mm-2739 === Subject New version has more stuff & more art and drama (for Two Culture's audience) on 1. Wheeler IT FROM BIT 3. Dark Matter Detectors as Much Ado About Nothing. 4. Oh say can you see by WMAP's early light? 5. Slam Dunk easy to follow animated explanations of Why Inflation? Dynamical instability of Dirac Vacuum. 6. How superconductors work - animated. 7. How Alcubierre's warp drive works - animated. etc. Great stuff below Saul-Paul I need to study it in airports and on planes Thursday. :-) You saw that Max Tegmark et-al do not buy the closed finite multiply-connected Universe. One cosmologist seems to think the low power in the WMAP quadrupole mode is a contingent artifact of the Earth's location? In any case this is a worthy issue - whatever the outcome. A high resolution foreground cleaned CMB map from WMAP http://arxiv.org/abs/astro-ph/0302496 http://www.hep.upenn.edu/~max/wmap.html If you got here from the New York Times soccer ball story or the story about the Nature paper suggesting that the Universe is a dodecahedron, you'll find our paper ruling it out here. The point is that the dodecahedron model predics matched circles exactly opposite each other in the sky just like a bagel (but six pairs instead of one), which our paper ruled out. See also the more thorough upcoming circle paper by Spergel, Starkman and Cornish. Soccer enthusiasts will note that a soccer ball is made up of pentagons and hexagons and isn't a dodecahedron. http://www.hep.upenn.edu/~max/wmap.html http://www.hep.upenn.edu/~angelica/topology.html ... all toroidal universes (cubes and rectangles) are also ruled out... Can the lack of symmetry in the COBE/DMR maps constrain the topology of the universe? http://xxx.lanl.gov/abs/astro-ph/9510109 Who is Indrid? He is getting more and more useful and relevant lately. He was probably one of Robert Anton Wilson's groupies in Berserkley is my guess? Obviously quite bright and a bit of a polymath. Hi Lou, Bob, et alia, The book *Homographies, Quaternions, and Rotations* by Patrick Du Val (Oxford, 1964) is indeed a wonderful book. It provides the most detailed pictures of the polyhedral fundamental domains --indicating the gluing of opposite faces with appropriate twists. Du Val's book is based on the quaternion nature of the action of the binary polyhedral groups in creating the spherical orbit spaces SU(2)/g, where g is a finite subgroup of SU(2), i.e., a cyclic group Zn, or a binary polyhedral group of type DDn, TD, OD, or ID. Thus I wish to present an aspect of this quaternion nature not covered by Du Val --i.e., the A-D-E classification of these finite subgroups of the unit-length quaternions, SU(2). First a note on my labelling: I use DDn (for the Dihedral double group) rather than the more common Dn (for binary dihedral group). It is a (possibly useful, and possibly confusing) coincidence that in the A-D-E classification of these groups, DDn goes with Dn as a Coxeter graph (and Lie algebra) label. Moreover, I use the Crystallographers names for the double polyhedral groups (rather than the more common mathematician's name of binary polyhedral groups). Just as SU(2) double covers SO(3) via the orbit space SU(2)/{+1,-1}, so the double polyhedral groups are double covers for the polyhedral groups, which are finite subgroups of SO(3). It is most interesting that the presentation formulas for each of the finite subgroups of SU(2) can be read directly from the Coxeter graph associated with this group. In every case except the very simple cyclic groups Zn, the famous (1843) formula of William Rowan Hamilton is generalized. In fact the D4 graph provides the structure of Hamilton's (8 element) quaternion group, which is the same as the dihedral double group of type DD4, actually the smallest of these groups. i^2 = j^2 = k^2 = ijk = -1. Here I will use R,S, and T instead to accord with Du Val and H.S.M. Coxeter's nomenclature (as in *Regular Polytopes* 3rd edition, 1973). Coxeter graph: Group label: Presentation: No. of elements: An: *--*--*-...-* (n nodes) Zn+1 R^n+1 = 1 (special case) n + 1 Dn: *--*--*-...-* (n nodes, DDn R^2 = S^2 = T^n-2 = -1 4(n-2) | n > 3) <2,2,n-2> * E6: *--*--*--*--* TD R^2 = S^3 = T^3 = -1 24 | <2,3,3> * E7: *--*--*--*--*--* OD R^2 = S^3 = T^4 = -1 48 | <2,3,4> * E8: *--*--*--*--*--*--* ID R^2 = S^3 = T^5 = -1 120 | <2,3,5> * Looking at these graphs and the group presentations especially as condensed in the form of Coxeter's label for these groups , it is easy to see how to read off the presentations from the graphs: simply count nodes along the three legs of the graphs, always starting with the node common to all three legs. This exercise in the classification of the subgroups of SU(2), suggests that there might be many other mathematical objects to classify. After all these graphs were first devised as part of the classification of crystallographic reflection groups (Coxeter groups). They were next used to classify Lie groups and Lie algebras (Dynkin diagrams). Then V.I. Arnold used them to classify a vast generalization of the Thom type catastrophe structures. Now more than 20 mathematical objects are classified by these truly magical (and very simple) graphs. The advantage is that these ojects must all be somehow related to each other -- though the relationships my lie very deep in the mathematical object underlying all the possible classifications provided by these A-D-E graphs, the study of which I call ADEX theory. Although, the above classification of finite subgroups of SU(2) was known to Coxeter (and others), it was not until 1979 that John Mckay found a truly deep connection between these finite groups and the A-D-E classified Lie algebras. This was presented at a Coxeter festscrift in 1979 and published in the proceedings under the title, *The Geometric Vein*, edited by Chandler Davis, Branko Grunbaum, and F.A. Sherk (Springer, 1981). This should be in any math library. And besides the great paper Representations and Coxeter Graphs by David Ford and John McKay, there is a wonderful paper by Patric Du Val, Crystallography and Cremona Transformations -- and many other goodies. The vision of the unification of disparate mathematical objects via the A-D-E Coxeter graphs is due largely to V.I. Arnold. His little book *Catastrophe Theory* (Springer, 1986). As he puts it: At first glance, functions, quivers, caustics, wave fronts and regular polyhedra have no connection with each other. But in fact, corresponding objects bear the same label not just by chance: for example, from the icosohedron one can construct the function x^2 + y^3 + z^5, and from it the diagram E8, and also the caustic and wave front of the same name. To easily checked properties of one of a set of associated objects correspond properties of the others which need not be evident at all. Thus the relations between all the A, D, E-classififications can be use for the simultaneous study af all simple objects, in spit of the fact that the origin of many of these relations (for example, of the connections between functions and quivers) remains and unexplained manifestation of the mysterious unity of all things. There is now material on the web on this subject. In particular you can download John McKay's A Rapid Introduction to ADE Theory at: http://math.ucr.edu/home/baez/ADE.html classifications. Notes on Hyperspace which covers some of this same material with application to unified field theory and string theory. All for now. Saul-Paul ---------- === Subject: RE: Dodecahedron & Sagan's Contact You are quite right. I misspelled the reference but you found the right book. The reason it is to be receommended is that it discusses the tiling of the three dimensional sphere by 120 dodecahedra from which one obtains the Dodecahedral space as an orbit space of the three sphere under the subgroup of the quaternions corresponding to three space symmetries of the dodecahedron. Indidentally, if we let a.b and axb be the dot and cross products of vectors in three space, then the quaternion product of these is ab = -a.b + axb This is the best way to line up the quaternions in the form that people generalize when they go to Clifford algebras. Note that ab is in four space wth scalar part -a.b and vector part axb. The nice miracle is that ab is associative while axb is not. Very best, Lou I couldn't find the reference you gave below, but I found a reference: Du Val, P. Homographies, Quaternions, and Rotations. Oxford, England: Oxford University Press, 1964 which is, no doubt, the same thing. But I can't find it on amazon.com. I found it in the local (U. of Rochester) library. However, I now prefer Geometric Algebra over just the quaternions. The quaternions are a sub-space of the 3-dimensional Geometric Algebra multivector space. See, for example, the paper Imaginary Numbers are not Real - the Geometric Algebra of Spacetime at http://www.mrao.cam.ac.uk/~clifford/introduction/intro/intro.html and the new book Geometric Algebra for Physicists by Chris Doran, Anthony Lasenby. The geometric algebra has an invertible product, which the dot, cross and exterior products, by themselves are not. (Its a shame that the vector cross product became popular....which is only usable in 3-d vector space. in the mathematical physics books.) Bob Gray -----Original Message----- === Subject: Re: Dodecahedron & Sagan's Contact Bob Gray, Hugh Matlock, and Lynnclaire Dennis may not have been on the mailing list of this discussion. It is worth noting the resonance of an image of the universe as Dodecahedral Manifold and Lynnclaire's geometry. We have previously raised the question of the relationship of Dodec space and The Pattern (as I prefer to call what is now called Mereon). Since Dodec space is built (the five fold cylic branched covering) from a trefoil knot and the Pattern is an interrelationship of the trefoil knot and the regular solids, it is tantalizing to think that there must be a deep relationship here. The key book for a lot of this geometry is Quaternions Homographies and Rotations by Patrick du Val. In any case, this idea/evidence for Universe as a Dodec manifold raises the question once again! Best, Lou K. Jack & Stan, Plato in the *Timaeus*, allows Timaeus in dialogue with Socrates to describe God's construction of the world by way of triangular elements. Midway through his long dissertation, Timaeus describes the construction the tetrahedron, the cube the octahedron, and the icosahedron. He does not describe the construction of the dodecahedron, but merely says: There was yet a fifth combination which God used in the delineation of the universe with figures of animals. [Transltion by Benjamin Jowett.] There is only a hint here that he is describing a 12 sided solid, since there were traditionally 12 signs of the zodiac. Probably the exact construction of the dodecahedron was, in Plato's day, a Pythagorean secret. Each pentagonal face of the dodecahedron has the star pentagon (called the pentagram) consisting of the 5 diagonals, and this structure was sacred to the Pythagoreans as a symbol of their brotherhood. Construction of this pentagram could have employed the fact that each of the 5 diagonal lines cuts two other diagonals as a golden section (which in the terminology of the Greeks was in mean and extreme ratio. Book XIII of Euclid's *Elements* deals in great detail with the construction of the 5 Platonic solids. It is believed that this book was written by Theaetetus (c. 380 B.C.), and was included by Euclid (c. 300 B.C.) as the last chapter of his *Elements*. (Cf., B.L. van der Waerden,*Science Awakening*, pp, 172-173). In any case the scolium No.1 to Euclid's Book XIII, says that this book concerns the 5 so-called Platonic figures, which however do not belong to Plato, 3 of the 5 being due to the Pythagoreans, namely the cube, the pyramid, and the dodecahedron, while the octahedron and the icosahedron are due to Theaetetus. I would guess that Theatetus constructed these figures as duals to the cube and the dodecahedron, the pyramid (i.e., the tetrahedron) being self-dual. Proposition 1 of Euclid's Book XIII is a description of the golden section: If a straight line be cut in extreme and mean ratio, the square on the greater segment added to the half of the whole is five times the square of the half. This construction of the golden section is used over and over again throughout the entire Book XIII. Proposition 6: If a rational straight line be cut in extreme and mean ratio, each of the segments is the irrational straight line called apotome. Proposition 8: If in an equilateral and equiangular pentagon straight lines subtend two angles taken in order, they cut one another in extreme and mean ratio, and their greater segments are equal to the side of the pentagon. Proposition 17: To construct a dodecahedron and comprehend it in a sphere, like the aforesaid figures, and to prove that the side of the dodecahedron is the irrational straight line called apotome. [Translated from the Greek by Sir Thomas L. Heath] Of course, the great geometrical master of these objects is H.S.M. t = 2 cos (pi/5) = [(5^1/2) + 1]/2 = 1.6180339887... [using t as a replacement for Coxeter's tau symbol] which is the positive root of the quadratic equation x^2 - x -1 = 0 Writing this equation as x = 1 + (1/x), we see that t = 1 + 1/1+1/1+1/1+/+ ... It is well known that, of all regualr continued fractions, this converges slowest. Its nth convergent is fn+1/fn, where f1, f2,...are the Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, 21, 34, 55,... Since t^n-1 + t^n = t^n = 1, the integral powers of t are given by the formulae 2t^n = fn(5^1/2) + (fn-1 + fn+1) [n odd] 2t^-n = fn(5^1/2) + (fn-1 + fn+1) [n odd] 2t^n = (fn-1 + fn+1) + fn(5^1/2) [n even] 2t^-n = (fn-1 + fn+1 - fn(5^1/5) [n even] e.g., t^3 = (5^1/2) + 2, t^-6 = 9 - 4(5^1/2) [On page 52, Coxeter points out that the twelve vertices of the icosahedron can be obtained by dividing the twelve edges of an octahedron according to the golden section. Then he goes on to provide the Cartesian coordinates for the vertices of the 5 Platonic solids. Cube: (edge = 2): (+/-1, +/-1, +/-1) Tetrahedron: (edge = 2(2^1/2)): (1,1,1), (1,-1,-1), (-1,1,-1), (-1,-1,1) Octahedron: (edge = 2^1/2): (+/-1,0,0), (0,+/-1,0), (0,0,+/-1) Icosahedron: (edge = 2): (0,+/-t,+/-1), (+/-1,0,+/-t), (+/-t,+/-1,0) Dodecahedron: (edge = 2/t): (0,+/-t^-1,+/-t), (+/-t,0,+/-t^-1) (+/-t^-1,+/-t,0), (+/-1,+/-1,+/-1) (where the last set of vertices are the vertices of one of the 5 cubes inscribed in the dodecahedron). There is much more use of the golden section throughout this book, which deals with higher dimensional regular figures. The 4th dimension is especialy rich, and the golden section plays a big role in Coxeter's description of these 4D polytopes. as is evident from his detailed tables on pp. 298-304. (1985), p. 222: The diagrams were published in an eight-volume 'coffe table' book set that was soon reprinted worldwide. All over the planet people tried to figure out the pictures. The dodecahedron and the quasi-biological forms were especially evocative. Many clever suggestions were made by the public and carefully sifted by the Argus team. Many harebrained interpretations were also widely available, especially in weekly newspapers. Whole new industries developed--doubtless unforeseen by those who devised the Message--dedicated to using the diagrams to bilk the public. The Ancient and Mystical Order of the Dodecahedron was announced. The Machine was a UFO. The Machine was Ezekiel's Wheel. An angel revealed the meaning of the Messge and the diagrams to a Brazilian businessman, who distributed--at first at his own expense--his interpretation worldwide. With so many enigmatic diagrams to interpret, it was inevitable that many religions would recognize some of their iconography in the Message from the stars. And on page 310 the secrecy of the project was mentioned: The ancient Pythagoreans, who first discovered the dodecahedron, had declared its very existence a secret, and the penalties for disclosure were severe. So perhaps it was only fitting that this house-sized dodecahedron, halfway around the world and 2,600 years later, was known only to a few. Vagay and Eda were deep in the arcana of gravitational physics--twistors, renormalization of ghost propagators, time-like Killing vectors, non-Abelian gauge invariance, goedesic refocussing, eleven-dimensional Kaluza-klein treatments of supergravity, and of course, Eda' own quite different superunification. I say: write Eda backwards and you get adE, which suggests A-D-E Coxeter graphs:-) Coxeter graphs: Lie group: SU(2) Finite subgroup: An: *--*--*- ...-* (n nodes) SU(n+1) Cyclic (n+1) Dn: *--*--*- ...-* (n nodes) SO(2n) Dihedral double | * E6: *--*--*--*--* E(27) Tetrahedral double | * E7: *--*--*--*--*--* E(56) Octahedral double | * E8: *--*--*--*--*--*--* E(248) Dodecahedral double | * (There's that Dodecahedron again :-) Since these graphs classify (in the mathematical & not the security sense) at least 20 different mathematical objects of great utility in unified field theory and especially string theory, the study of these graphs (which I call ADEX theory) is indeed a different approach to superunification. For example the gauge groups are within the A-D-E classification of Lie groups (and Lie algebras). Sagan also mentions the Penrose twistors. It happens that gravitational instantons are closely related to twistors (Cf. Ward & Wells, *Twistor Geometry and Field Theory, 1990). These gravitational instantons are classified by the A-D-E graphs via the finite subgroups g of SU(2). Such a gravitationa instanton is the orbit space C^2/g. This is a 4-d compact space with a 3-d membrane between two hemispheres like the equator. This membrane provides tunneling between the two parts of the instanton. The structure of this membrane is the orbit space SU(2)/g. Note that in the case of g being the dodecahedral double group, there is the Oct. 8 *Nature* cover paper proposing this as the structure of the cosmological space! Also the 11-D supergravity theory invokes the E7 group as summetry group, with the 7-d torus subgroup as the hidden dimensions. Moreover this 11-d structure is included as a 6th substructure in the M-theory superunification of the 5 competing superstring theories. See the cover story of the November issue of Scientific American, which is an interview with Brian string theory to be aired Oct. 28 and Nov. 4. All for now. Saul-Paul ---------- Tenen , Michael E. Brandt, Ph.D. === Subject: Re: CONTACT - Carl Sagan - Welcome to The Machine. Hey Saul-Paul is there any connection of Golden Mean to Dodecahedron as Indrid Cold below suggests or is that New Age Cargo Cult Ka Ka? Jack, All of the Platonics have golden mean proportions built in. There are three intersecting golden rectangles that form an icosahedron, which is of course dual to the dodecahedron. There's nothing wrong with the golden proportion in this situation. Hi Everyone, Picking up on Bob Gray's comment about Geometric Algebra.... One of the implications of the Geometric Algebra approach (it seems to me) would be that the same structures would appear at different scales. So I think we could expect to find geometric symmetries at the planetary scale as well as at the cosmological scale. While earlier this year I had presented (statistical) evidence of great circle alignment and dodecahedral symmetry in the locations of sacred sites around the world, a more detailed look is leading me away from models that are symmetric in 3D to models that are projections of higher dimensional structures. The Planetary Hologram My current hypothesis is that information from a regular 4D geometry (such as the 24 cell or the Poincare Homology Sphere) is projected onto the surface of a 3D sphere, via a holographic projection. The nodes would correspond to various places on the surface of the sphere and the lines connecting the nodes would show up as great circle arcs. I have collected a database of about 1500 sites and devised several algorithms to search for such circles. It turns out that there is a lot of evidence for such arcs. 1. There are several arcs which contain 20 sites within 5km of a perfect great circle. (The same algorithm running with randomly positioned sites can finds few such alignments of even 10 sites.) 2. Admitting circles with as few as 4 sites per circle generates models with around 220 circles that can fit over 99% of the data. A system of 220 great circles of 10km width will cover, in total, about 15% of the earth's surface, so much of the sacred space on earth is concentrated on less than about 1/7 of the available surface. 3. The models found give a dramatic reduction in the degrees of freedom required to fit the data. We would expect to need 750 circles to perfectly fit 1500 randomly arranged points on a sphere. (Each pair of points determine a circle.) 4. Some sites lie on several of the great circles. Stonehenge, for example, lies on 15 circles, each of which contains 9 or more sites. I see all of this as evidence that the data has some sort of projective symmetry. One of the interesting things about the latest models is that many of the constituent great circle arcs (derived from positions of sacred sites) also align with the mesoscale geology of the planet, and the most ancient and stable rock formations. The beauty of the hyperdimensional models is that they resolve two problems with traditional sacred geometry as an explanation for planetary structure... first, that the earth does not look like a regular crystal, and second, that, over geologic time, things have moved around, which would disrupt any 3D symmetries. Starting with the observation that the oldest and hardest rock (and best preserved fossils) lie along great circles connecting certain crossing points, the holographic paradigm suggests a new theory for what powers plate tectonics. If what we are experiencing in the 3D world is a slow (i.e. geologic speed) rotation of a higher dimensional object, then the control points (i.e. the images of the nodes of the hyperdimensional object) will move around relative to the spin axis of the planet. (To get an idea of what higher dimensional rotations look like when projected to 3d, check out Michael Gibbs' 4D Polytope Viewer http://members.aol.com/jmtsgibbs/draw4d.htm ) Movement of the relative locations of the control points will cause the intermediate crust to stretch or compress, giving us the seafloor spreading and mountain building that we know from geology. Peace, Hugh P.S. Here is a link to Jeff Weeks' site, from which you can download a viewer for various sorts of spaces. I did not see a file for the dodecahedral space he is proposing, however. Computer Graphics in Curved Spaces http://geometrygames.org/CurvedSpaces/index.html P.P.S. And now, a few links I came across related to the Poincare Homology Sphere.... John Baez This Week's Finds in Mathematical Physics (Week 164) You may recall from week163 that the Poincare homology 3-sphere is a compact 3-manifold that has the same homology groups as the ordinary 3-sphere, but is not homeomorphic to the 3-sphere. I explained how this marvelous space can be obtained as the quotient of SU(2) = S3 by a 120-element subgroup - the double cover of the symmetry group of the dodecahedron. Even better, the points in S3 which lie in this subgroup are the centers of the faces a 4d regular polytope with 120 dodecahedral faces. That's pretty cool. But here's another cool way to get the Poincare homology sphere.. E8 is the biggest of the exceptional Lie groups. As I explained in week64, the Dynkin diagram of this group looks like this: o----o---o----o----o----o----o | | o Now, make a model of this diagram by linking together 8 rings: / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / Imagine this model as living in S3. Next, hollow out all these rings: actually delete the portion of space that lies inside them! We now have a 3-manifold M whose boundary dM consists of 8 connected components, each a torus. Of course, a solid torus also has a torus as its boundary. So attach solid tori to each of these 8 components of dM, but do it via this attaching map: (x,y) -> (y,-x+2y) where x and y are the obvious coordinates on the torus, numbers between 0 and 2pi, and we do the arithmetic mod 2pi. We now have a new 3-manifold without boundary... and this is the Poincare homology sphere. We see here a strange and indirect connection between E8 and the dodecahedron. This is not the only such connection! There's also the McKay correspondence (see week65) and a way of getting the E8 root lattice from the icosians (see week20). Are these three superficially different connections secretly just different views of the same grand picture? I'm not sure. I think I'd know the answer to part of this puzzle if I better understood the relation between ADE theory and singularities. But Diarmuid Crowley told me much more.... http://math.ucr.edu/home/baez/week164.html Ruth Lawrence The PSU(3) invariant of the Poincare homology sphere Abstract: Using the R-matrix formulation of the sl_3 invariant of links, we compute the coloured sl_3 generalised Jones polynomial for the is obtained. This takes complex number values at roots of unity. The result obtained is formally an infinite sum, independent of the order of the root of unity, which at roots of unity reduces to a finite sum. This form enables the derivation of the PSU(3) analogue of the Ohtsuki series for the Poincare homology sphere, which it was shown by Thang Le could be extracted from the PSU(N) invariants of any rational homology sphere. http://www.ma.huji.ac.il/~ruthel/papers/psu3phs.html David Gillman UCLA The best picture of Poincare's homology sphere (video lecture) http://www.pims.math.ca/science/2002/cascade/gillman/ === Subject: Solving odd-even recurrence relation I am trying to find a method for solving linear recurrence relations of the form: a_2n = F(a_n) # Even terms a_2n+1 = G(a_n) # Odd terms where F() and G() are linear functions. For example, a_2n = c * a_n + d * a_n-1 + b * a_n-2 + 1 a_2n+1 = k * a_n - 1 --john === Subject: Re: Solving odd-even recurrence relation > I am trying to find a method for solving linear recurrence relations of > the form: > a_2n = F(a_n) # Even terms > a_2n+1 = G(a_n) # Odd terms > where F() and G() are linear functions. > For example, > a_2n = c * a_n + d * a_n-1 + b * a_n-2 + 1 > a_2n+1 = k * a_n - 1 > --john Look into generating functions. However, I think your kind of problem probably leads to pretty messy stuff. === Subject: Re: Relation between theorem and Rationals/Irrationals === >Subject: Re: Relation between theorem and Rationals/Irrationals > I need to prove a theorem that states: > > If R is a number set and for any 2 numbers there exists a point of R > between them, then every number is a limit point of R. > >>... then every number of R? is a limit ... >> >>R = {0} union (1,2) satisfies the condition, >>yet 0 isn't a limit point of R. >> Do you mean {0} union ]1, 2]? >No, I said ]1,2[ , however ]1,2] , that is (1,2] , will suffice >when 0 is added. >>I think the OP left out the condition >> that the number set be open/closed. >I'm supposed to be psychic? >Heck no, are students in such a hurry or so lazy that they >won't bother to ask coherent questions? >Alas too many students need to learn how to state questions. >The rationals, which are neither open nor closed, >satisfies both the condition and the conclusion. >So premises other than 'open' could be divined. >As for closed, { 1/n | n in N } / {0} is closed set >with only one limit point, namely 0. So closed doesn't >suffice as 1/2, 1/3, ... aren't limit points of the set. >It's pointless to guess what poorly written posts intend. >If it's of worth, the OP will soon make amends. I'm sorry if I ruined your day but the actual theorem is something that my change it to try to trick anyone. I was just thinking about it and decided to ask my question. === Subject: Re: Relation between theorem and Rationals/Irrationals [...] |>It's pointless to guess what poorly written posts intend. I don't agree, not with all of them at least. At work we occasionally deal with clients who are a bit, shall we say, mediocre at explaining what they need. Deciding that they are not being articulate enough is not always a reasonable option. |>If it's of worth, the OP will soon make amends. | |I'm sorry if I ruined your day but the actual theorem is something that my |didn't change it to try to trick anyone. It's true that we get a lot of poorly written posts on sci.math, sometimes to the point that all we can do is to try to get some kind of clarification. Once in awhile people jump the gun a bit at deciding that a posting in too unclear. But don't let it bother you. The original posting was fine. The intent was clear enough, and it was easy for someone practiced in doing proofs of this kind to answer. Keith Ramsay === Subject: Re: Relation between theorem and Rationals/Irrationals === Subject: Re: Relation between theorem and Rationals/Irrationals Adjunct Assistant Professor at the University of Montana. >I need to prove a theorem that states: >If R is a number set and for any 2 numbers there exists a point of R between >them, then every number is a limit point of R. >Is this basically the same proof as proving that there is a rational number >between any 2 irrationals and there is an irrational between any 2 rationals? No, I don't think so... >Or am I way off base here? I guess I'm not too sure how I could easily make >the transition from that to this theorem or if it's possible at all. You are assuming that: (*) Given any two numbers, x and y, there is a number r in R such that x0, there is a number r in R such that |x-r|0 be any positive number. Then both x and x+e are numbers, so what can you conclude from (*)? And given that conclusion, can you show that there must be some r in R with |x-r| < e? Or, if you defined limit point as the limit of a sequence, then we want to show that for every number x, there exists a sequence r_1, r_2, r_3, ...., r_n, .... of elements of R such that lim(r_n) = x. So: for each n, consider the numbers x and x+(1/n). What can you conclude from (*) for those two numbers? How do you use that conclusion to construct the sequence r_1, r_2, r_3, ... ? It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: simple question about subspaces |I was wondering about the minimal required definition. I was told the |inclusion of the zero vector must be explicit to avoid a problem where |the empty set could be a subspace of V, but 0 * v = 0 implies directly |that 0 must be in S so why should it be required explicitly? It would be enough to require that the set has an element in it, because given that some v is an element of the set, the fact that the set is closed under scalar multiplication would imply that 0*v=0 is an element too. So we could in fact replace the condition that 0 is an element by the condition only that the set has an element in it. But we need a condition like this; the empty set is closed under all operations, because there are no elements which need to be operated on. It might be worth pointing out that this is a case of the convention, always followed in mathematics, that a statement saying each element of a set has a property is considered vacuously true when the set is empty. I wouldn't say that this is consistently followed in natural language, but I think it would cause needless complications and a lot of confusion if we were to depart from this convention. So for every element v of the empty set, 0*v is an element (because there aren't any elements of the empty set). I find it interesting to ask myself what advantages certain conventions give us. We could, after all, leave this condition out of the definition of vector subspace, and also perhaps modify the definition of vector space to permit the empty set as a vector. Instead of having the usual axioms for 0 and -, we could have (v+(-v))+w=w and so on. If we defined the span of a set of vectors as the minimal subspace containing them, as usual, then the span of {} would be {}, and the span of {0} would be {0}. A set of vectors is considered independent if removing any element shrinks the span, so {0} would count as independent. That would create a special case for the criterion that a set of vectors be independent. As usual, a set of vectors would be independent if none of them could be written as a (nonempty) linear combination of other elements of the set. But that would no longer be equivalent to the more symmetrical condition that there not exist a way to write 0 as a nontrivial linear combination of elements of the set. Thinking of the sum (or linear combination) of an empty set of vectors as being 0 seems simply to be the right way to set things up. Keith Ramsay === Subject: Re: simple question about subspaces > |I was wondering about the minimal required definition. I was told the > |inclusion of the zero vector must be explicit to avoid a problem where > |the empty set could be a subspace of V, but 0 * v = 0 implies directly > |that 0 must be in S so why should it be required explicitly? > It would be enough to require that the set has an element in it, because > given that some v is an element of the set, the fact that the set is closed > under scalar multiplication would imply that 0*v=0 is an element too. So we > could in fact replace the condition that 0 is an element by the condition > only that the set has an element in it. But we need a condition like this; > the empty set is closed under all operations, because there are no elements > which need to be operated on. We are accustomed to thinking about addition in a vector space as defined by a binary operation, +:VxV->V. But what we really need are two operations: 0:V +:VxV->V Here 0 is not an operation proper, since it takes no arguments (maybe it is called a 0-ary operation). The empty set is /not/ closed under both theese operations. Also, instead of using theese two operations we may use a operation which to a finite set of vectors assigns the sum of the vectors. If we call the operation Sum, then we have Sum{}=0 and Sum{v1, v2}= v1+v2. Again, the empty set is not closed under this operation. This can be generalised. The operation Sum can be defined for a monoid, in which case we must replace finite sets with finite lists, and in which case we should call it Prod instead of Sum. If e is the neutral element, we have Prod[]=e and Prod[a1, a2]=a1a2. The function Prod can also be defined for a category. In this case, Prod[] is ambigous since there are (in general) many identity arrows, but this ambiguity gets resolved when it is used in expressions such as (Prod[])(Prod[a1, a2, a3])=a1a2a3. If we want to define Prod for a ring without unity, then its arguments must be /non-empty/ lists. Alternatively, we could define Prod[] to be an element that is not part of the ring, which can be multiplied with elements of the ring but not added to them (Can a ring without unity always be extended to a ring with unity?). The reason why the empty set is not a subspace of a vector space is the same as why an arbitrary subset is not a subspace: it is not closed under addition and scalar multiplication. But the closure of any subset is a subspace, the empty set being no exception. Mattias === Subject: Re: simple question about subspaces Adjunct Assistant Professor at the University of Montana. >> |I was wondering about the minimal required definition. I was told the >> |inclusion of the zero vector must be explicit to avoid a problem where >> |the empty set could be a subspace of V, but 0 * v = 0 implies directly >> |that 0 must be in S so why should it be required explicitly? >> It would be enough to require that the set has an element in it, because >> given that some v is an element of the set, the fact that the set is closed >> under scalar multiplication would imply that 0*v=0 is an element too. So we >> could in fact replace the condition that 0 is an element by the condition >> only that the set has an element in it. But we need a condition like this; >> the empty set is closed under all operations, because there are no elements >> which need to be operated on. >We are accustomed to thinking about addition in a vector space as >defined by a binary operation, +:VxV->V. But what we really need are >two operations: >0:V >+:VxV->V >Here 0 is not an operation proper, since it takes no arguments (maybe >it is called a 0-ary operation). It is a perfectly fine operation! It is a nullary or zeroary operation, indeed. In general, if I is a set, then an I-ary operation on A is a function A^I -> A. Since 0 = {}, a 0-ary operation is a function A^{}->A; the set A^{} consists of all function from {} to A; the only function from {} to A is the empty function, so a zero-ary operation translates to a map from {emptyset} to A, which means a distinguished element of A. This is all part of General or Universal Algebra. But if you want to be technical, a vector space V over the field K really has more than those 2 operations. It has 3 operations associated to the abelian group structure of V, and then one operation for every element of K: +:V x V -> V binary operation (-): V -> V unary operation sending v to its additive inverse 0: V^{} -> V a nullary operation identifying the zero vector; and for each k in K, k: V->V a unary operation corresponding to scalar multiplication by k. If you are in a variety of algebras, then the empty algebra is an algebra in the variety if and only if the type has no nullary operations. So the empty set is not a vector space, since the type of vector spaces includes a nullary operation. It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: simple question about subspaces |If you are in a variety of algebras, then the empty algebra is an |algebra in the variety if and only if the type has no nullary |operations. So the empty set is not a vector space, since the type of |vector spaces includes a nullary operation. I considered mentioning that, plus the fact that one could eliminate the references to 0 in the definition by having the axiom (x+(-x))+y=y and so on. This would make the usual universal algebra definitions compatible with the definition of vector space and vector subspace being extended to permit the empty set. I think maybe the suggestion that addition really is an operation on finite multisets of elements serves to explain why such an approach works out less elegantly than the usual one. Keith Ramsay === Subject: Re: simple question about subspaces Adjunct Assistant Professor at the University of Montana. >|If you are in a variety of algebras, then the empty algebra is an >|algebra in the variety if and only if the type has no nullary >|operations. So the empty set is not a vector space, since the type of >|vector spaces includes a nullary operation. >I considered mentioning that, plus the fact that one could eliminate >the references to 0 in the definition by having the axiom (x+(-x))+y=y >and so on. Would that axiom not allow an underlying structure which is an inverse monoid but not a group? Hmmm... Haven't thought it through; perhaps the scalar multiplication takes care of that... 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: Core error, FEAR is a natural response :> :> :> :> :> :> :> What is the false implication? :> :> :> :> :> :> :> :> :> : The false implication is that b is a non-unit factor of 2 like :> :> : sqrt(2), or 2^{1/3}, when in fact, appropriately, it's a factor of 1. :> :> :> : Unfortunately the poster deleted out pertinent information, so I'll :> :> : make the effort to put it back in so that it makes sense to readers :> :> : without forcing them to go back to previous posts. :> :> :> : I gave the example of xy = 2, where x and y are algebraic integers, :> :> : where x=2a and y=b, so b is an algebraic integer, but 'a' is not, but :> :> : it is a member of what I call the object ring. :> :> :> : The object ring is a commutative ring that includes all numbers such :> :> : that -1 and 1 are the only members that are both a unit and an :> :> : integer, where no non-unit member is a factor of any two integers that :> :> : are coprime. :> :> :> : Notice that definition for the object ring excludes possibilities like :> :> : a=1/2 as then 2 would be a unit since 2(1/2) = 1. :> :> :> : So, in fact, in the object ring, ab=1 and b IS a unit, but because 'a' :> :> : is inappropriately excluded from the ring of algebraic integers by the :> :> : definition of algebraic integers as roots of monic polynomials with :> :> : integer coefficients, you have the false implication that b is some :> :> : other type factor of 2. :> :> :> : Now then, say you follow algebra, and then you find that you have 2 as :> :> : a factor of x, well you've been pushed out of the ring of algebraic :> :> : integers where x does NOT have 2 as a factor, even though x=2a. :> :> :> :> :> : And you can see several problems that popped up by that exclusion as b :> :> :> :> : is not a unit in the ring of algebraic integers, when it should be, :> :> :> :> : and x does NOT have 2 as a factor in the ring, though x=2a. :> :> :> :> :> :> :> :> So, what is the problem? I haven't seen anything that looks like :> :> :> :> a contradiction, only an assertion that something is not a unit :> :> :> :> which should be according to some unstated principle. What's the :> :> :> :> principle? :> :> :> :> : 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. :> :> :> :> :> :> What is behind the shoulds, i.e. what are the bad consequences of :> :> :> excluding 'a'? Before you were talking about being able to prove two :> :> :> contradictory results. :> :> :> :> :> :> Mike :> :> :> : The problem occurs because algebra does NOT recognize the arbitrary :> :> : definition, so using algebra you can find yourself in situations where :> :> : you prove that x has 2 as a factor, but then again, because of this :> :> : arbitrary definition, it does not, in the ring of algebraic integers. :> :> :> : Since the assumption is that algebraic integers is an appropriate :> :> : ring, it appears that you've proven two contradictory things: :> :> :> :> What do you mean by an appropriate ring? :> :> : One where you can't appear to prove contradictory things. :> :> : :> :> : 1. x has 2 as a factor :> :> :> : 2. x does NOT have 2 as a factor :> :> :> :> But aren't 1. and 2. actually: :> :> :> :> 1. x has 2 as a factor in the object ring :> :> : I'll remind of the example of 2 and 6 in the ring of evens to help :> : readers with context, as there 2 is not a factor of 6 because 3 isn't :> : even. :> :> : Now then, with my example, you have x=2a, where x is an algebraic :> : integer, but as 'a' is not, it doesn't have 2 as a factor, in the ring :> : of algebraic integers, which falsely implies, given that xy=2, where y :> : is an algebraic integer that x and y have non-unit factors of 2 in the :> : same sense as like if you have sqrt(2) as a factor for both. :> :> : But, in fact, *neither* has what you might call non-trivial factors of :> : 2 in the ring of algebraic integers, which I have to say as trivially :> : they have themselves as factors. :> :> : That is, given that xy=2, x is trivially a factor of 2, in the ring of :> : algebraic integers, as is y, when in fact x has a factor that IS 2, a :> : non-trivial factor, in an appropriate ring, that is, one which does :> : not allow contradictions. :> :> : It's worth noting that proving that x doesn't have a factor of 2, in :> : the ring of algebraic integers, requires going to the field of :> : algebraic numbers, which is itself, of course, not flawed. :> :> :> 2. x does NOT have 2 as a factor in the algebraic integers :> :> : And proving that requires going to the field of algebraic numbers, :> : which hasn't really been discussed at this point. :> :> : So then *in the ring of algebraic integer* you can appear to prove :> : *both* things, and only figure out that you've been pushed out of the :> : ring of algebraic integers, by going to the field of algebraic :> : numbers. :> I'm not following. You let x = 2a where x is an algebraic integer :> and a is not. So x does not have a factor of 2 in the algebraic :> integers. What is the contradictory proof that x *does* have a factor :> of 2 in the algebraic integers? : In my case I use a special polynomial P(m), which is special in that I : can rather easily factor it into non-polynomial factors to appear to : prove within the ring of algebraic integers that x has 2 as a factor. Oh. I was under the impression that this x=2a example was supposed to illustrate the issue without depending on the details of your argument. Under the stipulations of the example, x does *not* have a factor of 2 in the algebraic integers. Correct? I take this to be an immediate consequence of the definition of factor in the algebraic integers. But you claim that you have an argument which proves that x *does* have 2 as factor in the algebraic integers; that is, your argument proves something false. Therefore, you conclude... what? I still don't get it. It would seem that either the argument is incorrect or it is based on a false assumption. I assume you believe the argument is correct, so what is the false assumption? Mike === Subject: Re: Core error, FEAR is a natural response > :> :> > :> : :> :> :> What is the false implication? > :> :> : :> :> :> > :> :> : The false implication is that b is a non-unit factor of 2 like > :> :> : sqrt(2), or 2^{1/3}, when in fact, appropriately, it's a factor of 1. > :> :> : Unfortunately the poster deleted out pertinent information, so I'll > :> :> : make the effort to put it back in so that it makes sense to readers > :> :> : without forcing them to go back to previous posts. > :> :> : I gave the example of xy = 2, where x and y are algebraic integers, > :> :> : where x=2a and y=b, so b is an algebraic integer, but 'a' is not, but > :> :> : it is a member of what I call the object ring. > :> :> : The object ring is a commutative ring that includes all numbers such > :> :> : that -1 and 1 are the only members that are both a unit and an > :> :> : integer, where no non-unit member is a factor of any two integers that > :> :> : are coprime. > :> :> : Notice that definition for the object ring excludes possibilities like > :> :> : a=1/2 as then 2 would be a unit since 2(1/2) = 1. > :> :> : So, in fact, in the object ring, ab=1 and b IS a unit, but because 'a' > :> :> : is inappropriately excluded from the ring of algebraic integers by the > :> :> : definition of algebraic integers as roots of monic polynomials with > :> :> : integer coefficients, you have the false implication that b is some > :> :> : other type factor of 2. > :> :> : Now then, say you follow algebra, and then you find that you have 2 as > :> :> : a factor of x, well you've been pushed out of the ring of algebraic > :> :> : integers where x does NOT have 2 as a factor, even though x=2a. > :> :> :> :> : And you can see several problems that popped up by that exclusion as b > :> :> :> :> : is not a unit in the ring of algebraic integers, when it should be, > :> :> :> :> : and x does NOT have 2 as a factor in the ring, though x=2a. > :> :> :> : :> :> :> :> So, what is the problem? I haven't seen anything that looks like > :> :> :> :> a contradiction, only an assertion that something is not a unit > :> :> :> :> which should be according to some unstated principle. What's the > :> :> :> :> principle? > :> :> :> : 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. > :> :> : :> :> :> What is behind the shoulds, i.e. what are the bad consequences of > :> :> :> excluding 'a'? Before you were talking about being able to prove two > :> :> :> contradictory results. > :> :> : :> :> :> Mike > :> :> : The problem occurs because algebra does NOT recognize the arbitrary > :> :> : definition, so using algebra you can find yourself in situations where > :> :> : you prove that x has 2 as a factor, but then again, because of this > :> :> : arbitrary definition, it does not, in the ring of algebraic integers. > :> :> : Since the assumption is that algebraic integers is an appropriate > :> :> : ring, it appears that you've proven two contradictory things: > :> : :> :> What do you mean by an appropriate ring? > :> : One where you can't appear to prove contradictory things. > :> : > :> :> : 1. x has 2 as a factor > :> :> : 2. x does NOT have 2 as a factor > :> : :> :> But aren't 1. and 2. actually: > :> : :> :> 1. x has 2 as a factor in the object ring > :> : I'll remind of the example of 2 and 6 in the ring of evens to help > :> : readers with context, as there 2 is not a factor of 6 because 3 isn't > :> : even. > :> : Now then, with my example, you have x=2a, where x is an algebraic > :> : integer, but as 'a' is not, it doesn't have 2 as a factor, in the ring > :> : of algebraic integers, which falsely implies, given that xy=2, where y > :> : is an algebraic integer that x and y have non-unit factors of 2 in the > :> : same sense as like if you have sqrt(2) as a factor for both. > :> : But, in fact, *neither* has what you might call non-trivial factors of > :> : 2 in the ring of algebraic integers, which I have to say as trivially > :> : they have themselves as factors. > :> : That is, given that xy=2, x is trivially a factor of 2, in the ring of > :> : algebraic integers, as is y, when in fact x has a factor that IS 2, a > :> : non-trivial factor, in an appropriate ring, that is, one which does > :> : not allow contradictions. > :> : It's worth noting that proving that x doesn't have a factor of 2, in > :> : the ring of algebraic integers, requires going to the field of > :> : algebraic numbers, which is itself, of course, not flawed. > :> :> 2. x does NOT have 2 as a factor in the algebraic integers > :> : And proving that requires going to the field of algebraic numbers, > :> : which hasn't really been discussed at this point. > :> : So then *in the ring of algebraic integer* you can appear to prove > :> : *both* things, and only figure out that you've been pushed out of the > :> : ring of algebraic integers, by going to the field of algebraic > :> : numbers. > : :> I'm not following. You let x = 2a where x is an algebraic integer > :> and a is not. So x does not have a factor of 2 in the algebraic > :> integers. What is the contradictory proof that x *does* have a factor > :> of 2 in the algebraic integers? > : In my case I use a special polynomial P(m), which is special in that I > : can rather easily factor it into non-polynomial factors to appear to > : prove within the ring of algebraic integers that x has 2 as a factor. > Oh. I was under the impression that this x=2a example was supposed > to illustrate the issue without depending on the details of your > argument. Well, it's possible to operate from the more inclusive ring of objects, but I find myself trying to explain with rings and fields that people are more familiar with, and as 'a' is NOT in the ring of algebraic integers, you can't have x=2a in that ring, but you can in the field of algebraic numbers. However, *within* the ring of algebraic integers I can prove that x has 2 as a factor, which is a fascinating oddity, as in fact, it does not *in the ring of algebraic integers* so I've been pushed out of the ring. >Under the stipulations of the example, x does *not* have a > factor of 2 in the algebraic integers. Correct? It turns out though that *in the ring of algebraic integers* you can appear to prove that it does and find out that 'a' is not an algebraic integer by going to the field of algebraic numbers. That's where you can prove that the root of a non-monic primitive irreducible over Q cannot be the root of a monic primitive, and therefore can't be an algebraic integer. It's actually rather fascinating. >I take this to be an > immediate consequence of the definition of factor in the algebraic > integers. But you claim that you have an argument which proves that x > *does* have 2 as factor in the algebraic integers; that is, your argument > proves something false. Actually the argument is correct, but here's how it goes so you can see how things get screwed up. If you use only algebraic integers and do various algebraic operations that are legitimate in the ring of algebraic integers then *presumably* you'll get algebraic integers. However, I've found a way to get numbers that are NOT algebraic integers. If the ring of algebraic integers was not screwed up, that wouldn't happen. So assuming that the ring is not screwed up you further assume that x has 2 as a factor, but if you know the ring is screwed up--as I do--then you can say it *should* have it as a factor. >Therefore, you conclude... what? I still don't get > it. It would seem that either the argument is incorrect or it is based > on a false assumption. I assume you believe the argument is correct, > so what is the false assumption? > Mike The false assumption is that the ring of algebraic integers is a proper ring i.e. a ring that wouldn't allow such a contradiction. So using ring operations you find a result, which is that x has 2 as a factor, but it turns out that the ring if flawed. It's fun stuff. James Harris === Subject: Re: Core error, FEAR is a natural response :> :> :> : Now then, with my example, you have x=2a, where x is an algebraic :> :> : integer, but as 'a' is not, it doesn't have 2 as a factor, in the ring :> :> : of algebraic integers, which falsely implies, given that xy=2, where y :> :> : is an algebraic integer that x and y have non-unit factors of 2 in the :> :> : same sense as like if you have sqrt(2) as a factor for both. :> :> :> : But, in fact, *neither* has what you might call non-trivial factors of :> :> : 2 in the ring of algebraic integers, which I have to say as trivially :> :> : they have themselves as factors. :> :> :> : That is, given that xy=2, x is trivially a factor of 2, in the ring of :> :> : algebraic integers, as is y, when in fact x has a factor that IS 2, a :> :> : non-trivial factor, in an appropriate ring, that is, one which does :> :> : not allow contradictions. :> :> :> : It's worth noting that proving that x doesn't have a factor of 2, in :> :> : the ring of algebraic integers, requires going to the field of :> :> : algebraic numbers, which is itself, of course, not flawed. :> :> :> :> 2. x does NOT have 2 as a factor in the algebraic integers :> :> :> : And proving that requires going to the field of algebraic numbers, :> :> : which hasn't really been discussed at this point. :> :> :> : So then *in the ring of algebraic integer* you can appear to prove :> :> : *both* things, and only figure out that you've been pushed out of the :> :> : ring of algebraic integers, by going to the field of algebraic :> :> : numbers. :> :> :> :> I'm not following. You let x = 2a where x is an algebraic integer :> :> and a is not. So x does not have a factor of 2 in the algebraic :> :> integers. What is the contradictory proof that x *does* have a factor :> :> of 2 in the algebraic integers? :> :> : In my case I use a special polynomial P(m), which is special in that I :> : can rather easily factor it into non-polynomial factors to appear to :> : prove within the ring of algebraic integers that x has 2 as a factor. :> Oh. I was under the impression that this x=2a example was supposed :> to illustrate the issue without depending on the details of your :> argument. : Well, it's possible to operate from the more inclusive ring of : objects, but I find myself trying to explain with rings and fields : that people are more familiar with, and as 'a' is NOT in the ring of : algebraic integers, you can't have x=2a in that ring, but you can in : the field of algebraic numbers. : However, *within* the ring of algebraic integers I can prove that x : has 2 as a factor, which is a fascinating oddity, as in fact, it does : not *in the ring of algebraic integers* so I've been pushed out of the : ring. :>Under the stipulations of the example, x does *not* have a :> factor of 2 in the algebraic integers. Correct? : It turns out though that *in the ring of algebraic integers* you can : appear to prove that it does and find out that 'a' is not an algebraic : integer by going to the field of algebraic numbers. : That's where you can prove that the root of a non-monic primitive : irreducible over Q cannot be the root of a monic primitive, and : therefore can't be an algebraic integer. : It's actually rather fascinating. :>I take this to be an :> immediate consequence of the definition of factor in the algebraic :> integers. But you claim that you have an argument which proves that x :> *does* have 2 as factor in the algebraic integers; that is, your argument :> proves something false. : Actually the argument is correct, but here's how it goes so you can : see how things get screwed up. : If you use only algebraic integers and do various algebraic operations : that are legitimate in the ring of algebraic integers then : *presumably* you'll get algebraic integers. : However, I've found a way to get numbers that are NOT algebraic : integers. The only operations that are allowed in the ring of algebraic integers are addition and multiplication. Are you claiming that the algebraic integers are not closed under these operations? Mike === Subject: Re: Core error, FEAR is a natural response > > :> > :> :> : Now then, with my example, you have x=2a, where x is an algebraic > :> :> : integer, but as 'a' is not, it doesn't have 2 as a factor, in the ring > :> :> : of algebraic integers, which falsely implies, given that xy=2, where y > :> :> : is an algebraic integer that x and y have non-unit factors of 2 in the > :> :> : same sense as like if you have sqrt(2) as a factor for both. > :> :> : But, in fact, *neither* has what you might call non-trivial factors of > :> :> : 2 in the ring of algebraic integers, which I have to say as trivially > :> :> : they have themselves as factors. > :> :> : That is, given that xy=2, x is trivially a factor of 2, in the ring of > :> :> : algebraic integers, as is y, when in fact x has a factor that IS 2, a > :> :> : non-trivial factor, in an appropriate ring, that is, one which does > :> :> : not allow contradictions. > :> :> : It's worth noting that proving that x doesn't have a factor of 2, in > :> :> : the ring of algebraic integers, requires going to the field of > :> :> : algebraic numbers, which is itself, of course, not flawed. > :> :> :> 2. x does NOT have 2 as a factor in the algebraic integers > :> :> : And proving that requires going to the field of algebraic numbers, > :> :> : which hasn't really been discussed at this point. > :> :> : So then *in the ring of algebraic integer* you can appear to prove > :> :> : *both* things, and only figure out that you've been pushed out of the > :> :> : ring of algebraic integers, by going to the field of algebraic > :> :> : numbers. > :> : :> :> I'm not following. You let x = 2a where x is an algebraic integer > :> :> and a is not. So x does not have a factor of 2 in the algebraic > :> :> integers. What is the contradictory proof that x *does* have a factor > :> :> of 2 in the algebraic integers? > :> : In my case I use a special polynomial P(m), which is special in that I > :> : can rather easily factor it into non-polynomial factors to appear to > :> : prove within the ring of algebraic integers that x has 2 as a factor. > : :> Oh. I was under the impression that this x=2a example was supposed > :> to illustrate the issue without depending on the details of your > :> argument. > : Well, it's possible to operate from the more inclusive ring of > : objects, but I find myself trying to explain with rings and fields > : that people are more familiar with, and as 'a' is NOT in the ring of > : algebraic integers, you can't have x=2a in that ring, but you can in > : the field of algebraic numbers. > : However, *within* the ring of algebraic integers I can prove that x > : has 2 as a factor, which is a fascinating oddity, as in fact, it does > : not *in the ring of algebraic integers* so I've been pushed out of the > : ring. > :>Under the stipulations of the example, x does *not* have a > :> factor of 2 in the algebraic integers. Correct? > : It turns out though that *in the ring of algebraic integers* you can > : appear to prove that it does and find out that 'a' is not an algebraic > : integer by going to the field of algebraic numbers. > : That's where you can prove that the root of a non-monic primitive > : irreducible over Q cannot be the root of a monic primitive, and > : therefore can't be an algebraic integer. > : It's actually rather fascinating. > :>I take this to be an > :> immediate consequence of the definition of factor in the algebraic > :> integers. But you claim that you have an argument which proves that x > :> *does* have 2 as factor in the algebraic integers; that is, your argument > :> proves something false. > : Actually the argument is correct, but here's how it goes so you can > : see how things get screwed up. > : If you use only algebraic integers and do various algebraic operations > : that are legitimate in the ring of algebraic integers then > : *presumably* you'll get algebraic integers. > : However, I've found a way to get numbers that are NOT algebraic > : integers. > The only operations that are allowed in the ring of algebraic > integers are addition and multiplication. Are you claiming that > the algebraic integers are not closed under these operations? > Mike That's partly how mathematicians were fooled into thinking the ring is without problems as it IS closed under those operations; however, there is decomposition or factorization, even in the ring of algebraic integers. It's in decomposition that the problem arises as you can have an algebraic integer that results from multiplying an algebraic integer times a number from a more inclusive ring, like with my x=2a example, where 'a' is an object but not an algebraic integer, while x and, of course, 2 are both algebraic integers. James Harris === Subject: Re: Core error, FEAR is a natural response > That's partly how mathematicians were fooled into thinking the ring is > without problems as it IS closed under those operations; however, > there is decomposition or factorization, even in the ring of algebraic > integers. > It's in decomposition that the problem arises as you can have an > algebraic integer that results from multiplying an algebraic integer > times a number from a more inclusive ring, like with my x=2a example, > where 'a' is an object but not an algebraic integer, while x and, of > course, 2 are both algebraic integers. In other words, you just figured out, about years later than everyone else, that the algebraic integers are a ring and not a field. Harris, you are such a genius. And not that I want to upset you, but my prime counting program was about one thousand times faster than yours! That's what I call getting your ass kicked! === Subject: Re: Core error, FEAR is a natural response >> That's partly how mathematicians were fooled into thinking the ring is >> without problems as it IS closed under those operations; however, >> there is decomposition or factorization, even in the ring of algebraic >> integers. >> It's in decomposition that the problem arises as you can have an >> algebraic integer that results from multiplying an algebraic integer >> times a number from a more inclusive ring, like with my x=2a example, >> where 'a' is an object but not an algebraic integer, while x and, of >> course, 2 are both algebraic integers. >In other words, you just figured out, about years later than everyone >else, that the algebraic integers are a ring and not a field. No. He figured out that they are not a field. He has no clue whatever why they form a ring - the only reason he believes that is he's been told it's so by lying mathematicians. >Harris, >you are such a genius. >And not that I want to upset you, but my prime counting program was >about one thousand times faster than yours! That's what I call getting >your ass kicked! ************************ David C. Ullrich === Subject: Re: Core error, FEAR is a natural response :> :> : If you use only algebraic integers and do various algebraic operations :> : that are legitimate in the ring of algebraic integers then :> : *presumably* you'll get algebraic integers. :> :> : However, I've found a way to get numbers that are NOT algebraic :> : integers. :> The only operations that are allowed in the ring of algebraic :> integers are addition and multiplication. Are you claiming that :> the algebraic integers are not closed under these operations? :> Mike : That's partly how mathematicians were fooled into thinking the ring is : without problems as it IS closed under those operations; however, : there is decomposition or factorization, even in the ring of algebraic : integers. : It's in decomposition that the problem arises as you can have an : algebraic integer that results from multiplying an algebraic integer : times a number from a more inclusive ring, like with my x=2a example, : where 'a' is an object but not an algebraic integer, while x and, of : course, 2 are both algebraic integers. I assume you don't consider the case a=1/2 to be problematic. Why is there a problem when a is an object? And according to you, does x have a factor of 2 in the algebraic integers when a is an object, or not? === Subject: Re: Core error, FEAR is a natural response > > :> > :> : If you use only algebraic integers and do various algebraic operations > :> : that are legitimate in the ring of algebraic integers then > :> : *presumably* you'll get algebraic integers. > :> : However, I've found a way to get numbers that are NOT algebraic > :> : integers. > : :> The only operations that are allowed in the ring of algebraic > :> integers are addition and multiplication. Are you claiming that > :> the algebraic integers are not closed under these operations? > : : :> Mike > : That's partly how mathematicians were fooled into thinking the ring is > : without problems as it IS closed under those operations; however, > : there is decomposition or factorization, even in the ring of algebraic > : integers. > : It's in decomposition that the problem arises as you can have an > : algebraic integer that results from multiplying an algebraic integer > : times a number from a more inclusive ring, like with my x=2a example, > : where 'a' is an object but not an algebraic integer, while x and, of > : course, 2 are both algebraic integers. > I assume you don't consider the case a=1/2 to be problematic. Why > is there a problem when a is an object? And according to you, does > x have a factor of 2 in the algebraic integers when a is an object, > or not? Remember 1/2 in the ring means that 2 is a unit, and my definition of the object ring excludes any integers besides -1 and 1 from being units. And there isn't *necessarily* a problem with x=2a, just because 'a' is an object as for instance, with x=10, a=5 and it is an object. However, because of the focus on *monic* polynomials the ring of algebraic integers is small enough that there are numbers that fit in a ring where -1 and 1 are the only units, but they are not algebraic integers because they aren't the roots of monic polynomials with integer coefficients. It's just the perfect storm type kind of problem as it can hide for a long time, and in fact, the problem has sat there in mathematics for over a hundred years. Now mathematicians have to just suck it up, and show they can handle the truth. James Harris === Subject: Re: Core error, FEAR is a natural response Adjunct Assistant Professor at the University of Montana. >> >> :> >> :> : If you use only algebraic integers and do various algebraic operations >> :> : that are legitimate in the ring of algebraic integers then >> :> : *presumably* you'll get algebraic integers. >> >> :> : However, I've found a way to get numbers that are NOT algebraic >> :> : integers. >> :> :> The only operations that are allowed in the ring of algebraic >> :> integers are addition and multiplication. Are you claiming that >> :> the algebraic integers are not closed under these operations? >> :> :> :> Mike >> >> : That's partly how mathematicians were fooled into thinking the ring is >> : without problems as it IS closed under those operations; however, >> : there is decomposition or factorization, even in the ring of algebraic >> : integers. >> >> : It's in decomposition that the problem arises as you can have an >> : algebraic integer that results from multiplying an algebraic integer >> : times a number from a more inclusive ring, like with my x=2a example, >> : where 'a' is an object but not an algebraic integer, while x and, of >> : course, 2 are both algebraic integers. >> I assume you don't consider the case a=1/2 to be problematic. Why >> is there a problem when a is an object? And according to you, does >> x have a factor of 2 in the algebraic integers when a is an object, >> or not? >Remember 1/2 in the ring means that 2 is a unit, and my definition of >the object ring excludes any integers besides -1 and 1 from being >units. The integers satisfy your definition of object ring, yet they are not the object ring. Why? The algebraic integers satisfy every condition you give in the definition of object ring, yet they are not the object ring. Why? You have stated, but not proven in any way, that in the object ring, one and only one of (11-sqrt(105))/2 and (11+sqrt(105))/2 will be a unit in your ring; that is, that one and only one of the roots of 4x^2 - 11x + 1 will be in your object ring. Why only one? And which one? And how do you know which one? You cannot have both. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Core error, FEAR is a natural response >That's partly how mathematicians were fooled into thinking the ring is >without problems as it IS closed under those operations; however, >there is decomposition or factorization, even in the ring of algebraic >integers. >It's in decomposition that the problem arises as you can have an >algebraic integer that results from multiplying an algebraic integer >times a number from a more inclusive ring, like with my x=2a example, >where 'a' is an object but not an algebraic integer, while x and, of >course, 2 are both algebraic integers. And what problem arises from this? And why doesn't it apply to the even numbers or the ordinary rational integers? Consider: x = 2a where x is an integer (for instance 3). You can have the property that this is true but a is not an integer. Does this show there's a problem with the integers? - Randy === Subject: Re: Core error, FEAR is a natural response Discussion, linux) > That's partly how mathematicians were fooled into thinking the ring is > without problems as it IS closed under those operations; however, > there is decomposition or factorization, even in the ring of algebraic > integers. > It's in decomposition that the problem arises as you can have an > algebraic integer that results from multiplying an algebraic integer > times a number from a more inclusive ring, like with my x=2a example, > where 'a' is an object but not an algebraic integer, while x and, of > course, 2 are both algebraic integers. But, you've never answered the obvious question: is the ring E of even numbers flawed? What about N? R? -- However, you presuppose that certain numbers *are* prime ideals, ... when in fact ...* they are not... (Maybe I should look up 'prime ideals' but the effort doesn't seem to be worth it. I assume some poster will get excited ... if I messed up.) --James Harris === Subject: Re: Core error, FEAR is a natural response Hi James, I'm new to the discussion here... [...] > Actually the argument is correct, but here's how it goes so you can > see how things get screwed up. > If you use only algebraic integers and do various algebraic operations > that are legitimate in the ring of algebraic integers then > *presumably* you'll get algebraic integers. > However, I've found a way to get numbers that are NOT algebraic > integers. Could you provide a concrete example? I mean, this way you can show everyone that you're right... On the other side, if you can't do it, don't expect anyone to take your proof seriously. -- Edgar === Subject: Re: Core error, FEAR is a natural response > Well, it's possible to operate from the more inclusive ring of > objects, but I find myself trying to explain with rings and fields > that people are more familiar with, and as 'a' is NOT in the ring of > algebraic integers, you can't have x=2a in that ring, but you can in > the field of algebraic numbers. Nonsense. If a=1/2 then 'a' is neither in the ring of integers nor in the ring of algebraic integers. However, 2a which equals unity, is in both. You don't have to 'go anywhere' to know that 1 is an element of the integers or the algebraic integers. -- I came, I saw, I goofed.. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Factorial/Exponential Identity, Infinity I have an idea for a constant, or a variety of constants. It's about a sequence that is defined as a binary sequence representing a real number where the sequence element is one if the subset of the positive integers contains the element at that index, otherwise zero, for subsets of the naturals such as primes. .0110101000101000101000100000101000001000101000... Using Leroy Quet's cool ASCII summation sign: -- / 1 / 2^p -- p prime or oo -- / 1/2^x s(x) -- x = 1 where s(x)=1 if x is prime else s(x)=0. I should write a program to calculate values for iterations of that. One way to do that is to make sequences of the bits of length n as an integer and then divide that by the sequence 100... of length n+1. So the program using Java and java.math.BigInteger and java.math.BigDecimal could use an algorithm that returns for each integer greater than zero whether it is prime. With that then it sets the bit at the integer bit index into an array of zero bytes. This continues for a byte array of given sizes, then that is used to construct a BigInteger. Then another BigInteger is created from a zero byte array of length one greater than the prime bit sequence and its least significant bit is set. Then, a BigDecimal is returned as the quotient of the two. Then, I can put that into the Inverse Symbolic Calculator and see what it says it is. A decimal approximation, for primes less than 2500: 0.414682509851111660248109622154307708365774238137916977868245414488640960.. . The Inverse Symbolic Calculator says Primes coded in binary, as well it is. It also suggests the sequence as near being sum 1/2^x s(x) for s(x) being A005180(x), the orders of simple groups, which has some elements differing from the primes. http://www.cecm.sfu.ca/cgi-bin/isc/lookup?lookup_type=browse&page_no=0&numbe r=.4146825098511116 http://pi.lacim.uqam.ca/cgi-bin/lookup.pl?lookup_type=browse&page_no=0&numbe r=.4146825098511116 I notice also that a variety of other sequences coded in binary are also represented in the browse list there. Another thing to consider is not just the primes, but the primes unioned with one and/or zero, or other functions with range [0,1] over the domain of the positive integers. Another thing to consider is besides 1/2^x to have 1/3^x, etcetera. 1/y^x for y>1. One well known example of this kind of summation is that of 1/2^x for x=1 to infinity, which is equal to one. These sequences appear to each converge, but I am trying to think of cases where they would not. I don't know if they are interrelated to other known constants. One reason to consider that is this: if a different method could be determined that approximates or rather equals the prime-founded number, then it could be used to generate the sequence and determine which of the elements of N are prime at some differing cost compared to calculating each, for example as to how an arbitrary digit of Pi may be computed. That's wishful thinking, I might add. The ISC does not show a value derivable from other constants or functions of the same value as that. Adding .5 to consider the number one among the primes leads to sequence A008578, the primes and one, noting today 1 is no longer regarded as a prime. The number that is the sum of each 1/2^x for x composite is obviously enough one minus the sum for x prime. 0.585317490148888339751890377845692291634225761862083022131754585511359039 I got to thinking about this from the discussion about how an infinite binary sequence represents a given subset of the natural numbers, or in this case positive integers. I suppose the binary sequence could as well represent any subset of any infinite subset of the naturals or integers, or rationals, etcetera. Anyways I'm wondering if there is a way similar to the Bailey Borwein Plouffe method for determining an arbitrary digit of pi to determine the arbitrary digit of the irrational number represented by the binary sequence, a digit extraction algorithm. Wouldn't that be something. There's certainly a digit extraction algorithm for the primes coded in binary, just determine if the digit index is a prime number, primes are in P. http://www.sciencenews.org/sn_arc98/2_28_98/mathland.htm It might be a feature of these known digit extraction algorithms for irrational constants that the number is normal or some such thing. As an aside I want to note that the author notes that the probability of two integers selected at random being coprime is 6/pi^2. Squaring the circle indeed. One thing to consider is some different ordering of the elements. For example, in the example of coding the primes in binary, consider this alternate method: the index is for (2, 1, 4, 3, 6, 5, ...) or something. Then the idea is that it is coding the same subset of the naturals or rather Z+, but the values at various places would have differing significances. Coding these subsets as binaries might be a dead end, as it were, I can't think of a way to represent the sequence of primes as any other thing. There are of course some obvious equivalences for other sets coded in this manner, for example where .1010(10)... = 2/3, the odd integers coded in binary. I borrowed this book, I think it's really good, it has information about generating functions, Goulden and Jackson's Combinatorial Enumeration, Wiley, 1983. Still so far it has not told me what the decompositions, additive partition sequences, of numbers that are distinct under rotation are, for the fast k-subset enumeration algorithm, but it did explain the permutation notation that I saw on MathWorld, where for example a permutation of {1, 2, 3, 4, 5} could be (1, 3, 5) (2, 4) where 3 replaces 1, 5 replaces 3, and 1 replaces 5, and then 4 replaces 2 which replaces 4, (3, 4, 5, 2, 1). Anyways I think it's a good book and researching about it seems to show that it is a good reference for combinatorial enumeration. That and graph and matroid theory books are borrowed, from a library, a good complexity book or two. Another good book I have around here is called something like Hypergraphics: Visualizing Complex Relationships in Art, Science, and Technology, Brisson, ed., which is about neither Berge's nor Harary's hypergraph, I was hoping it would help me understand hypermatrices. Ross === Subject: Re: Rotated Hyperbolic Paraboloids Why not go to Doctor Math in http://mathforum.org/dr.math/ ? Hint: By rotating Hypar z=x*y through 45 deg on z-axis you get z = (x^2-y^2)/2 ; the zx plane sections are x^2/2 and the zy plane sections are -y^2/2 ..NOTE: the Hypar is anti-symmetric ! === Subject: How Euclid did not prove the Pythagorean Theorem's converse In http://www.oswego.edu/~baloglou/math/euclid-1.48.html I present a geometrical argument that *at the same time* proves the Pythagorean Theorem and illustrates how it fails for non-right angles (thus proving its converse): this geometrical argument is a straightforward extension of Euclid's proof of the Pythagorean Theorem, and argument for a right angle, see http://www.oswego.edu/~baloglou/math/euclid-1.47.html, to the case of an obtuse angle, the argument for an acute angle being similar. As you may (not) recall, Euclid's original proof of the converse of the Pythagorean Theorem employs the Pythagorean Theorem itself, something that I do not like that much*: my efforts to locate other proofs of the converse in the literature led nowhere -- save for 'my' proof above, that is! Comments on both other proofs of the Pythagorean Theorem's converse and the broader issue of employing A ---> B in the proof of B ---> A welcome... *to the point of being tempted to replace how by why in the title :-) baloglouAToswego.edu === Subject: Re: How Euclid did not prove the Pythagorean Theorem's converse > ... Euclid's original proof of the converse of the > Pythagorean Theorem employs the Pythagorean Theorem itself, something that > I do not like that much ... Why not? Seems perfectly reasonable to me. It's not like he's proving the inverse. ~ Chris === Subject: Re: How Euclid did not prove the Pythagorean Theorem's converse > ... Euclid's original proof of the converse of the > Pythagorean Theorem employs the Pythagorean Theorem itself, something that > I do not like that much ... > Why not? Seems perfectly reasonable to me. It's not like he's proving the inverse. > ~ Chris In other words, you do not accept ordinary logic. In that case, you better formulate the logical principles you do accept. For all anyone knows, it cannot be proved in your logic, whatever it is. On the other hand, it is likely that not much can be proved in your logic. I certainly will not waste time on it (more than I have). Designing a logic is not for the faint of heart, nor for the unsophisticated. === Subject: Re: Need help polygons by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9J2xb005121; >
>then you will need to no that a dodecagon has 12 sizes and also a regular 
polygon.and the degrees of a dodecagon is 360.tell me more when you gets 
mains.holla
>Robin Chapman                           256 256 256.
>Department of Mathematics                O hel, ol rite; 256; whot's
>University of Exeter, EX4 4QE, UK        12 tyms 256? Bugird if I no.
>rjc@maths.exeter.ac.uk                   2 dificult 2 work out.
>http://www.maths.ex.ac.uk/~r
jc/rjc.html  Iain M. Banks - Feersum Endjinn
>      http://www.dejanews.com/     
Search, Read, Post to Usenet
>
=== Subject: Re: Need help polygons yeah, I agree with Kastrup; 13-gon. I'm not sure about the Greek either, but was going by the usage for the 24-hedron, icosa-kai-tetra-hedron, or 20+4-hedron (which can be faceted either by trigona, tetragona or pentagona, all catalan solids, using those words as modifiers .-) the more well-known archimedean duals can, thus, be called icosakaitetra-astera. --les ducs d'Enron! === Subject: Re: Boole's Rule, ..., Weddle's Rule? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9J2xdv05171; Get the website translated textual into english and more people can read the work done in Vienna, especially people of non-speaking german abilities. === Subject: Re: show that 0.99999... equals 1? > 1/9 = 0.1111... > 2/9 = 0.2222... > ... > 8/9 = 0.8888... > 9/9 = 0.9999... > But, 9/9 = 1, so 0.9999... equals 1 > This is just some funny stuff that crept into my head. Now, is there anything wrong in this proof? If so, what is it? Nothing wrong with it, except that it's way too complicated, and in order to justify things like: 8/9 = 0.8888... you would have to delve into the definition of real numbers (look up Dedekind cuts). The definition of 0.9999... is the least number that is greater than or equal to all numbers of the form 0.999...9 (i.e., all finite portions of 0.999...). That's 1 (:-). David -- To make invalid valid, reverse the letters === Subject: Re: show that 0.99999... equals 1? that's not needed; you only have to rely upon the definition of long division (circa 1500s ... as in your last sentence .-) > order to justify things like: > 8/9 = 0.8888... > you would have to delve into the definition of real numbers (look up > Dedekind cuts). > The definition of 0.9999... is the least number that is greater than or > equal to all numbers of the form 0.999...9 (i.e., all finite portions of > 0.999...). That's 1 (:-). --les ducs d'Enron! === Subject: Re: show that 0.99999... equals 1? >The problem is the 0.9999... = 10 * 0.09999..., right? At least the rest >looks basic arithmetic to me. The problem is that you're assuming the convergence of the sequence. And once you prove that the sequence converges to a limit, and that the limit is unique, you've finished the proof anyhow. Doug === Subject: Re: show that 0.99999... equals 1? I'd guess that the procedure was introduced by Stevin (long division of decimals .-) > You first have to show that 1/9 =0.1111.... . The rest follows. --les ducs d'Enron! === Subject: Re: show that 0.99999... equals 1? actually, it may be a sort of a key to some of the dyscoveries of Leibniz and Fermat, since 0.9999...9999 + 0.0000...0001 = 1.0000...0000, by simple arithmetic. > this applies, no matter where the endless nines begin, --les ducs d'Enron! === Subject: Re: show that 0.99999... equals 1? CC'ed to FAQ author. In sci.math, Randy Poe >> 1/9 = 0.1111... >> 2/9 = 0.2222... >> ... >> 8/9 = 0.8888... >> 9/9 = 0.9999... >> But, 9/9 = 1, so 0.9999... equals 1 >> This is just some funny stuff that crept into my head. Now, is >> there anything wrong in this proof? > There's nothing wrong with the result you obtained. > proof as well as giving hints about how to show the result > more formally. > http://www.faqs.org/faqs/sci-math-faq/specialnumbers/0.999eq1/ There appears to be an error in this FAQ, which makes it look a bit strange: An even easier argument multiplies both sides of 0.9999... = 1 by 0.3333... = 1/3 . The result is 3 . It probably should be divide by instead of multiply by. > http://mathforum.org/dr.math/faq/faq.0.9999.html > - Randy -- #191, ewill3@earthlink.net -- use this addy instead of the posted one :-) for replies It's still legal to go .sigless. === Subject: Re: show that 0.99999... equals 1? >1/9 = 0.1111... >2/9 = 0.2222... >... >8/9 = 0.8888... >9/9 = 0.9999... >But, 9/9 = 1, so 0.9999... equals 1 >This is just some funny stuff that crept into my head. >Now, is there anything wrong in this proof? If so, >what is it? >> Informally, it's fine. >> Formally, though, you are assuming without proof >> a) 1/9 = 0.1111... >> b) 9=0.1111... = 0.9999... oops, s.b. b) 9x0.1111.. = 0.9999... >> and c) both 0.1111... and 0.9999... somehow represent >> specific numbers. > There are all geometric series with ratio < 1 (1/10 to be exact) so the > series of partial terms converges. You know that; I know that; the regulars here know that. It seemed to me that the OP most likely did not know that. But more to (my) point, it seemed likely the OP also didn't know enough mathematical infrastructure to see why his/her argument (without some context) is heuristic and not a real proof (and might not grok the mathematical point of view that that context is essential). So I pointed out that each step needs some support. === Subject: C being equal to the partial recursive functions set C is defined as being the smallest set containing the addition, multiplication, projection functions as well as the caracteristic function of identity (the function E(x,y) being equal to 1 if x = y, and to 0 either). C is closed by substitution and by minimisation (-operator). I have to show that C is the same set as the partial recursive function set F*. I was able to show that C is included in F*. Regarding the inclusion of F* in C, I was able to prove that: - The successor function is in C. - That the constant functions are in C. - I'm in trouble regarding the recursion principle. I was able to build a function being equal to what should be a function build with the recursion principle up to any integer, and being equal to zero aftern. But not to build one defined by the recursion principle for any integer. Any idea much appreciated! === Subject: Re: Minimum polynomials > Given a quartic monic polynomial with integer coefficients and with four > complex roots (two pairs of complex conjugate roots), > eg: x4-10x3+47x2-90x+81 > how do I find the min poly of |x|^2 (involving the complex absolute value) Well, consider the sextic g(x) whose roots are the uv, where u and v are distinct roots of your quartic. Unless you are unlucky this will have two positive real roots, the |u|^2 you seek. Of course, this sextic will probably still be irreducible over Q :-( -- 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: Minimum polynomials >>Given a quartic monic polynomial with integer coefficients and with four >>complex roots (two pairs of complex conjugate roots), >>eg: x4-10x3+47x2-90x+81 >>how do I find the min poly of |x|^2 (involving the complex absolute value) > Well, consider the sextic g(x) whose roots are the uv, where u and v are > distinct roots of your quartic. Unless you are unlucky this will > have two positive real roots, the |u|^2 you seek. Of course, this sextic > will probably still be irreducible over Q :-( So there is something special about my particular quartic. Its splitting field has Galois group the Klein four group. I wonder if that is typical of the other polynomials that I get from the same source. I'll mention this again in another place. I had wrongly assumed that this was always the case for quartics with two pairs of complex conjugate roots and which are monic and irreducible over Q. But I now can't see why that would be the case.....just thinking out loud. Bill. === Subject: Re: Minimum polynomials >Given a quartic monic polynomial with integer coefficients and with four >complex roots (two pairs of complex conjugate roots), >eg: x4-10x3+47x2-90x+81 >how do I find the min poly of |x|^2 (involving the complex absolute >value) >> Well, consider the sextic g(x) whose roots are the uv, where u and v are >> distinct roots of your quartic. Unless you are unlucky this will >> have two positive real roots, the |u|^2 you seek. Of course, this sextic >> will probably still be irreducible over Q :-( > So there is something special about my particular quartic. Its splitting > field has Galois group the Klein four group. I wonder if that is typical > of the other polynomials that I get from the same source. I'll mention > this again in another place. > I had wrongly assumed that this was always the case for quartics with > two pairs of complex conjugate roots and which are monic and irreducible > over Q. But I now can't see why that would be the case.....just thinking > out loud. Well, knowing that the polynomial has no real roots tells you that its Galois group has an element of shape 2^2, but if we know the quartic is irreducible then we know the Galois group is transitive, and so must have an element of shape 2^2 anyway :-) -- 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: Minimum polynomials OK, someone has pointed out the solution to me. I feel a bit stupid for not knowing it. I should. In a case like the one I mention, the cubic resolvent equation has a rational root, which is easily calculated. This leads to the evaluation I'm after, since that root is precisely the value aa' + bb' where the a's and b' are the conjugate pairs of the original roots of my quartic. Bill. === Subject: Re: Minimum polynomials Could you please give a few more details. I don't see why the root of the resolvent is equal to aa'+bb'. -Michael. > OK, someone has pointed out the solution to me. I feel a bit stupid for > not knowing it. I should. > In a case like the one I mention, the cubic resolvent equation has a > rational root, which is easily calculated. This leads to the evaluation > I'm after, since that root is precisely the value aa' + bb' where the > a's and b' are the conjugate pairs of the original roots of my quartic. > Bill. === Subject: Re: Minimum polynomials Adjunct Assistant Professor at the University of Montana. >Given a quartic monic polynomial with integer coefficients and with four >complex roots (two pairs of complex conjugate roots), >eg: x4-10x3+47x2-90x+81 >how do I find the min poly of |x|^2 (involving the complex absolute value) There may not a well defined such animal; surely you are also assuming that the quartic monic polynomial is irreducible over Q. Otherwise, any product of two irreducible quadratics with roots of different norm will give you problems: e.g. (x^2+4)(x^2+25) = x^4 + 29x^2 + 100 is a quartic, monic polynomial with integer coefficiennts, four complex roots, 2i, -2i, 5i, and -5i, two of them have |x|^2 = 4 so they have minimal polynomial x-4, and two of them have |x|^2 = 25, so they have minimal polynomial x-25. >for the example above it is: x2-27x+81 (yeah I can get the 81 OK) I'm not sure I understand the question. Do you mean, if r is any root of x^4 + 10x^3 + 47x^2 + 90x + 81, then |x|^2 satisfies y^2 - 27y + 81? That is, |x|^4 - 27|x|^2 + 81 = 0? So |x|^2 = (27 + sqrt(729 - 324))/2 = (27 + sqrt(405))/2 or |x|^2 = (27 - sqrt(405))/2? (And I then assume that one of the pairs will have one of them as norm, the other will have the second value?) I'm having a much harder time seeing how you got the 27 than the 81, to be honest... Your assumptions give that there are two complex numbers, r and s, such that x^4 + 10x^3 + 47x^2 + 90x + 81 = (x-r)(x-conj(r))(x-s)(x-conj(s)) where conj(a) denotes the complex conjugate of a. Then you have x^4 + 10x^3 +47x^2 + 90x + 81 = (x^2-2Re(r)x+|r|^2)(x^2 -2Re(s)x+|s|^2) So |r|^2*|s|^2 = 81. Re(r)|s|^2 + Re(s)|r|^2 = -45. |r|^2 + |s|^2 + 4Re(r)Re(s) = 47. Re(s)+Re(r) = -5. You would want (x-|r|^2)(x-|s|^2) = x^2 - (|r|^2+|s|^2)x + |r|^2|s|^2 to have real coefficients, presumably, then. So the constant term is the constant term of the original polynomial, and the trick is figuring out what |r|^2 + |s|^2 is equal to. It is equal to 47 - 4*Re(r)*Re(s). So we need to figure out what Re(r)*Re(s) is equal to. You are saying it is equal to 5 in this instance, and I'm not sure I see why... 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: Minimum polynomials > There may not a well defined such animal; surely you are also assuming > that the quartic monic polynomial is irreducible over Q. Otherwise, > any product of two irreducible quadratics with roots of different norm > will give you problems: e.g. > (x^2+4)(x^2+25) = x^4 + 29x^2 + 100 > is a quartic, monic polynomial with integer coefficiennts, four > complex roots, 2i, -2i, 5i, and -5i, two of them have |x|^2 = 4 so > they have minimal polynomial x-4, and two of them have |x|^2 = 25, so > they have minimal polynomial x-25. Yes. Sorry I should have made that clear. The `min poly' also refers to the original polynomial. It is of course not a min poly if it is reducible over Q. >>for the example above it is: x2-27x+81 (yeah I can get the 81 OK) > I'm not sure I understand the question. Do you mean, if r is any root > of x^4 + 10x^3 + 47x^2 + 90x + 81, Some minus signs appear to have gone missing here. No I meant what I originally asked. You appear to be changing between r and x here and later, too. then |x|^2 satisfies > y^2 - 27y + 81? That is, > |x|^4 - 27|x|^2 + 81 = 0? > So |x|^2 = (27 + sqrt(729 - 324))/2 = (27 + sqrt(405))/2 > or > |x|^2 = (27 - sqrt(405))/2? > (And I then assume that one of the pairs will have one of them as > norm, the other will have the second value?) Yes. I think that is what I am saying.... Yes, that sounds right. > I'm having a much harder time seeing how you got the 27 than the 81, > to be honest... Try those minus signs. Of course I got the 27 by numerical computation, not by algebra. > Your assumptions give that there are two complex numbers, r and s, > such that > x^4 + 10x^3 + 47x^2 + 90x + 81 = (x-r)(x-conj(r))(x-s)(x-conj(s)) > where conj(a) denotes the complex conjugate of a. Then you have > x^4 + 10x^3 +47x^2 + 90x + 81 = (x^2-2Re(r)x+|r|^2)(x^2 -2Re(s)x+|s|^2) > So |r|^2*|s|^2 = 81. Yes, that's how I got the 81. > Re(r)|s|^2 + Re(s)|r|^2 = -45. > |r|^2 + |s|^2 + 4Re(r)Re(s) = 47. > Re(s)+Re(r) = -5. Yes. > You would want (x-|r|^2)(x-|s|^2) = x^2 - (|r|^2+|s|^2)x + |r|^2|s|^2 to > have real coefficients, presumably, then. > So the constant term is the constant term of the original polynomial, > and the trick is figuring out what |r|^2 + |s|^2 is equal to. Yep. > It is equal to 47 - 4*Re(r)*Re(s). So we need to figure out what > Re(r)*Re(s) is equal to. > You are saying it is equal to 5 in this instance, and I'm not sure I > see why... Me either, but the information is present in the question wouldn't you agree, since we may use Ferarri's method, factor the quartic, then calculate whatever we like. But I am certain there is a way of doing it without first obtaining the factorisation. > 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: Google Calculator > Type in half a cup in teaspoons > or sqrt(3) * sqrt(5) * sqrt(7) Or: e^(pi*i) arctan(-1) in degrees 60 mph in km/h 1999 in roman numerals the speed of light in km per second the mass of the sun divided by the mass of the earth the answer to life, the universe and everything But of course you should never trust a finite-precision calculator. Try: cos(7)^2+sin(7)^2-1 (Compare it with cos(7)^2+sin(7)^2.) -- Karl Ove Hufthammer === Subject: Re: Google Calculator > Type in half a cup in teaspoons > or sqrt(3) * sqrt(5) * sqrt(7) > Or: > e^(pi*i) > arctan(-1) in degrees > 60 mph in km/h > 1999 in roman numerals > the speed of light in km per second > the mass of the sun divided by the mass of the earth > the answer to life, the universe and everything > But of course you should never trust a finite-precision > calculator. Try: > cos(7)^2+sin(7)^2-1 > (Compare it with cos(7)^2+sin(7)^2.) Right, but that sort of thing can be expected. Perhaps it should be noted that, if you type in -2^2 , it then displays -2^2 = 4 , whereas, if you type in 0-2^2 , it then displays 0-(2^2) = -4 . And, BTW, it nicely gives 0^0 = 1 . David === Subject: Re: Google Calculator > Perhaps it should be noted that, > if you type in -2^2 , it then displays -2^2 = 4 , Nothing particularly unusual about that: bash-2.05a$ bc bc 1.06 Copyright 1991-1994, 1997, 1998, 2000 Free Software Foundation, Inc. This is free software with ABSOLUTELY NO WARRANTY. For details type `warranty'. -2^2 4 And C/C++ also both adopt the convention of binding unary minus expressions very tightly. > whereas, > if you type in 0-2^2 , it then displays 0-(2^2) = -4 . Unary operator '-' and binary operator '-' are not the same operator, they needn't even use the same symbol (but always seem to in any grammer I've encontered). Even in the grammer of a language that evaluates '-2^2' to be -4, you'll see that unary '-' and binary '-' are different operators. It's just a question of the operator precedence of the unary one that varies. Phil -- Unpatched IE vulnerability: XSS in Unparsable XML Files Description: Cross-Site Scripting on any site hosting files that can be misrendered in MSXML Reference: http://sec.greymagic.com/adv/gm013-ie/ Exploit: http://sec.greymagic.com/adv/gm013-ie/ === Subject: Re: I am a freshman of CST,which book of maths should I read first? >I am a freshman of computer science and technology. >I came across a problem.That is I don't know which book should I read >first,which should I read second...... >I have three books right here.They are ,Maths>,. >If you know,please help me! >>Change your major to basket weaving. What do any of the newsgroups have to do >>with CS? > Piss off, racist. What managed to rattle your cage? Need someone to explain to you what computer science covers? -- Evil Arabs refer to Israel as the Zionist entity. Good Israelis refer to Israel as the Zionist enterprise. And people thought PC speech in the US was stupid. -- The Iron Webmaster, 2873 === Subject: Maths Challenge. Series-> Formula Can somone please translate this series in a formula X Y 1 13 4 20 5 25 9 37 12 46 14 52 19 67 20 70 24 82 29 97 30 100 34 112 39 127 47 150 49 157 65 205 http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: Maths Challenge. Series-> Formula >Can somone please translate this series in a formula >X Y >1 13 >4 20 >5 25 >9 37 >12 46 >14 52 >19 67 >20 70 >24 82 >29 97 >30 100 >34 112 >39 127 >47 150 >49 157 >65 205 Here is a formula that works for your first three points. You should have no problem extending it for as many points as you like. View with monospaced font. 13(x-4)(x-5) + 20(x-1)(x-5) + 25(x-1)(x-4) y = ---------------------------------------------- (x-4)(x-5) + (x-1)(x-5) + (x-1)(x-4) One of the consequences of this is you can define a formula which fits any given sequence and still lets you choose what the next value will be. <> === Subject: Re: Maths Challenge. Series-> Formula > Can somone please translate this series in a formula > X Y > 1 13 > 4 20 > 5 25 > 9 37 > 12 46 > 14 52 > 19 67 > 20 70 > 24 82 > 29 97 > 30 100 > 34 112 > 39 127 > 47 150 > 49 157 > 65 205 It fit very well to a straight line, by : y = 12656x/4213 + 40914/4213 The major discrepancy is for the pont (4, 20). -- Ignacio Larrosa Caestro A Corua (Espaa) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Maths Challenge. Series-> Formula Hey You could put them in excel, let excel make a graph, and then can excel make the formula from the graph... i don't have excel on this pc, im sorry > Can somone please translate this series in a formula > X Y > 1 13 > 4 20 > 5 25 > 9 37 > 12 46 > 14 52 > 19 67 > 20 70 > 24 82 > 29 97 > 30 100 > 34 112 > 39 127 > 47 150 > 49 157 > 65 205 News==---- > http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups > ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Four Colours and infinite number of countries I was reading the book of Robin Wilson about the four colour problem. And I had the question whether the four colour theorem also hold for a map with infinite number of countries. I thought that this was a major obstacle, but suddenly I realized that there is a rather simple proof (see below), comparable to proofs of other infinite graph problems. The interesting part is that the Euler's formula does not hold for maps with infinite number of countries. There is no guarantee that there is a pentagon in the map, or even a hexagon. Since the proof of the four colour problem heavily depends on Euler's formula, this is rather odd. And suggests that there must be a simpler proof for the four colour problem (this is speculation, an alternative proof that does not use Euler's formula might also be more complex). Still, an interesting observation. Lucas B. Kruijswijk The Netherlands ----------------------------------------------- Proof of the four colour problem with a connected map with infinite number countries, where each country has a finite number of neighbours. ----------------------------------------------- First give each country a number, starting by 1. This can be done as follows. Start by a certain country. Give this country number 1. Distribute the remaining numbering over the neighbours of this country. Now each neighbour get its own infinite sequence of numbers. Do the trick applied to the starting country, also on all neighbours. Any country that has already a number is skipped. This ensures that each country gets a number. It can easily be seen that if there is a path from a country to the starting country, this country gets a number and there is even an upper limit to the number (based on the number of neighbours of the countries on the path). Now, make a program 'prog' that colours the first n countries in four colours. Since this subset is finite and the theorem of four colours is true for finite maps, this is possible. The 'prog' program should give only 1 way of colouring and should do this in a deterministic was. If applied a second time, it should give the same colouring. With this 'prog' program we can start colouring the infinite map. We first look to country number 1. We look to the colouring of this country for each prog(n), where n goes to infinity. There must be at least one colour that will keep appearing. If not then there will be an upper limit for each colour and that leads to an impossibility. The first country is assigned the colour that keeps appearing. For the second country we do the same trick. However, we only consider prog(n) for numbers for which the colouring of the first country is the colour as we assigned. This is still an infinite number of numbers. We can repeat this process for each country and this completes the colouring of the infinite map. Lucas B. Kruijswijk The Netherlands === Subject: Re: Four Colours and infinite number of countries This is an example of what they call a compactness argument. Often used in combinatorial situations to go from finite to infinite. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Four Colours and infinite number of countries > This is an example of what they call a compactness argument. > Often used in combinatorial situations to go from finite to infinite. I am still intrigued by the result. The proof of the four colour problem is based on situations that are absolutely not true in the infinite case. Does this occurs also in other proofs where the compactness argument is used? Lucas B. Kruijswijk === Subject: Re: Four Colours and infinite number of countries > This is an example of what they call a compactness argument. > Often used in combinatorial situations to go from finite to infinite. > I am still intrigued by the result. > The proof of the four colour problem is based on situations that are > absolutely not true in the infinite case. Does this occurs also in > other proofs where the compactness argument is used? > Lucas B. Kruijswijk The fact that it is a standard type of argument doesn't change the fact that you discovered it yourself and it is certainly not obvious. I doubt very many people would have discovered it and it is very elegant. === Subject: Re: Four Colours and infinite number of countries > This is an example of what they call a compactness argument. > > Often used in combinatorial situations to go from finite to infinite. > I am still intrigued by the result. > The proof of the four colour problem is based on situations that are > absolutely not true in the infinite case. Does this occurs also in > other proofs where the compactness argument is used? > Lucas B. Kruijswijk > The fact that it is a standard type of argument doesn't change the > fact that you discovered it yourself and it is certainly not obvious. > I doubt very many people would have discovered it and it is very > elegant. But for me the idea that there is maybe another proof for the four colour problem (that maybe does not require computers) is more exciting than the proof of the four colour problem in infinity. It is very strange, that a proof that relies so heavily on parts that are not true in infinity, extends so easily to infinity. Lucas === Subject: Re: What is 0/0 ? 0/1= 0... 1/1= infinity,but 0/0= error! Faith Dorell schreef in bericht > error or 0 ? === Subject: Re: What is 0/0 ? > 0/1= 0... 1/1= infinity,but 0/0= error! > Faith Dorell schreef in bericht > error or 0 ? I think you mean 1/0 = infinity. 1/1 = 1. === Subject: Re: What is 0/0 ? > If 0/0 = 1 then 0/0 = 1*n for arbitrary n. Then 1 = 0/0 = n hence 1 = n. > That is the kind of grief one gets if 0/0 is defined. That's misleading -- unintentionally, I suppose. Rather That is the kind of grief one gets if 0/0 is defined to be a _nonzero_ real number, like your choice of 1. But defining 0/0 to be 0 should not lead to grief. David Cantrell > Therefore it is not defined. > Bob Kolker === Subject: Re: What is 0/0 ? > _nonzero_ real number, like your choice of 1. But defining 0/0 to be 0 > should not lead to grief. Your point is well taken. If 0/0 equals anything then it ought to be 0. Bob Kolker === Subject: Re: What is 0/0 ? > _nonzero_ real number, like your choice of 1. But defining 0/0 to be 0 > should not lead to grief. > Your point is well taken. If 0/0 equals anything then it ought to be 0. Perhaps it's interesting to note: was the first Indian mathematician to have considered division by zero. He stated that 0/0 = 0; in modern times, he has often been derided, unjustly IMO, for having done so. David === Subject: Re: What is 0/0 ? >> _nonzero_ real number, like your choice of 1. But defining 0/0 to be 0 >> should not lead to grief. >> Your point is well taken. If 0/0 equals anything then it ought to be 0. >Perhaps it's interesting to note: >was the first Indian mathematician to have considered division by zero. >He stated that 0/0 = 0; in modern times, he has often been derided, >unjustly IMO, for having done so. Nope. The answer should be null rather than a number. If I'm dividing nothing into no piles, I had better not get an answer that is an element on the number line. /BAH Subtract a hundred and four for e-mail. === Subject: Re: What is 0/0 ? >> _nonzero_ real number, like your choice of 1. But defining 0/0 to be 0 >> should not lead to grief. >Your point is well taken. If 0/0 equals anything then it ought to be 0. Nope. Not when I'm using a computer. I want a trap at the minimum at execution time; it's real useful to have it be flagged as an exception at compile time. Zeroes in data processing are dangerous critters. /BAH Subtract a hundred and four for e-mail. === Subject: Re: What is 0/0 ? In sci.math, jmfbahciv@aol.com : > _nonzero_ real number, like your choice of 1. But defining 0/0 to be 0 > should not lead to grief. >>Your point is well taken. If 0/0 equals anything then it ought to be 0. > Nope. Not when I'm using a computer. I want a trap at the minimum > at execution time; it's real useful to have it be flagged as an > exception at compile time. Zeroes in data processing > are dangerous critters. Especially when they lead to ships having to be towed into port. :-) > /BAH > Subtract a hundred and four for e-mail. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: What is 0/0 ? >In sci.math, jmfbahciv@aol.com > >: >> _nonzero_ real number, like your choice of 1. But defining 0/0 to be 0 >> should not lead to grief. >Your point is well taken. If 0/0 equals anything then it ought to be 0. >> Nope. Not when I'm using a computer. I want a trap at the minimum >> at execution time; it's real useful to have it be flagged as an >> exception at compile time. Zeroes in data processing >> are dangerous critters. >Especially when they lead to ships having to be towed into port. :-) Among other things. You wouldn't want an MRI to go berserk because it encountered a zero while it was pointing at you. Just think about all those autos that can't run without a chip in their innards. /BAH Subtract a hundred and four for e-mail. === Subject: Re: What is 0/0 ? >> _nonzero_ real number, like your choice of 1. But defining 0/0 to be 0 >> should not lead to grief. >Your point is well taken. If 0/0 equals anything then it ought to be 0. > Nope. Not when I'm using a computer. I'd say that depends on _exactly how_ you're using a computer. If you're using floating-point arithmetic, then what you say sounds right. And the IEEE standard's NaN is certainly better than giving 0.0, IMO. > I want a trap at the minimum > at execution time; it's real useful to have it be flagged as an > exception at compile time. Zeroes in data processing > are dangerous critters. OTOH: If you're using integer arithmetic, I'd suggest that 0/x = 0 might be reasonable for all integer x (including x = 0). If you're using a computer algebra system, I'd suggest that, if the zero in the numerator is the system's symbolic 0, then 0/x should simplify to that symbolic 0 regardless of x (in particular, even if x is only approximately 0.0). And of course there are other cases which I haven't covered. David === Subject: Re: What is 0/0 ? >> >> > _nonzero_ real number, like your choice of 1. But defining 0/0 to be 0 > should not lead to grief. >> >>Your point is well taken. If 0/0 equals anything then it ought to be 0. >> Nope. Not when I'm using a computer. >I'd say that depends on _exactly how_ you're using a computer. True. But an architect of that hard/software cannot possibly predict how anybody is going to the gear. So, there has to be a result that people would expect (so they can check for it or induce the condition). > .. If you're >using floating-point arithmetic, then what you say sounds right. Everytime I dealt with computer arithemtic, I had to relearn how it gets done; it made my hair hurt. So I can't judge intelligently whether the notion is limited to floating point. I've certainly seen programs use the index loop counter as part of a calculation. I would also include integer arithmetic. > ... And >the IEEE standard's NaN is certainly better than giving 0.0, IMO. Anything is better than a success return. Even powering down the system is better than allowing an answer go through which gives the owner of the bits a secure feeling that all answers are correct. >> I want a trap at the minimum >> at execution time; it's real useful to have it be flagged as an >> exception at compile time. Zeroes in data processing >> are dangerous critters. >OTOH: >If you're using integer arithmetic, I'd suggest that 0/x = 0 might be >reasonable for all integer x (including x = 0). Nope. That means that I'd have to put a test for zero in every bloody loop of a FORTRAN program if I used the loop counter in an arithmetic operation. hmmm...I guess that would depend on how the compiler counts its loops. I bet C is worse. It would also depend on how the bare iron did the division; IOW, how the wire was wrapped. >If you're using a computer algebra system, I'd suggest that, if the zero >in the numerator is the system's symbolic 0, then 0/x should simplify to >that symbolic 0 regardless of x (in particular, even if x is only >approximately 0.0). >And of course there are other cases which I haven't covered. You have to think about the complements and what is zero. IIRC, there's a huge difference between the results of a one's complement and a two's complement but I've forgotten why or how. Computer arithmetic was on my list of guy things which itemized things that I didn't do well at all; I would leave these things for those more adept than I. It saved us from shipping a lot of crap to the field :-). The point is that 0/0 doesn't have meaning; don't you guys call that undefined in the math biz? /BAH Subtract a hundred and four for e-mail. === Subject: Re: What is 0/0 ? > It's not possible to define 0/0 as anything real because doing so will > lead immediately to contradiction. No. Defining 0/0 to be 0 should not lead immediately to contradiction. > Assume 0/0 is defined as Z, where Z is in R. Now, > Z + Z = (0 + 0)/0 = Z <=> Z = 0, but lim(n->0, n/n) = 1 = Z, so 0 = 1. Of course that limit equals 1, but the claim then that 1 = Z is not justified. No contradiction. David Cantrell === Subject: Re: What is 0/0 ? > error or 0 ? I'll assume that you're asking about 0/0 as merely a numerical expression. Based on that assumption, I'll give you an answer below. But first, I must take a paragraph to discuss something else. Had you asked instead about limits in the form 0/0, the answer would have been that the limit form is properly called indeterminate. Several other respondents have already shown you why that limit form is indeterminate. But please note that their responses have no direct bearing on whether the numerical expression 0/0 should be defined or not. Those responses, dealing with limits, show what can happen to a quotient as its numerator and denominator _approach_ 0. But that is not directly related to the question of whether a quotient having its numerator and denominator _equal to_ 0 should be defined and, if so, how. So what about the numerical expression 0/0? 1. 0/0 is normally considered to be undefined. 2. If 0/0 were to be assigned a real value, the best choice would be 0, as I have argued before in this newsgroup. It might also be noted that 0/0 = 0 in the computer language J (which is an offspring of APL, in which, unfortunately, 0/0 = 1). 3. In some systems of extended-real interval arithmetic, 0/0 is taken to be [-oo, +oo], that is, the set of extended reals. Related to this idea: 0/0 is defined as an element (often denoted as _|_ ) in algebraic structures called wheels. According to the internationally accepted standard for floating-point arithmetic, 0/0 is defined as a floating-point object (NaN, acronym for Not a Number). David Cantrell === Subject: contour integral.....?? C:|z|=3 f(z) = [(z^2)*(z+i)] / [{(z-2i)^2}*(z-3i)] solve contour integral... --------------------------------- some time ago, i know that it can not applied residue. if so, how solved that?? if z=3e^(io) (0=theta) i think that very difficult expansion. how to solved it.......please.... === Subject: Re: contour integral.....?? > C:|z|=3 > f(z) = [(z^2)*(z+i)] / [{(z-2i)^2}*(z-3i)] > solve contour integral... It diverges, because of a simple pole on the contour of integration. You get the principal value by using half of the residue on the contour. > f := ((z^2)*(z+I)) / (((z-2*I)^2)*(z-3*I)); 2 z (z + I) f := -------------------- 2 (z - 2 I) (z - 3 I) > r2 := residue(f,z=2*I); r2 := -28 I > r3 := residue(f,z=3*I); r3 := 36 I > 2*Pi*I*(r2+r3/2); 20 Pi === Subject: Calculating value of a matrix. Hello everyone. I have a problem. Im calculating the inverse value of an matrix A with the itterative method. Now what Im having a problem with is a part of this problem. I get an matrix ||E|| = ||0.2 -0.3|| which is supposed to be equal to = 0.5 ||0.15 -0.1|| The problem is that I have no idea what the double lines on the sides mean and how I calulate the value 0.5. Is it the absolutevalue of the matrix? Please help since Im totally lost. Sincerly Kristofer Nordstrm === Subject: Re: Calculating value of a matrix. > Hello everyone. > I have a problem. Im calculating the inverse value of an matrix A > with the itterative method. > Now what Im having a problem with is a part of this problem. I get an > matrix > ||E|| = ||0.2 -0.3|| which is supposed to be equal to = 0.5 > ||0.15 -0.1|| > The problem is that I have no idea what the double lines on the sides > mean and how I calulate the value 0.5. Is it the absolutevalue of the > matrix? > Please help since Im totally lost. > Sincerly Kristofer Nordstrm Could be the norm of the matrix. Maple says: > with(linalg): A := matrix([[0.2,-0.3],[0.15,-0.1]]); [.2 -.3] A := [ ] [.15 -.1] > norm(A,1);norm(A,2);norm(A,infinity);norm(A,frobenius); .4 .3981937098 .5 .4031128874 Looks like the infinity norm. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Area of a focal sector of an ellipse by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9JBpFq07274; >parts of this thread into a couple of web pages. If you are interested, >please see http://www.whim.org/nebu la/math/kepler.html and its link to >Planetary Orbits. >Rob Johnson >rob@whim.org In your paper (http://www.whim.org/nebu la/math/kepler.html) you say >Inverting [5] and using [8] allows us to compute position from time. > a^2 sqrt(1-e^2) > A = --------------- (s - e sin(s)) [5] > 2 > e sin(s) > t = s + 2 atan( -------------------------- ) [8] > 1 + sqrt(1-e^2) - e cos(s) By inverting, I take it you mean taking the function inverse (ie.solve [5] for s, making S(A) )? I would greatly appreciate if you could show me how this is done. === Subject: Re: Area of a focal sector of an ellipse Rob may not be reading this group currently; I haven't seen any posts of his since July. So I'll reply in his stead. But email him if you want his own answer. > In your paper > you say >Inverting [5] and using [8] allows us to compute position from time. > a^2 sqrt(1-e^2) > A = --------------- (s - e sin(s)) [5] > 2 > e sin(s) > t = s + 2 atan( -------------------------- ) [8] > 1 + sqrt(1-e^2) - e cos(s) > By inverting, I take it you mean taking the function inverse (ie.solve > [5] for s, making S(A) )? I would greatly appreciate if you could show > me how this is done. Solving [5] for s is the classic problem of solving Kepler's equation, which cannot be done in closed form in terms of familiar functions. I highly recommend that you read _Solving Kepler's Equation Over Three Centuries_, Peter Colwell, 1993, Willmann-Bell. David Cantrell === Subject: finite groups Is it true that in every finite simple group G all isomorphic maximal subgroups are Aut(G)-conjugate? Anvita === Subject: Re: finite groups >Is it true that in every finite simple group G >all isomorphic maximal subgroups >are Aut(G)-conjugate? No, but it is not easy to find examples. I was about to give up, but then I found one! Believe it or not, the alternating group A_{253} has two maximal subgroups isomorphic to PSL(2,23). One of them has dihedral D_{24} as point stabilizer, and the other has S_4, so they cannot be conjugate in S_{253}. Derek Holt. === Subject: Re: finite groups >Is it true that in every finite simple group G >all isomorphic maximal subgroups >are Aut(G)-conjugate? > No, but it is not easy to find examples. I was about to give up, but > then I found one! > Believe it or not, the alternating group A_{253} has two maximal subgroups > isomorphic to PSL(2,23). One of them has dihedral D_{24} as point > stabilizer, and the other has S_4, so they cannot be conjugate in > S_{253}. Hi Derek, I have one question about your example. The dihedral subgroup D_{24} of PSL(2,23) is extended to a subgroup of PGL(2,23) of the same index 253. This means that PGL(2,23) also has a permutational representation of degree 253. Could it be that the first of your subgroups is not maximal but a subgroup of index 2 in a larger subgroup isomorphic to PGL(2,23)? Anvita === Subject: Re: finite groups Anvita a crit dans le message de > Is it true that in every finite simple group G > all isomorphic maximal subgroups > are Aut(G)-conjugate? No, consider the group of order 16 with presentation: G = {a,b; a^4 = b^4 = 1, ba=a^-1b} It is the semi-direct product of Z/4Z by Z/4Z with morphism b |--> (a |->a^-1) G has 3 maximal subgroups, all isomorphic to Z/4Z x Z/2Z, but one of them say H (generated by ) has its cyclic subgroups of order 4: and normal in G The other two (say K_1= and K_2=) have their cyclic subgroups of order 4: and for K_1 and ,< a^3b> for K_2 not normal in G. Then H and K_1 subgroups can't be automorph i.e. no automorphism of G can send the first onto the second. Alain === Subject: Re: finite groups Alain Debreil a crit dans le message de > Anvita a crit dans le message de > Is it true that in every finite simple group G > all isomorphic maximal subgroups > are Aut(G)-conjugate? > No, consider the group of order 16 with presentation: > G = {a,b; a^4 = b^4 = 1, ba=a^-1b} > It is the semi-direct product of Z/4Z by Z/4Z with morphism b |--> (a > |->a^-1) > G has 3 maximal subgroups, all isomorphic to Z/4Z x Z/2Z, > but one of them say H (generated by ) has its cyclic subgroups of > order 4: and normal in G > The other two (say K_1= and K_2=) have their cyclic > subgroups of order 4: and for K_1 and ,< a^3b> for K_2 not > normal in G. > Then H and K_1 subgroups can't be automorph i.e. no automorphism of G can > send the first onto the second. Apologizes I miss reading simple group in the original post Please forget my answer. Alain === Subject: Re: Quick Math Guide to core error issues James, You post so much material that I don't have the time or desire to read everything and I lose track of your position and arguments. Some time ago you had an organised website with definitions and cross- reference and you have posted several summaries and histories on this newsgroup which would be easier to refer to if they were permanently available. I, for one, would be pleased if you would resurrect your website and put the information back. Here are just some suggestions as to what you could put there and how it could be organised. *Definitions* - Algebraic Integers and Algebraic Numbers An explanation of what an algebraic integer is , some examples and their general properties would be useful. I vaguely recall that you had some software that would take a monic quadratic equation with integer coefficients and factorise it into linear terms with algebraic integer coefficients. This could usefully go here as a Java applet. - Your 'Object Ring' and what numbers are in it and what aren't. I *think* you said 2^sqrt(3) was in it I *think* you said that it wasn't wholly contained in the complex numbers An explanation of why these numbers are in it would be useful - What an incomplete ring is. - Some expressions don't seem to translate across the Atlantic and I am thinking of 'has a factor of 5' in particular. To me, 'has a value of 42' means '42 is a value' so 'has a factor of 5' should mean '5 is a factor' but you don't seem to mean this all time. Personally, I prefer the active '5 divides N' or perhaps 'N is divisible by 5' to the passive 'N has 5 as a factor'. *History* I think you have written at least two histories to this newsgroup and a web site would be a much better place to put them. You have made several accusations of lying and you could use the website to substantiate them by pointing to documentation. *Points of Contention* Your Viewpoint I lose track of the arguments and how they are settled or even if they are. Your executive summary below is a good start but that is all it is. Hotlinks to expansions, with proofs, would be better. Java documentation is a fairly good example if the analogy isn't taken too far - at the top of the documentation for a class is an explanation of what this class is for and a list of the methods with the types of their parameters and each has a hotlink to an explanation of that method. One could go further and look at the code but it should never be necessary. Similarly, a mathematical exposition can have definitions, lemmas, propositions and theorems. The proof of a lemma or similar can cite consequences from the statement of another but mustn't ever refer to the proof - this would be as bad as code jumping into the middle of a subroutine or the documentation of a method referring to the code of another method. I suggest an exposition of your work would have the same structure with an executive summary at the top and cascading expansions of each point. (A good book contains a series of chapters. Each chapter starts out by saying what it is going to say, then says it in detail, and finally ends by summarising what it has said) Your previous website made a point of being terse but this isn't necessary. Other Viewpoints It is not at all an easy thing to do, but if you could try to explain, in detail, other people's standpoints as well as your own then this would be useful. I'm thinking in particular of the factorisation of certain polynomials in which you say that one root is coprime to a prime factor of the constant term of the polynomial and other people have shown explicit factorisations of the polynomial in which this is not true. I didn't follow your rebuttal. Dead Viewpoints It would take great courage but if you could also document points where you are no longer holding a former position then this also would interesting. I'm thinking in particular of whether a ring must be closed under an infinite summation and whether Z[pi] is isomorphic to the reals. *Simplifications* The polynomial that you use is very complicated with several parameters. Is all the complication necessary to the proof and does it add to it? Is the 'u' term necessary at all? You have resorted to actually substituting values in order to explain - why not go back and give the whole argument again but use a polynomial which is as simple as possible. In fact, I don't always follow your reasoning but perhaps it would be more obvious if you used a quadratic polynomial instead of a trinomial or explained why the argument doesn't work for quadratics. *Style* I, too, like long sentences (my second choice for a book to take if I were to be stranded on a desert island would be the collected works of Jane Austen) and you are much better at writing them than I am but they can be difficult for other people to follow. Take this as an example: . I already have as the polynomial is . . P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3f) . . and your attempts at confusing the issue by trying to push up x, . don't help you here, as the g's as factors of m, necessarily have . a constant term with respect to m, and your claim that it can vary . with m, is nonsensical on its face. It makes sense and I can understand it but I have to work at it. I suggest that shorter sentences would make it easier for the reader. Penny Hassett > For those of you trying to keep up with the mathematical facts in the > discussions about the error in core mathematics from a problem with a > definition, this post will outline the important ones quickly and > succinctly. > 1. First the problematic definition: > Algebraic integers are defined to be roots of monic polynomials with > integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where > monic refers to the leading coefficient. > My assertion is that the over hundred year old definition excludes > numbers that have to be included to keep from having contradiction > i.e. mathematical inconsistency. > 2. The important tool I use is a polynomial: > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) > The form of the polynomial allows me to factor P(m) into > non-polynomial factors, and the factorization with those factors is > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > where the a's are roots of the following cubic: > a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m). > 3. Dispute centers around what happens when I divide P(m) by f^2, > which you'll note is a factor of the polynomial in the ring of > algebraic integers. > 4. Mathematicians have argued that f^2 divides off as a function of m > because if they concede that it divides off independent of m, then I > can show that only two of the roots of > a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) > have f as a factor. > 5. However, it turns out that if you go to the field of algebraic > numbers you can prove that for *certain* values of m and f, the roots > of the cubic do not have f as a factor *in the ring of algebraic > numbers* which is the inconsistency. > That is, for the math to be consistent, two of the roots *should* have > f as a factor as long as m and f are algebraic integers, but while I > can show they do for a particular values like m=1, f=sqrt(2), there > are other values you can show they do not *in the ring of algebraic > integers* which results from the definition and its focus on monic > polynomials. > Note: In the ring of algebraic integers you can't see the problem but > have to go to the field of algebraic numbers as from within the ring > of algebraic integers it appears that only two of the roots have a > factor that is f. > James Harris === Subject: Re: Quick Math Guide to core error issues > James, > You post so much material that I don't have the time or desire to > read everything and I lose track of your position and arguments. > Some time ago you had an organised website with definitions and cross- > reference and you have posted several summaries and histories on this > newsgroup which would be easier to refer to if they were permanently > available. I, for one, would be pleased if you would resurrect your > website and put the information back. > Here are just some suggestions as to what you could put there and > how it could be organised. > *Definitions* > - Algebraic Integers and Algebraic Numbers > An explanation of what an algebraic integer is , some > examples and their general properties would be useful. > I vaguely recall that you had some software that would > take a monic quadratic equation with integer coefficients > and factorise it into linear terms with algebraic integer > coefficients. This could usefully go here as a Java applet. > - Your 'Object Ring' and what numbers are in it and what aren't. > I *think* you said 2^sqrt(3) was in it > I *think* you said that it wasn't wholly contained in the > complex numbers > An explanation of why these numbers are in it would be useful > - What an incomplete ring is. > - Some expressions don't seem to translate across the Atlantic and > I am thinking of 'has a factor of 5' in particular. To me, 'has a > value of 42' means '42 is a value' so 'has a factor of 5' should > mean '5 is a factor' but you don't seem to mean this all time. > Personally, I prefer the active '5 divides N' or perhaps 'N is > divisible by 5' to the passive 'N has 5 as a factor'. > *History* > I think you have written at least two histories to this newsgroup > and a web site would be a much better place to put them. You have > made several accusations of lying and you could use the website > to substantiate them by pointing to documentation. > *Points of Contention* > Your Viewpoint > I lose track of the arguments and how they are settled or even > if they are. Your executive summary below is a good start but that > is all it is. Hotlinks to expansions, with proofs, would be better. > Java documentation is a fairly good example if the analogy isn't > taken too far - at the top of the documentation for a class is > an explanation of what this class is for and a list of the methods > with the types of their parameters and each has a hotlink to an > explanation of that method. One could go further and look at the > code but it should never be necessary. > Similarly, a mathematical exposition can have definitions, lemmas, > propositions and theorems. The proof of a lemma or similar can cite > consequences from the statement of another but mustn't ever refer to > the proof - this would be as bad as code jumping into the middle of > a subroutine or the documentation of a method referring to the > code of another method. > I suggest an exposition of your work would have the same structure > with an executive summary at the top and cascading expansions of > each point. (A good book contains a series of chapters. Each chapter > starts out by saying what it is going to say, then says it in detail, > and finally ends by summarising what it has said) Your previous > website > made a point of being terse but this isn't necessary. > Other Viewpoints > It is not at all an easy thing to do, but if you could try to > explain, > in detail, other people's standpoints as well as your own then this > would be useful. I'm thinking in particular of the factorisation > of certain polynomials in which you say that one root is coprime > to a prime factor of the constant term of the polynomial and other > people have shown explicit factorisations of the polynomial in > which this is not true. I didn't follow your rebuttal. > Dead Viewpoints > It would take great courage but if you could also document points > where you are no longer holding a former position then this also > would interesting. I'm thinking in particular of whether a ring > must be closed under an infinite summation and whether Z[pi] is > isomorphic to the reals. > > *Simplifications* > The polynomial that you use is very complicated with several > parameters. Is all the complication necessary to the proof and does > it add to it? Is the 'u' term necessary at all? You have resorted > to actually substituting values in order to explain - why not go > back and give the whole argument again but use a polynomial which > is as simple as possible. In fact, I don't always follow your > reasoning but perhaps it would be more obvious if you used a > quadratic polynomial instead of a trinomial or explained why the > argument doesn't work for quadratics. > *Style* > > I, too, like long sentences (my second choice for a book to take > if I were to be stranded on a desert island would be the collected > works of Jane Austen) and you are much better at writing them than > I am but they can be difficult for other people to follow. > Take this as an example: > . I already have as the polynomial is > . P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3f) > . and your attempts at confusing the issue by trying to push up x, > . don't help you here, as the g's as factors of m, necessarily have > . a constant term with respect to m, and your claim that it can vary > . with m, is nonsensical on its face. > It makes sense and I can understand it but I have to work at it. > I suggest that shorter sentences would make it easier for the > reader. > Penny Hassett For those of you trying to keep up with the mathematical facts in the > discussions about the error in core mathematics from a problem with a > definition, this post will outline the important ones quickly and > succinctly. 1. First the problematic definition: Algebraic integers are defined to be roots of monic polynomials with > integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where > monic refers to the leading coefficient. My assertion is that the over hundred year old definition excludes > numbers that have to be included to keep from having contradiction > i.e. mathematical inconsistency. 2. The important tool I use is a polynomial: P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) The form of the polynomial allows me to factor P(m) into > non-polynomial factors, and the factorization with those factors is P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where the a's are roots of the following cubic: a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m). 3. Dispute centers around what happens when I divide P(m) by f^2, > which you'll note is a factor of the polynomial in the ring of > algebraic integers. 4. Mathematicians have argued that f^2 divides off as a function of m > because if they concede that it divides off independent of m, then I > can show that only two of the roots of a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) have f as a factor. 5. However, it turns out that if you go to the field of algebraic > numbers you can prove that for *certain* values of m and f, the roots > of the cubic do not have f as a factor *in the ring of algebraic > numbers* which is the inconsistency. That is, for the math to be consistent, two of the roots *should* have > f as a factor as long as m and f are algebraic integers, but while I > can show they do for a particular values like m=1, f=sqrt(2), there > are other values you can show they do not *in the ring of algebraic > integers* which results from the definition and its focus on monic > polynomials. Note: In the ring of algebraic integers you can't see the problem but > have to go to the field of algebraic numbers as from within the ring > of algebraic integers it appears that only two of the roots have a > factor that is f. James Harris What a breath of fresh air you are on this NG! Respectfully, John === Subject: Re: Quick Math Guide to core error issues > James, > You post so much material that I don't have the time or desire to > read everything and I lose track of your position and arguments. > Some time ago you had an organised website with definitions and cross- > reference and you have posted several summaries and histories on this > newsgroup which would be easier to refer to if they were permanently > available. I, for one, would be pleased if you would resurrect your > website and put the information back. I like you Penny Hasset and appreciate your commentary which is why I'm posting in a thread where I don't need to post as it's a quide. My problem is that I don't want to use MSN Groups, and I don't feel like going to another website provider. I *am* willing to allow someone else to host my work as long as they give me complete editorial control. Otherwise, it's easier for me to just post rather than try to maintain a website. > Here are just some suggestions as to what you could put there and > how it could be organised. > *Definitions* > - Algebraic Integers and Algebraic Numbers > An explanation of what an algebraic integer is , some > examples and their general properties would be useful. > I vaguely recall that you had some software that would > take a monic quadratic equation with integer coefficients > and factorise it into linear terms with algebraic integer > coefficients. This could usefully go here as a Java applet. Oh yeah, after Arturo Magidin tried to make a big deal out of some crap, I figured out what he was doing, which wasn't much more than a rather simple search for a factorization in algebraic integers. It was fun, but not that much fun. You see, I've found and dropped more mathematics than most people discover in a lifetime, as I'm a thrill seeker. You know, an adrenaline junkie. > - Your 'Object Ring' and what numbers are in it and what aren't. > I *think* you said 2^sqrt(3) was in it > I *think* you said that it wasn't wholly contained in the > complex numbers > An explanation of why these numbers are in it would be useful Actually *you* are in it Penny Hasset, as the object ring is rather large. You see, you are a mathematical object which I can prove using some rather basic logic and Goedel's proof. I like it that you're in Britain. If you're willing to advise me, I'm willing to toe a line. After all, it is math, but by myself I tend to be over the top. I need organization. > - What an incomplete ring is. It's a ring where you can have contradictions *within* the ring, which is the problem with algebraic integers. I can explain everything, but I'm a discoverer. I'm an artist. I'm NOT organized for this other stuff, like trying to convince people. I'm an artist. > - Some expressions don't seem to translate across the Atlantic and > I am thinking of 'has a factor of 5' in particular. To me, 'has a > value of 42' means '42 is a value' so 'has a factor of 5' should > mean '5 is a factor' but you don't seem to mean this all time. > Personally, I prefer the active '5 divides N' or perhaps 'N is > divisible by 5' to the passive 'N has 5 as a factor'. Hey Penny Hasset, if you can help me, and help me overcome these objections to the extent that I can make some money here, I'll pay you $250,000 US from any one math prize that I win that exceeds that amount. Since I'm a black male in America, as it has a rather stupendous history of racism, I should be able to pay that amount as well as $100,000 US as previously offered to a person or group with a machine proof of the core error, without much trouble, since I turn the world upside down. > *History* > I think you have written at least two histories to this newsgroup > and a web site would be a much better place to put them. You have > made several accusations of lying and you could use the website > to substantiate them by pointing to documentation. Oh, that's part of my fun. Unfortunately for me I have a tendency to scare people away when they figure out just how much I know and what I can do. Luckily for me, mathematicians are arrogant *and* dumb. So they're a perfect combination for someone like me, who otherwise gets kind of lonely. > *Points of Contention* > Your Viewpoint > I lose track of the arguments and how they are settled or even > if they are. Your executive summary below is a good start but that > is all it is. Hotlinks to expansions, with proofs, would be better. > Java documentation is a fairly good example if the analogy isn't > taken too far - at the top of the documentation for a class is > an explanation of what this class is for and a list of the methods > with the types of their parameters and each has a hotlink to an > explanation of that method. One could go further and look at the > code but it should never be necessary. > Similarly, a mathematical exposition can have definitions, lemmas, > propositions and theorems. The proof of a lemma or similar can cite > consequences from the statement of another but mustn't ever refer to > the proof - this would be as bad as code jumping into the middle of > a subroutine or the documentation of a method referring to the > code of another method. > I suggest an exposition of your work would have the same structure > with an executive summary at the top and cascading expansions of > each point. (A good book contains a series of chapters. Each chapter > starts out by saying what it is going to say, then says it in detail, > and finally ends by summarising what it has said) Your previous > website > made a point of being terse but this isn't necessary. I agree. Let's get started. I can make you rich, if you aren't rich already. If you are rich, I'll make you powerful. If you're already powerful, hell, why not just do it? > Other Viewpoints > It is not at all an easy thing to do, but if you could try to > explain, > in detail, other people's standpoints as well as your own then this > would be useful. I'm thinking in particular of the factorisation > of certain polynomials in which you say that one root is coprime > to a prime factor of the constant term of the polynomial and other > people have shown explicit factorisations of the polynomial in > which this is not true. I didn't follow your rebuttal. > Dead Viewpoints > It would take great courage but if you could also document points > where you are no longer holding a former position then this also > would interesting. I'm thinking in particular of whether a ring > must be closed under an infinite summation and whether Z[pi] is > isomorphic to the reals. > > *Simplifications* > The polynomial that you use is very complicated with several > parameters. Is all the complication necessary to the proof and does > it add to it? Is the 'u' term necessary at all? You have resorted > to actually substituting values in order to explain - why not go > back and give the whole argument again but use a polynomial which > is as simple as possible. In fact, I don't always follow your > reasoning but perhaps it would be more obvious if you used a > quadratic polynomial instead of a trinomial or explained why the > argument doesn't work for quadratics. > *Style* > > I, too, like long sentences (my second choice for a book to take > if I were to be stranded on a desert island would be the collected > works of Jane Austen) and you are much better at writing them than > I am but they can be difficult for other people to follow. > Take this as an example: > . I already have as the polynomial is > . P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3f) > . and your attempts at confusing the issue by trying to push up x, > . don't help you here, as the g's as factors of m, necessarily have > . a constant term with respect to m, and your claim that it can vary > . with m, is nonsensical on its face. > It makes sense and I can understand it but I have to work at it. > I suggest that shorter sentences would make it easier for the > reader. > Penny Hassett I like it Penny Hassett, and I'm willing to do some work, but not much as I'm the engine that drive everything anyway. You prepare to do some work--and make no mistake you WILL work very hard--and I'll try to give you success. After all, I'm already one of the most powerful men on the planet. With your help, I can get organized and maybe do some good in this world. Email me if you're interested, all offers are rescinded if you do not. James Harris === Subject: Re: Quick Math Guide to core error issues >>James, >>You post so much material that I don't have the time or desire to >>read everything and I lose track of your position and arguments. >>Some time ago you had an organised website with definitions and cross- >>reference and you have posted several summaries and histories on this >>newsgroup which would be easier to refer to if they were permanently >>available. I, for one, would be pleased if you would resurrect your >>website and put the information back. > I like you Penny Hasset and appreciate your commentary which is why > I'm posting in a thread where I don't need to post as it's a quide. > My problem is that I don't want to use MSN Groups, and I don't feel > like going to another website provider. > I *am* willing to allow someone else to host my work as long as they > give me complete editorial control. > Otherwise, it's easier for me to just post rather than try to maintain > a website. http:www.sphosting.com is easy to use, and allows file uploading. Of course, if you aren't willing to do any work, by all means keep everyone somewhat confused and off-balance. Perhaps you could post a weekly update? -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Quick Math Guide to core error issues > > I like it Penny Hassett, and I'm willing to do some work, but not much > as I'm the engine that drive everything anyway. > You prepare to do some work--and make no mistake you WILL work very > hard--and I'll try to give you success. > After all, I'm already one of the most powerful men on the planet. > With your help, I can get organized and maybe do some good in this > world. > Email me if you're interested, all offers are rescinded if you do not. > James Harris of you material up on a web-site then I'm willing to advise on the format and style by way of the sci.math newsgroup but that's all. PS. Lest you feel I am being dishonest when you find out later, let me say that it will be obvious to many people in Britain that I am using a nom-de-keyboard. === Subject: Re: Quick Math Guide to core error issues > I like it Penny Hassett, and I'm willing to do some work, but not much > as I'm the engine that drive everything anyway. You prepare to do some work--and make no mistake you WILL work very > hard--and I'll try to give you success. After all, I'm already one of the most powerful men on the planet. With your help, I can get organized and maybe do some good in this > world. Email me if you're interested, all offers are rescinded if you do not. James Harris > of you material up on a web-site then I'm willing to advise on the > format and style by way of the sci.math newsgroup but that's all. last couple of days, partly out of EXTREME FRUSTRATION at my situation. I've found that I can have fun with postings, which makes me feel better. Still I was sincere about the $250k but am now relieved that you declined. Oh yeah, I've taken your advice though as I'm using only m as a variable in my recent postings as I'm *really* ready to finish things up. > PS. Lest you feel I am being dishonest when you find out later, let > me say that it will be obvious to many people in Britain that > I am using a nom-de-keyboard. Oh hey, I'd started calling you my money penny too. Kind of like a James thing, you know, Bond, James Bond. Oh well, maybe someday you'll have reason to give me your name, but it's not a big deal. In any event, unlike with Nora Baron, I won't put quotes around your name. James Harris === Subject: Re: Quick Math Guide to core error issues === >Subject: Re: Quick Math Guide to core error issues >Message-id: <3F942CFC.408188F5@nospamm.dabsol.co.uk> >> >> I like it Penny Hassett, and I'm willing to do some work, but not much >> as I'm the engine that drive everything anyway. >> You prepare to do some work--and make no mistake you WILL work very >> hard--and I'll try to give you success. >> After all, I'm already one of the most powerful men on the planet. >> With your help, I can get organized and maybe do some good in this >> world. >> Email me if you're interested, all offers are rescinded if you do not. >> James Harris >of you material up on a web-site then I'm willing to advise on the >format and style by way of the sci.math newsgroup but that's all. >PS. Lest you feel I am being dishonest when you find out later, let >me say that it will be obvious to many people in Britain that >I am using a nom-de-keyboard. And do you live near the Archers, Penny? Marvin Marvin Sebourn osugeography@aol.com === Subject: Re: Quick Math Guide to core error issues > I'm willing to do some work, but not much I knew it! You're just too lazy! === Subject: Re: Quick Math Guide to core error issues putting aside the question of wether there is *any* real JSH, I still have to question this one's entitlement to the name, when he goes over the top and says that a correspondent can be proved to be in his newfound ring of what ever. back to teh question of the real one: if he's not being paid to do this, or has a pension that is allowing him to make fun of Whitey (?), then it really is pretty strange. note on Anglo-american history: slavery was a British institution; that's why they supported the Confederacy (along with the New York Times etc.) with ships & materiel, and actually organized the Civil War. (this makes for a good, revisionist question: What was our 3rd war with Great Britain?) [see http://tarpley.net] > I'm posting in a thread where I don't need to post as it's a quide. > An explanation of what an algebraic integer is , some > examples and their general properties would be useful. > I vaguely recall that you had some software that would > take a monic quadratic equation with integer coefficients > and factorise it into linear terms with algebraic integer > coefficients. This could usefully go here as a Java applet. > You see, I've found and dropped more mathematics than most people > Actually *you* are in it Penny Hasset, as the object ring is rather > large. > You see, you are a mathematical object which I can prove using some > rather basic logic and Goedel's proof. > I like it that you're in Britain. > I lose track of the arguments and how they are settled or even > if they are. Your executive summary below is a good start but that > is all it is. Hotlinks to expansions, with proofs, would be better. > If you're already powerful, hell, why not just do it? > It would take great courage but if you could also document points > where you are no longer holding a former position then this also > would interesting. I'm thinking in particular of whether a ring > must be closed under an infinite summation and whether Z[pi] is > isomorphic to the reals. > . I already have as the polynomial is > . > . P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3f) > . > . and your attempts at confusing the issue by trying to push up x, > . don't help you here, as the g's as factors of m, necessarily have > . a constant term with respect to m, and your claim that it can vary > . with m, is nonsensical on its face. --UN HYDROGEN (sic; Methanex (TM) reformanteurs) ECONOMIE?... La Troi Phases d'Exploitation de la Protocols des Grises de Kyoto: (FOSSILISATION [McCainanites?] (TM/sic))/ BORE/GUSH/NADIR @ http://www.tarpley.net/aobook.htm. Http://www.tarpley.net/bushb.htm (content partiale, below): 17 -- L'ATTEMPTER de COUP D'ETAT, 3/30/81 === Subject: Re: Quick Math Guide to core error issues > [...] > note on Anglo-american history: > slavery was a British institution; The British slave traders bought their slaves from Arab and African trade slavers, so it wasn't a British institution it was an international one. Slavery persists to this day of course. > [...] -- G.C. === Subject: Re: Quick Math Guide to core error issues > For those of you trying to keep up with the mathematical facts in the > discussions about the error in core mathematics from a problem with a > definition, this post will outline the important ones quickly and > succinctly. > 1. First the problematic definition: > Algebraic integers are defined to be roots of monic polynomials with > integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where > monic refers to the leading coefficient. > My assertion is that the over hundred year old definition excludes > numbers that have to be included to keep from having contradiction > i.e. mathematical inconsistency. This is sheer idiocy. A definition cannot lead to a contradiction. A definition, that is not correctly understood by a wannabe maths genius like yourself, followed by some ridiculously confused attempts at proving things, can lead a sufficiently stupid person to thinking there are contradictions. But that is _your_ problem, and not a problem of the definition. === Subject: Re: Quick Math Guide to core error issues In sci.physics, James Harris > For those of you trying to keep up with the mathematical facts in the > discussions about the error in core mathematics from a problem with a > definition, this post will outline the important ones quickly and > succinctly. > 1. First the problematic definition: > Algebraic integers are defined to be roots of monic polynomials with > integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where > monic refers to the leading coefficient. > My assertion is that the over hundred year old definition excludes > numbers that have to be included to keep from having contradiction > i.e. mathematical inconsistency. And these numbers are ... what? Presumably, you can produce a counterexample. > 2. The important tool I use is a polynomial: > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) > The form of the polynomial allows me to factor P(m) into > non-polynomial factors, and the factorization with those factors is > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > where the a's are roots of the following cubic: > a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m). Pedant point: ITYM a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) = 0. > 3. Dispute centers around what happens when I divide P(m) by f^2, > which you'll note is a factor of the polynomial in the ring of > algebraic integers. That it is, for what it's worth. However, you've not gotten around the f^(2/3) problem yet. I posit that a perfectly valid transformation of your cubic is Q(m) = P(m) / f^2 = (b_1 x + uf^(1/3))(b_2 x + uf^(1/3))(b_3 x + uf^(1/3)) where b_{i} = a_{i} / f^(2/3). In fact, that's probably what you'd end up with anyway! :-) I, however, make no claims regarding the b_{i} being integers, algebraic or otherwise. I'm not sure what can be deduced therefrom. It's worth noting that f^(1/3) is an algebraic integer if f is. Another transformation is R(m) = P(m) / f^3 = (c_1 x + u)(c_2 x + u) (c_3 x + u) although in this case we have a problem, as not all the c's are algebraic integers; there's a missing factor of 1/f in there somewhere. This factor can be added in: Q(m) = f R(m) = (c_1 x + u) (c_2 x + u) (c_3 fx + uf) but it could equally easily be added in: Q(m) = (c_1 f^(1/3) x + u f^(1/3)) (c_2 f^(1/3) x + u f^(1/3)) (c_3 f^(1/3)x + uf^(1/3)) or Q(m) = (c_1 x + u) ( c_2 f^(1/2) x + u f^(1/2) ) (c_3 f^(1/2) x + u f^(1/2)) I just don't know at this point. > 4. Mathematicians have argued that f^2 divides off as a function of m > because if they concede that it divides off independent of m, then I > can show that only two of the roots of > a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) > have f as a factor. If one sets f = 2, m = 1 we get a^3 + 9a^2 - 28 which has as one root a_1 = -2. Factoring, we get a^3 + 9a^2 - 28 = (a + 2) (a^2 + 7a - 14) so the other two roots are a_x = (-7 sqrt(49 + 56)) / 2 = -7/2 sqrt(105) / 2. Only one of these roots (-2) is divisible by 2. The other two roots -7/2 sqrt(105) / 2 are such that, if we set b_x = a_x/2, or a_x = 2b_x, we get b_x = -7/4 sqrt(105) / 4. What equation of integer coefficients does the b's satisfy? That's simple enough; substituting 2b for a, we get 4b^2 + 14b - 14 = 0 or 2b^2 + 7b - 7 = 0. Clearly, the b's are not algebraic integers, and therefore two of the original a's are not divisible by 2, for this particular setting of f and m. This is a counterexample to your original proposition. > 5. However, it turns out that if you go to the field of algebraic > numbers you can prove that for *certain* values of m and f, the roots > of the cubic do not have f as a factor *in the ring of algebraic > numbers* which is the inconsistency. > That is, for the math to be consistent, two of the roots *should* have > f as a factor as long as m and f are algebraic integers, but while I > can show they do for a particular values like m=1, f=sqrt(2), there > are other values you can show they do not *in the ring of algebraic > integers* which results from the definition and its focus on monic > polynomials. I take it you want to include -7/4 sqrt(105) / 4 in the ring of algebraic integers? Please clarify. > Note: In the ring of algebraic integers you can't see the problem but > have to go to the field of algebraic numbers as from within the ring > of algebraic integers it appears that only two of the roots have a > factor that is f. 2/3 can be divided by 3 (the result being 2/9). Did you have a point here? > James Harris -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Quick Math Guide to core error issues > In sci.physics, James Harris > ... > 5. However, it turns out that if you go to the field of algebraic > numbers you can prove that for *certain* values of m and f, the roots > of the cubic do not have f as a factor *in the ring of algebraic > numbers* which is the inconsistency. > > That is, for the math to be consistent, two of the roots *should* have > f as a factor as long as m and f are algebraic integers, but while I > can show they do for a particular values like m=1, f=sqrt(2), there > are other values you can show they do not *in the ring of algebraic > integers* which results from the definition and its focus on monic > polynomials. > I take it you want to include > -7/4 sqrt(105) / 4 > in the ring of algebraic integers? No, he can not add both. Their sum is -7/2, and adding both would make 2 a unit. He wants to add only one, but he will not tell us which one. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Quick Math Guide to core error issues In sci.physics, Dik T. Winter : > In sci.physics, James Harris > > ... > 5. However, it turns out that if you go to the field of algebraic > numbers you can prove that for *certain* values of m and f, the roots > of the cubic do not have f as a factor *in the ring of algebraic > numbers* which is the inconsistency. That is, for the math to be consistent, two of the roots *should* have > f as a factor as long as m and f are algebraic integers, but while I > can show they do for a particular values like m=1, f=sqrt(2), there > are other values you can show they do not *in the ring of algebraic > integers* which results from the definition and its focus on monic > polynomials. I take it you want to include > -7/4 sqrt(105) / 4 > in the ring of algebraic integers? > No, he can not add both. Their sum is -7/2, and adding both would make > 2 a unit. He wants to add only one, but he will not tell us which one. It's a package deal. :-) And he gets 4 for the price of 2; the reciprocals need to be added as well, as unit * unit = unit and unit / unit = unit. In fact, a lot more will be dragged in by this inclusion. But the original definition discriminates against this number (and for good reason). -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Quick Math Guide to core error issues > For those of you trying to keep up with the mathematical facts in the > discussions about the error in core mathematics from a problem with a > definition, this post will outline the important ones quickly and > succinctly. > 1. First the problematic definition: > Algebraic integers are defined to be roots of monic polynomials with > integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where > monic refers to the leading coefficient. > My assertion is that the over hundred year old definition excludes > numbers that have to be included to keep from having contradiction > i.e. mathematical inconsistency. The definition just defines a set of numbers. The definition itself cannot produce a contradiction. Contradictions are produced when one 'theorem' contradicts another theorem. If you think you have a contradiction here, what is the known theorem in algebraic number theory which is being contradicted? (See below at {###] for my speculation on this.) > 2. The important tool I use is a polynomial: > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) > The form of the polynomial allows me to factor P(m) into > non-polynomial factors, and the factorization with those factors is > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > where the a's are roots of the following cubic: > a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m). ... and the a's are therefore algebraic integers (this cubic is monic). > 3. Dispute centers around what happens when I divide P(m) by f^2, > which you'll note is a factor of the polynomial in the ring of > algebraic integers. > 4. Mathematicians have argued that f^2 divides off as a function of m > because if they concede that it divides off independent of m, then I > can show that only two of the roots of >[*] a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) > have f as a factor. No - we don't argue that it divides off as a function of m. We argue essentially that f^2 is distributed among the factors (ai*x + u*f) in a way which depends on m. For example, if m is such that the polynomial [*] that you give above is reducible, then one factor is relatively prime to f. But if m is such that [*] is irreducible, then *none* of the factors are relatively prime to f. In fact in the irreducible case, ALL of the factors (ai*x + u*f) are divisible by f^{2/3}. See my post of Oct 18 in the thread Finishing argument - core error proven for a proof of this. > 5. However, it turns out that if you go to the field of algebraic > numbers you can prove that for *certain* values of m and f, the roots > of the cubic do not have f as a factor *in the ring of algebraic > numbers* which is the inconsistency. The first part is true: it happens whenever [*] is irreducible, which is true for most values of m. But it does not result in a contradiction. The factorization is different when the polynomial [*] is irreducible than when it is not. The twain do not meet (i.e., [*] is either irreducible or it is not) so there is no inconsistency. > That is, for the math to be consistent, two of the roots *should* have > f as a factor as long as m and f are algebraic integers, but while I > can show they do for a particular values like m=1, f=sqrt(2), there > are other values you can show they do not *in the ring of algebraic > integers* which results from the definition and its focus on monic > polynomials. f = sqrt(2) is of no interest here. You original concern, relevant not only to your claims in Advanced Polynomial Factorization and Core error but also to your proof of Fermat's last theorem, dealt with f a a prime > 3 and m an integer relatively prime to f. Our counterexamples to your argument are restricted to the latter conditions. But even if one wanted to generalize for formal academic reasons: how f^2 distributes among the factors (a1*x + u*f) differs as described above for different combinations of m and f. Proving something about the form of the factorization for one combination does not prove it for others, as you inexplicably seem to believe. Similarly proving that for m = 0, f^2 distributes into the 3 factors as f, f, and 1, tells you nothing about cases for which m <> 0. I am surprised to see that you have mentioned your erroneous belief to the contrary, for the thousandth time. > Note: In the ring of algebraic integers you can't see the problem but > have to go to the field of algebraic numbers as from within the ring > of algebraic integers it appears that only two of the roots have a > factor that is f. A bizarre statement. In the field of algebraic numbers, every number has f as a factor! This is of no interest at all. The whole point of what you have been doing is lost if you decide to start talking about factorizations in a field. Nora B. > James Harris [###] Your result, if true, would contradict one of the following theorems: 1. Roots of non-monic primitive irreducible polynomial with integer coefficients cannot be algebraic integers. 2. The set of algebraic numbers constitutes a ring. I think you are refusing to say that your result contradicts 1. because you have gone on record (in 2002) as accepting that 1. is a correct theorem. You have not thought much about 2., which is also a theorem and one which is moderately difficult to prove. You have chosen to say that your result follows from an error in the definition of algebraic integers because you know that would lead to the conclusion that mathematics is inconsistent (which you and the rest of us abhor), OR that your own proof is wrong. And your emotional state is such that you cannot possibly accept the latter conclusion. But saying that your result is a consequence of an erroneous definition makes no sense at all on any scale. N.B. === Subject: Re: Quick Math Guide to core error issues Adjunct Assistant Professor at the University of Montana. [.snip.] >> 4. Mathematicians have argued that f^2 divides off as a function of m >> because if they concede that it divides off independent of m, then I >> can show that only two of the roots of >>[*] a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) >> have f as a factor. > No - we don't argue that it divides off as a function of >m. We argue essentially that f^2 is distributed among the >factors (ai*x + u*f) in a way which depends on m. For example, >if m is such that the polynomial [*] that you give above is >reducible, then one factor is relatively prime to f. This is not true in general either. If m=1 and f=2, one factor is a multiple of f, and the other two are multiples of proper factors of f; the polynomial [*] in that case factors as a product of a linear and an irreducible quadratic. What we have said is that: > But if >m is such that [*] is irreducible, then *none* of the factors >are relatively prime to f. But there have been no general conclusions about the reducible case in general, other than it may indeed be the case that one of the factors is coprime to f. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Quick Math Guide to core error issues ... > 1. First the problematic definition: > > Algebraic integers are defined to be roots of monic polynomials with > integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where > monic refers to the leading coefficient. > > My assertion is that the over hundred year old definition excludes > numbers that have to be included to keep from having contradiction > i.e. mathematical inconsistency. > The definition just defines a set of numbers. The definition > itself cannot produce a contradiction. Contradictions are > produced when one 'theorem' contradicts another theorem. > If you think you have a contradiction here, what is the > known theorem in algebraic number theory which is being > contradicted? (See below at {###] for my speculation on this.) ... > Your result, if true, would contradict one of the > following theorems: > 1. Roots of non-monic primitive irreducible polynomial with > integer coefficients cannot be algebraic integers. > 2. The set of algebraic numbers constitutes a ring. You mean algebraic integers here. > I think you are refusing to say that your result contradicts > 1. because you have gone on record (in 2002) as accepting that > 1. is a correct theorem. You have not thought much about 2., > which is also a theorem and one which is moderately difficult to > prove. Actually he is also on record accepting that 2 is true, the last time was not so long ago. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Quick Math Guide to core error issues > For those of you trying to keep up with the mathematical facts in the > discussions about the error in core mathematics from a problem with a > definition, this post will outline the important ones quickly and > succinctly. > 1. First the problematic definition: > Algebraic integers are defined to be roots of monic polynomials with > integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where > monic refers to the leading coefficient. > My assertion is that the over hundred year old definition excludes > numbers that have to be included to keep from having contradiction > i.e. mathematical inconsistency. > The definition just defines a set of numbers. The definition > itself cannot produce a contradiction. Contradictions are > produced when one 'theorem' contradicts another theorem. > If you think you have a contradiction here, what is the > known theorem in algebraic number theory which is being > contradicted? (See below at {###] for my speculation on this.) > 2. The important tool I use is a polynomial: > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) > The form of the polynomial allows me to factor P(m) into > non-polynomial factors, and the factorization with those factors is > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > where the a's are roots of the following cubic: > a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m). > ... and the a's are therefore algebraic integers (this cubic > is monic). > 3. Dispute centers around what happens when I divide P(m) by f^2, > which you'll note is a factor of the polynomial in the ring of > algebraic integers. > 4. Mathematicians have argued that f^2 divides off as a function of m > because if they concede that it divides off independent of m, then I > can show that only two of the roots of >[*] a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) > have f as a factor. > No - we don't argue that it divides off as a function of > m. We argue essentially that f^2 is distributed among the > factors (ai*x + u*f) in a way which depends on m. For example, > if m is such that the polynomial [*] that you give above is > reducible, then one factor is relatively prime to f. But if > m is such that [*] is irreducible, then *none* of the factors > are relatively prime to f. In fact in the irreducible case, > ALL of the factors (ai*x + u*f) are divisible by f^{2/3}. > See my post of Oct 18 in the thread Finishing argument - core > error proven for a proof of this. > 5. However, it turns out that if you go to the field of algebraic > numbers you can prove that for *certain* values of m and f, the roots > of the cubic do not have f as a factor *in the ring of algebraic > numbers* which is the inconsistency. > The first part is true: it happens whenever [*] is irreducible, > which is true for most values of m. But it does not result > in a contradiction. The factorization is different when the > polynomial [*] is irreducible than when it is not. The twain > do not meet (i.e., [*] is either irreducible or it is not) > so there is no inconsistency. > That is, for the math to be consistent, two of the roots *should* have > f as a factor as long as m and f are algebraic integers, but while I > can show they do for a particular values like m=1, f=sqrt(2), there > are other values you can show they do not *in the ring of algebraic > integers* which results from the definition and its focus on monic > polynomials. > f = sqrt(2) is of no interest here. You original concern, relevant > not only to your claims in Advanced Polynomial Factorization and Core > error but also to your proof of Fermat's last theorem, dealt with f a > a prime > 3 and m an integer relatively prime to f. Our counterexamples > to your argument are restricted to the latter conditions. But even if one > wanted to generalize for formal academic reasons: how f^2 distributes among > the factors (a1*x + u*f) differs as described above for different > combinations of m and f. Proving something about the form of the > factorization for one combination does not prove it for others, as > you inexplicably seem to believe. > Similarly proving that for m = 0, f^2 distributes into the 3 > factors as f, f, and 1, tells you nothing about cases for which > m <> 0. I am surprised to see that you have mentioned your erroneous > belief to the contrary, for the thousandth time. > Note: In the ring of algebraic integers you can't see the problem but > have to go to the field of algebraic numbers as from within the ring > of algebraic integers it appears that only two of the roots have a > factor that is f. > A bizarre statement. In the field of algebraic numbers, every > number has f as a factor! This is of no interest at all. The > whole point of what you have been doing is lost if you decide to > start talking about factorizations in a field. > Nora B. > James Harris > [###] > Your result, if true, would contradict one of the > following theorems: > 1. Roots of non-monic primitive irreducible polynomial with > integer coefficients cannot be algebraic integers. > 2. The set of algebraic numbers constitutes a ring. > I think you are refusing to say that your result contradicts > 1. because you have gone on record (in 2002) as accepting that > 1. is a correct theorem. You have not thought much about 2., > which is also a theorem and one which is moderately difficult to > prove. You have chosen to say that your result follows from > an error in the definition of algebraic integers because you > know that would lead to the conclusion that mathematics is > inconsistent (which you and the rest of us abhor), OR that > your own proof is wrong. And your emotional state is such > that you cannot possibly accept the latter conclusion. But > saying that your result is a consequence of an erroneous > definition makes no sense at all on any scale. > N.B. I want to make a note about James' dividing off claim. If you divide something off, you change the equation. Let me give you an example. Say we have x^2-x-12=x-4. If I solve this for x, I get x^2-2x-8=0 => (x-4)(x+2)=0 => x=4 or x=-2. Well, both solutions check, but let's do this a little differently, which is wrong. I see that this can be written as (x-4)(x+3)=x-4. If I divide both sides by x-4, I get x+3=1, so x=-2. Well, this is right, but I'm missing a solution, the case when x-4=0. So instead of dividing something off, you should factor it out so you don't change anything. David Moran === Subject: Re: Quick Math Guide to core error issues > 3. Dispute centers around what happens when I divide P(m) by f^2, > which you'll note is a factor of the polynomial in the ring of > algebraic integers. > 4. Mathematicians have argued that f^2 divides off as a function of m > because if they concede that it divides off independent of m, then I > can show that only two of the roots of > a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) > have f as a factor. Yup. > 5. However, it turns out that if you go to the field of algebraic > numbers you can prove that for *certain* values of m and f, the roots > of the cubic do not have f as a factor *in the ring of algebraic > numbers* which is the inconsistency. This is bogus. As the algebraic numbers form a field, f is a factor of *everything*. You can not prove that something is not a factor of something else in a field. So if you prove that in the algebraic numbers something (!= 0) is not a factor of something else you only prove that the algebraic numbers do not form a field. In the ring of algebraic integers, indeed, you can prove that for certain values of m and f, the roots of the cubic do not have f as a factor *in that ring*. But that is not a inconsistency. > Note: In the ring of algebraic integers you can't see the problem but > have to go to the field of algebraic numbers as from within the ring > of algebraic integers it appears that only two of the roots have a > factor that is f. That is a fata morgana. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Quick Math Guide to core error issues > That is a fata morgana. What does that mean? I've encountered the phrase before only as the title of a Werner Herzog film, the one that starts with a long sequence of airliners landing, shot from a great distance with a telephoto lens (I think). Gib === Subject: Re: Quick Math Guide to core error issues > That is a fata morgana. > What does that mean? I've encountered the phrase before only as the > title of a Werner Herzog film, the one that starts with a long sequence > of airliners landing, shot from a great distance with a telephoto lens > (I think). It is even in the Webster I own (though I doubt the ethymology given). It is a mirage. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Quick Math Guide to core error issues > > That is a fata morgana. > > What does that mean? I've encountered the phrase before only as the > > title of a Werner Herzog film, the one that starts with a long sequence > > of airliners landing, shot from a great distance with a telephoto lens > > (I think). > It is even in the Webster I own (though I doubt the ethymology given). It > is a mirage. Ah, I think I knew that once. It fits with that opening sequence - the planes are landing on a runway in the desert, and as the sun heats the tarmac the optical distortions increase. Gib === Subject: Re: Quick Math Guide to core error issues now, you've done it; sent me on at least two, intermiable webhunts! http://www.biblecodedigest.com/ --les ducs d'Enron! === Subject: Re: Quick Math Guide to core error issues |A definition cannot introduce a contradiction, since a definition is |simply shorthand for an expression. If instead of defining algebraic |integers we simply talked about roots of monic polynomials with |integer coefficients, there would be absolutely no change in |mathematics (except that all number theory books would be longer). actually they would be shorter, if we abbreviated it to rompwic. so will that remove the inconsistency, james, if all the mathematicians in the world start calling them rompwics instead of algebraic integers? -- [e-mail address jdolan@math.ucr.edu] === Subject: Re: Quick Math Guide to core error issues > For those of you trying to keep up with the mathematical facts in the > discussions about the error in core mathematics from a problem with a > definition Firstly, it is more of a diatribe by JSH, repeated in many guises, than a discussion, since JSH ignores most reasonable responses and merely repeats his refuted arguments. Secondly, JSH is the only one claiming an error in core mathematics. The rest of the world seems quite satisfied with the reliability of of mathematics, core and all. The error JSH alleges obviously being, to anyone who has followed JSH's sequence of diatribes, that others do not see JSH as having the genius he sees in himself. A common problem among those of little talent and excessive ego. === Subject: Re: Quick Math Guide to core error issues > For those of you trying to keep up with the mathematical facts in the > discussions about the error in core mathematics from a problem with a > definition, this post will outline the important ones quickly and > succinctly. > 1. First the problematic definition: > Algebraic integers are defined to be roots of monic polynomials with > integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where > monic refers to the leading coefficient. Forget about the definition. Forget about the algebraic integers. Let's call them roots of monic polynomials with integer coefficients instead, ok? Can you clearly formulate what's wrong with roots of monic polynomials with integer coefficients? === Subject: Re: Quick Math Guide to core error issues > For those of you trying to keep up with the mathematical facts in the > discussions about the error in core mathematics from a problem with a > definition, this post will outline the important ones quickly and > succinctly. > 1. First the problematic definition: > Algebraic integers are defined to be roots of monic polynomials with > integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where > monic refers to the leading coefficient. > My assertion is that the over hundred year old definition excludes > numbers that have to be included to keep from having contradiction > i.e. mathematical inconsistency. > 2. The important tool I use is a polynomial: > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) Yes, and then you claim you can find the terms which are independent of m by setting m=0. This is a blunder. Setting m=0 merely *evaluates* P(m) for the case m=0. Any multi-variable expression can be simplified by setting one or more of the variables to a constant value, of course, but whether you set m=0, m=1, m=I, m=Pi, or any other constant value, you have only succeeded in performing an evaluation of P(m) for that value. Naturally 'm' will not appear in the resulting expression because it has been replaced by a constant. This says nothing whatsoever about the properties of the original expression. For example, suppose we have: Q(m) = 2*(m-1) + 1 Does setting m=0 remove the terms dependent on 'm'? Is Q(m) dependent on 'm'? (Answer: yes, it is a function of 'm'. It has a distinct value for every value of 'm'.) What if: Q(m) = 2^m + 1 does setting m=0 remove the terms dependent on 'm'? Suppose: Q(m) = a^m + b*m + c/m + d*cos(m) + e*exp(m) What about this case? Can we set m=0 to remove terms dependent on 'm'? If not, why don't you explain what you mean when you state that you can find terms independent of 'm' by placing m=0? Apparently you either misunderstand the process of evaluating a multivariable expression by substitution, or you are simply misrepresenting your argument. -- A fool and his proof are soon refuted. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Quick Math Guide to core error issues > For those of you trying to keep up with the mathematical facts in the > discussions about the error in core mathematics from a problem with a > definition, this post will outline the important ones quickly and > succinctly. > 1. First the problematic definition: > Algebraic integers are defined to be roots of monic polynomials with > integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where > monic refers to the leading coefficient. > My assertion is that the over hundred year old definition excludes > numbers that have to be included to keep from having contradiction > i.e. mathematical inconsistency. > 2. The important tool I use is a polynomial: > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) Lost me again. You were doing fine in #1. Now please explain in simple English where you're going with this, before piling on the equations. What kind of number is excluded? If you start with the algebraic integers and apply standard ring operations of multiplication and addition, do you end up with something that's not an algebraic integer? Or is the extra number the result of some operation other than multiplication and addition? Before diving into the equations, just say where you're going. === Subject: Re: OLD HP CALUCLATORS > >> Once upon a time, the battery died and thought that I'd >> just use this other calculator until I had time to get >> a new one. I could not function. RPN is so imbedded >> into my thinking that I could not use a regular calculator. >> I ended up doing the stuff on paper and the task of getting >> a new battery jumped to number 1 priority. >> >> It's the same with me. I use a slide rule and/or paper and pencil if >> an RPN calc isn't available. (Though my 48GX, 41CX and 16C are usually >> within arm's reach.) > >My laptop is often within arm's reach and I use an RPN calculator on it, >although I do also have a physical one somewhere. I, too, can not >function with an algebraic-notation calculator, despite using algebraic >notation often enough in various programming languages. > I don't see any relationship between those other kinds of calculations > and algebra or programming languages. It's because of my familiarity > of programming that I can't use those other calculators. > RPN is completely algebraic to me. When calculating any equation, > you do the insides first with a heirarchy of operation. > Those non-RPN calculators are left to right with no push down list. > You mean the ones with algebraic notation and no parentheses? (Yes, they > really exist.) Or with the (seemingly always) insufficiently deep nesting. > I've exceeded 9 levels of parentheses with algebraic (left-to-right) entry, > but I vaguely remember *once* exceeding a 4-level stack on an RPN > calculator. I couldn't even tell you the circumstances. I might be > misremembering -- or remembering an error I made. > I get around the algebraic problems by (drum roll, please) working from the > inside out and either using stored memory locations or writing down the > intermediate results. I also claim that anyone who can actually use an > algebraic calculator for difficult calculations (like mortgage payment > calculations) does the equivalent (well, maybe not, since it can be done > with about 5 levels of parentheses). But after a while, you forget which > level of parentheses you're in, so you have to be making notes of some sort > on a sheet of paper. > Oh, I forgot algebraic notation but no algebraic hierarchy. So 1+2*3=9. > Aaaaaauuuuuuuggggggghhhhhhhhh! Usually with no parentheses. Just curious, I don't have a calculator with parentheses, but if you have one, could you test this out and tell how many seconds and how many keystrokes it takes to calculate this hypothetical thing: (9+7)(8-3)(5+4) ------------------ + (6-4)(3-1) (9+2)(9-4)(7+3) ? Of course you're not allowed to use your knowledge of simple arithmetic ;-) With RPN (and without practicing first) it this takes 39 keystrokes and approx. 13 seconds. And without using memory other than the implicit stack. How does this go with an algebraic with parentheses? By the way, I just have spent some time trying to do this with Windows' silly calculator - in Scientific Mode ;-) Calculating the top part of the fraction made me end up with a memory value of 720 and 20 keystrokes. Then I started with the 9 + 2 = of the bottom part of the fraction. Having 11 on the display and 720 in memory, I needed 9 keystrokes to divide the display by the memory value: / MR = MS 1 / MR = MS So , the fraction (left term) requires 20 + 4 + 9 + 9 + 9 = 51 keystrokes. And then I realized I got stuck: If I want to calculate and add the second term, I need a second memory location. AAARG! You probably understand why I spent quite some time writing my HP55 With A Vengeance ;-) Dirk Vdm === Subject: Re: OLD HP CALUCLATORS >> > Once upon a time, the battery died and thought that I'd > just use this other calculator until I had time to get > a new one. I could not function. RPN is so imbedded > into my thinking that I could not use a regular calculator. > I ended up doing the stuff on paper and the task of getting > a new battery jumped to number 1 priority. > > It's the same with me. I use a slide rule and/or paper and pencil if > an RPN calc isn't available. (Though my 48GX, 41CX and 16C are usually > within arm's reach.) >> >>My laptop is often within arm's reach and I use an RPN calculator on it, >>although I do also have a physical one somewhere. I, too, can not >>function with an algebraic-notation calculator, despite using algebraic >>notation often enough in various programming languages. >> I don't see any relationship between those other kinds of calculations >> and algebra or programming languages. It's because of my familiarity >> of programming that I can't use those other calculators. >> RPN is completely algebraic to me. When calculating any equation, >> you do the insides first with a heirarchy of operation. >> Those non-RPN calculators are left to right with no push down list. >You mean the ones with algebraic notation and no parentheses? (Yes, they >really exist.) Or with the (seemingly always) insufficiently deep nesting. >I've exceeded 9 levels of parentheses with algebraic (left-to-right) entry, >but I vaguely remember *once* exceeding a 4-level stack on an RPN >calculator. I couldn't even tell you the circumstances. I might be >misremembering -- or remembering an error I made. >I get around the algebraic problems by (drum roll, please) working from the >inside out and either using stored memory locations Except stored memory locations is a poor-man's RPN. A real calculator will do the data push as part of the operation rather than leave it as a specific action. > ..or writing down the >intermediate results. This is also doing the calculation by RPN. Just because the calculator doesn't make easy doesn't mean the the RPN process isn't done. > ..I also claim that anyone who can actually use an >algebraic calculator for difficult calculations (like mortgage payment >calculations) does the equivalent (well, maybe not, since it can be done >with about 5 levels of parentheses). Ah!! Yes. The calculation process is RPN. HP calculators are designed to not interfere with the process. Those others are designed to interrupt the process by forcing the user to write it down in some manner. There exist operating systems that also get in the way of accomplished work; there exist OSes who know when to stay out of the way and when to assist. > . .But after a while, you forget which >level of parentheses you're in, so you have to be >making notes of some sort on a sheet of paper. >Oh, I forgot algebraic notation but no algebraic hierarchy. So 1+2*3=9. >Aaaaaauuuuuuuggggggghhhhhhhhh! Usually with no parentheses. And every time a result has to be pushed onto a piece of paper, the chances of error quadruple. Data should only be written once; that way you have to check it only twice. /BAH Subtract a hundred and four for e-mail. === Subject: Re: OLD HP CALUCLATORS >I can't believe I've written this much about this. for this class? and the Why do I hafta do things by hand when I have a calculator? questions from students. I can explain what I know about how a calculator can be useful and how it can be useless, but I concede my viewpoint is quite limited. Others' experiences can be informative. dave === Subject: Re: OLD HP CALUCLATORS >>I can't believe I've written this much about this. >for this class? and the Why do I hafta do things by hand when I have a >calculator? questions from students. I can explain what I know about >how a calculator can be useful and how it can be useless, but I concede >my viewpoint is quite limited. Others' experiences can be informative. There exists a collection (I don't know if it's a web site or a newsgroup format) of stories about the dangers of too much calculator reliance. I can't remember the guy's name but he did ask me if he could collect a story I told. Ask about it in alt.folklore.computers. Somebody over there will know; they know everything or knows somebody who does ;-). /BAH Subtract a hundred and four for e-mail. === Subject: One interesting Partial differential equation I am doing Ph. D in astrophyics. In my work, i came across the following equation du = f in $omega$ u = 0 on the boundary (*) where i) f is a differential k+1 -form in $omega$, coefficients in L^p ($omega$), 1 I am doing Ph. D in astrophyics. In my work, i came across the > following equation > du = f in $omega$ > u = 0 on the boundary (*) > where i) f is a differential k+1 -form in $omega$, coefficients in > L^p ($omega$), 1 ii) $omega$ is an open, bounded subset in R^n > I need some result regarding the existence and regularity of the > solution in the appropriate sobolev spaces for the above equation in > my work. > My questions are : 1. Is there any necessary and sufficient condition > known regarding the existence of solution, ie. a differential k- form > in the appropriate sobolev spaces, satisfying the equation? > 2. what happens if we allow some non zero boundary condition? > Can anybody help me to solve the problem? Unless I'm misreading the ascii, this is known as the Neumann problem. I would look for that in the index of a reference on PDE's or integral equations. LH === Subject: Re: Finding the roots of Q(X)=X^9+... > Have a look at > http://www.library.cornell.edu/nr/bookcpdf/c9-5.pdf > Let phi=(1+sqrt(5))/2 and > Q(X)= X^9 - (4*phi^2)X^7 + a(3)X^6 + ... +a(9) . > It is known that all roots x_1,x_2,...,x_9 of Q(X) are real, and > moreover that Q(X) > 0 for all x in ( phi/3 , infty) . > It's possible to find the roots of Q(X) ? How ? Hi Roland, thank you for the interesting reference. However, by means of my own ,,computer I found x_1= - 8*phi/3 , x_2=x_3=...=x_9= phi/3 . === Subject: Re: Finding the roots of Q(X)=X^9+... Hi Alex, Because you did not defined a(i), I cannot check your results but I fear you used Newton-Raphson method without tricks avoiding to find several time the same root. The eigenvalue method is slower but more robust and provides you with all the roots, real or imaginary while using linear method easy to implement since eigenvalue search are present in most of numerical libraries (Lapack, Jama,.....). Roland > Have a look at > http://www.library.cornell.edu/nr/bookcpdf/c9-5.pdf > Let phi=(1+sqrt(5))/2 and > Q(X)= X^9 - (4*phi^2)X^7 + a(3)X^6 + ... +a(9) . > It is known that all roots x_1,x_2,...,x_9 of Q(X) are real, and > moreover that Q(X) > 0 for all x in ( phi/3 , infty) . > It's possible to find the roots of Q(X) ? How ? > Hi Roland, > thank you for the interesting reference. > However, by means of my own ,,computer I found > x_1= - 8*phi/3 , x_2=x_3=...=x_9= phi/3 . === Subject: Re: Finding the roots of Q(X)=X^9+... > Hi Alex, > Because you did not defined a(i), I cannot check > your results but I fear you used Newton-Raphson > method without tricks avoiding to find several time > the same root. Hi Roland, there is only one polynomial of form Q(X)=X^9 - (4*phi^2)X^7 +a(3)X^6+...+a(9) having properties: i) all roots are real , ii) Q(X) > 0 for all X in (phi/3,infty) . More exactly Q(X)=(X+ 8*phi/3)*(X-phi/3)^8= X^9 -(4*phi^2)X^7+.... . === Subject: Equivalent Binomial Random Variables Through experimentation I have come across with the fact that it is possible to observe some sort of equivalence among independent binomial random variables with mutually distinct parameters n and p. For example: X1(n1= 17 ; p1 = 1 - exp(5/100)); X2(n2 = 325; p2 = 1 - exp(5/2000)); X3(n3 = 1619; p3= 1 - exp(5/10000)). If we plot the probability mass function for X1, X2 and X3 we will see they are very close to each other. The differences will be concentrated in the respective tails where the probability values may be disregarded. Does anybody know why this happens? Is there any property related to the values of the parameters? A formal proof ? Would anybody point me to a good reference on this? Carlo. === Subject: Re: Equivalent Binomial Random Variables > Through experimentation I have come across with the fact that > it is possible to observe some sort of equivalence among independent > binomial random variables with mutually distinct parameters n and p. > For example: > X1(n1= 17 ; p1 = 1 - exp(5/100)); > X2(n2 = 325; p2 = 1 - exp(5/2000)); > X3(n3 = 1619; p3= 1 - exp(5/10000)). I assume you mean p = exp(#) - 1 > If we plot the probability mass function for X1, X2 and X3 we will > see they are very close to each other. The differences will be > concentrated in the respective tails where the probability values may > be disregarded. > Does anybody know why this happens? Is there any property related > to the values of the parameters? A formal proof ? Would anybody point > me to a good reference on this? Consider binomial RVs with p = 1/n for n = 10, 100, 1000, ... . What are their means and variances? How about the more general case p = m/n, m > 0, n = M, 10M, 100M, ..., where M is any integer > m. What is the distribution of the limiting case? === Subject: Re: Equivalent Binomial Random Variables > Through experimentation I have come across with the fact that > it is possible to observe some sort of equivalence among independent > binomial random variables with mutually distinct parameters n and p. > For example: > X1(n1= 17 ; p1 = 1 - exp(5/100)); > X2(n2 = 325; p2 = 1 - exp(5/2000)); > X3(n3 = 1619; p3= 1 - exp(5/10000)). > I assume you mean p = exp(#) - 1 If we plot the probability mass function for X1, X2 and X3 we will > see they are very close to each other. The differences will be > concentrated in the respective tails where the probability values may > be disregarded. > I am sorry Ray. I forgot the negatival signal. In fact I meant the parameters p are: p1 = 1 - exp(-5/100)); p2 = 1 - exp(-5/2000)); p3= 1 - exp(-5/10000)). All other remaining statements remain true. > Does anybody know why this happens? Is there any property related > to the values of the parameters? A formal proof ? Would anybody point > me to a good reference on this? > Consider binomial RVs with p = 1/n for n = 10, 100, 1000, ... . > What are their means and variances? > How about the more general case p = m/n, m > 0, n = M, 10M, 100M, ..., > where M is any integer > m. > What is the distribution of the limiting case? === Subject: Re: Equivalent Binomial Random Variables > I am sorry Ray. I forgot the negatival signal. In fact > I meant the parameters p are: > p1 = 1 - exp(-5/100)); > p2 = 1 - exp(-5/2000)); > p3= 1 - exp(-5/10000)). > All other remaining statements remain true. And my point also remains unchanged: that you should start by considering what happens to the mean and variance when you increase n and decrease p in a way that leaves n*p constant, which is what your values approximate. === Subject: Re: Equivalent Binomial Random Variables > I am sorry Ray. I forgot the negatival signal. In fact > I meant the parameters p are: > p1 = 1 - exp(-5/100)); > p2 = 1 - exp(-5/2000)); > p3= 1 - exp(-5/10000)). > All other remaining statements remain true. > And my point also remains unchanged: that you should start by considering > what happens to the mean and variance when you increase n and decrease p > in a way that leaves n*p constant, which is what your values approximate. Unless I misunderstood your point, the mean and variance will remain unchanged. That is, X1, X2 and X3 will have approximately the same mean and variance, but is enough to let me state that they are equivalent? If not, what are the conditions in order to state so? === Subject: Re: Equivalent Binomial Random Variables > Through experimentation I have come across with the fact that > it is possible to observe some sort of equivalence among independent > binomial random variables with mutually distinct parameters n and p. > For example: > X1(n1= 17 ; p1 = 1 - exp(5/100)); > X2(n2 = 325; p2 = 1 - exp(5/2000)); > X3(n3 = 1619; p3= 1 - exp(5/10000)). > I assume you mean p = exp(#) - 1 If we plot the probability mass function for X1, X2 and X3 we will > see they are very close to each other. The differences will be > concentrated in the respective tails where the probability values may > be disregarded. Does anybody know why this happens? Is there any property related > to the values of the parameters? A formal proof ? Would anybody point > me to a good reference on this? > Consider binomial RVs with p = 1/n for n = 10, 100, 1000, ... . > What are their means and variances? > How about the more general case p = m/n, m > 0, n = M, 10M, 100M, ..., > where M is any integer > m. > What is the distribution of the limiting case? I am sorry Ray. I forgot the negatival signal. In fact I meant the parameters p are: p1 = 1 - exp(-5/100)); p2 = 1 - exp(-5/2000)); p3= 1 - exp(-5/10000)). All other remaining statements remain true. === Subject: Re: Quanta and Cakes Mr. Heller mistake is, that he suppose that only two cells |l> and |r> Total number of all possible configurations is classicaly Z = N + N * (N - 1). (there is N of such states) second N * (N - 1) stands for Now N/2 of the states is in the right and N/2 is the left part of the box. L = N/2 + N/2 * (N/2 -1) same for right and one left, one right: LR = N * N/2 If N=2 (only two cells are possible) you obtain for the probabilities: P(L) = 1/4 P(R) = 1/4 P(LR)= 1/2 Fault of this approach (and this is the point of Mr. Heller's paper) is that I've summed the same states twice; since state (a(1),b(2)) is indistinguishable from the state (a(2),b(1)) (a, b mark some cells), it should be treated only once in sums of all possible states. Than you have Z = N + (N * (N - 1))/2 L = N/2 + (N/2 * (N/2 -1))/2 LR = (N * N/2)/2 and indeed for N=2 P(L) = 1/3 P(R) = 1/3 P(LR)= 1/3 (like V/(h*h*h), h is Planck constant, V is box volume). In fact you can make limit N -> infinity. You'll than get P(L) = 1/4 P(R) = 1/4 P(LR)= 1/2 regardless if classical or quantum approach is used. So you can safely continue with cakes and quanta discussion, except example with the box. Palo >> Hi >> Partly right. The point of the thread is that although > probability of ll is 1/4 > probability of rr is 1/4 > probability of lr is 1/2 >> are the probabilities that physicists would have expected to find, >> the probabilities that they have found are: > probability of ll is 1/3 > probability of rr is 1/3 > probability of lr is 1/3 >> I'm affraid that you are not right. They will find the above >> probabilities (1/4, 1/4, 1/2). I agree, that you cannot distiguish between the >> three configurations have the same probablity. They actually do not. >> The question then arises, what is it about quanta that is >> responsible for the difference between classical statistics >> and quantum statistics. Some physicists have attributed >> this difference to what they refer to as a 'lack of individuality' >> in quanta. >> Within the quantum statistics, the states which differ only by exchange >> counted only once into the statistical sums. >> states are numbered and you work with the ocupation numbers of >> particular state. >> all. >> Did I missed some point? >> Palo > lot of space to the 1/3,1/3,1/3-1/2/1/4/1/4 question. > --John -- +--------------------------------------------------------------+ | Pavol Domin | | | | Telephone: (56) (32) 654 507 | | Fax: (56) (32) 79 76 56 | | E-mail: pavol.domin@usm.cl | | | | Present address: | | Departamento de Fisica, Universidad Tecnica | | Federico Santa Maria, Casilla 110-V, | | Valparaiso, Chile. | +--------------------------------------------------------------+ === Subject: Re: Quanta and Cakes > bmu5p6$qp94u$1@ID-186372.news.uni-berlin.de>... Theres a misspelling: The authors name is Teller, not Heller! [Tellers homepage: http://www-philosophy.ucdavis.edu/teller/] PH P.S.: In the Usenet its customary to put ones own text BELOW the quoted one! === Subject: Re: Quanta and Cakes >> bmu5p6$qp94u$1@ID-186372.news.uni-berlin.de>... > There´s a misspelling: > The author´s name is Teller, not Heller! OK, it came to me like Heller, I'm sorry. > [Teller´s homepage: http://www-philosophy.ucdavis.edu/teller/] Oh the man is professor. But still, his example with the box with two some better example, with only two quantum states available (I can not find no realistic one). Palo > PH > P.S.: > In the Usenet it´s customary to put one´s own text BELOW the quoted one! === Subject: Re: Quanta and Cakes > Mr. Heller mistake is, that he suppose that only two cells |l> and |r> > Total number of all possible configurations is classicaly > Z = N + N * (N - 1). > (there is N of such states) second N * (N - 1) stands for > Now N/2 of the states is in the right and N/2 is the left part of the box. > L = N/2 + N/2 * (N/2 -1) > same for right and one left, one right: > LR = N * N/2 > If N=2 (only two cells are possible) you obtain for the probabilities: > > P(L) = 1/4 > P(R) = 1/4 > P(LR)= 1/2 > Fault of this approach (and this is the point of Mr. Heller's paper) is > that I've summed the same states twice; since state (a(1),b(2)) is > indistinguishable > from the state (a(2),b(1)) (a, b mark some cells), it should be treated only > once in sums of all possible states. > Than you have > Z = N + (N * (N - 1))/2 > L = N/2 + (N/2 * (N/2 -1))/2 > LR = (N * N/2)/2 > > and indeed for N=2 > P(L) = 1/3 > P(R) = 1/3 > P(LR)= 1/3 > (like V/(h*h*h), h is Planck constant, V is box volume). > In fact you can make limit N -> infinity. You'll than get > P(L) = 1/4 > P(R) = 1/4 > P(LR)= 1/2 > regardless if classical or quantum approach is used. > So you can safely continue with cakes and quanta discussion, except > example with the box. > Palo Doesn't the question still arise why the probabilities are 1/3,1/3,1/3 for the case that Heller is discussing, rather than 1/2,1/4,1/4? John >> Hi >> Partly right. The point of the thread is that although > probability of ll is 1/4 > probability of rr is 1/4 > probability of lr is 1/2 >> are the probabilities that physicists would have expected to find, >> the probabilities that they have found are: > probability of ll is 1/3 > probability of rr is 1/3 > probability of lr is 1/3 >> I'm affraid that you are not right. They will find the above >> probabilities (1/4, 1/4, 1/2). I agree, that you cannot distiguish between the >> three configurations have the same probablity. They actually do not. >> The question then arises, what is it about quanta that is >> responsible for the difference between classical statistics >> and quantum statistics. Some physicists have attributed >> this difference to what they refer to as a 'lack of individuality' >> in quanta. >> Within the quantum statistics, the states which differ only by exchange >> counted only once into the statistical sums. >> states are numbered and you work with the ocupation numbers of >> particular state. >> all. >> Did I missed some point? >> Palo lot of space to the 1/3,1/3,1/3-1/2/1/4/1/4 question. > --John > -- > +--------------------------------------------------------------+ > | Pavol Domin | > | | > | Telephone: (56) (32) 654 507 | > | Fax: (56) (32) 79 76 56 | > | E-mail: pavol.domin@usm.cl | > | | > | Present address: | > | Departamento de Fisica, Universidad Tecnica | > | Federico Santa Maria, Casilla 110-V, | > | Valparaiso, Chile. | > +--------------------------------------------------------------+ === Subject: Convergence of infinite sum; can anyone solve this? I am trying to prove the convergence of the following series; The limit as N approaches infinity of; (1/(ln(N)) times (sum from 1 to N) of |mu(N)|/N, where |mu(N)| is the absolute value of the Mobius function, and ln(N) is the natural log of N. |mu(N)| is 1 whenever N is square-free, and 0 otherwise. Bob Adams === Subject: Re: Convergence of infinite sum; can anyone solve this? Liz > I am trying to prove the convergence of the following series; > The limit as N approaches infinity of; > (1/(ln(N)) times (sum from 1 to N) of |mu(N)|/N, > where |mu(N)| is the absolute value of the Mobius function, and ln(N) > is the natural log of N. > |mu(N)| is 1 whenever N is square-free, and 0 otherwise. We know that (1/(ln(N)) times (sum from 1 to N) of 1/N has a limit at infinity, from the def'n of Euler's constant. So, we are down to showing that this has a limit: (1/(ln(N)) times (sum from 1 to N) of k(N)/N where k(N)=1 if N is divisible by a square, else k(N)=0. But that expression must go to zero, because sum(1/N^2) is finite. LH === Subject: Re: Convergence of infinite sum; can anyone solve this? >Liz >> I am trying to prove the convergence of the following series; >> The limit as N approaches infinity of; >> (1/(ln(N)) times (sum from 1 to N) of |mu(N)|/N, Presumably meaning 1/ln(N) sum_{n=1}^N |mu(n)|/n >> where |mu(N)| is the absolute value of the Mobius function, and ln(N) >> is the natural log of N. >> |mu(N)| is 1 whenever N is square-free, and 0 otherwise. >We know that >(1/(ln(N)) times (sum from 1 to N) of 1/N >has a limit at infinity, from the def'n of Euler's constant. So, we are down >to showing that this has a limit: >(1/(ln(N)) times (sum from 1 to N) of k(N)/N >where k(N)=1 if N is divisible by a square, else k(N)=0. >But that expression must go to zero, because sum(1/N^2) is finite. No, that expression doesn't go to 0. For example, k(4 j) = 1 for any positive integer j, so 1/ln(N) sum_{n=1}^N k(n)/n >= 1/ln(N) sum_{j=1}^floor(N/4) 1/(4j) ~= ln(N/4)/(4 ln(N)) -> 1/4 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: Convergence of infinite sum; can anyone solve this? > Liz > I am trying to prove the convergence of the following series; > The limit as N approaches infinity of; > (1/(ln(N)) times (sum from 1 to N) of |mu(N)|/N, > where |mu(N)| is the absolute value of the Mobius function, and ln(N) > is the natural log of N. > |mu(N)| is 1 whenever N is square-free, and 0 otherwise. > We know that > (1/(ln(N)) times (sum from 1 to N) of 1/N > has a limit at infinity, from the def'n of Euler's constant. So, we are down > to showing that this has a limit: > (1/(ln(N)) times (sum from 1 to N) of k(N)/N > where k(N)=1 if N is divisible by a square, else k(N)=0. > But that expression must go to zero, because sum(1/N^2) is finite. > LH Actually, |mu(n)| is 1 if n is NOT divisible by a square. It is easy to numerically show that it seems to converge, but I would like to see a proof. Bob === Subject: Re: Convergence of infinite sum; can anyone solve this? Liz > Larry Hammick > Liz > I am trying to prove the convergence of the following series; > > The limit as N approaches infinity of; > > (1/(ln(N)) times (sum from 1 to N) of |mu(N)|/N, > where |mu(N)| is the absolute value of the Mobius function, and ln(N) > is the natural log of N. > > |mu(N)| is 1 whenever N is square-free, and 0 otherwise. > We know that > (1/(ln(N)) times (sum from 1 to N) of 1/N > has a limit at infinity, from the def'n of Euler's constant. So, we are down > to showing that this has a limit: > (1/(ln(N)) times (sum from 1 to N) of k(N)/N > where k(N)=1 if N is divisible by a square, else k(N)=0. > Actually, |mu(n)| is 1 if n is NOT divisible by a square. True, but I'm just looking at the difference between the two sequences. We get home either way. > But that expression must go to zero, because sum(1/N^2) is finite. See Bob Israel's correction to this one:) LH === Subject: Re: Convergence of infinite sum; can anyone solve this? >> Liz >> I am trying to prove the convergence of the following series; >> >> The limit as N approaches infinity of; >> >> (1/(ln(N)) times (sum from 1 to N) of |mu(N)|/N, >> where |mu(N)| is the absolute value of the Mobius function, and ln(N) >> is the natural log of N. >> > Actually, |mu(n)| is 1 if n is NOT divisible by a square. It is easy > to numerically show that it seems to converge, but I would like to see > a proof. The limit ought to be 1/zeta(2) = 6/pi^2. Let f(s) = sum_{n=1}^infinity |mu(n)|/n^s, convergent for Re(s) > 1. Then f(s) = zeta(s)/zeta(2s). So let g(s) = f(s) - zeta(s)/zeta(2). This extends to a holomorphic function on Re(s) > 1/2. If we put g(s) = sum b_n/n^s then we need to show that sum_{n=1}^N b_n/n = o(log N). There must be some Dirichlet series/Mellin transform trick for this (the sum of b_n/n ought to be convergent) but I don't see it right now :-( -- 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: Convergence of infinite sum; can anyone solve this? >> Liz >> I am trying to prove the convergence of the following series; >> >> The limit as N approaches infinity of; >> >> (1/(ln(N)) times (sum from 1 to N) of |mu(N)|/N, >> where |mu(N)| is the absolute value of the Mobius function, and ln(N) >> is the natural log of N. >> Actually, |mu(n)| is 1 if n is NOT divisible by a square. It is easy > to numerically show that it seems to converge, but I would like to see > a proof. > The limit ought to be 1/zeta(2) = 6/pi^2. > Let f(s) = sum_{n=1}^infinity |mu(n)|/n^s, convergent for Re(s) > 1. > Then f(s) = zeta(s)/zeta(2s). So let > g(s) = f(s) - zeta(s)/zeta(2). This extends to a holomorphic function > on Re(s) > 1/2. If we put g(s) = sum b_n/n^s then we need to show that > sum_{n=1}^N b_n/n = o(log N). There must be some Dirichlet series/Mellin > transform trick for this (the sum of b_n/n ought to be convergent) > but I don't see it right now :-( proof, please let me know; I am stuck and cannot proceed furthur until this is put to rest. Bob Adams === Subject: Re: Strange Random Variables Problem > For example it's clear (I hope) that we can pick a sequence of > constants such that X_n/a_n converges to zero in expectation: i.e., > the real series E[X_n/a_n] converges to zero. That's one kind of > dominance. > I guess one problem I have is that the above claim is not clear (the claim > about expectation). Why is this true? Already answered and proposed as unecessary by others more accomplished. One of my quandaries about your question may have been one of yours: How much structure is implied in the assertion X_n is a random variable? I would have liked to assume each X_n has a mean and variance, but existence of the variance is not guaranteed, possibly not even the mean. The necessary and sufficient condition for your proof is apparently that the probability in the tails can be made as small as we like for sufficiently large cut-offs. This allows us to pick the a_n so that the total probability that {X_n/a_n} will poke outside some pre-defined convergent sequence can be made finite, hence (by some lemma) this poking will occur almost surely only a finite number of times. Hence there is almost surely a last poking beyond which our sequence is bounded by a convergent sequence, hence is convergent. Simple when experts show you how. :-) === Subject: Re: Strange Random Variables Problem I haven't looked at this sort of stuff in a while, but I think that this might work. Pick a_n so that P(x_n/a_n<1/n)<1/2^n. The probability that x_n/a_n exceeds 1/n for some n>N is bounded by the sum from n=N+1 to infinity of 1/2^n, which approaches 0 as N-> infinity. Therefore the sequence x_n/a_n will converge to 0 with probability 1. (If you're familiar with the Borel-Cantelli lemmas, this is basically an application of the first of them. We carefully chose our a_n so that the events {x_n/a_n<1/n} had a finite total sum of probabilities. By the 1st B-C lemma, almost surely only a finite number of these events will occur, which is sufficient for convergence.) === Subject: Re: Strange Random Variables Problem >>Show that for each sequence of random variables {X_n : n = 1,2,...} there >>exist a sequence of constants {a_n : n = 1,2,...} such that X_n/a_n >>converges to zero with probability one. Note : There is no hypothesis >>whatsoever on the random variables. >>How can I prove this? This seems like a strange assertion to me that lacks >>intuition. >> >First, I don't see why (as another poster mentioned) we would need the >X_n's defined on the same probability space. The resulting random >sequence lives on the infinite product of the probability spaces of >the X_n's, which we can construct without needing the individual >spaces to be the same. The definition of almost sure convergence (= convergence with probability 1) requires the r.v.'s be defined on the same probability space. We need to show that the set {s: X_n(s)/a_n -> 0} contains a set of probability measure 1. By putting the r.v.'s on the product space, you have made them independent; i.e., you added a hypothesis not included in the problem. >Now, for each n, choose a_n so large that P(|X_n| > a_n/n) < 2^{-n}. >By the Borel-Cantelli Lemma, the probability that |X_n| > a_n/n for >infinitely many n is zero. Thus, with probability one, |X_n/a_n| <= >1/n for all n sufficiently large, i.e. X_n/a_n --> 0. Now this works. As an intermediate step, the reader might note that U{{|X_n/a_n| >= 1/k i.o.} : k in N} is a subset of {|X_n/a_n| >= 1/n i.o.}. If this is homework, Steve, you should cite your sources in submitted work. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Strange Random Variables Problem >>Show that for each sequence of random variables {X_n : n = 1,2,...} there >>exist a sequence of constants {a_n : n = 1,2,...} such that X_n/a_n >>converges to zero with probability one. Note : There is no hypothesis >>whatsoever on the random variables. >> >>How can I prove this? This seems like a strange assertion to me that lacks >>intuition. >> >> >First, I don't see why (as another poster mentioned) we would need the >X_n's defined on the same probability space. The resulting random >sequence lives on the infinite product of the probability spaces of >the X_n's, which we can construct without needing the individual >spaces to be the same. > The definition of almost sure convergence (= convergence with > probability 1) requires the r.v.'s be defined on the same probability > space. We need to show that the set {s: X_n(s)/a_n -> 0} contains a > set of probability measure 1. By putting the r.v.'s on the product > space, you have made them independent; i.e., you added a hypothesis not > included in the problem. >Now, for each n, choose a_n so large that P(|X_n| > a_n/n) < 2^{-n}. >By the Borel-Cantelli Lemma, the probability that |X_n| > a_n/n for >infinitely many n is zero. Thus, with probability one, |X_n/a_n| <= >1/n for all n sufficiently large, i.e. X_n/a_n --> 0. > Now this works. As an intermediate step, the reader might note that > U{{|X_n/a_n| >= 1/k i.o.} : k in N} is a subset of {|X_n/a_n| >= > 1/n i.o.}. > If this is homework, Steve, you should cite your sources in submitted work. No problemo. THanks for the help. > -- > Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Strange Random Variables Problem >> > >> >>Show that for each sequence of random variables {X_n : n = 1,2,...} >> >> >>there >> >>exist a sequence of constants {a_n : n = 1,2,...} such that X_n/a_n >>converges to zero with probability one. Note : There is no hypothesis >>whatsoever on the random variables. >> >>[...] >> >For example it's clear (I hope) that we can pick a sequence of >constants such that X_n/a_n converges to zero in expectation: i.e., >the real series E[X_n/a_n] converges to zero. That's one kind of >dominance. > >>I guess one problem I have is that the above claim is not clear (the claim >>about expectation). Why is this true? Because then you are just dealing with a sequence of numbers. For example, take a_n = n EX_n if EX_n <> 0, a_n = 1 otherwise. Actually, you need E|X_n| finite for sufficiently large n for this to work. >> >It doesn't matter. It's a fact that |X_n| is finite almost surely. It >follows that P(|X_n| > a) -> 0 as a -> infinity. Hence you can >pick a_n such that P(|X_n| > a_n/n} < ___, and if you fill in the >blank properly that shows that X_n -> in probability. So now you have shown convergence in probability. The original question requires almost sure convergence. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Strange Random Variables Problem >>[...] >>It doesn't matter. It's a fact that |X_n| is finite almost surely. It >>follows that P(|X_n| > a) -> 0 as a -> infinity. Hence you can >>pick a_n such that P(|X_n| > a_n/n} < ___, and if you fill in the >>blank properly that shows that X_n -> in probability. >> >So now you have shown convergence in probability. The original question >requires almost sure convergence. Uh, that was a typo. Change that paragraph to this: >>It doesn't matter. It's a fact that |X_n| is finite almost surely. It >>follows that P(|X_n| > a) -> 0 as a -> infinity. Hence you can >>pick a_n such that P(|X_n| > a_n/n} < ___, and if you fill in the >>blank properly that shows that X_n -> almost surely. (It's true that some ways of filling in the blank would give convergence in probability but not almost surely. Other ways of filling in the blanks give almost sure convergence. Hint: If sum(P(E_n)) is finite then it follows that ___ .) ************************ David C. Ullrich === Subject: Re: Strange Random Variables Problem > Show that for each sequence of random variables {X_n : n = 1,2,...} there > exist a sequence of constants {a_n : n = 1,2,...} such that X_n/a_n > converges to zero with probability one. Note : There is no hypothesis > whatsoever on the random variables. Use the Chebyshev's inequality: For a random variable X with mean m and variance s^2, we have Prob[|X-m|>=ks]<=1/k^2. Choose the sequence {b_n} in such a way that X_n<=b_n with prob 1. Then choose {a_n} so that lim(b_n/a_n)=0 as n goes to infinity. === Subject: Re: Boolean Algebra - Arithmetic Relationship > ... > I'd like to recommend some reading. For logic generally, see Tarski An > Introduction to Logic and the Methodology of the Deductive Sciences > OUP. (Dreadful title, best book.) For recursion theory I'm at a bit of > a loss, how about Boolos & Jeffrey Computability and Logic CUP? And for set theory P Suppes Axiomatic Set Theory Dover which shows how the real numbers are defined in set theory. Also E Landau Foundations in terms of ordered n-tuples of real numbers. -- G.C. === Subject: Re: Boolean Algebra - Arithmetic Relationship > Mathematics seems like an entirely chaotic enterprise at the > macrosopic level, it hard to believe that anyone could exercise > leadership to influence it in a particular direction. > Is there some master reducibilty graph that mathematicians use to > show when one symbology is reducible to another? > Do mathematicians reduce or do they build up? Sometimes they > uncover the commonality between two apparently different branches of > mathematics--but that's not reducing it's finding a common foundation. > Category theory is a good example. > This is rather disheartening, I was hoping that reducibility might > lead to a hierarchical organization of mathematic symbology. > I was under the impression that some symbologies where, in fact, > conceptual subsets of others. Or in other words, everything that can > be proven in symbology X can be proven in symbology Y, but Y can prove > even more. I don't know what you mean by symbology but if you mean deductive system then X = propositional calculus and Y = first order logic, would be an example. But it seems to me that this Y is built up on this X. (Non conservative extension is the jargon.) This may be of interest to you: http://www.hep.upenn.edu/~max/toe.html . See Figure 1. Relationships between various basic mathematical structures. there. -- G.C. === Subject: Re: Boolean Algebra - Arithmetic Relationship > Mathematics seems like an entirely chaotic enterprise at the > macrosopic level, it hard to believe that anyone could exercise > leadership to influence it in a particular direction. > Is there some master reducibilty graph that mathematicians use to > show when one symbology is reducible to another? > > Do mathematicians reduce or do they build up? Sometimes they > uncover the commonality between two apparently different branches of > mathematics--but that's not reducing it's finding a common foundation. > Category theory is a good example. > This is rather disheartening, I was hoping that reducibility might > lead to a hierarchical organization of mathematic symbology. > I was under the impression that some symbologies where, in fact, > conceptual subsets of others. Or in other words, everything that can > be proven in symbology X can be proven in symbology Y, but Y can prove > even more. > I don't know what you mean by symbology One of my dictionaries says symbology /smbldi/ n. M19. [Irreg. f. SYMBOL n.1 + -LOGY.] The branch of knowledge that deals with the use of symbols; gen. the use of symbols, symbolism; symbols collectively.symbological a. M19. --------------------------------------------------------- Excerpted from Oxford Talking Dictionary Copyright 1998 The Learning Company, Inc. All Rights Reserved. and I can't relate that to what you're writing. -- G.C. === Subject: Re: Boolean Algebra - Arithmetic Relationship > I don't know what you mean by symbology > One of my dictionaries says > symbology /smbldi/ n. M19. [Irreg. f. SYMBOL n.1 + -LOGY.] The branch of > knowledge that deals with the use of symbols; gen. the use of symbols, > symbolism; symbols collectively.symbological a. M19. > and I can't relate that to what you're writing. Indeed symbology was a poor choice of words on my part. I meant a deductive system. (Although, I am interested in how certain symbolic arrangements assist in performing certain operations) I guess I was taking the symbolic representation of some logical construct to imply we are using a particular deductive system. In that way, I guess I was assuming the terms symbology and deductive system to be sematically synonymous. As you can see I'm fairly handicapped in the jargon department. I've been thinking about. Are there certain things common to all deductive systems? 1)An Explicit Symbolic Grammar 2)Certain fundamental assumptions/premises 3)A set of operations to manipulate a logical construct from one form into another. I'm also curious about proofs. As each proof is made, can it be added to the current set of operations as a de facto new operation? Thus adding a new tool the the mathematician's arsenal. Does every deductive system have an infinite set of proofs? I'm told Godel's Proof shows that our system of aritmetic cannot be proven consistent. Someday we might find out that 2+2=63 or something and our whole system will collapse. (Is this a fair characterization?) Has there even been a case where some primitive society used a deductive system for a while that was eventually determined to be inconsistent? Is it possible to construct an inconsistent deductive system? If so, why do all our current systems seem to pass the test? Given man's track record of scientific/technological/mathematical progress, I would assume he would have bungled up his deductive systems a few times before getting it right. However, I'm unaware of any evidence of this sort of thing happening. Once again thank you for answering all my questions. -Steve === Subject: Re: Boolean Algebra - Arithmetic Relationship > I don't know what you mean by symbology > One of my dictionaries says > symbology /smbldi/ n. M19. [Irreg. f. SYMBOL n.1 + -LOGY.] The branch of > knowledge that deals with the use of symbols; gen. the use of symbols, > symbolism; symbols collectively.symbological a. M19. > and I can't relate that to what you're writing. > Indeed symbology was a poor choice of words on my part. I meant a > deductive system. (Although, I am interested in how certain symbolic > arrangements assist in performing certain operations) I guess I was > taking the symbolic representation of some logical construct to imply > we are using a particular deductive system. In that way, I guess I > was assuming the terms symbology and deductive system to be > sematically synonymous. As you can see I'm fairly handicapped in the > jargon department. > I've been thinking about. > Are there certain things common to all deductive systems? > 1)An Explicit Symbolic Grammar > 2)Certain fundamental assumptions/premises > 3)A set of operations to manipulate a logical construct from one form > into another. A language/logic usually has: (1) A set of symbols (usually infinitely many). (2) Rules for forming strings of symbols that are deemed well-formed. (3) The positing of some of those well formed formulae, known as axioms. (4) Rules for deducing well formed formulae from the axioms and from previously deduced well formed formulae. Such deduced well formed formulae are known as theorems (of the logic). The deduction is known as a proof of the theorem. > I'm also curious about proofs. As each proof is made, can it be added > to the current set of operations as a de facto new operation? Thus > adding a new tool the the mathematician's arsenal. Yes, as each theorem, in effect, has the status of an axiom and may be used in subsequent proofs. > Does every deductive system have an infinite set of proofs? Yes. I suppose that one could construct systems with only a finite number of proofs but they'd be very artificial. > I'm told Godel's Proof shows that our system of aritmetic cannot be > proven consistent. Someday we might find out that 2+2=63 or something > and our whole system will collapse. (Is this a fair characterization?) Almost. Neither arithmetic, nor anything else, A say, can be proved consistent without appealing to a stronger system, S say. But if one feels the need for a consistency proof of A, what faith would one have in a stronger system S? > Has there even been a case where some primitive society used a > deductive system for a while that was eventually determined to be > inconsistent? Yes. I take it that our society is a primitive one. A deductive system of Quine's was shown to be inconsistent, and one of Church's that I think was never in the public eye. An especially famous case was Russell finding an inconsistency in a system of Frege's. > Is it possible to construct an inconsistent deductive system? If so, Yep. Add as an axiom any logically false well formed formula, say P and not-P (but it depends on the language, see above). > why do all our current systems seem to pass the test? Given man's > track record of scientific/technological/mathematical progress, I > would assume he would have bungled up his deductive systems a few > times before getting it right. However, I'm unaware of any evidence > of this sort of thing happening. Look up the cases mentioned above. > Once again thank you for answering all my questions. > -Steve -- G.C. === Subject: Re: Boolean Algebra - Arithmetic Relationship > [...] > why do all our current systems seem to pass the test? Given man's > track record of scientific/technological/mathematical progress, I > would assume he would have bungled up his deductive systems a few > times before getting it right. In my previous reply I neglected to mention that a number of logicians framed their rules of substitution wrongly: Frege, Hilbert, Russell. If _they_ can make mistakes, what hope have the rest of us got? [...] -- G.C. === Subject: Re: Boolean Algebra - Arithmetic Relationship > [...] > why do all our current systems seem to pass the test? Given man's > track record of scientific/technological/mathematical progress, I > would assume he would have bungled up his deductive systems a few > times before getting it right. > In my previous reply I neglected to mention that a number of logicians > framed their rules of substitution wrongly: Frege, Hilbert, Russell. If > _they_ can make mistakes, what hope have the rest of us got? > [...] So it would be fair to conjecture that the ancients didn't bungle up arithmetic or geometry several times because they based their deductive systems directly on natural phenomenon, which has a solid track record of not contradicting itself? Furthermore, is it fair to say all deductive systems we use today rest on the assumption that they are not contradictory. And that for these deductive systems to apply to empirical reality we must make another (more presumptuous) assumption that reality will obey the laws of the given deductive model. Strange it seems, to avoid contradiction in our systems we base them off of natural phenonmenon. Yet to apply them, we have to assume 'nature obeys the rules' and not the other way around. -Steve === Subject: Re: Boolean Algebra - Arithmetic Relationship > > > [...] > why do all our current systems seem to pass the test? Given man's > track record of scientific/technological/mathematical progress, I > would assume he would have bungled up his deductive systems a few > times before getting it right. > In my previous reply I neglected to mention that a number of logicians > framed their rules of substitution wrongly: Frege, Hilbert, Russell. If > _they_ can make mistakes, what hope have the rest of us got? > [...] > So it would be fair to conjecture that the ancients didn't bungle up > arithmetic or geometry several times because they based their > deductive systems directly on natural phenomenon, which has a solid > track record of not contradicting itself? > Furthermore, is it fair to say all deductive systems we use today rest > on the assumption that they are not contradictory. Yes, or almost yes, because from a contradiction one can prove anything: p ------ p or q not-p --------------- q Here p, not-p are the contradictory statements and q is any statement that you wish to deduce (or, indeed, any that you wish not to deduce). The above deduction can be formalized in any standard I'm sure that someone will deny one of the rules used above (such as p / p or q which is probably denied by relevant logicians) and someone may do so in support of some logic of contradictions. > And that for these > deductive systems to apply to empirical reality we must make another There are also deductive systems that are not intended to apply to reality. > (more presumptuous) assumption that reality will obey the laws of the > given deductive model. > Strange it seems, to avoid contradiction in our systems we base them > off of natural phenonmenon. Yet to apply them, we have to assume > 'nature obeys the rules' and not the other way around. > -Steve -- G.C. === Subject: Re: Boolean Algebra - Arithmetic Relationship Where do axioms come from? Are these simply highly reliable patterns that emerge out of empirical reality? Can axiom be thought of as a pattern or template, that given one form of truth can imply another? Can an axiom be modeled therefore as an edge in a directed graph? Given some truth (state) and an applicable axiom it moves me to another truth (state)? If this is true, doesn't it imply that all transformations can be modeled by a Turing machine? -Steve === Subject: JSH: So much fun Most of my life I've had this label: smart. But here on this newsgroup people made fun of me, and kept trying to argue with me. I could come out here and not be at the top of the heap but be considered on the bottom. I had people to play with. Usually, when people figure out what I can do they quit playing with me. They get tired of being beaten. But so many of you kept playing, and I've had a lot of fun. Now though I wonder if it's about over, so while I have the chance, I'll have even more fun. Now I call you names and curse. It's now much more fun. How long will you still play with me? Let's see. James Harris === Subject: Re: JSH: So much fun I don't know if this will be read, but... > Most of my life I've had this label: smart. This is probably true of many people on this newsgroup. Note that I've had two labels for much of my life: smart, no common sense. Some people would consider the second label more significant than the first. > But here on this newsgroup people made fun of me, and kept trying to > argue with me. I could come out here and not be at the top of the > heap but be considered on the bottom. And you still have more faith in being smart than in other people having knowledge through training. > I had people to play with. > Usually, when people figure out what I can do they quit playing with > me. You can toss a lot of symbols around and insult people. > They get tired of being beaten. > But so many of you kept playing, and I've had a lot of fun. We kept hoping you would choose to learn. > Now though I wonder if it's about over, so while I have the chance, > I'll have even more fun. > Now I call you names and curse. This is usually considered a negative reflection on your maturity and (possibly) intelligence. > It's now much more fun. > How long will you still play with me? Let's see. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: JSH: So much fun > Most of my life I've had this label: smart. This is more a reflection of the people who you've interacted with most of your life. ~ Chris === Subject: Re: JSH: So much fun >> Most of my life I've had this label: smart. >This is more a reflection of the people who you've interacted with >most of your life. Remember, you can't have a smart-ass without smart. === Subject: Re: JSH: So much fun >> Most of my life I've had this label: smart. >This is more a reflection of the people who you've interacted with >most of your life. > Remember, you can't have a smart-ass without smart. Corollary 1: ass is a necessary condition for smart-ass, but not for smart. === Subject: Re: So much fun James this is GOD. Actually, the label read smart-ass. The ass part of the label fell off, so by the time you could read (which I believe, was about the age of 16), you though it just said smart. Sorry about that oversight on my part, was meaning to get around to fixing it and all, and if you'll kindly report back to me, I'll add back on the ass part and rectify (if you'll pardon the pun) matters. === Subject: Re: JSH: So much fun >Most of my life I've had this label: smart. Then you should have no problem learning math by taking a math course or reading a math book. Smart means you're a good learner. The problem is when you think smart means you have knowledge without the benefit of learning. Sad news for you, James: the world is full of smart people. You aren't the only one. You knew that, right? - Randy === Subject: Re: JSH: So much fun > Most of my life I've had this label: smart. Now *there's* a truth-in-packaging error! Talk about false advertising! > But here on this newsgroup people made fun of me, and kept trying to > argue with me. Generally this happened when you were totally wrong, and were defending your errors as passionately as most decent persons defend the truth. > I could come out here and not be at the top of the > heap but be considered on the bottom. Whatever you're in, it's certainly some kind of heap. > I had people to play with. You should have stayed in your sandbox. At least you'd have been in appropriate company. > Usually, when people figure out what I can do they quit playing with > me. That's true of any kid that flings manure. > They get tired of being beaten. No, they get tired of being the undeserved targets of a raging cyber-crank. > But so many of you kept playing, and I've had a lot of fun. Fascinating (or should I say odd). It almost never sounded like you were having fun. > Now though I wonder if it's about over, so while I have the chance, > I'll have even more fun. Is this another one of your famous tricks where you promise to go away, but keep coming back? > Now I call you names and curse. Now? What do you mean Now? You've been doing that since the first person pointed out one of your errors. > It's now much more fun. So you delight in cursing and calling people names. Don't you think this reveals more about you than it does about them? > How long will you still play with me? Let's see. > James Wacky Harris If you want to play, like a little child, fine. But keep in mind that there are others who have grown up and have acquired some interest in the pursuit of truth. -- I know you are, but what am I? -- PeeWee Herman. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: So much fun > Now I call you names and curse. > Now? What do you mean Now? You've been doing that since the first > person pointed out one of your errors. Now he's going ONLY curse. How sad and pathetic. === Subject: Re: JSH: So much fun >Most of my life I've had this label: smart. >But here on this newsgroup people made fun of me, and kept trying to >argue with me. I could come out here and not be at the top of the >heap but be considered on the bottom. Yup. That's because most of the people here know something about math, and most of the others have some interest in _learning_ something about math. >I had people to play with. >Usually, when people figure out what I can do they quit playing with >me. >They get tired of being beaten. >But so many of you kept playing, and I've had a lot of fun. Sorry to break the news, but you haven't beaten anyone here. >Now though I wonder if it's about over, so while I have the chance, >I'll have even more fun. >Now I call you names and curse. Good plan. That's _certain_ to convince people that you're right about everything. >It's now much more fun. >How long will you still play with me? Let's see. >James Harris ************************ David C. Ullrich === Subject: Re: JSH: So much fun > Most of my life I've had this label: smart. Then why do you act so stupid? You insult people who are trying to offer genuine assistance. You are apparently unable to explain your argument in simple declarative English sentences. You make mistakes, apologise, then make more mistakes and issue more apologies, over and over for years on end. This is not smart behavior. === Subject: JSH: My lapdogs One of the funnier things in retrospect is how David Ullrich became so upset when I said years back that he'd acted as my lapdog in an instance, and in his anger he became my lapdog so many more times. I learned a lot from that exchange years ago. As I watched the newsgroup react to Ullrich's posts, even when I'd slyly inform that I was using him and the newsgroup, I learned more. Eventually, I knew how to use my little lapdog pack of Ullrich, Magidin, Winter, and Nora Baron. With them I could dominate the newsgroup at will to the point that from what I heard at times 25% of the traffic had to do with my little machinations. I've continually learned about you all and now I can wreak havok at will, as unfortunately I'm something of a social anarchist who has powerful mathematical tools at my disposal. Yup, so I LOVE that fascinating definition error that's over a hundred years old as its like a flame that draws so many moths. I mean, look at who I have so far: Barry Mazur, Andrew Granville, and Ralph McKenzie, among others. I can continue to use it, and continue to wreak social havok upon math society, and use my lapdogs to help me do it. As I said so many years ago, I will bring down the math establishment. And you know what? I'll enjoy myself doing it. It's just taking so damn long, but then again, math society has such a high position now that I guess it makes sense that it may take some time to draw it down to earth and raze it to the ground. James Harris === Subject: Re: JSH: My lapdogs > I mean, look at who I have so far: Barry Mazur, Andrew Granville, and > Ralph McKenzie, among others. e-mail? So you also have George Bush? - Randy === Subject: Re: JSH: My lapdogs > I've continually learned about you all and now I can wreak havok at > will, as unfortunately I'm something of a social anarchist who has > powerful mathematical tools at my disposal. Creating a high volume of posts is not a mathematical tool. > Yup, so I LOVE that fascinating definition error that's over a hundred > years old as its like a flame that draws so many moths. > I mean, look at who I have so far: Barry Mazur, Andrew Granville, and > Ralph McKenzie, among others. > I can continue to use it, and continue to wreak social havok upon math > society, and use my lapdogs to help me do it. > As I said so many years ago, I will bring down the math establishment. You overestimate the accuracy and significance of your work. The vast majority of the math establishment is probably completely ignorant of your work. > And you know what? I'll enjoy myself doing it. > It's just taking so damn long, but then again, math society has such a > high position now that I guess it makes sense that it may take some > time to draw it down to earth and raze it to the ground. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: JSH: My lapdogs > As I said so many years ago, I will bring down the math establishment. .--. .- -. __/ __ .--. / '----//o o----' ( _ ) ,____` = /`____, // `;'---' ` / / |-o // | | | | | | | |-o | | `--| |/ / '._ |-o | / '|| | .' | |-o |-` | | | |_/ _____| jsh | | | | | | '-.|.-'/ ] _|_ [ / | / / (___/` `___) _ _ === Subject: Re: My lapdogs > One of the funnier things in retrospect is how David Ullrich became so > upset when I said years back that he'd acted as my lapdog in an > instance, and in his anger he became my lapdog so many more times. > I learned a lot from that exchange years ago. > As I watched the newsgroup react to Ullrich's posts, even when I'd > slyly inform that I was using him and the newsgroup, I learned more. > Eventually, I knew how to use my little lapdog pack of Ullrich, > Magidin, Winter, and Nora Baron. > With them I could dominate the newsgroup at will to the point that > from what I heard at times 25% of the traffic had to do with my little > machinations. > I've continually learned about you all and now I can wreak havok at > will, as unfortunately I'm something of a social anarchist who has > powerful mathematical tools at my disposal. > Yup, so I LOVE that fascinating definition error that's over a hundred > years old as its like a flame that draws so many moths. > I mean, look at who I have so far: Barry Mazur, Andrew Granville, and > Ralph McKenzie, among others. > I can continue to use it, and continue to wreak social havok upon math > society, and use my lapdogs to help me do it. > As I said so many years ago, I will bring down the math establishment. > And you know what? I'll enjoy myself doing it. > It's just taking so damn long, but then again, math society has such a > high position now that I guess it makes sense that it may take some > time to draw it down to earth and raze it to the ground. > James Harris Why not stick to the math? I have a little tip for you, Instead of using the term dividing off, you should use the term factoring out. I'll show you why, if you haven't seen my example in another thread. Say we have the equation x^2-x-6=x-3 and we want to solve this for all x. We know that the graph of this is a parabola with a line touching it twice. Let's solve this and see what our x values are. If we get 0 on the right, we get x^2-2x-3=0, which is the same as (x-3)(x+1)=0 and obviously x=3 and x=-1. But let's say we divide off the common factor. I can rewrite the original equation as (x-3)(x+2)=x-3. If I divide both sides by x-3, I get x+2=1, so x=-1. Well, this is right, but what about the other solution? I lost it when I divided x-3 off. Therefore you shouldn't use the term dividing off. David Moran === Subject: Re: My lapdogs > One of the funnier things in retrospect is how David Ullrich became so > upset when I said years back that he'd acted as my lapdog in an > instance, and in his anger he became my lapdog so many more times. > I learned a lot from that exchange years ago. > As I watched the newsgroup react to Ullrich's posts, even when I'd > slyly inform that I was using him and the newsgroup, I learned more. > Eventually, I knew how to use my little lapdog pack of Ullrich, > Magidin, Winter, and Nora Baron. > With them I could dominate the newsgroup at will to the point that > from what I heard at times 25% of the traffic had to do with my little > machinations. > I've continually learned about you all and now I can wreak havok at > will, as unfortunately I'm something of a social anarchist who has > powerful mathematical tools at my disposal. James, am a little confused. It this post: === ] Subject: JSH: Ok, I'm a loser ] ] It's finally settled in that I'm just some pathetic loser. If I ] weren't so pathetic I'd just go away gracefully, but I'll send one ] more post, or who am I kidding, my patheticness is so great that I'll ] probably post yet again. ] ] I'm disgusting. I'm just a pile of . I should just die like so ] many of you have said. I hate myself. I despise this life. I'm ] nothing but a sick joke to be made fun of by those of you who have ] real educations. People who actually know something, when I know ] nothing. I'm just nothing. ] ] If I hadn't been such a disgusting human being I'd have come to this ] realization years ago instead of wasting your time. ] ] My life is nothing. I know nothing. I'm worth nothing. I'm just ] . ] ] Please forgive me. All your attacks were justified. ] ] James Harris you said that you are a pathetic loser and a pile of . Now you are saying that you are a social anarchist. So which, would you say, is the real you? -- Clive Tooth http://www.clivetooth.dk === Subject: Re: My lapdogs === >Subject: Re: My lapdogs >Message-id: >> One of the funnier things in retrospect is how David Ullrich became so >> upset when I said years back that he'd acted as my lapdog in an >> instance, and in his anger he became my lapdog so many more times. >> I learned a lot from that exchange years ago. >> As I watched the newsgroup react to Ullrich's posts, even when I'd >> slyly inform that I was using him and the newsgroup, I learned more. >> Eventually, I knew how to use my little lapdog pack of Ullrich, >> Magidin, Winter, and Nora Baron. >> With them I could dominate the newsgroup at will to the point that >> from what I heard at times 25% of the traffic had to do with my little >> machinations. >> I've continually learned about you all and now I can wreak havok at >> will, as unfortunately I'm something of a social anarchist who has >> powerful mathematical tools at my disposal. >James, am a little confused. It this post: === >] Subject: JSH: Ok, I'm a loser >] It's finally settled in that I'm just some pathetic loser. If I >] weren't so pathetic I'd just go away gracefully, but I'll send one >] more post, or who am I kidding, my patheticness is so great that I'll >] probably post yet again. >] I'm disgusting. I'm just a pile of . I should just die like so >] many of you have said. I hate myself. I despise this life. I'm >] nothing but a sick joke to be made fun of by those of you who have >] real educations. People who actually know something, when I know >] nothing. I'm just nothing. >] If I hadn't been such a disgusting human being I'd have come to this >] realization years ago instead of wasting your time. >] My life is nothing. I know nothing. I'm worth nothing. I'm just >] . >] Please forgive me. All your attacks were justified. >] James Harris >you said that you are a pathetic loser and a pile of . Now you are >saying that you are a social anarchist. >So which, would you say, is the real you? You mean somewhere there's a social anarchist who _isn't_ a pathetic loser pile of ? >-- >Clive Tooth >http://www.clivetooth.dk -- Mensanator Ace of Clubs === Subject: Re: JSH: My lapdogs >One of the funnier things in retrospect is how David Ullrich became so >upset when I said years back that he'd acted as my lapdog in an >instance, and in his anger he became my lapdog so many more times. >I learned a lot from that exchange years ago. Yes, it was funny, in retrospect. And I learned a lot from it as well. What I learned was not to think of certain posters on usenet as human beings. Was a very valuable lesson. >As I watched the newsgroup react to Ullrich's posts, even when I'd >slyly inform that I was using him and the newsgroup, I learned more. >Eventually, I knew how to use my little lapdog pack of Ullrich, >Magidin, Winter, and Nora Baron. Yup. You're incredibly skilled at using all of us to demonstrate that you're a total moron. It's really quite impressive. >With them I could dominate the newsgroup at will to the point that >from what I heard at times 25% of the traffic had to do with my little >machinations. >I've continually learned about you all and now I can wreak havok at >will, as unfortunately I'm something of a social anarchist That's a common misspelling: it's actual pathetic loser, not social anarchist. >who has >powerful mathematical tools at my disposal. Again, the correct spelling is who wouldn't recognize a powerful mathematical tool if it ran up to me and shouted 'Look, I'm a powerful mathematical tool!'. >Yup, so I LOVE that fascinating definition error that's over a hundred >years old as its like a flame that draws so many moths. >I mean, look at who I have so far: Barry Mazur, Andrew Granville, and >Ralph McKenzie, among others. Right. When you send people unsolicited email that counts as drawing them like moths to a flame. You're saving the rejection letters you get from journals, savoring them as more moths you've drawn to that flame, right? >I can continue to use it, and continue to wreak social havok upon math >society, and use my lapdogs to help me do it. >As I said so many years ago, I will bring down the math establishment. >And you know what? I'll enjoy myself doing it. >It's just taking so damn long, but then again, math society has such a >high position now that I guess it makes sense that it may take some >time to draw it down to earth and raze it to the ground. You know, if you ever sober up and try to get back to what you stupidly call math people are going to remind you of this little episode. >James Harris ************************ David C. Ullrich === Subject: Re: JSH: My lapdogs > That's a common misspelling: it's actual pathetic loser, not > social anarchist. Pathetic loser? You should talk! David Ullrich on Formalizing C3,C4 in ZFC C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A) Classification C4 AyAx[Az(z in y <-> z in x) -> {(set y & set x) <-> y=x}] (Equi-membered classes are identical iff these are sets.) It's clear to anyone with half a clue that your reasoning, if correct, can easily be formalized in ZFC. (Um: I should have said with some predicate in place of 'in' and also some predicate in place of '=', instead of just pointing out that your 'in' would not correspond to the 'in' in ZFC.) This is because there's nothing non-standard about your _reasoning_, all that's non-standard is your _axioms_ regarding 'in' and '='. In case you don't have half a clue, the part of the formalization corresponding to those two axioms might be C3 EyAx[ni(x, y) <-> Et(ni(x,t)) & A] (with y not free in A) Classification C4 AyAx[Az(ni(z,y) <-> ni(z,x)) -> {(S(y) & S(x)) <-> I(x,y)}] (Equi-membered classes are identical iff these are sets.) A white Ph.D. professor can babble with impunity, while a black amateur mathematician is judged by standards quite different. === Subject: Re: JSH: My lapdogs > That's a common misspelling: it's actual pathetic loser, not > social anarchist. > Pathetic loser? You should talk! > David Ullrich on Formalizing C3,C4 in ZFC > C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A) > Classification > C4 AyAx[Az(z in y <-> z in x) -> {(set y & set x) <-> y=x}] > (Equi-membered classes are identical iff these are sets.) > It's clear to anyone with half a clue that your reasoning, if > correct, can easily be formalized in ZFC. (Um: I should have > said with some predicate in place of 'in' and also some > predicate in place of '=', instead of just pointing out that > your 'in' would not correspond to the 'in' in ZFC.) This > is because there's nothing non-standard about your > _reasoning_, all that's non-standard is your _axioms_ > regarding 'in' and '='. > In case you don't have half a clue, the part of the formalization > corresponding to those two axioms might be > C3 EyAx[ni(x, y) <-> Et(ni(x,t)) & A] (with y not free in A) > Classification > > C4 AyAx[Az(ni(z,y) <-> ni(z,x)) -> {(S(y) & S(x)) <-> I(x,y)}] > (Equi-membered classes are identical iff these are sets.) > A white Ph.D. professor can babble with impunity, while a black amateur > mathematician is judged by standards quite different. A black Ph.D mathematics professor can babble with impunity also. Since if you have noticed yet, all mathematics professor are amatuers. Since just all of them moonlight as physicsts and Lawyers. Or as Einstone would have said. Play may play with dice, but he certainly doesn't play with the Fourier stooges of probability theory. === Subject: Re: JSH: My lapdogs > With them I could dominate the newsgroup at will to the point that > from what I heard at times 25% of the traffic had to do with my little > machinations. I agree that you are the best Usenet troll I've ever seen. === Subject: Re: JSH: My lapdogs > One of the funnier things in retrospect is how David Ullrich became so > upset when I said years back that he'd acted as my lapdog in an And you became upset about something that he _didn't_ say--that was funny. > instance, and in his anger he became my lapdog so many more times. > I learned a lot from that exchange years ago. > As I watched the newsgroup react to Ullrich's posts, even when I'd > slyly inform that I was using him and the newsgroup, I learned more. > Eventually, I knew how to use my little lapdog pack of Ullrich, > Magidin, Winter, and Nora Baron. Why scare quotes around Nora Baron's name? -- G.C. === Subject: Re: JSH: My lapdogs Nora Badboobda BaroN is a ... what's that word? on the wayside, I think, the statistic to which JSH refered to are just the local ones -- what ever is the criterium for that; I don't hink taht it hoolds for all of sci.math feeds. I'm dysappointed that I didn't make it into the list of Good Lapdogs; do I have another assignation? > Why scare quotes around Nora Baron's name? --UN HYDROGEN (sic; Methanex (TM) reformanteurs) ECONOMIE?... La Troi Phases d'Exploitation de la Protocols des Grises de Kyoto: (FOSSILISATION [McCainanites?] (TM/sic))/ BORE/GUSH/NADIR @ http://www.tarpley.net/aobook.htm. Http://www.tarpley.net/bushb.htm (content partiale, below): 17 -- L'ATTEMPTER de COUP D'ETAT, 3/30/81 === Subject: Re: JSH: My lapdogs Discussion, linux) >> One of the funnier things in retrospect is how David Ullrich became so >> upset when I said years back that he'd acted as my lapdog in an > And you became upset about something that he _didn't_ say--that was > funny. >> instance, and in his anger he became my lapdog so many more times. >> I learned a lot from that exchange years ago. >> As I watched the newsgroup react to Ullrich's posts, even when I'd >> slyly inform that I was using him and the newsgroup, I learned more. >> Eventually, I knew how to use my little lapdog pack of Ullrich, >> Magidin, Winter, and Nora Baron. > Why scare quotes around Nora Baron's name? Because, after months of correspondence with Nora Baron, James was told by a third party that her name is a palindrome. Now, of course, one *might* think that big Mega Foundation brains can notice such obvious word games all by themselves, but in any case, the incident left James so badly shaken that he has been adding extra punctuation since. -- Jesse Hughes We will run this with the same kind of openness that we've run Windows. Steve Ballmer, speaking about MS's new .Net project. === Subject: Re: JSH: My lapdogs > Because, after months of correspondence with Nora Baron, James was > told by a third party that her name is a palindrome. Now, of course, > one *might* think that big Mega Foundation brains can notice such > obvious word games all by themselves, but in any case, the incident > left James so badly shaken that he has been adding extra punctuation > since. Which still maintains the--shall we say--palindromicity of the name. Rick === Subject: Re: JSH: My lapdogs > >> Because, after months of correspondence with Nora Baron, James was > told by a third party that her name is a palindrome. Now, of course, > one *might* think that big Mega Foundation brains can notice such > obvious word games all by themselves, but in any case, the incident > left James so badly shaken that he has been adding extra punctuation > since. Which still maintains the--shall we say--palindromicity of the name. > Rick Hey Rick! No lesser a JSH critic than yourself--who is (to boot) a math professor--just *might* have something to say about which of the following is a *really* real proof. 1. AxAy[Az(z in x <-> z in y) -> x=y] 2. Ax(x=x) 1 C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A) C4 AyAx[Az(z in y <-> z in x) -> {(set y & set x) <-> y=x}] 1) EyAx(x in y <-> Et(x in t) & ~(x in x)) C3 2) Ax(x in r <-> Et(x in t) & ~(x in x)) 1,EI 3) r in r <-> (Et(r in t) & ~(r in r)) 2,UI 4) ~Et(r in t) 3 5) ~(set r) 4 6) ~(r=r) 5,C4 7) Ex~(x=x) 6,EG 8) ~Ax(x=x) 7 The right-wing definition of a proof (right-or-wrong, rule-of-law) definition is that a proof is a logically correct argument that establishes the truth of a given statement. If you are a right-winger (in Devlin's sense), which of the above is a logically-correct-argument-that- establishes-the-truth-of-its-last-line? (Take the background logic to be first-order logic *without* identity.) And supposing you are a left-winger (in Devlin's sense), what would it take to 'convince' you that the second of these arguments is sound? --John Correy PS Provide a *really* good answer and I won't assign this to Ullrich as homework. === Subject: Re: Altered Strong law help? > Suppose the random variables X_i, i = 1,2,... are i.i.d. Show > that if E((X_1)^2) < infinity and if M_n = max(X_1,X_2,...,X_n), then > (M_n)/squareroot(n) converges to zero with probability one. > I thought about altering the strong law of large numbers assuming second > moments, then using a subsequence trick, but am getting nowhere. > Can anyone give a proof for this please? > Marc Since M_n >= X_1, P[0 <= liminf M_n/sqrt{n}] = 1. Now, note that E|X_1|^2 = int_0^infty {P[|X_1|^2 > x] dx} >= sum_{n=1}^infty P[|X_n|^2 > n]. If X_1 has finite variance, then the Borel-Cantelli Lemma shows that, with probability one, |X_n| <= sqrt{n} for all n sufficiently large. Thus, P[limsup M_n/sqrt{n} <= 1] = 1. Similar work shows P[limsup M_n/sqrt{n} <= 1/k] = 1 for all k; thus, taking intersections, P[0 <= liminf M_n/sqrt{n} <= limsup M_n/sqrt{n} <= 0] = 1, i.e. M_n/sqrt{n} --> 0 a.s. === Subject: Re: measure theory and integration by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9JEVMT16480; >>does anybody know where i can find (some internet adresses) solved >>problems in measure theory and integration. my homework are exercises >>from w. rudin's book real and complex analysis (chapters 1,2,3,6,7 >>and 8). i hope somebody will help me. >The best help I can give you is to say that you should do the >problems yourself. Copying them from a website will be of no >help to you. The second best help I can give you is to say to >go ask your professor for assistance. The third is to ask for >hints on specific problems after you've tried them yourself. >--Dan Grubb Rudin is an idiiot ! Your professor is a bigger idiot !! Scoundrels who ought not to be teachning Mathematics have been allowed to teach Mathematics. That's the problem. Look at this: http://home.earthlink.net/~vadulam/ === Subject: Re: measure theory and integration > Rudin is an idiiot ! Your professor is a bigger idiot !! > Scoundrels who ought not to be teachning Mathematics have been allowed to > teach Mathematics. That's the problem. Actually, the problem is that any idiot with a computer can scream all he wants on Usenet. === Subject: Re: measure theory and integration >> Rudin is an idiiot ! Your professor is a bigger idiot !! >> Scoundrels who ought not to be teachning Mathematics have been allowed to >> teach Mathematics. That's the problem. >Actually, the problem is that any idiot with a computer can scream all he >wants on Usenet. Ah, you exist. Check the thread cauchy sequence - it's actually a question about something in Axler et al. I'm curious whether my guess regarding what's in the book that the OP is not telling us is correct... ************************ David C. Ullrich === Subject: Re: measure theory and integration > Look at this: > http://home.earthlink.net/~vadulam/ Why? -- 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: Complex Var Book : Good or Bad as Text I'm considering using Donald Sarason's book, Notes on Complex Function Theory, next year for a grad math Complex Var I course. This is my first time teaching complex vars at the grad level, and although I'm tempted to use Conway, Sarason has been the standard text used for the course in the last 3 years. I'm curious as to experiences other people might have had using this book, and what they saw as the advantages and disadvantages of this text. Jim === Subject: tangential quadrilateral If a circle can be inscribed in a convex quadrilateral, R, in the xy plane, then R is called a tangential quadrilateral. It is well known, and not hard to show, that if R is a tangential quadrilateral, then a + c = b + d, where a and c, and b and d denote the lengths of opposite sides of R. It is noted in John W. Harris and Horst Stocker: Handbook of Mathematics and Compuitational Science, Springer, New York, 1998, but not proved, that a + c = b + d is also sufficient for R to be a tangential quadrilateral. I might be able to grind out a proof myself, but I was wondering if anyone knows if the details of a proof have been published anywhere. === Subject: Re: tangential quadrilateral > If a circle can be inscribed in a convex quadrilateral, R, in the xy > plane, then R is called a tangential quadrilateral. It is well known, > and not hard to show, that if R is a tangential quadrilateral, then a + > c = b + d, where a and c, and b and d denote the lengths of opposite > sides of R. It is noted in John W. Harris and Horst Stocker: Handbook of > Mathematics and Compuitational Science, Springer, New York, 1998, but > not proved, that a + c = b + d is also sufficient for R to be a > tangential quadrilateral. I might be able to grind out a proof myself, > but I was wondering if anyone knows if the details of a proof have been > published anywhere. This seems somehow dual to Ptolemy's theorem on cyclic quadrilaterals. === Subject: Re: tangential quadrilateral find a proof for the apposite case, with vertices on the circle, a cyclic tetragon (sik; quadrilateral). > If a circle can be inscribed in a convex quadrilateral, R, in the xy > plane, then R is called a tangential quadrilateral. It is well known, > and not hard to show, that if R is a tangential quadrilateral, then a + > c = b + d, where a and c, and b and d denote the lengths of opposite > sides of R. It is noted in John W. Harris and Horst Stocker: Handbook of > Mathematics and Compuitational Science, Springer, New York, 1998, but > not proved, that a + c = b + d is also sufficient for R to be a > tangential quadrilateral. I might be able to grind out a proof myself, > but I was wondering if anyone knows if the details of a proof have been > published anywhere. --UN HYDROGEN (sic; Methanex (TM) reformanteurs) ECONOMIE?... La Troi Phases d'Exploitation de la Protocols des Grises de Kyoto: (FOSSILISATION [McCainanites?] (TM/sic))/ BORE/GUSH/NADIR @ http://www.tarpley.net/aobook.htm. Http://www.tarpley.net/bushb.htm (content partiale, below): 17 -- L'ATTEMPTER de COUP D'ETAT, 3/30/81 === Subject: Re: tangential quadrilateral > If a circle can be inscribed in a convex quadrilateral, R, in the xy > plane, then R is called a tangential quadrilateral. It is well known, > and not hard to show, that if R is a tangential quadrilateral, then a + > c = b + d, where a and c, and b and d denote the lengths of opposite > sides of R. It is noted in John W. Harris and Horst Stocker: Handbook of > Mathematics and Compuitational Science, Springer, New York, 1998, but > not proved, that a + c = b + d is also sufficient for R to be a > tangential quadrilateral. I might be able to grind out a proof myself, > but I was wondering if anyone knows if the details of a proof have been > published anywhere. Let A, B, C and D the vertices in order. Then suppose that AB + CD = BC+ AD If a circle can be inscribe in it, the centre must be at the intersection point of the four bisectrices of the angles. It is suffice to see that three bisectrices intersect at one point. If the quadrilateral is a rhombus, it's clear. Then suppose that two consecutive sides have distinct lengths. By example, AB > BC. Then CD < AD. Then take E on AB, as BE = BC. The triangle BCE is then isosceles. Similarly, take F on AD, as DF = DC. The triangle DFC is also isosceles. But then, the triangle AEF is also isosceles, because AE = AB - BC = AD - CD = AF. Then the bisectrix of angle B is the mediatrix of side CE, the bisectrix of D is the mediatrix of CF and the bisectrix of A is the mediatrix of EF. Then, the three bisectrices intersect at circumcentre of the triangle CFE, and the quadrilateral is circunscriptible. -- Ignacio Larrosa Caestro A Corua (Espaa) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: tangential quadrilateral Ignacio Larrosa Caestro escribi > If a circle can be inscribed in a convex quadrilateral, R, in the xy > plane, then R is called a tangential quadrilateral. It is well known, > and not hard to show, that if R is a tangential quadrilateral, then a + > c = b + d, where a and c, and b and d denote the lengths of opposite > sides of R. It is noted in John W. Harris and Horst Stocker: Handbook of > Mathematics and Compuitational Science, Springer, New York, 1998, but > not proved, that a + c = b + d is also sufficient for R to be a > tangential quadrilateral. I might be able to grind out a proof myself, > but I was wondering if anyone knows if the details of a proof have been > published anywhere. > Let A, B, C and D the vertices in order. > Then suppose that AB + CD = BC+ AD > If a circle can be inscribe in it, the centre must be at the intersection > point of the four bisectrices of the angles. It is suffice to see that three > bisectrices intersect at one point. > If the quadrilateral is a rhombus, it's clear. Then suppose that two > consecutive sides have distinct lengths. By example, AB > BC. Then CD < AD. > Then take E on AB, as BE = BC. The triangle BCE is then isosceles. > Similarly, take F on AD, as DF = DC. The triangle DFC is also isosceles. > But then, the triangle AEF is also isosceles, because AE = AB - BC = AD - CD > = AF. > Then the bisectrix of angle B is the mediatrix of side CE, the bisectrix of > D is the mediatrix of CF and the bisectrix of A is the mediatrix of EF. > Then, the three bisectrices intersect at circumcentre of the triangle CFE, > and the quadrilateral is circunscriptible. I forgot the cite: 'Acerca de la Demostracin en Geometra' (Coleccin 'Lecciones Populares de Matemticas'), A.I. Fetsov. Edit. MIR (Mosc) -- Ignacio Larrosa Caestro A Corua (Espaa) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Reliable Statistical Relationship I've got a quick stats question about determining if the results of a survey that I have conducted are valid. I sent out 24,000 surveys and got a response of 779. I know that I've received enough responses to make the survey valid, however, I can't for the life of me remember how to determine if the results of a question have enough responses to determine what confidence level it is significant at? For example: I received only 8 responses from Native Americans, I know that I can't generalize to the entire population from such a small response. However other questions had a 98% response rate, how do I say that it is significant at say the .001 or .01 or .05 level for my cross-tabs? Any help is GREATLY APPRECIATED!!! === Subject: Tensor analysis Hello! Can u please advice me a site, where i can find some lectures and tasks on tensor analysis. Best wishes, Serge. === Subject: Examples of 3D Fractals I have been experimenting with 3d fractals in Pov-Ray a while back. I do understand the theory behind it (the ones with Quaternions anyway - it would be kind of sad if I didn't understand that :) ). The funny thing is that I've spent like a year trying to make a 3-dimensional algebra space to use an iterative formula in (spending ages looking for a 'third imaginary number'), and apparently, I discovered some time ago, you need 4 dimensions to do it, using the quaternial i,j,k equation. I know there is another type of 4-dimensional complex number that is used in an iteration formula - hyper-complex numbers? I could be wrong about that term it has been a while and I never investigated those thoroughly. Anyway, my point is that I can make absolutely no visual sense whatsoever of the 3d fractals that Pov-Ray outputs, they look as pseudo-randomly as anything I've ever seen. Well not really but still, they share nothing with their 2d counterparts in terms of topological convergance. Does anybody have a link to illustrations on the different kinds of iterations that are used in these 3d fractals (well, I mean, slices of 4d fractals), and what models they yield to? Or also about the real 4d - with directions, -- Quaternion === Subject: Re: Examples of 3D Fractals >I have been experimenting with 3d fractals in Pov-Ray a while back. I do I'm not 100% sure I understand what you mean. I guess you refer to quaternion/whatsoever Julia/Mandelbrot objects: if this is so, just keep in mind that (complex) analytical dynamics and fractals are in fact two different things, the link between them being that (e.g.) Julia sets are actually fractal sets. But do *not* think that fractal = iterate something and see what happens! >understand the theory behind it (the ones with Quaternions anyway - it >would be kind of sad if I didn't understand that :) ). The funny thing is >that I've spent like a year trying to make a 3-dimensional algebra space to >use an iterative formula in (spending ages looking for a 'third imaginary >number'), and apparently, I discovered some time ago, you need 4 dimensions >to do it, using the quaternial i,j,k equation. I know there is another type >of 4-dimensional complex number that is used in an iteration formula - >hyper-complex numbers? I could be wrong about that term it has been a while AFAIK hypercomplex numbers is an old term to designate *any* 4-dimensional (real algebra that realizes a) complex extension. Indeed the second kind of POVray 4-dimensional complex numbers are/were[*] called hypercomplex in the docs. POVray's hypercomplex numbers were simply complex numbers with complex real and imaginary part, that is, couples of complex numbers to be multiplied like this: (a,b)(z,w):=(az-bw,aw+bz). This idea, though appealing, doesn't really bring anything new, since *that* multiplication it is not substantially different from this one: (a,b)(z,w):=(az,bw). So the quadratic iteration x|->x^2+c with these numbers is just the same as a couple of *parallel* iterations of the corresponding complex function. >Anyway, my point is that I can make absolutely no visual sense whatsoever of >the 3d fractals that Pov-Ray outputs, they look as pseudo-randomly as >anything I've ever seen. Well not really but still, they share nothing with I certainly don't think so. They have definitely a recognizable look: in both cases they seem to present the classical jerkiness of fractal objects only in one direction, and to be smooth in the other ones (say, you see typical trails of the cross section's edges). While this is obvious as for the second kind of numbers, keeping into account above's remark, I don't *think* it is so trivial for quaternions too. But the visual evidence supports the claim, that hopefully can be stated in mathematically proper terms too... Anyway I hope you're aware that what you see are 3-dimensional slices of the 4d objects! IIRC some treatment of these algorithms is done in The Science Of Fractal Images, and maybe in more recent and more in-depth >their 2d counterparts in terms of topological convergance. ?? >Does anybody have a link to illustrations on the different kinds of >iterations that are used in these 3d fractals (well, I mean, slices of 4d >fractals), and what models they yield to? Or also about the real 4d - with >directions, IMHO, a much more interesting hypercomplex algebra to be investigated is that of real 2x2 matrices, that can be thought of as C^2 with a suitable multiplication. [*] I'v not been used POVray for quite a long time. I don't remember if by any chance these objects were in the 3.0 release, some features of which became later deprecated. Michele -- > Comments should say _why_ something is being done. Oh? My comments always say what _really_ should have happened. :) - Tore Aursand on comp.lang.perl.misc === Subject: Re: Tensor analysis > Hello! > Can u please advice me a site, where i can find some lectures and > tasks on tensor analysis. > Best wishes, > Serge. Google. Look for Sean Carrol's lecture notes in pdf format. They are excelllent. Bob Kolker === Subject: Re: Tensor analysis > Hello! > Can u please advice me a site, where i can find some lectures and > tasks on tensor analysis. > Best wishes, > Serge. > Google. Look for Sean Carrol's lecture notes in pdf format. They are > excelllent. And Stefan Waner's notes at http://people.hofstra.edu/faculty/Stefan_Waner/diff_geom/tc.html http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/pdfs/DiffGeom.pdf Dirk Vdm === Subject: Re: Tensor analysis > And Stefan Waner's notes at > http://people.hofstra.edu/faculty/Stefan_Waner/diff_geom/tc.html > http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/pdfs/DiffGeom.pdf I concur. This also, is a lovely set of nets. Use the PDF version. It is easier to navigate in. Bob Kolker === Subject: Re: Question about tensor products Let me try to define S *_R N in this case, ignoring all that has gone before. I will use * for tensor product for ease of typing. Begin with the set (not abelian group) S x N. It has elements (s,n), s in S and n in N. Form the free abelian group F on this set. Its elements are finite sums (s_1,n_1) + (s_2,n_2) + ... with the obvious addition. Let A be the subgroup generated by all elements of one of the forms (s_1+s_2,n) - (s_1,n) - (s_2,n) (s,n_1+n_2) - (s,n_1) - (s.n_2) (sr,n) - (s,rn) for all s,s_1,s_2 in S, n,n_1,n_2 in N and r in R. Define S *_R N = F/A and denote the element (s,n) + A by s*n. Every element of S * N is a finite sum s_1*n_1 + s_2*n_2 + ... Moreover, on account of the nature of A, it is true that (s_1+s_2)*n = s_1*n + s_2*n s*(n_1+n_2) = s*n_1 + s*n_2 sr*n = s*rn You should work out why these equations hold. Further, define s(s_1*n_1 + s_2*n_2 + ...) = ss_1*n_1 + ss_2*n_2 + ... and check that this makes S *_R N into an S-module. That's all there is to it. More generally, if M is a right R-module and N is a left R-module, you can define the abelian group M *_R N. Whenever M is a left S, right R-bimodule, then M *_R N will be a left S-module. Similarly, if N is a left R, right T-bimodule, then M *_R N is a right T-module and if both happen, then M *_R N is a left S, right T-bimodule. Hope this helps. === Subject: statistical propability, just might work Two things you need to know up front: 1. You CAN NOT make $50,000 in a month or even six months by following the instructions found in this letter, or similar ones found on the net. 2. You CAN make $1,500 to $3,000 or more within a couple of months, with which you can buy a new computer, pay some bills, or go away on vacation. O.K. I won't lie to you and give you a list of success stories. Truth is, this is my first time trying this. I've seen letters like this come and go and I've ignored every one up until now. I've got 6 bucks in my back pocket and I've decided I'm going to find out if this thing really works... once and for all. So here I am. I figure my odds of at least getting my money back are far better than playing the lottery. In fact, the odds that I won't get my money back are very- very low. And all I have to do is post this message to some newsgroups and send 6 envelopes through the mail. I look at it like this. If everything fails and I don't get ANY money through the mail, isn't it worth $6 to know if the systems really works? It's that simple. If not for all the money I should be getting, then I'm just doing this to see if it really works. I'll leave the math to you, but when you finally figure out how much money you could make (even in the worst case), I don't see how you can turn this down. I promise, you will make much more than the original $6 you put into it. So now I'm asking you. Will you please give it a try. It's only $6 plus the cost of 6 stamps. A total of $8.42. If you decide to try it, one of those five dollars you use will be mine, just as future one dollar bills will be yours. And remember, that's just the first phase. After that your name moves up on the list and the amount of money you receive grows exponentially until you're pushed off the top. Come on. Give it a try just for chance if nothing else. All I had to do is write this little letter you just read and post it to the newsgroups. Everything below this point is copied from the letter I received. Now, all I do is sit back and wait. Easy. So, how does this work? Simple. You have probably seen this before. What it is, is a PERFECTLY LEGAL (see Title 18,h sections 1302 NS 1341 of Postal Lottery Laws) program of collecting names and addresses for a mailing list. Basically, you will start your own small company that collects and stores mailing lists. NOTE: You can sell these mailing lists to other companies and make even more money. My goal is to make money, not to deluge those who send me the money with piles of junk mail. STEP ONE - Print this letter. Save it, in case you want to try it again at a later date. STEP TWO - Send a real mail letter to each of the following people, reading simply Please add me to your mailing list, and containing your full name and mailing address, and $1.00 in cash. 1) Aaron Groves 4333 Angelina Dr.Plano, TX 75704 2)Lucy Crespo 135 Stoddard Dr. apt.223 Meriden, CT 06451 3) Sid McFarland 168 N. Lancastor apt. 1 Athens, OH 45701 4)Erika Malave 2606 Academy Dr.apt. 402 Huntsville, AL 35811 5) Kor Davis 809 N.11th Manhattan, KS 66502 6) Asa Black 435 Allen St. apt. 102 New Britian, CT 06053 STEP THREE - Edit this letter to read the way you want it to. Include some information about yourself. DO NOT LIE! I have explained that this is my first time trying this, but if it's not your first time, then don't say that. Just tell us exactly why you decided to try it. When you get to the list of names of above, remove the top name (#1), and move the others up. Then add your name to the bottom of the list (#5). In this way, you can earn money the investments of others after you, as you are helping those that came before. Remember to write these names down or make a copy of the file or print it out before you erase them so that you will know who to send your $1 to. STEP FOUR - Post your letter to as many internet newgroups or local BBS numbers as you wish. You can even send copies of it snail-mail, but this is not as considerate. When you post it on a newsgroup, the people who want to read it can, but they are not forced to receive it as junk mail. Also, when you post it, give it a title that will draw people to it. I have used the title Making Money by Reading Newsgroups. STEP FIVE - When the money starts coming in (and it may take two to three weeks, don't panic and don't give up), save all the letters you receive in a box or closet somewhere. Store the names and addresses in your computer, if you wish, to make your mailing list easier to handle, but always KEEP THE ORIGINAL LETTERS. This way, you will have proof that your business is legal and legitimate, in case you are ever asked. STEP SIX - Use the money you receive as you wish. You have earned it. If you need more (we all do, don't we?) then try again. Just do not forget to send the money to the people on your list! They are just like you, and nobody makes anything if this letter regenerates without any money changing hands. I hope you decide to give it a try, and I hope it gives you the same results I expect to get very soon. Asa Black === Subject: Re: Cross product distributive proof > .... > Could someone please give me a proof for distributive law of the cross > product: > a x (b + c) = a x b + a x c > But there's a catch! I am assuming the we have defined the cross > product as (|a||b|sinO)n, etc.... When teaching this several years ago, I used a proof slightly simplified from that of Seymour Schuster, Elementary Vector Geometry, (Wiley, 1962), pp.139-140. The trick is to work ahead (without assuming the distributive law) to the scalar triple product. You need to show that its dot and cross can be interchanged, and also to have proved the distributive law for the _dot_ product (which can also be a little delicate). After that, here's the proof. Please pretend every variable is in bold type. :-) Let d = a x (b + c) - a x b - a x c so it is required to prove that d = 0. d^2 = d.d = d.(a x (b + c) - a x b - a x c) = d . a x (b + c) - d . a x b - d . a x c = d x a . (b + c) - d x a . b - d x a . c = d x a . (b + c) - d x a . (b + c) = 0. Therefore d = 0, so a x (b + c) = a x b + a x c. Then you can multiply by -1 and use the anticommutative law to get the other distributive law. Schuster didn't use d^2 but I think it makes the proof clearer. Ken Pledger. === Subject: Re: Can I get parameters of an ellipse from conic section? > Ellipse is produced if a plane intersects only one nappe of a cone. > Is there a way to get the parameters of the ellipse as a function of > the angle of the plane with the axis of the cone? > .... D.M.Y. Sommerville, Analytical Conics, 3rd edition, Chapter IV. Ken Pledger === Subject: coding gain hi can anyone give the definition of coding gain in coding theory I check several books but can't find it === Subject: Re: coding gain >hi can anyone give the definition of coding gain in coding theory >I check several books but can't find it Here's a start at an explanation. Communications occur in the presence of noise. The noise disturbs or corrupts the signal as it is sent through a wire or through radio waves. Errors are unavoidable. That is, if there is noise, there is always some chance of errors. The best one can do is to try to minimize the rate at which errors occur. One way of reducing the error rate is to have a more powerful signal, which will be less affected by noise than a weak signal. Another way of reducing the rate at which errors occur is to use a more powerful code. Suppose code C when using 10 watts of power leaves the message receiver with 1 bit in 1000 in error, and that code A+ when using 1 watt of power also leaves the message receiver with 1 bit in 1000 in error. Obviously, code A+ is better than code C. Because code A+ uses a fraction of 1/10 of the power used by code C, for the same error rate, if we take -10*log_10(1/10), we get: 10, and engineers would say that code A+ has a 10 decibel or 10 dB coding gain relative to code C. Also, with complex codes, it can be time-consuming to do the best kind of decoding for a given code R. An alternative is to do less-than-optimal decoding, in less time, at the expense of some negative coding gain , i.e. the error rate is higher with the faster but sub-optimal decoding method... The code R is always the same here, what is changing is the decoding algorithm used... A 10 dB coding gain is large or excellent ... Here's a web page address that discusses coding gains of turbo codes: http://www.xenotran.com/turbo_tech.html David Bernier === Subject: Re: coding gain should? 0`l@G0P / ]  - 4 : k    0 v   C ] w          ! , B M c y    b    E    ^      ! 0 ? N d s          >  1 C U g y - o H @ s  L l    , ^    I   ! + G g     ;    V   ( o   ?    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X' XD Xt X X X X Y YQ Y Y Z Z/ Zl Zr Z Z Z [E [K [h [ [ E     ] ] ]8 ]= ]E ]K ]h ]t ] ^ ^Q ^f ^n ^ ^ ^ ^ ^ _ _ _ _ ` `Z ` ` ` ` a' a- aP ag a a a bE b b c- c4 c[ c` cf c c c ed by noise than a weak signal. Another way of reducing the rate at which errors occur is to use a more powerful code. Suppose code C when using 10 watts of power leaves the message receiver with 1 bit in 1000 in error, and that code A+ when using 1 watt of power also leaves the message receiver with 1 bit in 1000 in error. Obviously, code A+ is better than code C. Because code A+ uses a fraction of 1/10 of the power used by code C, for the same error rate, if we take -10*log_10(1/10), we get: 10, and engineers would say that code A+ has a 10 decibel or 10 dB coding gain relative to code C. Also, with complex codes, it can be time-consuming to do the best kind of decoding for a given code R. An alternative is to do less-than-optimal decoding, in less time, at the expense of some negative coding gain , i.e. the error rate is higher with the faster but sub-optimal decoding method... The code R is always the same here, what is changing is the decoding algorithm used... A 10 dB coding gain is large or excellent ... Here's a web page address that discusses coding gains of turbo codes: http://www.xenotran.com/turbo_tech.html David Bernier === Subject: Re: coding gain