defined, please ignore it. > Suppose we have B boys, G girls. B=G. > The boys b1 and b3 are neighbors of b2, b2 and b4 are neighbors of b3, > etc. > Same for the girls: g1 and g3 are neighbors of g2, g2 and g4 are > neighbors of g3, etc. > We want to find a matching such that it is stable, AND if the boy m(i) > is matched to the girl g(j), > then the neighbors of m(i) (that is, m(i-1) and m(i+1) ) must match to > the neighbors of g(j) (that is, g(j-1) or g(j+1) ). > Anyone can think of a good algo? === Subject: Re: Q wrt a number-theoretic generating function > I'm reading some introductory material about Dirichlet's generating > functions of arithmetical functions, but could not find anything about > the obvious generating function > Sum_p p{-x}, > where the sum is extended over all primes. Has it been studied? Is it > an independent function or can it be expressed in terms Riemann's > zeta? Anything else interesting about it? References? > Michele By the way, I might as well add this: exp(sum{p=primes} 1/px) = sum{k=1 to oo} A(k)/kx, where A(k) = product{p=primes} 1/(a(p,k))! , and where each a(p,k) is a nonnegative integer such that pa(p,k) is the highest power of the prime p which divides k. (I think...) Leroy Quet === Subject: Re: FUNctions/Continued-Fraction Puzzle UGGG! I used the wrong word (twice). Reposting with 2 solutions replaced with proofs. Sorry. --- >... > But, anyway, my solution is below the replied-to message. > (This might be actually trivial. But it does not seem to be with the > little thought I have given it. In any case, perhaps I should not have > cross-posted this to rec.puzzles {if I should have even posted it to > sci.math}; but what the...) > For all real x > 1, and for some function of x, y(x); > where each y is a real, y =y(x), based on x: it is so that: f([x; x2, x3, x4,...,xm]) = [y; y2, y3, y4,...,ym], for EVERY positive integer m; where: [x; x2, x3, x4,...,xm] is the continued-fraction 1 > x + ------------------ ; > 1 > x2 + -------------- > 1 > x3 + --------- > .... > + 1/xm and [y; y2, y3, y4,...,ym] is also a continued-fraction (obviously); and [x; x2, x3, x4,...,xm] converges to X; and f(w) is a real -> real function, such that f'(X) exists and is finite nonzero. So, what are the possible f(w)'s, given all of the conditions above?? > First, by the way, f'(X) is the (1st) derivative of f(w) at w = X, in > case this is not obvious. I should mention that f can equate to an infinite number of functions > if it need not be analytic. If it need by analytic, however, there are > a finite number of possible functions that can equal f(w). > (So, find the set of analytic f(w)'s.) This puzzle seems to be more difficult than I first assumed. > I will wait until Friday, at least, to post the answer if no one else > posts the solution before that. .... > ...the solution: > I get that the only possible analytic f is: > f(w) = w. > Proof: > limit{m -> oo} (x/y)(2m-1) = 1. > So, x must = y. And, consequently, f(w) must = w. > * earlier result at: I highly suspect my PROOF is far from the simplest. Is there a PROOF which is any simpler, even trivial? (Perhaps my result itself, that f(w) = w is the ONLY analytic function, is wrong.) Leroy Quet === Subject: Re: FUNctions/Continued-Fraction Puzzle >... > But, anyway, my solution is below the replied-to message. > (This might be actually trivial. But it does not seem to be with the > little thought I have given it. In any case, perhaps I should not have > cross-posted this to rec.puzzles {if I should have even posted it to > sci.math}; but what the...) > For all real x > 1, and for some function of x, y(x); > where each y is a real, y =y(x), based on x: it is so that: f([x; x2, x3, x4,...,xm]) = [y; y2, y3, y4,...,ym], for EVERY positive integer m; where: [x; x2, x3, x4,...,xm] is the continued-fraction 1 > x + ------------------ ; > 1 > x2 + -------------- > 1 > x3 + --------- > .... > + 1/xm and [y; y2, y3, y4,...,ym] is also a continued-fraction (obviously); and [x; x2, x3, x4,...,xm] converges to X; and f(w) is a real -> real function, such that f'(X) exists and is finite nonzero. So, what are the possible f(w)'s, given all of the conditions above?? > First, by the way, f'(X) is the (1st) derivative of f(w) at w = X, in > case this is not obvious. I should mention that f can equate to an infinite number of functions > if it need not be analytic. If it need by analytic, however, there are > a finite number of possible functions that can equal f(w). > (So, find the set of analytic f(w)'s.) This puzzle seems to be more difficult than I first assumed. > I will wait until Friday, at least, to post the answer if no one else > posts the solution before that. .... > ...the solution: > I get that the only possible analytic f is: > f(w) = w. > Proof: > limit{m -> oo} (x/y)(2m-1) = 1. > So, x must = y. And, consequently, f(w) must = w. > * earlier result at: I highly suspect my solution is far from the simplest. Is there a solution which is any simpler, even trivial? (Perhaps my result itself, that f(w) = w is the ONLY analytic function, is wrong.) Leroy Quet === Subject: Re: Euler books > where can I find english or german translations of some of euler's > works in the internet? > I can't afford to buy those. I know for instance that Euler's opera > omnia is very expensive. This would be a _really_ good place to start: http://math.dartmouth.edu/~euler/welcome.html This Google search will help you find others: HTH xanthian. -- === Subject: Re: Euler books > where can I find english or german translations of some of euler's > works in the internet? Oh, Internet. Try the Gutenberg Project, which has online full text of many classic books. One problem with many English translations of Euler is that they are so recent (published by Springer Verlag, in some cases I know) that they are still under copyright for the English text. The original Latin text of Euler's works is, of course, long since in the public domain. Hope this helps! -- Karl M. Bunday Christ has set us free. Galatians 5:1 Learn in Freedom (TM) http://learninfreedom.org/ kmbunday AT earthlink DOT net (preferred email address) === Subject: Re: Euler books > they are still under copyright for the English text. The original Latin text > of Euler's works is, of course, long since in the public domain. Euler was among the last scholars to write his papers in Latin. During the 19-th century and later most papers were published in national languages. Bob Kolker === even see it) === Subject: f(x+1) = f(x) + 1/f(x) What is the family of real -> real analytic functions f(x), such that: f(x+1) = f(x) + 1/f(x) for all real x? Leroy Quet === Subject: Re: f(x+1) = f(x) + 1/f(x) >What is the family of real -> real analytic functions f(x), such that: >f(x+1) = f(x) + 1/f(x) >for all real x? There cannot be any such function on all reals. If we let g=f2, the equation yields g(x+1) = g(x) + 2 + 1/g(x), so g(x+1) > g(x) + 2. As g is non-negative, one cannot go far in the negative direction. If we look at the equation for g and ignore its positivity, it cannot be continuous for the same reason; it must be 0 somewhere. If g starts at some x_0, the asymptotics of g are g(x) ~ 2x + C + ln(x+C/2)/2 + O(ln(x)/x), where C can be a periodic function of the fractional part of x. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: f(x+1) = f(x) + 1/f(x) > What is the family of real -> real analytic functions f(x), such that: > f(x+1) = f(x) + 1/f(x) > for all real x? > Leroy Quet It can't have any roots since then the relation is undefined. It also has to diverge to the right since lim f(x) = lim f(x+1) = lim f(x) + lim 1/f(x) ==> lim 1/f(x) = 0 ==> lim f(x) = +/-infinity where the limit is taken as x-->+infinity That's all I got right now. Have a tolerable existence. Eli -- === Subject: Re: f(x+1) = f(x) + 1/f(x) What is the family of real -> real analytic functions f(x), such that: >> f(x+1) = f(x) + 1/f(x) >> for all real x? >> Leroy Quet > It can't have any roots since then the relation is undefined. It also has > to diverge to the right since > lim f(x) = lim f(x+1) = lim f(x) + lim 1/f(x) > ==> lim 1/f(x) = 0 ==> lim f(x) = +/-infinity > where the limit is taken as x-->+infinity > That's all I got right now. > Have a tolerable existence. Eli > -- Both your statements are also true for all derivatives of f? .... Quaternion === Subject: Re: Factorial/Exponential Identity, Infinity Besides two, there actually is another integer where ths sum of it and itself is equal to the product of it and itself: zero. A binary number that is normal to base 2 has equal probability of a given element being a zero or a one, right? That is to say, statistically any contiguous finite subsequence is expected to have equal numbers of ones and zeros. Is that the same thing as its sequence having zero-density of one half? In any even length contiguous subsequence of (01)..., half the values are one and half zero. The current state of opinion appears to be that almost all real numbers are absolutely normal, that is, normal to each integer base greater than one. Then, there are uncountably many abnormal numbers. In the unit interval, then, what if normal numbers are almost half of them, with the rational numbers with zero-density of one half rounding the proportion to exactly one half? Bizarre, no? Then again, the only normal numbers thus far contrived are not shown to appear in the wild, I hear Sierpinski designed a construction of one. Yet, there is evidence that pi, for example, is not abnormal to base ten, eg, in the first million decimal digits each digit appears around 1/10 of the time. MathPages' Is e normal?: http://www.mathpages.com/home/kmath519.htm . Many (most?) sequences with equal densities of zeros and ones are not normal to base two. They don't contain every other expected sequence, ie being normal in base four, eight, etcetera. A number normal to base two is normal to base 2x for positive integer x. I read this quote of Euler off the Internet the other day and thought it was pretty good, Euler's rather caustic: Notable enough, however, are the controversies over the series 1 - 1 + 1 - 1 + 1 - ... whose sum was given by Leibniz as 1/2, although others disagree. ... Understanding of this question is to be sought in the word sum; this idea, if thus conceived -- namely, the sum of a series is said to be that quantity to which it is brought closer as more terms of the series are taken -- has relevance only for convergent series, and we should in general give up the idea of sum for divergent series. - L. Euler I think the sum is zero. Notice that Euler has the idea of sums of divergent series. Anyways, I got to thinking about the factorizations of ((sum n)x - sum(nx)) / s(n+1, n-x+1). I noticed the patterns of some of the factors, for example how for five values of x in a row the possible rational function, or ratio of polynomials, if it exists, that goes to x!, has the given factor. Then again, I'm still trying to determine an efficient method to determine a type of additive partitioning of a number, for use in enumeration. As well, I'd like to read more about the consideration of hypermatrices. I started this thread because I had the notion that half the sequences had densities of one half. Yet we have seen that n!/(n/2)!2 2n evaluates to a different asymptotic expression than that, yet now I read that multitudinous numbers are normal. Ross === Subject: Re: Factorial/Exponential Identity, Infinity > Besides two, there actually is another integer where ths sum of it and > itself is equal to the product of it and itself: zero. x*x = x + x being a quadratic equation, how many solutions did you expect to find? > The current state of opinion appears to be that almost all real > numbers are absolutely normal, that is, normal to each integer base > greater than one. Then, there are uncountably many abnormal > numbers. I do not see that your conclusion follows from your premise. > In the unit interval, then, what if normal numbers are almost half of > them, with the rational numbers with zero-density of one half rounding > the proportion to exactly one half? Since almost all and almost half are incompossible, why ask? > Bizarre, no? Your thought processes are indeed bizarre. > Then again, the only normal numbers thus far contrived are not shown > to appear in the wild, I hear Sierpinski designed a construction of > one. Yet, there is evidence that pi, for example, is not abnormal to > base ten, eg, in the first million decimal digits each digit appears > around 1/10 of the time. > MathPages' Is e normal?: http://www.mathpages.com/home/kmath519.htm > Many (most?) sequences with equal densities of zeros and ones are not > normal to base two. They don't contain every other expected sequence, > ie being normal in base four, eight, etcetera. A number normal to > base two is normal to base 2x for positive integer x. > I read this quote of Euler off the Internet the other day and thought > it was pretty good, Euler's rather caustic: > Notable enough, however, are the controversies over the series 1 - 1 > + 1 - 1 + 1 - ... whose sum was given by Leibniz as 1/2, although > others disagree. ... Understanding of this question is to be sought in > the word sum; this idea, if thus conceived -- namely, the sum of a > series is said to be that quantity to which it is brought closer as > more terms of the series are taken -- has relevance only for > convergent series, and we should in general give up the idea of sum > for divergent series. - L. Euler > I think the sum is zero. Notice that Euler has the idea of sums of > divergent series. Even Homer nods. === Subject: Integer-Alteration Game Each player takes turns coming up with algorithmic steps, one step at a time in sequence, each step giving a rule to transform an integer into another. (such as: multiply m by greatest prime p, where p divides m and is <= sqrt(|m|). If no such prime exists, leave m unchanged.) (or such as: m = floor(|m|/d(|m|)), where d(m) is number of positive divisors of m.) (or rules may involve meta-transformations, such as sending players to previous rules, depending somehow upon the value of the integer when reaching the step.) Anyway, players are encouraged to be creative when inventing rules! Each step is capable of transforming any integer, always giving an integer as output. After a predetermined number of steps have been created (the same number for each player), a random integer is generated somehow. Players then try to guess what the output intger will be. The algorithm is run using the random start-integer. The winner of the game is the player who comes closest to guessing the final output integer. (needs work....) Leroy Quet === Subject: Re: absolute value graph > Sketch the graph of f(x)=2abs(x) > ------ > 1+x2 > How would I start this question? > This function is even, i.e. f(-x) = f(x). Then sketch the graph of f(x) = > 2x/(1 + x2) for x >= 0. and reflect it on the OY axis. Not that the above is a bad answer, but here's another perspective. Near the zero of f[x] (@ x=0 obviously), the graph looks like an absolute value function - v-shaped. When x is big - in either direction, f[x] is positive and near zero - like a reciprocal function, in fact. Apparently then, for some x in between 0 and really big, your function f must have had a maximum value... hth, cdj === Subject: {Field Theory} Please help me understand this special case (was This could get confusing fast...) In my other thread I pointed out that in a field F of integers, the notation ab of two elements of that field is ambiguous, namely because it has the 3 possible and not necessarily equal interpretations: 1. ab = a * b where * is the multiplication of F 2. ab = (a + a + ... + a) (b times) 3. ab = (b + b + ... + b) (a times) Some people were replying to the effect that the 3 are in fact equal. And they are, when F is well behaved. But since it was confusing me so much, I sought a counterexample, and lo and behold I found one with little difficulty. This, I exhibit thus: let F as a set (not a field yet) simply equal Z, ie, all integers (do not confuse with Z_n) Now let p:N->Q be the mapping of N to Q which is used in Cantor's famous proof that the rationals are countable. For example, p(1)=1, p(2)=1/2, p(3)=2, p(4)=3, p(5)=1/3, etc. Now enhance p into a better function q defined by: for x > 0, q(x) = p(x) for x = 0, q(x) = q(0) = 0 for x < 0, q(x) = -p(-x) Finally let q' be the inverse of q. (I'll leave the existence proof details to you since it is all quite easy) Now associate with F the following addition and multiplication: a + b = q'[q(a)+q(b)] a * b = q'[q(a)q(b)] It is not hard to prove that F is thus turned into a field (with 1 as its unit and 0 as its zero) What, then, can we make of 5*3? Let us check the 3 possible interpretations: 1. 5*3 = f'[f(5)*f(3)] = f'[(1/3) * 2] = f'(2/3) = 7 2. 5*3 = 5 + 5 + 5 = f'[f(5)*3] = f'[1] = 1 3. 5*3 = 3 + 3 + 3 + 3 + 3 = f'[f(3)*5] = f'[10]. I didn't bother to write out Cantor's table all the way needed to calculate f'[10] but it is obviously way larger than 7. I have thus exhibited that the notation in question is indeed ambiguous. I am greatly muddled on the matter and am seeking clarification... Some have responded that there is no such thing as a field of integers... but this seems to make no sense to me. Yes, I can see how you can argue that Z_p is really a field of sets, not of integers, although I see nothing forbiding us from using {0,1,...,p-1} as the set and simply modifying the addition and multiplication so the modular arithmetic takes place there, and thus obtain a finite field of integers. Is it the case that the definition of a field is really some arcane, ethereal thing taught in Ph.D. level math and that Herstein's definition is just a handwaving gesture similar to an algebra 101 definition of the rationals? === Subject: Re: {Field Theory} Please help me understand this special case (was This could get confusing fast...) > In my other thread I pointed out that in a field F of integers, the One can indeed label the congruence classes of the finite field Z/(p) by integers, e.g. the congruence class representatives 0,1,2,...,p-1. The proper terminology here is a residue field of the ring of integers, or image field..., not field of integers. > notation ab of two elements of that field is ambiguous, namely > because it has the 3 possible and not necessarily equal > interpretations: > 1. ab = a * b where * is the multiplication of F > 2. ab = (a + a + ... + a) (b times) > 3. ab = (b + b + ... + b) (a times) > Some people were replying to the effect that the 3 are in fact equal. > And they are, when F is well behaved. But since it was confusing me Well-behaved field? That is not a mathematical term. In any case, the example I give below should help to alleviate any confusion here. > so much, I sought a counterexample, and lo and behold I found one with > little difficulty. This, I exhibit thus: > let F as a set (not a field yet) simply equal Z, ie, all integers (do > not confuse with Z_n) > Now let p:N->Q be the mapping of N to Q which is used in Cantor's > famous proof that the rationals are countable. For example, > p(1)=1, p(2)=1/2, p(3)=2, p(4)=3, p(5)=1/3, etc. > Now enhance p into a better function q defined by: > for x > 0, q(x) = p(x) > for x = 0, q(x) = q(0) = 0 > for x < 0, q(x) = -p(-x) > Finally let q' be the inverse of q. (I'll leave the existence proof > details to you since it is all quite easy) > Now associate with F the following addition and multiplication: > a + b = q'[q(a)+q(b)] > a * b = q'[q(a)q(b)] > It is not hard to prove that F is thus turned into a field (with 1 as > its unit and 0 as its zero) > What, then, can we make of 5*3? Let us check the 3 possible > interpretations: > 1. 5*3 = f'[f(5)*f(3)] = f'[(1/3) * 2] = f'(2/3) = 7 > 2. 5*3 = 5 + 5 + 5 = f'[f(5)*3] = f'[1] = 1 > 3. 5*3 = 3 + 3 + 3 + 3 + 3 = f'[f(3)*5] = f'[10]. You are somehow confused. For an enlightening example let's consider the isomorphism between the ring of Roman numerals and the ring of integers: V times III = q'(q(V) * q(III)) = q'(5*3) = q'(15) = XV V plus V plus V = q'(q(V) + q(V) + q(V)) = q'(5+5+5) = q'(15) = XV III plus...plus III = q'(q(III)+...+q(III)) = q'(3+3+3+3+3) = q'(15) = XV The roman numerals are simply new names for the integers, and the operations on them may be performed by translating the numerals to their standard names, performing the standard operations, then mapping the standard result to its Roman numeral. For further discussion of such isomorphisms see my prior post -Bill Dubuque > I didn't bother to write out Cantor's table all the way needed to > calculate f'[10] but it is obviously way larger than 7. > I have thus exhibited that the notation in question is indeed > ambiguous. I am greatly muddled on the matter and am seeking > clarification... > Some have responded that there is no such thing as a field of > integers... but this seems to make no sense to me. Yes, I can see > how you can argue that Z_p is really a field of sets, not of integers, > although I see nothing forbiding us from using {0,1,...,p-1} as the > set and simply modifying the addition and multiplication so the > modular arithmetic takes place there, and thus obtain a finite field > of integers. Is it the case that the definition of a field is really > some arcane, ethereal thing taught in Ph.D. level math and that > Herstein's definition is just a handwaving gesture similar to an > algebra 101 definition of the rationals? === Subject: Re: {Field Theory} Please help me understand this special case (was This could get confusing fast...) at 07:00 PM, snizpilbor@yahoo.com (Sniz Pilbor) said: >In my other thread I pointed Claimed. >field F of integers, The integers don't form a field. >and not necessarily equal interpretations: Incorrect. >Some people were replying to the effect that the 3 are in fact >equal. And they are, when F is well behaved. What do you mean by well behaved? If F is a filed then there are a well defined mappings *_1: ZxF -> F, *_2: FxZ -> F and I: Z -> F such that all three are equivalent once you make the functions explicit. >I found one Nope. It's not a counter example. You just got confused about what operator to apply when. >Now associate with F the following addition and multiplication: >a + b = q'[q(a)+q(b)] >a * b = q'[q(a)q(b)] You'll need to make everything explicit to see your error. Use distinct symbols for the operations in Z and in F. >It is not hard to prove that F is thus turned into a field (with 1 >as its unit and 0 as its zero) Only if p(1)=1. >What, then, can we make of 5*3? Nothing, without context. It might refer to any of your interpretations 1-3, all of which give the same result, or it might refer to a 4th interpretation, which gives a different result. >Some have responded that there is no such thing as a field of >integers... Let me rephrase that: it is not a standard Mathematical term, and you have not defined it. >I see nothing forbiding us from using {0,1,...,p-1} as the set or {0,1,3,4,...p}. Would that satisfy your definition of filed of integers? >simply modifying the addition and multiplication so the modular >arithmetic takes place there, If you chose different addition and multiplication operators, would you still call it a field of integers? >Is it the case that the definition of a field is really some arcane, >ethereal thing taught in Ph.D. level math No: what is the case is that when you introduce new nomenclature you are obligated to define it. Until you have a subject well and truly under you belt, introducing new nomenclature in a question is more likely to confuse the issue than to help. Introducing new nomenclature without a definition is guarantied to confuse the issue, since the readers are not likely to interpret it the same way that you do. -- spamtrap@library.lspace.org === Subject: Re: {Field Theory} Please help me understand this special case (was This could get confusing fast...) > In my other thread I pointed out that in a field F of integers, the > notation ab of two elements of that field is ambiguous, namely > because it has the 3 possible and not necessarily equal > interpretations: > 1. ab = a * b where * is the multiplication of F > 2. ab = (a + a + ... + a) (b times) > 3. ab = (b + b + ... + b) (a times) > Some people were replying to the effect that the 3 are in fact equal. > And they are, when F is well behaved. This has nothing to do with F being well beheved or not. It is a consequence of the way in which integers are defined within a ring R (with identity) in general: one uses the unique ringhomomorphism from Z to R. More generally (anticipating your example below) one can define the integers within a monoid M (= semigroup with identity) by using the unique monoid-homomorphism from N to M. > But since it was confusing me > so much, I sought a counterexample, and lo and behold I found one with > little difficulty. This, I exhibit thus: > let F as a set (not a field yet) simply equal Z, ie, all integers (do > not confuse with Z_n) > Now let p:N->Q be the mapping of N to Q which is used in Cantor's > famous proof that the rationals are countable. For example, > p(1)=1, p(2)=1/2, p(3)=2, p(4)=3, p(5)=1/3, etc. This is interesting but beside the point. Your p is not a homomorphism. Marc === Subject: Re: {Field Theory} Please help me understand this special case (was This could get confusing fast...) Some people were replying to the effect that the 3 are in fact equal. > And they are, when F is well behaved. > This has nothing to do with F being well beheved or not. > It is a consequence of the way in which integers are defined > within a ring R (with identity) in general: one uses the unique > ringhomomorphism from Z to R. > More generally (anticipating your example below) one can define the > integers within a monoid M (= semigroup with identity) by using > the unique monoid-homomorphism from N to M. replying? By integers I was meaning the normal integers, that is, 1, 2, 3, -1, -2, -3, 0, etc., encountered in basic high school algebra, not generalized integers of a ring. I apologize if this caused additional confusion. As for the ambiguity, in the post to which you replied I gave a specific example where the 3 interpretations of 3*5 all ended up being completely different. Your reply seems to be saying that no, no, they are all equal, whether or not F is well-behaved. Where, then, was my mistake? Was the F which I exhibited in my post not in fact a field, and if so why not, which axiom(s) of the field did it fail to uphold? The previous response to my post, whose writer seems to have actually read it through before replying, seemed to clear things up, but now the waters are rendered as muddy as ever. I sincerely want to understand what you are saying, but you must not begrudge me to remain skeptical when a counterexample is right in front of us which you have not explained. > This is interesting but beside the point. Your p is not a homomorphism. I never claimed that it was (why would it matter?). I was using it as a means of defining an addition and multiplication which would make F a field. I get the impression that you read just that far and then stopped. The original confusion of the ambiguity seems to be even greater when learned field theoreticians rise in abundance to say it doesn't exist, *blatantly ignoring the counterexample right there*. The original post in this thread (with some minor corrections): > In my other thread I pointed out that in a field F of integers, the > notation ab of two elements of that field is ambiguous, namely > because it has the 3 possible and not necessarily equal > interpretations: > 1. ab = a * b where * is the multiplication of F > 2. ab = (a + a + ... + a) (b times) > 3. ab = (b + b + ... + b) (a times) > Some people were replying to the effect that the 3 are in fact equal. > And they are, when F is well behaved. But since it was confusing me > so much, I sought a counterexample, and lo and behold I found one with > little difficulty. This, I exhibit thus: > let F as a set (not a field yet) simply equal Z, ie, all integers (do > not confuse with Z_n) > Now let p:N->Q+ be the mapping of N to Q+ (the positive rationals) > which is used in Cantor's > famous proof that the rationals are countable. For example, > p(1)=1, p(2)=1/2, p(3)=2, p(4)=3, p(5)=1/3, etc. > Now enhance p into a better function q defined by: > for x > 0, q(x) = p(x) > for x = 0, q(x) = q(0) = 0 > for x < 0, q(x) = -p(-x) > Finally let q' be the inverse of q. (I'll leave the existence proof > details to you since it is all quite easy) > Now associate with F the following addition and multiplication: > a + b = q'[q(a)+q(b)] > a * b = q'[q(a)q(b)] > It is not hard to prove that F is thus turned into a field (with 1 as > its unit and 0 as its zero) > What, then, can we make of 5*3? Let us check the 3 possible > interpretations: > 1. 5*3 = q'[q(5)*q(3)] = q'[(1/3) * 2] = q'(2/3) = 7 > 2. 5*3 = 5 + 5 + 5 = q'[q(5)*3] = q'[1] = 1 > 3. 5*3 = 3 + 3 + 3 + 3 + 3 = q'[q(3)*5] = q'[10]. > I didn't bother to write out Cantor's table all the way needed to > calculate q'[10] but it is obviously way larger than 7. > I have thus exhibited that the notation in question is indeed > ambiguous. I am greatly muddled on the matter and am seeking > clarification... > Some have responded that there is no such thing as a field of > integers... but this seems to make no sense to me. Yes, I can see > how you can argue that Z_p is really a field of sets, not of integers, > although I see nothing forbiding us from using {0,1,...,p-1} as the > set and simply modifying the addition and multiplication so the > modular arithmetic takes place there, and thus obtain a finite field > of integers. Is it the case that the definition of a field is really > some arcane, ethereal thing taught in Ph.D. level math and that > Herstein's definition is just a handwaving gesture similar to an > algebra 101 definition of the rationals? === Subject: Re: {Field Theory} Please help me understand this special case (was This could get confusing fast...) > In my other thread I pointed out that in a field F of integers, the > notation ab of two elements of that field is ambiguous, namely > because it has the 3 possible and not necessarily equal > interpretations: > 1. ab = a * b where * is the multiplication of F > 2. ab = (a + a + ... + a) (b times) > 3. ab = (b + b + ... + b) (a times) > Some people were replying to the effect that the 3 are in fact equal. > And they are, when F is well behaved. But since it was confusing me > so much, I sought a counterexample, and lo and behold I found one with > little difficulty. This, I exhibit thus: > let F as a set (not a field yet) simply equal Z, ie, all integers (do > not confuse with Z_n) > Now let p:N->Q be the mapping of N to Q which is used in Cantor's > famous proof that the rationals are countable. For example, > p(1)=1, p(2)=1/2, p(3)=2, p(4)=3, p(5)=1/3, etc. > Now enhance p into a better function q defined by: > for x > 0, q(x) = p(x) > for x = 0, q(x) = q(0) = 0 > for x < 0, q(x) = -p(-x) For this to give a bijection you probably meant that p is a bijection between N and the POSITIVE rationals. > Finally let q' be the inverse of q. (I'll leave the existence proof > details to you since it is all quite easy) > Now associate with F the following addition and multiplication: > a + b = q'[q(a)+q(b)] > a * b = q'[q(a)q(b)] > It is not hard to prove that F is thus turned into a field (with 1 as > its unit and 0 as its zero) This is correct, since the addition and multiplication is now that of the rationals. We simply decided to give them funny and unnatural names via q and q'. > What, then, can we make of 5*3? Let us check the 3 possible > interpretations: > 1. 5*3 = f'[f(5)*f(3)] = f'[(1/3) * 2] = f'(2/3) = 7 > 2. 5*3 = 5 + 5 + 5 = f'[f(5)*3] = f'[1] = 1 > 3. 5*3 = 3 + 3 + 3 + 3 + 3 = f'[f(3)*5] = f'[10]. Since in your field 1+1+1 is not equal 3 nor is 1+1+1+1+1 equal to 5, the computations 2. and 3. are incorrect, unless interpreted as below (when there is no reason to expect them to be equal to computation 1.). Nothing exciting about it. > I didn't bother to write out Cantor's table all the way needed to > calculate f'[10] but it is obviously way larger than 7. > I have thus exhibited that the notation in question is indeed > ambiguous. I am greatly muddled on the matter and am seeking > clarification... You seem to suffer from the following confusion: When the notation m*r, where m is an integer and r is an element of ring, is defined, it is done using ONLY the operations of the ring. Remember, only the ring addition and ring multiplication exist (at that moment). Let me add tags _R and _Z to denote that the element or operation is that of the ring R or of the integers Z. Observe that we actually have a third multiplication taking place between an element of a ring and an element of the ring, denote this by *_M (M stands for module, you will learn about these in your next algebra course). Multiplication by adding together copies of an element of the ring necessitates that there is an integer participating to the proceedings, i.e. this is the module multiplication. So, we define module multiplication as follows: 1) Start with defining multiplication by the integers 0 and 1: 0_Z *_M r_R = 0_R, 1_Z *_M r_R = r_R 2) Extend the definition (by induction) to larger positive integers so as to satisfy one of the Z-module axioms: (n_Z +_Z m_Z)*_M r_R= (n_Z *_M r_R) +_R (m_Z *_M r_R) 3) Extend the definition to negative integers by the usual trick of using the additive inverse (within the ring) Your algebra textbook then hopefully shows a list of natural looking consequencies of this definition of module multiplication, such as (n_Z *_Z m_Z) *_M r_R = n_Z *_M (m_Z *_M r_R). As a consequence of this definition we then end up with e.g. 3_Z *_M r_R = (1_Z +_Z 1_Z +_Z 1_Z)*_M r_R = (1_Z *_M r_R) +_R (1_Z *_M r_R) +_R (1_Z *_M r_R) = r_R +_R r_R +_R r_R as expected (here again the module multiplication axiom was used). Observe that of the various multiplications present: 1) *_Z applies, when both sides are integers 2) *_R applies, when both sides are ring elements 3) *_M applies, when one is integer and the other from the ring. Your example consisted of the following 3 different multiplications 1. 5_R *_R 3_R 2. 5_R *_M 3_Z 3. 5_Z *_M 3_R There is no reason, why these should be the same, when *_R is very unnatural as was the case here. You muddied the waters by making ring multiplication by 3 (really by 3_R) to be different from module multiplication by 3 (really by 3_Z). Algebra textbooks often denote all of these simply by *. You are correct in that, if you cook up something really weird (such as your example), then a confusion may result. When a confusion is a possibility, it is IMHO the responsibility of the cook to clearly indicate what is meant (e.g. by adding the subscripts to the operations as above). You may feel that it is a disservice to the students not to emphasize this distinction. I DO notify my freshman algebra students of these distinctions, but I won't unduly emphasize it and stick to the standard notation in class room usage. The reason to this is that it is part of their education to: A) be aware of these distinctions, B) be able to interpret a possibly ambiguous formula correctly from the context (this is necessary for them to be able to study more algebra on their own), C) be aware of the fact that usually the distinctions don't result in any changes (which is why the same notation is used). Good luck in your studies, Jyrki Lahtonen, Turku, Finland === Subject: 1/m +1/(m+1) +...+1/n close to x Let m be a fixed positive integer. Let x be a fixed positive real. Let n(m,x) be the highest positive integer such that 1/m + 1/(m+1) + 1/(m+2) + ...+ 1/n(m,x) <= x. A lot of questions can be asked about this. But what I am wondering now is, what is: E(m,x) = x - (1/m + 1/(m+1) + 1/(m+2) + ...+ 1/n(m,x)) asymptotical to? All this is highly related to the topic at: === Subject: Re: Two-planes in Four-Space > Let G(2,4) be the Grassmanian of 2-dimensional subspaces in R4. > I map G(2,4) -> RP2 as follows. Given a plane g, choose an > orthogonal basis e1, e2. Identifying R4 with the quaternions in > the obvious way, form the quaternion u = e1 * e2(-1). Then > u is a square root of -1, well defined up to a sign. The square > roots of -1 are naturally identified with the unit ball S2 sitting > in R3 sitting in R4 via (a,b,c) |-> (0,a,b,c), so u gives a > well-defined element of S2/(plus or minus 1) = RP2. One might add that the fiber F_u over the point in RP2 represented by the quaternion u has a natural group structure: Given two planes g1 and g2 in F_u, it follows that the set g1 g2 = {p q |p in g1 and q in g2} is also a plane in F_u. This fiber is naturally isomorphic to H*/(R+Ru)* . Another characterization of F_u is that it consists of all planes that are fixed by the 90 degree rotation represented by u. Now I repeat my two questions: 1) We now have a fibration G(2,4) -> RP2 in which the fibers are naturally isomorphic to the various groups H*/(R+Ru)*, where u ranges over square roots of -1 in the quaternions. Is there a standard name for this fibration? 2) Two planes g1 and g2 are in the same fiber F_u if and only if there exists a quaternion p such that p g1 = g2 (or equivalently, if and only if there exists a 90 degree rotation that fixes both g1 and g2). Is there a standard name for this equivalence relation? === Subject: Re: help me...my problem?? > sequence {An} > A1 = 1 > A2 = 8 > An = root (An-1 * An-2) (n=3,4,5....) > find lim An (n->infinite) > --------------------------------- > i want your warm advice. > help me...please === Subject: Re: help me...my problem?? Original Format Nice question ! Denote D=(0,infty). Suppose that f:(0,infty)--->f(D) is continuous and strictly monotonic. Let F: f(D) --->(0,infty) be its inverse function. Instead of geometric mean let us consider the ,,f-quasi mean M_f(a,b) of two positive numbers a,b, that is (*) M_f(a,b)=F((f(a)+f(b))/2) . For instance The last mean can be generalized as / ((ar+br)/2){1/r} for real r , r=/=0 , and MEAN_r(a,b):=/ lim_{r-->0}A_r(a,b)=G(a,b):= sqrt(ab) when r=0 . Let 0=1} defined by (1) A(n)=M_f(A(n-1),A(n-2)) , n=3,4,... , with A(1)=p , A(2)= q . By summing equalities 2*Y(k)=Y(k-1)+Y(k-2) , (k=3,4,...,n) we obtain 2*SUM_{k=3 to k=n}Y(k)=SUM_{k=2 to k=n-1}Y(k) + SUM_{k=1 to k=n-2}Y(k) that is (2) 2*Y(n) + Y(n-1) = Y(1) +2*Y(2) . Because min{a,b} =< M_f(a,b) =< max{a,b} for a,b >0, it's easy to prove that (X(n))_{n>=1}, and therefore also (Y(n))_{n>=1}, are convergent. Let LIM:=lim_{n-->infty}A(n), L:= lim_{n-->infty} Y(n)= f(LIM), or LIM=F(L) . and thus L = lim_{n-->infty}Y(n)=(Y(1)+2*Y(2))/3 . In conclusion LIM:= lim_{n--->infty}A(n)=F((Y(1)+2*Y(2))/3)= F( (f(A(1)+2*f(A(2))/3) . For instance , when r is a real number f(x)=xr (when r=/=0) , and f(x)= ln(x) when r=0 , we find for (A(n))_{n>=1} generated by A(1)=p , A(2)=q , 0infty} A(n)= Sqrt[3](64) =4 . Note: Let D=(0,infty), f:D--->f(D)) be a continuous strictly monotonic function and F: f(D)-->D its inverse function. An interesting question may be : Suppose as fixed the following numbers : k = a positive integer , k>=2 , p_1,p_2,...,p_k , (p_j >0 , j=1,2,...,k) , w_1,w_2,...,w_k are positiv numbers such that w_1+w_2+...+w_k = 1 . Furher let A_f(a_1,a_2,...,a_k):=F(w_1*f(a_1)+w_2*f(a_2)+...+w_k*f(a_k)) , a_j > 0 . Define the sequence (S(n))_{n>=1} by S(1)=p_1 , S(2)=p_2 ,...,S(k)=p_k , and S(n)= A_f(S(n-1),S(n-2),S(n-3),...,S(n-k)) , n in {k+1,k+2,...} . NEW QUESTION: what about the convergence of sequence (S(n))_{n>=1} ? === Subject: Re: JSH: About time > Now that I've revealed the odd and you could say esoteric error in > core mathematics with such a short, and rather simple argument, the > issue now is how long until mathematicians decide that they'd rather > have correct mathematics versus the *belief* that they had been > perfect in keeping error out of the collected body of work that is > called mathematics. > Really, I can not understand your work, not even the core error problem. > I apologize for replying in a public forum, but I've chosen, or been > guided, to work under certain constraints. The danger of e-mail is the > temptation to say different things to different people. It would be > better for me, professionally, to hide my support for you, but I refuse > to do it. Don't worry about it, as I find I'm liking your posts. Actually, I kind of got a kick out of some of your previous posts back when you were insulting me. My thinking is that you appreciate quality work when you see it. Now for those who wonder, being rather pissed at my current predicament, with *some* mathematicians--I have to remember not to characterize all of them--lying about my work or running away from it, I sent Jim Ferry and Keith Ramsay a testy email, where I informed them I considered them to be runners as well, and I would chase them down with the rest, by getting a computerized proof check of my work. And I also told Ferry to look for a job outside of mathematics. Now, I'm not so angry, and I think that maybe I shouldn't have sent that email, but then again, it all revolves around the question of whether or not Jim Ferry does understand the math. > My work is out there and rather easy to go over as can be seen at the > Hong Konk math site: See http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782 And I send people there because their allowal of the use of LaTeX > makes for a *much* better presentation, and given the *social* issues > I'm facing, I need all the help I can get. > The social issues. Indeed. What help could I be to you with the > mathematical issues? Even herd mathematicians like Magidin are better > at math than I. The fact is, you don't *need* any help with the > mathematical issues. Ah yes, rather prescient of you to emphasize need. It seems to me that things are about to get interesting. > I'm still working on your Plan. I've very busy right now, but I'm > finding ways to make the time for it. It will take several weeks > at least. As I began to conceive/receive it, I realized that it > would do no good if you had a Bush-like outlook on life. You seemed > annoyed by my detour into politics, but this is what I mean: you > being concerned with *mathematical* issues is like Bush trying to > build bigger bombs to fight terrorism. America is already powerful > enough, and doesn't need to become all-mighty so that all nations of > the world cower at the thought of incurring its wrath. According to > Bush, terrorists and the people who cheer for them are insane and evil. > According to Bush, a vast segment of the Islamic world is insane and evil. Bush is having issues with his father, so he reversed his father on everything. Bush Sr. said he wouldn't raise taxes and did. So Bush the younger lowers them repeatedly, even when it's cuckoo. Bush Sr. didn't invade Iraq beyond the UN resolutions, so Bush the younger flouts the UN and invades the country. Bush Sr. refuted Saddam Hussein for invading a country and Bush the younger emulated him by invading against international law. It's not as complicated as you put it Jim Ferry, as it's just a rather dysfunctional family playing out its issues on the world stage. It's happened before, I'm sure. > But George Bush is simpleminded, so perhaps we should forgive him his > simplistic viewpoint. Hatred for America is engendered by . . . America's > better teeth. Now here's what I want you to do . . . Only a suggestion, > mind you: I'm all for freedom and human rights and democracy (my ideas > are also better than yours), so do whatever you want. What? What are you > doing? Now I have to beat the crap out of you! Oh, boo hoo, you want > your natural resources back? Too late, I ate them. Ha ha. Hey! Don't > hit me there! Evil terrorist! I have a date with Brittany Spears tonight! Bush isn't simpleminded and in fact he's quite intelligent, but lazy. I think part of what he does is out of anger at his father, and the rest is just following along with Cheney, who appears to be, um, not quite sane. > I'm not saying that you've installed any puppet dictatorships, James, but > there are aspects to America's foreign policy arrogance that remind me of > you. Sure, it's a natural reaction to being better than everyone else. > Better at building weapons, better at doing math, easy to think better in > every way. Easy to wonder, I'm so great, but they all hate me . . . they > must be wrong. They must be evil. Well Bush and I seem to have similarities, which I've noticed to my chagrin. I've decided that it comes from my being brought up as a fundamentalist Christian, so I understand that rigid thinking. > This is why I doubt that working out your Plan is even worth the effort. > No one likes being told they have to change their behavior to achieve their > goals. My guide (or muse, or the Holy Spirit, or whatever it is that seems > to be sending this Plan to me) argues that because you so fervently want to > achieve your goal and because you realize that what you're doing isn't > working, that you are receptive to new ideas. But I think your attitude is > more like, Let Ferry write his little plan, and if I don't like it, I'll > just laugh at it. So, yes, I'm resisting having to do this pointless work, > but still the Holy Spirit (or guide, or muse, or whatever) is prodding me on. Uh oh. Maybe I need to help you out a bit, in case you're losing it. Human beings are in the object ring. You're just a pack of ideas, in one sense. As ideas, you can be infected by ideas, like in the movie The Matrix. It turns out that very complex ideas, like human beings, exist in other forms, and they can infect a human mind. Thing is, such infections are rather hard to fix, as it's not so easy like in that movie. I'll consider your future replies Jim Ferry, and I may cut you loose. My suggestion, see someone, like a priest (not a Catholic priest), or even better, a Zen master. > So please, just tell me, Oh, you want me to roll over and play Mr. Nice > Guy! No dice! and then I can forget this whole hassle. At first I was so > honored that the Holy Spirit was choosing to work through me . . . oh, it's > probably a load of rubbish anyway. Me wanting to be important and imagining > that I was getting divine inspiration. The whole thing just seems ridiculous. > Never mind. I understand the feeling, and I take you seriously. You need to talk to someone, and not on Usenet. > Some of you are now facing the reality of the human brain versus any > fantasy you might have had about being completely rational. Human > beings are NOT rational creatures but necessarily rely on social > forces to determine what they believe. You are creatures of society. You may have believed that your mathematical knowledge was based > completely on logic and rationality, but human beings don't work that > way; it's built-in to your wiring NOT to work that way. Some of you must learn to be more than human. > You know, I don't think genetics come into it. I'm in no position to > know, but here's what the little voice in my head says: > The chief difference between James Harris's and the Establishment's > mathematical systems lies not in the validity of each: rather it is > simply that the world has chosen to accept the Establishment viewpoint. > Reality is created by consciousness more than you know. James Harris > is attempting to create a new Reality to displace the old, but his > arguments simply *do not pertain* within the current Reality. The > current Reality is flawed, of course, by Goedelian indeterminancies, > and is ripe for replacement by something superior . . . Yop, you're infected Jim Ferry. > Yeah. Sounds like a load of crap. Like Langan's CTMU consciousness > creates the world crap. Forget it. > Mathematics is an absolute truth. Cbeck. That's Mathematics as opposed to mathematics. What most people call mathematics is a body of discoveries by people like me--discoverers. It can be flawed. But Mathematics is perfect. > The physical world is what it is. Check. The physical world is a finite bit of Mathematics. And like I said above, human beings are in the object ring. Yup, people, you are mathematical objects, just highly complicated ones. > I apologize for all my inconsistencies. I started writing this in one > frame of mind, and ended up in another. A little voice in my head > telling me how James Harris can conquer the (mathermatical) world? > People are going to start telling *me* to take my meds . . . Go to a Zen master. A good one can handle that mental virus you have. > You must learn to be truly rational, for the first times in your > lives. So it's about time, as I wait, and wonder, how many of you can handle > the truth. And how many of you prefer the fantasy which was the world you > believed in, which actually never existed, except in your > imaginations; your wishes for a nicer world, where your wishes matter. > James Harris > My imagination. Sigh. I'm sure that's all it was. Sorry for wasting > your time. Now that was interesting. James Harris === Subject: Re: JSH: About time > Now that I've revealed the odd and you could say esoteric error in > core mathematics with such a short, and rather simple argument, the > issue now is how long until mathematicians decide that they'd rather > have correct mathematics versus the *belief* that they had been > perfect in keeping error out of the collected body of work that is > called mathematics. > Really, I can not understand your work, not even the core error problem. > I apologize for replying in a public forum, but I've chosen, or been > guided, to work under certain constraints. The danger of e-mail is the > temptation to say different things to different people. It would be > better for me, professionally, to hide my support for you, but I refuse > to do it. > Don't worry about it, as I find I'm liking your posts. Actually, I > kind of got a kick out of some of your previous posts back when you > were insulting me. > My thinking is that you appreciate quality work when you see it. > Now for those who wonder, being rather pissed at my current > predicament, with *some* mathematicians--I have to remember not to > characterize all of them--lying about my work or running away from it, > I sent Jim Ferry and Keith Ramsay a testy email, where I informed them > I considered them to be runners as well, and I would chase them down > with the rest, by getting a computerized proof check of my work. > And I also told Ferry to look for a job outside of mathematics. > Now, I'm not so angry, and I think that maybe I shouldn't have sent > that email, but then again, it all revolves around the question of > whether or not Jim Ferry does understand the math. > My work is out there and rather easy to go over as can be seen at the > Hong Konk math site: See http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782 And I send people there because their allowal of the use of LaTeX > makes for a *much* better presentation, and given the *social* issues > I'm facing, I need all the help I can get. > The social issues. Indeed. What help could I be to you with the > mathematical issues? Even herd mathematicians like Magidin are better > at math than I. The fact is, you don't *need* any help with the > mathematical issues. > Ah yes, rather prescient of you to emphasize need. > It seems to me that things are about to get interesting. > I'm still working on your Plan. I've very busy right now, but I'm > finding ways to make the time for it. It will take several weeks > at least. As I began to conceive/receive it, I realized that it > would do no good if you had a Bush-like outlook on life. You seemed > annoyed by my detour into politics, but this is what I mean: you > being concerned with *mathematical* issues is like Bush trying to > build bigger bombs to fight terrorism. America is already powerful > enough, and doesn't need to become all-mighty so that all nations of > the world cower at the thought of incurring its wrath. According to > Bush, terrorists and the people who cheer for them are insane and evil. > According to Bush, a vast segment of the Islamic world is insane and evil. > Bush is having issues with his father, so he reversed his father on > everything. > Bush Sr. said he wouldn't raise taxes and did. So Bush the younger > lowers them repeatedly, even when it's cuckoo. > Bush Sr. didn't invade Iraq beyond the UN resolutions, so Bush the > younger flouts the UN and invades the country. > Bush Sr. refuted Saddam Hussein for invading a country and Bush the > younger emulated him by invading against international law. > It's not as complicated as you put it Jim Ferry, as it's just a rather > dysfunctional family playing out its issues on the world stage. > It's happened before, I'm sure. > But George Bush is simpleminded, so perhaps we should forgive him his > simplistic viewpoint. Hatred for America is engendered by . . . America's have > better teeth. Now here's what I want you to do . . . Only a suggestion, > mind you: I'm all for freedom and human rights and democracy (my ideas > are also better than yours), so do whatever you want. What? What are you > doing? Now I have to beat the crap out of you! Oh, boo hoo, you want > your natural resources back? Too late, I ate them. Ha ha. Hey! Don't > hit me there! Evil terrorist! I have a date with Brittany Spears tonight! > Bush isn't simpleminded and in fact he's quite intelligent, but lazy. > I think part of what he does is out of anger at his father, and the > rest is just following along with Cheney, who appears to be, um, not > quite sane. > I'm not saying that you've installed any puppet dictatorships, James, but > there are aspects to America's foreign policy arrogance that remind me of > you. Sure, it's a natural reaction to being better than everyone else. > Better at building weapons, better at doing math, easy to think better in > every way. Easy to wonder, I'm so great, but they all hate me . . . they > must be wrong. They must be evil. > Well Bush and I seem to have similarities, which I've noticed to my > chagrin. > I've decided that it comes from my being brought up as a > fundamentalist Christian, so I understand that rigid thinking. > This is why I doubt that working out your Plan is even worth the effort. > No one likes being told they have to change their behavior to achieve their > goals. My guide (or muse, or the Holy Spirit, or whatever it is that seems > to be sending this Plan to me) argues that because you so fervently want to > achieve your goal and because you realize that what you're doing isn't > working, that you are receptive to new ideas. But I think your attitude is > more like, Let Ferry write his little plan, and if I don't like it, I'll > just laugh at it. So, yes, I'm resisting having to do this pointless work, > but still the Holy Spirit (or guide, or muse, or whatever) is prodding me on. > Uh oh. Maybe I need to help you out a bit, in case you're losing it. > Human beings are in the object ring. > You're just a pack of ideas, in one sense. > As ideas, you can be infected by ideas, like in the movie The > Matrix. > It turns out that very complex ideas, like human beings, exist in > other forms, and they can infect a human mind. > Thing is, such infections are rather hard to fix, as it's not so easy > like in that movie. > I'll consider your future replies Jim Ferry, and I may cut you loose. > My suggestion, see someone, like a priest (not a Catholic priest), or > even better, a Zen master. > So please, just tell me, Oh, you want me to roll over and play Mr. Nice > Guy! No dice! and then I can forget this whole hassle. At first I was so > honored that the Holy Spirit was choosing to work through me . . . oh, it's > probably a load of rubbish anyway. Me wanting to be important and imagining > that I was getting divine inspiration. The whole thing just seems ridiculous. > Never mind. > I understand the feeling, and I take you seriously. You need to talk > to someone, and not on Usenet. > Some of you are now facing the reality of the human brain versus any > fantasy you might have had about being completely rational. Human > beings are NOT rational creatures but necessarily rely on social > forces to determine what they believe. You are creatures of society. You may have believed that your mathematical knowledge was based > completely on logic and rationality, but human beings don't work that > way; it's built-in to your wiring NOT to work that way. Some of you must learn to be more than human. > You know, I don't think genetics come into it. I'm in no position to > know, but here's what the little voice in my head says: > The chief difference between James Harris's and the Establishment's > mathematical systems lies not in the validity of each: rather it is > simply that the world has chosen to accept the Establishment viewpoint. > Reality is created by consciousness more than you know. James Harris > is attempting to create a new Reality to displace the old, but his > arguments simply *do not pertain* within the current Reality. The > current Reality is flawed, of course, by Goedelian indeterminancies, > and is ripe for replacement by something superior . . . > Yop, you're infected Jim Ferry. > Yeah. Sounds like a load of crap. Like Langan's CTMU consciousness > creates the world crap. Forget it. > Mathematics is an absolute truth. Cbeck. > That's Mathematics as opposed to mathematics. > What most people call mathematics is a body of discoveries by people > like me--discoverers. > It can be flawed. But Mathematics is perfect. > The physical world is what it is. Check. > The physical world is a finite bit of Mathematics. > And like I said above, human beings are in the object ring. > Yup, people, you are mathematical objects, just highly complicated > ones. > I apologize for all my inconsistencies. I started writing this in one > frame of mind, and ended up in another. A little voice in my head > telling me how James Harris can conquer the (mathermatical) world? > People are going to start telling *me* to take my meds . . . > Go to a Zen master. A good one can handle that mental virus you have. > You must learn to be truly rational, for the first times in your > lives. So it's about time, as I wait, and wonder, how many of you can handle > the truth. And how many of you prefer the fantasy which was the world you > believed in, which actually never existed, except in your > imaginations; your wishes for a nicer world, where your wishes matter. > James Harris > My imagination. Sigh. I'm sure that's all it was. Sorry for wasting > your time. > Now that was interesting. > James Harris James, What do you expect to accomplish with all these insults you're dishing out? Not like you'll answer me anyway. These insults are childish, grow up. David Moran === Subject: Re: JSH: About time >>Now that was interesting. >>James Harris > James, What do you expect to accomplish with all these insults you're > dishing out? Not like you'll answer me anyway. These insults are childish, > grow up. > David Moran ??? Were you and I reading the same post? James reply was insightful and fascinating. Sure, he and I disagree on a few points, but I don't understand how you could categorize it as insulting. It was actually quite helpful to me in my distress. Maybe James's and my upbringings are the root of some of our differences of opinion. James was brought up in a fundamentalist Christian household; my household was a jumble of various and/or no beliefs. My struggle is to find attunement with God in an empty, profane world. James probably has to work to shut out that oppressive, thundering, Old Testament Jehovah. Thus, whereas I welcome an inspiration of the Holy Spirit as fresh air, James seems to see such things as infections, mental diseases. Now that I see where you're coming from, James, let me say that I'm not oppressed by thunder, nor looking to be cured. Rather, I'm straining to hear something like far off music, like elven-song, and am amazed that I can still hear it if I try. I fear that it will abandon me if I fail to honor it, and that it would be a terrible shame to waste it. Though it merely moves through me toward you, when it moves through me, I am purified. I see now that it comes to me because you could never hear it, perhaps because of Jehovah's thunder, but more simply because of the hammer and tongs of your mathworks. === Subject: Re: JSH: About time >[...] >James, What do you expect to accomplish with all these insults you're >dishing out? It's an interesting question. Sometimes I think that he's actually going to get people to agree he's right this way: You're wrong. Idiot. Well, you're still wrong. Liar. No, agreeing you were right would be lying. Satan's going to kill you. Oh my, I just can't tolerate these insults! Ok, you're right. I mean it's hard to believe anyone could think that it's going to work out that way, but he does believe lots of things that it's hard to believe anyone would believe... >Not like you'll answer me anyway. These insults are childish, >grow up. >David Moran === Subject: Re: JSH: About time >>[...] >>James, What do you expect to accomplish with all these insults you're >>dishing out? >It's an interesting question. Sometimes I think that he's actually >going to get people to agree he's right this way: >You're wrong. >Idiot. >Well, you're still wrong. >Liar. >No, agreeing you were right would be lying. >Satan's going to kill you. >Oh my, I just can't tolerate these insults! Ok, you're right. >I mean it's hard to believe anyone could think that it's going >to work out that way, but he does believe lots of things that >it's hard to believe anyone would believe... But that's [bullying until somebody cried Uncle] is a successful tactic among children. MostW[sigh!]some people figure out this doesn't work to one's advantage when they attain adulthood. It is bothersome that elementary schooling encourages this kind of persistent behaviour; teachers aren't supposed to correct mistakes because that may bruise the little egos and not build their confidence. /BAH Subtract a hundred and four for e-mail. === Subject: Re: JSH: About time >[...] >James, What do you expect to accomplish with all these insults you're >dishing out? > It's an interesting question. Sometimes I think that he's actually > going to get people to agree he's right this way: > You're wrong. > Idiot. > Well, you're still wrong. > Liar. > No, agreeing you were right would be lying. > Satan's going to kill you. > Oh my, I just can't tolerate these insults! Ok, you're right. > I mean it's hard to believe anyone could think that it's going > to work out that way, but he does believe lots of things that > it's hard to believe anyone would believe... >Not like you'll answer me anyway. These insults are childish, >grow up. >David Moran > ************************ > David C. Ullrich Sometimes I wonder what his mathematical background is. He strikes me as someone who has had very little. If he actually has a physics degree, you would think that there'd be enough math there to set him straight; I'm a physics major myself (Meteorology). David Moran === Subject: Re: JSH: About time > Sometimes I wonder what his mathematical background is. He strikes me as > someone who has had very little. If he actually has a physics degree, you > would think that there'd be enough math there to set him straight; I'm a > physics major myself (Meteorology). You can see that he doesn't get it at all if you mention anything that isn't simple arithmetic or that he hasn't defined himself. I'd say his mathematical background is extremely limited. The fact that he is completely unwilling (incapable?) of learning makes things worse. === Subject: Re: JSH: About time >>[...] >>James, What do you expect to accomplish with all these insults you're >>dishing out? >> It's an interesting question. Sometimes I think that he's actually >> going to get people to agree he's right this way: >> You're wrong. >> Idiot. >> Well, you're still wrong. >> Liar. >> No, agreeing you were right would be lying. >> Satan's going to kill you. >> Oh my, I just can't tolerate these insults! Ok, you're right. >> I mean it's hard to believe anyone could think that it's going >> to work out that way, but he does believe lots of things that >> it's hard to believe anyone would believe... >>Not like you'll answer me anyway. These insults are childish, >>grow up. >>David Moran >> ************************ >> David C. Ullrich >Sometimes I wonder what his mathematical background is. None even though he might have received As in el-hi. I have a nephew who has been told all his young life that he knows a lot about computers. He knows nothing except how to point and click very fast. He has grown up thinking he knows it all just because he knew how to maniputate a mouse quickly. > .. He strikes me as >someone who has had very little. If he actually has a physics degree, you >would think that there'd be enough math there to set him straight; I'm a >physics major myself (Meteorology). My father claims he majored in history. His highest education level is 12th grade. When I went to high school, they also used the major word. It has nothing to do with the discipline associated with university terms. I suspect that some people may equate a high school major with a college-level degree. Did he really say physics degree or did you extrapolate that from him saying physics major? /BAH Subtract a hundred and four for e-mail. === Subject: Re: JSH: About time > My father claims he majored in history. His highest education level > is 12th grade. When I went to high school, they also used the > major word. It has nothing to do with the discipline associated > with university terms. I suspect that some people may equate > a high school major with a college-level degree. Did he really > say physics degree or did you extrapolate that from him saying > physics major? He claims to have a BS in Physics from Vanderbilt. I give him the benefit of the doubt on that. I can assure you that you can get a BS in physics without ever seeing math at a very abstract level. He also was apparently in gifted-talented programs as a youngster and had a lot of people telling him how smart he was. He seems unable to comprehend that a lot of the people in the newsgroups he frequents had the same life experiences, but managed to make the transition from big fish in little pond to little fish in big pond which, from all evidence, he did not. - Randy === Subject: Re: JSH: About time > He also was apparently in gifted-talented programs as a > youngster and had a lot of people telling him how smart > he was. He seems unable to comprehend that a lot of the > people in the newsgroups he frequents had the same life > experiences, but managed to make the transition from big fish > in little pond to little fish in big pond which, from > all evidence, he did not. Maybe he made a different transition. Out of the water . . . === Subject: Re: JSH: About time [...] James, What do you expect to accomplish with all these insults you're >dishing out? > It's an interesting question. Sometimes I think that he's actually > going to get people to agree he's right this way: > You're wrong. > Idiot. > Well, you're still wrong. > Liar. > No, agreeing you were right would be lying. > Satan's going to kill you. > Oh my, I just can't tolerate these insults! Ok, you're right. > I mean it's hard to believe anyone could think that it's going > to work out that way, but he does believe lots of things that > it's hard to believe anyone would believe... >Not like you'll answer me anyway. These insults are childish, >grow up. David Moran ************************ > David C. Ullrich >>Sometimes I wonder what his mathematical background is. >None even though he might have received As in el-hi. I have >a nephew who has been told all his young life that he >knows a lot about computers. He knows nothing except how >to point and click very fast. He has grown up thinking he >knows it all just because he knew how to maniputate a mouse >quickly. Presumably he doesn't know how to find usenet... >> .. He strikes me as >>someone who has had very little. If he actually has a physics degree, you >>would think that there'd be enough math there to set him straight; I'm a >>physics major myself (Meteorology). >My father claims he majored in history. His highest education level >is 12th grade. When I went to high school, they also used the >major word. It has nothing to do with the discipline associated >with university terms. I suspect that some people may equate >a high school major with a college-level degree. Did he really >say physics degree or did you extrapolate that from him saying >physics major? He definitely has a degree in physics (or at least that's the story). Every once in a while it's explained that this is why we should believe he's right. >/BAH >Subtract a hundred and four for e-mail. === Subject: Re: JSH: About time >>[...] >>James, What do you expect to accomplish with all these insults you're >>dishing out? >> It's an interesting question. Sometimes I think that he's actually >> going to get people to agree he's right this way: >> You're wrong. >> Idiot. >> Well, you're still wrong. >> Liar. >> No, agreeing you were right would be lying. >> Satan's going to kill you. >> Oh my, I just can't tolerate these insults! Ok, you're right. >> I mean it's hard to believe anyone could think that it's going >> to work out that way, but he does believe lots of things that >> it's hard to believe anyone would believe... >>Not like you'll answer me anyway. These insults are childish, >>grow up. >>David Moran >> ************************ >> David C. Ullrich >Sometimes I wonder what his mathematical background is. > None even though he might have received As in el-hi. I have > a nephew who has been told all his young life that he > knows a lot about computers. He knows nothing except how > to point and click very fast. He has grown up thinking he > knows it all just because he knew how to maniputate a mouse > quickly. > .. He strikes me as >someone who has had very little. If he actually has a physics degree, you >would think that there'd be enough math there to set him straight; I'm a >physics major myself (Meteorology). > My father claims he majored in history. His highest education level > is 12th grade. When I went to high school, they also used the > major word. It has nothing to do with the discipline associated > with university terms. I suspect that some people may equate > a high school major with a college-level degree. Did he really > say physics degree or did you extrapolate that from him saying > physics major? > /BAH > Subtract a hundred and four for e-mail. I remember him saying he had a physics degree, however, I could claim I have a math degree and you wouldn't know; that's the internet for you. By the way I don't have a math degree. David Moran === Subject: qual vs quan research boundary=----=_NextPart_000_0034_01C391E6.5E7AD640 --------------------------------------------------------------------- #1 - Is environmental dumping more likely to occur in impoverished neighborhoods compared to middle neighborhoods? (I think thit is quantative question) #2 - Do children raised in single parent families have greater academic difficulties than children raised in two parent families? (I think thit is quantative question) #3 - How does motivation lead to success in the workforce? (I think thit is qualatative question) === Subject: Re: qual vs quan research > following questions and have to state whether they are qualatative or > quantative. HEre are the questions and what I think they are. Can anyone > tell me if I'm correct? THANKS! > #1 - Is environmental dumping more likely to occur in impoverished > #neighborhoods compared to middle neighborhoods? (I think thit is > #quantative question) > #2 - Do children raised in single parent families have greater academic > #difficulties than children raised in two parent families? (I think thit > #is quantative question) > #3 - How does motivation lead to success in the workforce? (I think thit > #is qualatative question) -- I'm not speaking as an expert here but this is what I think. Third is definately qualitative. The first two could be considered quantitative because the contain comparative words like more but I consider them qualitative because they are really questions of cause. Nobody really cares whether more or less of something happens here or there by accident. Any satisfactory answer to these questions would have to suggest a reason why more or less of something happens here or there. Besides, for a quantitative question, I would expect something more specific than more or less. I would expect an equation or something. Have a tolerable existence. Eli === Subject: Re: Minimal Graph, Four Color Theorem I'd hesitate to ask you to go there, but it's Harris Steven James' 10-year program to prove the last theorem of Fermat. of course, it was actually his first one! he's wroking on the definition of divisibility, now. > I hesitate to ask, but what is HSJ? --les ducs de Enron! http://larouchepub.com === Subject: Re: Quadratic System Help from slick_shoes@punkass.com: >18=a(5)2+b(5)+c >69=a(12)2+b(12)+c >74=a(14)2+b(14)+c > I'm not sure how to go about solving this. In class, we've only solved >systems like this when on of the equations had x=0, so it rather easy to >solve for c. Can anybody point me in the right direction on this one? >slick_shoes My earlier response never entered the newgroup, so I try again: You have three equations and three unknown values, now a, b, and c. If you understand a little about matrices, then you already can figure out what..... but you are asking for help, so you probably know at most, parts of intermediate algebra. If this is so, then either use an equation to find expression of one variable in terms of the others and back substitute; or use elimination by adding or subtracting multiples of one equation to another. G C === Subject: Re: Quadratic System Help > from slick_shoes@punkass.com: >18=a(5)2+b(5)+c >69=a(12)2+b(12)+c >74=a(14)2+b(14)+c > I'm not sure how to go about solving this. One way is as a system of linear equations in a,b,and c as unknowns. 25*a + 5*b + c = 18 144*a + 12*b + c = 69 196*a + 14*b + c = 74 Another is by row reduction of the augmented matrix [[ 25 5 1 18 ] [ 144 12 1 69 ] [ 196 14 1 74 ]] === Subject: harmonic functions How do you prove: Suppose D is a connected domain and {f_n} is a sequence of non-negative harmonic functions on D. Then we have one the following: (1) The sequence {f_n} contains a subsequence diverging to infinitely pointwise on D, (2) The sequence {f_n} contains a subsequence converging uniformly on compact subsets of D. My intuition is that all that is needed is Harnack's inequality (in one form or another) and the Arzela-Ascoli theorem, but I can't seem to combine them in just the right way... === Subject: Re: harmonic functions >How do you prove: >Suppose D is a connected domain and {f_n} is a sequence of non-negative >harmonic functions on D. Then we have one the following: >(1) The sequence {f_n} contains a subsequence diverging to infinitely >pointwise on D, >(2) The sequence {f_n} contains a subsequence converging uniformly on >compact subsets of D. >My intuition is that all that is needed is Harnack's inequality (in one form >or another) and the Arzela-Ascoli theorem, but I can't seem to combine them >in just the right way... The fact that you seem to know what's needed to prove this makes it smell like homework - you can't quite figure out how to use the hint. So I'll give a few more hints: First, a relevant form of Harnack's inequality would be this: For any compact set K in D and any point p in D there is a constant c such that u(x) <= c u(p) for all x in K. Second, either the family is unbounded at some point or it isn't. Third, if K1, K2 are compact and K1 is contained in the interior of K2 then there exists c such that the sup of grad(u) over K1 is less than c times the sup of u over K2. (Note that bounds on grad(u) imply equicontinuity...) === Subject: inversion via lagrange theory hello, is it possible to invert something of a form say... f(t) = G(f(t)) + H(f(t),t) (G, H arbitrary functions, where as shown, H is explicitly dependant on t.) via say the lagrange inversion theorem to solve for f(t) ? (im not too familiar with the theorem, which is why i am asking...however on inspection of the theorem i would assume it is not possible....is there a theorem that allows for its solution?) cheers moth === Subject: Re: Value of PI using trigonometry and calculus yess, thats pure calculus way to prove !! in my earlier post same has been proved using geometry too. atul > pi = lim n*sin(180/n) > n->inf > lim(n->oo) n*sin 180/n = 180 lim(n->oo) (sin 180/n)/(180/n) > = 180 lim(x->0) (sin x)/x > = 180 provided x is in radians > I believe this thing is known and exists. > Basic calculus > lim(x->0) (sin x)/x = 1 === Subject: Re: Value of PI using trigonometry and calculus William, What I was saying, was correct. Finally I found it in one of the answers on Dr. Math website. It reads as follows : === Subject: Re: Pi and Polygons This method of finding successive approximations to Pi is one of the oldest methods known, because it is one of the easiest to understand, and you can draw nice pictures to explain it. If a regular polygon has n sides, then we can draw lines from its center to all the vertices, and these lines will divide the pie-shaped picture into n wedges. Each wedge's central angle will have a measure of 360/n degrees. So, we can draw this picture of one wedge, where C is at the center of the polygon: A /| / | / | / | / | C /__________| D | | | | | | B Segment AB here is one side of the original regular polygon. Since angle ACB is 360/n, angle ACD is 180/n. Therefore, if the length of AC is 1/2, the length of AD is sin(180/n)/2. Therefore, the length of AB is sin(180/n), and the perimeter of the entire polygon is n*sin(180/n). So, you are correct: when you let n go toward infinity, sin(180/n) will tend towards zero. Since we know that a circle whose radius is 1/2 has a circumference of Pi, the 0 and infinity balance each other in the limit. Of course, there is a practical problem with all of this. In order to calculate the sines, you need to know a thing or two about Pi. It is true that for some special angles like 30, 45, and 60 degrees (and their sums and differences, etc.) you can write down an explicit elementary expression for their sines, this is not true of some other angles like 180/11. So, how do we calculate the perimeters without knowing Pi already? We need to find some trick, or we need to find some other method entirely of approximating Pi. And that is where a great part of the glorious history of mathematics starts. For more about Pi, its role in history, and the various attempts to know it better, check out the excellent book _A History of Pi_ by Petr Beckmann. People have done some pretty clever things to get to know Pi. - Doctor Ken, The Math Forum http://mathforum.org/dr.math/ cheers, Atul > pi = lim n*sin(180/n) > n->inf > lim(n->oo) n*sin 180/n = 180 lim(n->oo) (sin 180/n)/(180/n) > = 180 lim(x->0) (sin x)/x > = 180 provided x is in radians > I believe this thing is known and exists. > Basic calculus > lim(x->0) (sin x)/x = 1 === Subject: JVC Anyone who knows the difference between the Jonker Volgenant (JV) and the Jonker Volgenant Castanion (JVC) Algorithm? === Subject: Re: Groups with 16 elements > I found an interesting program for doing things > with groups of order < 32 at http//math.ucsd.edu/~jwavrick . > program (or using telnet); [...] > ORDERS for Groups Number 35 and 38 > Group number 35 of Order 16 > 1 elements of order 1: A > 3 elements of order 2: C E G > 12 elements of order 4: B D F H I J K L M N O P > 0 elements of order 8: > 0 elements of order 16: > Both 35 and 38 have the same center; Z = {A E C G}. > This is the same distribution of orders as in Z_2 x Q, so > 35 or 38 is Z_2 x Q. > For 35 and 38, there are 6 = 12/2 elements of order 4, You mean 6 *cyclic subgroups* of order 4, not 6 elements. > and if we call these x_i; i=1,6, then let > x_12 = x_22, x_32 = x_42, x_52 = x_62, > which gives 3 elements of order 2, all of which are in the center. No. All elements of order 4 in Z_2 x Q square to the same element, so 2 of the 3 involutions in the center are not the square of an element of order 4. Likewise, only 2 of the 3 involutions in the center of the other non-abelian group with this distribution of element orders are the squares of elements of order 4. This group is the semidirect product Z_4 x| Z_4 with presentation . > I would like to look first at the 5 groups 34 thru 38 groups more > closely. (Note that none of these 5 have an element of order 8.) > Also, #36 and 37 have the same distribution of orders. > ORDERS for Group Number 36 > Group number 36 of Order 16 > 1 elements of order 1: A > 7 elements of order 2: C E G I K N P > 8 elements of order 4: B D F H J L M O > 0 elements of order 8: > 0 elements of order 16: > ORDERS for Group Number 37 > Group number 37 of Order 16 > 1 elements of order 1: A > 7 elements of order 2: C E G I K M O > 8 elements of order 4: B D F H J L N P > 0 elements of order 8: > 0 elements of order 16: > CENTER of Group Number 36 { A B C D } (cyclic--Z_4) > CENTER of Group Number 37 { A C E G } (Z_2 x Z_2) [...] > This shows the structure of 36, but I haven't figured out how to > write it concisely, as one does for D_8, e.g. This is the central product Z_4 * D_4 =~ Z_4 * Q that Derek mentioned. It has rank 3, and one easy presentation is . Group 37 has rank 2 with, for example, presentation . > I plan to look at the rest of the groups. I don't yet see > how one decides how many groups there are with each distribution of > orders. It didn't turn out quite like I thought--there are more > groups with no element of order 8 than I would have guessed. I know of no way to decide except to actually construct all of the groups of order 16 and count the elements of each order in each group. > There must be a systematic way of doing this--I know there are > papers, and I think some books, on groups of order 2n, but > I am not at a University, so I don't have the easy access > that I used to have. As I posted earlier in the thread, the systematic way is to look for all possible ways that 2 subgroups of order 8 can intersect, necessarily in a subgroup of order 4, such that one normalizes the other. You correctly started on this path when you looked at all the ways to have a cyclic subgroup of order 8. -- Jim Heckman === Subject: Re: JSH: $100,000 US offer, Abel Prize > Tee-hee. Nothing makes a pig happier than a roll in the wallow, as this tittering porker knows! http://www.shop4egifts.com/target.asp?item=/jpg/25112.jpg&width=314&height=4 00# > Heh-heh. Nothing makes a donkey happier than a rut in the barnyard, as this hee-hawing hoofer knows. http://www.rubylane.com/ni/shops/viperswife/iteml/GW-684 > David C. Ullrich === Subject: abstract math I was wondering if I could get some help on this problem. We know that multiplication of integers is commutative; i.e. ab=ba for all pairs of integers a and b. Prove that, for every natural number n, the product of n integers is independant of the order of the factors. === Subject: Re: abstract math >I was wondering if I could get some help on this problem. >We know that multiplication of integers is commutative; i.e. ab=ba for >all pairs of integers a and b. Prove that, for every natural number n, >the product of n integers is independant of the order of the factors. This requires associative as well. If f(a,b) = (a+b)2, this operation is commutative but not associative, and it is false, as f(1, f(1, 3)) = (1 + 10)2, and f(f(1,1), 3) = (4 + 3)2. For commutative and associative operations, it is an exercise in induction. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: abstract math > I was wondering if I could get some help on this problem. > We know that multiplication of integers is commutative; i.e. ab=ba for > all pairs of integers a and b. Prove that, for every natural number n, > the product of n integers is independant of the order of the factors. Hint: Given any permutation of ABC... normalize it by first moving A to the first place via transpositions then inductively doing the same on the tail after A, moving B to the 2nd place, C to the third place, etc. e.g. DCBA DCAB DACB ADCB now A is in normal place, work on rest DBC BDC now B is in normal place, work on rest CD now C is in normal place, so too is D ---- ABCD If you've studied permutation groups you'll recognize the relationship to the representation of a permutation as a product of transpositions. -Bill Dubuque === Subject: Re: abstract math >We know that multiplication of integers is commutative; i.e. ab=ba for >all pairs of integers a and b. Prove that, for every natural number n, >the product of n integers is independant of the order of the factors. I hate this type of homework: it is more difficult to neatly write down what it is that you have to prove [*] than to actually prove it. [*] For all natural numbers n, for all integers x_1,...,x_n, for all bijections f : {1,...,n} -> {1,...,n} x_1 * x_2 * ... * x_n = x_f(1) * x_f(2) * ... * x_f(n) and even then I should probably make clear where the parethesis are supposed to be in the products. Peter van Rossum === Subject: Re: abstract math > I was wondering if I could get some help on this problem. > We know that multiplication of integers is commutative; i.e. ab=ba for > all pairs of integers a and b. Prove that, for every natural number n, > the product of n integers is independant of the order of the factors. Mathematical Induction and the UFT. Bob Pease === Subject: Re: abstract math >> I was wondering if I could get some help on this problem. >> We know that multiplication of integers is commutative; i.e. ab=ba for >> all pairs of integers a and b. Prove that, for every natural number n, >> the product of n integers is independant of the order of the factors. > Mathematical Induction and the UFT. You don't need unique factorization. The property holds in any commutative ring, not just in a UFD. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: abstract math Amin schrieb im Newsbeitrag > I was wondering if I could get some help on this problem. > We know that multiplication of integers is commutative; i.e. ab=ba for > all pairs of integers a and b. Prove that, for every natural number n, > the product of n integers is independant of the order of the factors. Sorry for asking: But why would you want to prove this? I mean - I believe in it, that's enough for me. Don't want to offend you - I was just currious. -Gernot === Subject: Re: abstract math > Amin schrieb im Newsbeitrag > I was wondering if I could get some help on this problem. > We know that multiplication of integers is commutative; i.e. ab=ba > for > all pairs of integers a and b. Prove that, for every natural number > n, > the product of n integers is independant of the order of the > factors. > Sorry for asking: But why would you want to prove this? I mean - I > believe in it, that's enough for me. > Don't want to offend you - I was just currious. > -Gernot It is not so much offensive as indicative of potentially dangerous Snerdism. Would you post a message like this to a newsgroup like alt.religion.Christian? Sorry for asking: But why would you want to believe this? I mean - I believe in God, that's enough for me. Don't want to offend you - I was just currious. The CENTRAL reason for studying Math is to DO proofs. Whazzamatta you??? RJ Pease === Subject: Re: abstract math > The CENTRAL reason for studying Math is to DO proofs. I would just like to quote Laurent Schwartz : The reason for studying math is to discover some interesting properties, and to prove them just to be sure. Well, I would consider proofs as mere tools... Yann === Subject: Re: abstract math > Amin schrieb im Newsbeitrag > I was wondering if I could get some help on this problem. > We know that multiplication of integers is commutative; i.e. ab=ba > for > all pairs of integers a and b. Prove that, for every natural number > n, > the product of n integers is independant of the order of the > factors. > Sorry for asking: But why would you want to prove this? I mean - I > believe in it, that's enough for me. > Don't want to offend you - I was just currious. It's a standard approach in mathematics: Start from as few axioms as possible, and build things up. In this case, it shows you don't need to take equivalence of arbitrary permutation as an axiom, you just need to take commutation of pairs. And, I believe, associativity. To the OP: Try induction on n. Consider this at the induction step: (product of n-1 factors)*x Now, from your assumptions, you can do two thing without changing the product: - interchange x and the last element in (product of n-1 factors) - rearrange (product of n-1 factors) in any way you like. Can you convince yourself (and your teacher) that these two operations are sufficient to generate all permutations of n factors? Try it for 3 or 4 factors. If that's not sufficient, note that you can also do this operation: - regroup as (first factor)*(n-2 factors * x) and this: - regroup as (first factor)*(any permutation of n-2 factors and x) - Randy === Subject: Re: abstract math >Amin schrieb im Newsbeitrag >> I was wondering if I could get some help on this problem. >> We know that multiplication of integers is commutative; i.e. ab=ba for >> all pairs of integers a and b. Prove that, for every natural number n, >> the product of n integers is independant of the order of the factors. >Sorry for asking: But why would you want to prove this? I mean - I >believe in it, that's enough for me. >Don't want to offend you - I was just currious. This is only a guess. Beginning of school year = beginning of home work in class writing proofs. Can't see the answer immediately = ask the net for the answer. More advanced math work = believes only what can be proven. === Subject: Re: abstract math > I was wondering if I could get some help on this problem. > We know that multiplication of integers is commutative; i.e. ab=ba for > all pairs of integers a and b. Prove that, for every natural number n, > the product of n integers is independant of the order of the factors. Would an induction proof work? Bill === Subject: Intersection area of 2 circles assume 2 circles in 2D with given centers cx1, cy1 and cx2, cy2 and radiuses r1, r2. Now if dist(c1, c2) < r1+r2, the circles intersect. What I need is the intersection's center of mass (if 2d can have mass), and the area of the intersection. -- -Gernot In order to reply, revert my forename from: tonreG.Frisch.at.Dream-D-Sign.de@invalid.com ________________________________________ Looking for a good game? Do it yourself! GLBasic - you can do www.GLBasic.com === Subject: Re: Intersection area of 2 circles > assume 2 circles in 2D with given centers cx1, cy1 and cx2, cy2 and > radiuses r1, r2. > Now if dist(c1, c2) < r1+r2, the circles intersect. What I need is the > intersection's center of mass (if 2d can have mass), and the area of > the intersection. Draw the common chord and all the radii to its endpoints. The area of the intersection is a sector minus an isosceles triangle on each side of the chord. Add 'em up. I can't think of a much slicker way to do the CoM than the same construction. It's not very hard to work out the CoM of a sector[1] and the CoM of a triangle is its centroid, then use linearity. [1] if the sector has radius r and angle 2t then I make the CoM a distance (2 sin t)/(3t) from the centre. But that's just a quick guess. You are welcome :) Dave -- Remove the opinion on spam to reply. === Subject: Re: Intersection area of 2 circles >> assume 2 circles in 2D with given centers cx1, cy1 and cx2, cy2 and >> radiuses r1, r2. >> Now if dist(c1, c2) < r1+r2, the circles intersect. What I need is the >> intersection's center of mass (if 2d can have mass), and the area of >> the intersection. > Draw the common chord and all the radii to its endpoints. The area of the > intersection is a sector minus an isosceles triangle on each side of the > chord. Add 'em up. > I can't think of a much slicker way to do the CoM than the same > construction. It's not very hard to work out the CoM of a sector[1] and > the CoM of a triangle is its centroid, then use linearity. > [1] if the sector has radius r and angle 2t then I make the CoM a distance > (2 sin t)/(3t) from the centre. But that's just a quick guess. > You are welcome :) > Dave I'm not certain, but I would think that the center of mass would be the average of the centers of the two circles weighted by their areas, that is, ( (x1,y1)*pi*r12 + (x2,y2)*pi*r22 ) / ( pi*r12 + pi*r22 ) Have a tolerable existence. Eli -- === Subject: Re: Intersection area of 2 circles >> assume 2 circles in 2D with given centers cx1, cy1 and cx2, cy2 and >> radiuses r1, r2. >> Now if dist(c1, c2) < r1+r2, the circles intersect. What I need is the >> intersection's center of mass (if 2d can have mass), and the area of >> the intersection. > Draw the common chord and all the radii to its endpoints. The area of > the intersection is a sector minus an isosceles triangle on each side > of the chord. Add 'em up. > I can't think of a much slicker way to do the CoM than the same > construction. It's not very hard to work out the CoM of a sector[1] and > the CoM of a triangle is its centroid, then use linearity. > [1] if the sector has radius r and angle 2t then I make the CoM a > distance (2 sin t)/(3t) from the centre. But that's just a quick guess. > I'm not certain, but I would think that the center of mass would be the > average of the centers of the two circles weighted by their areas, that > is, ( (x1,y1)*pi*r12 + (x2,y2)*pi*r22 ) / ( pi*r12 + pi*r22 ) The center of mass of the intersection cannot be located as you thought. Consider, for example, a degenerate case when one disk is completely contained within another disk of larger radius. Then the center of mass of the intersection is always just the center of the smaller disk; it is independent of the center and radius of the larger disk. David Cantrell === Subject: Re: Intersection area of 2 circles assume 2 circles in 2D with given centers cx1, cy1 and cx2, cy2 and > radiuses r1, r2. > Now if dist(c1, c2) < r1+r2, the circles intersect. What I need is the > intersection's center of mass (if 2d can have mass), and the area of > the intersection. >> Draw the common chord and all the radii to its endpoints. The area of >> the intersection is a sector minus an isosceles triangle on each side >> of the chord. Add 'em up. >> I can't think of a much slicker way to do the CoM than the same >> construction. It's not very hard to work out the CoM of a sector[1] and >> the CoM of a triangle is its centroid, then use linearity. >> [1] if the sector has radius r and angle 2t then I make the CoM a >> distance (2 sin t)/(3t) from the centre. But that's just a quick guess. >> I'm not certain, but I would think that the center of mass would be the >> average of the centers of the two circles weighted by their areas, that >> is, ( (x1,y1)*pi*r12 + (x2,y2)*pi*r22 ) / ( pi*r12 + pi*r22 ) > The center of mass of the intersection cannot be located as you thought. > Consider, for example, a degenerate case when one disk is completely > contained within another disk of larger radius. Then the center of mass > of the intersection is always just the center of the smaller disk; it is > independent of the center and radius of the larger disk. > David Cantrell You're right. Oh, well. Have a tolerable existence. Eli -- === Subject: Re: Intersection area of 2 circles > assume 2 circles in 2D with given centers cx1, cy1 and cx2, cy2 and > radiuses r1, r2. > Now if dist(c1, c2) < r1+r2, the circles intersect. What I need is the > intersection's center of mass (if 2d can have mass), and the area of > the intersection. > Draw the common chord and all the radii to its endpoints. The area of the > intersection is a sector minus an isosceles triangle on each side of the > chord. Add 'em up. Or see formula (11) at . > I can't think of a much slicker way to do the CoM than the same > construction. It's not very hard to work out the CoM of a sector[1] and > the CoM of a triangle is its centroid, then use linearity. > [1] if the sector has radius r and angle 2t then I make the CoM a > distance (2 sin t)/(3t) from the centre. But that's just a quick guess. But of course that distance can't be independent of r. I suppose you intended to write (2 r sin t)/(3t); if so, your quick guess was correct. David Cantrell === Subject: Re: should Gauss's Law of Magnetism be: Integral B dot dA = q/p + q/p_t > So the new Gauss Law of Magnetism looks somewhat like this: > Integral B dot dA = q/p + q/p_t where t is temperature related. > This new law restores complete symmetry to the Maxwell Equations. It > dismisses the monopole idea completely. > It just says that we were just not advanced enough since the 1800s > when > the Gauss laws were built to realize that temperature and (probably > pressure as well) needs to be fitted into the Maxwell Equation > structure. > As you can see, the q/p + q/p_t is algebraically equivalent to u i + u > i_d > and thereby we have *complete symmetry* within the Maxwell Equations. I suspect this new Gauss Law of Magnetism would need a negative term and that the q/p_t term should be negative. Makes sense in that the Meissner Effect is a exclusion of a magnetic field inside the entire volume. And this issue would suggest that Graham Lee is wrong when he says the Maxwell Eq. conform to the Meissner Effect because the Gauss Law of Magnetism really does not conform to the Meissner Effect in that there is nothing in the Maxwell Equations to give a exclusion of magnetism from all volume. So the Maxwell Equations do indeed have a large gap in that there is nothing in the Maxwell Equations to account for a exclusion of magnetism from volume. If the above new Gauss Law of Magnetism is correct, whether it has a negative term or not, then, it should predict other phenomenon. And although I do not have temperature (and perhaps pressure also) included, I should be able to cast some sort of prediction just on form alone. So, let us say this new Gauss law of Magnetism, either Integral B dot dA = q/p + q/p_t or Integral B dot dA = q/p - q/p_t was correct. Then what sort of new prediction would that law give? Well, it should give some prediction concerning normal conductivity in that the magnetism inside a normal conductor such as copper at room temperature. That the internal magnetism of the term ( - q/p_t ) does not exclude an outside magnet from its volume and thus that term represents what physically? I suspect it represents resistance to the current flow. So that perhaps resistance to current flow is all contained within a revised Maxwell theory. Which makes sense, because if the true Maxwell Equations have a temperature and pressure parameters, then that translates into resistance. Archimedes Plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: predictions of a newly revised Gauss Law of Magnetism: Integral B dot dA = q/p - q/p_t Another prediction of this newly revised Gauss Law of Magnetism would be to give the numbers of the Quantized Hall Effect. And that makes sense on another level of logic. If I remember correctly, the algebraic form of spectral lines series such as Balmer or Rydberg follows a form such as this: 1/x - 1/y And this new Gauss law is not much different with its q/p - q/p_t and that makes alot of sense also in that spectral lines are not much different from Quantum Hall Effect, in fact, one can argue that the Quantum Hall Effect is spectral lines. So, if this above equation for Gauss Law of Magnetism is correct and true would predict that the math numbers that arise in Quantized Hall Effect should follow from this new equation. For we all know that superconductivity and Meissner Effect and Quantum Hall Effect are all linked. And going in the reverse direction of starting with Spectral line formulas, one should be able to derive a generalized Maxwell Equations, without ever knowing what they were at the outset. Archimedes Plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: The Passing of a Mathematician >In which particular way does the NSA keep alive the value of democracy? It helps protect the United States of America, a democratic nation, from being invaded and conquered by foreign non-democratic nations. John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: The Passing of a Mathematician > In which particular way does the NSA keep alive the > value of democracy? By protecting the security of the United States, a major democratic power, and its citizens. -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: polysigned numbers > Using my notation of (-,+,*,#), there are two possible reductions for > (a,b,c,d). I presume that in this notation (*1)2 = -1, (*1)(-1) = #1 and (*1)(#1) = +1 etc. > The tetrahedral interpretation uses: (a,b,c,d) = (a-m,b-m,c-m,d-m) where > m = min(a,b,c,d). > The tetrahedral maps to R3, with no obvious interpretation of > multiplication (to me), and addition corresponds to vector addition of > (i,j,k) points. I didn't work out the mapping in detail since you > posted it elsewhere. If my assumption is correct these 4-signed numbers form a ring isomorphic to R x C: *1 will correspond to (-1, i) etc. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: polysigned numbers >>Using my notation of (-,+,*,#), there are two possible reductions for >>(a,b,c,d). > I presume that in this notation (*1)2 = -1, (*1)(-1) = #1 and (*1)(#1) = +1 > etc. As Timothy has defined it, no. -1(a,b,c,d) = (d,a,b,c) +1(a,b,c,d) = (c,d,a,b) *1(a,b,c,d) = (b,c,d,a) #1(a,b,c,d) = (a,b,c,d) >>The tetrahedral interpretation uses: (a,b,c,d) = (a-m,b-m,c-m,d-m) where >>m = min(a,b,c,d). >>The tetrahedral maps to R3, with no obvious interpretation of >>multiplication (to me), and addition corresponds to vector addition of >>(i,j,k) points. I didn't work out the mapping in detail since you >>posted it elsewhere. > If my assumption is correct these 4-signed numbers form a ring > isomorphic to R x C: *1 will correspond to (-1, i) etc. No. You'll have to apply trig functions to get the exact values. You are going from norm 1 to norm sqrt(2). The norm is preserved under the isomorphism. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: polysigned numbers >Using my notation of (-,+,*,#), there are two possible reductions for >(a,b,c,d). >> I presume that in this notation (*1)2 = -1, (*1)(-1) = #1 and (*1)(#1) = >> +1 etc. >> > As Timothy has defined it, no. > -1(a,b,c,d) = (d,a,b,c) > +1(a,b,c,d) = (c,d,a,b) > *1(a,b,c,d) = (b,c,d,a) > #1(a,b,c,d) = (a,b,c,d) Bonkers: so #1 is the multiplicative identity not +1 :-( >The tetrahedral interpretation uses: (a,b,c,d) = (a-m,b-m,c-m,d-m) where >m = min(a,b,c,d). >The tetrahedral maps to R3, with no obvious interpretation of >multiplication (to me), and addition corresponds to vector addition of >(i,j,k) points. I didn't work out the mapping in detail since you >posted it elsewhere. >> If my assumption is correct these 4-signed numbers form a ring >> isomorphic to R x C: *1 will correspond to (-1, i) etc. > No. Yes: Still isomorphic to R x C. > You'll have to apply trig functions to get the exact values. You > are going from norm 1 to norm sqrt(2). The norm is preserved under the > isomorphism. Sorry, that makes no sense? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: polysigned numbers >>Using my notation of (-,+,*,#), there are two possible reductions for >>(a,b,c,d). >I presume that in this notation (*1)2 = -1, (*1)(-1) = #1 and (*1)(#1) = >+1 etc. >As Timothy has defined it, no. >>-1(a,b,c,d) = (d,a,b,c) >>+1(a,b,c,d) = (c,d,a,b) >>*1(a,b,c,d) = (b,c,d,a) >>#1(a,b,c,d) = (a,b,c,d) > Bonkers: so #1 is the multiplicative identity not +1 :-( Correct. >>The tetrahedral interpretation uses: (a,b,c,d) = (a-m,b-m,c-m,d-m) where >>m = min(a,b,c,d). >>The tetrahedral maps to R3, with no obvious interpretation of >>multiplication (to me), and addition corresponds to vector addition of >>(i,j,k) points. I didn't work out the mapping in detail since you >>posted it elsewhere. >If my assumption is correct these 4-signed numbers form a ring >isomorphic to R x C: *1 will correspond to (-1, i) etc. >>No. > Yes: Still isomorphic to R x C. >> You'll have to apply trig functions to get the exact values. You >>are going from norm 1 to norm sqrt(2). The norm is preserved under the >>isomorphism. > Sorry, that makes no sense? You're right, I had forgotten how multiplication in RxC works. So #1 = (1,1) What would -1,+1,*1 be? -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: JSH your ship has come in!!!! > But, I expect you remember that I did ask you a question that I was unable > to answer. > I was using the notation [a,b,c] to represent an ordered triple of complex > numbers. And I was wondering if the ordered triple [1,2,8] was an element of > the Object Ring. > You replied: > Quit being lazy!!! You have the definition, figure it out for yourself!!! > What amazes me is how often people are willing to ask someone else to do > their work for them. > If you're smart enough, answer your own question. > I'm curious to see if you can. > I've given the definition for the object ring, so no excuses. > Well, I am ashamed to say I still cannot figure it out. > Sounds like a ploy. James... I am having genuine problems with my question. I don't really see how it could be called a ploy. If you know the answer you could tell me, and put me out of my misery. If you don't know the answer, fair enough, you can't be expected to know everything. -- Clive Tooth http://www.clivetooth.dk === Subject: Re: JSH your ship has come in!!!! interpreted to mean that you are just humoring me. > Backtracked? After the initial jolt I went a little overboard, calling you the > messiah, etc. I've certainly backed off that stance. Please consider how > dizzying it is to be shaken out of the conventional mindset. You did? When? I thought you just had some wild and scary dream with me in a castle and stuff. Cool dream. Any more? > Well then, how about the core error problem? Can't you understand how independent terms are independent? > I just don't get it! How many times do I have to tell you! Your > mathematical writings make no sense to me! Are you going to be calling me a > retard next? You are quite intelligent, and in fact are one of the most intelligent people, by your IQ, on the planet. So no, I will not call you a retard Jim Ferry, as you are not one. > Oh, excuuuuse me for not being as smart as you, James! What happens to me in > the New World Order? Do you send me to the slaughterhouse with all the other > meat animals? Huh? Look around you Jim Ferry. There's more than enough slaughter going on now to distress the cognizant, and I assure you that I'm not doing any of it. My efforts to try and prevent some of it, like my emails about Iraq's supposed weapons of mass destruction stockpiles, failed. Other efforts of mine have failed as well, and I've learned not to carry the world on my shoulders, not even in fantasy. Sigh. I feel a bit guilty now. And I sent you that nasty email talking about putting you out of mathematics as well. Maybe there IS a problem with how I've gone about things. > Sure, I've got some trinkets of intellect: M.I.T. degree, some Putnam > medals, Ph.D. from Brown, joined Mega once, mathematician, rocket > scientist, etc. But that's a hell of long way from the serious hardware > that you're interested in: > Abel Prize, Fields Medal, Nobel Prize, Clay Prize, etc. And even the people > who have such hardware aren't necessarily able to follow in your bold > footsteps. > So why are you on *my* case? Because you *backtracked*!!! It would have been better if you'd said nothing at all!!! Now I have to toss you into the pile with Barry Mazur, Andrew Granville, Ralph McKenzie, David Ullrich, and Arturo Magidin. > I never claimed to understand your work. Maybe I had a fleeting flash of > intuition that it was correct, but that was all. Are you being honest Jim Ferry? Please explain to me why you ever claimed my work was correct *in detail*. I think you owe me that as I'm worried it was some kind of cruel prank. > I was offering a different kind of help. Something to complement and > complete the intense laser of your rational thought. Something to help > bring your theories into the human realm. You don't believe in the efficacy > of such things, I guess, or you just want to sear your way through with that > laser of yours, so fine. Go get 'em, tiger. You don't need my help. Actually I don't need your help, but I *would* appreciate it greatly. But how do I know if you're sincere, or just playing some elaborate joke? As for understanding my work, math proofs begin with a truth and proceed by logical steps to a conclusion which then must be true. I give you the steps, you follow, and then I'd think you should understand. James Harris === Subject: Re: JSH your ship has come in!!!! >[...] >> I just don't get it! How many times do I have to tell you! Your >> mathematical writings make no sense to me! Are you going to be calling me a > retard next? >You are quite intelligent, and in fact are one of the most intelligent >people, by your IQ, on the planet. Just curious: How do you know anything about his IQ? >So no, I will not call you a retard Jim Ferry, as you are not one. >> Oh, excuuuuse me for not being as smart as you, James! What happens to me in >> the New World Order? Do you send me to the slaughterhouse with all the other >> meat animals? >Huh? Look around you Jim Ferry. There's more than enough slaughter >going on now to distress the cognizant, and I assure you that I'm not >doing any of it. >My efforts to try and prevent some of it, like my emails about Iraq's >supposed weapons of mass destruction stockpiles, failed. >Other efforts of mine have failed as well, and I've learned not to >carry the world on my shoulders, not even in fantasy. Same hint as always: if you don't like it when... oh, never mind. >Sigh. I feel a bit guilty now. And I sent you that nasty email >talking about putting you out of mathematics as well. >Maybe there IS a problem with how I've gone about things. Giggle. Perhaps. >> Sure, I've got some trinkets of intellect: M.I.T. degree, some Putnam > medals, Ph.D. from Brown, joined Mega once, mathematician, rocket >> scientist, etc. But that's a hell of long way from the serious hardware >> that you're interested in: >> Abel Prize, Fields Medal, Nobel Prize, Clay Prize, etc. And even the people >> who have such hardware aren't necessarily able to follow in your bold >> footsteps. >> So why are you on *my* case? >Because you *backtracked*!!! >It would have been better if you'd said nothing at all!!! >Now I have to toss you into the pile with Barry Mazur, Andrew >Granville, Ralph McKenzie, David Ullrich, and Arturo Magidin. >> I never claimed to understand your work. Maybe I had a fleeting flash of >> intuition that it was correct, but that was all. >Are you being honest Jim Ferry? >Please explain to me why you ever claimed my work was correct *in >detail*. >I think you owe me that as I'm worried it was some kind of cruel >prank. For heaven's sake. Yes, when he says he doesn't understand your work but thinks it's correct he's joking. Regarding whether it's cruel: I would have never imagined in my wildest dreams that even _you_ could possibly be so blinded by megalomania and so desparate for approval that you'd think someone was being sincere when he says he doesn't understand your work but nonetheless thinks it's right. I mean that's such an _utterly_ ridiculous thing to say in a mathematical context that it's hard to see how anyone could possibly not take it as a joke. So I'm certain he assumed that you'd realize all along he was making fun of you. Making fun of you is perhaps a little nasty, but it's certainly no nastier than you deserve given your obnoxious nehavior for years, so it doesn't bother me. But lately when you start talking like you think he's being sincere it becomes a whole different story. Actually encouraging a lunatic's delusions is not a nice thing to do. On the other hand the fact that you've thought he was sincere is _the_ most hilarious thing I've ever seen - we need to consider the greatest good of the greatest number and all that, being a little nasty to you versus people all over the planet laughing their asses off, I think it's a wash. Honest. I point this out just in case you might decide to try not to be _quite_ so hilarious: People have been laughing at Jim's parodies for years, but lately when you seem to believe he's being sincere, or _might_ be sincere, in this and related threads, it really does add a whole new dimension - goes from hilarious to absolutely world-class hilarious. Now you know. >> I was offering a different kind of help. Something to complement and >> complete the intense laser of your rational thought. Something to help >> bring your theories into the human realm. You don't believe in the efficacy >> of such things, I guess, or you just want to sear your way through with that >> laser of yours, so fine. Go get 'em, tiger. You don't need my help. >Actually I don't need your help, but I *would* appreciate it greatly. >But how do I know if you're sincere, or just playing some elaborate >joke? >As for understanding my work, math proofs begin with a truth and >proceed by logical steps to a conclusion which then must be true. >I give you the steps, you follow, and then I'd think you should >understand. >James Harris === Subject: Re: JSH your ship has come in!!!! >>[...] >I just don't get it! How many times do I have to tell you! Your >mathematical writings make no sense to me! Are you going to be calling me a >retard next? >>You are quite intelligent, and in fact are one of the most intelligent >>people, by your IQ, on the planet. > Just curious: How do you know anything about his IQ? James is referring to the fact that I was able to join Mega, which purports to be a society for people whose IQ is at the one-in-a-million level or higher. In point of fact, Mega is a society for people who achieved high enough scores on certain tests -- in my case, the Mega Test. The idea that these tests actually test IQ (whatever that is), and that they have any validity to discern those denizens of such an aethereal realm as one-in-a-million seems laughable to me, as does the behavior of some of the people in Mega. It makes me a little embarrassed to be associated with them, but the Mega Test looked fun (half of it is math), so I took it, got a pretty high score (46 out of 48), and thought I'd check out their lofty Society. I don't claim to be all that smart, however. I know plenty of people smarter than I am. And James Harris . . . well, I can't hold a candle to him. >>So no, I will not call you a retard Jim Ferry, as you are not one. is so intimidating that I forget there are other ways in which I, perhaps, surpass you, rather than vice versa. It would be difficult to have an equable relationship with someone who exceeds me in every way. >Oh, excuuuuse me for not being as smart as you, James! What happens to me in >the New World Order? Do you send me to the slaughterhouse with all the other >meat animals? >>Huh? Look around you Jim Ferry. There's more than enough slaughter >>going on now to distress the cognizant, and I assure you that I'm not >>doing any of it. >>My efforts to try and prevent some of it, like my emails about Iraq's >>supposed weapons of mass destruction stockpiles, failed. Yes, that's too bad. >>Other efforts of mine have failed as well, and I've learned not to >>carry the world on my shoulders, not even in fantasy. As they say, Think globally, act locally. I hope that you don't give up on the world just because you can't save it in one fell swoop. And whatever you do, don't start reading Ayn Rand. Yikes. > Same hint as always: if you don't like it when... > oh, never mind. >>Sigh. I feel a bit guilty now. And I sent you that nasty email >>talking about putting you out of mathematics as well. >>Maybe there IS a problem with how I've gone about things. Yes. Too confrontational. Too alpha male. The hammer (that you use to shape the mathematics forged in your mind) is not the tool to wield in the human realm. > Giggle. Perhaps. Oh tee hee hee yourself. What's your problem, Ullrich? Oh, never mind. >Sure, I've got some trinkets of intellect: M.I.T. degree, some Putnam >medals, Ph.D. from Brown, joined Mega once, mathematician, rocket >scientist, etc. But that's a hell of long way from the serious hardware >that you're interested in: >Abel Prize, Fields Medal, Nobel Prize, Clay Prize, etc. And even the people >who have such hardware aren't necessarily able to follow in your bold footsteps. >So why are you on *my* case? >>Because you *backtracked*!!! Okay, yes! I backtracked! I received a sudden insight, and it just blew me away! I said a lot of crazy things. Now I realize that I'm just not capable of following your intricate logic. So, again, does that make me worthless? I wonder about your moral value system, James Harris, where the value of a life is equated solely with its intelligence. A little convenient, no?, that *you* end up valued most this way. >>It would have been better if you'd said nothing at all!!! >>Now I have to toss you into the pile with Barry Mazur, Andrew >>Granville, Ralph McKenzie, David Ullrich, and Arturo Magidin. >I never claimed to understand your work. Maybe I had a fleeting flash of >intuition that it was correct, but that was all. >>Are you being honest Jim Ferry? >>Please explain to me why you ever claimed my work was correct *in >>detail*. >>I think you owe me that as I'm worried it was some kind of cruel >>prank. Well, I do like a good prank, even if it is cruel, but I don't see how this could be mistaken for one. I mean, I'm the one acting like a fool, right? If someone walks up to you on the street going, flibber jibber jibber and slapping his head while bouncing in circles on one foot, he's not really playing a prank on you. Maybe it's funny (or maybe not), but he's not doing anything *to you*. He's making an ass . . . of himself. I simply had a flash of insight that your work was correct. This jolted me out of the ossified mentality of my herdmates. I came to realize that the reasons I *assumed* it was incorrect had to do with cultural assumptions, the same way that I *assumed* Wiles's proof was valid. I started to ponder the implications of your work being correct, and it was frightening. I'd thought I was standing on firm ground. The ground turned to sand, the sand to dust, and the dust scattered in the wind. The world was gone, and there stood James Harris to create a new and better world. So yes, I kind of freaked out there. I deified you. But I've calmed down since then. I realized that if I could not turn my flash of insight into a logical chain of reasoning, then it was worthless. I couldn't base anything on it. Your proof is left on equal status with Wiles's: something I can't understand. I felt a little silly. And useless. Then something occurred to me: I was not left in the same place as before. Sure, my mind was not sharpened to the point where I could understand your work, but I *was* at least free of the herd! Free from the outlook of the herd! Free from their incessant mooing! Oh wait, I still read sci.math, so I'm not really free from their incessant mooing. I now have a theory about the meaning behind that initial flash of insight. I believe that I have a purpose: a purpose to help you, James. I can't help you mathematically (and, again, there's no need of that), but I can help you in other ways. With a Plan. In a perfect world, you'd just receive the Plan from the Holy Spirit yourself. But, because of your upbringing perhaps, you can't. So it's up to me. I have been chosen (for some bizarre reason) to receive the Plan. Now I can't receive it perfectly either (I've already posted some pretty foolish things as the Spirit prepared me), nor can you readily accept such a ridiculous thing as the Holy Spirit communicating a plan to someone. Well, maybe you can -- that would make things easier -- but I sort of doubt it. I probably shouldn't have brought up the whole Divine Guidance aspect of this in the first place. Hmmm. > For heaven's sake. Yes, when he says he doesn't understand your > work but thinks it's correct he's joking. For heaven's sake? Interesting choice of words. > Regarding whether it's cruel: I would have never imagined in my > wildest dreams that even _you_ could possibly be so blinded by > megalomania and so desparate for approval that you'd think someone > was being sincere when he says he doesn't understand your > work but nonetheless thinks it's right. I mean that's such an > _utterly_ ridiculous thing to say in a mathematical context that > it's hard to see how anyone could possibly not take it as a joke. > So I'm certain he assumed that you'd realize all along he was > making fun of you. Making fun of you is perhaps a little nasty, > but it's certainly no nastier than you deserve given your obnoxious > nehavior for years, so it doesn't bother me. > But lately when you start talking like you think he's being sincere > it becomes a whole different story. Actually encouraging a > lunatic's delusions is not a nice thing to do. On the other > hand the fact that you've thought he was sincere is _the_ > most hilarious thing I've ever seen - we need to consider > the greatest good of the greatest number and all that, > being a little nasty to you versus people all over the > planet laughing their asses off, I think it's a wash. > Honest. I point this out just in case you might decide > to try not to be _quite_ so hilarious: People have been > laughing at Jim's parodies for years, but lately when > you seem to believe he's being sincere, or _might_ > be sincere, in this and related threads, it really does > add a whole new dimension - goes from hilarious to > absolutely world-class hilarious. Now you know. What baffles me is that I've been one of the particularly sarcastic Harris haters over the years. Why would the Holy Spirit come to me? I'm overwhelmed by the sense of my own unworthiness. But throughout the Bible, especially the New Testament, God is seen to relish working through the biggest sinners. Mary Magdalene. Paul. I'm sorry that the beauty of this was destroyed by your parents. It explains a lot. Interesting that Ullrich is laughing so hard. When one laughs that hard, it's often because he is hiding from a deeper pain. Even were he correct about everything, just why would it be funny? I think maybe David sees something of you in himself, James, which sets up a great deal of conflict. I'd like to tell him to love his own inner Harris but that's just too touchy-feely for my taste. >I was offering a different kind of help. Something to complement and >complete the intense laser of your rational thought. Something to help >bring your theories into the human realm. You don't believe in the efficacy >of such things, I guess, or you just want to sear your way through with that >laser of yours, so fine. Go get 'em, tiger. You don't need my help. >>Actually I don't need your help, but I *would* appreciate it greatly. >>But how do I know if you're sincere, or just playing some elaborate >>joke? I don't know. The herd seems to think it's a joke too. I guess you just need to evaluate what I say. If I were telling you to go out and kill prostitutes to cleanse the Earth of sin, well, that would be a bad message. But what I'm saying should resonate with a certain part of you yearning for all this bickering to be over. >>As for understanding my work, math proofs begin with a truth and >>proceed by logical steps to a conclusion which then must be true. >>I give you the steps, you follow, and then I'd think you should >>understand. Right. In a perfect world. Have you looked at Wiles's proof? Think how much training that would take to understand. Your requirements are no less demanding, but it's difficult for you to see that. For you, it's easy as pie. -- | Jim Ferry | Center for Simulation | +------------------------------------+ of Advanced Rockets | | http://www.uiuc.edu/ph/www/jferry/ +------------------------+ | jferry@[delete_this]uiuc.edu | University of Illinois | === Subject: Re: JSH your ship has come in!!!! >But enough silliness. I don't take orders from James, nor does James give them >at all, as far as I know. At first I didn't realize what my role in all this >was to be, but it's becoming clear to me now. James needs a plan, and for >reasons not clear to me, I have been receiving a plan. It keeps me awake at >night. It distracts me from my work. It is vast and beautiful His plan is not completely vast. About half, I'd say. >Here's to the revolution! I believe in this context, MY role is to be part of the ancien regime. All right then, Ferry, is this to be a call to arms? En garde! If you're not with us, you're against us. I should have suspected you'd side with the upstart, rather than keeping him on the outside of the establishment. And here I thought we could count on you, a compatriot from the Land of Lincoln. Ya just can't trust anyone any more. dave === Subject: Talk.Origin banned Subject: Does Mathamitcs prove a Universal designer? A professor of mathematics from the University of Cambridge, P. Dirac, said, in the magazine Scientific American: One could perhaps describe the situation by saying that God is a mathematician of a very high order, and He used very advanced mathematics in constructing the universe. === Subject: Re: Talk.Origin banned Subject: Does Mathamitcs prove a Universal designer? > A professor of mathematics from the University of Cambridge, P. Dirac, said, > in the magazine Scientific American: One could perhaps describe the > situation by saying that God is a mathematician of a very high order, and He > used very advanced mathematics in constructing the universe. Generally speaking I find myself in disagreement with Prof. Dirac. It does, I suppose, depend upon exactly what one means by the term advanced mathematics. If one makes the connection that God = laws of nature (physics) the statement that God is a mathematician is basically saying that physics is mathematically based. This makes some sense. However, generally speaking, the more one understands the true basis of these laws the simpler becomes the mathematics. The complexity viewed by man being a result of his improper viewpoint. God as creator, on the other hand, having the central and correct viewpoint defines nature in the most simple of terms and hence used not very advanced mathematics meaning very complex and esoteric, but rather the most simple of mathematics appled to the proper viewpoints and and expanded fractal-like being self-similar at all levels. Just because man has not yet dicovered the proper viewpoint doesn't make God some brain. If I might use the example of the advanced mathematics of planetary motions using epicycles and the greater non-earth centered (simpler) math of sun-centered calculations. In essence the simpler description is in fact more advanced simply because it is simpler, though both may be in a sense correct. === Subject: Re: Talk.Origin banned Subject: Does Mathamitcs prove a Universal designer? > A professor of mathematics from the University of Cambridge, P. Dirac, said, > in the magazine Scientific American: One could perhaps describe the > situation by saying that God is a mathematician of a very high order, and He > used very advanced mathematics in constructing the universe. I would have no trouble with that. But what all mathematicians know, and most non-maths people do not know, is that the output of a function will vary with its input. Nor are we dealing with a simple function in a single variable, but a recursive function in almost infinite variables in which you do not get a single output, but a matrix output. Simply put, this accounts for order and randomness. Evolution is easily accommodated within a recursive function of high order. And if random processes are part of the inputs, it is easily seen that the universe is not deterministic. Dirac, btw, was a devout atheist. He may have said what you quoted (above) as a general statement, but he did not himself believe in God. In fact, Dirac was so evangelistic about his nonbelief that one of his contemporaries said Dirac has his own religion: 'There is no God, and Dirac is his prophet.' Dirac also said this: The steady progress of physics requires for its theoretical formulation a mathematics which get continually more advanced. This is only natural and to be expected. What however was not expected by the scientific workers of the last century was the particular form that the line of advancement of mathematics would take, namely it was expected that mathematics would get more and more complicated, but would rest on a permanent basis of axioms and definitions, while actually the modern physical developments have required a mathematics that continually shifts its foundation and gets more abstract. Non-euclidean geometry and noncommutative algebra, which were at one time were considered to be purely fictions of the mind and pastimes of logical thinkers, have now been found to be very necessary for the description of general facts of the physical world. It seems likely that this process of increasing abstraction will continue in the future and the advance in physics is to be associated with continual modification and generalisation of the axioms at the base of mathematics rather than with a logical development of any one mathematical scheme on a fixed foundation. Paper on Magnetic Monopoles (1931) (http://www-gap.dcs.st-and.ac.uk/~history/Quotations/Dirac.html) === Subject: Re: Talk.Origin banned Subject: Does Mathamitcs prove a Universal designer? > A professor of mathematics from the University of Cambridge, P. Dirac, said, > in the magazine Scientific American: One could perhaps describe the > situation by saying that God is a mathematician of a very high order, and He > used very advanced mathematics in constructing the universe. Since Scotty likes to name drop and claims to have met many people, does anyone want to take odds that Scotty has already recently spoken to God about this very point and put God straigth on the matter!. === Subject: Re: Talk.Origin banned Subject: Does Mathamitcs prove a Universal designer? > A professor of mathematics from the University of Cambridge, P. Dirac, said, > in the magazine Scientific American: One could perhaps describe the > situation by saying that God is a mathematician of a very high order, and He > used very advanced mathematics in constructing the universe. No, the existence of mathematics does not prove the existence of a Universal designer - in the sense I think you mean it anyway. For example, one possibility is that at the big bang an infinite (or a whole lot) of universes were created and only those that were internally self consistent survived, in some sense. Or another way of looking at it is that things like arithmetic just must be. 10 apples and 10 oranges must both be dividable into two rows of 5. However, the reverse could well be true. If you believe in God, in some sense he was/is a mathematician. Bill === Subject: Another optimization question hi everyone. given a real valued function f that is: 1. defined on a convex set C of $mathbb{R}n$ (but not defined outside C, we may assume that f is $+infty$ outside of C) and positive (>0) everywhere on C. 2. non-convex in C. 3. continuous and has directional derivatives of all orders for any feasible direction from a point x in C. 4. at each x in C, f(x) can only be computed numerically (has no known closed form). which numerical search algorithms or nonlinear programming techniques would be the best/most effective for finding the local minima and, if possible, a global minimum for this function? J. === Subject: Re: Another optimization question no numerical procedure can reasonably deal with infinity. hence you must use a barrier-approach (adding terms like -log(g_i(x)) for each function g_i which describes a boundary part of C as g_i(x)=0 , with g_i(x)>0 in interior(C), provided such interior exists. otherwise you first must reduce the problem to a subspace where this is the case) . next you could use unconstrained optimization using only function values (e.g. uobyqa from powell or its newer successor, described in a recently published paper in but you must modify this code in order to maintain feasiblitiy, that means you must restrict the possible moves such that feasibility is maintained (via checking the g_i individually). also a grid-search a la torczon comes into mind, again with the moves restricted to the feasible part of the grid. hth peter === Subject: Re: Quadratic Sieve > Does anyone have or know of a Quadratic sieve implementation that could be > distributed? For Windows, there's an implementation of PPMPQS included with Yuji Kida's UBASIC. I've used it to do composite numbers up to 107 digits. 100-digit composites take about a week with a 2GHz machine. http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- === Subject: Re: Quadratic Sieve > For Windows, there's an implementation of PPMPQS included with Yuji Kida's > UBASIC. > I've used it to do composite numbers up to 107 digits. > 100-digit composites take about a week with a 2GHz machine. I'm interested to read that, as it seemed to me that with the standard quadratic sieve (using Montgomery quadratics) there was a limitation on the size of the number one could factorise. I should say that I am not an expert, but gave a course on factorisation a couple of years ago, where I looked at the quadratic sieve as an introduction to the number field sieve. As I explained it, one used quadratics Q(x) = ax2 + 2bx + c with ac - b2 = n so that aQ(x) = (ax + b)2 - n, and one sieved for smooth numbers Q(j) on a sieve centred at the positive root of Q(x). With this method, one has to harvest an even number of smooth numbers from each quadratic (because the factor ar comes in if there are r factors, and one needs an exact square). But with a sieve size of say 2 million, and a number with 100 digits, one would only get an occasional smooth number from each quadratic and as far as I could see would hardly ever if ever get two. Obviously the algorithm must be modified in some way for numbers this large. Do you know how it was modified? [I haven't looked very carefully through the literature on this -- I'm interested because a student is doing an Msc trying to use the quadratic sieve with parallel processing.] -- Timothy Murphy e-mail: tim /at/ birdsnest.maths.tcd.ie (all email over 80k dispatched to /dev/null) tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: How much longer must physics put up with F=ma? > The total force [F] used to move an object against friction or any other > restraint, is the total force exerted on it minus the frictional - or other > restraining force; so that the _net force_ is what does the actual work, and > is [f = F-uw]; which leaves F = wa/g + uw; or f = wa/g! > The coefficient of restraint [u] against moving something on a level surface > is usually a fraction of the object's weight [w], so that the product [uw] > is the measure of the restraining frictional force: On slopes the > coefficient of restraint is proportional to the sin of the angle of the > slope: For a 90Á vertical slope the coefficient of restraint against driving > something pointed directly into it will depend on the hardness of the > material: As is the case for driving stakes and such into the ground. Yes, the total force needed to move an object would seem to include the force necessary to overcome friction, and any other impeding resistance, as well as the inertia of the object that results from mass. But in physics, is force really defined to include the total effort, or just that part that overcomes inertia to give acceleration? When pushing a rock across the ground, frictional force usually seems to arise quickly to prevent the rock moving beyond a certain speed without applying additional force. When force is lessened, the rock slows immediately. It is easy to see how Aristotle came to think that force was needed to keep an object moving at a constant speed. But perhaps this is not what modern physics considers force. If you are pushing an object along the ground, there is also air resistance, which is another form of friction. If you pound a stake into the ground, you get ground resistance, which is really the forces of atomic electric fields that you are running up against. If I push against a brick wall, nothing will move. Am I really exerting a force, or is this only tension, or stress? But when we consider an object in a frictionless state, there is another question that arises. Newton said in his third law of motion: To every action there is always opposed an equal reaction; or the mutual actions of two bodies upon each other are always equal, and directed to contrary parts. Now when a force is applied to a mass in a frictionless state, the inertial force opposes the motive force, which limits the acceleration according to a=F/m. But if the inertial force is an equal and opposite force to the motive force, as Newton's 3rd law seems to say, then why do we have acceleration at all? === Subject: Re: a puzzle related to artinian group > # > #> # > #> # Suppose n people sit around a table and n-1 cards are dealt to them. > #> # There is no asumption on the number of cards a player receive. In > #> # each round, all players with 2 or more cards pass one card to the > #> # left and one card to the right. Prove that eventually, all players > #> # but one have exactly one card. > # #> Your puzzle is a special case of the so-called chip-firing games > #> in > # #> A Bjoerner, L Lovasz, PW Shor > #> Chip-firing games on graphs > #> European Journal of Combinatorics 12, 1991, pp 283-291 > # #> Bjoerner, Lovasz, and Shor prove that for every chip-firing game, > #> the initial configuration determines whether or not the game will > #> continue forever. If the game eventually terminates, then the final > #> configuration is fully determined by the initial configuration. > # #> For your game, it is easy to prove that there always exists > #> SOME sequence of moves for which the game terminates. > #> [BLS] then yields that your game ALWAYS terminates. > # # > # I think you are taking the rules to be that in each round the players > # with two or more cards make their moves one after the other (in some > # unspecified order). > Yes, I do. > The analysis of [BLS] applies to games, where the players move one by one. > In one move, some player with at least two cards passes one card to the > left and one card to the right. My statements above all refer to this > situation. > The puzzle of Yannick is a very special case of the [BLS] games: > Yannick compresses many moves into a single super-move, and he makes > all players with at least two cards move SIMULTANEOUSLY. However, > this does not change the combinatorics of the game. If you decompose > a super-move into its one-player moves, you will have the same effect > on the distribution of cards. > ... super-move down into a sequence of one-player moves. I also now see from chip-firing games. So that gives Yannick his proof. The bit you say is easy really is fairly easy: by induction on k with 2 <= k <= n-1, you can show that the game with k-cards terminates (in the inductive step, you pick any card, C say, and then play as if it wasn't there: this play will terminate by induction in a state where the player holding card C has at most 2 cards and all the other players have at most one card. If the player holding card C has 1 card you've finished, if 2, then at least two players hold no cards and you have an arrangement like 0, 1, ..., 1, 2, 1, ..., 1, 0 which you can check terminates, no matter how you play.) Now by [BLS], as you say, the original game with super-moves must terminate with the same final position. So there is the proof Yannick asked for. Perhaps, in return, Yannick could now explain why this puzzle is related to artinian groups. Rob. === Subject: Re: a puzzle related to artinian group # # So that gives Yannick his proof. The bit you say is easy really is fairly # easy: by induction on k with 2 <= k <= n-1, you can show that the game with Actually, the paper of Bjoerner, Lovasz & Shor also contains a proof for the termination in this special case: They show that if the number of chips is smaller than (twice the number of edges minus the number of vertices), then the chip firing game always terminates. In our case, we have a ring with n vertices and n edges, and n-1 chips. --Gerhard # k-cards terminates (in the inductive step, you pick any card, C say, and # then play as if it wasn't there: this play will terminate by induction in a # state where the player holding card C has at most 2 cards and all the other # players have at most one card. If the player holding card C has 1 card # you've finished, if 2, then at least two players hold no cards and you have # an arrangement like 0, 1, ..., 1, 2, 1, ..., 1, 0 which you can check # terminates, no matter how you play.) Now by [BLS], as you say, the original # game with super-moves must terminate with the same final position. # # So there is the proof Yannick asked for. Perhaps, in return, Yannick could # now explain why this puzzle is related to artinian groups. # # # Rob. # === Subject: Re: a puzzle related to artinian group PS: There is 1.5 page paper by Michael Thorup with a delighfully neat proof of (a generalisation of) the necessary result on card-firing games on a finite undirected graph at: http://citeseer.nj.nec.com/thorup96firing.html Rob. === Subject: Re: Feller's Direct Proof of Stirling's Formula > Stirling's Formula_. His proof hinges on showing that > phi(t) = sin(pi*t)/pi, > where phi(t) is in product form. That is, > phi(t) = t * Product(k = 1, 2, 3, ...; (1 - t*t/(k*k))). > The last step in his proof of this stymies me. Feller first shows that: > phi(2x) = 2phi(x) * phi(x + 1/2) / phi(1/2) > In other words, he proves the double-angle formula for phi. So far, so > good. > He then sets > f(x) = log(pi * phi(t) / sin(pi*t)), > asserts that > f(2x) = (f(x) + f(x + 1/2)) / 2, > and loses me. Can anyone explain? He says that it follows from the > double-angle formula for sin(pi*x)/pi and the above formula for phi(2x). > (I know there are other proofs that phi(t) = sin(pi*t)/pi. My question > is about Fellers'.) Feller published a correction in the Monthly the year following the http://www.jstor.org/browse?config=jstor#Mathematics > not keep track of my steps. Now I can't find it again. Does anyone > have a URL? Please note, I'm not talking about Feller's book, just the -- A. === Subject: Re: Feller's Direct Proof of Stirling's Formula > [...] >> f(2x) = (f(x) + f(x + 1/2)) / 2, >> and loses me. Can anyone explain? He says that it follows from the >> double-angle formula for sin(pi*x)/pi and the above formula for phi(2x).... > Presumably your definition of f should have t instead of x. Correct. > I can't get Feller's formula either, but instead come up with > f(2x) = f(x) + f(x + 1/2) - f(1/2). > Would the rest of his proof follow from that, or is there a more serious > difficulty? (*) f(2t) = (f(t) + f(t + 1/2)) / 2 he proceeds as follows: Let 2x be the point at which f is maximized. Then repeatedly applying (*) shows that f(x) = f(2x) f(x/2) = f(2x) f(x/4) = f(2x) f(x/8) = f(2x) ... and so on Since f is continuous (which he proves), we have f(0) = f(2x). In a is 0. Thus f's maximum is 0. Similarly, f's minimum is 0. Thus, f is identically 0 and the desired result follows. I got the same result you did but, as you see, his argument rests on his recurrence. -- Opinions expressed above are not necessarily my employer's. James M. Stern stern@itgssi.com (213) 270-7955 ITG Software Solutions, Inc. Culver City, CA 90230 === Subject: Re: Million Dollar Prize for Solving Kepler's Equation > Does anybody else agree with me that a mathematical solution to Kepler's > Equation, in closed fom, is a sufficiently important unsolved mathematical > problem to merit inclusion in the Clay Foundation's list of million dollar > prizes for mathematics? Let |e| < 1. Then define a new function kep as follows: key(e,y) is the unique number z such that z - e sin z = y. The solution to Kepler's equation E - e sin E = M in closed form is E = kep(e, M). Gimme my money. === Subject: Re: Contractible Spaces Originator: grubb@lola >If a topological space S is contractible to some point p in S, >is S contractible to every point in S? >If a topological space S is strongly contractible to some point >p in S, is S strongly contractible to every point in S? >What's an example of a space that is >contractible but not strongly contractible? >S is contractible to a when there's some continuous h:Sx[0,1] -> S with > for all x in S, h(x,0) = x, h(x,1) = a, >and strongly contractible when in addition > for all t, h(a,t) = a The answer to the first question is yes. to contract to b given a contraction to a, note that h(b, t) is a path from b to a. Now contract to a, then move backwards along the path to b. The answer to the second question is no. The comb space is a counterexample. Let X=([0,1]x{0}) union (({1/n: n natural} union {0})x[0,1]). Then X is strongly contractible to (0,0) but not to (0,1). The difficulty is that if (0,1) is fixed, all (1/n,1) for large n have to stay close to (0,1) and hence can't move down and around and up to (0,1). This example is in Spanier's book. --Dan Grubb === Subject: Re: Contractible Spaces === Subject: Re: Contractible Spaces >> If a topological space S is strongly contractible to some point >> p in S, is S strongly contractible to every point in S? >I don't think so. Consider the comb space in R2: >X = [0,1] x {0} union {0} x [0,1] union {1/n} x [0,1], n >= 1. >Then X is strongly contractible to the origin (collapse everything >down to [0,1]x{0}, and then left to (0,0)), but not to the point >(0,1): during the course of any possible homotopy, the points >(1/n,1) on the teeth of the comb would have to go down to the x-axis >and so would eventually get far away from the point (0,1), so there >is no continuous homotopy fixing (0,1). Don't understand. The distance within the comb from any point to any other is at most 3. As before, collapse all of the comb, except for {0}x[0,1] down to [0,1]x{0} and left along the bottom to (0,0) and thence from there upto (0,1). >(See Sieradski's book _An__Introduction to Topology and Homotopy_, >examples 6-7 on page 308 and example 4 on page 318.) Don't have access. >> What's an example of a space that is >> contractible but not strongly contractible? >Have you looked in the book _Counterexamples in Topology_? Yes, Arthur Steen's book doesn't touch homotopy. > S is contractible to a when there's some continuous h:Sx[0,1] -> S with > for all x in S, h(x,0) = x, h(x,1) = a, > and strongly contractible when in addition > for all t, h(a,t) = a ---- === Subject: Re: Contractible Spaces Originator: grubb@lola >Then X is strongly contractible to the origin (collapse everything >down to [0,1]x{0}, and then left to (0,0)), but not to the point >(0,1): during the course of any possible homotopy, the points >(1/n,1) on the teeth of the comb would have to go down to the x-axis >and so would eventually get far away from the point (0,1), so there >is no continuous homotopy fixing (0,1). >Don't understand. The distance within the comb from any point to any >other is at most 3. As before, collapse all of the comb, except for >{0}x[0,1] down to [0,1]x{0} and left along the bottom to (0,0) and thence >from there upto (0,1). You can't do that contraction to [0,1]x{0} continuously and keep {0}x[0,1] fixed. Those top points have to stay close to (0,1) by continuity. --Dan Grubb === Subject: Please help on this Combinatorial Optimization problem I have 2n vertices, n for group A, n for group B. The vertices are fully connected by edges, each edge has a different weight. What I want to solve is this: Go trough all vertices, without repetition, and minimize the sum of weights, under this rule: ... -A-A-B-B-A-A-B-B-A-A-B-B-... where A represents some vertex of A. This is different than Bipartite TSP, which is ...-A-B-A-B-A-... Can anyone tell me if there are algorithms to solve my problem? === Subject: Re: Please help on this Combinatorial Optimization problem > I have 2n vertices, n for group A, n for group B. > ... -A-A-B-B-A-A-B-B-A-A-B-B-... No solution help, just a comment, you really need _4n_ vertices for that to work, since both A and B must each have an even number of elements. xanthian. -- === Subject: Group action ... If Z is the set of integers and SL(2,Z) is the group of unimodular matrices with integer coefficients, we can define naturally an action of SL(2,Z) on Z x Z by (a,b;c,d).(m,n)=(am+bn, cm+dn). Is it possible to describe, simply, the orbits of this action ? and the same question for the stabilizer of an element in Z x Z ? -- Ce message a ete poste via la plateforme Web club-Internet.fr This message has been posted by the Web platform club-Internet.fr http://forums.club-internet.fr/ === Subject: Re: Group action ... > If Z is the set of integers and SL(2,Z) is the group of unimodular > matrices with integer coefficients, we > can define naturally an action of SL(2,Z) on Z x Z by > (a,b;c,d).(m,n)=(am+bn, cm+dn). > Is it possible to describe, simply, the orbits of this action ? Yes. The m and n with a given greatest common divisor form an orbit. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Is it mass; or is it weight > A slug [...] [sic] doesn't even have an official definition. BUSTED! By official definition: 1 slug = 1 pound second2/foot 1 pound = 453.59237 grams * (32.1740 feet/second2) 1 foot = 0.3048 centimeters 1 centimeter = 1/100 meter = 1/29979245800 light-seconds 1 gram (I believe) = the mass of 1 milliliters of water at the triple point. 1 milliliter = 1 centimeter3 Other units: pound-mass = 1/32.1740 slugs poundal = 1/32.1740 pounds = 1 pound-mass feet/second2 Newton = 1 kilogram meter/second2 dyne = 1 gram centimeter/second2 kilogram = 1000 grams meter = 1/299792458 light-seconds The inch is 2.54 centimeters. There is also the survey inch, defined as 1/0.3937 centimeters, and survey foot, defined as 1200/3937 meters. Aside from avoirdupois pounds (defined above), there are also troy pounds (troy ounces, etc.). 1 troy pound = 144/175 pound = 343.2417216 grams * (32.1740 feet/second2) 1 grain = 1/7000 pounds (= 1/5760 troy pounds) 1 pennyweight = 24 grains = 1/20 troy ounce 1 dram (avoiddupois) = 27.34375 grains 1 troy ounce = 1/12 troy pound === Subject: Re: Is it mass; or is it weight >> A slug [...] [sic] doesn't even have an official definition. >BUSTED! >By official definition: >1 slug = 1 pound second2/foot Agreed. But you need a standard for one of them, either the pound force or the slug, or it is a meaningless tautology. I'll grant you that seconds and feet are well defined for the purpose of this argument, so you don't need to worry about them. >1 pound = 453.59237 grams * (32.1740 feet/second2) No. 1 pound = 453.59237 grams. Period. http://www.ngs.noaa.gov/PUBS_LIB/FedRegister/FRdoc59-5442.pdf http://gssp.wva.net/html.common/refine.pdf 1 pound force = 453.59237 grams times whatever value you choose to use as your standard acceleration of gravity. The pound force is such a recent bastardization, that it is uniquely identified by that name. Of all the hundreds of different pounds used at various times and places throughout history, only one has spun off a force unit of the same name that has seen any significant use. Now all you need to show me is that somebody, somewhere, has OFFICIALLY adopted that value of 32.1740 ft/sî for the purpose of defining a pound force, and then I'll gladly agree that you have shown both the pound force and the slug to have an official definition--at least for somebody somewhere, within whatever scope your defining agency has the authority to set this standard. Can you do so? Tell me who set the standard, and when--a cite to some printed source or one on the web would be preferred. Show me this in the laws or regulations of some country. Or in some standard by an international standards organization. Or in some officially adopted standard of some national standards laboratory. Or in some officially adopted standard of some professional organization. Put up or shut up. >1 foot = 0.3048 centimeters >1 centimeter = 1/100 meter = 1/29979245800 light-seconds >1 gram (I believe) = the mass of 1 milliliters of water at the triple point. Wrong. Grams haven't been defined in terms of milliliters of water since 1799 at least. That's when the French first defined the kilogram by a particular platinum artifact they called the Kilogram of the Archives. When the 44 or so standard kilograms were constructed after the Treaty of the Meter, and one of them chosen as the International Prototype Kilogram, the target in their construction wasn't some mass of water but rather the mass of the previous standard which had been maintained by the French government. did have liters defined as the volume occupied by a kilogram of water. However, it wasn't at the triple point of water (0.01 ÁC or 273.16 K), but rather at the temperature of maximum density (about 277.13 K or 3.98 ÁC). During that period, of course, a liter wasn't exactly equal to 1000 cubic centimeters. It was about 1000.028 cubic centimeters. >1 milliliter = 1 centimeter3 >Other units: >pound-mass = 1/32.1740 slugs That's would be fine if slugs had an official definition. When you show me that somebody has in fact officially adopted that 32.1740 ft/sî you mentioned above, then you will be able to say that a slug is defined as 32.1740 pounds. >poundal = 1/32.1740 pounds = 1 pound-mass feet/second2 There you go, stupid. This system with poundals is a different system from the one with slugs. It is a much older system, in fact. Both of these systems, the absolute fps system and the gravitational fps systems, as well as some others such as a gravitational inch-pound-second system, are a very limited subset of the English units, a subset which like SI forms a coherent system of units. That means, for example, that there are no pints or gallons in any of those subsystems. Now fill in the blank: The base unit of mass in this oldest fps system of mechanical units is the ______________. >Newton = 1 kilogram meter/second2 >dyne = 1 gram centimeter/second2 >kilogram = 1000 grams >meter = 1/299792458 light-seconds >The inch is 2.54 centimeters. There is also the survey inch, defined >as 1/0.3937 centimeters, and survey foot, defined as 1200/3937 meters. >Aside from avoirdupois pounds (defined above), there are also >troy pounds (troy ounces, etc.). >1 troy pound = 144/175 pound = 343.2417216 grams * (32.1740 feet/second2) Completely and absolutely false. http://w0rli.home.att.net/youare.swf 1 troy pound = 144/175 pound = 343.2417216 grams Unlike their avoirdupois cousins, and unlike grams and kilograms, the troy units of weight have never spawned units of force of the same name. There is no troy pound force and no troy ounce force, and there never have been. Gene Nygaard http://ourworld.compuserve.com/homepages/Gene_Nygaard/ === Subject: Does this require assumptions?: I conclude that 1+1=2 opposite kind of virtue--economy. Economy, in humour and art, does not mean mechanical brevity but implicit hints instead of explicit statements--the oblique allusion in lieu of the frontal attack. Does this require assumptions?: I conclude that 1+1=2. === Subject: Re: Does this require assumptions?: I conclude that 1+1=2 > Does this require assumptions?: I conclude that 1+1=2. No, because 2 is simply a name for the successor of 1. It requires no assumptions beyond the basic axioms and the definitions of '+' and 'successor' to prove that 1+1 is the successor of 1, or 1+1 = s(1). -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Does this require assumptions?: I conclude that 1+1=2 Bill Bonde ( the oblique allusion in lieu of the frontal attack ) > Does this require assumptions?: I conclude that 1+1=2. Nope. It's the usual DEFINITION of 2. you must. of course assume the definition of + , your universe of discourse and principal existential axioms. Bob Pease === Subject: Re: How long must physics put up w/f=ma? In sci.physics, Gene Nygaard : >>In sci.physics, tadchem >>: > The answer to the question How long must physics put up w/f=ma? is: >> As long a F = ma describes reality And just what does that have to do with *w/f=ma*? >>I think the 'w/' in this case is an abbreviation for 'with', >>not part of the formula proper. Of course writing it that >>way isn't exactly helpful. > Of course it isn't helpful. > But if you haven't figured out yet, Shead is an idiot. He'll never > change. Yet we other idiots keep replying to a thread whose subject > asks about w/f=ma? > It might be understandable that after all these years, Shead still > hasn't the foggiest idea what a factorial is, or the symbol for this > operation, a symbol he keeps throwing into his formulas. But he damn > sure out to be held to a standard of knowing that a slash / is a > symbol for a mathematical operation. Don't be blaming this confusion > on anyone other than the culprit, Dense Donny Shead. > Donald Shead is also a crackpot who keeps using ft as a symbol for > the multiplication of force by time--and is too stupid to figure out > why people don't understand what he is saying. I was under the impression that he was using 'ft' as an abbreviation for 'foot' (since he highly prefers English units, for some bizarre reason). As for being an idiot -- I prefer not to make judgements on the person, but certainly Shead's postings are consistent with being an intractable quack, unable to see that SI, for all of its warts (one of them being the requirement of an existent of the mass-artifact of 1 kg; another being the liter/litre controversy) doesn't have the conversion problems and/or other difficulties of contemporary English/Imperial units: 1 km = 1,000 m = 1,000,000 mm and that's more or less that, all throughout SI, regardless of the item being measured. (Some artifacts, however, still exist: 1 cal = 4.180 J, for instance. I'm not sure where 'cal' came from apart from its relationship to water (1 cal heats 1 gm water 1 degree C) and it may not be, strictly speaking, part of SI. There are also some issues on the electromagnetic side of the issues, which I'm not that familiar with. Of course this is a far cry from the grains/oz/lbs [pick a flavor!], the mil/in/ft/rod/mile/furlough, or the dry vs wet quart issues perambulating all over Imperial measurements. Even pints are different.) I base this primarily on his non-response to the elementary observation that a mass artifact (the aforementioned 1 kg item) is far easier to handle than an apparatus that can be proven to generate a 1 Newton (or, for those so inclined, 1 lb-force) force. (Arguably, the simplest method to reliably produce this force would rely on the ideal gas law, and require precise temperature measurements.) Nor is he treading with the crowd on his notions regarding inertia, mass, and force. I don't see why it makes any difference when writing F = ma a = F/m m = F/a assuming scalar quantities. (In vector quantities, the third must be tossed, for obvious reasons; F and a are inherently vectors.) Use whichever one is appropriate for the problem solution. Side issue: I'll admit it is a pity we're not willing to spend money on the gigantic conversion efforts necessary to bring our roadway and speedometer systems consistent with Europe, though. :-/ But oh well. :-) > Gene Nygaard > http://ourworld.compuserve.com/homepages/Gene_Nygaard/ -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Naive User Stories Followups. In sci.physics, Richard Henry <0zPib.33608$La.24164@fed1read02>: >> There are subtler effects, however. In princple you can get kT ln 2 >> joules of useful energy by randomizing a bit that starts out in a >> known state. And of course energy has mass. So perhaps a blank >> diskette, or one that contains the digits of pi or some other >> determinate pattern, weighs slightly more than one that contains >> random data. >> How, pray tell, does one distinguish the digits of pi from random data? > In what radix? 10, 16, 2, pi, etc? > In what representation? Binary, ascii, BCD, etc? Does it matter? Given any finite digit sequence, pi embeds it. Somewhere. :-) (At least, such is my understanding.) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Naive User Stories > Followups. > In sci.physics, Richard Henry > : >> There are subtler effects, however. In princple you can get kT ln 2 >> joules of useful energy by randomizing a bit that starts out in a >> known state. And of course energy has mass. So perhaps a blank >> diskette, or one that contains the digits of pi or some other >> determinate pattern, weighs slightly more than one that contains >> random data. >> How, pray tell, does one distinguish the digits of pi from random data? > In what radix? 10, 16, 2, pi, etc? > In what representation? Binary, ascii, BCD, etc? > Does it matter? Given any finite digit sequence, pi embeds it. > Somewhere. :-) > (At least, such is my understanding.) I'm afraid I don't understand your understanding. === Subject: Re: Jordan measure Any takers??? -- === Subject: Re: How to calculate the total coverage area of a few circles? > late. The method is very skillful. I believe it has a excellent efficiency. > Thaks again. _ It's true that Arcs (the list of arcs between x=xleft and x=xright) can be arranged so that each even/odd pair of entries in Arcs gives bounds of a contiguous area, bounded below by the first arc, above by the second arc, and left and right by xleft and xright. But the step Sort Arcs [e] so both ends are in non-decreasing y-order (which is always possible because of the Event list construction). In one pass, flag arcs that are inside other circles, and in another pass, remove the flagged arcs is wrong and (for example) fails where an overlap of two circles fills between two other circles; for example, with circles of radius 13 at (13,31), (19,13), (25,47), and (33,29). The only way around this I have thought of is O(n2), and leads to a worst case complexity of O(n4) and an average complexity of O(n3). I will post a program later illustrating it. -jiw === Subject: Re: How to calculate the total coverage area of a few circles? 3QLpj-NoP*NzsIC,boYU]bQ]H'y<#4ga3$21: late. The method is very skillful. I believe it has a excellent efficiency. > Thaks again. _ > It's true that Arcs (the list of arcs between x=xleft and x=xright) > can be arranged so that each even/odd pair of entries in Arcs gives > bounds of a contiguous area, bounded below by the first arc, above by > the second arc, and left and right by xleft and xright. But the step > Sort Arcs [e] so both ends are in non-decreasing y-order (which is > always possible because of the Event list construction). In one pass, > flag arcs that are inside other circles, and in another pass, remove > the flagged arcs is wrong and (for example) fails where an overlap of > two circles fills between two other circles; for example, with circles > of radius 13 at (13,31), (19,13), (25,47), and (33,29). The only way > around this I have thought of is O(n2), and leads to a worst case > complexity of O(n4) and an average complexity of O(n3). I will > post a program later illustrating it. The method of Edelsbrunner that I referred to earlier gives you complexity O(n log n). Basically you get one term in the sum for every feature of the power diagram of the circles (Voronoi diagram of their centers in the case that all the circles have the same radius). -- David Eppstein http://www.ics.uci.edu/~eppstein/ Univ. of California, Irvine, School of Information & Computer Science === Subject: Re: Fun and Mental; Raisin for High; Achievements of Shoes >You mean, you still don't realize that the above is a fake and that >>there is no such institution as The Lovenstein Institute of Scranton >>Pennsylvania. Even the Guardian which, at the time, jumped >>enthusiastically on this crap, had to retract and admit that they were >>taken for a ride. I suggest you contact them. >I wonder if this is the reason for liberals to work on dumbing >down public schools? Might be, might be:-) > Even people who appear to have an ability >to think use this rumor of Bush dumbness to get a Democrat >elected next year. With a stress on appear. Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: Fun and Mental; Raisin for High; Achievements of Shoes there is no such institution as The Lovenstein Institute of Scranton >Pennsylvania. Even the Guardian which, at the time, jumped >enthusiastically on this crap, had to retract and admit that they were >taken for a ride. I suggest you contact them. I wonder if this is the reason for liberals to work on dumbing down public schools? Even people who appear to have an ability to think use this rumor of Bush dumbness to get a Democrat elected next year. /BAH Subtract a hundred and four for e-mail. === Subject: Re: Fun and Mental; Raisin for High; Achievements of Shoes >You mean, you still don't realize that the above is a fake and that >>there is no such institution as The Lovenstein Institute of Scranton >>Pennsylvania. Even the Guardian which, at the time, jumped >>enthusiastically on this crap, had to retract and admit that they were >>taken for a ride. I suggest you contact them. >I wonder if this is the reason for liberals to work on dumbing >down public schools? 'Scuse me? I happen to be a knee-jerk liberal, and I have no interest at all in dumbing down schools. > Even people who appear to have an ability >to think use this rumor of Bush dumbness to get a Democrat >elected next year. Every time I've ever seen a picture of him next to a senior advisor, I get the impression of a 4-year-old on career day, visiting the grown ups. - Randy === Subject: Re: Proving the Four Color Theorem > Given what you know about graphs and vertex coloring, which do you > think would be the easiest task?. > 1. Prove that a 5-chroma planar graph cannot exist. > 2. Prove that all planar graphs are 4-colorable. > 3. Prove that a 5-chroma graph cannot be planar. This is as meaningful a question as asking which of the following is the easiest task? 1. Prove 4 + 5 = 9. 2. Prove 9 - 5 = 4. 3. Prove 9 - 4 = 5. J === Subject: Re: Proving the Four Color Theorem >> Given what you know about graphs and vertex coloring, which do you >> think would be the easiest task?. >> 1. Prove that a 5-chroma planar graph cannot exist. >> 2. Prove that all planar graphs are 4-colorable. >> 3. Prove that a 5-chroma graph cannot be planar. > This is as meaningful a question as asking which of the following is > the easiest task? > 1. Prove 4 + 5 = 9. > 2. Prove 9 - 5 = 4. > 3. Prove 9 - 4 = 5. Which is it? :-) -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Proving the Four Color Theorem > Given what you know about graphs and vertex coloring, which do you > think would be the easiest task?. > 1. Prove that a 5-chroma planar graph cannot exist. > 2. Prove that all planar graphs are 4-colorable. > 3. Prove that a 5-chroma graph cannot be planar. They are all the same problem. Therefore, they are equally hard/easy. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Naive Q: Set theory, logic - which comes first? <1xUcb.14254$O85.6040@pd7tw1no> <3f79e264$7$fuzhry+tra$mr2ice@news.patriot.net> <3f82ebb8$3$fuzhry+tra$mr2ice@news.patriot.net> that its truth has been established beyond all doubt. No. That is not either a Mathematical concept or an achievable goal. >The integers have more properties than can be captured by any axiom >system. And what would those be? How does one distinguish an axiom system in which those proerties are true from one in which they are not? >You said that there are statements about natural numbers that can >neither be proven nor disproven, The term Natural Numbers usually refers to the axiom system known as the Peano Postulates. >which may or may not be true, but >does not follow from what G.9adel showed. It certainly does, unless you are using the term natural numbers as some vague metaphysical concept having nothing to do with any axiom system. What do *you* mean by natural numbers? >G.9adel did show that in a given >formal system for natural numbers there are statements about natural >numbers that can neither be proven nor disproven. Which means that if you stick to Mathematics my statement is true. It doesn't matter which axiom system you pick, as long as it includes Peanos' Postulates among its consequences, there will be statements that can neither be proven nor disproven. -- spamtrap@library.lspace.org === Subject: Re: Homeomorphism and boundaries <3f85d5c4$21$fuzhry+tra$mr2ice@news.patriot.net> <3f89fd84$48$fuzhry+tra$mr2ice@news.patriot.net> at 01:53 PM, magidin@math.berkeley.edu (Arturo Magidin) said: >Then you mapped (0,2pi) to C by sending x to exp(ix), and that map is >also not an open map. Of course, if your domain changed to [0,2pi], >compact, then it would be a closed continuous mapping, but it would >not be injective. It's the range I should have changed, not the domain. Specifically, I should have stated that the map was an injection into the unit circle whose image was missing one point. Or, equivalently, given the map as the natural injection of the open interval into its one-point compactification. -- spamtrap@library.lspace.org === Subject: Re: Fundamental Reason for High Achievements of Jews at 11:59 PM, Noah Roberts said: >Hell, during WW2 my country commited similar horrific acts right >here on our own soil - Japanese concentration camps. What we did to the Nissei was shameful, but it wasn't remotelcy close to what Germany or even Japan did. -- spamtrap@library.lspace.org === Subject: Re: Fundamental Reason for High Achievements of Jews > What do you think will happen when all these > bitter Black, Latino and Redneck trained warriors, > get back to America, after wasting many months > in a foreign land, and they begin to take stock??? > They might make the Nazi look like choir boys, > because these guys are not as sophisticated > as the Germans were/are. > Tom Potter > OOOOOOahahaha, Tom Potter the (mostly self) righteous > dude, who sees himself as the true epidemy of MORALS, > nonchalantly describes Blacks, Latinos and Rednecks as > not as sophisticated as Germans.......AHAhahahhaa..... > AHAHAHAHAh.........ahahahahah........ahahahahah.. > Potter, ....spoken like a true Jew........welcome into the fold. > ahahahahaha.........ahahahahanson Sophisticated: Having or appealing to those having worldly knowledge and refinement and savoir faire Maybe Harry Hyena hanson is right, and that when the Americans return from the Middle East, they will have more worldly knowledge, but I dare say that they will not be as refined. But the question is: how will they react to the fact that their buddies have been killed and maimed, and that they have been deprived of the joys of being with their friends and families, etc. when they begin to digest all this worldly knowledge, and see that others instigated the events that lead to their misfortune, and profited from it? -- Tom Potter http://tompotter.us === Subject: Re: Fundamental Reason for High Achievements of Jews >>Actually, the masses have been brainwashed >>to give the German people a bad rap, >>for trying to defend themselves from the Bolsheviks. Mr. Potter's attempt to trivialize Nazi's is rather sickening, > but I must also say that giving German people a bad rap is > guilty of the same mentality. it wasn't just the Germans who > murdered Jews during WW2....... > ....Several groups trying to completely whipe out each other > in ongoing wars and mass genocides [today]. > Hitler's manipulations could be just as effective today, > and in fact I have feared such a thing happening in my country > ever since 9/11. Hell, during WW2 my country commited similar > horrific acts right here on our soil - Japanese concentration camps. > When you mix a fearful there is no bottom to how bad it can get. > The human capacity for great evil is very great, and it lies within > each and it lies within each of us no matter what our culture > or belief systems stipulate. Mass murder and wanton cruelty > are unfortunately very human traits. > This is none of the fundamental aspects that Tom does not see, does > not comprehend and much less accept. But, Noah, it is even worse than > what you describe, re wanton cruelty. Take all the wanton cruelty > arising from connections with politics/culture/religion magically away > and you will still see a substantial permanent residue of people who > demand and PAY to receive cruel treatment and men and women who > get PAID and enjoy torturing others. As you say, Noah, wanton cruelty > is unfortunately a very human trait ....and it is are here to stay..... > and cruelty hurts just the same whether it comes from hormonal or > institutional aberrations. > ......And now comes Tom's response to Noah, > Noah who did NOT suggest that ALL Jewish people instigate..... > It is interesting to see that Noah Roberts > seems to be suggesting that all Jewish people > instigate conflict and war, for power and riches. > Now, that Tom believes that he has placed the turd successfully > onto Noah quickly washes his own dirty hands with strident > backpedaling (for the moment): > All Jews are not instigators of conflict and war for profit, > any more than all Muslims are terrorists, > or all Italians are Mafia members, > or all Gypsies are fortune tellers. > And of course, all instigators of conflict and war > for power and riches are not Jews, as can be > seen by the examples of Bush, Blair and Rumfeld. > Now, him feeling clean and righteous, Tom is ratcheting > up the bar again from NOT ALL Jews to MOST were Jews > .....ahahahaha...ahahahaha...... > The problem is, that most of the instigators > of the class wars of the 1900's were Jews, > and most of the instigators of the religious wars > of the 2000's are Jews, and it seems to me, > that the good, moral Jews, who believe > that all folks are entitled to life, liberty, > and the pursuit of happiness, should take > an aggressive stand against the war instigators, > as it will bite all Jews in the ass, just as it did > in the class wars. > The Bolsheviks, who were mostly Jews , > caused a lot of problems for Germany and the > world during the 1900's, and over one million > of them migrated to Israel after the native Russians > regained control of their government, and as can be seen, > they are desperately trying to get a world religious war > going, so they can get back in the chips. > Tom, now back in business with MOST of the Jews as the culprit > instigators, the time has come for him to explain who and > why one ought to react harshly against ALL Jews.......... > ahahahaha......ahahahaha......and that one should not be > surprised (which, all fun aside, is a possible real scenario): > It is not surprising that many Germans reacted harshly > against all Jews, good and bad, and it will not be > surprising if many Americans react the same way, > if America is pulled into a war that costs many > Blacks, Latinos, and Rednecks their lives, limbs, > liberties and fortunes. > Most people are intimidated by the character assassination blitz > put out by a small immoral Jewish element when anyone suggests > that the motives and actions of some Jews are less than noble, > but the truth needs to be told to prevent an enormous amount of > suffering by Jews and the peoples of the entire world. > In the 1900's, the German leaders and people > were also intimidated by the Bolsheviks, > because they were not only engaging in character assassination, > the Bolsheviks did anything to silence opposition, including > beatings and real assassinations. This is why some Germans formed > strong arm groups to protect themselves and their leaders > for being physically attacked by the Bolsheviks. > AHAHAHahhahah........ahahahaha.....IAROTFLMAO....... > Tom must have seen Hitler's propaganda ca. 1940's films > Der ewige Jude being a Symphonie des Ekels und des Grauens, > and in Jud S.9fss in which Hitler depicted the Jews to be like > rats, but Potter makes another, novel leap and pits worms against > rats........ahahahah.....What Potter does not see is that in his world > in a clash between worms and rodents, the rats will win hands > down. .....Unless of course the worms infect the rats, and > cause rat extermination by internal systems collapse....ahahahaha... > Even intimidated and brainwashed worms turn > when the hits the fan, and there are a couple of billion > intimidated and brainwashed worms, and the > is beginning to hit the fan. > What do you think will happen when all these > bitter Black, Latino and Redneck trained warriors, > get back to America, after wasting many months > in a foreign land, and they begin to take stock??? > They might make the Nazi look like choir boys, > because these guys are not as sophisticated > as the Germans were/are. > Tom Potter > OOOOOOahahaha, Tom Potter the (mostly self) righteous > dude, who sees himself as the true epidemy of MORALS, > nonchalantly describes Blacks, Latinos and Rednecks as > not as sophisticated as Germans.......AHAhahahhaa..... > AHAHAHAHAh.........ahahahahah........ahahahahah.. > Potter, ....spoken like a true Jew........welcome into the fold. > ahahahahaha.........ahahahahanson === Subject: Re: Fundamental Reason for High Achievements of Jews > Harry Hyena hanson has lost the ability > to make a rational, intelligent, MORAL > response to my messages. > The only MORAL response to your messages is to > condemn you [Tom Potter] for the ignoramabus and bigot > that you are. You don't like Jews. > Bob Kolker > Easy, Bob, easy. It is not a crime for anyone not to like Jews. > It is anyone's fundamental right to like or dislike whatever they wish. > There is neither a law nor obligation which says that you should > or must be liked. The luxury of being liked must be earned. > As a matter of fact this maybe one of the core-problems that Jews > suffer from: Their bloated and loudly advertised need to be liked. > May that be as is, what makes me ROTFL about > Peeping Tom Crackpotter is that he has the chutzpah to use the > word MORAL, when in fact he is simple quite angry at me because > I refused to join him in: > utters terrorist threats against the President of the United States, > & calls and recruits for the violent overthrow of the government: 888888888888 - in Potter's own words - 888888888888 > [Tom Potter:] > Hopefully, hanson will join with me,.......... > ... and begin to slaughter the instigators of conflict and war. > ... most Americans will turn against the Bush's.... > ... and join in the blood fest. In fact,..... > ... Bush and his friends, associates, and family and property > ... will be pissed on and destroyed, like Saddam's statues. > 8888888888888888888888888888888888888888888888 [Tom Potter] > Frankly, I'd like to see ..... the American government to investigate > and try to make a case against me...., Tom Potter. ...... > ************************* > Ahahahahaha........ahahahahaha..... > Peeping Tom Crackpotter, the MORAL man, ahahahaha.....hahahaha > ahahahaha.....hahahanson As can be seen, by Harry Hyena hanson's post, the bull machine is on. Like to see bull in motion? Just expose the people who defend and support the instigators of conflict and war, and lo!!!! The bull machine turns on. -- Tom Potter http://tompotter.us === Subject: Re: Fundamental Reason for High Achievements of Jews > As can be seen, Bush is willing partner > in the Bolshevik game. -- Tom The Bolsheviks are dead or hiding in the closet > or under your bed. --- Bob It is interesting to see that Bob Kolker and > Harry Hyena hanson see anti-Bolsheviks under > their beds, and they it is driving them mad. > I wonder why???? > Tom Potter Ahahahaha.......ahahaha.......so, it is interesting to/for you > to check under other peoples beds, Tom. Hmm, that then > is a new revelation about your character, you PEEPING TOM. > But your behavior here is not very surprising as it belongs > into the same level of your apparently very wide ranging > prurient interests, like when you bragged a while ago that a > two bit whore hit you over the head with her earnings,...... > a bag full of pennies. ..... man, you are hanging out in very > strange places, Tom. It looks more like it's YOU who is driving > mad from peeping and cavorting with hookers at the age of 71. > AHhahahahhaha.......ahahahanson > Yep! > Harry Hyena hanson has lost it. (That is, if he ever had it.) > Harry's problem is that he gets all bent out of shape, > when I point out that his beloved Bolsheviks > migrated to Israel and New York,from where > they are instigating the religious wars. -- Tom > Ahahahaha....AHAHAHAH......They are beloved to you, alright. > But I never said that I liked them. So, to make you happy for > the week, Tom, here's a battle song from the Landsers of WWII > on their march against the Bolsheviks. Sang your German soldier: > --- Hundertzehn Patronen in der Tasche, > --- and der Seite das Gewehr, > --- in der Hand die Handgranate, > --- Bolscheviki komm daher! > Isn't this balsam for your scrotal soul, Tom? > Tom, if the Bolsheviks, the object & subjects of your fetish, were > not there, then people like you will find or invent other groups for > the goal of their visions. There are/will always be people with > agendas that instigate, just like yours. All have the same goal, > power and riches, they just cloak it differently, like i.e. you do. > Or have you lost the ability to discriminate and see that you are > no different then the ones you attack, save the odor and tint? > Say, where are you from hanson, > and what nations are you a citizen of????? > Tom Potter > See, ....you are at it again, .....you are PEEPING again, > like a senile Peeping Tom who can't see the obvious any > longer. Consider an apology to your 2 bit whore and then > have her cleanse your plumbing. It may re-normalize you. > But don't be a cheap , Potter, pay her more that 2 bits. > or she may hit you over the head again........ahahahaha > AHAHAHAH.....hahahahaha.........ahahahahanson I am pleased to see that Harry Hyena hanson is beginning to understand the parable about the whore who hit the person who called her a two bit whore, over the head with a bag of quarters. As intelligent, rational, MORAL folks know, and as can be seen in the news, on the street, at work, at play, and in the news groups, call some people's hyping of conflict and war bull, and they dump all kinds of bull on you. Exposing the war for profit game is like turning on the tap of a high pressure bull faucet. Watch out for the quarters! Watch out for the bull! -- Tom Potter http://tompotter.us === Subject: Re: Fundamental Reason for High Achievements of Jews > I prefer to give any interested parties the benefit > of my experiences to see if it has any bearing > on their life. liberty and happiness. > Tom Potter This is too much....you wanna dole out benefits now , Tom? > I doubt that you'll get many takers, the way you turned out. > especially if you need to see if it has any bearing on their life > admitting here and now that you are a ing PEEPING TOM ...... > Ahahahaha........ahahahanson > I think that Harry Hyena hanson has lost it. > (That is, if he ever had it.) > Note his desperate efforts to negate my messages: > the way you turned out. > PEEPING TOM > This is too much > ahhahaha........ahahahahahaha > Tom-Tom, Tom, this is not a negation, it is an EVALUATION > of your message. Your complaint looks like a bitter reaction > to the fact that you haven't **found** it yet, > weven after 71 years...... > ahahahahaha......ahahahaha.......ahahahahah If you don't think I've **found** it take a look at my web site, and take a look at my recent pictures that show me doing my natural thing. (Enjoying life.) Why don't you post a lot of pictures of you engaged in your regular activities?? A picture is worth a thousand words. (Or really about 100,000 bytes in JPG.) I'd be more than happy to compare my life, accomplishments, and morality to the people who support and defend the people who instigate conflict and war, for power and riches. The point being, that morality, and commitment to life, liberty, and happiness for ALL people, rather than selfishness and greed, brings happiness. Spread the word. -- Tom Potter http://tompotter.us === Subject: Re: Fundamental Reason for High Achievements of Jews >> You are an utter ass, Tom Potter. Blame the victim is a totally played >> out strategy. You know this so you attempt to surround it with large >> doses of unprovable accusations to make the victim appear to be an >> incredibly effective plotter. >> Aside from all that, the Bolsheviks are dead and gone. Communism died >> (no tears shall we shed). Red fascism mostly died with Stalin and the >> Soviet Union is gone, and let us rejoice over that. So Bolshevism at >> this stage is a boogey man and a boojum. >> Now there are collectivists, socialists and people who want to >> redistribute what is not theirs, but this is not particularly Jewish. >> Politically savvy folk have been playing Lord Bountiful with other >> people's wealth and property since God invented dirt. >The Bolsheviks are dead and gone, >they migrated to Israel and New York, yes I hear they all live in a cold water flat in Harlem. >from where they are instigating the religious wars of the 2000's, >as the loot from their instigation of the class wars of the 1900's >is almost gone. cold water flats in NYC are expensive so the loot must be gone by now. >No doubt Robert J. Kolker knows that Communism, >like Socialism, Republics, Democracies, Monarchies, etc. >are just forms of governments, and they can be good or bad, >and all have their strong and weak points. quite true and the idea is to get rid of the ones that don't work isn't it. >Forms of government don't instigate conflict and war for >power and riches, the Bolsheviks do, and they have >had a long history of doing so. Look out! I just saw one run under your bed!!!! >As can be seen, I don't Blame the victims, >I blame the people who instigate conflict and war >for power and riches. Like Bush? >The Blacks, Latino's, Rednecks, Muslims, etc. >who sacrifice their lives, limbs, liberties and money, >to enrich these immoral instigators, like Bush? (hey, this is alt.politics.bush... gotta keep on topic) THOM >are the victims >Tom Potter http://tompotter.us === Subject: Re: Fundamental Reason for High Achievements of Jews >(hey, this is alt.politics.bush... gotta keep on topic) No. This crap is cross-posted to: - soc.college.admissions - sci.math - sci.physics - sci.chem - comp.lang.c - alt.politics.bush And nothing of your drivel is topical in (at very least) one of these groups. Are you able to spot? And now stop polluting c.l.c. -- Irrwahn -- Irrwahn (irrwahn33@freenet.de) === Subject: Re: Fundamental Reason for High Achievements of Jews >> You are an utter ass, Tom Potter. Blame the victim is a totally played >> out strategy. You know this so you attempt to surround it with large >> doses of unprovable accusations to make the victim appear to be an >> incredibly effective plotter. >Aside from all that, the Bolsheviks are dead and gone. Communism died >(no tears shall we shed). Red fascism mostly died with Stalin and the >Soviet Union is gone, and let us rejoice over that. So Bolshevism at >this stage is a boogey man and a boojum. >Now there are collectivists, socialists and people who want to >redistribute what is not theirs, sorry but socialists aren't into that. Modern socialism is a balance of public and private enterprise with an adiquate social safety net. THOM >but this is not particularly Jewish. >Politically savvy folk have been playing Lord Bountiful with other >people's wealth and property since God invented dirt. >Bob Kolker === Subject: Re: Fundamental Reason for High Achievements of Jews > sorry but socialists aren't into that. Modern socialism is a balance > of public and private enterprise with an adiquate social safety net. That means stealing from one and giving to other via taxes. Taxation is Theft (because it is enforced at gun point and is not voluntary like charity). The adequate social safety is hardley ever adequate, is almost never safe and is very unsocial to those who are taxed to pay for it. Bob Kolker === Subject: Re: Fundamental Reason for High Achievements of Jews >> sorry but socialists aren't into that. Modern socialism is a balance >> of public and private enterprise with an adiquate social safety net. >That means stealing from one and giving to other via taxes. Taxation is >Theft Stupidity alert! >(because it is enforced at gun point and is not voluntary like >charity). The adequate social safety is hardley ever adequate, is almost >never safe and is very unsocial to those who are taxed to pay for it. >Bob Kolker === Subject: Re: Fundamental Reason for High Achievements of Jews >> The German people did not advocate >> Do unto others, before they do unto you. >> as the poster suggests. >> What they did was respond as best they could >> to a greedy, immoral attempt to take over, >> and rape their nation. >> Ah, that explains the gas chambers. It was all self defense! >> Dude, you are a moron. >> If FDR had joined with the German people, >> rather than selling out to the Bolsheviks, >> WWII would have lasted two months, >> and there would have been no nuclear weapons, >> no Cold War, no Korean War, no Vietnam, and no >> instigation of religious wars. >> You are pretty silly too. >> bob >> - who would have been put on a train if he'd been living in Germany >> during the round-ups (one grandparent is all it took). >First of all, >the gas chambers have been vastly hyped, Why? isn't 6 million enough for you? Shall we throw in the SS Einzotgruppe that roamed around Russia just shooting Jews rather than gasing them? >and have become the a key part of an over all strategy, >as can be seen by looking at the facts, >and by taking an intelligent, rational, unbiased look at the facts. >Secondly, there is no doubt that people >use tit for tat as a social strategy, and no doubt >some Germans took tit for tat to extreme levels, >just as Israel is taking tit for tat to extreme levels today. yes they have been suckered into that haven't they by the Palistianians. >Most Germans recognized that their plight, >and the destruction of their nation, >had been brought about because many Jews >in Germany and around the world, had supported >the class war instigations of the Bolsheviks, Wrong again, the Bolsheviks were long gone by the 20's and replaced by Stalinists. >and had waged a propaganda war against >Germany and the nations who took action >to control the Bolshevik terrorists, >and they resented this. wrong again. >It is interesting to note that and that after the death of FDR, Truman, Hoover, and others, >who recognized the Bolshevik threat did a 180, >and began to counter them, just as the German people had >tried to do. ZZZZZZZZZZZZZZzzzzzzzzzzzzzz... I'm sorry did you say something? Hoover, Hoover... wasn't that the dude that threw us into the Great Depression? >As can be seen, Bush is a willing partner of the Bolsheviks, >who are instigating the religious wars of the 2000's, >to get back into the chips, as the loot from their >class wars is rapidly running out. >If Bush had not gotten into bed with Sharon >and the Bolsheviks, that bed must be in a grave yard since the Bolsheviks have been gone an awful long time. >there would have been no 911, >no war with Afghanistan and Iraq, Why, right wing wackos aren't the result of Bolshevikism or communism, they are the result of right wing wackoism. >no movement of America toward a police and military state, >no weakening of America's relationships with most nations on Earth, >no weakling of the United nations, NATO, the G8, etc. >no dollar flight, no meltdown of the economy, etc. no jelly in my peanut butter and jelly sandwich no rust or decay no All in the Family or Married with Children on TV no Gypsie fortune tellers no.... _________________ (please insert your favorite predjudice) >Also note, that thanatos@coldmail.nu >uses the standard tactic of people who support, >or have been brainwashed by Bolshevik propaganda, >of attacking the messenger. >They are masters of character and real assassination. >These are their major weapons. yah, that Alexander Korinski is a real terrorist. I'll bet he'se hiding out with Bin Laden or Saddam. THOM >Tom Potter http://tompotter.us === Subject: Re: Fundamental Reason for High Achievements of Jews >> The German people did not advocate >> Do unto others, before they do unto you. >> as the poster suggests. >> What they did was respond as best they could >> to a greedy, immoral attempt to take over, >> and rape their nation. >> Ah, that explains the gas chambers. It was all self defense! >> Dude, you are a moron. >> If FDR had joined with the German people, >> rather than selling out to the Bolsheviks, >> WWII would have lasted two months, >> and there would have been no nuclear weapons, >> no Cold War, no Korean War, no Vietnam, and no >> instigation of religious wars. >> You are pretty silly too. >> bob >> - who would have been put on a train if he'd been living in Germany >> during the round-ups (one grandparent is all it took). >First of all, >the gas chambers have been vastly hyped, > Why? isn't 6 million enough for you? Shall we throw in the SS > Einzotgruppe that roamed around Russia just shooting Jews rather than > gasing them? >and have become the a key part of an over all strategy, >as can be seen by looking at the facts, >and by taking an intelligent, rational, unbiased look at the facts. >Secondly, there is no doubt that people >use tit for tat as a social strategy, and no doubt >some Germans took tit for tat to extreme levels, >just as Israel is taking tit for tat to extreme levels today. > yes they have been suckered into that haven't they by the > Palistianians. Thom makes an interesting point when he suggests that the Germans were suckered into tit for tat responses against all Jews because of the class war instigations of a few Jews, that were so harmful to the German people, and the entire world. No doubt the religious war instigations of the Bolsheviks will be far more disastrous to mankind, and the tit for tat responses will be much greater against innocent people. If FDR had joined with Germany, Spain, Japan, and other nations, in going after the Bolshevik terrorists, rather than backing them, WWII would has lasted two months, and there would have been no millions of deaths, no Cold War, no Korea, no Vietnam , no nuclear weapons, and no instigation of global religious war. And if Bush had gone after the Bolshevik instigators, rather than selling out to them and Sharon, there would have been no 9/11, no movement of America toward a police and military state, no loss of freedoms to Americans, no danger to American's at home and abroad, no collapse of the economy, no shattering of the national budget, no war against the Iraqi people, with its' attendant cost, depletion of non-renewable resources, environmental damage, etc, no loss of trust by most nations of the world, no hatred of America by hundreds of millions of Muslims, no raid on Social Security, no freeze of Civil Service employees salaries, no bankruptcy of major airlines, no dollar flight, no meltdown of the dollar, no meltdown of America, no weakening of the United Nations, no weakening of NATO, the G8, no independent European GPS system, no loss of trillion dollar markets by Boeing, G.E. and other American firms, etc. -- Tom Potter http://tompotter.us === Subject: Re: Fundamental Reason for High Achievements of Jews > Hoover, Hoover... wasn't that the dude that threw us into the Great > Depression? Actually not. A lot of things were happening beyond the control of Hoover or anyone else. The world was on a binge during the 20's and the bubble went pop. So no matter who was in charge there would have been a recession. Puting the Smoot-Hawley tarrif in place in 1931 did not help. Many people think the stock market crash caused the depression. Not so. The recession was underway before that. Stock market crashes do not cause depressions. We had a real sharp crash in 1986 and no depression resulted. By the way, T. Potter sees Bolshies under the bed, behind the drapes, in the woodshed and inhabiting the shadows. He equates Bolshiveks with Jews. Never mind the the Bolshies that did the most damage were the gentiles, Lenin and Stalin. Poor Trotsky, aka Lev Bronshteyn (who was not very nice, btw) did only a fraction of the damage done by the other two. It was that ex seminary student Joe Stalin, a Groozysian thug, who forced farm collectivization on his country and caused at least 7.5 millions to starve. Stalin also created the Gulag. But Potter will blame that on the Jews too. Look at all those Jewish-Bolshivek villains. Beria, Kruschev who spilled the blood of their countrymen. Potter will blame all the bad weather on the Jews too. Bob Kolker === Subject: Re: Fundamental Reason for High Achievements of Jews > Hoover, Hoover... wasn't that the dude that threw us into the Great > Depression? > Actually not. A lot of things were happening beyond the control of > Hoover or anyone else. The world was on a binge during the 20's and the > bubble went pop. So no matter who was in charge there would have been a > recession. Puting the Smoot-Hawley tarrif in place in 1931 did not help. > Many people think the stock market crash caused the depression. Not so. > The recession was underway before that. Stock market crashes do not > cause depressions. We had a real sharp crash in 1986 and no depression > resulted. > By the way, T. Potter sees Bolshies under the bed, behind the drapes, in > the woodshed and inhabiting the shadows. He equates Bolshiveks with > Jews. Never mind the the Bolshies that did the most damage were the > gentiles, Lenin and Stalin. Poor Trotsky, aka Lev Bronshteyn (who was > not very nice, btw) did only a fraction of the damage done by the other > two. It was that ex seminary student Joe Stalin, a Groozysian thug, who > forced farm collectivization on his country and caused at least 7.5 > millions to starve. Stalin also created the Gulag. But Potter will blame > that on the Jews too. Look at all those Jewish-Bolshivek villains. > Beria, Kruschev who spilled the blood of their countrymen. > Potter will blame all the bad weather on the Jews too. It is interesting to see that Robert J. Kolker continues to suggest that ALL Jews are instigators of conflict and war for profit. I assert that all Jews are not instigators of conflict, any more than all Muslims are terrorists, or all Gypsies are fortune tellers. And as can be seen in the cases of Bush, Blair and Rumfeld, all instigators of conflict and war for power and riches are not Jewish. Robert J. Kolker does bring up a good point when he points out that it was a long up hill struggle for the native Russians to regain control of their country after the Bolshevics assassinated the native Russian leaders, and the Royal family, and co-opted the Russian government. Although Stalin, who had a Jewish wife, was a front man for the Bolshevics, he was motivated by self-survival, and had enough power to eliminate any one who was a threat to him. Robert J. Kolker brings up another good point when he suggest that FDR, who inherited a recession, and turned it into a long depression, by meddling with the economy in order to cater to the people who were instigating class warfare in America, and all over the world. Note that there was a more drastic stock market crash when Reagan was president, and as he let market forces make the necessary corrections, a year or two later, there was prosperity, rather than depression. And Robert J. Kolker makes another good point when he points out that Stalin, like FDR and many other international leaders, were forced by the class war instigations, to make drastic changes to his nations market system, with disastrous results. Of course, the historical data shows that the Bolshevics were responsible for co-opting the food from the area of Russia where millions starved, as punishment for anti-Semitic activities in that area. It amazes me, how so much spin has been put on history, and even current events, right in front of people's eyes, and that anyone buys the spin. It goes to show you that Pavlov was right. -- Tom Potter http://tompotter.us === Subject: Re: Fundamental Reason for High Achievements of Jews >> If thine enemy smite thee on thy cheek, tear his head off and down >> his neck. [...] A real man >> kills his enemies, takes his land and cattler, rapes his women, enslaves >> his children and contemplates how good life can be. >> In this evil world, if you >> are not on top, some other part is and is humping you. The price of >> liberty is efficiently killing the enemies of liberty. The rule is very >> simple. Support and help your friends. Kill your enemies. That is the >> only way to survive. >> Do unto others, before they do unto you. >> Fascinating, even if it's off topic. >> Robert Kolker, who claims to be Jewish, advocates a philosophy >> of life which sounds a lot like what we normally would associate >> with Nazis and supremacists. >> If someone claimed the Jews believe what Kolker claims to >> believe, that person would surely be labelled an anti-Semite. >> And yet, Kolker seems to be serious, and has received some support >> from others in this thread. >> What gives? >Actually, the masses have been brainwashed >to give the German people a bad rap, >for trying to defend themselves from >the Bolsheviks.who were instigating >conflict and war, for power and riches in the 1900's. Please give examples. Stalin was no angel but except for his defeat in Finland never tried what the right wing Hitler did. All his crimes were internial Stalin did not invade Germany in June 1941, Hitler Invaded Russia, Stalin did however cooperate with Hitler in the occupation of Poland. >As the Bolsheviks.were assassinating leaders and breaking >up the meetings of political rivals, Please give examples including dates. For the most part it was the Nazi's that that broke up and attacked rivals. >a few nations, >such as Germany, Italy, Spain, Japan, and Turkey >turned to strong leaders to counteract the Bolshevik menace, >which sought to co-opt their governments, >as they had the Russian government. Russia didn't have a government, it had a dictatorship run by the Tzars. Next you need to learn the difference between the Trioika Communists and the Bolsheviks. The Bolshevik Government under Alexander Korenski was over thown by the Troika (and keep in mind that Lenin was sent to Russia by the Kiaser to take power and knock Russia out of the war so why not Blame Willheilm for the left wing mess. >As can be seen by their contributions to society, >and their relationships with folks all over the world, >the German people are probably the most intelligent, >most moral, and most productive people on Earth. Yes they produced 6 million bodies in the gas chambers and in some cases produced consumer goods from the bodies like wigs and lamp shades made from human skin, plus they recycled all the gold in teeth and the like into the Reich's bank. >The Bolsheviks, who have a history of instigating >conflict and war for profit, excuse me that you have that wrong. Its the Nazi's and republicans that do stuff like for money and profit. >have a vested interest >in tagging their critics and adversaries, with negative boilerplate >as this keeps their scam alive. >Note that the Bolsheviks.instigated the class wars of the 1900's, >and after the native Russians regained control of their nation, >they migrated to Israel and New York, >from where they are instigating the religious wars of the 2000's, >to get back into the chips. are you drunk? >The German people did not advocate >Do unto others, before they do unto you. >as the poster suggests. >What they did was respond as best they could >to a greedy, immoral attempt to take over, >and rape their nation. By Hitler, the Nazi's and the rich industrialists that backed them. >If FDR had joined with the German people, >rather than selling out to the Bolsheviks, sorry but the Bolsheviks were long gone by 1940. >WWII would have lasted two months, >and there would have been no nuclear weapons, Except German ones >no Cold War, no Korean War, no Vietnam, OK. lets look at VN. Ho Chi Minn was a nationalists who even in the 30's was tryiung to get the French out of his country. When the Japs invaded the OSS gave him aid and he sided with us but when we broke our promises to him and gave SEA back to France to bribe them to join the UN, Ho turned to the only people left, the commies. Now given that, why do you think Ho Chi Minn and other Vietnamese Nationalists would not have fought for independence from the French??? > and no >instigation of religious wars. Yah right the Koran and bible would have gone puff and disappeared. >As can be seen, >Bush has also sold out to the Bolsheviks hard to do since the Bolsheviks haven't existed for 80+ years and the USSR for 13+ years. Bush sold out to the New World Order. THOM >Tom Potter http://tompotter.us === Subject: Re: Fundamental Reason for High Achievements of Jews >> If thine enemy smite thee on thy cheek, tear his head off and down >> his neck. [...] A real man >> kills his enemies, takes his land and cattler, rapes his women, enslaves >> his children and contemplates how good life can be. >> In this evil world, if you >> are not on top, some other part is and is humping you. The price of >> liberty is efficiently killing the enemies of liberty. The rule is very >> simple. Support and help your friends. Kill your enemies. That is the >> only way to survive. >> Do unto others, before they do unto you. >> Fascinating, even if it's off topic. >> Robert Kolker, who claims to be Jewish, advocates a philosophy >> of life which sounds a lot like what we normally would associate >> with Nazis and supremacists. >> If someone claimed the Jews believe what Kolker claims to >> believe, that person would surely be labelled an anti-Semite. >> And yet, Kolker seems to be serious, and has received some support >> from others in this thread. >> What gives? >Actually, the masses have been brainwashed >to give the German people a bad rap, >for trying to defend themselves from >the Bolsheviks.who were instigating >conflict and war, for power and riches in the 1900's. > Please give examples. Stalin was no angel but except for his defeat > in Finland never tried what the right wing Hitler did. All his crimes > were internial > Stalin did not invade Germany in June 1941, Hitler Invaded Russia, > Stalin did however cooperate with Hitler in the occupation of Poland. >As the Bolsheviks.were assassinating leaders and breaking >up the meetings of political rivals, > Please give examples including dates. For the most part it was the > Nazi's that that broke up and attacked rivals. It is interesting to see that Thom does not comprehend, that Germany, and many nations, tried to control the class war instigations of the Bolsheviks by many peaceful means, and by making treaties to oppose the instigation of conflict and war, and when all measures failed, they went after the Bolsheviks in the nations they were using as bases of operations, just as America went after real terrorists, and imagined terrorists in Iraq. Note that the Germans were welcomed in most nations as their local leaders had been bought or intimidated by the Bolsheviks and were unable to maintain the social, financial and cultural integrity of their nations. Even France had little opposition to the Germans. If FDR had joined with Germany, Spain, Japan, and other nations, in going after the class war instigators, rather than backing them, WWII would has lasted two months, and there would have been no millions of deaths, no Cold War, no Korea, no Vietnam , no nuclear weapons, and no instigation of global religious war. Bush has also sold out to the religious war instigators, and brought much harm to America and the world. And it is interesting to see that Thom does not know that the Bolsheviks were assassinating leaders, and assaulting people, and breaking up meetings, and that the Black Shirts and Nazi were reactions to this. I suggest that he read some factual, concurrent history of those times, rather than allow himself to be brainwashed by present day spin motivated more by agenda, than by facts or a respect for mankind. -- Tom Potter http://tompotter.us === Subject: Re: Fundamental Reason for High Achievements of Jews > Yes they produced 6 million bodies in the gas chambers and in some > cases produced consumer goods from the bodies like wigs and lamp > shades made from human skin, plus they recycled all the gold in teeth > and the like into the Reich's bank. Twelve million. The Nazis killed untermenschen other than Jews. Stalin killed 7.5 million Ukranians and only God knows how many others he worked to death in the Gulags. Stalin was a Monster of the same magnitude as Hitler. The Chinese Communists win the prize however. They liquidated 40 to 60 million class enemies to acheive their revolution. Nazi-ism was racial insanity but Lenninist Communistm was sanitized by Western intellectuals and still was death and destruction. Pinko-stinko fellow travelers and useful idiots put a good face on Stalin's doings. Bob Kolker === Subject: Re: Fundamental Reason for High Achievements of Jews ... > Stalin killed 7.5 million Ukranians and only God knows how many others > he worked to death in the Gulags. Stalin was a Monster of the same > magnitude as Hitler. > The Chinese Communists win the prize however. They liquidated 40 to 60 > million class enemies to acheive their revolution. Are you objecting to this? I think it was you who said that it is OK to kill all his enemies. I wonder how a brainless individual like you manages to write. And now, *please* stop posting to clc. Jirka === Subject: Re: Fundamental Reason for High Achievements of Jews >... >> Stalin killed 7.5 million Ukranians and only God knows how many others >> he worked to death in the Gulags. Stalin was a Monster of the same >> magnitude as Hitler. >> The Chinese Communists win the prize however. They liquidated 40 to 60 >> million class enemies to acheive their revolution. >Are you objecting to this? I think it was you who said that it is OK to >kill all his enemies. I wonder how a brainless individual like you manages >to write. It's all a matter of probability: let a bunch of bonobos play around with computer terminals for a suitable amount of time, and sooner or later something like what you are replying to is likely to find its way to usenet. >And now, *please* stop posting to clc. Seconded. -- Irrwahn === Subject: Re: Fundamental Reason for High Achievements of Jews : It's all a matter of probability: let a bunch of bonobos play around : with computer terminals for a suitable amount of time, and sooner or : later something like what you are replying to is likely to find its : way to usenet. And just what makes you think that this does not already occur on a daily basis? ----- Richard Schultz schultr@mail.biu.ac.il Department of Chemistry, Bar-Ilan University, Ramat-Gan, Israel Opinions expressed are mine alone, and not those of Bar-Ilan University ----- an optimist is a guy/ that has never had/ much experience === Subject: Re: Fundamental Reason for High Achievements of Jews >: It's all a matter of probability: let a bunch of bonobos play around >: with computer terminals for a suitable amount of time, and sooner or >: later something like what you are replying to is likely to find its >: way to usenet. >And just what makes you think that this does not already occur on a >daily basis? Nothing, I don't think so, and I didn't say so. In fact, it merely seems to occur even more frequently. :) Just as I said: it's a matter of probability... > an optimist is a guy/ that has never had/ much experience True enough. Irrwahn -- The generation of random numbers is too important to be left to chance. === Subject: Re: Fundamental Reason for High Achievements of Jews > Yes they produced 6 million bodies in the gas chambers and in some > cases produced consumer goods from the bodies like wigs and lamp > shades made from human skin, plus they recycled all the gold in teeth > and the like into the Reich's bank. > Twelve million. The Nazis killed untermenschen other than Jews. > Stalin killed 7.5 million Ukranians and only God knows how many others > he worked to death in the Gulags. Stalin was a Monster of the same > magnitude as Hitler. > The Chinese Communists win the prize however. They liquidated 40 to 60 > million class enemies to acheive their revolution. > Nazi-ism was racial insanity but Lenninist Communistm was sanitized by > Western intellectuals and still was death and destruction. Pinko-stinko > fellow travelers and useful idiots put a good face on Stalin's doings. It is interesting to see how many people have been conditioned to blame Communism for events orchestrated by a group of people who instigate conflict and war, for power and riches. Communism, like Socialism, Monarchies, Democracies, etc. are just forms of government, and they can be good or bad, and ALL forms of government have their good and bad points. Of course, as can be seen by historical examples, the major weakness of a Democracy is that it can be bought by special interest groups, or manipulated by demogogues. As can be seen by the historial evidence, the Bolshevics were the major instigators of the class wars of the 1900's, the colonial wars of the 1800's, and they are right in there instigating the religious wars of the 2000's, and naturally when wars are instigated, bad things happen. The instigation of conflict and war, for power and riches is the stock in trade of Bolsheviks, much as fortune telling is the stock in trade of Gypsies. They netted trillions of dollars from their class wars, and they expect to net far more from their religious wars. The shame of it all is, that most of the Bolsheviks, and after these conflicts rage for a while, the combatants eventually turn against the good Jews. Can you imagine what will happen in America, when hundreds of thousands of bitter Blacks, Latino's, and Rednecks return from the religious wars, and who they will blame for their wounds, the deaths of their buddies, their loss of liberty, separation from their children and spouses, and their financial and social misfortunes, and the degeneration of their nation???? -- Tom Potter http://tompotter.us === Subject: Re: Fundamental Reason for High Achievements of Jews >> AHhahahhaha......Bob, are you having a bout of genocidal >> upwellings again? As a Jew, you are supposed to be better >> and more lovable then the Muslims...and 5.5 times smarter. >> You said so yourself!. >> Now, you've become like them, a Jewish Homicide bomber, >> a Wailing-Wall-Martyr....a Sui-Holocauster........ >> ahahahhha............ahahahahanson >There is nothing wrong with killing people who have harmed you, are will >harm you eventually. We send our soldiers to die regularly. It is >business as usual. I just want a high body count. what kind of stupid are you ? Bruce ----------------------------------------------------------------------- It was so much easier to blame it on Them. It was bleakly depressing to think that They were Us. If it was Them, then nothing was anyone's fault. If it was Us, what did that make Me ? After all, I'm one of Us. I must be. I've certainly never thought of myself as one of Them. No-one ever thinks of themselves as one of Them. We're always one of Us. It's Them that do the bad things. <=> Terry Pratchett. Jingo. === Subject: Limit point compacness help? Is a limit point compact subspace of a Hausdorff space necessarily closed? Well since I can't seem to prove this and since if this is true, the proof would look nothing like the proof that a compact subspace of a Hausdorff space is closed, I am trying to come up with a counterexample. There is an example of a limit point compact space that isn't compact in my Munkres book which is : Let Y consist of two points and give Y the topology consisting of Y and the empty set. Then X = Z_+ x Y is limit point compact but not compact. (Z_+ is the positive integers). Can I use this to show a counterexample? I guess the question is does X sit inside of a Hausdorff space making X open? I don't think it does because X itself is not Hausdorff since the open sets consist of unions of {n} x Y and thus if Y = {a,b}, then 1 x a and 1 x b do not have disjoint neighborhoods. Thoughts? Steve === === === Subject: Re: Is ...9999.9999... = 0 ? This is a much more intelligent question than most of the responses indicate. The simplest answer is that ...999.999... is not a real number, because the series sum[i=0..inf]{9*10i} does not converge over the real numbers. However, there are situations in which number representations like ...999.999... do make sense. For example, in the binary two's-complement notation used by all digital computers, the number representation ...1111 corresponds to -1. The reason for that is that, as with your example, adding 1 to ...1111 causes all digits to be wiped out by carries. This can be similarly extended to ten's complement arithmetic, where ...999 equals -1 and, equivalantly, ...999.999... equals 0. So the question you've asked shows some great intuition along these lines. To be precise, constructs like two's complement and ten's complement are number representations (as are decimal numbers and Roman numerals), rather than number systems (as are the real numbers and the integers). So when you talk about ...999.999... you are talking about a scheme for encoding real numbers into written symbols, and not about a fundamental property of real numbers themselves. However, there exists a number system in which concepts like ...999 are more fundemantal: the p-adic numbers (for any prime number p, for example for p=7 there exist the 7-adic numbers). There, instead of repeating decimals like 0.272727.., you can have numbers like ..27272727.0. p-adic numbers are of theoretical interest and have some counterintuitive properties, but they aren't very applicable to everyday use. === Subject: Re: Is ...9999.9999... = 0 ? Brian Quincy Hutchings scribbled the following: > there's no opposite of a projection, but > two or more of them is a perspectivity, > viz painting (Brunelleschi et al). > the real meet (sik) of this is just in ten's complimentation, Spelling nitpick: (sic), not (sik). > in the base of ten. or nine's complimentation. that is, > when you subtract a larger number from a smaller, > this is whta you get, til you do the complimentation > to get the negative number. -- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ To err is human. To really louse things up takes a computer. - Anon === Subject: Re: Is ...9999.9999... = 0 ? Chapman (I guess) had me correctly: I use sik when I make a mistake on purpose, and sic when others make a mistake, on prupose or not. > viz painting (Brunelleschi et al). > the real meet (sik) of this is just in ten's complimentation, > Spelling nitpick: (sic), not (sik). --Dec.2000 'WAND' Chairman Paul O'Neill, reelected to Board. Newsish? http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac === Subject: Re: Is ...9999.9999... = 0 ? > Brian Quincy Hutchings scribbled the following: >> there's no opposite of a projection, but >> two or more of them is a perspectivity, >> viz painting (Brunelleschi et al). >> the real meet (sik) of this is just in ten's complimentation, > Spelling nitpick: (sic), not (sik). >> in the base of ten. or nine's complimentation. that is, >> when you subtract a larger number from a smaller, >> this is whta you get, til you do the complimentation >> to get the negative number. Annd another spelling nitpick: complementation not complimentation. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Is ...9999.9999... = 0 ? Robin Chapman scribbled the following: >> Brian Quincy Hutchings scribbled the following: > there's no opposite of a projection, but > two or more of them is a perspectivity, > viz painting (Brunelleschi et al). > the real meet (sik) of this is just in ten's complimentation, >> Spelling nitpick: (sic), not (sik). > in the base of ten. or nine's complimentation. that is, > when you subtract a larger number from a smaller, > this is whta you get, til you do the complimentation > to get the negative number. > Annd another spelling nitpick: complementation not complimentation. Ignoring the nitpick on the first word of your reply, who says it wasn't intentional? Perhaps you have to say something like: Please, O most illustrious holy one, the beacon of wisdom, emperor for eternity, may I get the 'negative' number?. -- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ The truth is out there, man! Way out there! - Professor Ashfield === Subject: Mathematical Olympics I've always wondered about Mathematical Olympics. Don't the judges have to be smarter than any of the contestants, to be able to judge them accordingly? In normal Olympics, the judges can be any skinny little wimps off the street, as they can merely watch the athletes without competing themselves. But mathematics is different. Unlike athletics, it's not plain to see who is better than who. You have to be a mathematician yourself. There's *one* thing that suggest that the judges don't have to be the smartest ones after all - in certain theoretical calculations, checking whether something has been done is far easier than actually doing it. For example consider the simple array sort. Actually sorting the array can't be done faster than O(nlog n) time, this has been proven. But checking if the array has been sorted can be done in O(n) time with a simple algorithm any child could come up with. Maybe this can be expanded to more complicated calculations? -- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ Roses are red, violets are blue, I'm a schitzophrenic and so am I. - Bob Wiley === Subject: OLD HP CALUCLATORS I am looking for some old HP calculators like HP 41CV, HP 41CX, HP 71B, HP 15C, HP 16C, HP 67 and any others in the 1980's era...If you have one you would like to sell please email me at === Subject: Re: OLD HP CALUCLATORS > I am looking for some old HP calculators like HP 41CV, HP 41CX, HP > 71B, HP 15C, HP 16C, HP 67 and any others in the 1980's era...If you > have one you would like to sell please email me at Here's a good source of older HP calculators: http://www.hpmuseum.org/cgi-sys/cgiwrap/hpmuseum/adforum.cgi -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: OLD HP CALUCLATORS I've got an old 97 but it's not a wreck so I'm not selling. Sorry > I am looking for some old HP calculators like HP 41CV, HP 41CX, HP > 71B, HP 15C, HP 16C, HP 67 and any others in the 1980's era...If you > have one you would like to sell please email me at === Subject: Re: OLD HP CALUCLATORS > I am looking for some old HP calculators like HP 41CV, HP 41CX, HP > 71B, HP 15C, HP 16C, HP 67 and any others in the 1980's era...If you > have one you would like to sell please email me at I have an HP55 and no, I'm not selling. Bought it in 1975 and it still works :-) http://www.dotpoint.com/xnumber/hp55.htm Dirk Vdm === Subject: differential equation Hi! I have to solve this differential equation: r'' = k/(|r|2) where r is a 2-d vector. Marco === Subject: Re: differential equation >Hi! I have to solve this differential equation: >r'' = k/(|r|2) >where r is a 2-d vector. What is k? (My guess would be that k was a constant, but if so this can't be exactly the equation you need to solve, because the left side is a vector and the right side is a scalar...) > Marco === Subject: Re: how to prove that this is a measurable function >>Show that >>1) (x,y) |-> x+y is measurable; >>2) if f: X -> Y, g: Y -> Z are measurable, then so is g op f >>(composition); and >>3) if f, g: X -> R are measurable, so is fg (product). >> >2) is false. The composition of Lebesgue measurable (LM) functions need not >be LM . However, compositions work for Borel measurable functions, so one >could work with that. Another approach: X_B(y) is clearly LM on R2 and >X_A(x+y) is the characteristic function of T(R x A), where T is a linear >transformation on R2. > > How could it be false? I am talking of abstract measurable spaces > here. Given a set A in the sigma field on Z, isn't (g o f)-1 (A) > = f-1 (g-1 (A)) in the sigma field on X? Is the definition of LM > only that the inverse maps Borel sets to Lebesgue sets? The problem with saying a function f is measurable with this definition is that continuous functions need not be measurable when both sigma-algebras are the collection of Lebesgue measurable sets. In fact, let g be the 'Cantor staircase' or 'devil's staircase' and let f(x)=x+g(x). Then f is strictly increasing, continuous, and takes a set of measure 0, C, to a set of positive measure f(C). Now let A be any non-measurable subset of f(C). Then f(A) is measurable, but f-1 (f(A)) is not. The point is that, while your definition is perfectly good when dealing with Borel sets, or generic sigma-algebras, it has unpleasant effects when dealing with Lebesgue measurable sets. In particular, the composition of a measurable function and a continuous one need not be measurable. However, this is the situation the OP wanted. What is needed is that inverse images of sets of measure 0 via the continuous function have to again be of measure 0. This does happen for the OP. --Dan Grubb === Subject: Re: how to prove that this is a measurable function There's a problem in a real analysis book I'm reading that asks to show that (x,y) |-> f(xy) is measurable in R2 if f(x) is measurable in R. For that, I want to show that k{-1}(E) is lebesgue measurable if E is measurable in R. I seem to be unable to show this... Any ideas? Any help appreciated, cheers, nojb. How can I prove that the function g(x,y) = X_A(x+y) X_b(y) is lebesgue > measurable in R2 where A, Bsubset R are measurable sets and X_A, X_B > are the charactersitic functions? > If A and B are Borel measurable subsets of R, then the proof you request > is easy: both (x,y) --> x+y and (x,y) --> y are continuous maps of R2 > into R, hence they are both Borel measurable. Therefore so are the > compositions (x,y) --> X_A(x+y) and (x,y) --> X_B(y) Borel measurable, > hence so is their product. > If A and B are Lebesgue measurable, then so is your function. This is > because both of the maps (x,y) --> x+y and (x,y) --> y have the property > that the inverse image (with respect to either map) of a Lebesgue > measurable set is Lebesgue measurable. The same composition argument as > before finishes the matter. === Subject: Re: how to prove that this is a measurable function > There's a problem in a real analysis book I'm reading that asks to > show that (x,y) |-> f(xy) is measurable in R2 if f(x) is measurable > in R. If E is a Borel set in R of measure 0, use Fubini's theorem to show that {(x,y) : xy lies in E} has measure 0 in R2. This implies f(xy) is a.e. equal to a Borel measurable function on R2, hence is Lebesgue measurable on R2. === Subject: Re: how to prove that this is a measurable function >2) is false. The composition of Lebesgue measurable (LM) functions need >not >be LM . However, compositions work for Borel measurable functions, so one >could work with that. Another approach: X_B(y) is clearly LM on R2 and >X_A(x+y) is the characteristic function of T(R x A), where T is a linear >transformation on R2. > >How could it be false? I am talking of abstract measurable spaces >>here. Given a set A in the sigma field on Z, isn't (g o f)-1 (A) >>= f-1 (g-1 (A)) in the sigma field on X? Is the definition of LM >>only that the inverse maps Borel sets to Lebesgue sets? >> >If f : R -> R, then a standard definition is that f is LM iff >f(-1)[(a,oo)] is LM for all a in R. > If so, the answer to my latter question is affirmative. gave. > In general > measurable spaces, a funciton is measurable iff its inverse map sets > from one sigma field into the other. Contrary to your declaration, my > original claim, > 2) if f: X -> Y, g: Y -> Z are measurable, then so is g op f > (composition) > is true. In that context, yes you are of course right. The OP was asking about LM functions, and compositions of those can give headaches. Yes, we can use your 2) in this context if we're careful to specify that the map T(x,y) = x + y is more than just LM; T(-1)(E) should be LM for every LM E contained in R. === === Subject: Re: why is Hamel dimension well-defined? > [[ This message was both posted and mailed: see > the To, Cc, and Newsgroups headers for details. ]] > Would someone please point me to a proof of this theorem? Any two Hamel > bases of a vector space have the space have the same cardinality. This is > driving me crazy. Have a tolerable existence. Eli There is a proof in Real and Abstract Analysis by Hewitt & Stromberg. My favorite one is in Jacobson's Basic Algebra II. It uses the idea of an abstract dependence relation. The nice thing about it is that it can be generalized to 'countable dependence relation' and obtain that the dimension of a Hilbert space is well-defined also. Jacobson uses it to obtain results about completely reducible modules. --Dan Grubb === Subject: Re: Rotate Quaternion numbers? > Is it possible, using quaternion numbers, to rotate 3 dimensions > around the fourth. A bit like rotating x and y around the z axis. Or > am i being stupid? > I have an object that exists in 4D Quaternion space. I can view it in > 3D (b using a simple ray-tracing technique) and i can rotate in 3D > around either of the other axis. Any ideas? The (continuous) automorphisms of the quaternions (these are the maps that preserve the structures, addition and multiplication as well as the topology) correspond to the 3D rotations, i.e. the group SO(3). Given an arbitrary nonzero quaternion u, we can construct an automorphism in defining the map q |--> u*q*(u-1) where u-1 is the inverse of u. Because any real number commutes with all quaternions, the the u deletes out when q is real; so the whole real axis is mapped onto itself. And the other three (imaginary) dimensions are rotated around the axis given by u (because u itself is also mapped to itself, so it is a fixed point) by an angle which is given by 2*arctan(Im(u)/Re(u)) (sketch of proof: calculate the trace of the above mapping as an endomorphism of a real vector space and compare to the trace 1+2*cos(phi) of a rotation of angle phi in 3D). Given two nonzero quaternions u1 and u2, they induce the same rotation if and only if their quotient is a real number. === Subject: Re: Rotate Quaternion numbers? I'm still not 100% fully understanding though. Perhaps it's the Math syntax: q' is the rotated Quaternion number yes? q is the original? That leaves a and b. Are these the 2 angles you mentioned? I need to think about this a lot more! Mat > Hello everybody, Is it possible, using quaternion numbers, to rotate 3 dimensions > around the fourth. A bit like rotating x and y around the z axis. Or > am i being stupid? > Not at all, but as you probably reaslise it's a bit more complicated in > 4D. In 4D rotations take place about planes. the simplest rotation in 4D > fixes a plane (e.g. the x-y plane) and rotates in the other two > dimensions, so a 90 degree rotation takes z -> w and w -> -z. > A more general 4D rotation taks place about two orthogonal planes and > rotates about both of them simultaneously through two different angles. To > completely specify the rotation you need to give one plane (the second is > uniquely defined by being orthogonal to the first) and both angles, + > senses of rotation if it is not obvious from the way the plane(s) are > defined. > This has a number of consequences. Unlike in 3D (or 5D or 7D) a general 4D > rotation has no fixed point except the origin. Also above 3D rotations are > not simple, meaning that for rotation R it is not generally true that Rt > == I for some t. > You can use quaternions to rotate in 4D. The formula is > q' = aqb > The pair (a, b) of unit quaternions specifies the rotation, multiplication > is just quaternion multipilication, and the pairing is unique up to > replacing (a, b) with (-a, -b). You can of course use special orthogonal > matrices, but they are in my opinion the least user friendly way to get > things done in 4D. > John === Subject: Re: Rotate Quaternion numbers? > I'm still not 100% fully understanding though. > Perhaps it's the Math syntax: > q' is the rotated Quaternion number yes? > q is the original? > That leaves a and b. Are these the 2 angles you mentioned? No. a and b are unit quaternions. The formula q' = a * q * b gives a 4D rotation, from q to q'. '*' is quaternion multiplication. It is obvious that this is a linear transformation, and the properties of quaternion multiplication and the fact that a and b are unit quaternions mean it preseves lengths and so is orthogonal. It's less obvious that all 4D rotations can be written this way, and that for each rotation the pair of rotations that generates it is unique, up to multiplying both quaternions by -1. I'm not sure how useful this method of generating 4D rotations is: it's mathematical compactness and symmetery suggests there's more to it than meets the eye, but I've not come across anything more about it than given above. I mentioned it as your post metioned 'using quaternion numbers' to do rotations in 4D, and this one way to do so. John === Subject: Re: Core error, FEAR is a natural response >: Here's a demonstration of how you can get a problem, using xy=2, where >: x and y are algebraic integers. Consider x=2a, and y = b, so b is an >: algebraic integer, but 'a' is not, so x does not technically have 2 as >: a factor in the ring of algebraic integers, as that requires *both* >: factors be algebraic integers, while 'a' is not one. So b *should* be >: a unit, but because ab=1, it is NOT a unit, because the unit >: definition would require that both 'a' and b be algebraic integers. >I don't understand how you are deciding what should and should >not be in the algebraic integers. >>Here's a simpler example, consider 2 and 6 in the ring of evens, but 2 >>is NOT a factor of 6 because 3 is not even, understand? > I'm going to join with the others in asking what point > you meant to make here regarding what should be in > the ring of algebraic integers. > You have the situation that 2a = 6. Yet a is not in the > ring of evens. Clearly 2 is a factor of 6, but 2 is not > a factor of 6 in the ring of evens. Are you saying > that a should be in the ring of evens, that there's > an error in the definition of the ring, that the ring > is incomplete? > What is the should test? The fundamental problem is that should is not a word with a generally accepted meaning in mathematics. JSH follows his own inscrutable rules, not those of mathematics. Gib === Subject: Re: Core error, FEAR is a natural response >I think many of you *should* be terrified beyond the capacity for >rational thought, which is what I'm seeing. Is this the same should as in your proof, or do the asterisks change the logical meaning? >[...]*belief*[...]*selectively*[...]*has* I see a definite pattern of usage here. Could you clarify the asterisk operator for us? === Subject: Re: Core error, FEAR is a natural response linux) > I see a definite pattern of usage here. Could you clarify the asterisk > operator for us? The asterisk operator is a standard means of emphasis (boldface) on Usenet. Some newsreaders (like Gnus) even render the boldface and omit the asterisks. Maybe you knew that, but maybe not. -- When I am grown to man's estate I shall be very proud and great, and tell the other girls and boys not to meddle with my toys. --Robert Louis Stevenson === Subject: Re: Core error, FEAR is a natural response is this the Real JSH, at MSN? anyway, I saw the problem with it, the first time you'd stated it: it's nonsensical, because you've simply exchanged the even numbers for the units; that is, it's just a matter of labelling: 2 is your unit, and 4 is your first even mumber, 6 is odd. or are the evens really a ring, since they add, subtract and multiply? there's a very thin line between trembling in fear, and shaking with laughter, dood. as with I could, but I don't think it's worth the effort, and it wouldn't really show you anything. maybe you should move on to greener pastures; I know, I said, I would! I mean, I'd spend a lot less time in sci.math, if i didn't indulge in this ridiculous pasttime. > Here's a simpler example, consider 2 and 6 in the ring of evens, but 2 > is NOT a factor of 6 because 3 is not even, understand? > Similarly, because 'a' above is not an algebraic integer, then though > b should be a unit, it's not because the *definition* for unit > requires that both 'a' and b be in the ring of algebraic integers, so > b is not a unit, which means it is a non-unit factor of 2, which gives > a false implication, since x = 2a, like with 6, 2(3) = 6. > I don't understand this definition. Could you specify a numerical > value for a, to make the example more concrete? > Well, I could, but I don't think it's worth the effort, and it > wouldn't really show you anything. > As for the definition it suffices to note that it doesn't allow 'a' to > be 1/2, as that would mean that 2 is a unit, since 2(1/2) = 1. > Given that xy=2, with x=2a, y=b, with b an algebraic integer, x an > algebraic integer, while 'a' is not, it *should* be that y does not > share non-unit factors with 2, since x *should* have 2 as a factor, > but in the ring of algebraic integers, because the definition > arbitrarily excludes 'a', neither of those is the case. --les ducs de Buffet; vote NONE OF THE BELOW on Trickier Dick Cheney's California Recall & e-Dereg! http://larouchepub.com === Subject: Re: Core error, FEAR is a natural response >What I'm facing is a situation where the math--rather basic >algebra--says I'm right. > Of course, you believe it's right. And if other people believed it > was right, then your explanation of their actions might be plausible. > But it seems to me that they *don't* believe you're right; rather they > believe you are mistaken, confused, deluded or however they think of > it. Then isn't there a rather more obvious explanation for their > actions? It's MATH. It's not about belief. A math proof begins with a truth and proceeds by logical steps to a conclusion which then MUST be true! That's what makes this situation obscene as I can step through the proof from begining to end, and mathematicians simply decide to avoid or deny. They run away. Remember besides posting on newsgroups I've spoken to a mathematician in-person and explained it step-by-step, and he was quite skeptical and challenging as to the math, though otherwise the conversation was congenial. I handled every challenge, explained every step, and showed how it flowed logically from the previous. The proof is in fact that--a proof. There is no mathematical basis for challenging it at this point, and as it is a short proof, there is no RATIONAL way that mathematicians can avoid its conclusion. It's mathematicians who are acting on belief, not me. After all, who needs belief, when you have algebra? James Harris === Subject: Re: Core error, FEAR is a natural response What I'm facing is a situation where the math--rather basic >algebra--says I'm right. Of course, you believe it's right. And if other people believed it > was right, then your explanation of their actions might be plausible. > But it seems to me that they *don't* believe you're right; rather they > believe you are mistaken, confused, deluded or however they think of > it. Then isn't there a rather more obvious explanation for their > actions? > It's MATH. It's not about belief. A math proof begins with a truth > and proceeds by logical steps to a conclusion which then MUST be true! The Harris kind of proof proceeds by steps whose meaning can only be inferred by guessing, has the odd error in it, plus some f***ing big gap right in the middle where things get interesting, and ends with a conclusion that only you could belief. === Subject: Re: Core error, FEAR is a natural response > The Harris kind of proof proceeds by steps whose meaning can only be > inferred by guessing, has the odd error in it, plus some f***ing big gap > right in the middle where things get interesting, and ends with a > conclusion that only you could belief. A concise description of Harristotelian logic... Gib === Subject: Re: Core error, FEAR is a natural response > It's MATH. It's not about belief. A math proof begins with a truth > and proceeds by logical steps to a conclusion which then MUST be true! Show me the logical step that produces the word *should*. I know how to prove formal statements A is B and A is not B with formal logic. But how to prove A should be B with logic? > I handled every challenge, explained every step, and showed how it > flowed logically from the previous. The proof is in fact that--a > proof. Explained every step? Really? And how exactly A should be B followed from the previous? === Subject: Re: Core error, FEAR is a natural response > It's MATH. It's not about belief. A math proof begins with a truth > and proceeds by logical steps to a conclusion which then MUST be true! > Show me the logical step that produces the word *should*. > I know how to prove formal statements A is B and A is not B with formal logic. > But how to prove A should be B with logic? Readers should note that I have given the proof step-by-step and it doesn't have should in it at all. In fact, besides writing a paper, which I received email confirmation yesterday is still under review at a mainstream math journal, I've given repeated explanations, in detail. Rather than deal with the actual math argument, posters pick and choose from various posts I've made explaining various aspects of the definition problem currently sitting at the heart of mathematics in an area called algebraic number theory. It has been there for over a hundred years, so necessarily it has its subtleties. > I handled every challenge, explained every step, and showed how it > flowed logically from the previous. The proof is in fact that--a > proof. > Explained every step? Really? > And how exactly A should be B followed from the previous? And again, I point out that posters pick and choose from discussions where I explain, rather than going to the posts that actually do step through the argument carefully, and in detail. It'd be one thing if it were just a Usenet problem, but my experience with Professor McKenzie of Vanderbilt University, where I explained everything in-person and handled objections as he brought them up, yet he still sought to run away by claiming it was out of his area, highlights that it's a systemic problem. Mathematicians *can* decide to ignore a correct math argument, and keep doing so, despite realizing that the negative consequences if they are caught are huge. You should think about that reality. Basically, clearly certain mathematicians know they can lie to you about mathematics. James Harris === Subject: Re: Core error, FEAR is a natural response > It's MATH. It's not about belief. A math proof begins with a truth > and proceeds by logical steps to a conclusion which then MUST be true! > Show me the logical step that produces the word *should*. > I know how to prove formal statements A is B and A is not B with formal logic. > But how to prove A should be B with logic? > Readers should note that I have given the proof step-by-step and it > doesn't have should in it at all. In fact, besides writing a paper, > which I received email confirmation yesterday is still under review at > a mainstream math journal, I've given repeated explanations, in > detail. All your short explanations of the error in core are using extremely imprecise and consufing word should. You even use asterisks, like that - *should*. So can you give the short explanation without that non-math word? > Rather than deal with the actual math argument, posters pick and > choose from various posts I've made explaining various aspects of the > definition problem currently sitting at the heart of mathematics in an > area called algebraic number theory. > It has been there for over a hundred years, so necessarily it has its > subtleties. > I handled every challenge, explained every step, and showed how it > flowed logically from the previous. The proof is in fact that--a > proof. > Explained every step? Really? > And how exactly A should be B followed from the previous? > And again, I point out that posters pick and choose from discussions > where I explain, rather than going to the posts that actually do step > through the argument carefully, and in detail. > It'd be one thing if it were just a Usenet problem, but my experience > with Professor McKenzie of Vanderbilt University, where I explained > everything in-person and handled objections as he brought them up, yet > he still sought to run away by claiming it was out of his area, > highlights that it's a systemic problem. > Mathematicians *can* decide to ignore a correct math argument, and > keep doing so, despite realizing that the negative consequences if > they are caught are huge. > You should think about that reality. > Basically, clearly certain mathematicians know they can lie to you > about mathematics. > James Harris === Subject: Re: {Field Theory} This notation could get very confusing very fast... > I am reading Herstein's Abstract Algebra and teaching myself field > theory. I came to the following and it gave me a bit of trouble (I am > paraphrasing it for people who don't own that book): > If F is a finite field with q elements, then viewing F simply as an > abelian group under its addition, +, we have from group theory that > qx=0 for all x in F. > Now after a bit of consideration I came to the conclusion that, yes, > this is true, if we go by the notation (which Herstein did not > explicitly define) that qx, where q is a counting number and x is an > abstract object, means x+x+x+...+x, q times. Yes, Herstein could have made this more explicit in the discussion on groups. The point is that xn for a multiplicative group is written n*x for an additive group. > In terms of fields > this takes a little bit of chewing before swallowing since now we have > not one but two very different kinds of multiplication, each of which > uses the same symbol. This happens often. For example, when a field K contains a subfield F, you can view K as a vector space with F as scalars. Then, f*k could either be field multiplication in K or scalar multiplication in vector space K. > This is still unambiguous when the elements of our field are abstract > things, or at least anything BUT counting numbers. But if our field > is in fact counting numbers, it brings up a very gaping problem, at > least as I see things (although, true, I am very much a newbie!) [cut] > I am probably just overlooking something obvious, and am sure I will > be corrected! I don't think you are overlooking anything. Yes, the above notation can lead to ambiguity. If that happens, then you have to use a more precise, explicit notation to distinguish the field multiplication from the group repetition operation. Such sloppiness is very common in mathematics. -- Bill Hale === Subject: Re: {Field Theory} This notation could get very confusing very fast... at 09:48 PM, snizpilbor@yahoo.com (Sniz Pilbor) said: >In terms of fields >this takes a little bit of chewing before swallowing since now we >have not one but two very different kinds of multiplication, each of >which uses the same symbol. Correct; it is quite common in Mathematics to use the same symbol for related notions and to epect the reader to identify by context which is meant. >This is still unambiguous when the elements of our field are >abstract things, It's unambiguous across the board. >But if our field is in fact counting numbers, What do you mean by counting numbers? Integers? They don't form a field. >To be specific, suppose F is a field of counting numbers, There is no such thing. What is the multiplicative inverse of 2? Now, perhaps you meant the ring of integers, but there is still no ambiguity. >What, then, are we to make of ab? It is ambiguous No it isn't. >has the three possible meanings: No. If you apply the distributive law you will see that all three expressions have the same value. >1. a mulitplied by b by the multiplication operation of F (1+1+...1 (a times))*(1+1+...1 (b times)) >2. a+a+a+...+a (b times), (1+1+...1 (a times))*(1+1+...1 (b times)) >3. b+b+b+...+b (a times) (1+1+...1 (a times))*(1+1+...1 (b times)) >Was Herstein's choice of symbols merely a bad fluke? No. It was normal and consistent. >else it seems that the study of fields of integers There are none. Perhaps you mean rings. >would soon be rendered hopelessly ambiguous No. >and context-dependant. Yes, but it is a convenient and harmless context dependence, and can easily be removed if your willing to use more cumbersome expressions. >I am probably just overlooking something obvious, See above. > When you say 2 = 1+1, by your + are you talking about >normal addition, or the field's addition? It doesn't matter, because you get the same result either way. There is only one nondegenerate homomorphism of Z into a ring or field, so it is convenient to use the same symbol for the integer 1 and the identity of the ring or field. >In a very well behaved field, What is a well behaved field? The relevant porperties apply to *ALL* fields. >but in a field of integers What is that? >whose addition and multiplication have nothing to do with normal >addition and multiplication, It doesn't matter, as long as they satisfy the requirements for a field. >it becomes ambiguous. Take your field F. It has a unique additive identity, all it O, and a unique multiplicative Identity, call it I. There is a unique nondegenerate homomorphism E: Z->F, since E(0) must be O, E(1) must be I and all other values are determined by the fact that it is a homomorphism. Now if q is a positive integer and b is in F, b+b+...b (q times) is b*I+b*I+...b*I (q times). Now apply the distributive law and you get b*(I+I+...I (q times)). But (I+I+...I (q times)) is G(q). So there is no ambiguity in overloading the symbols. -- spamtrap@library.lspace.org === Subject: Re: {Field Theory} This notation could get very confusing very fast... > As F is a field, it has an multiplicative identity usually notated 1. > Thus 1x = x is no problem. If you consider 2 = 1+1, etc., > then '2'x = (1+1)x = 1x + 1x = x+x = 2x, etc. > Thus you've your situation. > When you say 2 = 1+1, by your + are you talking about > normal addition, or the field's addition? In the above all was within the field except for = 2x where I use the notation (2)x = x+x where (2) is the integer 2 and '2' = 1+1 in field. By convention nx = x+x+..+x unto n x's but xn isn't used where n is integer and x is group element. === Subject: Re: JSH: Harris's Big Fat Blunder > James Harris posted a new version of his proof of a core > error on October 12. He made a simple algebra mistake > which he has since acknowledged. > The interesting thing about this is that, when the algebra > mistake is corrected, it leads immediately to a proof that > him main conclusions are wrong. ... JSH will ignore your post. Gib === Subject: Re: JSH: Harris's Big Fat Blunder Nora Badboobda Baron, you stole that line from Captain Arnie! oh, wait; that was Peter Townsend. nevermind. PS: just as I've been saying (for years, in hte larger aspect), Captain Arnie has immediately proposed complete dereg of energy, along with green supplies, no offshore drilling, and not happy with the President on the Clean Air Act. at least, he didn't come out against nuclear power. Where's Warren? > the ball game is over: because in that case, b1 and > b2 CANNOT be algebraic integers. --les ducs de Buffet; vote NONE OF THE BELOW on Trickier Dick Cheney's California Recall & e-Dereg! http://larouchepub.com === Subject: Simple combinations/Permutations question I'm not much of a mathmatician, but was reviewing the definitions of combinations and permutations today. permutations is that with permutations, order matters. The question I have is if thisis the case, why do we talk about a safe(e.g. in a bank) as having a combination ? Shouldn't it be a permutation ? To open a bank safe, order matters, so why don't we say the permutation to the safe ? Is the phrase combination to the safe mathmatically incorrect ? Any help in enlightenting me in this would most likely ensure I sleep soundly tonight. -NH === Subject: Re: Simple combinations/Permutations question Visiting Assistant Professor at the University of Montana. >I'm not much of a mathmatician, but was reviewing the definitions of >combinations and permutations today. >permutations is that with permutations, order matters. >The question I have is if thisis the case, why do we talk about a >safe(e.g. in a bank) as having a combination ? Because you are talking about different things. When we say a safe has a combination, we are not talking about mathematics, and we are not talking about the ->technical<- meaning of combinations and permutations. Just as when you talk about adding someone to the group you do not mean that you will put a big + sign between that person and the group, and then come up with a number... > Shouldn't it be a >permutation ? To open a bank safe, order matters, so why don't we say >the permutation to the safe ? Is the phrase combination to the safe >mathmatically incorrect ? The phrase has no mathematical meaning in the first place, so it is neither correct nor incorrect. Combination to the safe is an English sentence, not a mathematical statement. It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Variation on an impossible problem >Allowing only the use of the + operator, is it possible to use each of >the digits 0 1 2 3 4 5 6 7 8 9 once only to make a total of 1000? >One can combine digits to create sums like 210 + 345 + 6 + 78 + 9 = >648 etc. >>No. sum(i,i=0..9) = 0 mod 9, whereas 1000 = 1 mod 9. > I could possibly form must be divisible be nine? I can see that if all > such sums must be divisible by nine then 1000 cannot be a possible > sum. > Moreover, I can see that sum(i,i=0..9) = 0 mod 9 but how does this > generalise to all possible sums formed? > Mitch. Note that every number N with digits d(n)n, d(n-1), ..., d(1), d(0): N = d(n) d(n-1) d(n-2) ... d(1) d(0) can be written as follows: N = sum(10(k) d(k)), where the index runs from 0 to n. Since 10 is congruent to 1 modulo 9, one also has 10k congruent to 1 modulo 9, and thus N is congruent to sum(d(k)) modulo 9. Any sum of numbers thus obtained (from the digits 0, ... ,9) is congruent to the sum of the individual numbers, and each individual is congruent to the sum of its digits. Thus, the sum of all the numbers is congruent to the sum of the full collection of digits used. Dale. === Subject: help me....my teacher....my problem is~~ let |z-10i | =6 let x : argument of z find max * min of 6cos(x) + 8sin(x) -------------------------- i wait your hot-advice...please... thank in advance. === Subject: Re: help me....my teacher....my problem is~~ > let |z-10i | =6 A circle radius 6 centred at 10i > let x : argument of z > find max * min of 6cos(x) + 8sin(x) 6 and 8, eh? sin and cos (at right angles) eh? Combine those to get 10sin(x+a) for some a. Then you just want to find the max and the min. If (x+a) stays within (-Pi/2, Pi/2), then the the min and max occur at extremal values of x, as 10sin(x+a) is monotone increasing in that range. If however, (x+a) exceeds either or both those bounds then the max and/or min will be +10 or -10 respectively. Phil -- Unpatched IE vulnerability: window.open search injection Description: cross-domain scripting, cookie/data/identity theft, command execution Reference: http://safecenter.net/liudieyu/WsFakeSrc/WsFakeSrc-Content.HTM Exploit: http://safecenter.net/liudieyu/WsFakeSrc/WsFakeSrc-MyPage.htm === Subject: Re: help me....my teacher....my problem is~~ > If (x+a) stays within (-Pi/2, Pi/2), then the the min > and max occur at extremal values of x, as 10sin(x+a) > is monotone increasing in that range. > If however, (x+a) exceeds either or both those bounds then > the max and/or min will be +10 or -10 respectively. ------------------ i am not understand. i think that interval of x is less than pi. addition explanation ...please === Subject: signed graph = directed graph ? A signed graph is just an ordinary graph with each of its edges labeled with either a + or a -. ... Those of you who are really quick may have noticed that this labeling is basically just labeling each edge with an element of GF(2). What is the difference between signed graph and directed graph? === Subject: Re: Finishing argument, core error proven > For me there have been two perspectives as I work to figure out how to > explain the definition problem in mathematics with LOTS of opposition, > and I wonder about mathematicians so dedicated to attacking an > argument that is clearly correct. I remind of that as I present what should finish their ability to > distract, as I've seen a strange and dedicated effort to ignore the > actual math, and simply toss up just about anything rather than face > the truth. All variables are in the ring of algebraic integers unless otherwise > stated. Let P(m) = f2((m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f) and let R(m) = (m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f so P(m) = f2 R(m). Now consider P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where the a's are given by the following cubic: a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). Then it must be true that the following factorization exists R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where the b's are given by the following cubic: b3 + 3(-1+mf2)b2 - (m3 f4 - 3m2 f2 + 3m), > should be > b3+...+ 3(-1+mf2)b2+...- (m3 f4 - 3m2 f2 + 3m) > where a_3 = b_3, and at m=0, b_3 = 3. > It's not true in general that a_3 = b_3. Um, here's one spot where I got it wrong, when I was right the first time, as pointed out in this thread on sci.math by the poster Arturo Magidin. And in fact, in general, a_3 = b_3. The cubic I gave was wrong though, again, as the correct cubic is b3+ (...)b2 + (...)b+...- (m3 f4 - 3m2 f2 + 3m) which may seem strange, but consider m=1, f=sqrt(2), gives b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 where you can see what's hidden in general. James Harris === Subject: Re: Finishing argument, core error proven > For me there have been two perspectives as I work to figure out how to > explain the definition problem in mathematics with LOTS of opposition, > and I wonder about mathematicians so dedicated to attacking an > argument that is clearly correct. I remind of that as I present what should finish their ability to > distract, as I've seen a strange and dedicated effort to ignore the > actual math, and simply toss up just about anything rather than face > the truth. All variables are in the ring of algebraic integers unless otherwise > stated. Let P(m) = f2((m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f) and let R(m) = (m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f so P(m) = f2 R(m). Now consider P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where the a's are given by the following cubic: a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). Then it must be true that the following factorization exists R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where the b's are given by the following cubic: b3 + 3(-1+mf2)b2 - (m3 f4 - 3m2 f2 + 3m), > should be > b3+...+ 3(-1+mf2)b2+...- (m3 f4 - 3m2 f2 + 3m) > where a_3 = b_3, and at m=0, b_3 = 3. > It's not true in general that a_3 = b_3. Um, here's one spot where I got it wrong, when I was right the first time, as pointed out in this thread on sci.math by the poster Arturo Magidin. And in fact, in general, a_3 = b_3. The cubic I gave was wrong though, again, as the correct cubic is b3+ (...)b2 + (...)b+...- (m3 f4 - 3m2 f2 + 3m) which may seem strange, but consider m=1, f=sqrt(2), gives b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 where you can see what's hidden in general. James Harris === Subject: Re: Finishing argument, core error proven > [.snip.] >> This is not correct. Note that a1/f = b1, or a1 = f*b1. >> Putting this into the equation above that the a's >> satisfy and simplifying, one obtains >> f*b13 + 3*(-1 + m*f2)*b12 - (m3*f4 - 3*m2*f2 + 3*m). >> Clearly b2 is a root of the same polynomial. Since >> b3 = a3, b3 satisfies a different equation: >> b33 + 3*(-1+mf2)b32 - f2(m3 f4 - 3m2 f2 + 3m) = 0. >Um, Nora Baron do you accept that there exists a SINGLE cubic which >has the b's as it's roots? >That is, ONE cubic with the roots b_1, b_2, and b_3? > Um, James Harris, since b3=a3, but b2 is not equal to a2 and b1 is > not equal to a1, they cannot be roots of the same cubic UNLESS the > cubic that defines the a's is reducible. It's not true that a_3 = b_3 in general, which I noted in a follow-up post. What is true is that at m=0, a_3 = b_3 = 3. Now then, there MUST exist a *single* cubic that has b_1, b_2 and b_3 as its roots, and in fact, at m=0, I can give that single cubic as b3 - 3b2, whose roots correctly give b_1 = b_2 = 0, and b_3 = 3. No other primitive cubic exists with the same roots. The SINGLE cubic is important as it can't be non-monic. If it were non-monic, then how would it give a monic at m=0? Well you may try to find some function of m that behaves that way, but then you have to reconcile that with R(m) = (m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where clearly, there's no way that a non-monic will work. For readers confused here, consider that a non-monic is something like P(x) = 2x + 1, where the leading coefficient is not 1 or -1. My point is that the SINGLE cubic defining the b's MUST be monic. > If the cubic that defines the a's is irreducible, then any cubic with > integer coefficients that has a3 as a root must have a1 and a2 as > roots, and that's all the roots, and none of them are equal to b1 or > to b2. So in order to be able to have a cubic that has a3 as a root, > and also b1 and b2, then the original cubic defining a1, a2, and a3 > must have a linear factor corresponding to a3, which means that a3 > must be an integer. It is true that the SINGLE cubic defining the b's can't have all algebraic integers coefficients for all m, though at m=0, it is b3 - 3b2, so it CAN for certain values of m. In general the defining cubic is b3+...+ 3(-1+mf2)b2 +...- (m3 f4 - 3m2 f2 + 3m) where in general certain terms are inexpressible and they are not, in general, algebraic integers. James Harris === Subject: Re: Finishing argument, core error proven Visiting Assistant Professor at the University of Montana. Cc: >> [.snip.] > This is not correct. Note that a1/f = b1, or a1 = f*b1. > Putting this into the equation above that the a's > satisfy and simplifying, one obtains f*b13 + 3*(-1 + m*f2)*b12 - (m3*f4 - 3*m2*f2 + 3*m). Clearly b2 is a root of the same polynomial. Since > b3 = a3, b3 satisfies a different equation: b33 + 3*(-1+mf2)b32 - f2(m3 f4 - 3m2 f2 + 3m) = 0. >>Um, Nora Baron do you accept that there exists a SINGLE cubic which >>has the b's as it's roots? >>That is, ONE cubic with the roots b_1, b_2, and b_3? >> Um, James Harris, since b3=a3, but b2 is not equal to a2 and b1 is >> not equal to a1, they cannot be roots of the same cubic UNLESS the >> cubic that defines the a's is reducible. >It's not true that a_3 = b_3 in general, which I noted in a follow-up >post. And unless you are in a specific family of cases, you must have a_3=b_3 in every other case. >What is true is that at m=0, a_3 = b_3 = 3. >Now then, there MUST exist a *single* cubic that has b_1, b_2 and b_3 >as its roots, and in fact, at m=0, I can give that single cubic as Sigh. Given any three COMPLEX numbers, there exists a cubic that has those three complex numbers as roots. Namely, if you have the complex numbers a, b, and c, then the single cubic that has those three complex numbers as roots is (x-a)(x-b)(x-c). That's basic precalculus stuff. The point, however, is that any old cubic is not good enough. If you want to conclude stuff about the integrality of b_1, b_2, and b_3 (e.g., whether or not they are algebraic integers) then you need the cubic to be monic and to have algebraic integer (or better yet, integer) coefficients. In any case when b_3=a_3 (which is every case in which a_1x is not equal to -uf and a_2x is not equal to -uf), you have b_3=a_3 but b_2 different from a_1 and a_2, and b_1 different from a_1 and a_2 (assuming f is not a unit, which you do). IF the polynomial that defines the a's is irreducible over Q, then ANY polynomial with integer coefficients that has a_3 as a root must be a multiple of the polynomial that defines the a's. And therefore, it is either of degree strictly larger than 3, or else it is a scalar multiple of the polynomial that defines the a's and has exactly a_1, a_2, and a_3 as roots. Therefore, in all such cases, it is impossible for there to be a single cubic WITH INTEGER COEFFICIENTS which has b_1, b_2, and b_3 as roots. This is simple, basic, precalculus algebra of polynomials. >The SINGLE cubic is important as it can't be non-monic. >If it were non-monic, then how would it give a monic at m=0? Let f(x) = (2m+1)x3 - 3x2 +1. It's a single cubic. It gives a monic cubic when m=0. Are you claiming that it is ALWAYS monic? [.rest deleted.] 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 critize. A great many people are staggered to this extend, 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 === Subject: Re: Finishing argument, core error proven >For me there have been two perspectives as I work to figure out how to >explain the definition problem in mathematics with LOTS of opposition, >and I wonder about mathematicians so dedicated to attacking an >argument that is clearly correct. I remind of that as I present what should finish their ability to >distract, as I've seen a strange and dedicated effort to ignore the >actual math, and simply toss up just about anything rather than face >the truth. All variables are in the ring of algebraic integers unless otherwise >stated. Let P(m) = f2((m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f) and let R(m) = (m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f so P(m) = f2 R(m). Now consider P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where the a's are given by the following cubic: a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). Then it must be true that the following factorization exists R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where the b's are given by the following cubic: b3 + 3(-1+mf2)b2 - (m3 f4 - 3m2 f2 + 3m), > This is not correct. Note that a1/f = b1, or a1 = f*b1. > Putting this into the equation above that the a's > satisfy and simplifying, one obtains f*b13 + 3*(-1 + m*f2)*b12 - (m3*f4 - 3*m2*f2 + 3*m). Clearly b2 is a root of the same polynomial. Since > b3 = a3, b3 satisfies a different equation: b33 + 3*(-1+mf2)b32 - f2(m3 f4 - 3m2 f2 + 3m) = 0. Um, Nora Baron do you accept that there exists a SINGLE cubic which > has the b's as it's roots? That is, ONE cubic with the roots b_1, b_2, and b_3? If so, then your objection fails as provably that cubic must be > non-monic. Do you understand? No, no, no. You made a simple mistake in algebra. Check it > over again. One clue is that your equation for the b's > is different from your equation for the a's, even though > b3 = a3. > As for b1 and b2: just take your equation for the a's > (which is correct) and make the substitution a = f*b, and > do the algebra. > Another clue: note that YOUR equation for the b's is monic. > If it were correct, you could conclude immediately, without > having to think about R(0) etc., that the b's were all > algebraic integers. Easier than you thought, eh? But the > correct equation for b1 and b2 however is *non-monic* ! That's not possible. Readers should note that the b's vary with the value of m, and at m=0, you have b3 - 3b2 as the SINGLE cubic which correctly gives the value for ALL of the b's, as then you have b_1 = b_2 = 0, and b_3 = 3. Now Nora Baron needs that SINGLE cubic to be non-monic for some values of m, but that leads to a problem, there's no mathematical way for the cubic to be non-monic, and have b3 - 3b2, which is clearly monic, at m=0. Those of you who think m=0 is a special case can have fun trying to find some function of m, such that the cubic will be non-monic, as needed for *some* values. Now then Nora Baron is correct in noting that if the cubic had all algebraic integer coefficients, it would have algebraic integer roots, so guess what? It's coefficients, in general, are not all algebraic integers, which is why you need ... to in any way display the cubic in general. James Harris === Subject: Re: Finishing argument, core error proven ... > Now Nora Baron needs that SINGLE cubic to be non-monic for some > values of m, but that leads to a problem, there's no mathematical way > for the cubic to be non-monic, and have b3 - 3b2, which is clearly > monic, at m=0. f_0(m).b3 + f_1(m).b2 + f_2(m).b + f_3(m), where the f_i(m) are algebraic integer functions of m and f_0(0) = 1, f_1(0) = -3, f_2(0) = 0 and f_3(0) = 0? I think that is the situation you have. -- === Subject: Re: Finishing argument, core error proven > ... > Now Nora Baron needs that SINGLE cubic to be non-monic for some > values of m, but that leads to a problem, there's no mathematical way > for the cubic to be non-monic, and have b3 - 3b2, which is clearly > monic, at m=0. > f_0(m).b3 + f_1(m).b2 + f_2(m).b + f_3(m), > where the f_i(m) are algebraic integer functions of m and f_0(0) = 1, > f_1(0) = -3, f_2(0) = 0 and f_3(0) = 0? Which would require that f_3(m)/f_0(m) = -(m3 f4 - 3m2 f2 + 3m) so then, does anyone else besides this poster accept that possibility? The poster deleted out important information, which I'll replace. R(m) = (m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) > I think that is the situation you have. Well then do you believe that f_3(m)/f_0(m) can equal what it does and ever have f_0(m) NOT be a factor of f_3(m)? Readers, this post is a checkmate post. Oh yeah, I also need to correct my previous posts as I've apparently given the wrong cubic for the b's as verified by a test. The proper cubic is b3 + (...)b2 + (....)b -(m3 f4 - 3m2 + 3m) and I found out that my cubic was wrong by using m=1, f=sqrt(2), which gives the following cubic for the b's: b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 Those of you with sufficient mathematical sophistication will realize this is a checkmate post, and will hopefully not force me to spell it all out in detail before conceding to the math. I don't know why any of you decided to fight the math anyway. James Harris === Subject: Re: Finishing argument, core error proven > ... > Now Nora Baron needs that SINGLE cubic to be non-monic for some > values of m, but that leads to a problem, there's no mathematical way > for the cubic to be non-monic, and have b3 - 3b2, which is clearly > monic, at m=0. > > f_0(m).b3 + f_1(m).b2 + f_2(m).b + f_3(m), > > where the f_i(m) are algebraic integer functions of m and f_0(0) = 1, > f_1(0) = -3, f_2(0) = 0 and f_3(0) = 0? > Which would require that > f_3(m)/f_0(m) = -(m3 f4 - 3m2 f2 + 3m) > so then, does anyone else besides this poster accept that possibility? Oh, yes, for each integer m. But are f_1(m)/f_3(m) and f_2(m)/f_3(m) algebraic integers for all m? Consider the cubic (x-b1)(x-b2)(x-b3), obviously this is a monic cubic with constant term -b1.b2.b3 = -(m3.f4 - 3m2.f2 + 3m). However we do not know whether the coefficients of b2 and b are algebraic integer for all m. But for each m they are algebraic numbers. So choose an arbitrary m: m0. Say the coefficient for b2 is c2 and the coefficient for b is c1. We can write c2 and c1 as the quotient of an algebraic integer and a normal integer. Take r0 the lowest common multiple of the two normal integers. Multiply the polynomial by r0, then r0.(b-b1)(b-b2)(b-b3) has algebraic integer coefficients for m=m0. Define the f_i(m0) accordingly. Do that for each integer m and you get the given functions f_i(m). Note that the r's depend on m! > I think that is the situation you have. > Well then do you believe that f_3(m)/f_0(m) can equal what it does and > ever have f_0(m) NOT be a factor of f_3(m)? O, yes for each integer m, f_0(m) is a factor of f_3(m). So what? > Readers, this post is a checkmate post. O. Why? This is not sufficient to show that the b's are algebraic integers. To show that you must also have that f_0(m) is for each m a factor (in the algebraic integers) of f_1(m) and f_2(m). To show that you must have that the coefficients of (b-b1)(b-b2)(b-b3) are algebraic integers, and to show that you must have that b1, b2 and b3 are algebraic integers. Looks a bit circular to me. -- === Subject: Re: Finishing argument, core error proven > ... > Now Nora Baron needs that SINGLE cubic to be non-monic for some > values of m, but that leads to a problem, there's no mathematical way > for the cubic to be non-monic, and have b3 - 3b2, which is clearly > monic, at m=0. f_0(m).b3 + f_1(m).b2 + f_2(m).b + f_3(m), where the f_i(m) are algebraic integer functions of m and f_0(0) = 1, > f_1(0) = -3, f_2(0) = 0 and f_3(0) = 0? Which would require that > f_3(m)/f_0(m) = -(m3 f4 - 3m2 f2 + 3m) > so then, does anyone else besides this poster accept that possibility? > Oh, yes, for each integer m. But are f_1(m)/f_3(m) and f_2(m)/f_3(m) > algebraic integers for all m? Consider the cubic (x-b1)(x-b2)(x-b3), > obviously this is a monic cubic with constant term -b1.b2.b3 = > -(m3.f4 - 3m2.f2 + 3m). However we do not know whether the > coefficients of b2 and b are algebraic integer for all m. But for each > m they are algebraic numbers. So choose an arbitrary m: m0. Say the > coefficient for b2 is c2 and the coefficient for b is c1. We can write > c2 and c1 as the quotient of an algebraic integer and a normal integer. > Take r0 the lowest common multiple of the two normal integers. Multiply > the polynomial by r0, then > r0.(b-b1)(b-b2)(b-b3) > has algebraic integer coefficients for m=m0. Define the f_i(m0) > accordingly. Do that for each integer m and you get the given functions > f_i(m). Note that the r's depend on m! Well readers, did anyone see what's wrong with this poster's position? He suggests > f_0(m).b3 + f_1(m).b2 + f_2(m).b + f_3(m), where the f_i(m) are algebraic integer functions of m and f_0(0) = 1, > f_1(0) = -3, f_2(0) = 0 and f_3(0) = 0? But the poster fails to point out that his position requires that for certain values of m and another important variable I call f, that his f_0(m) *must* have f as a factor, so it's basically a step function. Is that the right word step function? Here's important information left out by the poster, including where the b's fit into the big picture: P(m) = f2((m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f) P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). R(m) = (m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) Here reducibility over rationals of the cubic defining the a's a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m) has been given as an issue, and the posters position requires that his f_0(m) vary such that it's 1 if that cubic is reducible over rationals, and f, if it's irreducible over rationals. > I think that is the situation you have. Well then do you believe that f_3(m)/f_0(m) can equal what it does and > ever have f_0(m) NOT be a factor of f_3(m)? > O, yes for each integer m, f_0(m) is a factor of f_3(m). So what? I'll direct a question to the poster now. Well, then, what is f_0(m) if a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m) is irreducible over Q? > Readers, this post is a checkmate post. > O. Why? This is not sufficient to show that the b's are algebraic > integers. To show that you must also have that f_0(m) is for each m > a factor (in the algebraic integers) of f_1(m) and f_2(m). To show that > you must have that the coefficients of (b-b1)(b-b2)(b-b3) are > algebraic integers, and to show that you must have that b1, b2 and b3 > are algebraic integers. Looks a bit circular to me. The poster has no room to run and has walked into the checkmate. All that's left is for the poster to concede (or run away) that his position requires that f_0(m) behave as described. Checkmate. James Harris === Subject: Re: Finishing argument, core error proven > ... >b3 + 3(-1+mf2)b2 - (m3 f4 - 3m2 f2 + 3m), This is not correct. Note that a1/f = b1, or a1 = f*b1. > Putting this into the equation above that the a's > satisfy and simplifying, one obtains f*b13 + 3*(-1 + m*f2)*b12 - (m3*f4 - 3*m2*f2 + 3*m). Clearly b2 is a root of the same polynomial. Since > b3 = a3, b3 satisfies a different equation: b33 + 3*(-1+mf2)b32 - f2(m3 f4 - 3m2 f2 + 3m) = 0. Um, Nora Baron do you accept that there exists a SINGLE cubic which > has the b's as it's roots? > Nora Baron *only* state that the single cubic you give is not one which > has the b's as its roots. Um, do *you* accept that there exists a SINGLE cubic which has the b's as its roots? Readers, I'm emphasizing this point because I know that the characteristics of that single cubic must be such as to end the debate. Now, this poster and others would like to throw TWO cubics at you because you can't tell from TWO cubics where for one only two of the roots give you b_1 and b_2, while for the other only one of the roots gives you b_3. So what good are their cubics, when I can give ONE which brings b_1, b_2 and b_3 together? I'm emphasizing the *single* cubic because it must be monic. > If so, then your objection fails as provably that cubic must be > non-monic. > There is such a cubic indeed. Possibly it is non-monic, more likely though, > it does not have integer coefficients. It can't be non-monic as then you wouldn't get b_1=b_2=0, b_3=3 at m=0, which requires that the cubic, at m=0, MUST be b3 -3b2 = 0, as no other primitive cubic exists that will give the appropriate roots. So at a minimum you MUST concede the cubic at m=0. Do you concede it Dik T. Winter? > ... Not correct, as noted above. And as I noted above, this poster seems to be lost on the idea that > given b_1, b_2 and b_3 there must exist a *single* cubic for which > they are the roots, as this poster tried to give TWO, which is rather > interesting nonsense. > I thought it is *your* obsession that there must exist a *single* cubic > for which they are the roots. Good for you, there is such a polynomial, > alas, most likely not one with integer coefficients. It has integer coefficients at m=0, as then it's just b3 - 3b2 and readers note the dancing. I'm emphasizing this single cubic because its very important, as if it's conceded that it's monic in general, then the debate is over. > It doesn't, except when m = 0. For example, if m = 1 > and f = 5 and u = 1, the CORRECT equation for b1 and b2 > becomes 5*b3 + 72*b2 - 553 = 0. a primitive irreducible *non-monic* polynomial with integer > coefficients. This implies that b1, for example, > cannot be an algebraic integer. Since b1 = a1/f = a1/5, > this implies that a1 does not have a factor of f = 5. But must there not exist a SINGLE cubic for which b_1, b_2 and b_3 are > roots? > There is: (x - b1)(x - b2)(x - b3). But, does it have integer coefficients, > or rational coefficients? Not when f = 5 and u = 1, as shown above. But is it monic? > Then why are you the one giving two cubics in one case, when you > didn't try to give two cubics for the a's? > Because the a's satisfy a single cubic *with integer coefficients*. > The b's do *not* so in general. The important question for the SINGLE cubic for which the b's are roots is, IS IT MONIC? > Actually with a *single* cubic, the algebra shows quite clearly that I > can't be wrong, which is probably why you tossed out two cubics, as if > no one would realize that a single cubic will suffice for three > values. > Pray give the single cubic with b1, b2 and b3 as roots. And *show* that > the coefficients are integer. The coefficients aren't integers, and in fact, they're not even algebraic integers, in general, though they may be algebraic integers for specific values of m and f. But you see, that's the point, the ring of algebraic integers isn't big enough. Considering that SINGLE cubic proves it. James Harris === Subject: Re: Finishing argument, core error proven > ... >b3 + 3(-1+mf2)b2 - (m3 f4 - 3m2 f2 + 3m), > > This is not correct. Note that a1/f = b1, or a1 = f*b1. > Putting this into the equation above that the a's > satisfy and simplifying, one obtains > > f*b13 + 3*(-1 + m*f2)*b12 - (m3*f4 - 3*m2*f2 + 3*m). > > Clearly b2 is a root of the same polynomial. Since > b3 = a3, b3 satisfies a different equation: > > b33 + 3*(-1+mf2)b32 - f2(m3 f4 - 3m2 f2 + 3m) = 0. > > Um, Nora Baron do you accept that there exists a SINGLE cubic which > has the b's as it's roots? > > Nora Baron *only* state that the single cubic you give is not one which > has the b's as its roots. > Um, do *you* accept that there exists a SINGLE cubic which has the b's > as its roots? Yes, I do, and I think everyone does. (x-b1)(x-b2)(x-b3) serves just fine. It is a cubic in b, but as the b's depend on m, it is not necessarily a cubic in m. > Readers, I'm emphasizing this point because I know that the > characteristics of that single cubic must be such as to end the > debate. Ah. > So what good are their cubics, when I can give ONE which brings b_1, > b_2 and b_3 together? > I'm emphasizing the *single* cubic because it must be monic. Pray show. (I think you mean a cubic polynomial?) > If so, then your objection fails as provably that cubic must be > non-monic. > > There is such a cubic indeed. Possibly it is non-monic, more likely though, > it does not have integer coefficients. > It can't be non-monic as then you wouldn't get b_1=b_2=0, b_3=3 at > m=0, which requires that the cubic, at m=0, MUST be b3 -3b2 = 0, as > no other primitive cubic exists that will give the appropriate roots. > So at a minimum you MUST concede the cubic at m=0. At m=0 it is b2.(b-3). > Do you concede it Dik T. Winter? So, yup, no problem. > I thought it is *your* obsession that there must exist a *single* cubic > for which they are the roots. Good for you, there is such a polynomial, > alas, most likely not one with integer coefficients. > It has integer coefficients at m=0, as then it's just > b3 - 3b2 Oh, you noted that already. > But must there not exist a SINGLE cubic for which b_1, b_2 and b_3 are > roots? > > There is: (x - b1)(x - b2)(x - b3). But, does it have integer coefficients, > or rational coefficients? Not when f = 5 and u = 1, as shown above. > But is it monic? As written, yes it is monic. No problem. But it does not necessarily have integer coefficients, so what do you conclude from it being monic? > Then why are you the one giving two cubics in one case, when you > didn't try to give two cubics for the a's? > > Because the a's satisfy a single cubic *with integer coefficients*. > The b's do *not* so in general. > The important question for the SINGLE cubic for which the b's are > roots is, IS IT MONIC? No. The important question is: is it monic and does it have integer coefficients? Otherwise you can not conclude anything about the b's being algebraic integers. > Actually with a *single* cubic, the algebra shows quite clearly that I > can't be wrong, which is probably why you tossed out two cubics, as if > no one would realize that a single cubic will suffice for three > values. > > Pray give the single cubic with b1, b2 and b3 as roots. And *show* that > the coefficients are integer. > The coefficients aren't integers, and in fact, they're not even > algebraic integers, in general, though they may be algebraic integers > for specific values of m and f. They can be made algebraic integers for *all* values of m and f. That is not the problem. Let's take (x-b1)(x-b2)(x-b3); all b's are algebraic numbers, so all coefficients are algebraic numbers. Each algebraic number can be written as the quotient of an algebraic integer and a normal integer. Do so. Multiply by the LCM of the denominators and you have a polynomial with algebraic integer coefficients of which the three b's are roots. The question is: is that polynomial monic? The answer is: in general, no. > But you see, that's the point, the ring of algebraic integers isn't > big enough. Big enough for what? For b1, b2 and b3 all three being algebraic integers? Yes, that is right. This simply means that your argument that they are algebraic integers fails. They may be objects in some other ring (the algebraic numbers for instance), but your proof hinge on the concept of co-primeness, and when you use that concept you have to show what ring you are using, and what definition of co-primeness you are using and a whole lot of other things. Your definition of object ring is incomplete as it is impossible to determine whether numbers are in it or not. -- === Subject: Re: Finishing argument, core error proven > ... >b3 + 3(-1+mf2)b2 - (m3 f4 - 3m2 f2 + 3m), This is not correct. Note that a1/f = b1, or a1 = f*b1. > Putting this into the equation above that the a's > satisfy and simplifying, one obtains f*b13 + 3*(-1 + m*f2)*b12 - (m3*f4 - 3*m2*f2 + 3*m). Clearly b2 is a root of the same polynomial. Since > b3 = a3, b3 satisfies a different equation: b33 + 3*(-1+mf2)b32 - f2(m3 f4 - 3m2 f2 + 3m) = 0. Um, Nora Baron do you accept that there exists a SINGLE cubic which > has the b's as it's roots? Nora Baron *only* state that the single cubic you give is not one which > has the b's as its roots. Um, do *you* accept that there exists a SINGLE cubic which has the b's > as its roots? > Yes, I do, and I think everyone does. (x-b1)(x-b2)(x-b3) serves just fine. > It is a cubic in b, but as the b's depend on m, it is not necessarily a > cubic in m. The question is to the solution for the b's, just like earlier in the post a solution for the a's is given as they are roots of the cubic a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). Readers should note the need for me to put *back* in math that is needed to properly evaluate various claims, as this poster left them out. My assessment of that behavior of selectively deleting out key bits of information is that it is done to try and make a convincing argument against the facts. In my experience, posters seem adept at producing posts calculated to convince other readers on the newsgroup that I'm wrong, by referring selectively to information. It's basically posting sleight-of-hand. Here's more pertinent information: P(m) = f2((m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f) P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) R(m) = (m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) > Readers, I'm emphasizing this point because I know that the > characteristics of that single cubic must be such as to end the > debate. > Ah. The reason is that posters like this one have worked to convince readers of mathematically false things, and I can show that their assertions would force the cubic that has the b's for roots to be *selectively* non-monic, where if they are correct, the cubic has to be non-monic if a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). is irreducible over Q, and monic if it's not. That is, you'd need a function of m that would equal 1, when a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m) is reducible over Q, and f, when it is not, and that function would be the leading coefficient of the cubic defining the b's. > So what good are their cubics, when I can give ONE which brings b_1, > b_2 and b_3 together? I'm emphasizing the *single* cubic because it must be monic. > Pray show. (I think you mean a cubic polynomial?) The cubic cannot, in general, be given without using ..., as then it is b3 + (...)b2 + (...)b - (m3 f4 - 3m2 f2 + 3m). However, for certain values of m and f, the terms are algebraic integers, and the cubic is easily displayed. For instance, for m=1, f=sqrt(2), it is b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 and note that here a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m), gives a3 + 3a2 - 2, which IS reducible, but intriguingly, isn't even reducible to linear terms over Q. The answer is that posters arguing against the cubic defining the b's being in general monic, are just wrong, and reducibility over Q just determines whether or not you can easily *see* that they are wrong. Of course, also for m=0, and any f, it is, as I've mentioned before b3 - 3b2, and a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m), gives a3 - 3a2. > If so, then your objection fails as provably that cubic must be > non-monic. There is such a cubic indeed. Possibly it is non-monic, more likely though, > it does not have integer coefficients. It can't be non-monic as then you wouldn't get b_1=b_2=0, b_3=3 at > m=0, which requires that the cubic, at m=0, MUST be b3 -3b2 = 0, as > no other primitive cubic exists that will give the appropriate roots. So at a minimum you MUST concede the cubic at m=0. > At m=0 it is b2.(b-3). > Do you concede it Dik T. Winter? > So, yup, no problem. What's important about that readers is it forces the poster to either accept that the cubic is monic in general, which means accepting the core problem, or claim that there's a *function* of m, which varies in some odd way. > I thought it is *your* obsession that there must exist a *single* cubic > for which they are the roots. Good for you, there is such a polynomial, > alas, most likely not one with integer coefficients. It has integer coefficients at m=0, as then it's just b3 - 3b2 > Oh, you noted that already. > But must there not exist a SINGLE cubic for which b_1, b_2 and b_3 are > roots? There is: (x - b1)(x - b2)(x - b3). But, does it have integer coefficients, > or rational coefficients? Not when f = 5 and u = 1, as shown above. But is it monic? > As written, yes it is monic. No problem. But it does not necessarily > have integer coefficients, so what do you conclude from it being monic? Well, no, it doesn't necessarily have integer coefficients, like I showed before with f=sqrt(2), m=1, where b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 but readers can see just how misleading Nora Baron was in posts showing TWO cubics, where one must exist. Now for certain values, the cubic has algebraic integer coefficients, while for others, it does not because the ring of algebraic integers doesn't include certain numbers that it should, which leads to a problem in core as you can have the appearance of two proofs contradicting each other. > Then why are you the one giving two cubics in one case, when you > didn't try to give two cubics for the a's? Because the a's satisfy a single cubic *with integer coefficients*. > The b's do *not* so in general. The important question for the SINGLE cubic for which the b's are > roots is, IS IT MONIC? > No. The important question is: is it monic and does it have integer > coefficients? Otherwise you can not conclude anything about the b's > being algebraic integers. That's bogus, as I can show with b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1. Now this poste may *want* a function as the leading coefficient of the general b cubic such that it magically equals f when a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m) is irreducible over Q, and switches to 1, when it is reducible, but there's no logical basis for that position. It's simply some person *wishing* something they cannot prove. > Actually with a *single* cubic, the algebra shows quite clearly that I > can't be wrong, which is probably why you tossed out two cubics, as if > no one would realize that a single cubic will suffice for three > values. Pray give the single cubic with b1, b2 and b3 as roots. And *show* that > the coefficients are integer. The coefficients aren't integers, and in fact, they're not even > algebraic integers, in general, though they may be algebraic integers > for specific values of m and f. > They can be made algebraic integers for *all* values of m and f. That is not > the problem. I split up a paragraph to point out the usual tactic used by such posters as this one. Make a statement, which they cannot prove, and then keep talking. It's like a fast talking technique where you say something false, and hope to snow readers by moving on before they catch you. In fact, the coefficients provably cannot be algebraic integers for all values of m and f, like for instance, with m=1, f=sqrt(5), they are not. >Let's take (x-b1)(x-b2)(x-b3); all b's are algebraic numbers, Here's the second sneak, as yes, the b's are in fact algebraic numbers, even when they're not algebraic integers, but the poster has moved to a field. > so all coefficients are algebraic numbers. Each algebraic number can be > written as the quotient of an algebraic integer and a normal integer. Here's the next point of the attempt at snowing the reader, as the poster is pushing the position that the b's can be written like, say, x/f, where x is some algebraic integer that doesn't have f as a factor, but possibly shares non-unit factors with f. > so. Multiply by the LCM of the denominators and you have a polynomial with > algebraic integer coefficients of which the three b's are roots. The question > is: is that polynomial monic? The answer is: in general, no. Now notice, given all of that, you'd think that there'd be some *logical* argument, but instead you have the poster simply asserted the desired conclusion. It's this type of poster that gives me fits as Nora Baron and Arturo Magidin both engage in this behavior. Some readers just jump to their conclusion, and apparently assume they proved it, when in fact, they simply use tactics. Later I'm accused of not handling their objections, when they never give mathematically valid objections, but instead use various techniques to *appear* to give valid objections. > But you see, that's the point, the ring of algebraic integers isn't > big enough. > Big enough for what? For b1, b2 and b3 all three being algebraic integers? > Yes, that is right. This simply means that your argument that they are > algebraic integers fails. They may be objects in some other ring (the > algebraic numbers for instance), but your proof hinge on the concept of > co-primeness, and when you use that concept you have to show what ring > you are using, and what definition of co-primeness you are using and a > whole lot of other things. Your definition of object ring is incomplete > as it is impossible to determine whether numbers are in it or not. So then, some reader comes along--skimming--doesn't bother to see all the places where the poster cheated, and assumes that someone has proven me wrong, when in fact, nothing of the kind has happened. I'll make my reply and the poster just does the same thing. Readers skim and think I'm just refusing to acknowledge the truth. The thread becomes unmanageably huge and I make a new thread, and readers claim I'm running away. Later, the poster comes back with the same techniques, never actually managing to refute a single point of my argument mathematically. James Harris === Subject: Re: Finishing argument, core error proven Visiting Assistant Professor at the University of Montana. [.snip.] >> Um, do *you* accept that there exists a SINGLE cubic which has the b's >> as its roots? >> Yes, I do, and I think everyone does. (x-b1)(x-b2)(x-b3) serves just fine. >> It is a cubic in b, but as the b's depend on m, it is not necessarily a >> cubic in m. >The question is to the solution for the b's, just like earlier in the >post a solution for the a's is given as they are roots of the cubic >a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). >Readers should note the need for me to put *back* in math that is >needed to properly evaluate various claims, as this poster left them >out. Will you quit your whining already? If people don't trim, you whine that their posts are too long; if they trim, you whine that they trimmed. If they make multiple replies to keep them short, you whine that they make multiple replies; if they make a single reply, you whine that they address different issues and are trying to snow people. And if anybody ->dares<- to say that you should stop grandstanding like you do above, you whine that they are not addressing the math. Well, neither are you: you are just whining. If someone left something out, and you think it is important, then bring it back up and mention it was there already. No need to start whining and telling readers what they should or should not note. >My assessment of that behavior of selectively deleting out key bits of >information is that it is done to try and make a convincing argument >against the facts. My assessment of your behavior is that you are far more interested in appearances than you are in substance. That is why you keep inserting these Appeals to the Gallery in the middle of your posts. [.snip.] >Here's more pertinent information: >P(m) = f2((m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f) >P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) >R(m) = (m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f >R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) So: R(m) = P(m)/f2. Assuming that P(m) is not zero, we have: R(m) = P(m)/f2 = (1/f)(a_1x+uf)(1/f)(a_2x+uf)(a_3x+uf) = (b_1x + u)(b_2x+u)(b_3x+uf) and you have defined b_1=a_1/f, b_2=a_2/f. So b_1x+u = (a_1x+uf)/f, and b_2x + u + (a_2x+uf)/f. Since P(m) is not zero, we can cancel terms and get b_3x + uf = a_3x + uf. So b_3x = a_3 x. Since x is not zero (it is coprime to f, which is not a unit), we have that b_3=a_3. So, UNLESS P(m)=0, we must have b_3=a_3. That means that b_3 is a root of the cubic you mention above: a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). There are two cases: CASE 1. The cubic is irreducible over Q; that is, x3 + 3(-1+mf2)x2 - f2(m3f4 - 3m2f2+3m) has no rational roots; that is, none of a1, a2, a3 are rational numbers (and being monic, they would necessarily be integers). and CASE 2. The cubic irreducible over Q; that is, it has at least one, possibly three, rational roots. That means, exactly one of a1, a2, a3 is an integer, or all three are integers. In CASE 2, which is NOT the case m=0, or the case u=1, f=sqrt(2), m=1 that you've used as well, we have the following: b3 is a root of G(x) = x3 + 3(-1+mf2)x2 - f2(m3f4 - 3m2f2+3m). Assume F(x) is ANY polynomial with rational coefficients that has b3 as a root. Then we can divide F(x) by G(x), to get F(x) = G(x)*Q(x) + r(x) where r(x) and Q(x) are polynomials with rational coefficients, and r(x)=0 or else r(x) has degree strictly less than the degree of G. Since F(x) has b3 as a root, then 0=F(b3)=G(b3)Q(b3) + r(b3) = r(b3), so r(x) also has b3 as a root. If r(x) is not zero, then we can divide G(x) by r(x), and we have G(x) = r(x)Q'(x) + s(x), with s(x) equal to 0 or with strictly smaller degree than r(x), and therefore either constant or degree 1. Then 0 = G(b3) = r(b3)Q'(b3) + s(b3) = s(b3), so s(b3)=0; that means either that r(x) divides G(x) (if s is constant), which is impossible because G(x) was assumed to be irreducible; or else s(b3)=0 and s is a polynomial of degree 1, which means that b3 is a rational number, which is also impossible since G(x) was assumed irreducible. The conclusion is that r(x) must be the constant zero, which means the original F(x) must be a multiple of G(x), which means that either F(x) is the zero polynomial, is a constant multiple of G(x), or else it has strictly larger degree than G(x) (i.e., it is of degree greater than 3). Therefore: If F(X) is ANY cubic polynomial with rational coefficients which has b3 as a root, then it must be a constant multiple of G(x), and its roots are a1, a2, and a3=b3. (Assuming the polynomial that defines the a's is irreducible over Q). So, if there is a cubic polynomial with rational coefficients that defines the b's, it must mean that we have {b1,b2} = {a1, a2}. Which means, either b1=a1 and b2=a2, or else b1=a2 and b2=a1. The first is impossible, since b1=a1/f, so a1=a1/f means a1=0, which would mean the original G(x) had a rational root, impossible. The latter is also impossible, for if b1=a2 and b2=a1, then a1/f = a2, so a1 = f(a1/f) = f*b1= f*a2 = f*(f*b2) = f2*b2 = f2*a1, so a1=0 and again we have a contradiction. Therefore, if the polynomial defining the a's is irreducible over Q, then there is NO polynomial with rational coefficients which is (a) cubic; and (b) has b1, b2, and b3 as roots. Therefore, IF there is a single cubic polynomial with rational coefficients that has b1, b2, and b3 as roots, then the original polynomial was reducible over Q. Which, surprise surprise, is exactly the two cases that you manage to handle. >> Readers, I'm emphasizing this point because I know that the >> characteristics of that single cubic must be such as to end the >> debate. >> Ah. >The reason is that posters like this one have worked to convince >readers of mathematically false things, and I can show that their >assertions would force the cubic that has the b's for roots to be >*selectively* non-monic, where if they are correct, the cubic has to >be non-monic if >a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). >is irreducible over Q, and monic if it's not. ->YOU<- can show? So you're not just repeating what they proved already? I always knew you were not above claiming works of others as your own. >That is, you'd need a function of m that would equal 1, when >a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m) >is reducible over Q, and f, when it is not, and that function would be >the leading coefficient of the cubic defining the b's. Yes, there is a cubic which has exactly b1, b2, and b3 as roots. However, in the case where P(m) is not zero and G(x) is irreducible over Q, this polynomial (which is a scalar multiple of (x-b1)(x-b2)(x-b3)) does NOT have integer coefficients. It does not even have RATIONAL coefficients; to get rational coefficients you need to go to a polynomial of larger degree. But, setting that aside, (1) What do you mean, reducible over Q, and f? What does it mean to be reducible over f? (2) What's wrong with a function which is 1 for some values of m but not 1 for others? Surely you've seen plenty of those? >> So what good are their cubics, when I can give ONE which brings b_1, >> b_2 and b_3 together? >> I'm emphasizing the *single* cubic because it must be monic. >> Pray show. (I think you mean a cubic polynomial?) >The cubic cannot, in general, be given without using ..., as then it >b3 + (...)b2 + (...)b - (m3 f4 - 3m2 f2 + 3m). It cannot? If we cannot see the cubic, then how do you know the cubic has integer coefficients? >However, for certain values of m and f, the terms are algebraic >integers, and the cubic is easily displayed. For instance, for m=1, >f=sqrt(2), it is >b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 >and note that here >a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m), gives >a3 + 3a2 - 2, >which IS reducible, but intriguingly, isn't even reducible to linear >terms over Q. x3+3x2 - 2 = (x+1)(x2+2x-2) Why is that intriguing? In this case, your cubic factors into a linear times an irreducible quadratic; you get lucky that your linear term has constant term coprime to f, so you get the other two factors sharing f2 in equal portions. You did not get so lucky when u=1, f=2, m=1, where the linear term did not avoid the value of 2; then you had a3 + 3(-1+4)a2 - 4(16 - 3(4) + 3) a3 + 9a2 - 28 = (a+2)(a2+7a-14) in that case, the factors of 2 did NOT factor out as you hoped they would, because both the linear term and the irreducible quadratic contributed to the total factor of 22 = 4. >The answer is that posters arguing against the cubic defining the b's >being in general monic, are just wrong, and reducibility over Q just >determines whether or not you can easily *see* that they are wrong. Nonsense. In the case u=1, f=2, m=1, the value of a3=b3=-2, and the other two values are b1 = a1/2 = (-7-sqrt(105))/4 b2 = a2/2 = (-7+sqrt(105))/4 Neither is an algebraic integer: they are roots of (x-b1)(x-b2) = x2 - (b1+b2)x + b1*b2 = x2 - (-14/4)x + (49-105)/16 = x2 + (7/2)x + (56/16) = x2 + (7/2)x + (7/2) so they are roots of the irreducible non-monic 2x2 + 7x + 7. Hence they are not algebraic integers. They cannot be the roots of any monic polynomial with algebraic integer coefficients, let alone a cubic. In fact, any cubic that has them as a roots would be a scalar multiple of (x-b1)(x-b2)(x-b3) = (x2 + (7/2)x+(7/2))(x+2) = x3 + (7/2)x2 + (7/2)x + 2x2 + 7x + 7 = x3 + (11/2)x2 + (21/2)x + 7 and if we clear denominators, we have 2x3 + 11x2 + 21x + 14, so no monic cubic with integer coefficients has all three as roots (it would have to be a scalar multiple of the above; since 2 and 11 are coprime, there is no algebraic integer other than units that divides all coefficients, so you cannot get rid of the 2 in the leading term). In case you argue that I just assigned the b's wrong, here are the two other possibilities (up to exchanging b1 and b2): CASE 2: b1 = -2/2, b2 = (-7-sqrt(105))/4, b3 = (-7+sqrt(105))/2 Then any cubic having all three as roots is a scalar multiple of (x-b1)(x-b2)(x-b3) = (x+1)(x+(7+sqrt(105))/4)(x+(7-sqrt(105))/2) = (x+1)(x2 + ((21-sqrt(105))/4)x -(56/8)) = x3 + ((21-sqrt(105))/4)x2 - 7x + x2 + ((21-sqrt(105))/4)x - 7 = x3 + [(25-sqrt(105))/4]x2 - [(7+sqrt(105))/4]x - 7 and clearing denominators, we have 4x3 + (25-sqrt(105))x2 - (7+sqrt(105))x - 28. Nonmonic with algebraic integer coefficients. It is not hard to see that this is irreducible over Q[sqrt(105)], and so none of its roots are algebraic integers. And of course, there is no cubic with INTEGER coefficients that will work, either. Finally, CASE 3: b1 = -2/2, b2 = (-7+sqrt(105))/4, b3 = (-7-sqrt(105))/2 Then any cubic having all three as roots is a scalar multiple of (x-b1)(x-b2)(x-b3) = (x+1)(x+(7-sqrt(105))/4)(x+(7+sqrt(105))/2) = (x+1)(x2 + ((21+sqrt(105))/4)x -(56/8)) = x3 + ((21+sqrt(105))/4)x2 - 7x + x2 + ((21+sqrt(105))/4)x - 7 = x3 + [(25+sqrt(105))/4]x2 - [(7-sqrt(105))/4]x - 7 and clearing denominators we have 4x3 + (25-sqrt(105))x2 - (7-sqrt(105))x - 28 again nonmonic, irreducible over Q[sqrt(105)], and none of its roots are algebraic integers. No cubic with INTEGER coefficients will work. So we can SEE that you are wrong. >Of course, also for m=0, and any f, it is, as I've mentioned before >b3 - 3b2, >and >a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m), gives >a3 - 3a2. Which is reducible into three linear terms, and reducibility of the polynomial is NECESSARY before there can be a cubic with integer coefficients that defines the b's (though not sufficient, as seen just above). [.snip.] >What's important about that readers is it forces the poster to either >accept that the cubic is monic in general, which means accepting the >core problem, or claim that there's a *function* of m, which varies >in some odd way. Yeah, some really odd way. Like, for example, (2m+1) which has value 1 with m=0 but not when m is not zero. Or a function which is 1 at m=0 and at m=1, but not anywhere else? Something REALLY odd like (2m-1)2? [.snip.] >> There is: (x - b1)(x - b2)(x - b3). But, does it have integer coefficients, >> or rational coefficients? Not when f = 5 and u = 1, as shown above. >> But is it monic? >> As written, yes it is monic. No problem. But it does not necessarily >> have integer coefficients, so what do you conclude from it being monic? >Well, no, it doesn't necessarily have integer coefficients, like I >showed before with f=sqrt(2), m=1, where >b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 You are using values of f that are not integers. But here, you get algebraic integers. However, as seen above, the MONIC version does not necessarily have ALGEBRAIC integer coefficients either. You are being misleading here, since you do not clarify just what it is you are claiming. Which of the following is your claim? POSSIBLE CLAIM 1: There is a cubic polynomial, whose coefficients are functions of u, f, m, with the property that for every algebraic integer value of u, f, and m, the polynomial is monic, has algebraic integer coefficients, and has b1, b2, and b3 as roots. POSSIBLE CLAIM 2: For every algebraic integer value of u, f, and m, there is a cubic monic polynomial with algebraic integer coefficients that has b1, b2, and b3 as roots. POSSIBLE CLAIM 3: For every algebraic integer value of u, f, and m, there is a monic cubic polynomial which has b1, b2, and b3 as roots, but we make no claim about the coefficients of that polynomial ? POSSIBLE CLAIM 3 gives you nothing; only from possible claims 1 or 2 could you conclude that b1, b2, and b3 are algebraic integers. And the situation you are in is 3. >but readers can see just how misleading Nora Baron was in posts >showing TWO cubics, where one must exist. You're wrong. For any value of m for which P(m) is not zero, and the polynomial a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). is irreducible over Q, you cannot have a single cubic with rational coefficients that has b1, b2, and b3 as roots. (NOTE THAT in order to be irreducible over Q, the coefficients MUST BE rational numbers; your example with f=sqrt(2) does not satisfy these conditions, since the polynomial DOES NOT have rational coefficients; the corresponding claim would have to be that the resulting polynomial is irreducible over Q[sqrt(2)] ). [.snip.] >Now for certain values, the cubic has algebraic integer coefficients, >while for others, it does not because the ring of algebraic integers >doesn't include certain numbers that it should, which leads to a >problem in core as you can have the appearance of two proofs >contradicting each other. Ah, yes. So your cubic doesn't have the right numbers, so once again, things that ->should<- be there aren't there. But why should they be there? Because you ->want<- them to be, not for any other reason. Well, tough. [.snip.] >> The important question for the SINGLE cubic for which the b's are >> roots is, IS IT MONIC? >> No. The important question is: is it monic and does it have integer >> coefficients? Otherwise you can not conclude anything about the b's >> being algebraic integers. >That's bogus, as I can show with >b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1. No, it's not bogus. Here, your original polynomial defining the a's DOES NOT have rational coefficients. Here, the question must become wether it is monic and irreducible over Q[sqrt(2)]. And this is all one BIG red herring. You are always assuming f is a prime integer, except when it is convenient for you to distract with other values. [.snip.] >It's simply some person *wishing* something they cannot prove. You mean, like there should be a cubic, even though you cannot prove there is one? [.snip.] >>Let's take (x-b1)(x-b2)(x-b3); all b's are algebraic numbers, >Here's the second sneak, as yes, the b's are in fact algebraic >numbers, even when they're not algebraic integers, but the poster has >moved to a field. There is no second sneak. This is basic precalculus, which even you must have gone over in your gifted program. [.snip.] >> so all coefficients are algebraic numbers. Each algebraic number can be >> written as the quotient of an algebraic integer and a normal integer. >Here's the next point of the attempt at snowing the reader, as the >poster is pushing the position that the b's can be written like, say, >x/f, where x is some algebraic integer that doesn't have f as a >factor, but possibly shares non-unit factors with f. That's a lie, James. Nowhere is he saying or implying that the denominator will be ->f<-; but EVERY algebraic number can be written as a quotient of an algebraic integer and a regular integer, though this expression NEED NOT BE IN LEAST TERMS. Nowhere is it assumed that it is in least terms, it is unnecessary for the argument. So you are erecting strawmen and lying. [.snip.] >It's this type of poster that gives me fits as Nora Baron and Arturo >Magidin both engage in this behavior. Ad hominem. >Some readers just jump to their conclusion, and apparently assume they >proved it, when in fact, they simply use tactics. Yeah, that's what you do. Should be multiples. That's jumping to a conclusion. When asked to justify, instead of explaining the should, you insult, you whine, and eventually explain SOME OTHER THING NOBODY COMPLAINED ABOUT. And then, having committed a strawman and a red herring, you claim to have handled the objection, because you addressed ->something<-... just not what people were asking and/or complaining about. [.snip.] 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 critize. A great many people are staggered to this extend, 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 === Subject: Re: Finishing argument, core error proven there's got to be some sort of *job* that could employ such skills in whining. seriously, it seems like about half of JSH's output is grandstanding, but it could be argued that you're doing the same. I mean, how many times have I seen the math reduced to a simple problemma in the definition of (say) divisibility -- often the first chapter of elementary books on numbertheory -- -- and you've also deomnstrated such, as I recall -- but you make these long & tedious corrections? I can't follow them, thus the quotes, nor do I want to bother with most of JSH's explanations, since I'm quite satisfied with the shorter, unanswered refutations. I'm really glad, though, that everyone takes him on his word, and his apparent pursuit of aggrandizement, since the idea that he *is* being paid to do this is just too much (at least *one*, two, much) to bear. [massive deletives impleted] > And if anybody ->dares<- to say that you should stop grandstanding > like you do above, you whine that they are not addressing the math. --Dec.2000 'WAND' Chairman Paul O'Neill, reelected to Board. Newsish? http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac === Subject: Re: Finishing argument, core error proven > [.snip.] >> Um, do *you* accept that there exists a SINGLE cubic which has the b's >> as its roots? >> Yes, I do, and I think everyone does. (x-b1)(x-b2)(x-b3) serves just fine. >> It is a cubic in b, but as the b's depend on m, it is not necessarily a >> cubic in m. >The question is to the solution for the b's, just like earlier in the >post a solution for the a's is given as they are roots of the cubic >a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). >Readers should note the need for me to put *back* in math that is >needed to properly evaluate various claims, as this poster left them >out. > Will you quit your whining already? If people don't trim, you whine > that their posts are too long; if they trim, you whine that they > trimmed. If they make multiple replies to keep them short, you whine > that they make multiple replies; if they make a single reply, you > whine that they address different issues and are trying to snow > people. And here readers should notice that Arturo Magidin starts right at the beginning, not with mathematics, but with a protest. Beyond that consider the abhorrent *length* of this person's post. I'll leave it all in this time, but I want readers to pay attention to this particular poster's tactics. > And if anybody ->dares<- to say that you should stop grandstanding > like you do above, you whine that they are not addressing the math. > Well, neither are you: you are just whining. If someone left something > out, and you think it is important, then bring it back up and mention > it was there already. No need to start whining and telling readers > what they should or should not note. The previous poster left out information pertinent to his own comments, without which, readers were left guessing. Given that poster's assumption that he'd get the benefit of the doubt, since SO many posters keep saying I'm wrong, though they're incapable of proving it, it's not surprising that the poster would leave out pertinent mathematics, pertinent to his own post, without fear of getting caught. Which is why I'm taking extra effort to highlight the tactics of these posters. >My assessment of that behavior of selectively deleting out key bits of >information is that it is done to try and make a convincing argument >against the facts. > My assessment of your behavior is that you are far more interested in > appearances than you are in substance. That is why you keep inserting > these Appeals to the Gallery in the middle of your posts. > [.snip.] I'm simply noting the behavior, especially given that certain posters--who basically just say I'm wrong--are given the benefit of the doubt. As I've watched what happens when I refute these posters mathematically, I can see that it doesn't pay for me to just hope that readers are catching them. >Here's more pertinent information: >P(m) = f2((m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f) >P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) >R(m) = (m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f >R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) > So: R(m) = P(m)/f2. Assuming that P(m) is not zero, we have: > R(m) = P(m)/f2 = (1/f)(a_1x+uf)(1/f)(a_2x+uf)(a_3x+uf) > = (b_1x + u)(b_2x+u)(b_3x+uf) > and you have defined b_1=a_1/f, b_2=a_2/f. So b_1x+u = (a_1x+uf)/f, > and b_2x + u + (a_2x+uf)/f. Since P(m) is not zero, we can cancel > terms and get > b_3x + uf = a_3x + uf. > So b_3x = a_3 x. > Since x is not zero (it is coprime to f, which is not a unit), we have > that b_3=a_3. Hey, that's correct, which means I had it right the first time, but later second guessed myself about a_3 and b_3. > So, UNLESS P(m)=0, we must have b_3=a_3. > That means that b_3 is a root of the cubic you mention above: > a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). > There are two cases: > CASE 1. The cubic is irreducible over Q; that is, > x3 + 3(-1+mf2)x2 - f2(m3f4 - 3m2f2+3m) > has no rational roots; that is, none of a1, a2, a3 are > rational numbers (and being monic, they would necessarily be > integers). > and > CASE 2. The cubic irreducible over Q; that is, it has at least one, > possibly three, rational roots. That means, exactly one of > a1, a2, a3 is an integer, or all three are integers. > In CASE 2, which is NOT the case m=0, or the case u=1, f=sqrt(2), m=1 > that you've used as well, we have the following: > b3 is a root of G(x) = x3 + 3(-1+mf2)x2 - f2(m3f4 - 3m2f2+3m). > Assume F(x) is ANY polynomial with rational coefficients that has b3 as > a root. Then we can divide F(x) by G(x), to get > F(x) = G(x)*Q(x) + r(x) > where r(x) and Q(x) are polynomials with rational coefficients, and > r(x)=0 or else r(x) has degree strictly less than the degree of G. > Since F(x) has b3 as a root, then 0=F(b3)=G(b3)Q(b3) + r(b3) = r(b3), > so r(x) also has b3 as a root. If r(x) is not zero, then we can > divide G(x) by r(x), and we have > G(x) = r(x)Q'(x) + s(x), with s(x) equal to 0 or with strictly smaller > degree than r(x), and therefore either constant or degree 1. Then > 0 = G(b3) = r(b3)Q'(b3) + s(b3) = s(b3), so s(b3)=0; that means either > that r(x) divides G(x) (if s is constant), which is impossible because > G(x) was assumed to be irreducible; or else s(b3)=0 and s is a > polynomial of degree 1, which means that b3 is a rational number, > which is also impossible since G(x) was assumed irreducible. > The conclusion is that r(x) must be the constant zero, which means the > original F(x) must be a multiple of G(x), which means that either F(x) > is the zero polynomial, is a constant multiple of G(x), or else it has > strictly larger degree than G(x) (i.e., it is of degree greater than > 3). > Therefore: If F(X) is ANY cubic polynomial with rational coefficients > which has b3 as a root, then it must be a constant multiple of G(x), > and its roots are a1, a2, and a3=b3. (Assuming the polynomial that > defines the a's is irreducible over Q). > So, if there is a cubic polynomial with rational coefficients that > defines the b's, it must mean that we have {b1,b2} = {a1, a2}. Which > means, either b1=a1 and b2=a2, or else b1=a2 and b2=a1. > The first is impossible, since b1=a1/f, so a1=a1/f means a1=0, which > would mean the original G(x) had a rational root, impossible. The > latter is also impossible, for if b1=a2 and b2=a1, then a1/f = a2, so > a1 = f(a1/f) = f*b1= f*a2 = f*(f*b2) = f2*b2 = f2*a1, so a1=0 and > again we have a contradiction. > Therefore, if the polynomial defining the a's is irreducible over Q, > then there is NO polynomial with rational coefficients which is (a) > cubic; and (b) has b1, b2, and b3 as roots. That can be the case even when it is reducible over Q, as I've pointed out by giving the defining cubic when m=1, f+sqrt(2), which is b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 as readers can actually see that only two of the coefficients, the leading and last, are rational. So far Arturo Magidin hasn't done anything but take up space. > Therefore, IF there is a single cubic polynomial with rational > coefficients that has b1, b2, and b3 as roots, then the original > polynomial was reducible over Q. > Which, surprise surprise, is exactly the two cases that you manage to > handle. Now the surprise surprise here really doesn't fit in context. Readers can guess what the purpose of the poster was in putting that phrase there. >> Readers, I'm emphasizing this point because I know that the >> characteristics of that single cubic must be such as to end the >> debate. >> Ah. >The reason is that posters like this one have worked to convince >readers of mathematically false things, and I can show that their >assertions would force the cubic that has the b's for roots to be >*selectively* non-monic, where if they are correct, the cubic has to >be non-monic if >a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). >is irreducible over Q, and monic if it's not. > ->YOU<- can show? So you're not just repeating what they proved > already? > I always knew you were not above claiming works of others as your > own. Now readers should consider how long this poster has gone on without actually saying anything. My assessment of this particular poster is that he has learned that most sci.math readers don't read in detail, but skim, and they are *especially* likely to skim a long post. Skimming along, they may feel that Arturo Magidin has said a lot, when he hasn't said anything as of yet, as I've pointed out. Now then, mathematically his position would require that the cubic defining the b's vary such that if the cubic defining the a's is reducible over Q, then that cubic is monic, but illogically, if the cubic defininng the a's is irreducible over Q, then his position requires a leading coefficient of f. Now if this poster were honest, then he'd simply acquiesce to the truth. However, instead, look at the LENGTH of his post!!! >That is, you'd need a function of m that would equal 1, when >a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m) >is reducible over Q, and f, when it is not, and that function would be >the leading coefficient of the cubic defining the b's. > Yes, there is a cubic which has exactly b1, b2, and b3 as > roots. However, in the case where P(m) is not zero and G(x) is > irreducible over Q, this polynomial (which is a scalar multiple of > (x-b1)(x-b2)(x-b3)) does NOT have integer coefficients. It does not > even have RATIONAL coefficients; to get rational coefficients you need > to go to a polynomial of larger degree. Which still isn't saying anything. Again I give the cubic for the b's for m=1, f=sqrt(2), as it is b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 and readers can see that, yeah, it has some irrational coefficients. However, it is monic. > But, setting that aside, > (1) What do you mean, reducible over Q, and f? What does it mean to > be reducible over f? This poster also has a habit of trying to find little ways to distract. Like here he broke up a paragraph where I noted that if the cubic defining the a's is NOT irreducible over Q then somehow supposedly the cubic defining the b's has a leading coefficient with a factor of f, so I wasn't talking about reducibility over f. The posters tactics are rather transparent, but readers typically come in believing that I'm wrong, so this poster can get away with distractions. Just remember readers, the correct math should be *more* concise, and direct. But notice how certain posters, like this one and the poster that goes by Nora Baron have posts that get longer, and longer, and longer. I find it rather annoying, as they keep playing tricks. > (2) What's wrong with a function which is 1 for some values of m but > not 1 for others? Surely you've seen plenty of those? Now you see the delivery from the previous setup, as this poster carefully hides the full position that when the cubic defining the a's is reducible over Q the equation would be monic, but if its irreducible, by his claims, it'd have a leading coefficient with a factor that is f. So you get the question, seemingly rational, but its purpose is to distract rather than be mathematically precise. And notice how LONG this poster has gone on here already and the length of this post!!! >> So what good are their cubics, when I can give ONE which brings b_1, >> b_2 and b_3 together? >> I'm emphasizing the *single* cubic because it must be monic. >> Pray show. (I think you mean a cubic polynomial?) >The cubic cannot, in general, be given without using ..., as then it >is >b3 + (...)b2 + (...)b - (m3 f4 - 3m2 f2 + 3m). > It cannot? If we cannot see the cubic, then how do you know the cubic > has integer coefficients? I didn't say it did. An example, again is b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 which results from m=1, f=sqrt(2). >However, for certain values of m and f, the terms are algebraic >integers, and the cubic is easily displayed. For instance, for m=1, >f=sqrt(2), it is >b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 >and note that here >a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m), gives >a3 + 3a2 - 2, >which IS reducible, but intriguingly, isn't even reducible to linear >terms over Q. > x3+3x2 - 2 = (x+1)(x2+2x-2) > Why is that intriguing? In this case, your cubic factors into a > linear times an irreducible quadratic; you get lucky that your linear > term has constant term coprime to f, so you get the other two factors > sharing f2 in equal portions. You did not get so lucky when u=1, f=2, > m=1, where the linear term did not avoid the value of 2; then you had > a3 + 3(-1+4)a2 - 4(16 - 3(4) + 3) > a3 + 9a2 - 28 = (a+2)(a2+7a-14) > in that case, the factors of 2 did NOT factor out as you hoped they > would, because both the linear term and the irreducible quadratic > contributed to the total factor of 22 = 4. It's intriguing because it shows that the cubic doesn't need to be fully reducible over Q, which makes it even more obvious how quirky the assertions of the poster are. Now then, as to the poster's assertion about factors, I ask that the poster give the cubic defining the b's, the SINGLE cubic. The answer is that the cubic defining the b's in that case do not have all algebraic integer coefficients. It's not complicated, but it's not the answer this poster will want to admit, without possibly trying to move to the field of algebraic numbers, as this poster needs a cubic with algebraic integer coefficients, and a leading coefficient with a factor of f. >The answer is that posters arguing against the cubic defining the b's >being in general monic, are just wrong, and reducibility over Q just >determines whether or not you can easily *see* that they are wrong. > Nonsense. In the case u=1, f=2, m=1, the value of a3=b3=-2, and the > other two values are > b1 = a1/2 = (-7-sqrt(105))/4 > b2 = a2/2 = (-7+sqrt(105))/4 > Neither is an algebraic integer: they are roots of > (x-b1)(x-b2) = x2 - (b1+b2)x + b1*b2 > = x2 - (-14/4)x + (49-105)/16 > = x2 + (7/2)x + (56/16) > = x2 + (7/2)x + (7/2) > so they are roots of the irreducible non-monic > 2x2 + 7x + 7. Hence they are not algebraic integers. They cannot be > the roots of any monic polynomial with algebraic integer coefficients, > let alone a cubic. > In fact, any cubic that has them as a roots would be a scalar multiple > of (x-b1)(x-b2)(x-b3) = (x2 + (7/2)x+(7/2))(x+2) > = x3 + (7/2)x2 + (7/2)x + 2x2 + 7x + 7 > = x3 + (11/2)x2 + (21/2)x + 7 > and if we clear denominators, we have > 2x3 + 11x2 + 21x + 14, > so no monic cubic with integer coefficients has all three as roots (it > would have to be a scalar multiple of the above; since 2 and 11 are > coprime, there is no algebraic integer other than units that > divides all coefficients, so you cannot get rid of the 2 in the > leading term). Reader should note that I've never claimed a cubic with integer coefficients for all values of m and f. This poster is adept at making up requirements, not given, and presenting them as if they are important. Here the poster is presenting an issue around having integer coefficients, when my point is that for values of m and f, coefficients are not even algebraic integers, let alone, integers. Let's get to the end as this post is long enough anyway. Read along and I'll give final comments there. > In case you argue that I just assigned the b's wrong, here are the two > other possibilities (up to exchanging b1 and b2): > CASE 2: b1 = -2/2, b2 = (-7-sqrt(105))/4, b3 = (-7+sqrt(105))/2 > Then any cubic having all three as roots is a scalar multiple of > (x-b1)(x-b2)(x-b3) = (x+1)(x+(7+sqrt(105))/4)(x+(7-sqrt(105))/2) > = (x+1)(x2 + ((21-sqrt(105))/4)x -(56/8)) > = x3 + ((21-sqrt(105))/4)x2 - 7x > + x2 + ((21-sqrt(105))/4)x - 7 > = x3 + [(25-sqrt(105))/4]x2 - [(7+sqrt(105))/4]x - 7 > and clearing denominators, we have > 4x3 + (25-sqrt(105))x2 - (7+sqrt(105))x - 28. Nonmonic with > algebraic integer coefficients. It is not hard to see that this is > irreducible over Q[sqrt(105)], and so none of its roots are algebraic > integers. And of course, there is no cubic with INTEGER coefficients > that will work, either. > Finally, > CASE 3: b1 = -2/2, b2 = (-7+sqrt(105))/4, b3 = (-7-sqrt(105))/2 > Then any cubic having all three as roots is a scalar multiple of > (x-b1)(x-b2)(x-b3) = (x+1)(x+(7-sqrt(105))/4)(x+(7+sqrt(105))/2) > = (x+1)(x2 + ((21+sqrt(105))/4)x -(56/8)) > = x3 + ((21+sqrt(105))/4)x2 - 7x > + x2 + ((21+sqrt(105))/4)x - 7 > = x3 + [(25+sqrt(105))/4]x2 - [(7-sqrt(105))/4]x - 7 > and clearing denominators we have > 4x3 + (25-sqrt(105))x2 - (7-sqrt(105))x - 28 > again nonmonic, irreducible over Q[sqrt(105)], and none of its roots > are algebraic integers. No cubic with INTEGER coefficients will > work. So we can SEE that you are wrong. >Of course, also for m=0, and any f, it is, as I've mentioned before >b3 - 3b2, >and >a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m), gives >a3 - 3a2. > Which is reducible into three linear terms, and reducibility of the > polynomial is NECESSARY before there can be a cubic with integer > coefficients that defines the b's (though not sufficient, as seen just > above). > [.snip.] >What's important about that readers is it forces the poster to either >accept that the cubic is monic in general, which means accepting the >core problem, or claim that there's a *function* of m, which varies >in some odd way. > Yeah, some really odd way. Like, for example, (2m+1) which has value 1 > with m=0 but not when m is not zero. > Or a function which is 1 at m=0 and at m=1, but not anywhere else? > Something REALLY odd like (2m-1)2? > [.snip.] >> There is: (x - b1)(x - b2)(x - b3). But, does it have integer coefficients, >> or rational coefficients? Not when f = 5 and u = 1, as shown above. >> But is it monic? >> As written, yes it is monic. No problem. But it does not necessarily >> have integer coefficients, so what do you conclude from it being monic? >Well, no, it doesn't necessarily have integer coefficients, like I >showed before with f=sqrt(2), m=1, where >b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 > You are using values of f that are not integers. But here, you get > algebraic integers. However, as seen above, the MONIC version does not > necessarily have ALGEBRAIC integer coefficients either. > You are being misleading here, since you do not clarify just what it > is you are claiming. > Which of the following is your claim? > POSSIBLE CLAIM 1: There is a cubic polynomial, whose coefficients are > functions of u, f, m, with the property that for every algebraic > integer value of u, f, and m, the polynomial is monic, has algebraic > integer coefficients, and has b1, b2, and b3 as roots. > POSSIBLE CLAIM 2: For every algebraic integer value of u, f, and m, > there is a cubic monic polynomial with algebraic integer coefficients > that has b1, b2, and b3 as roots. > POSSIBLE CLAIM 3: For every algebraic integer value of u, f, and m, > there is a monic cubic polynomial which has b1, b2, and b3 as roots, > but we make no claim about the coefficients of that polynomial > POSSIBLE CLAIM 3 gives you nothing; only from possible claims 1 or 2 > could you conclude that b1, b2, and b3 are algebraic integers. > And the situation you are in is 3. >but readers can see just how misleading Nora Baron was in posts >showing TWO cubics, where one must exist. > You're wrong. For any value of m for which P(m) is not zero, and the > polynomial > a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). > is irreducible over Q, you cannot have a single cubic with rational > coefficients that has b1, b2, and b3 as roots. > (NOTE THAT in order to be irreducible over Q, the coefficients MUST > BE rational numbers; your example with f=sqrt(2) does not satisfy > these conditions, since the polynomial DOES NOT have rational > coefficients; the corresponding claim would have to be that the > resulting polynomial is irreducible over Q[sqrt(2)] ). > [.snip.] >Now for certain values, the cubic has algebraic integer coefficients, >while for others, it does not because the ring of algebraic integers >doesn't include certain numbers that it should, which leads to a >problem in core as you can have the appearance of two proofs >contradicting each other. > Ah, yes. So your cubic doesn't have the right numbers, so once > again, things that ->should<- be there aren't there. But why should > they be there? Because you ->want<- them to be, not for any other > reason. > Well, tough. > [.snip.] >> The important question for the SINGLE cubic for which the b's are >> roots is, IS IT MONIC? >> No. The important question is: is it monic and does it have integer >> coefficients? Otherwise you can not conclude anything about the b's >> being algebraic integers. >That's bogus, as I can show with >b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1. > No, it's not bogus. Here, your original polynomial defining the a's > DOES NOT have rational coefficients. Here, the question must become > wether it is monic and irreducible over Q[sqrt(2)]. > And this is all one BIG red herring. You are always assuming f is a > prime integer, except when it is convenient for you to distract with > other values. > [.snip.] >It's simply some person *wishing* something they cannot prove. > You mean, like there should be a cubic, even though you cannot prove > there is one? > [.snip.] >>Let's take (x-b1)(x-b2)(x-b3); all b's are algebraic numbers, >Here's the second sneak, as yes, the b's are in fact algebraic >numbers, even when they're not algebraic integers, but the poster has >moved to a field. > There is no second sneak. This is basic precalculus, which even you > must have gone over in your gifted program. > [.snip.] >> so all coefficients are algebraic numbers. Each algebraic number can be >> written as the quotient of an algebraic integer and a normal integer. >Here's the next point of the attempt at snowing the reader, as the >poster is pushing the position that the b's can be written like, say, >x/f, where x is some algebraic integer that doesn't have f as a >factor, but possibly shares non-unit factors with f. > That's a lie, James. Nowhere is he saying or implying that the > denominator will be ->f<-; but EVERY algebraic number can be written > as a quotient of an algebraic integer and a regular integer, though > this expression NEED NOT BE IN LEAST TERMS. Nowhere is it assumed that > it is in least terms, it is unnecessary for the argument. > So you are erecting strawmen and lying. > [.snip.] >It's this type of poster that gives me fits as Nora Baron and Arturo >Magidin both engage in this behavior. > Ad hominem. >Some readers just jump to their conclusion, and apparently assume they >proved it, when in fact, they simply use tactics. > Yeah, that's what you do. Should be multiples. That's jumping to a > conclusion. When asked to justify, instead of explaining the should, > you insult, you whine, and eventually explain SOME OTHER THING NOBODY > COMPLAINED ABOUT. And then, having committed a strawman and a red > herring, you claim to have handled the objection, because you > addressed ->something<-... just not what people were asking and/or > complaining about. > [.snip.] Now then, if you read all the way to here then you're probably a rare reader unless you just jumped to the end. For most readers taking the time to carefully go through such a long post is a major investment, and I think Arturo Magidin knows it is. So he makes overlong posts where he goes on and on and on, basically relying on you finally accepting that simply his act of disagreement means that I'm wrong. James Harris === Subject: Re: Finishing argument, core error proven Visiting Assistant Professor at the University of Montana. [.snip.] >>Here's more pertinent information: >>P(m) = f2((m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f) >>P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) >>R(m) = (m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f >>R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) >> So: R(m) = P(m)/f2. Assuming that P(m) is not zero, we have: >> R(m) = P(m)/f2 = (1/f)(a_1x+uf)(1/f)(a_2x+uf)(a_3x+uf) >> = (b_1x + u)(b_2x+u)(b_3x+uf) >> and you have defined b_1=a_1/f, b_2=a_2/f. So b_1x+u = (a_1x+uf)/f, >> and b_2x + u + (a_2x+uf)/f. Since P(m) is not zero, we can cancel >> terms and get >> b_3x + uf = a_3x + uf. >> So b_3x = a_3 x. >> Since x is not zero (it is coprime to f, which is not a unit), we have >> that b_3=a_3. >Hey, that's correct, which means I had it right the first time, but >later second guessed myself about a_3 and b_3. IF P(m) is not zero, you can conclude this. >> So, UNLESS P(m)=0, we must have b_3=a_3. >> That means that b_3 is a root of the cubic you mention above: >> a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). >> There are two cases: >> CASE 1. The cubic is irreducible over Q; that is, >> x3 + 3(-1+mf2)x2 - f2(m3f4 - 3m2f2+3m) >> has no rational roots; that is, none of a1, a2, a3 are >> rational numbers (and being monic, they would necessarily be >> integers). >> and >> CASE 2. The cubic irreducible over Q; that is, it has at least one, >> possibly three, rational roots. That means, exactly one of >> a1, a2, a3 is an integer, or all three are integers. Drat. This should read The cubic is reducible over Q. >> In CASE 2, which is NOT the case m=0, or the case u=1, f=sqrt(2), m=1 >> that you've used as well, we have the following: And this should be In CASE 1. >> b3 is a root of G(x) = x3 + 3(-1+mf2)x2 - f2(m3f4 - 3m2f2+3m). >> Assume F(x) is ANY polynomial with rational coefficients that has b3 as >> a root. Then we can divide F(x) by G(x), to get >> F(x) = G(x)*Q(x) + r(x) >> where r(x) and Q(x) are polynomials with rational coefficients, and >> r(x)=0 or else r(x) has degree strictly less than the degree of G. >> Since F(x) has b3 as a root, then 0=F(b3)=G(b3)Q(b3) + r(b3) = r(b3), >> so r(x) also has b3 as a root. If r(x) is not zero, then we can >> divide G(x) by r(x), and we have >> G(x) = r(x)Q'(x) + s(x), with s(x) equal to 0 or with strictly smaller >> degree than r(x), and therefore either constant or degree 1. Then >> 0 = G(b3) = r(b3)Q'(b3) + s(b3) = s(b3), so s(b3)=0; that means either >> that r(x) divides G(x) (if s is constant), which is impossible because >> G(x) was assumed to be irreducible; or else s(b3)=0 and s is a >> polynomial of degree 1, which means that b3 is a rational number, >> which is also impossible since G(x) was assumed irreducible. >> The conclusion is that r(x) must be the constant zero, which means the >> original F(x) must be a multiple of G(x), which means that either F(x) >> is the zero polynomial, is a constant multiple of G(x), or else it has >> strictly larger degree than G(x) (i.e., it is of degree greater than >> 3). >> Therefore: If F(X) is ANY cubic polynomial with rational coefficients >> which has b3 as a root, then it must be a constant multiple of G(x), >> and its roots are a1, a2, and a3=b3. (Assuming the polynomial that >> defines the a's is irreducible over Q). >> So, if there is a cubic polynomial with rational coefficients that >> defines the b's, it must mean that we have {b1,b2} = {a1, a2}. Which >> means, either b1=a1 and b2=a2, or else b1=a2 and b2=a1. >> The first is impossible, since b1=a1/f, so a1=a1/f means a1=0, which >> would mean the original G(x) had a rational root, impossible. The >> latter is also impossible, for if b1=a2 and b2=a1, then a1/f = a2, so >> a1 = f(a1/f) = f*b1= f*a2 = f*(f*b2) = f2*b2 = f2*a1, so a1=0 and >> again we have a contradiction. >> Therefore, if the polynomial defining the a's is irreducible over Q, >> then there is NO polynomial with rational coefficients which is (a) >> cubic; and (b) has b1, b2, and b3 as roots. >That can be the case even when it is reducible over Q, as I've pointed >out by giving the defining cubic when m=1, f+sqrt(2), which is >b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 THIS POLYNOMIAL IS NOT REDUCIBLE OVER Q. For a polynomial to be reducible over Q, the very FIRST thing it must satisfy is that its coefficients must be RATIONAL NUMBERS. This polynomial does NOT have rational numbers, and therefore this polynomial is NOT reducible over Q. In addition, you will note that your reply is a complete waste of space. I explained exactly why, when the polynomial defining the a's is IRREDUCIBLE, then there can be no cubic. And then you reply that That can be the case even when it is reducible. So, you address cases that I was not talking about, as if they were relevant. Not to mention that your example is nonsense and inapplicable. >as readers can actually see that only two of the coefficients, the >leading and last, are rational. Which means it is not reducible over Q. It is not even DEFINED over Q. >So far Arturo Magidin hasn't done anything but take up space. No, I have PROVEN that your implicit claim that there is a cubic with rational coefficients defining the b's is false whenever the cubic defining the a's is irreducible over Q. Your example is not applicable, since the polynomial defining the a's is not defined over Q. case), it does not mean that I have simply 'wasted space. >> Therefore, IF there is a single cubic polynomial with rational >> coefficients that has b1, b2, and b3 as roots, then the original >> polynomial was reducible over Q. >> Which, surprise surprise, is exactly the two cases that you manage to >> handle. >Now the surprise surprise here really doesn't fit in context. Yes, it does. The two cases you manage to handle by providing explicitly a cubic for b1, b2, and b3 which is defined over Q are cases where the polynomial defining the a's is reducible. Just as I predicted. [.snip.] >Now readers should consider how long this poster has gone on without >actually saying anything. That's a lie. I just haven't said anything you ->understood<-. >My assessment of this particular poster is that he has learned that >most sci.math readers don't read in detail, but skim, and they are >*especially* likely to skim a long post. No, YOU skim a lot, and that's why you say so much nonsense. If you writing irreducible when I clearly meant reducible, and you might have seen the proof that you cannot have a cubic with integer coefficients defining b1, b2, and b3 when the polynomial defining the a's is irreducible over Q (which requires it to have rational coefficients). [.snip.] >Skimming along, they may feel that Arturo Magidin has said a lot, when >he hasn't said anything as of yet, as I've pointed out. No, you haven't pointed out, you have simply ->lied<- about. That's how you handle objections. By lying about them. [.snip.] >Now if this poster were honest, then he'd simply acquiesce to the >truth. The truth is you don't know what you are talking about, as you amply demonstrate in this post. I've acquiesced to that truth already. >However, instead, look at the LENGTH of his post!!! Quit your whining, sonny. [.snip.] >>That is, you'd need a function of m that would equal 1, when >>a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m) >>is reducible over Q, and f, when it is not, and that function would be >>the leading coefficient of the cubic defining the b's. >> Yes, there is a cubic which has exactly b1, b2, and b3 as >> roots. However, in the case where P(m) is not zero and G(x) is >> irreducible over Q, this polynomial (which is a scalar multiple of >> (x-b1)(x-b2)(x-b3)) does NOT have integer coefficients. It does not >> even have RATIONAL coefficients; to get rational coefficients you need >> to go to a polynomial of larger degree. >Which still isn't saying anything. Aren't you reading? IF P(m) is not zero, AND G(x) is irreducible over Q, then no polynomial with rational coefficients and degree 3 can have all of b1, b2, and b3 as roots. >Again I give the cubic for the b's for m=1, f=sqrt(2), as it is >b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 >and readers can see that, yeah, it has some irrational coefficients. Again, I note that your example is a big fat red herring. This polynomial is not even defined over Q; I stated clearly: THE POLYNOMIAL DOES NOT EVEN HAVE RATIONAL COEFFICIENTS when P(m) is not zero and G(x) is irreducible over Q. The reason why your example is just a waste of space is: this polynomial does not have rational coefficients, so it would SUPPORT my statement above (that the polynomial WILL NOT have rational coefficients), not show it is false as you imply. When m=1 and b=sqrt(2), and u=1, the polynomial defining the a's is a3 + 3(-1+mf2)a2-f2(m3 f4 - 3m2 f2 + 3m) = a3 + 3(-1+2)a2 - 2(4 - 6 + 3) = a3 +3a2 - 2 So there are two reasons why your example SUPPORTS my points, you ninny: (a) I talked about no cubic with rational coefficients existing. So what do you do? You give a cubic that does not have rational coefficients. (b) I specify that I am talking about the case where the polynomial defining the a's is irreducible. IN this case, the polynomial is REDUCIBLE: a3+3a2-2 = (a+1)(a2 + 2a - 2), so your example cannot In short, it was a big fat red herring and a big whopping lie on your part. >Like here he broke up a paragraph where I noted that if the cubic >defining the a's is NOT irreducible over Q then somehow supposedly the >cubic defining the b's has a leading coefficient with a factor of f, >so I wasn't talking about reducibility over f. What does reducibility over f mean? There is NO SINGLE CUBIC that defines the b's. There is an infinite number of cubics, all constant multiples of (x-b1)(x-b2)(x-b3), which is the UNIQUE MONIC POLYNOMIAL WITH COMPLEX COEFFICIENTS that is cubic and has b1, b2, and b3 as roots. When the original polynomial is irreducible over Q (which requires it to be DEFINED over Q), then NO multiple of this polynomial has rational coefficients, let alone integer ones. Your statement above is also a lie. Nobody is saying that if the polynomial defining the a's is reducible over Q then the cubic defining the b's has a leading coefficient with a factor of f. What has been said is exactly the opposite: IF THE CUBIC DEFINING THE a's IS IRREDUCIBLE OVER Q, then there is NO cubic polynomial with rational coefficients that defines the b's. NONE AT ALL. And since any scalar multiple of a polynomial has the same roots as the polynomial, it makes no sense to talk about the leading coefficient of a polynomial with given roots. So you can stop your lies and your distractions, and you appeals to the gallery. Stick to what is actually written, not what you invent to try to make your point. [.snip.] >> (2) What's wrong with a function which is 1 for some values of m but >> not 1 for others? Surely you've seen plenty of those? >Now you see the delivery from the previous setup, as this poster >carefully hides the full position that when the cubic defining the a's >is reducible over Q the equation would be monic, but if its >irreducible, by his claims, it'd have a leading coefficient with a >factor that is f. You are lying. No such position has been given. The ONLY thing I stated here is that if the polynomial defining the a's is IRREDUCIBLE over and P(m) is not zero, then there is NO CUBIC WITH RATIONAL COEFFICIENTS that can have b1, b2, and b3 as roots. I never said anything about an equation being monic or not monic, and I certainly never made a claim as stupid as the leading coefficient has a factor that is f when you have not specified which the infinite number of scalar multiples of a polynomial you might refer to. You are either skimming or you are lying. In either case, your (already non-existent) credibility takes a dip. [.snip.] >And notice how LONG this poster has gone on here already and the >length of this post!!! Quit yer whining, sonny. [.snip.] >b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 >which results from m=1, f=sqrt(2). >>However, for certain values of m and f, the terms are algebraic >>integers, and the cubic is easily displayed. For instance, for m=1, >>f=sqrt(2), it is >>b3 + (1+sqrt(2))b2 + (sqrt(2)-1)b - 1 >>and note that here >>a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m), gives >>a3 + 3a2 - 2, >>which IS reducible, but intriguingly, isn't even reducible to linear >>terms over Q. >> x3+3x2 - 2 = (x+1)(x2+2x-2) >> Why is that intriguing? In this case, your cubic factors into a >> linear times an irreducible quadratic; you get lucky that your linear >> term has constant term coprime to f, so you get the other two factors >> sharing f2 in equal portions. You did not get so lucky when u=1, f=2, >> m=1, where the linear term did not avoid the value of 2; then you had >> a3 + 3(-1+4)a2 - 4(16 - 3(4) + 3) >> a3 + 9a2 - 28 = (a+2)(a2+7a-14) >> in that case, the factors of 2 did NOT factor out as you hoped they >> would, because both the linear term and the irreducible quadratic >> contributed to the total factor of 22 = 4. >It's intriguing because it shows that the cubic doesn't need to be >fully reducible over Q, which makes it even more obvious how quirky >the assertions of the poster are. No, it makes nothing obvious about it. It is nothing but a red herring. >Now then, as to the poster's assertion about factors, I ask that the >poster give the cubic defining the b's, the SINGLE cubic. Sigh. ANY cubic defining the b's will be a constant multiple of (x-b1)(x-b2)(x-b3). That's BASIC FREAKING PRECALCULUS, so even YOU must have read about it at some point. But this polynomial is useless for determining whether b1, b2, and b3 are algebraic integers, because you do not know that it has algebraic integer coefficients UNLESS You already know that b1, b2, and b3 are algebraic integers. However, whenever the polynomial defining the a's is IRREDUCIBLE OVER Q, there is NO cubic polynomial that has the b's as roots AND has rational coefficients. ALL the constant multiples of (x-b1)(x-b2)(x-b3) will have NON rational coefficinets. The monic, and all the other ones. >The answer is that the cubic defining the b's in that case do not have >all algebraic integer coefficients. It's not complicated, but it's >not the answer this poster will want to admit, without possibly trying >to move to the field of algebraic numbers, as this poster needs a >cubic with algebraic integer coefficients, and a leading coefficient >with a factor of f. Why would I not want to admit it? IF THE POLYNOMIAL DEFINING THE b'S DOES NOT HAVE ALL ALGEBRAIC INTEGER COEFFICIENTS, AND IS MONIC, THEN THE b'S ARE NOT ALGEBRAIC INTEGERS. It would ->prove my point<-, you silly goose, it would ->prove you are wrong<-. You are so confused that you don't even know what you are arguing for any more. [.snip.] >>The answer is that posters arguing against the cubic defining the b's >>being in general monic, are just wrong, and reducibility over Q just >>determines whether or not you can easily *see* that they are wrong. >> Nonsense. In the case u=1, f=2, m=1, the value of a3=b3=-2, and the >> other two values are >> b1 = a1/2 = (-7-sqrt(105))/4 >> b2 = a2/2 = (-7+sqrt(105))/4 >> Neither is an algebraic integer: they are roots of >> (x-b1)(x-b2) = x2 - (b1+b2)x + b1*b2 >> = x2 - (-14/4)x + (49-105)/16 >> = x2 + (7/2)x + (56/16) >> = x2 + (7/2)x + (7/2) >> so they are roots of the irreducible non-monic >> 2x2 + 7x + 7. Hence they are not algebraic integers. They cannot be >> the roots of any monic polynomial with algebraic integer coefficients, >> let alone a cubic. >> In fact, any cubic that has them as a roots would be a scalar multiple >> of (x-b1)(x-b2)(x-b3) = (x2 + (7/2)x+(7/2))(x+2) >> = x3 + (7/2)x2 + (7/2)x + 2x2 + 7x + 7 >> = x3 + (11/2)x2 + (21/2)x + 7 >> and if we clear denominators, we have >> 2x3 + 11x2 + 21x + 14, >> so no monic cubic with integer coefficients has all three as roots (it >> would have to be a scalar multiple of the above; since 2 and 11 are >> coprime, there is no algebraic integer other than units that >> divides all coefficients, so you cannot get rid of the 2 in the >> leading term). >Reader should note that I've never claimed a cubic with integer >coefficients for all values of m and f. James Harris should note that his claim is empty unless he can say something about the cubic. GIVEN ANY 3 COMPLEX NUMBERS, there is a cubic that has those three numbers as roots and no other complex number as root. Saying that there is a cubic that has b1, b2, and b3 as roots is as empty as saying that b1, b2, and b3 are numbers. It's useless. [.Ad hominem personal attacks and no math deleted.] >Now then, if you read all the way to here then you're probably a rare >reader unless you just jumped to the end. I love the way you accuse everyone of your own shortcomings. You lie, so you accuse everyone of lying. You don't read what people write, so you accuse people of not reading what others write. You misunderstand, so you accuse people of misunderstanding. And here, you jumped to the end without comprehending, so you lie and appeal to the gallery. 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 critize. A great many people are staggered to this extend, 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 === Subject: Re: Finishing argument, core error proven Visiting Assistant Professor at the University of Montana. >> [.snip.] > Um, do *you* accept that there exists a SINGLE cubic which has the b's > as its roots? Yes, I do, and I think everyone does. (x-b1)(x-b2)(x-b3) serves just fine. > It is a cubic in b, but as the b's depend on m, it is not necessarily a > cubic in m. >>The question is to the solution for the b's, just like earlier in the >>post a solution for the a's is given as they are roots of the cubic >>a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). >>Readers should note the need for me to put *back* in math that is >>needed to properly evaluate various claims, as this poster left them >>out. >> Will you quit your whining already? If people don't trim, you whine >> that their posts are too long; if they trim, you whine that they >> trimmed. If they make multiple replies to keep them short, you whine >> that they make multiple replies; if they make a single reply, you >> whine that they address different issues and are trying to snow >> people. >And here readers should notice that Arturo Magidin starts right at the >beginning, not with mathematics, but with a protest. Clearly, the answer is No, I will not quit my whining. And, here, readers, should note that James Harris starts right at the beginning, not with mathematics, but with a protest. === Subject: Re: Finishing argument, core error proven > For me there have been two perspectives as I work to figure out how to > explain the definition problem in mathematics with LOTS of opposition, > and I wonder about mathematicians so dedicated to attacking an > argument that is clearly correct. I remind of that as I present what should finish their ability to > distract, as I've seen a strange and dedicated effort to ignore the > actual math, and simply toss up just about anything rather than face > the truth. All variables are in the ring of algebraic integers unless otherwise > stated. Let P(m) = f2((m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f) and let R(m) = (m3 f4 - 3m2 f2 + 3m) x3 - 3(-1+mf2 )x u2 + u3 f so P(m) = f2 R(m). Now consider P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where the a's are given by the following cubic: a3 + 3(-1+mf2)a2 - f2(m3 f4 - 3m2 f2 + 3m). Then it must be true that the following factorization exists R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where the b's are given by the following cubic: b3 + 3(-1+mf2)b2 - (m3 f4 - 3m2 f2 + 3m), > should be > b3+...+ 3(-1+mf2)b2+...- (m3 f4 - 3m2 f2 + 3m) > So, James Harris, how about apologizing to Nora Baron. After all, James > Harris, YOU made a mistake, not Nora Baron. That's an interesting question. > And in general you would come off a lot less pitiful if you would stop > attempting to discredit those who present clear cogent math with personal > attacks. It makes you appear as a pompous ass, which I suppose if we were > to meet you in person you are not. Really, it would serve your cause to > just address the math with math. Most of the time it's clear that you > cannot. So you resort to (a) insults, (b) pontificating, (c) simply > repeating the original discredited post, or (d) abandoning the thread and > starting over. I like arguing. And I like going over my math results, but what I find with certain posters like Nora Baron in particular is that they cheat. Now I like objections that I've handled to not come back up. I like it when basic algebra isn't challenged. I like it when other people play fair. > I suspect after an insulting parting shot we'll see (d) next. > KeithK You sound quite emotional. Math isn't about emotion, but cold, hard logic. Now if you think that Nora Baron is being treated badly by me, then, hey, you have the right to that opinion. But I think the poster can take care of her or himself, as since the poster uses a pseudonym, you can't be sure Nora Baron is female. Remember, this is Usenet and the Internet. People are often not who you think they are. James Harris === Subject: Re: Finishing argument, core error proven > I like it when basic algebra isn't challenged. Your track record proves otherwise. You delight in challenging basic algebra. > I like it when other people play fair. You reserve the right to play unfairly for yourself. Furthermore, you like it best when someone accepts your erroneous arguments. > Now if you think that Nora Baron is being treated badly by me, then, > hey, you have the right to that opinion. > But I think the poster can take care of her or himself, as since the > poster uses a pseudonym, you can't be sure Nora Baron is female. > Remember, this is Usenet and the Internet. People are often not who > you think they are. > James Harris We still don't know if James Harris is your real name, or an anagram for Mr. J. Harie Ass. Wacky, isn't it? But, hey, that's just basic math. Yup, yup, yup! -- 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 === Subject: Re: Big Bertha Thing blogs Big Bertha Thing progress Cosmic Ray Series Possible Real World System Constructs http://web.onetel.com/~tonylance/progress.html Access page to 6K Web page Astrophysics net ring access site Newsgroup Reviews including uk.transport The Progress of Discontent by T. Warton (Written at Oxford in 1746) Half-hours With the Best Authors The Chandos Classics Edited by Charles Knight Published by Frederick Warne and Co. 1890 Big Bertha Thing besides Some people think that Big Bertha is less suitable for the Have they heard the long name for CP Conf? Perspective. They think that is a better place to put Big Bertha. They must be besides themselves, which proves the thing. (C) Copyright Tony Lance 1998 To comply with my copyright, please distribute complete and free of charge. Tony Lance pobox47@big-bertha-thing.com Big Bertha Thing testament New Testament Rosary 1. The Angel Tells of the Virgin Birth. 2. The Blessing of John the Baptist in the Womb by the Unborn Christ.(13 weeks) 3. The Birth of Our Lord. 4. The Purification of Our Lady and the Presentation of Our Lord. 5. The Finding in the Temple. 6. The Baptism of Our Lord in the Jordan. (Luke 3) 7. The Wedding at Cana. (John 2) 9. The Transfiguration of Christ with Moses and Elias. (Matthew 17) 10. The Last Supper. (Matthew 26) 11. The Agony in the Garden. 12. The Scourging at the Pilar. 13. The Crowning With Thorns. 14. The Carrying of the Cross. 15. The Crucifixion. 16. The Resurrection. 17. The Ascension into Heaven. 18. The Coming Down of the Holy Ghost. 19. The Assumption into Heaven. 20. The Crowning of Our Lady Queen of Heaven. (1st five Joyfull, 2nd five Luminous, 3rd five Sorrowful, 4th five Glorious.) Old Testament Rosary by Tony Lance 1. The Promise of Hannah. (I Kings 1) Birth of Samuel. 2. The Three Angels Visit Abraham. (Genesis 18) Birth of Isaac. 3. The Betrothal of Ruth. (Ruth 3) Stranger and great-grandmother of David. 4. The Offering up of Isaac by Abraham. (Genesis 22) 5. The Finding of Joseph in Egypt by Jacob. (Genesis 46) 7. The Feast in the Temple for the Rechabites. (Jeremias 35) 8. The Covenant of Moses on Mount Sinai. (Exodus 19) 9. The Forty Day Walk of Elias. (III Kings 19) 10. The Writing on the Wall. (Daniel 5) The Fall of Empires. 11. The Water Dungeon of Jeremias. (Jeremias 38) The Babylonian Transmigration for 70 years. 12. The Deriding of David by the Kinsman of Saul. (II Kings 16) The Fall of Jerusalem. 14. The Ark is Carried for Forty Years. (Numbers 14) The Entry into the Promised Land. 15. The Execution of Jonah. (Jonah 1) The City of Nineveh Wears Sackcloth and Ashes. 16. The Raising to Life by the Tomb of Eleusius. (IV Kings 13) 17. The Taking up of Elias in the Fiery Chariot. (IV Kings 2) 18. The Spirit of Moses Comes Down on the Judges. (Numbers 11) The Law of Precident Cases. 19. The Slaying of Holofernes by Judith. (Judith 13) The Treasure to the 12 tribes. 20. The Crowning of Esther Queen of Susan. (Esther 2) The Decree of Pogrom is reversed. Rosary Intentions for each decade of St. Louis Marie de Montfort Luminous Mysteries by Tony Lance Lord through thy Holy Mother that we may:- 1. know true humility. 2. have charity one for another. 3. love poverty. 4. have purity of body and soul. 5. obtain Divine Wisdom. 6. be free from Original Sin.(Wet head saying 'I baptise you in the name of the Father and of the Son and of the Holy Ghost. Amen') 7. give grace one to another. 8. enlighten one another. 9. obtain Union with Christ.(3 Nights of Soul) St.John of the Cross Battle vices and deny 5 physical senses. (Pictures, icons and Rock and Roll.) Battle ghosts of old vices and deny spiritual senses. (Old sins plague you for second time.) Visions, Locutions and tears. All else gone except the love of God. (Last of all the memory is perfected. It gets worse first.) 10. break bread together. 11. have perfect contrition. 12. know mortification. 13. have contempt for the world. 14. have patience in crosses. 15. obtain repentance for sinners, perserverance in grace for the just and reief for the Holy Souls in Purgatory. 16. love thee and be fervent in they service. 17. love Heaven our true home. Annual Preparation. 12 days seek Contempt for World, 6 days each know self, Mary and Christ. 19. know the Holy Ghost. 20. obtain perserverance in grace for the just and the crown of Heavenly Glory. NB. Douai-Rheims cross reference links. The Sermon on the Mount. Mark 1,14:15 Matthew 4,12:17 Matthew 5,3 Luke 6,20 NB. Parallels between old and new testament rosary event readings. They had no wine, yet drank it. Others had wine and drank not. One was crowned with thorns and crucified. Another robed as a king and decreed to hang. One was scourged at the pillar. Another derided at the head of a whole column of soldiers. One was found with the wise men. Another was found, as the wisest man in Egypt. Elizabeth and Rachel were barren and both bore child. One was crucified. and descended into Hell (Limbo), for three days. Another walked the plank and was swallowed by a whale, for three days. One was born in a stable. Another betrothed in a barn. One was crowned Queen of Heaven. Another crowned Queen of an empire. One rose up into Heaven. Another was taken up in a fiery chariot. One was baptised in the Jordan. Another was saved from the bullrushes. That is 10 out of 20 direct parallels. The other 10 are more nuanced, but who can tell? NB. The Original Divine Office of the Church was Simply to Read Book Of Psalms Once a Week. (Divisions into days and hours can be the 15 minute segments as follows.) 3 by 7 Each psalm is followed by the prayer. Glory by to the Father and to the Son and to the Holy Ghost, as it was in the beggining, is now and ever shall be, world without end. Amen. 1-9 10-17 18-24 (25-32 33-37 38-44) 45-53 54-60 61-67 (68-72 73-77 78-84) 85-89 90-98 99-104 (105-109 110-117 118-118) 119-131 132-139 140-150 NB. On meditation and contemplation. You have to dig the field before you can watch the flowers grow. Both are problem solving. Dread Visions Six Harms By Tony Lance (Based on the writings of St.John of the Cross; the Mystic Doctor of the Church.) Faith fails, replaced by a vision sensible. What we see is more real, than God invisible. Visions turn the spirit from its flight Heavenward. Had Mary Magdallen touched the feet of the Lord. Then grounded in faith, would she never be. How can we hold this vision dearer than Thee. In nakedness of spirit, hold ourselves renouncible. Grace fails, imperceptible forsaken for perceptible. The gift becomes a right, which denies the gift. Decide the vision not from God, with Satan left. Seeking after visions is a sinfull occasion. Turning into diabolical, that divine vision. With the Devil disguised, as an angel of light, And man too ignorant, to even take fright. Grace cannot be rejected; into the soul infused. The taint of evil can be rejected, by vision refused. === Subject: math and selfstudie Hi there. I've got a question, it's fairly newbe, so bare with me. I want to do math as a selfstudie , for fun and soothing a little fascination/obsession.(well , everybody needs a hobby , eh?) What are actually good books or websites on calculus and algebra one can recommend? (Difficulty level is not important, I have a notion that most things can be learned given time, motivation and persperation). I want to studie the proofs of calculus and algebra theorems and I'm searching for a textbook/site where those proofs are as exact and precise as can be, as close to the original proof as the person who thougt of it. If some one knows such books, I'd be a happy lad. Bye. === Subject: Re: math and selfstudie > Hi there. > I've got a question, it's fairly newbe, so bare with me. > I want to do math as a selfstudie , for fun and soothing a little > fascination/obsession.(well , everybody needs a hobby , eh?) > What are actually good books or websites on calculus and algebra one > can recommend? (Difficulty level is not important, I have a notion > that most things can be learned given time, motivation and > persperation). > I want to studie the proofs of calculus and algebra theorems and I'm > searching for a textbook/site where those proofs are as exact and > precise as can be, as close to the original proof as the person who > thougt of it. > If some one knows such books, I'd be a happy lad. > Bye. if I may recommend a strategy, look at the book: How to Solve It by Polya. I think it gives a good perspective on the art of proof. There are other very good books on the subject too, but I'd start there. It is not so much the topic (math), but the classes of methods used in doing proofs. Get an wonderful orientation in those and I think that will go a very long way in helping you in this new hobby. HTH, Flip