mm-3159 === Subject: Re: Analytical solution > im searching for an analytical solution of the following equation: > a^p + b^p = k > a,b,k.... known > p = ? > Does an analytical solution exist? What does it looks like? Not likely. If it were needed sufficiently often in applications, someone would call it, for example, Bernd's function, p=bernd(a,b,k). Something like this happened to Lambert's W-function, which has many applications in the study of difference-differential equations, and elsewhere. (It defines y=W(x) by y*exp(y)=x.) Theorem, it could be proved that this bernd function is analytic in all three variables, with the exception of a small set where it would have a singularity. Then someone else would classify the singular set, thus The trouble is: this function seems to remain a curiosity item, with no wide applications. Unless someone shows the world such applications. The other trouble: the casual use of analytical (vaguely, expressible with the use of finitely many applications of algebraic and standard transcendental operations) is different from analytic (locally expressible as a power series). The meaning of analytical depends on the user's current list of standard transcendentals, so it can change during the user's lifetime; does one allow Gamma function, Bessel functions, etc.? === Subject: Re: norms again > Some days ago I was asking about the fact that the maximum norm and the > taxicab norm could not descend from an inner product. > Now I want to ask: *only* the euclidean norm comes from an inner product? If > yes, what it is so? No! Pick your favorite positive definite quadratic form and take its square root............................................. === Subject: Re: Which model will be more accurate? >Herman, >I think that what you're saying is not correct. The correlation >amongst each independent variable with the dependent variable is 0.30. >The correlations between each of the independent variables with each >other is very low (~0). If you sum up the correlations as you say, >then you get a R^2 of 1.50, which is impossible. If the R^2 are each .30, and the correlations between the independent variables are 0, this would indeed be impossible. Now if the correlations are .3, the R^2 would be .09, and you would get an overall R^2 of .45. Small non-negative correlations between the independent variables can make substantial differences in the results. >>Which multivariate statistical model will yield a more accurate model: >>Model I: There are 5 independent variables, but the dependent variable >>is only loosely correlated with any one of those independent variables. >> Moreover, there is very little correlation between all 5 independent >>variables. The R^2 for each model, when plotted individually against >>the dependent variable is 0.30. >>Model II: There is only 1 independent variable to predict the >>dependent variable, and the R^2 is 0.90. >> With this information, it is impossible to decide. >> If the five predictor variables in the first model >> are uncorrelated with each other, the R^2 for the >> combined model is the sum of the R^2 for the >> individual models. If not, it would have to be >> computed. >> -- >> This address is for information only. I do not claim that these views >> are those of the Statistics Department or of Purdue University. >> Herman Rubin, Department of Statistics, Purdue University >> hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Which model will be more accurate? >>Herman, >>I think that what you're saying is not correct. The correlation >>amongst each independent variable with the dependent variable is 0.30. >>The correlations between each of the independent variables with each >>other is very low (~0). If you sum up the correlations as you say, >>then you get a R^2 of 1.50, which is impossible. >If the R^2 are each .30, and the correlations between the >independent variables are 0, this would indeed be impossible. If you have six random variables X_1,...,X_6 such that the correlation of X_i with X_j is 0 for 2 <= i < j and c for i = 1 < j, then c^2 <= 1/5. This is because the matrix [ 1 c c c c c ] [ c 1 0 0 0 0 ] [ c 0 1 0 0 0 ] [ c 0 0 1 0 0 ] [ c 0 0 0 1 0 ] [ c 0 0 0 0 1 ] has eigenvalues 1, 1 + c sqrt(5) and 1 - c sqrt(5). But the correlation matrix of random variables must be positive semidefinite. Perhaps the original problem should have stated R^2 = 0.2. Then the total R^2 would end up as 1, and X_1 is perfectly predicted by X_2 to X_6, namely (if everything is normalized to mean 0 and variance 1) X_1 = (X_2 + ... + X_6)/sqrt(5) almost surely. On the other hand, if the correlations for 2 <= i < j are d instead of 0 (where 0 < d < 1), then the requirement becomes c^2 <= (1+4d)/5. So c^2 = 0.3 would be allowed if d >= 1/8. === Subject: Re: OEIS does it exist? Today I cannot get to OEIS. Is this a local or temporary > problem? Or has OEIS dropped of the face of the internet? > -- > Jim Buddenhagen > According to the seqfan-mailing list they're installing > a new serverprogram. It was announced just to wait. > Gottfried Helms It seems to be on-line again. === Subject: Re: Simple Graph theory.. Please help! > Let G be a graph of order n and let k be a positce integer with k prove that if the min degree of gG >= (n+k-2)/2 Then G is k-connected. Suppose you have a set X with size < k which disconnects G (i.e., GX breaks G into two graphs H and K). Try counting edges from X to X, X to H and K, edges with both ends in H, and edges with both ends in K. You should end up with fewer than the minimum degree condition guarantees. === Subject: Re: Graph theory > How do you prove this theorem: > If G is a connected graph with order n and is k-edge-connected. > Show that if > min degree of vertices in G, d >= (n-1)/2 then k=d. Use the hint in your other thread, except with X a set of edges. === Subject: Re: simple problem.. > Going to get a headache solving these theorems... > Anyway... > Suppose a graph,G is k-connected, you add a new vertice That's _vertex_ (singular form). I'll let the typo below go. > v and connect > it to any k vertices in G. Show that the resultant grapg is also > k-connected > I wish it's as trivial as it seems :) Try proof by contradiction. Choose a set X of < k vertices, and suppose G+v X is not connected. (1) Can v be in X? (2) Can X be the set of neighbors of v? (3) If v is not in X, then does X separate G? (v must be in one of the components of G+v X, by (1) and (2).) === Subject: Re: Ellipse inside a triangle > What the maximum area of an ellipse inside a triangle I think Jim's answer is correct. Area of ellipse / Area of triangle = pi/sqrt(27) If b and h are the base and height dimension of the triangle then max area of the ellipse would be h*b*pi/6/sqrt(3) === Subject: Re: Ellipse inside a triangle > What the maximum area of an ellipse inside a triangle I have got the following useful link === Subject: Re: Ellipse inside a triangle Jim Ferry escribi.97: >> What the maximum area of an ellipse inside a triangle > For an equilateral triangle, the inscribed circle is the inscribed > ellipse of maximal area. This area is pi/sqrt(27) times the area of > the triangle. > For a general triangle the maximal area of an inscribed ellipse is > also pi/sqrt(27) times the area of the triangle. This can be seen > easily by considering affine transformations. > An affine transformation takes every point x to a new point Ax+b where > A is a non-singular matrix and b is a vector. An affine > transformation Ax+b multiplies the area of every object by det(A) (or > the volume, if we're in 3-d, etc.) Furthermore, for any two > triangles there exists an affine transformation which takes one to > the other. Therefore, any ellipse in a arbitrary triangle has an > area of det(A) times an ellipse in some fixed equilateral triangle, > so the area of the former ellipse is maximized precisely when the > area of the latter ellipse is. This is the Steiner inner-ellipse of the triangle. Naturally, it is tangent totriangle sides in its midpoints. Its centre is the centroid of the triangle and its foci are the zeroes of the derivative of the polynomial of third degree that has the vertices as roots. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Ellipse inside a triangle <3q304dFd4d15U1@individual.net> So the maximum area circle in a scalene triangle less in area than that of a maximum ellipse. Right? === Subject: Re: Ellipse inside a triangle Narasimham escribi.97: > So the maximum area circle in a scalene triangle less in area than > that of a maximum ellipse. Right? Yes, the Steiner inner-ellipse is the biggest ellipse inscribed in the triangle, and circles are special ellipses. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Space is clearly something not nothing You cannot see all the interstellar photons en route because they cannot be seen at an angle or side-on.!! Thats the way photons are longitudinally polarized.!! You MUST look exactly UPstream to see them, at all.!! VERY sincerely u c, ```Brian >><>><>><>><>>< | I just dont understand why people persist in saying space > | is nothing when it clearly isnt. > It is clearly nothing or it would not be clear. insert ..see top. > | How can nothing expand ? > It can't. > Androcles. === Subject: Re: Space is clearly something not nothing <433C566B.502F@nf.sympatico.ca | I just dont understand why people persist in saying space > | is nothing when it clearly isnt. > It is clearly nothing or it would not be clear. Does cyberspace means also 'nothingness' to you? === Subject: Re: Space is clearly something not nothing | > | I just dont understand why people persist in saying space | > | is nothing when it clearly isnt. No, I didn't write that. You are clearly a delusional phuckwit if you think I did. Androcles === Subject: Re: modern math is failure -- back to the drawing board <1111929.1127989672891.JavaMail.jakarta@nitrogen.mathforum.org True only if, of course, Mark measures truth in the mythology of Mark's mind. What do you mean if? If you believe there is any other truth but what is in your mind, you are deluding yourself. === Subject: Re: modern math is failure -- back to the drawing board > True only if, of course, Mark measures truth in the > mythology of Mark's mind. > What do you mean if? If you believe there is any > other truth but > what is in your mind, > you are deluding yourself. An ironic statement, not what we usually mean by delusion. No doubt you find comfort in solipsism, as others find it in religion. Tom === Subject: Re: modern math is failure -- back to the drawing board <1111929.1127989672891.JavaMail.jakarta@nitrogen.mathforum.org> True only if, of course, Mark measures truth in the mythology of Mark's mind. > What do you mean if? If you believe there is any other truth but > what is in your mind, > you are deluding yourself. I thought deluding yourself involved believing whatever is in your mind is true. -- you, but just remember, I'm the guy who proved Fermat's Last Theorem in just a bit over 6 years [...] My standards are kind of high. --James Harris, founding a new mathematical school === Subject: Re: modern math is failure -- back to the drawing board >True only if, of course, Mark measures truth in the mythology of Mark's mind. >>What do you mean if? If you believe there is any other truth but >>what is in your mind, >>you are deluding yourself. > I thought deluding yourself involved believing whatever is in your > mind is true. Truth is that which, when you stop believing in it, doesn't go away. Han de Bruijn === Subject: Re: modern math is failure -- back to the drawing board <1111929.1127989672891.JavaMail.jakarta@nitrogen.mathforum.org> <87mzlvo8i1.fsf@phiwumbda.org> <603d5$433cfb49$82a1e3ad$2617@news2.tudelft.nl>True only if, of course, Mark measures truth in the mythology of Mark's mind. >What do you mean if? If you believe there is any other truth but > what is in your mind, you are deluding yourself. >> I thought deluding yourself involved believing whatever is in your >> mind is true. > Truth is that which, when you stop believing in it, doesn't go away. Thrilling. If I stop believing in cheese, it doesn't go away. Therefore, cheese is truth. Okay, probably a good percentage of the Dutch agree that cheeses (along with yogurt, bikes, and croquets) *are* truth, so that's a bad example. Let me pick something else. Say, fuzzy hats. -- I don't know why I live in a world with so many supposed mathematicians who are all so dumb AND rude. Why oh why couldn't someone like Gauss or Dedekind still be around? Shoot, I'd even take someone like Hardy at this point. -- James S Harris compromises === Subject: Re: modern math is failure -- back to the drawing board > True only if, of course, Mark measures truth in the mythology of Mark's mind. >> What do you mean if? If you believe there is any other truth but >> what is in your mind, >> you are deluding yourself. >I thought deluding yourself involved believing whatever is in your >mind is true. It might be worse if you believed that whatever is in Mark's mind is true. === Subject: Re: modern math is failure -- back to the drawing board <1111929.1127989672891.JavaMail.jakarta@nitrogen.mathforum.org> <87mzlvo8i1.fsf@phiwumbda.org> True only if, of course, Mark measures truth in the mythology of Mark's mind. > What do you mean if? If you believe there is any other truth but > what is in your mind, > you are deluding yourself. >>I thought deluding yourself involved believing whatever is in your >>mind is true. > It might be worse if you believed that whatever is in Mark's mind is > true. I doubt anyone, Mark included, falls for his little fantasies. -- Naomi Klein reports that Microsoft is helping develop e-government in Iraq, 'which Dees admits is a little ahead of the curve, since there is no g-government in Iraq--not to mention functioning phones lines.' -- as reported in The Register, http://www.theregister.co.uk === Subject: Re: modern math is failure -- back to the drawing board <1111929.1127989672891.JavaMail.jakarta@nitrogen.mathforum.org> <87mzlvo8i1.fsf@phiwumbda.org> What do we care what you think? === Subject: Re: modern math is failure -- back to the drawing board <1111929.1127989672891.JavaMail.jakarta@nitrogen.mathforum.org> <87mzlvo8i1.fsf@phiwumbda.org What do we care what you think? I'm just fascinated by this situation. You said that T.H. Ray is deluded, because he doesn't believe that whatever he believes is true. Thus, it must be the case that whatever he believes *is* true (since one is deluded regarding X iff he believes wrongly regarding X). So whatever he believes *is* true, and therefore he cannot be deluded about anything (since he has the proper beliefs about everything). It's a cute paradox. Something like T.H. is an omniscient being who believes he is not omniscient. You're not a millionaire financial mathematician. You're a philosopher, baby! -- Rob Enderle [predicts] that Longhorn will provide 'vast improvements in security.' We can cheer this happy prospect, but at the same time we must ignore the snide laughs of Macintosh users who have yet to encounter a virus... -- New York Times === Subject: Re: Octave Software - Null function Basically, a singular value decomposition of a matrix A results in three matrices U, S and V such that U and V are unitary, S is diagonal and A = U * S * V' or A * V = U * S. So by choosing those columns of V such that A * V = 0, you get a unitary basis for the null space of A. This is exactly what Matlab and Octave do. This is probably a shoddy explanation, but any good textbook on linear algebra should have a better one. === Subject: Strongly balanced tournaments A tournament is an oriented complete graph on n vertices. Call a tournament balanced if all n vertices have equal in- and out-degrees. Clearly a balanced tournament must have odd n. Balanced tournaments exist for all positive odd n: place the vertices on a circle and direct edges from each vertex to each of the next (n-1)/2 on the circle. Define a strongly balanced tournament to be a balanced tournament such that for each vertex v, the graph induced by the successors of v is itself strongly balanced (where the trivial n=1 tournament is considered strongly balanced). This requires n to be of the form 2^k-1. Question: For every n of the form 2^k-1 does there exist a strongly balanced tournament? For example, for n=7 this reduces to asking whether there exists a tournament such that for each vertex v the graph induced by the successor-set of v is a directed 3-cycle. Such a tournament does indeed exist. I don't know about n=15, though. -Jim Ferry Metron, Inc. f rr @m tsc .c m e y e i o === Subject: Re: Strongly balanced tournaments I realized that some of the terminology I was using is non-standard. Cf. Sloane sequence A096368 and the following site that it links to: http://cs.anu.edu.au/~bdm/data/digraphs.html Balanced tournaments already had a name, namely regular tournaments. What I was calling the graph induced by the successor-set of v is more concisely and canonically called the out-neighborhood of v. The link above gives, for example, the three non-isomorphic balanced tournaments on 7 vertices. I haven't seen anything about strongly bal..., I mean strongly regular tournaments, however. -Jim Ferry Metron, Inc. f rr @m tsc .c m e y e i o === Subject: Cartan Forms & Warp Drive Star Gate Lectures 1 & 2 Lecture 2 Ex 4 Mechanical Model of a Phase Singularity Simplest case. Imagine a plane. Pick an origin O. Use polar coordinates, (r,theta) for arbitrary moving point P. Pick a point 0' with fixed coordinates (a, chi). Draw a circle of radius b < a centered at O' with coordinates (b,phi) Let point P move around this circle whose center O' is displaced from origin O. Obviously when a =/= 0 the total theta angle integral of the 1-form dtheta swept out in one complete circuit round the circle is ZERO. Basically theta oscillates. Note that the angle theta depends on the angles chi and phi. Half of the movement is clockwise and then counter-clockwise for dtheta on successive semicircles as P winds around the circumference of the displaced circle. This is most easily seen intuitively all at once when O' is vertical compared to O (on y-axis ordinate). Note what happens when you move the circle to different locations on the plane. Draw tangents from O to the circle in different locations. In contrast, when a = 0, or alternatively, b > a the total angle integral of dtheta is 2pi. Homework Problem Use trigonometry to make an algebraic proof. ------------------- For 3D flat metric, the Hodge * is with the right-hand rule convention *dx/dy = dz *dy/dz = dx *dz/dx = dy Left-hand rule is *dy/dx = dz *dz/dy = dx *dx/dz = dy Parity transformation interchanges left and right hand rules in 3D. (x,y,z) -> (-x,-y,-z) SU(2)hypercharge breaks parity symmetry and it also may be the origin of inertia and gravity. p-forms |p) = (p!)^-1Fuv ... dx^u/dx^v ... p-factors, p =< n = dim of manifold. Important formula |p)/|q) = (-1)^(pq)|q)/|p) The exterior product / of forms is a parallelepiped in the co-tangent n-dim space of constant phase wave fronts in contrast to the tangent For R^3 A = Axdx + Aydy + Azdz F = dA = ( Az,y - Ay,z)dy/dz + (Az,x - Ax,z)dz/dx + (Ax,y - Ay,x)dx/dy 2-form independent of metric *F = *dA = ( Az,y - Ay,z)dx + (Az,x - Ax,z)dy + (Ax,y - Ay,x)dz * dual 1-form in 3D manifold with a metric specified. Note, if A = df F = dA = d^2f = 0 Therefore ( Az,y - Ay,z) = 0 etc , is ordinary partial derivative i.e. mixed second order partial derivatives of the 0-form f commute in that case. However, in the case of a phase-singularity, there is some kind of region in the manifold where the mixed partials of the 0-form Goldstone phase of the local macro-quantum coherent vacuum order parameter Higgs field in our primary application to physics of this formalism do not commute. This is a topological defect in the vacuum manifold G/H, where I write A = 'd'f d'd' =/= d^2 = 0 due to multiply-connected manifolds F = dA =/= 0 e.g. non-integrable anholonomic multi-valued gauge transformation of Hagen Kleinert AKA Flux without flux see also the related idea of the nonlocal Bohm-Aharonov effect using Feynman amplitude Wilson loop operators. In 3 space d|0) is gradient of a function, i.e. scalar field d|1) is curl of a vector field d|2) is divergence of a vector field B = Bxydx/dy + Byzdy/dz + Bzxdz/dx dB = (Bxy,z + Byz,x + Bzx,y)dx/dy/dz Static 4D Metrics without gravimagnetism (non-rotating spacetimes) & without gravity waves (c = 1 convention) here (curved metric) = g = -dt^2 + 3^g The toy model wormhole is of this form. We need positive dark zero point energy density with negative pressure to keep the wormhole open. There is no event horizon in this wormhole. It's not a black hole! A metric allows the symmetric inner product { , }. Classical energy density of the EM field in the absence of sources is (1/2)[{E,E} + {B,B}] The Lagrangian density is (1/2)[{E,E} - {B,B} E = (Ftx, Fty, Ftz) electric field B = (Fyz, Fzx, Fxy) magnetic field F = B + E/dt F & B are 2-forms E is a 1-form We need a classical EM stress-energy density tensor T to compute T ~ &(Dynamical Action)/&(metric) & is functional derivative of classical Lagrangian field theory (not w = (pressure/energy density) Note, the above is classical without any quantum zero point fluctuations. w = +1/3 for classical far-field radiation with only 2 transverse polarizations. For example, the cosmic black body radiation has w = +1/3 http://www-conf.slac.stanford.edu/ssi/2005/lec_notes/Kolb1/kolb1new_Page_05_ jpg.htm It's wrong to use w = +1/3 for vacuum zero point energy that bends spacetime absolutely. This is an error in SED used by HRP. The Casimir force is not important for metric engineering. Equivalence principle + local Lorentz invariance imply w = -1 for all kinds of zero point energy (isotropically distributed). That is the ZPF stress-energy diagonal is for pressure P (P,-P,-P,-P) i.e. Trace is -2P If we stick in plates or somehow break the rotational symmetry (rotating superconducting disks that phase lock to the vacuum Goldstone phase?) Then (-P, +aP, + bP, + cP), the trace is now (1 - a - b - c)P This is quintessence and it can perhaps be done with the Shipov torsion field. Lecture 1 on Cartan Forms I am using John Baez's Ch 4 of Gauge Fields, Knots and Gravity for the standard ideas. All the local physical observables in classical gauge force field theories are examples of Eli Cartan's differential forms, e.g., Au, Fuv, ju. The integrals of forms over manifolds are premetrical until we define a Hodge * operation taking a p-form to a N-p form for N-dim manifolds. The p-forms are very much like Bishop BerkeleyÍs ñghosts of departed quantities.î They ñare neither finite, nor .83 infinitely small, nor yet nothing.î The 1-forms are dual to tangent vector fields on the manifold. A vector picture of quantum theory. The 1-form (AKA ñcotangent vectorî) is like a stack of wave fronts (AKA ñlittle hyperplanesî) of small extent as in Fig. 1 p. 45 (Baez) also in MTWÍs ñGravitation.î ñThe bigger df is, the more tightly packed the hyperplanes are.î Given a Cartesian coordinate basis of tangent vector fields {,u} and a dual basis of 1-forms {dx^u}, then duality here is dx^u,v = 1v^u = Kronecker delta NxN identity matrix. Given a 1-form df and any vector field v, the directional real number df(v) ñcounts how many little hyperplanes in the stack df the vector v crosses.î Linearity is built in as a postulate. The Cartan forms are invariants of local coordinate LNIF transformations Diff(N). Diff(N) is what you get when you locally gauge the global ND translation group. In 1915 GR the Cartan forms are also invariant under the local LIF Lorentz transformations O(1,3). In general this would be O(N) pre-signature. That is TNxO(N). / is the exterior product. Obviously we have a kind of quasi-algebra equivalent to dissecting an N-1 simplex or ñbraneî also giving partially ordered (non-Boolean?) lattices with the 0-form on bottom and the N-form on the top. N = 1 (dx), i.e. 1 N = 2 (dx, dy, dx/dy = -dy/dx), i.e. 3 N = 3 (dx,dx,dz, dx/dy, dx/dz, dy/dz, dx/dy/dz), i.e 7 N = 4 (dx,dy,dz,dt, dx/dy, dx/dz, dy/dx, dt/dx, dt/dy,dt/dz, dx/dy/dz, dt/dx/dy, dt/dx/dz, dt/dy/dz, dt/dx/dy/dz), i.e. 15 If we include the 0-form we have 2, 4, 8, 16, i.e. 2^N elements in the quasi-algebra that suggests the Clifford Algebras. There are obviously N!/p!(N-p)! p-forms in N space. This is also like an information space of N c-Bit Shannon Boolean strings. Obviously there will be some kind of matrix representation. For example N = 2 should correspond to the 3 Paul 2x2 spin matrices with the unit matrix. Therefore, there is a connection to U(1)xSU(2) here. N = 3 should have something to do with the 8 SU(3) matrices, and N = 4 obviously connects with the Dirac algebra and possibly U(4) especially when we complexify each real number space-time dimension and even go beyond that to quaternions & octonians. Classical gauge force theories include Maxwell's U(1) electromagnetic theory, Yang-Mills theories of the SU(2) weak and SU(3) strong forces of the leptons and quarks in the standard model and Einstein's theory of gravity (General Relativity, 1915 AKA GR) provided you do not work at the symmetric metric tensor level guv(x), but work at the square root 1-form tetrad e level. Note that Einstein's local equivalence principle is simply (curved metric ) = e(flat metric)e where e is the Einstein-Cartan 1-form tetrad field. You can write e = 1 + B B = curvature tetrad field Since the forms are local frame invariant this decomposition is objective. Global Special Relativity 1905 AKA SR is when B = 0 everywhere-when. Note that the (curved) metric has linear in B elastic terms and nonlinear quadratic in B plastic terms (H. Kleinert), i.e., (curved metric) = (flat metric) + 1(flat metric)B + B(flat metric)1 + B(flat metric)B The B^2 terms show that the gravity field is self-interacting like the SU(2) & SU(3) gauge fields, but unlike the U(1) Maxwell EM field. The Cartan exterior derivative operator d on forms generalizes the gradient, curl and divergence. Together with its dual boundary operator & on co-forms, there is a generalization of Stokes & Gauss's theorems to N-dimensional manifold integrations with multiple-connectivity (e.g. wormholes). The p-dim form |p) is the thing integrated. The dual co-form (p| is the manifold on which the integral is done. I use a variation on the Dirac bra-ket notation. The basic integration theorem, is like the adjoint operation in quantum theory, i.e. (&(p+1)|p) = (p+1|d|p) The two identities d^2 = 0 &^2 = 0 are analogous to the antisymmetric Pauli exclusion principle in quantum field theory where a^2 = 0 a*^2 = 0 a* creates a fermion, a destroys a fermion. However, we use the notations ïdÍ and ï&î partially introduced by John Ex. 1: ïdthetaÍ = (xdy [CapitalEth] ydx)/(x^2 + y^2) for the polar angle ñthetaî where dr = (xdy + ydx)/(x^2 + y^2) x = rcos(theta) y = rsin(theta) The 1-form ïdthetaÍ above is closed, but not exact. In effect this means dÍdÍ =/= 0 ONLY when the integral is over a non-bounding co-form (AKA non-bounding cycle). Therefore, for this particular example, there is phase (theta) ambiguity at the origin r = 0. When the closed loop integral (&Í2|ÍdÍthetaÍ) = (2Í|dÍdthetaÍ) =/= 0 encircles the origin r = 0 it does not vanish. Note that if the closed loop integral of ïdthetaÍ does not encircle the branch point r = 0, it will vanish. In this sense, ïdthetaÍ is closed, but not exact and &2Í is not a true boundary because of the ñholeî at r = 0. Note that the co-form (2Í| is the area enclosed by the loop &Í2 minus the ñholeî at r = 0. If we extend this to cylindrical coordinates, then we have a vortex core string provided we have a local U(1) complex order parameter PSI(r,theta,z) such that PSI(0,theta,z) = 0 PSI(r,theta,z) = PSI(r, theta + 2pi, z) for equilibrium ñstationary statesî when the closed system relaxes expelling excess flux in the Meissner effect. In that case, (&Í2|ÍdÍthetaÍ) = (2Í|dÍdthetaÍ) = 2piN N = +-1, +-2 .83. N = winding number around the string vortex core on the z-axis. To review, the rigorous theorem is (&(p+1)|p) = (p+1|dp) Where (&(p+1)| is a true boundary, which means (&^2(p+1)| = 0 When |p) is exact, that means |p) = |d(p-1)) and |dp) = |d^2(p-1)) = 0 However, when the topology of the co-form manifold is multiply connected we can have closed p-manifolds, AKA ñnon-bounding p-cyclesî, (&Í(p+1)| that are not true boundaries together with non-exact p-forms |dÍ(p-1)) such that (&Í(p+1)|dÍ(p-1) = (p+1Í|ddÍ(p-1)) =/= 0 The non-bounding p-cycles are p-dim wormhole mouths or ñStar Gate Portalsî that are ñThrough The Looking Glassî Darkly as it were, down the Rabbit Hole in Hyperspace. Ex. 2: Consider the 3D space-like metric of a static spherically symmetric non-rotating uncharged wormhole Star Gate is (3-metric) = dr^2 + f(r)^2(dtheta^2 + sin^2thetadphi^2) Where f(r) is the wormhole shape function. Each wormhole mouth looks like a closed spherical surface of radius R where R = f(r*) df(r*)/dr = 0 d^2f(r*)/dr^2 > 0 This closed S2 surface is not a complete boundary (&3| enclosing a 3-space because it has a twin wormhole mouth somewhere-when else perhaps in a parallel universe next door in hyperspace. Therefore, all wormhole mouths for actual time travel to distant places in negligible proper time for the traveler are really (&Í3| not (&3|. Furthermore, the curved tetrad field B = (hG/c^3)^1/2ÍdÍ(Goldstone Phase) is not exact, i.e. in a 1-D loop around the wormhole mouth S2 surface with the multiply-connected quasi StokeÍs theorem (1Í|B) = (&Í3|dB) =/= 0 This is the curved tetrad flux through the closed loop that ñcutsî the spherical wormhole mouth. Ex. 3: Given in cylindrical coordinates the vortex string along the z-axis ïdÍtheta = (xdy [CapitalEth] ydx)/(x^2 + y^2) For any closed loop (1Í| = (&Í2| around the z-axis (1Í|ÍdÍtheta) = (2Í|dÍdÍtheta) =/= 0 i.e. Nonlocal Bohm-Aharonov ñFlux without fluxî Given the above wormhole 3-metric define a Hodge * operation, with the non-exact 2-form 3*ÍdÍtheta = (xdy/dz + ydz/dx + zdx/dy)/f(r)^3 Where now we have a multiply-connected quasi-divergence Gauss theorem (3Í|d3*ÍdÍtheta) = (&Í3|3*ÍdÍtheta)=/= 0 when (&Í3| is a wormhole portal. There is now a radial 3*ÍdÍtheta flux through the wormhole closed surface in addition to the ïdÍtheta circulation around a closed loop that cuts the wormhole closed surface that is not a complete boundary. === Subject: Re: Cartan Forms & Warp Drive Star Gate Lectures 1 & 2 Now to comment on something. > Lecture 2 JW Jack Sarfatti: In my opinion, you are the smartest in the world and it's a pleasure to be with you on the Internet. === Subject: Re: Open Sets & Subspace Topologies > <16436630.1128018318895.JavaMail.jakarta@nitrogen.math > forum.org>, > Narcoleptic Insomniac >**This is a homework exercise so only hints please** >If anyone could take a look at my reasoning and let >me know if this seems correct I'd appreciate it. >Consider the subspace Y = [-1, 1] of R where R has >the usual topology T. Then the subspace topology T_y >is defined as the collection {[-1, 1] / K : K is in T}. >It's clear to me that it is possible for a set to be >open in Y but not be open in R. >Okay, now let C be the set C = {x : 1/2 <= |x| < 1} >which can be written as C = (-1, -1/2] / [1/2, 1). I >know that C is not open in R since C is not an >element of T (that is to say, C cannot be expressed >as a union nor as an intersection of open sets in >R). Moreover, C is not open in Y either, since it is >not in the collection T_y. > Well, surely you would have to justify that last > part. How do you know there is no weird open set in R > which when you intersect it with Y just happens to > yields C? >Now consider the set B = {x : 1/2 < |x| <= 1} which >can expressed as B = [-1, -1/2) / (1/2, 1]. This set >is not open in T for the same reasons that C was not >open in T. However, B is open in T_y since the set >[-1, -1/2) can be seen as the intersection [-1, 1] / >(-n, -1/2) where n > 1. The argument is similar for >(1/2, 1] and thus the union (which is B) is open in >T_y. >These aren't formal proofs, > I'm unclear what statement you are trying to establish > in any case. Are you trying to show that it is > possible for a subset of Y to be open in Y, but not > open when you consider it a subspace of R? Or > somethign else? Are you trying to prove that B is open > in Y but not in R, and that C is not open in either R > or Y? Well, these two sets B and C are just a few of the sets from a homework exercise. The objective of the exercise is to find out if these various sets (like B and C) are open in Y and/or open in R (if they're even open at all). The exercises are in the section dealing with subspace topology. So yeah, there isn't a general statement that needs proof; the problems are just intended for us to become familiar with open sets with respect to subspaces. >but I just wanted to see if >the reasoning and logic I'm using seems correct. > If you are trying to show that B is open in Y but not > in R, then I would suggest using a specific value of n > rather than your unspecified n above; just exhibit the > open sets you want, explicitly and with nothing left > in the air. If you are also trying to show that C is > not open in Y, then I would say you have a bit more > work to do. You need to explain why there is no open > set U in R such that U / Y = C. You should be able > to do this easily, but noting what the fact that 1/2 > and -1/2 have to be in U implies for the intersection. This last part with the set C is my main concern and the reason I decided to begin this thread. You're right, I definately have to justify my intuition on this, and I should be able to do this easily. Unfortunately I'm just not seeing it at the moment. Kyle === Subject: Re: Open Sets & Subspace Topologies days. My association with the Department is that of an alumnus. [.snip.] >>Consider the subspace Y = [-1, 1] of R where R has >>the usual topology T. Then the subspace topology T_y >>is defined as the collection {[-1, 1] / K : K is in T}. [...] >>Okay, now let C be the set C = {x : 1/2 <= |x| < 1} >>which can be written as C = (-1, -1/2] / [1/2, 1). I >>know that C is not open in R since C is not an >>element of T (that is to say, C cannot be expressed >>as a union nor as an intersection of open sets in >>R). Moreover, C is not open in Y either, since it is >>not in the collection T_y. [...] Me: >> If you are also trying to show that C is >> not open in Y, then I would say you have a bit more >> work to do. You need to explain why there is no open >> set U in R such that U / Y = C. You should be able >> to do this easily, but noting what the fact that 1/2 >> and -1/2 have to be in U implies for the intersection. >This last part with the set C is my main concern and the >reason I decided to begin this thread. You're right, I >definately have to justify my intuition on this, and I >should be able to do this easily. Unfortunately I'm just >not seeing it at the moment. If U is an open set in R such that U / Y = C, then U must contain 1/2. And since it is open, U must contain an open interval centered at 1/2. And therefore, the intersection with Y would... -- 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: Open Sets & Subspace Topologies > <12699718.1128029947541.JavaMail.jakarta@nitrogen.math > forum.org>, > Narcoleptic Insomniac > [.snip.] >>Kyle: >>Consider the subspace Y = [-1, 1] of R where R has >>the usual topology T. Then the subspace topology T_y >>is defined as the collection {[-1, 1] / K : K is in T}. > [...] >>Okay, now let C be the set C = {x : 1/2 <= |x| < 1} >>which can be written as C = (-1, -1/2] / [1/2, 1). >>I know that C is not open in R since C is not an >>element of T (that is to say, C cannot be expressed >>as a union nor as an intersection of open sets in >>R). Moreover, C is not open in Y either, since >>it is not in the collection T_y. > [...] >> Arturo: >> If you are also trying to show that C is >> not open in Y, then I would say you have a bit >> more work to do. You need to explain why there is no >> open set U in R such that U / Y = C. You should be >> able to do this easily, but noting what the fact >> that 1/2 and -1/2 have to be in U implies for the >> intersection. >This last part with the set C is my main concern and the >reason I decided to begin this thread. You're right, I >definately have to justify my intuition on this, and I >should be able to do this easily. Unfortunately I'm just >not seeing it at the moment. > If U is an open set in R such that U / Y = C, then U > must contain 1/2. And since it is open, U must contain > an open interval centered at 1/2. And therefore, the > intersection with Y would... ..either be an open interval of the form (1/2, b) where 1/2 < b < 1 or an interval of the form (1/2, 1]. The same goes for the -1/2 which would only yeild an open interval of the form (a, -1/2) where -1 < a < -1/2 or an interval of the form [-1, -1/2). This is why the set B was open in Y but the set C is not. Okay, I think I see Kyle === Subject: Re: Open Sets & Subspace Topologies days. My association with the Department is that of an alumnus. >> <12699718.1128029947541.JavaMail.jakarta@nitrogen.math >> forum.org>, >> Narcoleptic Insomniac >> [.snip.] >Kyle: Consider the subspace Y = [-1, 1] of R where R has >the usual topology T. Then the subspace topology T_y >is defined as the collection {[-1, 1] / K : K is in T}. >> [...] >> >Okay, now let C be the set C = {x : 1/2 <= |x| < 1} >which can be written as C = (-1, -1/2] / [1/2, 1). >I know that C is not open in R since C is not an >element of T (that is to say, C cannot be expressed >as a union nor as an intersection of open sets in >R). Moreover, C is not open in Y either, since >it is not in the collection T_y. >> [...] > Arturo: If you are also trying to show that C is > not open in Y, then I would say you have a bit > more work to do. You need to explain why there is no > open set U in R such that U / Y = C. You should be > able to do this easily, but noting what the fact > that 1/2 and -1/2 have to be in U implies for the > intersection. >>This last part with the set C is my main concern and the >>reason I decided to begin this thread. You're right, I >>definately have to justify my intuition on this, and I >>should be able to do this easily. Unfortunately I'm just >>not seeing it at the moment. >> If U is an open set in R such that U / Y = C, then U >> must contain 1/2. And since it is open, U must contain >> an open interval centered at 1/2. And therefore, the >> intersection with Y would... >..either be an open interval of the form (1/2, b) where >1/2 < b < 1 or an interval of the form (1/2, 1]. Yikes... The intersection with Y would contain an open interval centered at 1/2 (since Y itself also contains an open interval centered at 1/2, and the intersection of two open intervals centered at 1/2 is an open interval centered at 1/2). As C does NOT contain an open interval centered at 1/2, we cannot have U/Y = C for any set U which is open in R. -- 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: Open Sets & Subspace Topologies > Yikes... Yikes with a captial F! I don't know why I was looking at the intersection Y / C instead of Y / U! ~Kyle === Subject: Re: 4 spheres inside a tetrahedron >> On 28 Sep 2005 21:02:31 -0700, >> jmorriss@idirect.com >>The only condition I can think of is that the >> smallest sphere not fall >>through the gap where the first three meet. >> Now that you state it that way, it seems obvious. >> Ok, so the question then is this: >> Given 3 mutually tangent spheres with radii >> r1<=r2<=r3, what is the >> smallest radius r4 such that a sphere of radius r4 >> can be placed >> mutually tangent to the other 3? In other words, >> find r4 such that any >> sphere of radius less than r4 could fall right >> through without >> touching (in basketball terminology -- swish). >> OK, the centres of the three spheres lie in some >> plane, and the >> tightest >> fit for the fourth sphere will be when its centre is >> in the same plane. >> Looking at that plane, we can rephrase the question >> as: given three >> mutually tangent circles with radii r1 <= r2 <= r3, >> what is the radius >> of a fourth circle tangent to the other three? >> Using coordinates, setting up equations and using a >> Groebner basis >> with Maple's help, I get >> (r1^2*r3^2+r2^2*r1^2-2*r2*r1^2*r3-2*r2^2*r1*r3-2*r1*r2 >> *r3^2+r2^2*r3^2)*r4^2 >> -2*r1*r2*r3*(r1*r2+r2*r3+r1*r3)*r4+r2^2*r1^2*r3^2 = 0 >> which has solutions >> r4 = >> (r1*r3+r1*r2+r2*r3+2*(r2^2*r1*r3+r2*r1^2*r3+r1*r2*r3^2 >> )^(1/2))/ >> (r1^2*r3^2+r2^2*r1^2-2*r2*r1^2*r3-2*r2^2*r1*r3-2*r1*r2 >> *r3^2+r2^2*r3^2)*r3*r1*r2, >This value can be simplified to >r4=r1*r2*r3/((r1*r2+r1*r3+r2*r3)-2*(r1*r2*r3*(r1+r2+r3))^(1/2)) OK. >This value is always negative No. For example, with r3 = r2 it is (2 r1 + r2 + 2 sqrt(r1^2 + 2 r1 r2)) r1/(r2 - 4 r1) which could be either positive or negative depending on how r2 relates to 4 r1. The cases with denominator 0 correspond to the three circles being tangent to a straight line. Positive and negative values correspond to the three circles being outside or outside the fourth circle (with radius |r4|). >> and >> r4 = >> (r1*r3+r1*r2+r2*r3-2*(r2^2*r1*r3+r2*r1^2*r3+r1*r2*r3^2 >> )^(1/2))/ >> (r1^2*r3^2+r2^2*r1^2-2*r2*r1^2*r3-2*r2^2*r1*r3-2*r1*r2 >> *r3^2+r2^2*r3^2)*r3*r1*r2 >This value can be simplified to >r4=r1*r2*r3/((r1*r2+r1*r3+r2*r3)+2*(r1*r2*r3*(r1+r2+r3))^(1/2)) >This value is always positive OK. Your form is better in that it works in the case where the three circles are tangent to a straight line. === Subject: Cumbersome system of equations help I am trying to show that for any floretions x, y there exists a third floretion such that ExEy = Ez. x and y have 16 components each (x = x[0]'i + x[1]'j + ... x[15]e with real numbers x[i]) and E = 'i + i' + 'ii' + 'jj' + 'kk' + 'jk' + kj' + e Formulated explicitely (This is very long- my sincere apologies if I made a mistake at some point. I initially tried to use the Maple solve command but I'm a Maple newcomer and am having trouble getting Maple to say anything in the case at hand.): I need to know if real numbers z[0], z[1], ..., z[15] can be chosen such that (Note: some of the u's are in fact equal- for ex. u6 and u7. Furthermore, had set up the problem such that we were trying to prove xEyE = zE it would be u6 and u8 which would have been equal): u0 = z[15] - z[6] - z[3] - z[14] + z[12] - z[8] + z[7] + z[0] u1 = -z[2] - z[11] + z[13] - z[4] - z[10] - z[5] - z[9] + z[1] u2 = z[1] - z[13] - z[11] + z[9] - z[5] + z[10] - z[4] + z[2] u3 =-z[6] + z[15] - z[0] - z[12] + z[14] + z[7] - z[8] + z[3] u4 = -z[9] - z[5] + z[10] - z[1] - z[13] - z[11] - z[2] + z[4] u5 = -z[10] + z[4] - z[9] + z[11] - z[2] - z[1] + z[13] + z[5] u6 = z[3] + z[0] + z[15] + z[8] + z[7] - z[14] - z[12] + z[6] u7 = z[3] + z[0] + z[15] + z[8] + z[7] - z[14] - z[12] + z[6] u8 = z[12] + z[14] + z[7] + z[6] + z[15] - z[0] - z[3] + z[8] u9 = z[4] - z[10] - z[5] + z[2] - z[11] + z[13] - z[1] + z[9] u10 = z[5] + z[9] + z[4] - z[13] - z[1] + z[2] + z[11] + z[10] u11 = -z[13] + z[1] - z[2] + z[5] - z[9] - z[4] + z[10] + z[11] u12 = z[15] - z[6] - z[3] - z[14] + z[12] - z[8] + z[7] + z[0] u13 = z[11] + z[2] + z[1] - z[10] - z[4] + z[9] + z[5] + z[13] u14 =-z[6] + z[15] - z[0] - z[12] + z[14] + z[7] - z[8] + z[3] u15 = z[12] + z[14] + z[7] + z[6] + z[15] - z[0] - z[3] + z[8] ***AND************** (in MAPLE Format) u0 := x[2]*y[11]+ x[2]*y[9]+ x[2]*y[5]+ x[2]*y[2]- x[11]*y[11]- x[11]*y [9]- x[11]*y[5]- x[11]*y[2]+ x[13]*y[1]+ x[13]*y[13]- x[13]*y[10]- x[13] *y[4]+ x[4]*y[1]+ x[4]*y[13]- x[4]*y[10]- x[4]*y[4]- x[10]*y[1]- x[10]*y [13]+ x[10]*y[10]+ x[10]*y[4]- x[5]*y[11]- x[5]*y[9]- x[5]*y[5]- x[5]*y[ 2]+ x[9]*y[11]+ x[9]*y[9]+ x[9]*y[5]+ x[9]*y[2]- x[1]*y[1]- x[1]*y[13]+4 *x[1]*y[10]+ x[1]*y[4]+ x[3]*y[6]- x[3]*y[15]- x[3]*y[7]+ x[3]*y[8]+ x[0 ]*y[6]- x[0]*y[15]- x[0]*y[7]+ x[0]*y[8]+ x[15]*y[0]+ x[15]*y[12]- x[15] *y[14]- x[15]*y[3]+ x[8]*y[0]+ x[8]*y[12]- x[8]*y[14]- x[8]*y[3]+ x[7]*y [0]+ x[7]*y[12]- x[7]*y[14]- x[7]*y[3]- x[14]*y[6]+ x[14]*y[15]+ x[14]*y [7]- x[14]*y[8]- x[12]*y[6]+ x[12]*y[15]+ x[12]*y[7]- x[12]*y[8]+ x[6]*y [0]+ x[6]*y[12]- x[6]*y[14]- x[6]*y[3]; u1 := - x[6]*y[4]+ x[3]*y[11]+ x[3]*y[9]+ x[3]*y[5]+ x[3]*y[2]- x[14]*y[ 11]- x[14]*y[9]- x[14]*y[5]- x[14]*y[2]- x[12]*y[11]- x[12]*y[9]- x[12]* y[5]- x[12]*y[2]+ x[8]*y[1]+ x[8]*y[13]- x[8]*y[10]- x[8]*y[4]+ x[7]*y[1 ]+ x[7]*y[13]- x[7]*y[10]- x[7]*y[4]+ x[0]*y[11]+ x[0]*y[9]+ x[0]*y[5]+4 *x[0]*y[2]+ x[2]*y[7]- x[2]*y[6]+ x[2]*y[15]- x[2]*y[8]- x[11]*y[7]+ x[ 11]*y[6]- x[11]*y[15]+ x[11]*y[8]- x[13]*y[12]+ x[13]*y[14]- x[13]*y[0]+4 *x[13]*y[3]- x[4]*y[12]+ x[4]*y[14]- x[4]*y[0]+ x[4]*y[3]+ x[10]*y[12]- x[10]*y[14]+ x[10]*y[0]- x[10]*y[3]- x[5]*y[7]+ x[5]*y[6]- x[5]*y[15]+ x [5]*y[8]+ x[9]*y[7]- x[9]*y[6]+ x[9]*y[15]- x[9]*y[8]+ x[1]*y[12]- x[1]* y[14]+ x[1]*y[0]- x[1]*y[3]+ x[15]*y[1]+ x[15]*y[13]- x[15]*y[10]- x[15] *y[4]+ x[6]*y[1]+ x[6]*y[13]- x[6]*y[10]; u2 := x[6]*y[2]- x[3]*y[1]+ x[3]*y[13]- x[3]*y[10]+ x[3]*y[4]+ x[14]*y[ 1]- x[14]*y[13]+ x[14]*y[10]- x[14]*y[4]+ x[12]*y[1]- x[12]*y[13]+ x[12] *y[10]- x[12]*y[4]- x[8]*y[11]+ x[8]*y[9]- x[8]*y[5]+ x[8]*y[2]- x[7]*y[ 11]+ x[7]*y[9]- x[7]*y[5]+ x[7]*y[2]- x[0]*y[1]+ x[0]*y[13]- x[0]*y[10]+ x[0]*y[4]+ x[2]*y[12]+ x[2]*y[14]- x[2]*y[0]- x[2]*y[3]- x[11]*y[12]- x[11]*y[14]+ x[11]*y[0]+ x[11]*y[3]- x[13]*y[7]- x[13]*y[6]- x[13]*y[15]- x[13]*y[8]- x[4]*y[7]- x[4]*y[6]- x[4]*y[15]- x[4]*y[8]+ x[10]*y[7]+ x [10]*y[6]+ x[10]*y[15]+ x[10]*y[8]- x[5]*y[12]- x[5]*y[14]+ x[5]*y[0]+ x [5]*y[3]+ x[9]*y[12]+ x[9]*y[14]- x[9]*y[0]- x[9]*y[3]+ x[1]*y[7]+ x[1]* y[6]+ x[1]*y[15]+ x[1]*y[8]- x[15]*y[11]+ x[15]*y[9]- x[15]*y[5]+ x[15]* y[2]- x[6]*y[11]+ x[6]*y[9]- x[6]*y[5]; u3 := - x[2]*y[11]- x[2]*y[9]- x[2]*y[5]- x[2]*y[2]+ x[11]*y[11]+ x[11]* y[9]+ x[11]*y[5]+ x[11]*y[2]+ x[13]*y[1]+ x[13]*y[13]- x[13]*y[10]- x[13 ]*y[4]+ x[4]*y[1]+ x[4]*y[13]- x[4]*y[10]- x[4]*y[4]- x[10]*y[1]- x[10]* y[13]+ x[10]*y[10]+ x[10]*y[4]+ x[5]*y[11]+ x[5]*y[9]+ x[5]*y[5]+ x[5]*y [2]- x[9]*y[11]- x[9]*y[9]- x[9]*y[5]- x[9]*y[2]- x[1]*y[1]- x[1]*y[13]+ x[1]*y[10]+ x[1]*y[4]+ x[3]*y[6]- x[3]*y[15]- x[3]*y[7]+ x[3]*y[8]+ x[ 0]*y[6]- x[0]*y[15]- x[0]*y[7]+ x[0]*y[8]- x[15]*y[0]- x[15]*y[12]+ x[15 ]*y[14]+ x[15]*y[3]- x[8]*y[0]- x[8]*y[12]+ x[8]*y[14]+ x[8]*y[3]- x[7]* y[0]- x[7]*y[12]+ x[7]*y[14]+ x[7]*y[3]- x[14]*y[6]+ x[14]*y[15]+ x[14]* y[7]- x[14]*y[8]- x[12]*y[6]+ x[12]*y[15]+ x[12]*y[7]- x[12]*y[8]- x[6]* y[0]- x[6]*y[12]+ x[6]*y[14]+ x[6]*y[3]; u4 := x[6]*y[4]+ x[3]*y[11]+ x[3]*y[9]+ x[3]*y[5]+ x[3]*y[2]- x[14]*y[ 11]- x[14]*y[9]- x[14]*y[5]- x[14]*y[2]- x[12]*y[11]- x[12]*y[9]- x[12]* y[5]- x[12]*y[2]- x[8]*y[1]- x[8]*y[13]+ x[8]*y[10]+ x[8]*y[4]- x[7]*y[1 ]- x[7]*y[13]+ x[7]*y[10]+ x[7]*y[4]+ x[0]*y[11]+ x[0]*y[9]+ x[0]*y[5]+4 *x[0]*y[2]- x[2]*y[7]+ x[2]*y[6]- x[2]*y[15]+ x[2]*y[8]+ x[11]*y[7]- x[ 11]*y[6]+ x[11]*y[15]- x[11]*y[8]- x[13]*y[12]+ x[13]*y[14]- x[13]*y[0]+4 *x[13]*y[3]- x[4]*y[12]+ x[4]*y[14]- x[4]*y[0]+ x[4]*y[3]+ x[10]*y[12]- x[10]*y[14]+ x[10]*y[0]- x[10]*y[3]+ x[5]*y[7]- x[5]*y[6]+ x[5]*y[15]- x [5]*y[8]- x[9]*y[7]+ x[9]*y[6]- x[9]*y[15]+ x[9]*y[8]+ x[1]*y[12]- x[1]* y[14]+ x[1]*y[0]- x[1]*y[3]- x[15]*y[1]- x[15]*y[13]+ x[15]*y[10]+ x[15] *y[4]- x[6]*y[1]- x[6]*y[13]+ x[6]*y[10]; u5 := - x[6]*y[2]+ x[3]*y[1]- x[3]*y[13]+ x[3]*y[10]- x[3]*y[4]- x[14]*y [1]+ x[14]*y[13]- x[14]*y[10]+ x[14]*y[4]- x[12]*y[1]+ x[12]*y[13]- x[12 ]*y[10]+ x[12]*y[4]+ x[8]*y[11]- x[8]*y[9]+ x[8]*y[5]- x[8]*y[2]+ x[7]*y [11]- x[7]*y[9]+ x[7]*y[5]- x[7]*y[2]+ x[0]*y[1]- x[0]*y[13]+ x[0]*y[10] - x[0]*y[4]- x[2]*y[12]- x[2]*y[14]+ x[2]*y[0]+ x[2]*y[3]+ x[11]*y[12]+4 *x[11]*y[14]- x[11]*y[0]- x[11]*y[3]+ x[13]*y[7]+ x[13]*y[6]+ x[13]*y[15] + x[13]*y[8]+ x[4]*y[7]+ x[4]*y[6]+ x[4]*y[15]+ x[4]*y[8]- x[10]*y[7]- x[10]*y[6]- x[10]*y[15]- x[10]*y[8]+ x[5]*y[12]+ x[5]*y[14]- x[5]*y[0]- x[5]*y[3]- x[9]*y[12]- x[9]*y[14]+ x[9]*y[0]+ x[9]*y[3]- x[1]*y[7]- x[1] *y[6]- x[1]*y[15]- x[1]*y[8]+ x[15]*y[11]- x[15]*y[9]+ x[15]*y[5]- x[15] *y[2]+ x[6]*y[11]- x[6]*y[9]+ x[6]*y[5]; u6 := - x[2]*y[1]+ x[2]*y[13]- x[2]*y[10]+ x[2]*y[4]+ x[11]*y[1]- x[11]* y[13]+ x[11]*y[10]- x[11]*y[4]+ x[13]*y[11]- x[13]*y[9]+ x[13]*y[5]- x[ 13]*y[2]+ x[4]*y[11]- x[4]*y[9]+ x[4]*y[5]- x[4]*y[2]- x[10]*y[11]+ x[10 ]*y[9]- x[10]*y[5]+ x[10]*y[2]+ x[5]*y[1]- x[5]*y[13]+ x[5]*y[10]- x[5]* y[4]- x[9]*y[1]+ x[9]*y[13]- x[9]*y[10]+ x[9]*y[4]- x[1]*y[11]+ x[1]*y[9 ]- x[1]*y[5]+ x[1]*y[2]- x[3]*y[0]+ x[3]*y[12]+ x[3]*y[14]- x[3]*y[3]- x[0]*y[0]+ x[0]*y[12]+ x[0]*y[14]- x[0]*y[3]+ x[15]*y[6]+ x[15]*y[15]+ x [15]*y[7]+ x[15]*y[8]+ x[8]*y[6]+ x[8]*y[15]+ x[8]*y[7]+ x[8]*y[8]+ x[7] *y[6]+ x[7]*y[15]+ x[7]*y[7]+ x[7]*y[8]+ x[14]*y[0]- x[14]*y[12]- x[14]* y[14]+ x[14]*y[3]+ x[12]*y[0]- x[12]*y[12]- x[12]*y[14]+ x[12]*y[3]+ x[6 ]*y[6]+ x[6]*y[15]+ x[6]*y[7]+ x[6]*y[8]; u7 := - x[2]*y[1]+ x[2]*y[13]- x[2]*y[10]+ x[2]*y[4]+ x[11]*y[1]- x[11]* y[13]+ x[11]*y[10]- x[11]*y[4]+ x[13]*y[11]- x[13]*y[9]+ x[13]*y[5]- x[ 13]*y[2]+ x[4]*y[11]- x[4]*y[9]+ x[4]*y[5]- x[4]*y[2]- x[10]*y[11]+ x[10 ]*y[9]- x[10]*y[5]+ x[10]*y[2]+ x[5]*y[1]- x[5]*y[13]+ x[5]*y[10]- x[5]* y[4]- x[9]*y[1]+ x[9]*y[13]- x[9]*y[10]+ x[9]*y[4]- x[1]*y[11]+ x[1]*y[9 ]- x[1]*y[5]+ x[1]*y[2]- x[3]*y[0]+ x[3]*y[12]+ x[3]*y[14]- x[3]*y[3]- x[0]*y[0]+ x[0]*y[12]+ x[0]*y[14]- x[0]*y[3]+ x[15]*y[6]+ x[15]*y[15]+ x [15]*y[7]+ x[15]*y[8]+ x[8]*y[6]+ x[8]*y[15]+ x[8]*y[7]+ x[8]*y[8]+ x[7] *y[6]+ x[7]*y[15]+ x[7]*y[7]+ x[7]*y[8]+ x[14]*y[0]- x[14]*y[12]- x[14]* y[14]+ x[14]*y[3]+ x[12]*y[0]- x[12]*y[12]- x[12]*y[14]+ x[12]*y[3]+ x[6 ]*y[6]+ x[6]*y[15]+ x[6]*y[7]+ x[6]*y[8]; u8 := - x[2]*y[1]+ x[2]*y[13]- x[2]*y[10]+ x[2]*y[4]+ x[11]*y[1]- x[11]* y[13]+ x[11]*y[10]- x[11]*y[4]- x[13]*y[11]+ x[13]*y[9]- x[13]*y[5]+ x[ 13]*y[2]- x[4]*y[11]+ x[4]*y[9]- x[4]*y[5]+ x[4]*y[2]+ x[10]*y[11]- x[10 ]*y[9]+ x[10]*y[5]- x[10]*y[2]+ x[5]*y[1]- x[5]*y[13]+ x[5]*y[10]- x[5]* y[4]- x[9]*y[1]+ x[9]*y[13]- x[9]*y[10]+ x[9]*y[4]+ x[1]*y[11]- x[1]*y[9 ]+ x[1]*y[5]- x[1]*y[2]+ x[3]*y[0]- x[3]*y[12]- x[3]*y[14]+ x[3]*y[3]+ x[0]*y[0]- x[0]*y[12]- x[0]*y[14]+ x[0]*y[3]+ x[15]*y[6]+ x[15]*y[15]+ x [15]*y[7]+ x[15]*y[8]+ x[8]*y[6]+ x[8]*y[15]+ x[8]*y[7]+ x[8]*y[8]+ x[7] *y[6]+ x[7]*y[15]+ x[7]*y[7]+ x[7]*y[8]- x[14]*y[0]+ x[14]*y[12]+ x[14]* y[14]- x[14]*y[3]- x[12]*y[0]+ x[12]*y[12]+ x[12]*y[14]- x[12]*y[3]+ x[6 ]*y[6]+ x[6]*y[15]+ x[6]*y[7]+ x[6]*y[8]; u9 := x[6]*y[2]+ x[3]*y[1]- x[3]*y[13]+ x[3]*y[10]- x[3]*y[4]- x[14]*y[ 1]+ x[14]*y[13]- x[14]*y[10]+ x[14]*y[4]- x[12]*y[1]+ x[12]*y[13]- x[12] *y[10]+ x[12]*y[4]- x[8]*y[11]+ x[8]*y[9]- x[8]*y[5]+ x[8]*y[2]- x[7]*y[ 11]+ x[7]*y[9]- x[7]*y[5]+ x[7]*y[2]+ x[0]*y[1]- x[0]*y[13]+ x[0]*y[10]- x[0]*y[4]+ x[2]*y[12]+ x[2]*y[14]- x[2]*y[0]- x[2]*y[3]- x[11]*y[12]- x[11]*y[14]+ x[11]*y[0]+ x[11]*y[3]+ x[13]*y[7]+ x[13]*y[6]+ x[13]*y[15]+ x[13]*y[8]+ x[4]*y[7]+ x[4]*y[6]+ x[4]*y[15]+ x[4]*y[8]- x[10]*y[7]- x [10]*y[6]- x[10]*y[15]- x[10]*y[8]- x[5]*y[12]- x[5]*y[14]+ x[5]*y[0]+ x [5]*y[3]+ x[9]*y[12]+ x[9]*y[14]- x[9]*y[0]- x[9]*y[3]- x[1]*y[7]- x[1]* y[6]- x[1]*y[15]- x[1]*y[8]- x[15]*y[11]+ x[15]*y[9]- x[15]*y[5]+ x[15]* y[2]- x[6]*y[11]+ x[6]*y[9]- x[6]*y[5]; u10 := x[6]*y[4]- x[3]*y[11]- x[3]*y[9]- x[3]*y[5]- x[3]*y[2]+ x[14]*y[ 11]+ x[14]*y[9]+ x[14]*y[5]+ x[14]*y[2]+ x[12]*y[11]+ x[12]*y[9]+ x[12]* y[5]+ x[12]*y[2]- x[8]*y[1]- x[8]*y[13]+ x[8]*y[10]+ x[8]*y[4]- x[7]*y[1 ]- x[7]*y[13]+ x[7]*y[10]+ x[7]*y[4]- x[0]*y[11]- x[0]*y[9]- x[0]*y[5]-4 *x[0]*y[2]- x[2]*y[7]+ x[2]*y[6]- x[2]*y[15]+ x[2]*y[8]+ x[11]*y[7]- x[ 11]*y[6]+ x[11]*y[15]- x[11]*y[8]+ x[13]*y[12]- x[13]*y[14]+ x[13]*y[0]-4 *x[13]*y[3]+ x[4]*y[12]- x[4]*y[14]+ x[4]*y[0]- x[4]*y[3]- x[10]*y[12]+ x[10]*y[14]- x[10]*y[0]+ x[10]*y[3]+ x[5]*y[7]- x[5]*y[6]+ x[5]*y[15]- x [5]*y[8]- x[9]*y[7]+ x[9]*y[6]- x[9]*y[15]+ x[9]*y[8]- x[1]*y[12]+ x[1]* y[14]- x[1]*y[0]+ x[1]*y[3]- x[15]*y[1]- x[15]*y[13]+ x[15]*y[10]+ x[15] *y[4]- x[6]*y[1]- x[6]*y[13]+ x[6]*y[10]; u11 := - x[6]*y[2]- x[3]*y[1]+ x[3]*y[13]- x[3]*y[10]+ x[3]*y[4]+ x[14]* y[1]- x[14]*y[13]+ x[14]*y[10]- x[14]*y[4]+ x[12]*y[1]- x[12]*y[13]+ x[ 12]*y[10]- x[12]*y[4]+ x[8]*y[11]- x[8]*y[9]+ x[8]*y[5]- x[8]*y[2]+ x[7] *y[11]- x[7]*y[9]+ x[7]*y[5]- x[7]*y[2]- x[0]*y[1]+ x[0]*y[13]- x[0]*y[ 10]+ x[0]*y[4]- x[2]*y[12]- x[2]*y[14]+ x[2]*y[0]+ x[2]*y[3]+ x[11]*y[12 ]+ x[11]*y[14]- x[11]*y[0]- x[11]*y[3]- x[13]*y[7]- x[13]*y[6]- x[13]*y[ 15]- x[13]*y[8]- x[4]*y[7]- x[4]*y[6]- x[4]*y[15]- x[4]*y[8]+ x[10]*y[7] + x[10]*y[6]+ x[10]*y[15]+ x[10]*y[8]+ x[5]*y[12]+ x[5]*y[14]- x[5]*y[0] - x[5]*y[3]- x[9]*y[12]- x[9]*y[14]+ x[9]*y[0]+ x[9]*y[3]+ x[1]*y[7]+ x [1]*y[6]+ x[1]*y[15]+ x[1]*y[8]+ x[15]*y[11]- x[15]*y[9]+ x[15]*y[5]- x[ 15]*y[2]+ x[6]*y[11]- x[6]*y[9]+ x[6]*y[5]; u12 := x[2]*y[11]+ x[2]*y[9]+ x[2]*y[5]+ x[2]*y[2]- x[11]*y[11]- x[11]* y[9]- x[11]*y[5]- x[11]*y[2]+ x[13]*y[1]+ x[13]*y[13]- x[13]*y[10]- x[13 ]*y[4]+ x[4]*y[1]+ x[4]*y[13]- x[4]*y[10]- x[4]*y[4]- x[10]*y[1]- x[10]* y[13]+ x[10]*y[10]+ x[10]*y[4]- x[5]*y[11]- x[5]*y[9]- x[5]*y[5]- x[5]*y [2]+ x[9]*y[11]+ x[9]*y[9]+ x[9]*y[5]+ x[9]*y[2]- x[1]*y[1]- x[1]*y[13]+ x[1]*y[10]+ x[1]*y[4]+ x[3]*y[6]- x[3]*y[15]- x[3]*y[7]+ x[3]*y[8]+ x[ 0]*y[6]- x[0]*y[15]- x[0]*y[7]+ x[0]*y[8]+ x[15]*y[0]+ x[15]*y[12]- x[15 ]*y[14]- x[15]*y[3]+ x[8]*y[0]+ x[8]*y[12]- x[8]*y[14]- x[8]*y[3]+ x[7]* y[0]+ x[7]*y[12]- x[7]*y[14]- x[7]*y[3]- x[14]*y[6]+ x[14]*y[15]+ x[14]* y[7]- x[14]*y[8]- x[12]*y[6]+ x[12]*y[15]+ x[12]*y[7]- x[12]*y[8]+ x[6]* y[0]+ x[6]*y[12]- x[6]*y[14]- x[6]*y[3]; u13 := - x[6]*y[4]- x[3]*y[11]- x[3]*y[9]- x[3]*y[5]- x[3]*y[2]+ x[14]*y [11]+ x[14]*y[9]+ x[14]*y[5]+ x[14]*y[2]+ x[12]*y[11]+ x[12]*y[9]+ x[12] *y[5]+ x[12]*y[2]+ x[8]*y[1]+ x[8]*y[13]- x[8]*y[10]- x[8]*y[4]+ x[7]*y[ 1]+ x[7]*y[13]- x[7]*y[10]- x[7]*y[4]- x[0]*y[11]- x[0]*y[9]- x[0]*y[5]- x[0]*y[2]+ x[2]*y[7]- x[2]*y[6]+ x[2]*y[15]- x[2]*y[8]- x[11]*y[7]+ x[ 11]*y[6]- x[11]*y[15]+ x[11]*y[8]+ x[13]*y[12]- x[13]*y[14]+ x[13]*y[0]-4 *x[13]*y[3]+ x[4]*y[12]- x[4]*y[14]+ x[4]*y[0]- x[4]*y[3]- x[10]*y[12]+ x[10]*y[14]- x[10]*y[0]+ x[10]*y[3]- x[5]*y[7]+ x[5]*y[6]- x[5]*y[15]+ x [5]*y[8]+ x[9]*y[7]- x[9]*y[6]+ x[9]*y[15]- x[9]*y[8]- x[1]*y[12]+ x[1]* y[14]- x[1]*y[0]+ x[1]*y[3]+ x[15]*y[1]+ x[15]*y[13]- x[15]*y[10]- x[15] *y[4]+ x[6]*y[1]+ x[6]*y[13]- x[6]*y[10]; u14 := - x[2]*y[11]- x[2]*y[9]- x[2]*y[5]- x[2]*y[2]+ x[11]*y[11]+ x[11] *y[9]+ x[11]*y[5]+ x[11]*y[2]+ x[13]*y[1]+ x[13]*y[13]- x[13]*y[10]- x[ 13]*y[4]+ x[4]*y[1]+ x[4]*y[13]- x[4]*y[10]- x[4]*y[4]- x[10]*y[1]- x[10 ]*y[13]+ x[10]*y[10]+ x[10]*y[4]+ x[5]*y[11]+ x[5]*y[9]+ x[5]*y[5]+ x[5] *y[2]- x[9]*y[11]- x[9]*y[9]- x[9]*y[5]- x[9]*y[2]- x[1]*y[1]- x[1]*y[13 ]+ x[1]*y[10]+ x[1]*y[4]+ x[3]*y[6]- x[3]*y[15]- x[3]*y[7]+ x[3]*y[8]+ x[0]*y[6]- x[0]*y[15]- x[0]*y[7]+ x[0]*y[8]- x[15]*y[0]- x[15]*y[12]+ x[ 15]*y[14]+ x[15]*y[3]- x[8]*y[0]- x[8]*y[12]+ x[8]*y[14]+ x[8]*y[3]- x[7 ]*y[0]- x[7]*y[12]+ x[7]*y[14]+ x[7]*y[3]- x[14]*y[6]+ x[14]*y[15]+ x[14 ]*y[7]- x[14]*y[8]- x[12]*y[6]+ x[12]*y[15]+ x[12]*y[7]- x[12]*y[8]- x[6 ]*y[0]- x[6]*y[12]+ x[6]*y[14]+ x[6]*y[3]; u15 := - x[2]*y[1]+ x[2]*y[13]- x[2]*y[10]+ x[2]*y[4]+ x[11]*y[1]- x[11] *y[13]+ x[11]*y[10]- x[11]*y[4]- x[13]*y[11]+ x[13]*y[9]- x[13]*y[5]+ x[ 13]*y[2]- x[4]*y[11]+ x[4]*y[9]- x[4]*y[5]+ x[4]*y[2]+ x[10]*y[11]- x[10 ]*y[9]+ x[10]*y[5]- x[10]*y[2]+ x[5]*y[1]- x[5]*y[13]+ x[5]*y[10]- x[5]* y[4]- x[9]*y[1]+ x[9]*y[13]- x[9]*y[10]+ x[9]*y[4]+ x[1]*y[11]- x[1]*y[9 ]+ x[1]*y[5]- x[1]*y[2]+ x[3]*y[0]- x[3]*y[12]- x[3]*y[14]+ x[3]*y[3]+ x[0]*y[0]- x[0]*y[12]- x[0]*y[14]+ x[0]*y[3]+ x[15]*y[6]+ x[15]*y[15]+ x [15]*y[7]+ x[15]*y[8]+ x[8]*y[6]+ x[8]*y[15]+ x[8]*y[7]+ x[8]*y[8]+ x[7] *y[6]+ x[7]*y[15]+ x[7]*y[7]+ x[7]*y[8]- x[14]*y[0]+ x[14]*y[12]+ x[14]* y[14]- x[14]*y[3]- x[12]*y[0]+ x[12]*y[12]+ x[12]*y[14]- x[12]*y[3]+ x[6 ]*y[6]+ x[6]*y[15]+ x[6]*y[7]+ x[6]*y[8]; === Subject: Re: Cumbersome system of equations help Correction: > I am trying to show that for any floretions x, y > there exists a third floretion z such that ExEy = Ez. === Subject: A little extra money for college students for more info email me for an immediate response: lukejea@coastalnow.net or http://www.feederfund.net/index.asp?refcode=Luke101 --------------------------------------------------------------------------- If you just can't wait to make some extra money you can make a little extra money by surfing the internet. http://www.studiotraffic.com/login.php?refid=657092&username=luke101 Math and Computer science students are the best === Subject: Re: Transcendence degree of field extensions > I know this is a basic question, but I'm having trouble seeing that > the transcendence degree of a field extension is well defined, > i.e., that any two transcendence bases have the same cardinality. > (A transcendence basis of K/F is defined as a set {a1,a2,...,an} of > elements in K that are algebraically independent over F and such > that K is algebraic over F(a1,a2,...,an).) Not sure if this is kosher, but can't one consider the vector space of derivatives, or the dual space of differentials, over the space F(a_1,...,a_n)? Isn't the dimension of this space equal to the degree of transendence. (A basis is given by d/da_1....,d/da_n, or dually by da_1,...,da_n .) -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: Transcendence degree of field extensions the transcendence degree of a field extension is well defined, > i.e., that any two transcendence bases have the same cardinality. > (A transcendence basis of K/F is defined as a set {a1,a2,...,an} of > elements in K that are algebraically independent over F and such > that K is algebraic over F(a1,a2,...,an).) > Not sure if this is kosher, > but can't one consider the vector space of derivatives, > or the dual space of differentials, over the space F(a_1,...,a_n)? > Isn't the dimension of this space equal to the degree of transendence. > (A basis is given by d/da_1....,d/da_n, or dually by da_1,...,da_n .) > -- > Timothy Murphy > e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie > tel: +353-86-2336090, +353-1-2842366 > s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland You probably can do that. But here is how it is usually done (assuming the transcendence degree is finite). Show that if T is a set of generating transcendentals and I is a set of independent transcendentals, then there is a subset J of T of the same cardinality as I such that T - J cup I is also a set of generating transcendentals. (Sound familiar?) The argument is much the same kind of argument as used when proving the uniqueness of basis number for vector spaces. You carry out the replacement one at a time, taking care not to discard any element of I at any time, which is where the independence of I is used. The infinite case is much harder (as is the proof for linear bases) and requires care. Mostly, in dealing with infinite dimensional vector spaces, people just ignore the problem since the main consequence (that a linear transformation is 1-1 iff it is onto) is false in any case. But both of these basis numbers are unique; it is just much harder to prove and I have forgotten how. === Subject: Re: Motion > The cause when CV exists wrt other objects is the fact that they are > all moving at the same speed and in the same direction. If you are moving with constant velocity wrt to another object, then that object is, by definition, at rest with respect to you. > Whatever > causes them to be in such a state can vary, but in our case above, the > fact that we are at rest wrt to the Earth is the cause of our CV wrt > the Earth. If we are at rest with respect to Earth, we are not movign at constyant velocity wrt the Earth. > For a sole object free of any external forces, CV is caused by the > inherent force which I claim exists in every body and which is the > total of the momentum of the mass and the energy it has due to its > motion. Can you describe an experiment you could perform that would establish whether a sole object is moving at constant velocity as opposed to being at rest? http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- === Subject: Re: Motion <1127684424_23799@spool6-east.superfeed.net> <1127751351_25367@spool6-east.superfeed.net> <1128036566_34385@spool6-east.superfeed.net The cause when CV exists wrt other objects is the fact that they are > all moving at the same speed and in the same direction. > If you are moving with constant velocity wrt to another object, then that > object is, by definition, at rest with respect to you. Yes, correct. > Whatever > causes them to be in such a state can vary, but in our case above, the > fact that we are at rest wrt to the Earth is the cause of our CV wrt > the Earth. > If we are at rest with respect to Earth, we are not movign at constyant > velocity wrt the Earth. Why not? We are moving at the same speed and in the same direction as the Earth, and that constitutes constant velocity. > For a sole object free of any external forces, CV is caused by the > inherent force which I claim exists in every body and which is the > total of the momentum of the mass and the energy it has due to its > motion. > Can you describe an experiment you could perform that would establish > whether a sole object is moving at constant velocity as opposed to being > at rest? As I have mentioned before, it is not possible, with what we know about our universe, to do that because everything is in motion in the universe for a number of reasons, the most common one being that the universe is in a state of expansion. The only way for constant velocity to occur is as a comparison between two or more objects like your home and the surface of the Earth as they move through space together as one object. Newton's 1st law of motion states that a body having no external net forces working on it will achieve constant velocity by itself. However, because that body is necessarily caught up in the universal expansion process, it too is involved in it one way or another. Besides the obvious effect that everything is moving away from each other, Newton's sole body is in motion due to its own inherent force, as my model claims, or due to it needing no motive power, i.e., a force to move it, as we have always been taught. We have no way of proving Newton's 1st law, but there is no reason to think a body will not achieve CV when all net external forces are removed from it. We could send a spaceship out and shut off its motive power, then it should achieve CV if it is so that space is expanding instead of the visible matter in the universe moving away from each other. At this time we are not sure what is really happening. I tend to think now that space is what is expanding because we see galaxies move apart but yet maintain their original coordinates wrt each other. I cannot think how that can happen unless it is space doing the expansion and moving the objects apart and not the objects themselves moving apart under their net forces. Good question. === Subject: Re: Motion >> If we are at rest with respect to Earth, we are not movign at constyant >> velocity wrt the Earth. > Why not? We are moving at the same speed and in the same direction as > the Earth, and that constitutes constant velocity. But when you are co-moving with another object, you are at rest wrt that object. That's what ti means when a physicist using the expression at rest. There is no way to make it more meaningful unless you can establish a true frame of refenrence that is at rest. So far, no one has been able to do that. >> Can you describe an experiment you could perform that would establish >> whether a sole object is moving at constant velocity as opposed to >> being >> at rest? > As I have mentioned before, it is not possible, with what we know about > our universe, to do that because everything is in motion in the > universe for a number of reasons, the most common one being that the > universe is in a state of expansion. If it's not possible, based on we know, then perhaps it's not possible at all. There is no way to tell. Based on what we know now, there is no way to make the determination. > The only way for constant > velocity to occur is as a comparison between two or more objects like > your home and the surface of the Earth as they move through space > together as one object. Right. Now, take that sentence and replace the words constant velocity with no velocity. Here is what you get: The only way for no velocity to occur is as a comparison between two or more objects like your home and the surface of the Earth as they move through space together as one object. > We have no way of proving Newton's 1st law, We have no way of proving ANY law. What we can do, though, is try to find exceptions to laws. In which case we can disprove them. That is all. When, despite all attempts to find an exception, we fail, it boosts our confidence in a law. That is currently the status of Newton's 1st Law. http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- === Subject: Re: Motion > Yes, I keep talking about the relation between force and > momentum, whereas you keep correcting me that the topic > of discussion is the relation between force and momentum, > clearly a different thing. Sorry about that. No, you keep talking about a derivative of momentum > Because dp/dt = F is a derivative of momentum. > in a futile effort > to show that momentum is not a force. > Because dp/dt is not equal to p. Hey, that's my position, not yours! A derivative cannot be equal to the function from which it is derived! If you knew what a derivative is you would have known that. > And you can't grasp what's wrong with that picture. > Because your ignorance is impervious. > And you've said every physics textbook does so, all over > the place. Is that a fact or is that your opinion based on one of your far-out > inferences? Cite the post so that I can check is you're lying or not. > The exact quote was this: >> Sure we do, just read a physics book. It is rare where we don't >> use terms like force and momentum nterchangably. > So I read this as saying that you believe if I pick up any > physics textbook, I will find that force and momentum are > used interchangeably. It is rare that they do not. Therefore > it should be easy to find such a passage in any physics book. > You can no doubt provide a quote from a physics > textbook where the term momentum is used to mean a force, > right? Thereby proving me wrong when I insist that nowhere > in physics is momentum used to mean force? How would that show that nowhere in physics is it used that way? > Read it again. Nowhere in physics is MY position. You show a real problem in comprehension of the written word. I know that is your position - did not say it was mine! Your low comprehension level shows that you don't understand English well enough to carry on a conversation without you requiring explanations of the meaning of almost every sentence. > Your > position is that it's rare for momentum and force NOT to be > used equivalently. Keep your position straight. And from that you came to believe that I think p and dp/dt are the same thing when I said they are not. Reading comprehension problem - to the woodpile for you. See if we can unscramble your brain. > Finding such a passage in a physics textbook would prove > me wrong, because my claim is that such a passage doesn't > exist. Yours (do try to remember this) is that it does, and > such passages are common. Proved you wrong, didn't I? Gave you one but you still claim it does not prove you wrong. Which only proves nothing will prove you wrong. There is really no sense continuing this discussion with you, is there? > [Snip remaining rant. Same old story: Tom thinks p and dp/dt > are interchangeable]. No, you're the one who said that so you're the one who thinks that. === Subject: Re: Motion > Because dp/dt is not equal to p. > Hey, that's my position, not yours! A derivative cannot be equal to > the function from which it is derived! If that's your position, then can you explain this comment of yours? False logic. dp/dt is a derivative of p so if force is a derivative of p, it must by definition also be p. It sure looks to me like you're saying the derivative of p is also by definition p. Now as to this: >> Sure we do, just read a physics book. It is rare where we don't >> use terms like force and momentum nterchangably. > So I read this as saying that you believe if I pick up any > physics textbook, I will find that force and momentum are > used interchangeably. It is rare that they do not. Therefore > it should be easy to find such a passage in any physics book. > You show a real problem in comprehension of the written word. I know > that is your position - did not say it was mine! Did you say just read a physics book. It is rare where we don't use terms like force and momentum nterchangably.? Is your position that you did not say that? > Finding such a passage in a physics textbook would prove > me wrong, because my claim is that such a passage doesn't > exist. Yours (do try to remember this) is that it does, and > such passages are common. > Proved you wrong, didn't I? By the Tipler quote that says force is the time rate of change of momentum? Again, how does that prove that F = dp/dt is wrong? > Gave you one but you still claim it does > not prove you wrong. Yes, I really can't see how force is the time rate of change of momentum proves F = dp/dt is wrong. Nor can I see how work done on an object equals change in KE of the object contradicts force and momentum are not interchangeable. I really can't. - Randy === Subject: Re: Motion > Hey, that's my position, not yours! A derivative cannot be equal to > the function from which it is derived! If you knew what a derivative > is you would have known that. What, not even when it's the exponential function? === Subject: Re: Motion <3q34kfFd335dU2@individual.net Hey, that's my position, not yours! A derivative cannot be equal to > the function from which it is derived! If you knew what a derivative > is you would have known that. > What, not even when it's the exponential function? I was wondering if Tom would muddy the waters with that one, though I already thought through my response. It's mathematically possible, but not physically possible for p to be equal to k*exp(t) on the basis of units. You don't see things like exp(t) in physics. Arguments to exponentials are always unitless. They'd better be, or the outcome would depend on the units of t. - Randy === Subject: Re: Motion > I was wondering if Tom would muddy the waters with that one, > though I already thought through my response. It's > mathematically possible, but not physically possible for > p to be equal to k*exp(t) on the basis of units. I tend to try not to think about this, because I get all confused, but if I'm forced to think about it, I think that the 't' there is just a number, which we interpret as the number of seconds (hours, fortnights, whatever) which have elapsed. Clearly, in this case, 'k' would have to have units of momentum. > You don't see things like exp(t) in physics. Used to see it all the time when I did damped harmonic motion. (OK, it was usually exp(-t) then.) But I suspect that I'm jumping in on a discussion and adding even more cross-purposes than were already there. Sorry about that. === Subject: Re: Motion <3q34kfFd335dU2@individual.net> <3q4tehFd1c9tU1@individual.net I was wondering if Tom would muddy the waters with that one, > though I already thought through my response. It's > mathematically possible, but not physically possible for > p to be equal to k*exp(t) on the basis of units. > I tend to try not to think about this, because I get > all confused, but if I'm forced to think about it, > I think that the 't' there is just a number, which > we interpret as the number of seconds (hours, fortnights, > whatever) which have elapsed. Nope. Can't work unless there's a time constant in there which makes the argument of exp() unitless. However... see below. > Clearly, in this case, > 'k' would have to have units of momentum. Right. It's the momentum at time t=0. > You don't see things like exp(t) in physics. > Used to see it all the time when I did damped > harmonic motion. (OK, it was usually exp(-t) then. It's really exp(a*t) where a is a rate constant, or exp(t/Tau) where Tau is a time constant, and you could have a situation where Tau is 1. But Tau would only be 1 for a particular choice of units of t. Then you could say something like p = p0*exp(-t) if t is in fortnights. But the preferred expression, in my opinion, is one that doesn't depend on a particular choice of units: p = p0*exp(-t/Tau) where Tau = 1 fortnight or the equivalent in the units of t. > But I suspect that I'm jumping in on a discussion > and adding even more cross-purposes than were already > there. Sorry about that. Naah, Tom is getting boring and repetitive. And now it's getting weird since he's telling me that my positions are the ones he's been insisting on all along. So I welcome a new voice, and Tom can't get any more confused than he is already. - Randy === Subject: lim ( f(x)/x ) Please so me if there are any mistakes in the proof that follows and also how to extend it by not assuming f nondecreasing Given : 1. f (x) differentiable and nondecreasing over R 2. lim f ' (x) = 0 as x->infinity Show : lim ( f (x) / x ) -> 0 as x->infinity Proof : 2 => ( f(x) - f(xo) ) / (x - xo) < e if x > xo >N and x-xo < delta after multiplying both sides by (x-xo) we get (f(x) - f(xo) )< e * (x - xo) after adding f(xo) to both sides we get f(x) < e*(x-xo) + f (xo) = e*x - e*xo + f(xo) after dividing both sides by x we get 0 < f(x)/x < e - e*xo/x + f(xo)/x e can be made as small as we please and xo and f(xo) are finite real numbers therefore as x->infinity the RHS converges to 0 so f(x)/x converges to 0 === Subject: Re: lim ( f(x)/x ) > Please so me if there are any mistakes in the proof that follows and > also how to extend it by not assuming f nondecreasing > Given : > 1. f (x) differentiable and nondecreasing over R > 2. lim f ' (x) = 0 as x->infinity > Show : > lim ( f (x) / x ) -> 0 as x->infinity For an easier proof, think about L'Hopital's rule. > Proof : > 2 => ( f(x) - f(xo) ) / (x - xo) < e if x > xo >N and x-xo < delta > after multiplying both sides by (x-xo) we get > (f(x) - f(xo) )< e * (x - xo) > after adding f(xo) to both sides we get > f(x) < e*(x-xo) + f (xo) = e*x - e*xo + f(xo) > after dividing both sides by x we get > 0 < f(x)/x < e - e*xo/x + f(xo)/x > e can be made as small as we please and xo and f(xo) are finite real > numbers therefore > as x->infinity the RHS converges to 0 so f(x)/x converges to 0 === Subject: Re: where can I read the book Conics written by Apollonius online? You're very unlikely to find it scanned into the Web. (If you do, please let me know!) Even in book form, there's no complete translation of Apollonius into English. Much of his work is quite deep and difficult to read. The version by Sir Thomas Heath (1896 and reprints) is not an exact translation but a paraphrase using modern notation to help you if you're interested in just the mathematical ideas rather than careful scholarship. The translation by R. Catesby Taliaferro in Great Books of the Western World, Vol. 11, has only Books I to III, and also uses some modern symbols. I have a French translation which the experts say is very accurate, by Paul ver Eecke (1922 and reprints). You could try to find any of those books in a university library or on interloan. Ken Pledger. === Subject: Re: where can I read the book Conics written by Apollonius online? > www.alexandrialib.edu Now if only they had SCANNED all those OTHER scrolls before they burned... --Ron Bruck === Subject: Vector Cross Product in n-dimensions? I am trying to figure out how to calculate the cross product of 2 vectors of n-dimension. All the textbook references seem only give 3-d examples. Can any guru give me a hand on any tips or hints on this topic. (1) Is this possible or well-defined mathematically? (2) If yes, === Subject: Re: Vector Cross Product in n-dimensions? 09/29/2005 at 05:08 PM, mndeng@yahoo.com said: >I am trying to figure out how to calculate the cross product of 2 >vectors of n-dimension. All the textbook references seem only give >3-d examples. That's because the cross product makes sense in an oriented 3-space with metric but not in an arbitrary vector space. >Can any guru give me a hand on any tips or hints on this topic. Google for exterior algebra or wedge product. >(1) Is this possible or well-defined mathematically? Perhaps in a 7-dimensional vector space; certainly not in general. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Vector Cross Product in n-dimensions? > I am trying to figure out how to calculate the cross product of 2 > vectors of n-dimension. 'tis my belief that you can only do so in dimensions 1, 3 and 7. The three imaginary spaces Im C, Im H and Im O, with inner products given by dots and vector product given by u x v = (u.v - v.u)/2 are the only (up to isometric isomorphism) vector product algebras. C, H, O are the complex numbers, quaternions and octonions. Considering these as real algebras A: Im A = {x in A: x^2 in Re and x not in Re{0} } e being the algebra unit. A vector product algebra is an R-algebra with an inner product . and a product x such that u x v = -v x u, (u x v).w = u.(v x w). Massey, W S, Cross products of vectors in higher dimensional Euclidean spaces vol 90, pp 697-701. Walsh, B, The scarcity of cross products in Euclidean spaces vol 74, pp 188-194. -- I don't know who you are Sir, or where you come from, but you've done me a power of good. === Subject: Re: Vector Cross Product in n-dimensions?- Hermann Grassmann laid in his Ausdehnungslehre (theory of extensions) (1844; 1862) the foundations of what is nowadays known as exterior algebra and exterior calculus. These subjects belong to multilinear algebra. Two other keywords are altrnating and anitsymmetric. I guess that Grassmann got his ideas from Cauchy's theorem on development of determinants in terms of subdeterminants. An example of Cauchy's theorem that applies to the 3D case is the following: Let A = (a1, a2, a3), B = (b1, b2, b3), C = (c1, c2, c3) be three 3D vectors. Then Det (A, B, C) = a1 x (b2.c3 - b3.c2) + a2 x (b3.c1 - b1.c3) + a3 x (b1.c2 - b2.c1). This is the familiar box product of the vectors A, B, C. The 2x2 determinants are the components of the cross product (exterior product) B x C. Analogous developments can be set up in any number of dimensions. In the mathematics of the Theory of Relativity four-dimesional space (Minkowskian (3+1)D space rather than Euclidean 4D space) plays a main role. One encounters (in Max von Laue's terminology) Vierervektoren and Sechservektoren. Vierervektoren are the usual four-component vectors (a1, a2, a3, a4); Sechservektoren are so-called bivectors; let A and B be vectors, then A x B is defined as the six-component vector consisting of the determinants a1.b4-a4.b1, a2.b4-a4.b2, a3.b4-a4.b3, a2.b3-a3.b2, a3.b1-a1.b3, a1.b2-a2.b1. Likewise, a trivector (Vierer-pseudovektor(?) - I do not remember exactly the terminology) consists of the four 3x3 determinants that can be formed from the 3x4 components. One may speak of the exterior or cross product of three vectors. Finally, the common 4x4 determinant can be interpreted as the 4D box product of a vector and a trivector. I hope this helps somewhat. Ciao: Johan E. Mebius >I am trying to figure out how to calculate the cross product of 2 >vectors of n-dimension. All the textbook references seem only give 3-d >examples. Can any guru give me a hand on any tips or hints on this >topic. (1) Is this possible or well-defined mathematically? (2) If yes, === Subject: Re: Vector Cross Product in n-dimensions? > I am trying to figure out how to calculate the cross product of 2 > vectors of n-dimension. All the textbook references seem only give 3-d > examples. Can any guru give me a hand on any tips or hints on this > topic. (1) Is this possible or well-defined mathematically? (2) If yes, You can't find the cross product of two vectors in more than 3 dimensions; if you have n dimensions, you need n-1 vectors to 'cross' together to get another. If you have that many, then the usual way of writing the cross product as a determinant whose top row consists of the basis vectors and whose other rows are the components of the vectors still works. === Subject: Re: Vector Cross Product in n-dimensions? > I am trying to figure out how to calculate the cross product of 2 > vectors of n-dimension. All the textbook references seem only give 3-d > examples. Can any guru give me a hand on any tips or hints on this > topic. (1) Is this possible or well-defined mathematically? (2) If yes, In an n-dimensional space you can define a vector perpendicular to n-1 other vectors by using the Levi-Cevita symbol. This is a symbol with n indices, each index having values from 1 to n. If any two indices are equal its value is zero. If the indices form an even permutation of the n integers 1 to n, its value is +1. If the indices form an odd permutation, its value is -1. It's usually denoted by a Greek epsilon with n subscripts. So in three dimensions the cross product C of two vectors A and B is given by: C_i = Sum(j=1 to n;k=1 to n)eps_i_j_k*A_j*B_k You can show this gives: C_1 = A_2*B_3 - A_3*B_2 C_2 = A_3*B_1 - A_1*B_3 C_3 = A_1*B_2 - A_2*B_1 which is the cross product A X B written in terms of components. In four dimensions the vector product D of three vectors A, B, C is D_h = Sum(i=1 to n;j=1 to n;k=1 to n)eps_h_i_j_k*A_i*B_j*C_k I won't write that out in detail, It's easy to show that D is orthogonal to each of the vectors A, B, C by considering the scalar products such as Sum(h=1 to n)D_h*A_h Put in D_h from its definition, and use the definition of the Levi-Cevita symbol to show that the scalar product is zero. You can see how to proceed in any general n-dimensional space. Bob === Subject: Re: Vector Cross Product in n-dimensions? > C_i = Sum(j=1 to n;k=1 to n)eps_i_j_k*A_j*B_k > Put in D_h from its definition, and use the definition of the > Levi-Cevita symbol to show that the scalar product is zero. > You can see how to proceed in any general n-dimensional space. However, for n > 3 the n-th order cross product ceases to be a binary operation and therefore loses some of its algebraic character. Bob Kolker === Subject: Re: Vector Cross Product in n-dimensions? > C_i = Sum(j=1 to n;k=1 to n)eps_i_j_k*A_j*B_k > > Put in D_h from its definition, and use the definition of the > Levi-Cevita symbol to show that the scalar product is zero. > > You can see how to proceed in any general n-dimensional space. > However, for n > 3 the n-th order cross product ceases to be a binary > operation and therefore loses some of its algebraic character. > Bob Kolker I think much of that algebraic character generalizes to the vector product of n-1 vectors in an n-dimensional space. Let's denote the vector product in an n-dimensional space by [A, B, C, ...] (involves n-1 vectors in all) Then in 3 dimensions [A, B] = A X B (the usual cross product) In 4 dimensions: [A, B, C]_h = Sum(i=1 to n;j=1 to n;k=1 to n)eps_h_i_j_k*A_i*B_j*C_k an even permutation of the three vectors gives the original vector, while an odd permutation gives the negative of the original vector. [A, B, C] = -[A, C, B] = -[B, A, C] = +[B, C, A] = +[C, A, B] = -[C, B, A] In 3 dimensions: [A, B] = -[B, A] which is the same as A X B = -B X A. This is one type of generalization. In 3 dimensions we have a distributive law for the cross product. (A + A') X B = A X B + A' X B This also generalizes, which I'll illustrate in 4 dimensions. It's easy to show from the fundamental definition that: [A + A', B, C] = [A, B, C] + [A', B, C] [A, B + B', C] = [A, B, C] + [A, B', C] [A, B, C + C'] = [A, B, C] + [A, B, C'] The identity in 3 dimensions: A X (B X C) = B(A.C) - C(A.B) can be proved using an index sum on the product of two Levi-Civita symbols. So that would generalize in some manner. I haven't done that exercise myself. Anyway I do think there is a rich set of algebraic identities for vector products in n-dimensions. Bob === Subject: Re: Vector Cross Product in n-dimensions? > C_i = Sum(j=1 to n;k=1 to n)eps_i_j_k*A_j*B_k > > Put in D_h from its definition, and use the > definition of the > Levi-Cevita symbol to show that the scalar product > is zero. > > You can see how to proceed in any general > n-dimensional space. > However, for n > 3 the n-th order cross product > ceases to be a binary > operation and therefore loses some of its algebraic > character. Thats true, but why is a non-binary operation less algebraic? In A_infty spaces you have operations for any arity n and they are related algebraicly === Subject: Re: Vector Cross Product in n-dimensions? > I am trying to figure out how to calculate the cross product of 2 > vectors of n-dimension. All the textbook references seem only give 3-d > examples. Can any guru give me a hand on any tips or hints on this > topic. (1) Is this possible or well-defined mathematically? (2) If yes, Yes, it is possible in a sense, but not as a simple vector as in the 3 dimensional case. You need to understand that the cross product doesn't actually define a vector. It is a special case of the Clifford outer product and is referred to as a bivector (or axial vector). We can identify the bivector in 3 dimensions with a vector only because it is the Clifford dual of a vector. This won't work in n dimensions, since for dual objects, the sums of the grades must add up to n. For n = 3, a vector (grade 1) is dual to the bivector (grade 2). Thus, for arbitrary n, a bivector (the cross product) will be dual to an object of grade n-2. And beyond n = 3, there is no preferred direction for a representation of these objects since these higher order grades tend to be associated with areas, volumes, or even hypervolumes. Googling on Clifford algebra should be sufficient for followup material. Good luck. === Subject: Re: Vector Cross Product in n-dimensions? > I am trying to figure out how to calculate the cross product of 2 > vectors of n-dimension. All the textbook references seem only give 3-d > examples. Can any guru give me a hand on any tips or hints on this > topic. (1) Is this possible or well-defined mathematically? (2) If yes, > Yes, it is possible in a sense, but not as a simple vector as in the 3 > dimensional case. You need to understand that the cross product > doesn't actually define a vector. It is a special case of the Clifford > outer product and is referred to as a bivector (or axial vector). We > can identify the bivector in 3 dimensions with a vector only because it > is the Clifford dual of a vector. > This won't work in n dimensions, since for dual objects, the sums of > the grades must add up to n. For n = 3, a vector (grade 1) is dual to > the bivector (grade 2). Thus, for arbitrary n, a bivector (the cross > product) will be dual to an object of grade n-2. And beyond n = 3, > there is no preferred direction for a representation of these objects > since these higher order grades tend to be associated with areas, > volumes, or even hypervolumes. > Googling on Clifford algebra should be sufficient for followup > material. Good luck. Maybe it's just me, but I can't help thinking that someone who is looking in textbooks for cross products isn't going to derive much pleasure from Clifford algebras, not for a few years, anyway. Anyway, what are you really looking for in a cross product? One thing the cross product does for you in 3-d is it gives you a vector orthogonal to both the vectors you started with. Do you want a vector in n-dimensions, orthogonal to two given vectors? That's easy enough to arrange, but there are lots of them, and no good reason to prefer any one of them to all the others. In 3-d, all the vectors orthogonal to the given ones are scalar multiples of the cross product, but in n dimensions the vectors orthogonal to two given ones form an n - 2 dimensional space. [I'm sure this could all be stated in terms of Clifford algebras, but....] Maybe you had better decide what you want out of a cross product first, then it will be easier to see whether there is one and, if so, how to compute it. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Vector Cross Product in n-dimensions? > I am trying to figure out how to calculate the cross product of 2 > vectors of n-dimension. All the textbook references seem only give 3-d > examples. Can any guru give me a hand on any tips or hints on this > topic. (1) Is this possible or well-defined mathematically? (2) If yes, > > Yes, it is possible in a sense, but not as a simple vector as in the 3 > dimensional case. You need to understand that the cross product > doesn't actually define a vector. It is a special case of the Clifford > outer product and is referred to as a bivector (or axial vector). We > can identify the bivector in 3 dimensions with a vector only because it > is the Clifford dual of a vector. > > This won't work in n dimensions, since for dual objects, the sums of > the grades must add up to n. For n = 3, a vector (grade 1) is dual to > the bivector (grade 2). Thus, for arbitrary n, a bivector (the cross > product) will be dual to an object of grade n-2. And beyond n = 3, > there is no preferred direction for a representation of these objects > since these higher order grades tend to be associated with areas, > volumes, or even hypervolumes. I seem to recall that Cliford algebras provide something like a vector product in 7 dimensions as well as in 3 dimensions, but only fir those two. > > Googling on Clifford algebra should be sufficient for followup > material. Good luck. > Maybe it's just me, but I can't help thinking that someone who > is looking in textbooks for cross products isn't going to derive > much pleasure from Clifford algebras, not for a few years, anyway. > Anyway, what are you really looking for in a cross product? > One thing the cross product does for you in 3-d is it gives you > a vector orthogonal to both the vectors you started with. Do you > want a vector in n-dimensions, orthogonal to two given vectors? > That's easy enough to arrange, but there are lots of them, and > no good reason to prefer any one of them to all the others. In 3-d, > all the vectors orthogonal to the given ones are scalar multiples > of the cross product, but in n dimensions the vectors orthogonal > to two given ones form an n - 2 dimensional space. [I'm sure > this could all be stated in terms of Clifford algebras, but....] > Maybe you had better decide what you want out of a cross product > first, then it will be easier to see whether there is one and, > if so, how to compute it. === Subject: what area of math this problem belongs to? I am trying to work out some computer program to solve some math problem. To simplify the problem, I can describe it as the following: 1) A1 and B1 are positive integers and A1 + 2^n1 - 1 = B1. Here n1 is a constant. 2) Similarly we have A2 and B2, A2+2^n2-1=B2, n2 is also constant. 3) Similarly we have A3 and B3, A3+2^n3-1=B3, n3 is constant. 4) [A1,B1], [A2,B2] and [A3,B3] do not overlap 5) n1, n2, n3 are all constant and less than 10. Given the above, I need to i) Find A4 and B4, so that A4 + 2^n4 - 1 = B4 and [A4,B4] covers [A1,B1], [A2,B2] and [A3,B3]. If I can find A4, n4 and B4, then I am done; if I can not, then ii) Find [A4,B4] and [A5, B5], so that their union covers [A1,B1], [A2,B2] and [A3,B3]. If I can find them, then I am done; otherwise, iii) in the case I know the answer for the above problem is [A1,B1], [A2,B2] and [A3,B3]. Because the actual problem is much more complex than the above, so I can not just write a brute-force program. I need to solve this thing through some math calculation. I very very roughly remember when I was in college there are some either linear algebra or optimization theory class mentioning some algorithm for the problem like this. But that is over 10 years ago. Could you point me what exactly such problem belongs to? === Subject: Re: what area of math this problem belongs to? > I am trying to work out some computer program to solve some math >problem. To simplify the problem, I can describe it as the following: > 1) A1 and B1 are positive integers and A1 + 2^n1 - 1 = B1. Here n1 is >a constant. > 2) Similarly we have A2 and B2, A2+2^n2-1=B2, n2 is also constant. > 3) Similarly we have A3 and B3, A3+2^n3-1=B3, n3 is constant. > 4) [A1,B1], [A2,B2] and [A3,B3] do not overlap > 5) n1, n2, n3 are all constant and less than 10. > Given the above, I need to > i) Find A4 and B4, so that A4 + 2^n4 - 1 = B4 and [A4,B4] covers >[A1,B1], [A2,B2] and [A3,B3]. If I can find A4, n4 and B4, then I am >done; if I can not, then Must n4 be less than 10? A4 = min(A1,A2,A3) B4_guess = max(B1,B2,B3) n4 = ceil(log2(B4_guess+1-A4)) B4 = A4+2^n4-1 > ii) Find [A4,B4] and [A5, B5], so that their union covers [A1,B1], >[A2,B2] and [A3,B3]. If I can find them, then I am done; otherwise, > iii) in the case I know the answer for the above problem is [A1,B1], >[A2,B2] and [A3,B3]. > Because the actual problem is much more complex than the above, so I >can not just write a brute-force program. I need to solve this thing >through some math calculation. > I very very roughly remember when I was in college there are some >either linear algebra or optimization theory class mentioning some >algorithm for the problem like this. But that is over 10 years ago. > Could you point me what exactly such problem belongs to? Algorithms? --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: what area of math this problem belongs to? problem. To simplify the problem, I can describe it as the following: > 1) A1 and B1 are positive integers and A1 + 2^n1 - 1 = B1. Here n1 is >a constant. > 2) Similarly we have A2 and B2, A2+2^n2-1=B2, n2 is also constant. > 3) Similarly we have A3 and B3, A3+2^n3-1=B3, n3 is constant. > 4) [A1,B1], [A2,B2] and [A3,B3] do not overlap > 5) n1, n2, n3 are all constant and less than 10. > Given the above, I need to > i) Find A4 and B4, so that A4 + 2^n4 - 1 = B4 and [A4,B4] covers >[A1,B1], [A2,B2] and [A3,B3]. If I can find A4, n4 and B4, then I am >done; if I can not, then > Must n4 be less than 10? > A4 = min(A1,A2,A3) > B4_guess = max(B1,B2,B3) > n4 = ceil(log2(B4_guess+1-A4)) > B4 = A4+2^n4-1 > ii) Find [A4,B4] and [A5, B5], so that their union covers [A1,B1], >[A2,B2] and [A3,B3]. If I can find them, then I am done; otherwise, > iii) in the case I know the answer for the above problem is [A1,B1], >[A2,B2] and [A3,B3]. > Because the actual problem is much more complex than the above, so I >can not just write a brute-force program. I need to solve this thing >through some math calculation. > I very very roughly remember when I was in college there are some >either linear algebra or optimization theory class mentioning some >algorithm for the problem like this. But that is over 10 years ago. > Could you point me what exactly such problem belongs to? > Algorithms? > --Keith Lewis klewis {at} mitre.org > The above may not (yet) represent the opinions of my employer. Hi Keith, Sorry I made a mistake on the problem description. After the 5 conditions, I should have said this way, Given the above, I need to i) Find A4 and B4, so that A4 + 2^n4 - 1 = B4 and [A4,B4] covers [A1,B1], [A2,B2] BUT NOT [A3,B3]. If I can find A4, n4 and B4, then I am done; if I can not, then ii) Find [A4,B4] and [A5, B5], so that their union covers [A1,B1], [A2,B2] BUT NOT [A3,B3]. Actually in this case I know they are [A1,B1] and [A2,b2]. The problem is something like, [A1,B1]=[0,1], [A2,B2]=[4,5] and [A3,B3]=[6,7]. You can see that the only answer is [A1,B1] and [A2,B2]. For problem like this, a try-and-fail-and-adjust algorithm is good enough, but the real problem set is very large, I must have some math workout first. If you have some idea how to formalize this into some math equation like what you did in your reply, or point me where I should look for related solution, I would appreciate very much. === Subject: Re: Is there a name for these equations? >> Is there a name for the polynomial equations of the form >> x^4 + ax^2 + b = 0 ? >> Jose Carlos Santos >> >> Bicuadratic ('bicuadrada' in spanish) >biquadratic in English, but that would allow x^3 and x terms, too. > A degree-n polynomial (sometimes, I think, specifically > a monic degree-n polynomial) p(x) in which the coefficient > of x^{n-1} is 0 is classically called a depressed n-ic. > So x^4 + ax^2 + b is a depressed biquadratic with no > linear term, which isn't much of a name, I admit. The polynomial reciprocal to x^4 + a x^2 + b is b x^4 + a x^2 + 1, which is also depressed (except perhaps for the thing about being monic). So you could call it a monic depressed quartic with depressed reciprocal. Enough to induce an epsiode of monic depression. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Is there a name for these equations? On Fri, 30 Sep 2005 10:10:32 +1000, Gerry Myerson > Is there a name for the polynomial equations of the form x^4 + ax^2 + b = 0 ? > Jose Carlos Santos > > Bicuadratic ('bicuadrada' in spanish) >>biquadratic in English, but that would allow x^3 and x terms, too. >> A degree-n polynomial (sometimes, I think, specifically >> a monic degree-n polynomial) p(x) in which the coefficient >> of x^{n-1} is 0 is classically called a depressed n-ic. >> So x^4 + ax^2 + b is a depressed biquadratic with no >> linear term, which isn't much of a name, I admit. >The polynomial reciprocal to x^4 + a x^2 + b is >b x^4 + a x^2 + 1, which is also depressed (except perhaps >for the thing about being monic). >So you could call it a monic depressed quartic >with depressed reciprocal. >Enough to induce an epsiode of monic depression. Well, it can be cured by a formula, however the formula is quite radical and there are reports of hallucinatory side effects (such as claiming to see imaginary numbers). === Subject: Re: Is there a name for these equations? >Is there a name for the polynomial equations of the form > x^4 + ax^2 + b = 0 ? > ... > http://www.wordreference.com/definition/biquadratic and : We found no English translation for 'bicarr.8ee' in our French to English Dictionary seems there is *no* specific word in English for that ? -- philippe mail : chephip at free dot fr site : http://chephip.free.fr/ Warning ! inside architecture changed ! (change your bookmarks) === Subject: Re: Is there a name for these equations? G. A. Edgar escribi.97: >> Jos.8e Carlos Santos escribi.97: > Hi all, > Is there a name for the polynomial equations of the form > x^4 + ax^2 + b = 0 ? > Jose Carlos Santos >> Bicuadratic ('bicuadrada' in spanish) > biquadratic in English, but that would allow x^3 and x terms, too. In spanish 'bicuadrada' (biquadratic) and 'cu.87rtica' (quartic) are not equivalente. The first is a special case of the second, without terms in x^3 nor x. With that name it is studied in secondary school as an easy extension of the quadratic one. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Is there a name for these equations? <290920050753596231%edgar@math.ohio-state.edu.invalid> <3q30d1Fd2pg3U1@individual.net In spanish 'bicuadrada' (biquadratic) and 'cu.87rtica' (quartic) > are not equivalente. The first is a special case of the second, > without terms in x^3 nor x. I've always assumed that a biquadratic is a quartic whose Galois group is a power of two, i.e. can be reduced to a product of quadratics with coefficients in the field K[sqrt(k)] where k is in the field K of the original quartic/biquadratic. (But then some things I've always assumed have turned out to be wrong...) === Subject: Re: Romanian math olympiad problem I cannot thank you enough for your kindness and nice solution. Have a good day~ === Subject: Re: proofs of Infinitude Twin Primes, Perfect Numbers, Fermat primes; using Math Induction in conjunction with Euclid proof ... > So in the Math Induction to prove Infinitude of Twin Primes what I do > for case of 1 is show true that if you multiply twin primes together > and multiply by 2 then add and subtract 1, that you produce another new > twin prime. I assume > true for case N that if you have n Twin Primes and thus multiply 2n > primes together in a lot and then multiply that lot by 2 then add > subtract 1, a new twin > prime not on the list is produced. This is wrong. When you multiply twin primes together, multiply by 2 and add and subtract 1 you get a twin, but you do not know whether they are prime. They may be divisible by a number that is prime but not part of a twin prime. But let's see, start with the pairs (3,5) and (5,7). Your operation generates (1049,1051). Both prime, indeed. Doing it again with these new primes we get (1157623949,1157623951). The first is prime, the second is divisible by 11. By sheer luck 11 is part of the twin (11,13). But that is not the twin you generated. When you continue this way you will get at the pair (857994985397549,857994985397551) where the first is divisible by 47 and the second by 691, and they are not part of a pair. > Euclid's proof of the infinitude of primes has an advantage over yours, > in that if the product of the primes plus or minus 1 is not prime, > Euclid could still find a new prime. You will have the face the > possibility that at least one of the two numbers you produce is not a > prime. The proof might collapse then, because if you get new prime > factors, you have the burdon of showing those primes are twins. > Wrong. You do not fully understand the Euclid IP proof mechanism. > Euclid guaranteed to produce Twin Primes in his proof mechanism. > Suppose 3 was the only prime in existence. Consider P!+1 and P!-1. > Because 3 is the only prime in existence guarantees by that supposition > that 2 and 4 are primes not on the original list. Within that > Supposition Space Euclid not only guarantees P!+1 as a new prime not > on his original list but guarantees Twin Primes. It is that fact, that > if a Supposition Space has finite primes that P!+1 and P!-1 guarantees > Twin Primes. Yes? So if there are finitely many primes a twin prime can always be found, that contradicts the supposition that there are finitely many primes. It does *not* contradict that there are finitely many twin primes. > So you're proving the infinitude of primes of the form 2^n - 1 by > induction on N in the statement: > > P(N) = When you multiply any 2N primes of the form 2^n - 1 by each > other, multiply by 6, and add 1, you get a (new) prime of the form 2^n > - 1. > > I think you're going to have trouble proving even P(1), and actually > will have trouble showing the new number is of the form 2^n - 1. > > Suppose m and n are both at least 3. > Wrong, this is the supposition step of Math Induction, so I will have > no trouble at all. Perhaps. In induction you have to show two things: 1. When it is true for some N it is true for N+1 2. It is true for some value of N (like 1) N = 1 and m = n = 3. > The trouble starts when I have to show N+1 and to > weave together the N step with the Euclid supposition of IP. Not the problem starts when you have to show that it is valid for even a single N. -- 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: proofs of Infinitude Twin Primes, Perfect Numbers, Fermat primes; using Math Induction in conjunction with Euclid proof > So in the Math Induction to prove Infinitude of Twin Primes what I do > > for case of 1 is show true that if you multiply twin primes together > > and multiply by 2 then add and subtract 1, that you produce another new > > twin prime. I assume > > true for case N that if you have n Twin Primes and thus multiply 2n > > primes together in a lot and then multiply that lot by 2 then add > > subtract 1, a new twin > > prime not on the list is produced. > This is wrong. When you multiply twin primes together, multiply by 2 and > add and subtract 1 you get a twin, but you do not know whether they are > prime. They may be divisible by a number that is prime but not part of > a twin prime. But let's see, start with the pairs (3,5) and (5,7). Your > operation generates (1049,1051). Both prime, indeed. Doing it again Indeed, and that is my case #1 for a Math Induction. I need just one case sample for Math Induction. Then I go onto the supposition case of N and then have to show for N+1. Actually I am correct, Dik, not wrong, because in the proof of Euclid of Infinitude of Primes, he yields at the end not just one new prime on the finite list but Twin primes. What you are failing to consider is the space of actual numbers and the space of supposition in a reductio ad absurdum. Euclid guarantees yielding Twin primes because he is under a supposition of all the primes are finite. Whereas you, Dik, not under any supposition and just roaming through the numbers can find cases where twin primes A and twin primes B, mulitiplied together, multiply 2, add and subtract 1 can find dozens of undesirable results. Euclid in his IP, was under a supposition, and being under that supposition guaranteed that his final two numbers of Primes factorial add 1 and Primes factorial subtract 1 are Twin primes. > with these new primes we get (1157623949,1157623951). The first is > prime, the second is divisible by 11. By sheer luck 11 is part of the > twin (11,13). But that is not the twin you generated. When you continue > this way you will get at the pair (857994985397549,857994985397551) where > the first is divisible by 47 and the second by 691, and they are not part > of a pair. I am not running a Math Induction to prove every twin primes x and y when multiplied together, then multiply by 2, then add subtract 1 will yield a new twin primes. Dik, I am running Math Induction on a subset of Twin Primes that obeys what I want it to obey. A subset of the Twin Primes that when multiplied by 2, add subtract 1 yields a new twin prime. And that this subset of twin primes is infinite. Dik, you are thinking that the Math Induction is that every twin primes will yield a new twin prime by multiply by 2, add subtract 1, No, I am doing a Math Induction on a subset of the Twin Primes and discovering that it is infinite and thus Twin Primes are infinite. > > Euclid's proof of the infinitude of primes has an advantage over yours, > > in that if the product of the primes plus or minus 1 is not prime, > > Euclid could still find a new prime. You will have the face the > > possibility that at least one of the two numbers you produce is not a > > prime. The proof might collapse then, because if you get new prime > > factors, you have the burdon of showing those primes are twins. > > Wrong. You do not fully understand the Euclid IP proof mechanism. > > Euclid guaranteed to produce Twin Primes in his proof mechanism. > > Suppose 3 was the only prime in existence. Consider P!+1 and P!-1. > > Because 3 is the only prime in existence guarantees by that supposition > > that 2 and 4 are primes not on the original list. Within that > > Supposition Space Euclid not only guarantees P!+1 as a new prime not > > on his original list but guarantees Twin Primes. It is that fact, that > > if a Supposition Space has finite primes that P!+1 and P!-1 guarantees > > Twin Primes. > Yes? So if there are finitely many primes a twin prime can always be > found, that contradicts the supposition that there are finitely many > primes. It does *not* contradict that there are finitely many twin > primes. > > So you're proving the infinitude of primes of the form 2^n - 1 by > > induction on N in the statement: > > P(N) = When you multiply any 2N primes of the form 2^n - 1 by each > > other, multiply by 6, and add 1, you get a (new) prime of the form 2^n > > - 1. > > I think you're going to have trouble proving even P(1), and actually > > will have trouble showing the new number is of the form 2^n - 1. > > Suppose m and n are both at least 3. > > Wrong, this is the supposition step of Math Induction, so I will have > > no trouble at all. > Perhaps. In induction you have to show two things: > 1. When it is true for some N it is true for N+1 > 2. It is true for some value of N (like 1) > N = 1 and m = n = 3. > > The trouble starts when I have to show N+1 and to > > weave together the N step with the Euclid supposition of IP. > Not the problem starts when you have to show that it is valid for even > a single N. > -- > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 > home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ I will need to know all the various forms of Math Induction. For the case of Twin Primes, the Math Induction out of Peano axioms (standard induction) is what I used above (1) show true for case #1 (2) assume true for N (3) show true for N+1. Yours, Dik, is not formal enough. The trouble I sense in your argument Dik, and probably within Chris's objections is that you forgot to realize that my Math Induction is on a subset of the Twin Primes and not on the whole entire set of Twin Primes. So if I find the subset to be infinite leads to the obvious conclusion that all the Twin Primes are infinite. Dik, I am having trouble in finding a case #1 of Math Induction for of form (n^2)+1 are generated from even numbers for the n where the last digit is either 6,4 or 0. Dik, can you prove that it is impossible to multiply any two of these primes then multiply by 2, add or subtract 1 and end up with a new prime of that form? My difficulty is not a proof, but if proved impossible then my method says that primes of this form are FINITE. And obviously a huge vindication that this method is valid. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: proofs of Infinitude Twin Primes, Perfect Numbers, Fermat primes; using Math Induction in conjunction with Euclid proof ... > > The trouble starts when I have to show N+1 and to > > weave together the N step with the Euclid supposition of IP. > > Not the problem starts when you have to show that it is valid for even > a single N. ... > The trouble I sense in your argument Dik, and probably within Chris's > objections is that you forgot to realize that my Math Induction is on a > subset of the Twin Primes and not on the whole entire set of Twin > Primes. So if I find the subset to be infinite leads to the obvious > conclusion that all the Twin Primes are infinite. Yes, if you find that the subset of twin primes that generates a new twin prime is infinite, than, indeed, the number of twin primes is infinite. That is a triviality. It is the first part that is difficult. > Dik, I am having trouble in finding a case #1 of Math Induction for > of form (n^2)+1 are generated from even numbers for the n where the > last digit is either 6,4 or 0. Dik, can you prove that it is impossible > to multiply any two of these primes then multiply by 2, add or subtract > 1 and end up with a new prime of that form? > My difficulty is not a proof, but if proved impossible then my method > says that primes of this form are FINITE. And obviously a huge > vindication that this method is valid. Why do you not try it? For starters, for n^2 - 1. Suppose we have two numbers of that form: (p^2 - 1) and (q^2 - 1). Multiply together: (p.q)^2 - p^2 - q^2 + 1. Multiply by 2 and add or subtract 1: 2.(p.q)^2 - 2.p^2 -2.q^2 + 1 or 2.(p.q)^2 - 2.p^2 - 2.q^2 + 3. Under what conditions is one of the last two numbers of the form n^2 - 1? A quick test should reveal to you that this is probably never the case when both p and q are even. Now it is up to you to prove that. -- 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: proofs of Infinitude Twin Primes, Perfect Numbers, Fermat primes; using Math Induction in conjunction with Euclid proof > > The trouble starts when I have to show N+1 and to > > > weave together the N step with the Euclid supposition of IP. > > Not the problem starts when you have to show that it is valid for even > > a single N. > ... > > The trouble I sense in your argument Dik, and probably within Chris's > > objections is that you forgot to realize that my Math Induction is on a > > subset of the Twin Primes and not on the whole entire set of Twin > > Primes. So if I find the subset to be infinite leads to the obvious > > conclusion that all the Twin Primes are infinite. > Yes, if you find that the subset of twin primes that generates a new twin > prime is infinite, than, indeed, the number of twin primes is infinite. > That is a triviality. It is the first part that is difficult. Yes, I have to go back and put a big clear paragraph into the proof that when using Math Induction on Twin Primes or Perfect Numbers primes or Fermat primes that I am proving a subset of Twin Primes is infinite and a subset of Perfect Numbers primes is infinite and a subset of Fermat primes is infinite, hence the larger set of Twin primes, Perfect numbers primes, Fermat primes are infinite. > > Dik, I am having trouble in finding a case #1 of Math Induction for > > of form (n^2)+1 are generated from even numbers for the n where the > > last digit is either 6,4 or 0. Dik, can you prove that it is impossible > > to multiply any two of these primes then multiply by 2, add or subtract > > 1 and end up with a new prime of that form? > > My difficulty is not a proof, but if proved impossible then my method > > says that primes of this form are FINITE. And obviously a huge > > vindication that this method is valid. > Why do you not try it? For starters, for n^2 - 1. Suppose we have > two numbers of that form: (p^2 - 1) and (q^2 - 1). Multiply together: > (p.q)^2 - p^2 - q^2 + 1. Multiply by 2 and add or subtract 1: > 2.(p.q)^2 - 2.p^2 -2.q^2 + 1 or 2.(p.q)^2 - 2.p^2 - 2.q^2 + 3. > Under what conditions is one of the last two numbers of the form > n^2 - 1? A quick test should reveal to you that this is probably > never the case when both p and q are even. Now it is up to you to > prove that. > -- > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 > home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ So I have a super dooper tester: bring to me any form of primes such as (n^2)+1 or (n^2)-1 and if it is impossible to form the first Math Induction step by mulitiplying two prime members together, multiply by 2, add/subtract 1 and impossible to derive a new prime of that same identical form, means it is impossible to make a Math Induction on primes of that form. Which means the set is FINITE. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: proofs of Infinitude Twin Primes, Perfect Numbers, Fermat primes; using Math Induction in conjunction with Euclid proof So I have a super dooper tester: bring to me any form of primes such as > (n^2)+1 or (n^2)-1 and if it is impossible to form the first Math > Induction step by mulitiplying two prime members together, multiply by > 2, add/subtract 1 and impossible to derive a new prime of that same > identical form, means it is impossible to make a Math Induction on > primes of that form. Which means the set is FINITE. No, it only means this method doesn't prove the result, and the answer can be finite or infinite. A proof, one way or another, may exist which DOESN'T use math induction. You can't create a .bmp (graphics) file with a text editor, but that doesn't mean there are no such things as .bmp files. You REALLY need to take a course in proofs. === Subject: Re: proofs of Infinitude Twin Primes, Perfect Numbers, Fermat primes; using Math Induction in conjunction with Euclid proof >Here is a conjecture that you should be able to prove Chris: >Conjecture: given Euclid's IP proof that this proof is a method and >this method is guaranteed able to prove the question of whether a >subset of primes of the set of all primes is finite or infinite. In >other words, since Euclid's IP proof mechanism works for all the >primes, it must work as a proof mechanism for all subsets of the >primes. Perhaps there is a theorem already proven in mathematics that >confirms this conjecture. On the contrary, there is a theorem that denies it. If k and h are coprime then there is an infinity of primes of the form kn + h. This was first proved by Dirichlet, and proving it is not easy. However, for some special cases there is a straightforward euclidean proof, and Schur proved that there is always such a proof if h^2 - 1 is divisible by k. But Murty has proved that there can be no euclidean proof if h^2 - 1 is not divisible by k. For example, the euclidean approach cannot prove that the sequence 2, 7, 12, 17, 22, 27, ... contains an infinity of primes. -- John Roberts-Jones === Subject: Re: proofs of Infinitude Twin Primes, Perfect Numbers, Fermat primes; using Math Induction in conjunction with Euclid proof Conjecture: given Euclid's IP proof that this proof is a method and >this method is guaranteed able to prove the question of whether a >subset of primes of the set of all primes is finite or infinite. In >other words, since Euclid's IP proof mechanism works for all the >primes, it must work as a proof mechanism for all subsets of the >primes. Perhaps there is a theorem already proven in mathematics that >confirms this conjecture. On the contrary, there is a theorem that denies it. If k and h are coprime then there is an infinity of primes of the form kn + h. This was first proved by Dirichlet, and proving it is not easy. However, for some special cases there is a straightforward euclidean proof, and Schur proved that there is always such a proof if h^2 - 1 is divisible by k. But Murty has proved that there can be no euclidean proof if h^2 - 1 is not divisible by k. For example, the euclidean approach cannot prove that the sequence 2, 7, 12, 17, 22, 27, ... contains an infinity of primes. -- John Roberts-Jones Well I am glad you brought this up. Not that I believe you refuted my conjecture but that of the problem I ran into last night in trying to find the initial case of Math Induction for the primes of form (n^2)+1. You see, I cannot get a Math Induction initial case for these primes, leading me to the suspicion that these primes are finite. And of course, primes of form (n^2)-1 would have to be finite if primes of form (n^2)+1 is finite, which is a simple symmetry argument. So your above is not a counterexample to my Conjecture. A proof of my conjecture would look to the ideas that if a Method works in proving the Inifinitude of a set A, then given subsets of that set, call them B, C, D, E.... the same method must work in proving whether they are finite or infinite, provided some adaption is included (in my case, injecting Math Induction along with Euclid IP method). What this Conjecture, if true, says is that if someone wanted to prove Infinitude of Twin Primes, his/her best bet is to re-adapt the Euclid IP method for a guaranteed success. If they choice to look elsewhere for a method to prove Twin Primes, there is no guarantee that the method will work. If they use Euclid IP, then they are guaranteed that the method will work. For now it is difficult for me to get a initial Math Induction step of two primes of form (n^2) +1 multiply together, then multiply by 2 then either add or subtract 1 and yield a new prime of form (n^2)+1 (or likewise (n^2)-1). Difficult is not impossibility, perhaps I did not look hard enough or find it. Can someone demonstrate that it is impossible to get primes of form (n^2)+1, multiply together, then multiply by 2 then either add 1 or subtract 1 and yield a new prime of form (n^2)+1. If I cannot form a Math Induction step #1, then these primes are finite. They thin out until they are impossible to form primes beyond a certain distance. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: proofs of Infinitude Twin Primes, Perfect Numbers, Fermat primes; using Math Induction in conjunction with Euclid proof .......... > Well I am glad you brought this up. Not that I believe you refuted my > conjecture but that of the problem I ran into last night in trying to > find the initial case of Math Induction for the primes of form (n^2)+1. > You see, I cannot get a Math Induction initial case for these primes, > leading me to the suspicion that these primes are finite. And of > course, primes of form (n^2)-1 would have to be finite if primes of > form (n^2)+1 is finite, which is a simple symmetry argument. ........... I am convinced that the number of primes of the form n^2 - 1 is finite. === Subject: Re: proofs of Infinitude Twin Primes, Perfect Numbers, Fermat primes; using Math Induction in conjunction with Euclid proof Well I am glad you brought this up. Not that I believe you refuted my > conjecture but that of the problem I ran into last night in trying to > find the initial case of Math Induction for the primes of form (n^2)+1. > You see, I cannot get a Math Induction initial case for these primes, > leading me to the suspicion that these primes are finite. And of > course, primes of form (n^2)-1 would have to be finite if primes of > form (n^2)+1 is finite, which is a simple symmetry argument. > ........... > I am convinced that the number of primes of the form n^2 - 1 is finite. Induction combined with the Euclid Infinitude of Primes proof requires the initial Math Induction step, but primes of form (n^2)+1 and (n^2)-1 cannot deliver those initial Math Induction step. So my method would prove these primes are finite but unfortunately my method does not seem to be able to point to where the last prime of those forms lie. At least I do not suspect my method can pinpoint the last prime. Caveat: unless we can measure the deviation from Math Induction initial step and thus tie that deviation to the primes of that form as they expand outward. For example in primes of form (n^2)+1 the last four such primes, if I have not made a mistake, are 7057, 8101, 8837, 12101 out to n=110. So if we use the Math Induction construction of multiply two primes of this form, multiply by 2, add/subtract 1 and find a numerical deviation from that construction. Scratch that above. I see no link between Math Induction and pinpointing the last prime of form (n^2)+1 or (n^2)-1 Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: proofs of Infinitude Twin Primes, Perfect Numbers, Fermat primes; using Math Induction in conjunction with Euclid proof scratch that, as in, one cannot find any *reason* for the twin primes to be finite? I keep on saying, induction is isomprphic to deduction; the proof is less than 3 pages in Mathematics magazine. > I am convinced that the number of primes of the form n^2 - 1 is finite. > Induction combined with the Euclid Infinitude of Primes proof requires > the initial Math Induction step, but primes of form (n^2)+1 and (n^2)-1 > cannot deliver those initial Math Induction step. So my method would > prove these primes are finite but unfortunately my method does not seem > to be able to point to where the last prime of those forms lie. At > least I do not suspect my method can pinpoint the last prime. > Caveat: unless we can measure the deviation from Math Induction initial > step and thus tie that deviation to the primes of that form as they > expand outward. For example in primes of form (n^2)+1 the last four > such primes, if I have not made a mistake, are 7057, 8101, 8837, 12101 > out to n=110. So if we use the Math Induction construction of multiply > two primes of this form, multiply by 2, add/subtract 1 and find a > numerical deviation from that construction. > Scratch that above. I see no link between Math Induction and > pinpointing the last prime of form (n^2)+1 or (n^2)-1 --Trickier Dick's Obnoxico? http://larouchepub.com/other/2005/3237energy_heist.html === Subject: Re: proofs of Infinitude Twin Primes, Perfect Numbers, Fermat primes; using Math Induction in conjunction with Euclid proof > The other thread was getting too confusing and cluttered and the > possibility that primes of form (n^2)+1 are finite. So let me preserve > and start afresh with the progress of Twin Primes, Perfect Numbers and > Fermat Primes here. PROOF OF THE INFINITUDE OF TWIN PRIMES So in the Math Induction to prove Infinitude of Twin Primes what I do > for case of 1 is show true that if you multiply twin primes together > and multiply by 2 then add and subtract 1, that you produce another new > twin prime. I assume > true for case N that if you have n Twin Primes and thus multiply 2n > primes together in a lot and then multiply that lot by 2 then add > subtract 1, a new twin > prime not on the list is produced. Finally I must show for case N+1 of > Math Induction that for 2n+2 primes multiplied together and multiplied > by 2 add and subtract 1 yields a new twin prime. This is where Euclid's > proof method enters. > Euclid's proof of the infinitude of primes has an advantage over yours, > in that if the product of the primes plus or minus 1 is not prime, > Euclid could still find a new prime. You will have the face the > possibility that at least one of the two numbers you produce is not a > prime. The proof might collapse then, because if you get new prime > factors, you have the burdon of showing those primes are twins. > Wrong. You do not fully understand the Euclid IP proof mechanism. I have doubts about whether you do; see (2) below. > Euclid guaranteed to produce Twin Primes in his proof mechanism. I don't think so. That would mean that Euclid solved the Twin Primes Conjecture long ago. > Suppose 3 was the only prime in existence. Consider P!+1 and P!-1. > Because 3 is the only prime in existence guarantees by that supposition > that 2 and 4 are primes not on the original list. (1) 3! is not 3. (However, the product of the elements in {3} _is_ 3. When you say Consider P!+1 and P!-1, this is confusing, because P! is the product of the elements in {1,2,...,P}. (If you want to talk about the product of the elements in a set S, use prod(S) -- which I will use for the rest of the post -- or some other such notation.) (2) 4 is not prime, but has a prime factor not in {3}. This is in accordance with Euclid's proof, where the product of the (alleged) set S of prime numbers plus 1 is prime OR has a prime factor not in S. Euclid's number thus need not be a prime, which you are (falsely) claiming. > Within that > Supposition Space Euclid not only guarantees P!+1 as a new prime No he doesn't. P!+1 for the set S = {3,5} is 3*5+1 = 16, which is not prime. There are many more examples. > not on his original list but guarantees Twin Primes. No he doesn't, since prod(S) + 1 and/or prod(S) - 1 need not be primes. > It is that fact, that > if a Supposition Space has finite primes that P!+1 and P!-1 guarantees > Twin Primes. No, not by any stretch of the definition of P! is this true. The calculations aren't difficult. However, you have proven the long-conjectured Twin Integers Conjecture, that there are an infinite number of n's such that n and n+2 are integers. > What I have ventured in doing is combining that Euclid IP mechanism > with Math Induction and to weave and blend and mix the N Supposition > step of Math Induction with the fact that Euclid IP guarantess Twin > Primes, that combining the N supposition step of Math Induction proves > Infinitude of Twin Primes. > I may not have the best form of expression of the Math Induction with > suppose N and show N+1. For there are many versions of Math Induction > including Fermat's Infinite Descent. > The important thing in all of this is not that I have proven Twin > Primes, Perfect Numbers and Fermat Primes, [...] In fact, you have NOT proven ANY of these. > Here is a conjecture that you should be able to prove Chris: > Conjecture: given Euclid's IP proof that this proof is a method and > this method is guaranteed able to prove the question of whether a > subset of primes of the set of all primes is finite or infinite. In > other words, since Euclid's IP proof mechanism works for all the > primes, it must work as a proof mechanism for all subsets of the > primes. Such a conjecture has nothing to do with twin primes, or primes of a given form, because Euclid only had to prove that his new number prod(S)+1 _leads to_ a prime. I suspect what you're asking is whether S can be _any_ set of primes, and the answer is yes. He didn't have to show an additional requirement that this new prime is of a certain form (2^n-1, for instance). This extra work is necessary and very difficult to do. For instance, if you let S = {2^2 - 1, 2^3 - 1} = {3, 7}, and let Euclid's number be N = prod(S)+1, then N = (3 * 7) + 1 = 22 = 2 * 11. The new prime not in S is either 2 or 11. (Note that prod(S)-1 isn't prime either, being 2^2 * 5.) Euclid's proof can stop here; all he needs is a new prime, period. But you have to show that 2 or 11 can be written as 2^n - 1 for some integer n. This is actually not the case, so Euclid wins and you lose. > The important thing about this conjecture is that it saves time in that > anyone wanting to prove a set of primes is finite or infinite merely > has to fiddle around with Euclid's IP in order to answer his/her > question. And saves the bloke from hunting around for some other > method. > And I think I can say that P!+1 and P!-1 is that new twin > primes. > ... for an infinite number of P's. Your statement is NOT true for all > P. If > P = 5, P!-1 = 119 = 7*17, and P!+1 = 121 = 11^2. In fact, looking at > the prime factors of P! +/- 1 (7, 11, 17), you don't get any new twin > primes, like your proof claims would happen. > Then you fail to understand Reductio Ad Absurdum, Don't start this again. I KNOW Reductio Ad Absurdum, I have PUBLISHED PAPERS USING Reductio Ad Absurdum, I TEACH Reductio Ad Absurdum in classes. > that Euclid has a > supposition-space and this supposition space guarantees that for any > finite set of primes P!+1 and P!-1 are guaranteed Twin Primes. No. Do some calculations. Look at my example above. > If 3 is > the only prime in existence, under that supposition space, 2 and 4 are > necessarily Twin Primes. They're Twin Integers, but since one of them is not prime, they're not Twin Primes. > In my proof of the Infinitude of Twin Primes, the Infinitude of Perfect > Numbers, the Infinitude of Fermat Primes, I weave together the > Supposition Space of Euclid's IP proof with the Supposition Space of > the Math Induction step of suppose N to be true, then show N+1 true. > By weaving together those two supposition spaces, I conquer those > proofs. > In the proof of Twin Primes the supposition step of N in Math > Induction is sort of superfluous because Euclid's IP has all the twin > primes in addition to having the other primes so the Euclid IP > overtakes all the supposition step of Math Induction and easily gives > me the N+1 step. > It was easy for Euclid because all he was after was a prime. You are > after a prime with certain other properties, so you have to show you > get these new properties as well. That means more work for you, and I > wouldn't expect the extra work to be easy. > The extra work is embodied in the fact of combining Math Induction with > Euclid's IP proof mechanism. No, it isn't. The extra work is used to show that the numbers you get are twin primes, or of a certain form. If you combine math induction with Euclid's proof mechanism, you get the following: THEOREM. There are an infinite number of primes. Proof. Let P(N) be the statement There exists a set of N primes, and proceed with induction. P(1) is true, because {2} is a set of one prime. Now assume P(k) is true, that there is a set of k primes; now we need to show P(k+1), that there is a set of k+1 primes. Let S be a set of k primes (whose existence is guaranteed by the induction hypothesis), and let n be the product of the elements in S, added to 1. Then n has a prime factor p (perhaps equal to n), which is not in S. Then the set S union {p} is a set of k+1 primes. QED. ... which may be a paragraph longer than proof by contradiction (RAA). > Sure there is extra work because we want > to know if a subset of an infinite set is also infinite. So we have to > add a mechanism that describes and characterizes this subset from the > larger set of primes that it is embedded within. Math Induction is what > describes this subset and demarks it apart from the larger set of > primes for which it is embedded. Math Induction describes the subset > and Euclid's IP proof mechanism determines whether it is finite or > infinite. But the prime that Euclid's IP proof mechanism produces may or may not be a twin prime; all you know is that p is prime. Nothing in the IP shows that p+2 or p-2 is prime; in fact, it may not be true. (p may turn out to be 23, which isn't a twin prime to _anything_.) IF prod(S)+1 is prime and IF prod(S)-1 is prime, THEN these numbers are twin primes, but you have to prove they're both primes. And in a lot of cases, prod(S)+1 isn't prime. > Therefore the proof. QED One thing I am certain of. That since the method used by Euclid to > prove infinitude of primes, that the same method must be useable to > prove the infinitude of prime subsets. In other words, the method of > Euclid must be able to prove Infinitude of Twin Primes and Perfect > Numbers and Fermat primes. What is missing is only the > details. Has anyone proven a Conjecture that says -- since Euclid IP > proof is also a method that such a method must be useable to prove the > infinitude of a subset of primes of the infinite set of primes. A proof > may not be easy to find but we are guaranteed of finding a proof of the > infinitude of a subset of primes by using Euclid's method demonstrated > in proving the Infinitude of primes. PROOF OF THE INFINITUDE OF PERFECT NUMBERS > Let me start the proof of the Infinitude of Perfect Numbers using > Euclid's proof method for infinitude of primes. I need to prove the > infinitude of primes of form (2^n)-1 prime 3 > prime 7 > prime 31 > prime 127 > prime 8191 > # n prime (2^n-1) > 7 19 524287 > 8 31 2147483647 > 9 61 2305843009213693951 > 10 89 618970019642690137449562111 > 11 107 162259276829213363391578010288127 > 12 127 170141183460469231731687303715884105727 For the case 1 by Math Induction I need to show that if I multiply two > of these primes of form (2^n)-1 together and multiply by 6 and then add > by 1 that I get another prime of form (2^n)-1. > So you're proving the infinitude of primes of the form 2^n - 1 by > induction on N in the statement: > P(N) = When you multiply any 2N primes of the form 2^n - 1 by each > other, multiply by 6, and add 1, you get a (new) prime of the form 2^n > - 1. > I think you're going to have trouble proving even P(1), and actually > will have trouble showing the new number is of the form 2^n - 1. > Suppose m and n are both at least 3. > Wrong, this is the supposition step of Math Induction, so I will have > no trouble at all. The trouble starts when I have to show N+1 and to > weave together the N step with the Euclid supposition of IP. No; again, you don't understand what I've read, or you just skipped over it. I am looking for conditions on m and n so that (2^m-1)*(2^n-1)*6+1 is a prime of the form 2^p-1. Thus, I am starting with the set {2^m-1, 2^n-1}, which is in the N=1 case. If I _were_ working with the general case, I'd have a set with the following elements: 2^(n_1)-1, 2^(n_2)-1, ..., 2^(n_(2N-1))-1, 2^(n_(2N))-1, but I don't have any sets with that many elements in it. > I have to > weave together the N step that suppose n number of primes of form > (2^n)-1 multiply by 6 and add 1, and weave that into the newly formed > set of primes in IP of P! add 1. You see all the primes in the N > supposition are contained in Euclid's IP. Yes; they're all primes. > So I marshall together these > two suppositions in hopes that I can show the N+1 step of Math > Induction. > The hardest task of this proving is to find the case #1 step of Math > Induction. I have thoughtfully provided you with conditions for finding the #1 step (P(1)); one of the two primes in the set has to be 3. This is because if you have two primes of the form 2^n-1 which are at least 2^3-1 = 7, then their product cannot be written as 2^M-1 for any integer M. Hence, one of the two primes you're starting with must be 2^0 - 1, 2^1 - 1, or 2^2 - 1, i.e., 0, 1, or 3. Since 3 is the only prime in the list, this MUST be one of your primes in the 2-element set. > In the conjecture of whether the primes of form (n^2)+1, I > am still unable to show where two primes of this form multiply together > can yield a new prime of that same form. The product of two primes is not prime. > One of the reasons is that > only numbers ending in 6,4,0 can form a prime for form (n^2)+1, leaving > me with the impression that they are finite and not infinite. If it is > impossible to generate step #1 of Math Induction then the set under > consideration is finite. No, it just means you can't prove step #1. Maybe a proof exists which uses something other than math induction. (After all, Euclid's IP can be proven without induction.) > So I have asked whether any veteran mathematician has also found this > prime form of (n^2)+1 as highly suspicious of being a finite set. > (2^p - 1) = 6(2^m - 1)(2^n - 1) + 1 > 2^p - 1 = 6 * 2^(m+n) - 6*2^m - 6*2^n + 7 > 2^p = 6 * 2^(m+n) - 6*2^m - 6*2^n + 8 > Now, if m and n are at least 3, then let M = m-3 and N = n-3. Then M > and N > are nonnegative, and 2^M and 2^N are integers. > 2^p = 6 * 2^6 * 2^M * 2^N - 6 * 2^3 * 2^M - 6 * 2^3 * 2^N + 8 > 2^p = 8 (6 * 8 * 2^M * 2^N - 6 * 2^M - 6 * 2^N + 1) > The quantity inside parentheses is odd, being one more than twice > 3 * 8 * 2^M * 2^N - 3 * 2^M - 3 * 2^N. If this is to equal a power of > 2, then we must have > 6 * 8 * 2^M * 2^N - 6 * 2^M - 6 * 2^N + 1 = 1, so > 6 * 8 * 2^M * 2^N - 6 * 2^M - 6 * 2^N = 0 > 8 * 2^M * 2^N - 2^M - 2^N = 0 > 8 * 2^M * 2^N = 2^M + 2^N. > Now suppose N >= M. Then > 8 * 2^M * 2^N = 2^M + 2^N <= 2 * 2^N, and after cancelling, > 4 * 2^M < 1, > which is a contradiction, as 2^M is at least 1. A similar consequence > happens if M >= N. This means that your statement P(1) is false, when > you pick m and n which are at least 3. > I prefer to multiply by > 2,not 6 but so few of these primes of this form are known. So I get > what I want for the initial Math Induction step case #1. Because 3x7=21 > which when multiplied by 6 gives 126 then add 1 is the prime 127. > Re-reading your sentence above: > [...] I need to show that if I multiply two > of these primes of form (2^n)-1 together and multiply by 6 and then add > by 1 that I get another prime of form (2^n)-1. > Maybe what you _meant_ to say is that _you can find_ two primes of the > form > 2^n - 1 such that when you multiply them together, multiply the result > by 6, and add 1, then you get another prime of the form 2^n - 1. That > means your induction hypothesis would be > Q(N) = There exists a set of 2N distinct primes of the form 2^n - 1, > for all N. (If you can prove Q(N) for all N, then the set of primes of > the form 2^n - 1 must be infinite; if it's a finite number S, then you > can use your proof to show that Q((S+1)/2) or Q((S+2)/2) is true > (depending on whether S is odd or even), contradicting the finiteness > result. > As I said before, there are many forms of Math Induction including > Fermat's Infinite Descent. I may not have chosen the best and optimal > form of Math Induction for Infinitude of Twin Primes, Perfect Numbers > or Fermat Primes. I have chosen the standard Math Induction of suppose > N and show N+1. If the argument is easier to see with a different > version of Math Induction such as what Chris is doing above. Well, by > all means be my guest. It's the same version; it has just been written up using the Mathematics language. > Since you are unaware of the difference between this and P(1), I can > only guess what you intended. > Now for the Math Induction step where I suppose true for the case N. I > suppose that given n number of primes of form (2^n)-1 that when > multiplying the lot together and multiply by 6 and then add 1 will > produce a new prime of form (2^n)-1. Finally for the Math Induction > step of N+1, I must show true that for n+1 number of primes of form > (2^n)-1 when multiplied the lot and multiply by 6 and add 1 yields a > new prime of form (2^n)-1. > Now, _this_ suggests that you are trying to prove P(N+1) after all, > since you're saying n number of primes. > No, I am sticking with the standard version of Math Induction and later > fit each proof with the version of Math Induction that makes it flow > the easiest. What _is_ your induction hypothesis? I suspect it's the following: (1) There exists a set of 2N twin primes. > Here is where I transfer the data of > Euclid's Infinitude of Primes proof where P!+1 is prime and thus the > case for N+1 is satisfied. Reasoning: in Euclid IP there is a 6 in the > form of 2 and 3 and there are all the other primes of the N supposition > step and even more primes in the Euclid proof. > So you're going to have > P! = 2 * 3 * (product of 2N primes of the form 2^n - 1). > This is impossible unless P = 3. If P >= 4, then 4 is a divisor of P!, > but the right-hand side is 2 times an odd number, so 4 cannot divide > evenly into it. This method will thus fail. > (This is the sort of calculation that you should be doing yourself > before you do any posting about a possible proof.) > So, I excise all the > primes out of Euclid that is not in the Supposition of the Math > Induction step. And I end up with a prime P!+1 which has form (2^n)-1. > Again, if P and n are of any reasonable size, you cannot possibly have > P! + 1 = 2^n - 1, because then > P! + 2 = 2^n. > If P >= 4, then P! >= 24, and 2^n = P! + 2 >= 26. Thus n >= 5, which > means 4 (2^2) is a divisor of 2^n. > Thus P! will be a multiple of 4 and 2^n will be a multiple of 4. But if > you add 2 to a multiple of 4, you can't get a multiple of 4. > If P = 3, then P! + 2 = 8, and sure enough, n = 3 works. > If P = 2, then P! + 2 = 4, and n = 2 works. > If P = 0 or 1, then P! + 2 is odd, and no n works. > Consequently, if you have P! + 1 = 2^n - 1, then (n,P) is one of (3,3) > or (2,2). Again, your method will fail. > (This is the sort of calculation that you should be doing yourself > before you do any posting about a possible proof.) > QED the Perfect Numbers are infinite because > primes of form (2^n)-1 are infinite. > The official webpage of Mersenne Primes is > http://primes.utm.edu/mersenne/ . > The Lucas-Lehmer Test is how it is proven that 2^p - 1 is prime. > PROOF OF THE INFINITUDE OF FERMAT PRIMES: The only known Fermat primes are > F_0 = 3 (1) > F_1 = 5 (2) > F_2 = 17 (3) > F_3 = 257 (4) > F_4 = 65537 --- end quoting Mathworld --- Outline of proof: Case 1 of Math Induction, 3 x 5 x 17 = 255 then add 2 > and the result is a third new Fermat Prime. Assume true for case N > means multiply n Fermat Primes together add 2 and the new number is a > Fermat Prime. Finally show true for case N+1 of Math Induction to show > that n+1 Fermat Primes multiplied together then add 2 is a new Fermat > Prime. > If you mean multiply F_1, F_2, ..., F_(N+1) together and add 2, then > you won't get a prime, since F_5 is composite (641 divides evenly into > it), and > F_1 * F_2 * ... * F_n + 2 = F_(n+1). > If you mean multiply F_(i_1), F_(i_2), ..., F_(i_(N + 1)) together, for > any integer variables i_*, and add 2, then this includes the case > above, and you won't be guaranteed a prime. > If you mean to prove the statement > R(N) = There exist n Fermat primes, for all N, > then you'll have a tough time doing the product. > Here is where the Euclid's Infinitude of all primes enters the > picture and to use the fact that P!+2 is a new prime > Sorry, P! is divisible by 2 if P >= 2. There's no excuse for not having > caught _this one_. > in the form of > Fermat Primes in conjunction with the use of the case N supposition of > Math Induction. I excise out of Euclid proof of IP the prime 2 so that > the P!+2 is prime and all other primes in Euclid which are not in the N > supposition step of the Math Induction. Thus P!+2 must be a new prime > and of the form 2^(2^n)+1. > QED What I am doing here is repeating these proves and making changes > because the proofs are not crystal clear flowing in my head just yet. > The transfer of Euclid's IP with Math Induction supposition step of N > and N+1 has to be a smooth transition. [...] > If there is one, you haven't found it yet. > I have found proofs of these three conjectures, what I have not done is > made the proofs flow with crystal clarity. And much more, I have given > a conjecture that says my method out of necessity has to be a proof. In > other words, Chris, my conjecture proves that I have found a proof > method. The only thing that remains is to dot the i and cross the t. I have shown that your proofs, if they use the facts that have been posted, cannot work. If you look at the actual calculations and manipulation, you will see that they refute any proof which includes these facts. If you start with a finite set of twin primes, or primes of a certain form, then showing that there is another prime isn't enough. You have to show that the new prime is also a twin prime, or is also of that same form. You find a new prime and then stop, so the proof is incomplete. === Subject: Re: proofs of Infinitude Twin Primes, Perfect Numbers, Fermat primes; using Math Induction in conjunction with Euclid proof > Wrong. You do not fully understand the Euclid IP proof mechanism. > I have doubts about whether you do; see (2) below. > Euclid guaranteed to produce Twin Primes in his proof mechanism. > I don't think so. That would mean that Euclid solved the Twin Primes > Conjecture long ago. > Suppose 3 was the only prime in existence. Consider P!+1 and P!-1. > Because 3 is the only prime in existence guarantees by that supposition > that 2 and 4 are primes not on the original list. > (1) 3! is not 3. (However, the product of the elements in {3} _is_ 3. > When you say Consider P!+1 and P!-1, this is confusing, because > P! > is the product of the elements in {1,2,...,P}. (If you want to talk > about the product of the elements in a set S, use prod(S) -- which Chris, Euclid did not solve Twin Primes conjecture, although he produced Twin Primes via his method to solve infinitude of all-primes. There is a difference which you are well aware, of yielding Twin Primes and yielding Infinitude of Twin Primes. You can run the proof of IP (Euclid Infinitude of Primes) by going after P! +1 (where P! represents the multiplication set of primes only). You can run Euclid's IP by going after P!-1 and gain the same desired conclusion. Or, you can run Euclid's IP by going after both P!+1 and P!-1 simultaneously. What Euclid Method proves is that primes are infinite, by showing that both P!+1 and P!-1 are new primes not on any finite original list. Any finite original list of Euclid generates Twin primes not on that list. The maximum statement that the Euclid IP proof holds is that if we assume the primes are finite, then we can produce Twin Primes not on that list. I am to exploit that fact to its maximum by coupling it with Mathematical Induction. I could use your help, Chris, in finding the best form of Math Induction to weave together the Supposition step of N in Math Induction. As for your comments about P!, a sort of confusing borrowing of terminology, I agree. I found it rather cute at one time because it saves me from having to constantly write (multiply all the existing primes together). Perhaps a better terminology is the Primes-factorial which means multiply together all-and-only primes. I usually hold the notion that anyone reading my work, and cannot figure out what my terminology is, has no business in reading it in the first place. I do not want people who demand every sentence and word and phrase be a solid monolith of logic. Rather I desire those who read my work to look beyond the pettiness of errors of terms, errors of spelling etc etc. I want readers who focus on the big key ideas and issues under consideration. Chris, try to make your responses shorter. Try to focus on a key issue and not a line by line critique. I could use help in outlining the variety of Math Inductions, including Fermat Infinite Descent. Can you enumerate all the forms of Math Induction that you know of? Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: proofs of Infinitude Twin Primes, Perfect Numbers, Fermat primes; using Math Induction in conjunction with Euclid proof > [...] > Euclid guaranteed to produce Twin Primes in his proof mechanism. > I don't think so. That would mean that Euclid solved the Twin Primes > Conjecture long ago. > Suppose 3 was the only prime in existence. Consider P!+1 and P!-1. > Because 3 is the only prime in existence guarantees by that supposition > that 2 and 4 are primes not on the original list. > (1) 3! is not 3. (However, the product of the elements in {3} _is_ 3. > When you say Consider P!+1 and P!-1, this is confusing, because > P! > is the product of the elements in {1,2,...,P}. (If you want to talk > about the product of the elements in a set S, use prod(S) -- which > I > Chris, Euclid did not solve Twin Primes conjecture, No kidding. > although he > produced Twin Primes via his method to solve infinitude of all-primes. That is an unproven, and unprovable statement. In fact, it is provable that his method will not produce twin primes. If he had, he would have proven the Twin Prime Conjecture. He did not prove the Twin Prime Conjecture; therefore, his method does not produce twin primes. His method might produce one twin of a twin prime p, but you still need to prove that p+2 or p-2 is a prime number to conclude that it's part of a twin. > There is a difference which you are well aware, of yielding Twin Primes > and yielding Infinitude of Twin Primes. You can run the proof of IP > (Euclid Infinitude of Primes) by going after P! +1 (where P! represents > the multiplication set of primes only). If that's what you're using P! for, it's highly misleading. Use something like prod(P), which is what I use for the product of the elements in the set P. > You can run Euclid's IP by > going after P!-1 and gain the same desired conclusion. Or, you can run > Euclid's IP by going after both P!+1 and P!-1 simultaneously. What > Euclid Method proves is that primes are infinite, by showing that both > P!+1 and P!-1 are new primes not on any finite original list. No, it doesn't, because prod(P)+1 does NOT have to be a prime number. You don't seem to realize this, which means you don't understand Euclid's proof. Do the math; it's not hard. If P = {3, 5}, then prod(P)+1 = 3*5+1 = 16, and 16 is NOT a prime. (This is basic arithmetic here.) However, 16 DOES have a prime factor not in P (namely 2), which is precisely what Euclid concludes. If we take this same P, then prod(P)-1 = 3*5-1 = 14, which is NOT prime, either. 14 has a prime factor not in P, though. (Any prime factor of 14 is not in P, in fact.) You do not have twin primes, because the pairs (2,2) and (2,7) are NOT pairs of twin primes. Got that? Here it is one more time: prod(P)+/-1 HAS A PRIME FACTOR (A PRIME DIVISOR) not in P. This does NOT mean prod(P)+/-1 is prime! > Any > finite original list of Euclid generates Twin primes not on that list. A twin prime is a prime p such that p+2 is also prime. Euclid's method on the set {3,5} produces the prime 2. This is not a twin prime, because 4 is not prime. > The maximum statement that the Euclid IP proof holds is that if we > assume the primes are finite, then we can produce Twin Primes not on > that list. This involves proof beyond what Euclid did. Euclid could only guarantee a prime not in the set P of finite primes already. > I am to exploit that fact to its maximum by coupling it with > Mathematical Induction. > I could use your help, Chris, in finding the best form of Math > Induction to weave together the Supposition step of N in Math > Induction. I'm too busy with Goldbach's Conjecture, remember? (That was sarcasm.) I'm not at a point in time where I can spend time on results which have been open for centuries. What you're asking for involves a lot of trial and error, something which is usually not acknowledged in the final proof. If you have an actual mechanism that works _when tried on examples_, then I will think about making the argument rigorous. > As for your comments about P!, a sort of confusing borrowing of > terminology, I agree. I found it rather cute at one time because it > saves me from having to constantly write (multiply all the existing > primes together). Perhaps a better terminology is the Primes-factorial > which means multiply together all-and-only primes. Well, not ALL of the primes. The product of all the primes diverges to infinity. I assume you mean all of the primes in a finite set P. > I usually hold the > notion that anyone reading my work, and cannot figure out what my > terminology is, has no business in reading it in the first place. Then why publicize it in the first place? If people have to guess at what you mean by certain phrases, or what you mean by a term which has a standardized meaning, but you're using a slightly different meaning, they will not want to read your proof. The idea of mathematics is to communicate ideas of a non-verbal nature, and if you can't communicate an idea, you've failed. > I do > not want people who demand every sentence and word and phrase be a > solid monolith of logic. For most sentences, this will be unnecessary. But you have a habit of saying, for instance, (in your alleged GC proof) Contradiction by Chebyshev's Theorem, without actually stating what the contradiction _is_. I had to drag it out of you before I could state that Chebyshev's Theorem could not be used in the way you wanted it to be used. A proof should be able to stand on its own, and be precise where it needs to be. There have been a lot of false proofs of conjectures like the GC, the Twin Primes Conjecture, conjectures involving perfect numbers, etc., and it's usually at a place where an important detail is missing. So when proving these sorts of conjectures, you really do need to be careful and know whether it's okay to go from step 11 to step 12, or from step 52 to step 53, by saying it's obvious. The simplest way to determine whether a proof is okay is to do what you've done: post it in a public place, and let mathematicians read through it. If there's a place where more details need to be supplied, someone will ask about it. > Rather I desire those who read my work to look > beyond the pettiness of errors of terms, errors of spelling etc etc. I > want readers who focus on the big key ideas and issues under > consideration. If the idea can't be communicated properly, then the details are an afterthought. > Chris, try to make your responses shorter. Try to focus on a key issue > and not a line by line critique. Sometimes I want to talk about more than one thing in one of your posts, so this is really inevitable. > I could use help in outlining the variety of Math Inductions, including > Fermat Infinite Descent. Can you enumerate all the forms of Math > Induction that you know of? There are two basic forms of mathematical induction. Suppose you're trying to prove P(n) for all n >= N. Then you can do that by proving the following: (1) Prove P(N). (2) Assume that P(k) is true, where k >= N; then prove P(k+1) is also true. (I avoid n here, because otherwise it looks like you're assuming that P(n) is true, which makes you wonder if you're using circular logic. Using a letter other than n helps clear out these wonderings.) Infinite Descent works by doing the following: (1) Prove P(N). (2) Assume that if P(k) is false, where k >= N. Prove that P(k-1) is also false. Step (2) in the standard induction and in the infinite descent are logically equivalent. P -> Q is true iff NOT(Q) -> NOT(P). There is another form of math induction, where you prove the following: (1) Prove P(N). (2) Assume that P(N), P(N+1), ..., P(k) are true; then prove P(k+1) is also true. An example of another form, which isn't used as often, is the following: THEOREM. If n >= 2, then n can be written as the product of prime numbers. Proof. Let P(n) be the statement n can be written as the product of prime numbers. P(2) is easily seen to be true -- 2 is prime. Now assume that P(2), P(3), ..., P(k) are true; now we need to write k+1 as the product of prime numbers. If k+1 is prime, we're done. (Write k+1 as k+1.) Otherwise, k+1 is composite, so there are integers a and b between 2 and k such that a*b = k+1. The induction hypothesis says a and b can be written as the product of prime numbers. Multiplying these products together yields a prime factorization of k+1. Thus P(k+1) is true, whether k+1 is prime or not. By math induction, the theorem is true for all n >= 2. QED. === Subject: Re: proofs of Infinitude Twin Primes, Perfect Numbers, Fermat primes; using Math Induction in conjunction with Euclid proof (snip) > although he > produced Twin Primes via his method to solve infinitude of all-primes. > That is an unproven, and unprovable statement. In a sense that is true, for historically Euclid focused on just prodP+1. But his method yields prodP+1 and prodP-1 which are twin primes. > In fact, it is provable that his method will not produce twin primes. > If he had, he would have proven the Twin Prime Conjecture. He did not > prove the Twin Prime Conjecture; therefore, his method does not produce > twin primes. I did not say he did. I said Euclid proved Infinitude of Primes by a method that yields two new primes (twin primes) not on the original list. > His method might produce one twin of a twin prime p, but you still need > to prove that p+2 or p-2 is a prime number to conclude that it's part > of a twin. No, Chris. > There is a difference which you are well aware, of yielding Twin Primes > and yielding Infinitude of Twin Primes. You can run the proof of IP > (Euclid Infinitude of Primes) by going after P! +1 (where P! represents > the multiplication set of primes only). > If that's what you're using P! for, it's highly misleading. Use > something like prod(P), which is what I use for the product of the > elements in the set P. > You can run Euclid's IP by > going after P!-1 and gain the same desired conclusion. Or, you can run > Euclid's IP by going after both P!+1 and P!-1 simultaneously. What > Euclid Method proves is that primes are infinite, by showing that both > P!+1 and P!-1 are new primes not on any finite original list. > No, it doesn't, because prod(P)+1 does NOT have to be a prime number. > You don't seem to realize this, which means you don't understand > Euclid's proof. Do the math; it's not hard. Wrong Chris, you are doing too much math and arithmetic and not focused on the logic of Euclid IP. You are forgetting that Euclid is bound by a Assumption. The assumption that the primes are finite. Thus when Euclid near the end of the proof constructs both prodP+1 (historically he only constructed prodP+1) and constructs prodP-1, then, Chris the assumption binding the proof forces the conclusion that both prodP+1 and prodP-1 are primes and twin primes at that. Look at it this way, Chris, if the Universe of primes is only 3 and 7 and you assume these are the only two primes that exist, then prodP+1 and prodP-1 are 22 and 20. In this universe of assumption then both 20 and 22 are prime numbers because the only primes that exist are 3 and 7 and they do not divide into 20 and 22 evenly. And they are Twin Primes. Your trouble is that you forget you are under the assumption-space which guarantees both prodP+1 and prodP-1 are primes and twin-primes. If you do not understand this now, I am not going to further correct you. > If P = {3, 5}, then prod(P)+1 = 3*5+1 = 16, and 16 is NOT a prime. Again, you forget that Euclid was under the assumption of finite primes and indeed in this Universe of primes 16 is a prime. > (This is basic arithmetic here.) However, 16 DOES have a prime factor > not in P (namely 2), which is precisely what Euclid concludes. It maybe arithmetic but you have lost sense of the proof of Euclid. > If we take this same P, then prod(P)-1 = 3*5-1 = 14, which is NOT > prime, either. 14 has a prime factor not in P, though. (Any prime > factor of 14 is not in P, in fact.) You do not have twin primes, > because the pairs (2,2) and (2,7) are NOT pairs of twin primes. > Got that? Here it is one more time: prod(P)+/-1 HAS A PRIME FACTOR (A > PRIME DIVISOR) not in P. This does NOT mean prod(P)+/-1 is prime! And for the last time, you failed to follow a Reductio Ad absurdum. In a direct proof prodP +/- 1 may or may not be prime. In a Reductio where you are under the assumption of finite set of primes, these two primes are new primes and twin at that. Think about it Chris. (snip) > There are two basic forms of mathematical induction. Suppose you're > trying to prove P(n) for all n >= N. Then you can do that by proving > the following: > (1) Prove P(N). > (2) Assume that P(k) is true, where k >= N; then prove P(k+1) is also > true. > (I avoid n here, because otherwise it looks like you're assuming that > P(n) is true, which makes you wonder if you're using circular logic. > Using a letter other than n helps clear out these wonderings.) Any more forms? > Infinite Descent works by doing the following: > (1) Prove P(N). > (2) Assume that if P(k) is false, where k >= N. Prove that P(k-1) is > also false. > Step (2) in the standard induction and in the infinite descent are > logically equivalent. P -> Q is true iff NOT(Q) -> NOT(P). > There is another form of math induction, where you prove the following: > (1) Prove P(N). > (2) Assume that P(N), P(N+1), ..., P(k) are true; then prove P(k+1) is > also true. > An example of another form, which isn't used as often, is the > following: > THEOREM. If n >= 2, then n can be written as the product of prime > numbers. > Proof. Let P(n) be the statement n can be written as the product of > prime numbers. P(2) is easily seen to be true -- 2 is prime. > Now assume that P(2), P(3), ..., P(k) are true; now we need to write > k+1 as the product of prime numbers. > If k+1 is prime, we're done. (Write k+1 as k+1.) Otherwise, k+1 is > composite, so there are integers a and b between 2 and k such that > a*b = k+1. The induction hypothesis says a and b can be written as the > product of prime numbers. Multiplying these products together yields a > prime factorization of k+1. Thus P(k+1) is true, whether k+1 is prime > or not. > By math induction, the theorem is true for all n >= 2. QED. > Question: is there any math proof in existence that uses Math Induction to prove a set is infinite. I suspect mine are the first in history. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: proofs of Infinitude Twin Primes, Perfect Numbers, Fermat primes; using Math Induction in conjunction with Euclid proof you want to prove is the following: THEOREM ITP1. There EXISTS a sequence of primes p(1), p(2), ... such that: (1) p(1) and p(2) are twin primes, (2) p(2n+1) = p(1)p(2)...p(2n)-1 is prime, and (3) p(2n+2) = p(1)p(2)...p(2n)+1 is also prime. What people (including me) have been _thinking_ is that you want to prove: THEOREM IPT2. If P = {p(1), p(2), ..., p(n)} is ANY set of primes, then (1) p(2n+1) = p(1)p(2)...p(2n)-1 is prime, and (2) p(2n+2) = p(1)p(2)...p(2n)+1 is also prime. You are not doing well at communicating what's going on with your proof; if you had clarified this issue at the start, a lot of frustration could have been avoided. (Note that both Theorem IPT1 and IPT2 would imply an infinitude of twin primes, and sci.math posters have been telling you IPT2 is false.) The proof by math induction would go: Proof (outline). Let P(N) be the statement p(2N+1) and p(2N+2) are prime, for N >= 0. P(N) will be proven by induction on N. P(0) is true because ... ((You have to show p(1) and p(2) are.)) Now assume P(0), P(1), ..., P(k) are true. Then, if p(2k+1) = p(1)p(2)...p(2k)-1 and p(2k+2) = p(1)p(2)...p(2k)+1, then p(2k+2)-p(2k+1) = 2 by construction, and p(2n+1) and p(2n+2) are prime because ... ((the important part of the proof goes here; why neither of p(2n+1) and p(2n+2) is composite.)) Thus P(k+1) is also true. (After reflection): But even Theorem ITP1 is false. This is because 2 cannot be on the list of primes, because 2 is not a twin prime. That means all the p(i)'s are odd, but the product of odd numbers is odd, and when you add or subtract 1 from an odd number, you get an even number, which is not prime. (In particular, you'll end up with two even numbers, and both of these can't prime, because they can't both equal 2.) So the best chance of a proof based on what you want to do is to prove: THEOREM ITP3. There EXISTS a sequence of primes p(0), p(1), p(2), ... such that: (1) p(1) and p(2) are twin primes, (2) p(2n+1) = p(0)p(1)p(2)...p(2n)-1 is prime, (3) p(2n+2) = p(0)p(1)p(2)...p(2n)+1 is prime, and (4) p(0) = 2. Then, once you have all four conditions, you have a proof of the infinitude of primes. > (snip) > although he > produced Twin Primes via his method to solve infinitude of all-primes. > That is an unproven, and unprovable statement. > In a sense that is true, for historically Euclid focused on just > prodP+1. > But his method yields prodP+1 and prodP-1 which are twin primes. There are sets P such that neither of prod(P)+1 or prod(P)-1 is prime, and consequently, prod(P)+1 and prod(P)-1 are not twin primes. One example is {3,5}. But if you start with the right set P, then it's possible for this to happen. ((For those readers just joining us, prod(P) is my notation for the product of the elements of P.)) > In fact, it is provable that his method will not produce twin primes. > If he had, he would have proven the Twin Prime Conjecture. He did not > prove the Twin Prime Conjecture; therefore, his method does not produce > twin primes. > I did not say he did. I said Euclid proved Infinitude of Primes by a > method that yields two new primes (twin primes) not on the original > list. If you add if P is chosen correctly, I would agree. > His method might produce one twin of a twin prime p, but you still need > to prove that p+2 or p-2 is a prime number to conclude that it's part > of a twin. Again, if you add if P is chosen correctly, I would agree. > There is a difference which you are well aware, of yielding Twin Primes > and yielding Infinitude of Twin Primes. You can run the proof of IP > (Euclid Infinitude of Primes) by going after P! +1 (where P! represents > the multiplication set of primes only). > If that's what you're using P! for, it's highly misleading. Use > something like prod(P), which is what I use for the product of the > elements in the set P. > You can run Euclid's IP by > going after P!-1 and gain the same desired conclusion. Or, you can run > Euclid's IP by going after both P!+1 and P!-1 simultaneously. What > Euclid Method proves is that primes are infinite, by showing that both > P!+1 and P!-1 are new primes not on any finite original list. Again, if you add if P is chosen correctly, I would agree. > Wrong Chris, you are doing too much math and arithmetic and not focused > on the logic of Euclid IP. If so, then you aren't doing _enough_; you haven't even been telling us whether there exists a set P of primes such that ... are prime, or for all sets P of primes, ... are prime. There is a world of difference here, and this is one place where you need to be precise. The middle ground is the place where a proof is. In dealing with conjectures that have histories of false proofs, it is better to err on the side of too much math and arithmetic, because the ideas may not connect the way you think they do. > You are forgetting that Euclid is bound by a > Assumption. The assumption that the primes are finite. Thus when Euclid > near the end of the proof constructs both prodP+1 (historically he only > constructed prodP+1) and constructs prodP-1, then, Chris the assumption > binding the proof Here, the proof means your proof. As we both know, Euclid had nothing to do with twin primes. (This is the sort of sloppiness that makes people upset at what you say; they've been trained to avoid it and you haven't.) > forces the conclusion that both prodP+1 and prodP-1 > are primes and twin primes at that. Again, if you add if P is chosen correctly, I would agree. > Look at it this way, Chris, if the Universe of primes is only 3 and 7 > and you assume these are the only two primes that exist, then prodP+1 > and prodP-1 are 22 and 20. In this universe of assumption then both 20 > and 22 are prime numbers because the only primes that exist are 3 and 7 > and they do not divide into 20 and 22 evenly. And they are Twin Primes. Again, if you add if P is chosen correctly, I would agree. But it won't ALWAYS happen. > Your trouble is that you forget you are under the assumption-space > which guarantees both prodP+1 and prodP-1 are primes and twin-primes. Again, if you add if P is chosen correctly, I would agree. But you have to prove the claim that will work. For instance, if you prove THEOREM ITP3. There EXISTS a sequence of primes p(0), p(1), p(2), ... such that: (1) p(1) and p(2) are twin primes, (2) p(2n+1) = p(0)p(1)p(2)...p(2n)-1 is prime, (3) p(2n+2) = p(0)p(1)p(2)...p(2n)+1 is prime, and (4) p(0) = 2. ... then you have to show that p(googol) is prime, something which is probably beyond computation on any machine on Planet Earth at the will invariably be prime. (If you look up Chebyshev's Theorem, you'll find out he doesn't just state his result about primes; he also shows why it's actually true. Just because someone says something doesn't mean it's a Theorem and is automatically true; it's only a conjecture. If you don't prove something, beyond the shadow of a doubt, it's not a proof.) > If P = {3, 5}, then prod(P)+1 = 3*5+1 = 16, and 16 is NOT a prime. > Again, you forget that Euclid was under the assumption of finite primes > and indeed in this Universe of primes 16 is a prime. ... or 16 has a prime DIVISOR not in {3,5}. You must not have read the right version of Euclid's proof, because he never claims that prod(P)+1 ALWAYS WILL BE PRIME; only that by considering it, a prime not in P will be found. Find an introductory textbook on Number Theory -- not a popular math book -- and check it. > (This is basic arithmetic here.) However, 16 DOES have a prime factor > not in P (namely 2), which is precisely what Euclid concludes. > It maybe arithmetic but you have lost sense of the proof of Euclid. No; you've added something to Euclid's proof that wasn't there before; the fact that prod(P)+1 (and prod(P)-1) is prime. It was not there before, one reason being that it is not true. Euclid's proof is: (1) Assume a finite set P of primes; (2) Let N be a certain integer depending on P; (3) Any prime which divides into N cannot be in P; (4) contradiciton between (1) and (3). > And for the last time, you failed to follow a Reductio Ad absurdum. > [...] In a Reductio where > you are under the assumption of finite set of primes, these two primes > are new primes and twin at that. But wait a minute ... if these new numbers aren't in the finite set of prime numbers, they can't be primes now, can they? Let's back up here: (1) Let P be the set of ALL primes; then anything not in P, by definition, is not prime. We can show that 2 and 3 are prime. (P is infinite, but we don't know this yet.) (2) We want to know whether P is finite or infinite. Let's adopt the hypothesis that P is finite and see where it leads us. (3) Let N1 = prod(P)+1 and N2 = prod(P)-1. Since P has at least 2 elements, N1 and N2 are strictly bigger than any element of P; in particular, they are larger than the maximum element of P. (4) That means N1 and N2 are not in P, so N1 and N2 are not prime, after all. The point is that you (among lots of other people) seem to be confused proven. > [...] > Question: is there any math proof in existence that uses Math Induction > to prove a set is infinite. I suspect mine are the first in history. Yes; the direct proof of the infinitude of primes uses math induction. THEOREM. For every positive integer N, there exists a set of N prime numbers. Proof: Induction on N. Certainly the statement when N=1 is true; {2} is a set of one prime number. Now suppose that N=k is true, that there is a set of k primes, p(1), p(2), ..., p(k). Let E = p(1)p(2)...p(k) + 1. Since every integer >= 2 can be factored into primes, E has a prime divisor q, and this prime divisor q is different from p(1), ..., p(k). Thus {p(1),p(2), ...,p(k),q} is a set of k+1 primes. By math induction, the result follows. QED. COROLLARY: The set of primes is infinite. Proof: If the set of primes is finite, then it has M elements, for some integer M. But the Theorem states that there is a set of M+1 primes, contradicting the assumption that the set of primes is finite. === Subject: Re: proofs of Infinitude Twin Primes, Perfect Numbers, Fermat primes; using Math Induction in conjunction with Euclid proof you want to prove is the following: THEOREM ITP1. There EXISTS a sequence of primes p(1), p(2), ... such that: (1) p(1) and p(2) are twin primes, (2) p(2n+1) = p(1)p(2)...p(2n)-1 is prime, and (3) p(2n+2) = p(1)p(2)...p(2n)+1 is also prime. What people (including me) have been _thinking_ is that you want to prove: THEOREM IPT2. If P = {p(1), p(2), ..., p(n)} is ANY set of primes, then (1) p(2n+1) = p(1)p(2)...p(2n)-1 is prime, and (2) p(2n+2) = p(1)p(2)...p(2n)+1 is also prime. You are not doing well at communicating what's going on with your proof; if you had clarified this issue at the start, a lot of frustration could have been avoided. (Note that both Theorem IPT1 and IPT2 would imply an infinitude of twin primes, and sci.math posters have been telling you IPT2 is false.) I have the proofs of Infinitude of Twin Primes, of Perfect Numbers and also of Fermat primes. I have the overall framework of these proofs as a Math Induction coupled with Euclid's method of Infinitude of Primes proof. What I do not have at this moment are all the details inside the proofs. One such important detail was revealed to me last night in a reply to Dik Winter. The detail is that Math Induction works on a SUBSET of the Twin Primes, on a subset of the primes of form (2^n)-1 and a subset of the Fermat primes. If I show that a subset of twin primes or perfect number primes or fermat primes is infinite, clearly I have shown that the larger set is infinite. So, Chris, when I take two twin primes, multiply them together, multiply by 2 and then add/subtract 1 and end up with a new twin primes serves as my case #1 Math Induction step. What this ends up with is that I have proven a subset of Twin Primes is infinite. So that is probably why many are confused. Because I need not show that any and every pair of twin primes multiplied together, multiply 2, add/subtract 1 furnishes us with a new twin prime. And to further show you that the method is valid. This method proves that primes of form (n^2)+1 and (n^2)-1 are FINITE. Because it is impossible to start the Math Induction because it is impossible to have case #1 as Dik pointed out by the algebra in a different post to this thread. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: proofs of Infinitude Twin Primes, Perfect Numbers, Fermat primes; using Math Induction in conjunction with Euclid proof > you want to prove is the following: > THEOREM ITP1. There EXISTS a sequence of primes p(1), p(2), ... such > that: > (1) p(1) and p(2) are twin primes, > (2) p(2n+1) = p(1)p(2)...p(2n)-1 is prime, and > (3) p(2n+2) = p(1)p(2)...p(2n)+1 is also prime. > What people (including me) have been _thinking_ is that you want to > prove: > THEOREM IPT2. If P = {p(1), p(2), ..., p(n)} is ANY set of primes, then > (1) p(2n+1) = p(1)p(2)...p(2n)-1 is prime, and > (2) p(2n+2) = p(1)p(2)...p(2n)+1 is also prime. > You are not doing well at communicating what's going on with your > proof; if you had clarified this issue at the start, a lot of > frustration could have been avoided. (Note that both Theorem IPT1 and > IPT2 would imply an infinitude of twin primes, and sci.math posters > have been telling you IPT2 is false.) > I have the proofs of Infinitude of Twin Primes, of Perfect Numbers and > also of Fermat primes. I have the overall framework of these proofs as > a Math Induction coupled with Euclid's method of Infinitude of Primes > proof. What I do not have at this moment are all the details inside the > proofs. > One such important detail was revealed to me last night in a reply to > Dik Winter. The detail is that Math Induction works on a SUBSET of the > Twin Primes, on a subset of the primes of form (2^n)-1 and a subset of > the Fermat primes. If I show that a subset of twin primes or perfect > number primes or fermat primes is infinite, clearly I have shown that > the larger set is infinite. Yes, that's what I mean by There EXISTS an (infinite) sequence of primes. > So, Chris, when I take two twin primes, multiply them together, > multiply by 2 and then add/subtract 1 and end up with a new twin primes > serves as my case #1 Math Induction step. What this ends up with is > that I have proven a subset of Twin Primes is infinite. So that is > probably why many are confused. Because I need not show that any and > every pair of twin primes multiplied together, multiply 2, add/subtract > 1 furnishes us with a new twin prime. > And to further show you that the method is valid. This method proves > that primes of form (n^2)+1 and (n^2)-1 are FINITE. No, it doesn't; it just means that particular method doesn't work. It's like saying a cat can't give birth to an adult elephant, so adult elephants don't exist. Basic logic lesson: Let P be the statement: Math Induction proves result X, and Q be the statement: There is a proof of result X. Then P implies Q is true, but not P implies not Q is false. You're saying that not P implies not Q must be true. Nope, invalid. The form n^2 - 1 can be taken care of in a non-induction way: If p is a prime of the form n^2-1, then p = n^2 - 1 = (n+1)(n-1). This means n+1 = 1 or n-1 = 1; if n were any other value, you would have written p as the product of two numbers >= 2. This means n = 0 or n = 2. This makes p = 0^2 - 1 = -1 (which is impossible), or p = 2^2 - 1 = 3. So there is only one prime of the form n^2 - 1, namely 3. The n^2 + 1 case is harder. === Subject: Re: Direct Sums I guess my question is now then, if only given that V=R(T)+N(T), how does one prove that R(T)*intersection*N(T)={0} since apparently it is needed? I am only given that V=R(T)+N(T)... I understand IF I am given R(T)*intersection*N(T)={0}, then the proof follows easily... but am not sure where to being if I am not given this info, but still asked to prove the direct sum. I know that N(T) and R(T) are subspaces of the vector space, and have proved this already,but the definition of a direct sum requires both that V=W1+W2 AND W1*intersection*W2={0}. In my case, W1 and W2 are replaced by N(T) and R(T), obviously. So, if the definition requires both conditions, then how can I be asked to prove the direct sum being given only one condition? === Subject: Re: Direct Sums days. My association with the Department is that of an alumnus. >I guess my question is now then, >if only given that V=R(T)+N(T), >how does one prove that >R(T)*intersection*N(T)={0} since >apparently it is needed? It is needed to verify that V is the direct sum of R(T) and N(T), since it is part of the definition of direct sum. >I am only given that V=R(T)+N(T)... But you are also supposed to know certain theorems about nullspaces and ranges. In particular, there is something called the Dimension Theorem or the Rank-Nullity Theorem, which I expclitly mentioned earlier, that says: RANK-NULLITY THEOREM If V is a finite dimensional vector space, and T:V->W is any linear transformation, then dim(V) = dim(R(T)) + dim(N(T)). (The dimention of R(T) is known as the rank of T; the dimension of N(T) is known as the nullity of T; hence the Rank-Nullity name). I also mentioned the following, which you should PROVE: EXERCISE: Let V be a vector space, and let W1 and W2 be finite dimensional subsapces of V. Then dim(W1+W2) = dim(W1)+dim(W2)-dim(W1 intersect W2). Hint for proof: Start with a basis for W1 intersect W2, say {w1,...,wk}. Then find vectors {v(k+1),...,vn} and {z{k+1},....,zm} such that {w1,...,wk,v(k+1),...,vn} is a basis for W1, and {w1,...,wk,z(k+1),...,zm} is a basis for W2. Show that {w1,...,wk,v(k+1),...,vn,z(k+1),...,zm} is a basis for W1+W2. dim(N(T)+R(T)) = dim(V) (by hypothesis) = dim(N(T)) + dim(R(T)) (by Rank-Nullity Thm) and dim(N(T)+R(T)) = dim(N(T)) + dim(R(T)) - dim(N(T) intersect R(T)) (by exercise) So dim(N(T))+dim(R(T)) = dim(N(T))+dim(R(T))-dim(N(T)intersect R(T)) hence dim(N(T) intersect R(T)) = 0. What can you say about a space that has dimension 0? >So, if the definition requires both >conditions, then how can I be asked to >prove the direct sum being given only >one condition? Because, in the case of the subspace you are given, one condition turns out to be sufficient. OBVIOUSLY you need to use other facts about those two particular subspaces (such as the Rank-Nullity theorem above). I already indicated all of this in a previous response. -- 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: Direct Sums I'm sorry... I understand now. explainations. I have pieced it all === Subject: Re: Image of boundary under affine mappings of the image of a set A, is contained in the image of the mapping of >the boundary of the A for a continous mapping. > Certainly not true in general. What other assumptions are you not > telling us? You mentioned affine in the Subject line, but continuous > here. Which is it? If it's affine but you need to say continuous > because that's not automatic, you must be working in infinite-dimensional > spaces. > Besides William's example (which you might think of as a bit of a cheat > because the boundary of A is empty), you could take f: R^2 -> R^2 > given by f(x,y) = (x,0), with A = {(x,y): 0 < x < 1}. What am I to think of f and A? bd f(A) = bd (0,1)x{0} = [0,1] x {0} f(bd A) = f([0,1] x R) = [0,1] x {0} === Subject: Re: Image of boundary under affine mappings of the image of a set A, is contained in the image of the mapping of >the boundary of the A for a continous mapping. > Certainly not true in general. What other assumptions are you not > telling us? You mentioned affine in the Subject line, but continuous > here. Which is it? If it's affine but you need to say continuous > because that's not automatic, you must be working in infinite-dimensional > spaces. > Besides William's example (which you might think of as a bit of a cheat > because the boundary of A is empty), you could take f: R^2 -> R^2 > given by f(x,y) = (x,0), with A = {(x,y): 0 < x < 1}. > What am I to think of f and A? > bd f(A) = bd (0,1)x{0} = [0,1] x {0} > f(bd A) = f([0,1] x R) = [0,1] x {0} No, bd A = {0,1} x R so f(bd A) = {0,1} x {0}. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: derivative of a discrete fourier transform interpolation Hi everyone I'm trying to understand this problem which arises in a paper I've been reading. I'm quite new to fourier analyses, and I haven't been able to find anything in books/web which answers it. If x(s) is an (unknown) real function which is sampled at evenly spaced intervals denoted s0,s1,s2,s3...s(N-1), we get a known real sequence x0,x1,x2,...x(N-1). Now we make an interpolation of the original function. Define X0,X1,X2....X(N-1) as the discrete fourier transform of x0,x1...x(N-1). The Xj's are typically complex numbers. Now, I believe we can interpolate x(s) as x(s)= (1/N)*sum {n=0...(N-1)} Xn*exp(2i*pi*n*s/N) My problem arises when trying to calculate dx/ds using this interpolation. Differentiating term by term, we find dx/ds(s)= (1/N^2)*2i*pi*sum {n=0...(N-1)} n*Xn*exp(2i*pi*n*s/N) Now, it seems to me that dx/ds is typically not real valued, even though x(s) is real valued. 1. Is what I've said above correct?? 2. What is a better approach to calculating the derivatives. Gareth Davies === Subject: Re: derivative of a discrete fourier transform interpolation >I'm trying to understand this problem which arises in a paper I've been >reading. I'm quite new to fourier analyses, and I haven't been able to >find anything in books/web which answers it. >If x(s) is an (unknown) real function which is sampled at evenly spaced >intervals denoted s0,s1,s2,s3...s(N-1), we get a known real sequence >x0,x1,x2,...x(N-1). >Now we make an interpolation of the original function. Define >X0,X1,X2....X(N-1) as the discrete fourier transform of x0,x1...x(N-1). >The Xj's are typically complex numbers. Now, I believe we can >interpolate x(s) as >x(s)= (1/N)*sum {n=0...(N-1)} Xn*exp(2i*pi*n*s/N) >conjugate pairs. Not as a real-valued function of a continuous variable s, but for integer s. This is because if s is an integer, exp(2i pi s) = 1 and so conjugate(exp(2i pi n s/N)) = exp(2i pi (N-n) s/N). If you want a real-valued interpolation, replace exp(2i pi n s/N) by exp(2i pi (n-N) s/N) for n > N. Then x(s) is real-valued iff X(N-n) = conjugate(X(n)) for all n. >My problem arises when trying to calculate dx/ds using this >interpolation. Differentiating term by term, we find >dx/ds(s)= (1/N^2)*2i*pi*sum {n=0...(N-1)} n*Xn*exp(2i*pi*n*s/N) >Now, it seems to me that dx/ds is typically not real valued, even though >x(s) is real valued. The derivative of a real-valued function has to be real-valued. It will be if you make the change I suggested above. === Subject: Toeplitz Matrix, DFT, and rank Hello All, I recall reading a paper that stated that an N X N conjugate symmetric (Hermitian) Toeplitz matrix is nonsingular if none of elements of the discrete Fourier transform of the first row equal zero. I can't find the paper in my files, can some one provide a reference? Also interested in knowing if the rank of the matrix can be determine by the number of non-zero elements of the discrete Fourier transform of the first row. Scott === Subject: Re: Toeplitz Matrix, DFT, and rank >I recall reading a paper that stated that an N X N conjugate symmetric >(Hermitian) Toeplitz matrix is nonsingular if none of elements of the >discrete Fourier transform of the first row equal zero. You must not recall correctly. For example, the 3 x 3 Toeplitz matrix [ a b c ] [ b a b ] [ c b a ] has determinant (a-c)(a^2 + ac - 2 b^2), while the discrete Fourier transform has elements (a + w b + w^2 c)/sqrt(3), where w is a cube root of 1. For example if a=c=0, b<>0 the DFT has no zeros but the matrix is singular. Perhaps the paper was talking about cyclic Toeplitz matrices, where the eigenvalues are sqrt(N) times those discrete Fourier transform values. === Subject: Re: Toeplitz Matrix, DFT, and rank > Hello All, > I recall reading a paper that stated that an N X N conjugate symmetric > (Hermitian) Toeplitz matrix is nonsingular if none of elements of the > discrete Fourier transform of the first row equal zero. I can't find > the paper in my files, can some one provide a reference? Also > interested in knowing if the rank of the matrix can be determine by the > number of non-zero elements of the discrete Fourier transform of the > first row. This is certainly true for a cyclic matrix, which is special Toeplitz, since the eigenvalues are the DFT coefficients. rusty === Subject: Re: How to solve (2x + 3y)^3? <8293110.1128017866676.JavaMail.jakarta@nitrogen.mathforum.org>, > Just for laffs, the query as to how can one foil a trinomial with a > binomial: > (a + b + c)*(d + e) > We sing: first, outer, inner, mouter, minner, last > Here mouter is b*e and minner is b*d on the idea > that we're taking the middle of the trinomial with ... > I don't think this will catch on. I make it First, outer, penultimate (that's bd), second (that's be), inner, last, for a catchy mnemonic fopsil. Anyone know a good mnemonic for spelling mnemonic? No jokes about mnemonic depression, please, that's another thread. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: How to solve (2x + 3y)^3? >Just for laffs, the query as to how can one foil a trinomial with a binomial: > (a + b + c)*(d + e) >We sing: first, outer, inner, mouter, minner, last >Here mouter is b*e and minner is b*d on the idea >that we're taking the middle of the trinomial with ... >I don't think this will catch on. We can only hope... === Subject: Re: irreducible represenation of Lie algebra sl_3 > Is there for sl_3 something like to Clebsch-Gordon decomposition > formulae for tensor product of two irreducible representation of sl_2? > Let T_(a,b) and T_(c,d) be two irreducible represenation of sl_3, here > a,b,c,d are natural numbers. What is exact decomposition T_(a,b) x > T_(c,d)? Here x is tensor product. The tensor product of irreducible representations of sl_3 is, in general, not multiplicity-free. One way of talking about this algebraically is to look at the intertwining operators between T_(a,b) x T_(c,d) and T_(e,f); for some choices of the integers, these operators form a vector space of dimension greater than 1. (In the case of sl_2, the spa ce of intertwining operators between T_a x T_b and T_c is zero if T_c is not contained in the product, and 1 if it is. The Clebsch-Gordan coefficients are the matrix elements of a particular intertwining operator, with respect to convenient bases of the representations.) Therefore, for explicit decompositions like the Clebsch-Gordan decomposition, you need to find a way to resolve the multiplicity, that is, to choose a basis for the space of intertwining operators. There have been lots of papers over the years in the Journal of Mathematical Physics which flail away at the problem and, and some have claimed to solve it. -- Chris Henrich http://www.mathinteract.com God just doesn't fit inside a single religion. === Subject: Re: irreducible represenation of Lie algebra sl_3 >> Is there for sl_3 something like to Clebsch-Gordon decomposition >> formulae for tensor product of two irreducible representation of sl_2? >> Let T_(a,b) and T_(c,d) be two irreducible represenation of sl_3, here >> a,b,c,d are natural numbers. What is exact decomposition T_(a,b) x >> T_(c,d)? Here x is tensor product. > There have been lots of papers over the years in the Journal of > Mathematical Physics which flail away at the problem and, and some have > claimed to solve it. Surely that is a bit pessimistic. I think the irreducible representations of sl(n) correspond to the Young diagrams with <= n columns, and the rule for multiplying these can be found eg in Weyl's Classical Groups. -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: irreducible represenation of Lie algebra sl_3 >> Is there for sl_3 something like to Clebsch-Gordon decomposition >> formulae for tensor product of two irreducible representation of sl_2? >> Let T_(a,b) and T_(c,d) be two irreducible represenation of sl_3, here >> a,b,c,d are natural numbers. What is exact decomposition T_(a,b) x >> T_(c,d)? Here x is tensor product. > There have been lots of papers over the years in the Journal of > Mathematical Physics which flail away at the problem and, and some have > claimed to solve it. > Surely that is a bit pessimistic. > I think the irreducible representations of sl(n) > correspond to the Young diagrams with <= n columns, > and the rule for multiplying these > can be found eg in Weyl's Classical Groups. The irreducible representations of sl(n) do correspond to Young diagrams. Weyl shows this, and finds the characters of the representations. As for rules for multiplying representations, I do not recall that Weyl discusses them at all. Other people have - I think the leading result is called the Littlewood - Richardson rules. But these rules are a way of counting the multiplicity of an irreducible representation in a tensor product; they do not resolve the product in the very explicit, detailed way that I think the OP was asking for. -- Chris Henrich http://www.mathinteract.com God just doesn't fit inside a single religion. === Subject: Re: irreducible represenation of Lie algebra sl_3 >> I think the irreducible representations of sl(n) >> correspond to the Young diagrams with <= n columns, >> and the rule for multiplying these >> can be found eg in Weyl's Classical Groups. > The irreducible representations of sl(n) do correspond to Young > diagrams. Weyl shows this, and finds the characters of the > representations. > As for rules for multiplying representations, I do not recall that Weyl > discusses them at all. I could be wrong about that - someone has borrowed/stolen my copy of Weyl. However, the product of representations corresponds to the product of characters, which is easy enough to compute if n = 3. > Other people have - I think the leading result > is called the Littlewood - Richardson rules. But these rules are a > way of counting the multiplicity of an irreducible representation in a > tensor product; they do not resolve the product in the very explicit, > detailed way that I think the OP was asking for. I'm not sure what you mean. The product of representations _is_ the tensor product. The Littlewood-Richardson rules - which would not be very complicated if n = 3 - seem to me completely explicit. In any case, this _is_ the rule for multiplying representations, so if it is complicated that is life ... -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: irreducible represenation of Lie algebra sl_3 >> I think the irreducible representations of sl(n) >> correspond to the Young diagrams with <= n columns, >> and the rule for multiplying these >> can be found eg in Weyl's Classical Groups. > The irreducible representations of sl(n) do correspond to Young > diagrams. Weyl shows this, and finds the characters of the > representations. > > As for rules for multiplying representations, I do not recall that Weyl > discusses them at all. > I could be wrong about that - > someone has borrowed/stolen my copy of Weyl. > However, the product of representations corresponds > to the product of characters, > which is easy enough to compute if n = 3. > Other people have - I think the leading result > is called the Littlewood - Richardson rules. But these rules are a > way of counting the multiplicity of an irreducible representation in a > tensor product; they do not resolve the product in the very explicit, > detailed way that I think the OP was asking for. > I'm not sure what you mean. > The product of representations _is_ the tensor product. > The Littlewood-Richardson rules - > which would not be very complicated if n = 3 - > seem to me completely explicit. > In any case, this _is_ the rule for multiplying representations, > so if it is complicated that is life ... The OP wanted more than just the multiplicities of the irreducible components of a product. He wanted explicit matrix elements for the decomposition. In the case of sl_2, this is a solved problem. If we let T_a, where a in an integer >= 0, stand for the representation of degree a+1, then T_a x T_b decomposes into the sum of one copy each of T_{|a-b|}, T_{|a-b+1}, ..., T_{a+b}. (TeX notation.) For example, T_2 x T_1 = T_1 + T_2 + T_3. Because this decomposition is always multiplicity-free, it is a well-defined problem to compute the matrix of the map which embeds T_3 in T_2 x T_1. There is only one arbitrary choice of phase to be made. The general solution is called the Clebsch-Gordan coefficients, and can be found in texts on quantum mechanics and angular momentum. Wigner's book on group theory and quantum mechanics is a classic reference. The decomposition of products of representations of sl_3 is not multiplicity-free in general. For example, from old lecture notes I find that T_{1,1} x T_{1,1} = T_{3,0} + T_{2,2} + 2T_{1,1} + T_{0,3} + T_{0,0}. For the OP, this does not complete the solution. He needs explicit matrix elements for the operators that embed T_{a,b} into T_{1,1} x T_{1,1}. There is a one-dimensional space of embeddings of T_{2,2} into the product; so, a norm-preserving embedding is unique up to a choice of phase. But there is a two-dimensional space of embeddings of T_{1,1} into the product. Explicit Clebsch-Gordan coefficients are not defined, even in principle, until one has a way of choosing a basis of that two-dimensional space. I don't know if a satisfactory solution to this problem has been found. -- Chris Henrich http://www.mathinteract.com God just doesn't fit inside a single religion. === Subject: any tool in Matlab that can help me explore the relation between position of poles and the response? Hi all, I am looking for such a tool in Matlab that can use Graphic User Interface to help me explore and understand the relation between position of the poles and the response itself? For example, for function f(t), the Fourier Transform F(w) can be deemed as complex function F(z) when we change omega into variable z and I want to examine how does the distance from the poles to real axis, the distance from the poles to imaginary axis, and the distance from the poles to the origin affect the behavior of f(t)? Any general rules about these poles? === Subject: Re: any tool in Matlab that can help me explore the relation between position of poles and the response? 6y59pkR8)L]j8{eQmixxBq{y[uG13Ekt3@`{$F6?Hj%@)Y{V[~BT;SJ}c-,bth[~` ']|Y^+JDq5'>mb.3vRMu91|)ffD;27>$9k]`0H > Hi all, > I am looking for such a tool in Matlab that can use Graphic User > Interface to help me explore and understand the relation between > position of the poles and the response itself? > For example, for function f(t), the Fourier Transform F(w) can be > deemed as complex function F(z) when we change omega into variable > z and I want to examine how does the distance from the poles to real > axis, the distance from the poles to imaginary axis, and the distance > from the poles to the origin affect the behavior of f(t)? > Any general rules about these poles? Take a look at the documentation for the Signal Processing Toolbox. There are a number of GUI-based tools there. KP === Subject: Re: any tool in Matlab that can help me explore the relation between position of poles and the response? > Hi all, > I am looking for such a tool in Matlab that can use Graphic User > Interface to help me explore and understand the relation between > position of the poles and the response itself? > For example, for function f(t), the Fourier Transform F(w) can be > deemed as complex function F(z) when we change omega into variable > z and I want to examine how does the distance from the poles to real > axis, the distance from the poles to imaginary axis, and the distance > from the poles to the origin affect the behavior of f(t)? Somewhere on the WEB, there's a Java applet that does (at least most of) Jerry -- Engineering is the art of making what you want from things you can get. .88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N 210N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N21 0N.88N.88N.88N.88N === Subject: Re: any tool in Matlab that can help me explore the relation between position of poles and the response? I am looking for such a tool in Matlab that can use Graphic User > Interface to help me explore and understand the relation between > position of the poles and the response itself? > For example, for function f(t), the Fourier Transform F(w) can be > deemed as complex function F(z) when we change omega into variable > z and I want to examine how does the distance from the poles to real > axis, the distance from the poles to imaginary axis, and the distance > from the poles to the origin affect the behavior of f(t)? > Somewhere on the WEB, there's a Java applet that does (at least most of) > Jerry > -- > Engineering is the art of making what you want from things you can get. > ¿¿¿¿¿ ¿¿¿[OSl ash]¿¿¿¿[DownQuesti on]¿¿¿ ¿¿¿¿¿ ¿¿¿[OSl ash]¿¿¿¿[DownQuesti on]¿¿¿ ¿¿¿¿¿ ¿¿¿[OSl ash]¿¿¿¿[DownQuesti on]¿¿¿ ¿¿¿¿¿ ¿¿¿[OSl ash]¿¿¿¿[DownQuesti on]¿¿¿ ¿¿¿ matlab: spttool http://www.mathworks.com/company/newsletters/digest/september99/spttool/ lot's of pole-zero plotting programs such as the one at Mississppi state university 'Pole/zero analysis of linear systems': http://www.cavs.msstate.edu/hse/ies/publications/courses/ece 4773/ === Subject: Re: any tool in Matlab that can help me explore the relation between position of poles and the response? matlab: spttool http://www.mathworks.com/company/newsletters/digest/september99/spttool/ lot's of pole-zero plotting programs such as the one at Mississppi state university 'Pole/zero analysis of linear systems': http://www.cavs.msstate.edu/hse/ies/publications/courses/ece_4773/ Hi Lucy - http://users.ece.gatech.edu/~bonnie/education/LSLNR/ for general rules you could try googling root locus or maybe bode plot or bode diagrams Best of Luck - Mike === Subject: Re: any tool in Matlab that can help me explore the relation between position of poles and the response? > Somewhere on the WEB, there's a Java applet that does (at least most of) > Jerry > -- Maybe one of these: http://www.nst.ing.tu-bs.de/schaukasten/polezero/en_idx.html http://www.jhu.edu/~signals/explore/index.html http://www.thole.org/manfred/polezero/en_idx.html This are the first 3 results with Google search: poles functions response java applet === Subject: Tricky Math Problem - Phones Here is a math trick so unbelievable that it will stump you. Personally I would like to know who came up with this and where they had the time to figure this out. 1. Grab a calculator. (you won't be able to do this one in your head) 2. Key in the first three digits of your phone number (NOT the area code) 3. Multiply by 80 4. Add 1 5. Multiply by 250 6. Add the last 4 digits of your phone number 7. Add the last 4 digits of your phone number again. 8. Subtract 250 9. Divide number by 2 Do you recognize the answer? The pressure is outrageous. Everyone is picked apart and it's so superficial and not real. I'm not superskinny and not overweight. I'm just normal. -- Hilary Duff === Subject: Re: Tricky Math Problem - Phones In message <6rlpj1lmupuoik7h5j5vmkp6qb8ur6m5p5@4ax.com>, Ablang >Here is a math trick so unbelievable that it will stump you. >Personally I would like to know who came up with this and where they >had the time to figure this out. > 1. Grab a calculator. (you won't be able to do this one in your >head) > 2. Key in the first three digits of your phone number (NOT the >area code) > 3. Multiply by 80 > 4. Add 1 > 5. Multiply by 250 > 6. Add the last 4 digits of your phone number > 7. Add the last 4 digits of your phone number again. > 8. Subtract 250 > 9. Divide number by 2 > Do you recognize the answer? Why is it unbelievable? 81*250 = 20,250 etc -- Jeremy Boden === Subject: Re: Tricky Math Problem - Phones This trick has no ingenuity at all. It is just the observation that abcdefg = abc * 10000 + defg obfuscated by lots of noise. The first thing you should note is that steps 4 and 8 cancel eachother out-- those 2 steps are purely noise and have no purpose but to mystify people who don't realize what's going on. Mentally erasing those steps, it becomes clear that steps 3 and 5 really just say multiply by 20000. Steps 6 and 7 together just say add 2 times the last 4 digits. So what do we have at this point? 20000 * abc + 2 * defg. Dividing by 2 gives 10000 * abc + defg. In fact steps 3/5 and steps 7/9 cancel eachother out except that steps 3/5 coalesce into a single multiply by 10000, and the whole thing becomes 1. Grab a calculator 2. Enter first 3 digits 3. Multiply by 10000 4. Add last 4 digits If you want to impress whoever told you this trick, you could convolute it even further by adding more noise. For example, you could give it an occult aspect by changing step 4 to Add 666 and step 8 to Subtract 166600 and then add 100. > Here is a math trick so unbelievable that it will stump you. > Personally I would like to know who came up with this and where they > had the time to figure this out. > 1. Grab a calculator. (you won't be able to do this one in your > head) > 2. Key in the first three digits of your phone number (NOT the > area code) > 3. Multiply by 80 > 4. Add 1 > 5. Multiply by 250 > 6. Add the last 4 digits of your phone number > 7. Add the last 4 digits of your phone number again. > 8. Subtract 250 > 9. Divide number by 2 > Do you recognize the answer? > The pressure is outrageous. Everyone is picked apart and it's so superficial and not real. I'm not superskinny and not overweight. I'm just normal. > -- Hilary Duff === Subject: Re: Tricky Math Problem - Phones Trivial stuff really. Lot's of camouflage to hide something really simple in the basic operations. === Subject: Re: Tricky Math Problem - Phones > Here is a math trick so unbelievable that it will stump you. I doubt it. > Personally I would like to know who came up with this and where they > had the time to figure this out. > 1. Grab a calculator. (you won't be able to do this one in your > head) > 2. Key in the first three digits of your phone number (NOT the > area code) > 3. Multiply by 80 > 4. Add 1 > 5. Multiply by 250 > 6. Add the last 4 digits of your phone number > 7. Add the last 4 digits of your phone number again. > 8. Subtract 250 > 9. Divide number by 2 > Do you recognize the answer? If abc-defg is your phone number, (Where are you at, where you don't use area codes!?!?!) then here's what you get, step by step. 2. 100a + 10b + c 3. 8000a + 800b + 80c 4. 8000a + 800b + 80c + 1 5. 2,000,000a + 200,000b + 20,000c + 250 6. 2,000,000a + 200,000b + 20,000c + 1000d + 100e + 10f + g + 250 7. 2,000,000a + 200,000b + 20,000c + 2000d + 200e + 20f + 2g + 250 8. 2,000,000a + 200,000b + 20,000c + 2000d + 200e + 20f + 2g 9. 1,000,000a + 100,000b + 10,000c + 1000d + 100e + 10f + g, which is abcdefg in decimal notation. Only algebra. 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Fourier series is for boundary value problems,for example ,on a finite interval (like [-pi,pi].Laplace transforms and the closely related Fourier transform requires the functions to be defined on all of the real line.These are used to transform the differential equations to algebraic equations like Z(s)L(f)(s)=L(g)(s) where L is the Laplace transform and Z(s) is at least to Electrical Engineers as impedence transfer functions useful for desighning electrical networks.If one ptus iw for s (i complex number whose square is -1 ) one gets the Fourier transform which is more convenient sometimes.Hope this === Subject: Discontinuous Derivative What is an example of a function f:R -> R with a derivative f' defined on all of R, that is somewhere or other discontinous? Inquiring minds want to know. === Subject: Re: Discontinuous Derivative > What is an example of a function f:R -> R > with a derivative f' defined on all of R, > that is somewhere or other discontinous? > Inquiring minds want to know. A dramatic example is this: f(0)=0, f(x) = x^2 * sin(1/x^2) for all non-zero x Then f'(0) = 0, and f' is unbounded in every neighborhood of 0, so badly so that f' is not absolutely integrable over any neighborhood of 0. (It is used as an example showing that Kurzweil-Henstock integral (also called gauge integral) is a proper extension of Lebesgue integral. As for bigger sets of discontinuities: If f' exists on an interval then f' is of first Baire class (pointwise limit of continuous functions), and the set of points of continuity of f' is a set of second category: complement of a countable union of nowhere dense sets. (a big, certainly uncountable set).It can still have Lebesgue measure zero, but I do not remember how to give an example (of f). === Subject: Re: Discontinuous Derivative Does there exist a differentiable real-valued function on the unit interval whose derivative is continuous nowhere? If not, one whose derivative is discontinuous almost everywhere? -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: Discontinuous Derivative function on the unit interval whose derivative > is continuous nowhere? If not, one whose > derivative is discontinuous almost everywhere? No, any derivative is necessarily continuous at a dense in R set -- indeed, even a c-dense in R set -- of points. Even more, the points of discontinuity form a first category (i.e. meager) set of points, although it is possible for this set to have full measure in every interval. In fact, most derivatives in the sense of Baire category (sup norm) have the property that their points of continuity form a set of measure zero (or even a set of Hausdorff h-measure zero for any pre-assigned Hausdorff measure function h). HISTORICAL ESSAY ON CONTINUITY OF DERIVATIVES Dave L. Renfro === Subject: Re: Discontinuous Derivative > What is an example of a function f:R -> R > with a derivative f' defined on all of R, > that is somewhere or other discontinous? The integral of |t| from 0 to x; i.e., the function f(x) = {x^2, if x >= 0; -x^2, if x < 0}. === Subject: Re: Discontinuous Derivative > What is an example of a function f:R -> R > with a derivative f' defined on all of R, > that is somewhere or other discontinous? > The integral of |t| from 0 to x; i.e., the function > f(x) = {x^2, if x >= 0; -x^2, if x < 0}. f'(x) = 2x, if x > 0 = -2x, if x < 0. f'(0) = 0, ie f'(x) = |2x| How is f' discontinuous? === Subject: Re: Discontinuous Derivative >>What is an example of a function f:R -> R >>with a derivative f' defined on all of R, >>that is somewhere or other discontinous? > The integral of |t| from 0 to x; i.e., the function > f(x) = {x^2, if x >= 0; -x^2, if x < 0}. I think you lost either a 2 or a 1/2 somewhere. But in any case, surely that has continuous derivative everywhere. f'(x)= 2x if x>0, f'(x) = -2x if x<0, and f'(0)=0 by direction computation, so f'(x) is just 2|x|, which is continuous. === Subject: Re: Discontinuous Derivative >>What is an example of a function f:R -> R >>with a derivative f' defined on all of R, >>that is somewhere or other discontinous? > The integral of |t| from 0 to x; i.e., the function > f(x) = {x^2, if x >= 0; -x^2, if x < 0}. > I think you lost either a 2 or a 1/2 somewhere. > But in any case, surely that has continuous > derivative > everywhere. > f'(x)= 2x if x>0, f'(x) = -2x if x<0, > and f'(0)=0 by direction computation, so > f'(x) is just 2|x|, which is continuous. Not only that but in general the integral of a continuous function will have a continuous derivative. === Subject: Re: Discontinuous Derivative >What is an example of a function f:R -> R >with a derivative f' defined on all of R, >that is somewhere or other discontinous? f(x) = x^2 cos(1/x) -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: Discontinuous Derivative >> What is an example of a function f:R -> R >> with a derivative f' defined on all of R, >> that is somewhere or other discontinous? > f(x) = x^2 cos(1/x) Cute. So f'(x) = 2x cos(1/x) + sin(1/x) when x is not zero. f'(0) we get straight from the definition: lim_{h -> 0} h^2cos(1/h) / h = 0. But as x -> 0, f'(x) has no limit at all, so f' is not continuous at 0. As I said above, cute. === Subject: Re: ring of Cauchy sequences > Affirmation:Suppose that we have 2 Cauchy sequences a(n) and b(n) with > n from 1 to infinity ,with elements in a field. we can form a ring of > all Cauchy sequences where addition and multiplication is defined > componentwise.The ideal which consists of all sequences that converge to > 0 is maximal. We define the metric such that its image is some totally > ordered group G. My question : Is the affirmation true in general or we What's needed is a metric that provides for the addition and multiplication of two Cauchy sequences to be a Cauchy sequence. Dealing with a norm |.| instead of a metric will be happier. What's needed is |a + b| <= |a| + |b| |ab| <= |a| |b| Thus the metric for F is d(x,y) = |x - y| This imparts to the metric, translation invariance, d(a + x, b + x) = d(a,b) and scalar sizing d(ax, bx) = d(x,0) * d(a,b) Thus you see besides addition, you want multiplication, ie to use an ordered ring instead of an ordered group. Upon closer examination of the proof that the sum of two Cauchy sequences is Cauchy sequence, one spies that division by 2 is most helpful. Hence I recommend a totally ordered field, a minimal of Z[1/2], for the domain of the norm and/or metric. === Subject: =?iso-8859-1?q?Re:__Wanted_information_on_Poincar=E9_equation_:F(a*z)=3Da*F (z)*(1-_F(z))_?= <290920050949382944%edgar@math.ohio-state.edu.invalid> I propose iterated solutions for the given equations: 1Á) f(ax) = b*f(x)*(1-f(x)) , f(x) =(b*L*(1-L) )^[ln(x)/ln(a) + k] closed forms computable for b = -2 , 2, 4 ... 2Á) h(z^2) = h(z)^2 + c , h(z) = (L^2 +c)^[k*ln(z)] , [iterands] are built from Abel counting functions, k stand for invariants terms , in 1Á) when x -> a*x 2Á) z -> z^2 In both cases we iterate functions of L , Alain. === Subject: Re: The Pigeonhole Principle-1 pigeon 0 holes? Russell, that's the most awesome answer I've got out of asking I looked up Vacuous truths on Wikipedia..and they're soooo interesting. === Subject: Algebraic extensions Let a be an algebraic number of odd degree with respect to Q. Prove that a and a^2 have the same degree with respect to Q. === Subject: Re: Algebraic extensions > Let a be an algebraic number of odd degree with respect to Q. Prove that > a and a^2 have the same degree with respect to Q. Consider Q(a) as an extension of Q(a^2). -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: Algebraic extensions Timothy Murphy skrev: >>Let a be an algebraic number of odd degree with respect to Q. Prove that >>a and a^2 have the same degree with respect to Q. > Consider Q(a) as an extension of Q(a^2). === Subject: Re: Algebraic extensions > Let a be an algebraic number of odd degree with respect to Q. Prove that > a and a^2 have the same degree with respect to Q. I don't suppose it is homework... === Subject: Re: proving a function is continuous On 29 Sep 2005 08:50:48 -0700, Artur I'd like some hints tro prove that, if I is an [open] interval of R and f:I -R is bounded on I and satisfies f((x+y)/2) <= ((f(x) + f(y))/2 for >every x and y in I, then f is continuous. It's easier to prove a stronger statement: If x is in I then (*) lim sup_{y -> x} |f(x) - f(y)|/|x-y| < infinity. Suppose that (*) fails. Then there are y_n -> x where such that |f(x) - f(y_n)|/|x-y_n| tends to infinity. Assume that in fact (f(x) - f(y_n))/(x-y_n) tends to infinity (you get this or a limit of -infinity by taking a subsequence). Note that f(x+2h) - f(x+h) >= f(x+h) - f(x) for h > 0 and you can now show that f is unbounded. >Artur ************************ David C. Ullrich === Subject: Re: proving a function is continuous > On 29 Sep 2005 08:50:48 -0700, Artur Hello >>I'd like some hints tro prove that, if I is an [open] interval of R and >>f:I ->R is bounded on I and satisfies f((x+y)/2) <= ((f(x) + f(y))/2 for >>every x and y in I, then f is continuous. > It's easier to prove a stronger statement: If x is in I then > (*) lim sup_{y -> x} |f(x) - f(y)|/|x-y| < infinity. What is the definition for lim sup_(x->a) f(x)? I understand the sequential version. > Suppose that (*) fails. Then there are y_n -> x where such that > |f(x) - f(y_n)|/|x-y_n| tends to infinity. Assume that > in fact (f(x) - f(y_n))/(x-y_n) tends to infinity (you get > this or a limit of -infinity by taking a subsequence). > Note that f(x+2h) - f(x+h) >= f(x+h) - f(x) for h > 0 and > you can now show that f is unbounded. >>Artur > ************************ > David C. Ullrich === Subject: Re: proving a function is continuous I'm not sure, I'd never seen this concept before, but from what David If f is defined on a set D, has values on R and a is an accumulation point of D, then the extended real number lim sup ( x ->a) f(x) = infimum {lim sup f(x_n)}, where the infimum is taken over the collection of all sequences x_n in D -{a} that converges to a. Artur === Subject: Re: proving a function is continuous > Yes, you're right. The interval I must be open. Otherwise, the > statement is actually false. > If we have f((x+y)/2) <= ((f(x) + f(y))/2 for every x and y in I and f > is continuous, then f is convex, even it it's not bounded. No, it need not be convex. As I said, you need additional conditions to ensure convexity. Boundedness will ensure convexity, as will monotonicity, and (if memory serves), so does (Lebesgue) measurability. The counterexamples to your assertion must, therefore, be pretty wild: non-measurable and unbounded. I have never actually seen proofs of these results, but have read them in sources I have long since forgotten. It is true without further restrictions that f(rx + (1-r)y) <= r f(x) + (1-r) f(y) for all _rational_ r in (0,1), but extension to irrational r needs additional assumptions on f. > If f > satisfies such functional equation on I and is Lebesgue measurable, OK, so here you are restricting f a bit; your original posting did not include measurability. > then f is contnuous and, therefore, convex. > So, what I have to prove is: If f satisfies the given functional > equation and is bounded on an open interval I, then f is continuous (if > the statement istrue, of course) Unfortunately, I do not recall the sources of my assertions, and cannot look them up anymore, since I got rid of a numerous books and papers when I retired and moved across the continent. However, if you have access to a university library, a few days searching under the topic of convexity ought to work. RGV > Artur === Subject: Re: proving a function is continuous >> Yes, you're right. The interval I must be open. Otherwise, the >> statement is actually false. >> If we have f((x+y)/2) <= ((f(x) + f(y))/2 for every x and y in I and f >> is continuous, then f is convex, even it it's not bounded. >No, it need not be convex. As I said, you need additional conditions to >ensure convexity. Boundedness will ensure convexity, as will >monotonicity, and (if memory serves), so does (Lebesgue) measurability. >The counterexamples to your assertion must, therefore, be pretty wild: >non-measurable and unbounded. I have never actually seen proofs of >these results, but have read them in sources I have long since >forgotten. >It is true without further restrictions that f(rx + (1-r)y) <= r f(x) + >(1-r) f(y) for all _rational_ r in (0,1), You're sure you mean rational here, as opposed to dyadic rational? >but extension to irrational r >needs additional assumptions on f. >> If f >> satisfies such functional equation on I and is Lebesgue measurable, >OK, so here you are restricting f a bit; your original posting did not >include measurability. >> then f is contnuous and, therefore, convex. >> So, what I have to prove is: If f satisfies the given functional >> equation and is bounded on an open interval I, then f is continuous (if >> the statement istrue, of course) >Unfortunately, I do not recall the sources of my assertions, and cannot >look them up anymore, since I got rid of a numerous books and papers >when I retired and moved across the continent. However, if you have >access to a university library, a few days searching under the topic of >convexity ought to work. >RGV >> Artur ************************ David C. Ullrich === Subject: Re: =?UTF-8?Q?2n_=3D_P_+_Q_,_P_=E2=89=A0_Q?= In message <21592190.1127879743526.JavaMail.jakarta@nitrogen.mathforum.org>, Chou >There's a little problem confused me !! >If we assume GoldBach Conjecture will be correct , so any even number >2n can represent as P+Q ( P,Q is prime). >Is any even number except 6 be writted as 2n = P+Q , P!=Q for some P >, Q ?? ex: 10=5+5=3+7 >14=7+7=9+5 >22=11+11=5+17 >Would you find any counter-example except 6 for the statement above ?? >P.S Would you introduce me to relative paper ?? How about 2n = 4? -- Jeremy Boden === Subject: Optimization problem Hi I have prepared a program for electric relays coordination for n numbers of relays I have the following for n=3 x1=Max( 0.1 ,C12*x2+b12,C13*x2+b13) x2=Max(C21*x1+b21, 0.1 ,C23*x2+b23) x3=Max(C31*x1+b31,C32*x2+b32, 0.1 ) Range of x1,x2,x3 is 0.1 to 1.1 x_i has only two digits c_ij and b_ij are positive Actually n = 300, each raw contains 6 element of c_ij and b_ij as maximum I used iterations method, starting all x_i = 0.1 I have always convergent results Mathematically, is iterations methods applicable safely for such optimization problems === Subject: lim f(x)/x Please so me if there are any mistakes in the proof that follows and also how to extend it by not assuming f nondecreasing Given : 1. f (x) differentiable and nondecreasing over R 2. lim f ' (x) = 0 as x->infinity Show : lim ( f (x) / x ) -> 0 as x->infinity Proof : 2 => ( f(x) - f(xo) ) / (x - xo) < e if x > xo >N and x-xo < delta after multiplying both sides by (x-xo) we get (f(x) - f(xo) )< e * (x - xo) after adding f(xo) to both sides we get f(x) < e*(x-xo) + f (xo) = e*x - e*xo + f(xo) after dividing both sides by x we get 0 < f(x)/x < e - e*xo/x + f(xo)/x e can be made as small as we please and xo and f(xo) are finite real numbers therefore as x->infinity the RHS converges to 0 so f(x)/x converges to 0 === Subject: Re: lim f(x)/x > Please so me if there are any mistakes in the proof that follows and also > how to extend it by not assuming f nondecreasing > Given : > 1. f (x) differentiable and nondecreasing over R > 2. lim f ' (x) = 0 as x->infinity > Show : > lim ( f (x) / x ) -> 0 as x->infinity > Proof : > 2 => ( f(x) - f(xo) ) / (x - xo) < e if x > xo >N and x-xo < delta I have a few qustions here: 1) What is _x_? 2) What is _xo_? 3) What is _e_? 4) What is _N_? 5) What is _delta_? > after multiplying both sides by (x-xo) we get > (f(x) - f(xo) )< e * (x - xo) > after adding f(xo) to both sides we get > f(x) < e*(x-xo) + f (xo) = e*x - e*xo + f(xo) > after dividing both sides by x we get > 0 < f(x)/x < e - e*xo/x + f(xo)/x > e can be made as small as we please and xo and f(xo) are finite real numbers > therefore > as x->infinity the RHS converges to 0 so f(x)/x converges to 0 Where did you use the fact that f is nondecreasing? Jose Carlos Santos === Subject: Re: lim f(x)/x >> Please so me if there are any mistakes in the proof that follows and >> also >> how to extend it by not assuming f nondecreasing >> Given : >> 1. f (x) differentiable and nondecreasing over R >> 2. lim f ' (x) = 0 as x->infinity >> Show : >> lim ( f (x) / x ) -> 0 as x->infinity >> Proof : >> 2 => ( f(x) - f(xo) ) / (x - xo) < e if x > xo >N and x-xo < delta > I have a few qustions here: > 1) What is _x_? > 2) What is _xo_? > 3) What is _e_? > 4) What is _N_? > 5) What is _delta_? >> after multiplying both sides by (x-xo) we get >> (f(x) - f(xo) )< e * (x - xo) >> after adding f(xo) to both sides we get >> f(x) < e*(x-xo) + f (xo) = e*x - e*xo + f(xo) >> after dividing both sides by x we get >> 0 < f(x)/x < e - e*xo/x + f(xo)/x >> e can be made as small as we please and xo and f(xo) are finite real >> numbers >> therefore >> as x->infinity the RHS converges to 0 so f(x)/x converges to 0 > Where did you use the fact that f is nondecreasing? > Jose Carlos Santos N is an integer... x , xo are real numbers...e (as in epsilon) and delta are real numbers arbitrarily small. I'm sure you are familiar with the epsilon ,delta usage in the definition of continuity in most standard texts of Calculus... As for the nondecreasing part I think that maybe is not necessary but if I don't use it I probably have to use absolute values in the inequality... instead of 0<=f(x)-f(xo) <= e*(x-xo) use 0<=| f(x) - f(xo) | <= e* | x-xo | But the question remains is my proof a proper proof or just nonsense? === Subject: Re: lim f(x)/x > N is an integer... > x , xo are real numbers...e (as in epsilon) and delta are real numbers > arbitrarily small. > I'm sure you are familiar with the epsilon ,delta usage in the definition of > continuity in most standard texts of Calculus... I sure am. But you cannot just say that x , xo are real numbers and so *every* real number _x_? Or are you saying that it's true for *some* real number _x_. The same questions apply to _xo_ (BTW x0 or x_0 are better choices) and to all the other numbers that you've mentioned. > As for the nondecreasing part I think that maybe is not necessary but if I > don't use it I probably have to use > absolute values in the inequality... > instead of 0<=f(x)-f(xo) <= e*(x-xo) use 0<=| f(x) - f(xo) | <= e* | x-xo | > But the question remains is my proof a proper proof or just nonsense? It will be hard for me to answer that until you tell me more about all those numbers. And another thing: why did you write that 0 < f(x)/x? Jose Carlos Santos === Subject: adding values in hour:minute:second format with TI-85 Does the TI-85 calculator have a feature to add a long list of values in hour:minute:second format? Or degree-minute-second format? I can do it the long way, but was curious. === Subject: Re: adding values in hour:minute:second format with TI-85 > Does the TI-85 calculator have a feature to add a long list of values > in hour:minute:second format? Or degree-minute-second format? I can > do it the long way, but was curious. Keith Lewis is undoubtably correct that this question is best answered by consulting the manual. The FM can be downloaded from http://education.ti.com/us/product/tech/85/features/features.html Hope that helps -John Coleman === Subject: Re: adding values in hour:minute:second format with TI-85 On 30 Sep 2005 11:52:44 -0700, John Coleman >> Does the TI-85 calculator have a feature to add a long list of values >> in hour:minute:second format? Or degree-minute-second format? I can >> do it the long way, but was curious. >Keith Lewis is undoubtably correct that this question is best answered >by consulting the manual. The FM can be downloaded from >http://education.ti.com/us/product/tech/85/features/features.html >Hope that helps >-John Coleman Well, I've got a TI-89. I decided to look this up in the manual. I looked up time, difftime, deltatime, date, diffdate, and found nothing. I looked in both the quickfind section and alphabetical listing of functions. The guys who enjoy saying RTFM might suggest what it's under if they are the experts. So, what's the keyword for the manual to look this up? === Subject: Re: adding values in hour:minute:second format with TI-85 >Does the TI-85 calculator have a feature to add a long list of values >in hour:minute:second format? Or degree-minute-second format? I can >do it the long way, but was curious. You should RTFM for questions like this. If you don't have the FM, look for a button labelled (DMS) or (o'). Put in the degrees or hours, press once, put in minutes, press again, and put in seconds. If that doesn't work try it backwards. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: x^3-x=6*y^3 Has anyone reference to this Diophantine equation? or a proof that it has only the trivial solutions (0,0),(1,0),(-1,0)? === Subject: Re: x^3-x=6*y^3 > Has anyone reference to this Diophantine equation? or a proof that it has only the trivial solutions (0,0),(1,0),(-1,0)? x^3 - x = 6*y^3 is bi-rationally equiv to the elliptic curve Y^2 = X^3 + 36 via X = -6*y/x, Y = -6/x (i.e. it is genus 1 not 2). The curve Y^2 = X^3 + 36 is rank 1 with generator (X,Y) = (-3,3) which corresponds to (x,y) = (-2,-1) of your curve, and with torsion corresponding to (x,y)=(+-1,0) and (x,y) = (0,0) of your curve. This allows you to find all rational points and presumably analyse which are integral, but I didn't look further. -- Jim Buddenhagen === Subject: Re: x^3-x=6*y^3 > Has anyone reference to this Diophantine equation? > or a proof that it has only the trivial solutions > (0,0),(1,0),(-1,0)? The guy who said this has genus 2 is wrong, because for any given integer N in place of 6, a rational solution can be derived from a rational solution to any one of a finite number of equations of the form c.x^3 + d.y^3 = 2^f.b where N = b.c.d. In particular taking c = d = 1, the latter equation often has an infinite number of non-trivial rational solutions, which can be obtained by the Bachet form of the Mordell-Weill chord-tangent process, applied to x^3 + y^3 = N', i.e. using pairs of solutions (or one solution doubled) to leap-frog recursively to another. For the stated equation, with N = 6, any rational solutions can be derived from integer solutions to one of the following equal the cube of an integer: X^3 + 6.Y^3 X^3 + 12.Y^3 2.X^3 + 3.Y^3 4.X^3 + 3.Y^3 The first two have no non-trivial solutions, by the congruence conditions of Example 3.1 (on page 3) in http://emis.library.cornell.edu/journals/JIS/VOL6/Broughan/broughan25.pdf and I strongly suspect that neither of the other two does either. It's easy to prove this, by starting with the homogenized version X.(X^2 - Y^2) = 6.Z^3 for integers X, Y, Z (with reassigned X, Y we can assume coprime), which implies the following for integers a, b p, q, r, s with a.b = 6 and p, q squarefree: X, X^2 - Y^2 = b.p.q^2.r^3, a.p^2.q.s^3 Plugging the first in the second we conclude successively: p | Y so that p = 1 (as also p | X) q | s and hence q | Y so that q = 1 Thus b^2.r^6 - Y^2 = a.s^3, and as the factors b.r^3 +/- Y have GCD 2^e where e = 0 or 1, this implies there are integers c, d, u, v with c.d = a (and hence b.c.d = 6) so that: b.r^3 + Y, b.r^3 - Y = 2^e.c.u^3, 2^e.d.v^3 Finally, adding these gives the following, from where the cases listed above were derived: 2^(1-e).b.r^3 = c.u^3 + d.v^3 (Note that if e = 0 then c.d must be odd.) John R Ramsden (jhnrmsdn@yahoo.com.uk) ^ remove m from com to reply === Subject: Re: x^3-x=6*y^3 >Has anyone reference to this Diophantine equation? or a proof that it has only >the trivial solutions (0,0),(1,0),(-1,0)? What about (2,1) and (3,2)? --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: x^3-x=6*y^3 (3,2) does not work. In fact the only consecutive solutions meaning x=y+1, are (2,1) and (1,0). We only need postive solutions due to the symmetry of the curve. === Subject: Re: x^3-x=6*y^3 >>Has anyone reference to this Diophantine equation? or a proof that it has only >>the trivial solutions (0,0),(1,0),(-1,0)? > What about (2,1) and (3,2)? > --Keith Lewis klewis {at} mitre.org > The above may not (yet) represent the opinions of my employer. Substituting x = 3 and y = 2, I get 3^3 - 3 = 6 * (2^3) 24 = 48 Am I missing something? Matt === Subject: Re: x^3-x=6*y^3 That indeed is true but I have to look for it....looks non-trivial. === Subject: Re: x^3-x=6*y^3 (x = 2, y = 1) and (x = -2, y = -1) satisfy the equation too. Hence the solution set contains non-trivial solutions. === Subject: Re: x^3-x=6*y^3 Is there a general solution? Positives will suffy due to the symmetry. === Subject: Re: x^3-x=6*y^3 <6338129.1128086804739.JavaMail.jakarta@nitrogen.mathforum.org> Is there a general solution? What do you mean by general solution? It is a curve of genus 2, and hence has finitely many integer solutions by Siegel's theoem. It also has finitely many rational solutions by Falting's thm. === Subject: Re: x^3-x=6*y^3 <6338129.1128086804739.JavaMail.jakarta@nitrogen.mathforum.org It is a curve of genus 2, and hence has finitely many > integer solutions by Siegel's theoem. It also has > finitely many rational solutions by Falting's thm. Genus 1, because x^3 - x = N.y^3 is equivalent to x^2 - 1 = N.x^2.(y/x)^3, i.e. - N.(y/x)^3 + 1 = (1/x)^2 === Subject: Re: x^3-x=6*y^3 Now, do we know how many? or which nonnegative solutions are there besides (1,0) and (2,1)? BTW, by general solution, I meant a formula or description of all the finitely many solutions. === Subject: Was 'i shifted from an imaginary to an invisible number in 1797 by Caspar Wessel? Were unseen dimensions demonstrated in 1797 by Caspar Wessel? This question arises from 3 sources: 1 - Paul J Nahin An Imaginary Tale: The Story of {square-root} -1, Princeton, NJ, Princeton University Press, 1998. 2 - Previous physics papers regarding unseen dimensions: the sub-millimeter dimensions of Arkani-Hamed, Dimopoulos and Dvali and the warped geometry in extra dimensions of Randall and Sundrum. 3- The Earth's Magnetosphere - detectable but unseen torus with symmetry distorted by the solar wind. Wessel appears to have shifted i from an imaginary to an invisible (unseen) number. If square root of +1 is union of -1, +1 [-1 or +1] then square root of -1 is intersection of -1, +1 [-1 and +1]. Nahin credits Wessel, an applied mathematician working as a surveyor, as the first to demonstrate the existence of i (square root of -1) in polar form: i = 1 angle-operator 90 degrees (operator from electrical engineering). That is multiplying by i is geometrically, simply a rotation by 90 degrees in the counterclockwise (CCW) sense yielding the directed line segment of length one pointing straight up the vertical axis. The harmonic oscillators of the Steinmetz phasor equation (used by Feynman to develop his 'sum over paths' method) and the Schrodinger wave equation may by this description include an unseen dimension. Could the Magnetosphere be a detectable entity with unseen dimensions? If so, would this be related to the work of Arkani-Hamed or Randall or something else? I have no access to the work of geophysicists such as Gary Glatzmaier in this field, but find it difficult to comprehend that i would not be used. Is it time to change the name of i to invisible rather than imaginary that more accurately describes the nature of this extant number? === Subject: Re: Was 'i shifted from an imaginary to an invisible number in 1797 by Caspar Wessel? i=sqrt(-1) is a purely mathematical concept. Physicists (as well as scientists and engineers in many fields) use it. You don't redefine it in terms of physics things. === Subject: Does anyone have copies of Paul's Online Math Notes?? They're an excellent summary of a many math concepts. Normally they reside on a server at Lamar University at http://tutorial.math.lamar.edu. Due to damage caused by hurricane Rita, the university has no power and has closed down until the damage is repaired which may take a few months. Does anyone know if there is a mirror of his website or know of copies located somewhere? === Subject: The Ultimate Online Pharmaceutical boundary=----------56574AEC40556AB by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id j8UEMvL02411 for ; Fri, 30 Sep 2005 10:22:57 -0400 by support2.mathforum.org (8.12.10/8.12.10/The Math Forum, $Revision: 1.6 secondary) with SMTP id j8UEK6FU029721 for ; Fri, 30 Sep 2005 10:22:47 -0400 by cm218-255-36-195.hkcable.com.hk (Qmailv1) with ESMTP id 05CDCE57C3 for ; Fri, 30 Sep 2005 10:23:37 -0400 --------------------------------------------------------------------- Vlaygra $3.3 Leviltra $3.3 Ciawlis $3.7 Imittrex $16.4 Flomiax $2.2 Ultrwam $0.78 Viopxx $4.75 Amblien $2.2 Valhium $0.97 Xanahx $1.09 Somma $3 Meriddia $2.2 our site http://comexpand.com/?GFSMKDQV1AQBhCUVoZWFBGW3RYUEZbUlpDR14aWkNV ___ Online Pharmaceuticals === Subject: The Ultimate Online Pharmaceutical boundary=----------30EED1F86120925 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id j8UEXML03522 for ; Fri, 30 Sep 2005 10:33:22 -0400 by support2.mathforum.org (8.12.10/8.12.10/The Math Forum, $Revision: 1.6 secondary) with SMTP id j8UEWxF5029882 for ; Fri, 30 Sep 2005 10:33:12 -0400 by pc-65-11-86-200.cm.vtr.net (Qmailv1) with ESMTP id 25EB9FEC13 for ; Fri, 30 Sep 2005 22:38:43 -0700 --------------------------------------------------------------------- Get Cipzro, Avanxdia, Profzac and more ONLINE! 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Visit our website: http://wcominer.com === Subject: Re: Testable Predictions by HdB > We trust the theory when the theorems are appropriately proven. [ ... ] The following quote is from an abstract, written by Eckard Blumschein: > Any actual physical state is the sum of only past influences. Future data > do not matter yet. And here comes (part of) my (modified) response. This is not only an issue in your domain of interest. Numerical Methods, for example, have been suffering from a violation of this principle as well. The remedy, meanwhile, is well-known. It's called upwinding or upwind differencing. Here is a reference: http://farside.ph.utexas.edu/teaching/329/lectures/node107.html I have been working with the idea for many years, But it's difficult to find a lucid explanation on the web. The best way to become accustomed with the idea is to is to read the chapter Upwind Scheme in the book: S.V. Patankar; Numerical Heat Transfer and Fluid Flow; Hemisphere Publishing Company U.S.A. 1980. To be found, for sale, at: http://www.cfd-online.com/Books/show_book.php?book_id=8 I have copied some text and a picture from this book and placed it on the web, without permission: The notion of upwind is quite important in CFD (: Computational Fluid Dynamics). Thus it seems that numerical analysis provides more adequate picture of the arrow of time than the corresponding exact model. This is remarkable. And again: where have your theorems been at that time? Only to establish that Upwind is O(h)? While the _not-working_ numerical schemes were / are(!) advertized as the better O(h^2)? Han de Bruijn === Subject: Re: Testable Predictions by HdB > I have not done very much in differential equations (I was a linear > algebra man and special functions). But if you post that question > in sci.math.num-analysis you might get an asnwer. Off-hand I can > only say that, yes, there are restrictions I think. There _are_ restrictions. Your intuition is quite good! I have posted this in sci.math.num-analysis as ODE and tri-diagonal system thread. Breathlessly awaiting, as usual ... Han de Bruijn === Subject: Re: Testable Predictions by HdB > Apparently nobody had thought that exceptional cases can occur. [ ... ] > Computer graphics uses a form of numerical mathematics. That's why exceptional cases are abundant in computer graphics as well. A nice example is the contouring of a function: f(x,y) - c = 0 , where ALL of the special cases must be distinguished carefully. If it is done at linear triangles, there are 27 of them. They can be summarized as: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 <== <=< <=> <<= <<< <<> <>= <>< <>> 19 20 21 22 23 24 25 26 27 >== >=< >=> ><= ><< ><> >>= >>< > Why does the contouring routine not work at that triangle in the corner of the flow domain? Well, I never thought that could happen (: case 1). 0 | | | : boundary conditions are often zero | |____ 0 0 Han de Bruijn === Subject: Re: Testable Predictions by HdB > We trust the theory when the theorems are appropriately proven. Routines > are indeed tested, but when the result is not conform the theory the > reason is always that the theory is not correctly implemented. Or, what > also does occur, the routine is used in a situation that is not covered > by the theory. The following story demonstrates that the above is .. wishful thinking. Incompressible and irrotational (thus ideal) flow of an inviscid fluid is described by the following system of linear first (not second) order Partial Differential Equations (PDE's): du/dx + dv/dy = 0 dv/dx - du/dy = 0 Here: (x,y) = coordinates , (u,v) = velocity-components. There does _not_ exist a kind of natural variational principle for the above differential equations. Conventional Finite Element Methods, however, are very much dependent upon the existence of such principles. There _must_ be something to minimize (or to make stationary). In cases like the above, it seems, at first sight, that a Least Squares Finite Element approach offers a possible solution. That is because an L.S.FEMethod proceeds by constructing an alternative minimum principle: square the equations just as-they-are (!), add these squares together, integrate their sum over the area of interest and minimize the result, as a function of the unknowns. This is the approach as described in O.C. Zienkiewicz The Finite Element Method (1977) chapter 3.14.2. In Februari 1976, Dr. Th.E. Labrujere, the National Aerospace Laboratory in English: The Least Squares Finite Element Method [L.S.FEM] applied to 2-D Incompressible Flow around a Circular Cylinder. In this report, it was established that a straightforward application of the Least Squares Method, using linear triangular Finite Elements, quite unexpectedly, DOES NOT WORK !! And how did NLR discover this? Exactly! By executing a computer program! As far as I know, Labrujere has been the first and only person who dared to report this, while everywhere else in the world researchers continued claiming to be successful with this method. Didn't they have theorems to prevent them from doing so? In December 1976, Labrujere's problem was finally solved by de Vries, Labrujere himself and Norrie, at the mechanical Engineering Department of The University of Calgary, Alberta, Canada. The result is written down in their Report no.86: A Least Squares Finite Element Solution for Potential Flow. What do you think of this story? Clearly, it has been a computer program which made the flaws in the theory visible here. This is confirmation of my statement that what you have stated above is wishful thinking indeed. Han de Bruijn === Subject: Re: Testable Predictions by HdB > > [ .... ] An example is the theorem that (sqrt(5)+1)/2 is an invariant > > point, which can _never_ be tested by the computer program at hand. > Indeed, that is one of them. But also the theorem that (1-sqrt(5))/2 > is an invariant can not be tested by that program. Also the theorem > that those are the only two is not testable. And there are more. > A program that iterates x := 1/(x-1) and outputs x that always seems to > converge to a number that is close to (1-sqrt(5))/2 comes as a surprise Thus executing the program for several inputs _suggests_ that iterations always converge to that value. That is the least we can say. Right? Then the theory comes in and _proves_ that (1-sqrt(5))/2 is an invariant. Are we playing with words or what? Han de Bruijn === Subject: Re: Testable Predictions by HdB ... > A program that iterates x := 1/(x-1) and outputs x that always seems to > converge to a number that is close to (1-sqrt(5))/2 comes as a surprise > Thus executing the program for several inputs _suggests_ that iterations > always converge to that value. That is the least we can say. Right? Then > the theory comes in and _proves_ that (1-sqrt(5))/2 is an invariant. Are > we playing with words or what? But now you are doing something different. You start with the testing, the theory comes afterwards. But that does not meant the theory comes with testable predictions. More like the theory explains the results of the tests. -- 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: Testable Predictions by HdB > But now you are doing something different. You start with the testing, > the theory comes afterwards. But that does not meant the theory comes > with testable predictions. More like the theory explains the results > of the tests. However, if we reverse the time axis, the theory _predicts_ the results of the tests. Feels like a chicken and egg problem ... :-) Han de Bruijn === Subject: Re: Testable Predictions by HdB > > which basically does nothing else than [ x := 1/(x-1) ] , then I find it > > quite a _surprise_, yes, that such iterations bear a relationship with > > the beautiful mathematics of Fibonacci Numbers and the Golden Ratio: > Any mathematician that has looked shortly on Fibonacci numbers would > know immediately that there is a close connection. So to me it did > not come as a susprise at all. But indeed, mathematical theorems can > give surprises, also for mathematicians. To me, it came as a complete surprise. This Aha Erlebnis is exemplified in my paper on the first page: See the pattern?. And it kept me going. But in order to establish that I am not the only one who experiences to be surprised by things which are trivial in the eyes of the experienced, I have a challenge for you which is similar to the Fibonacci Iterations. Of course, you are free to take it or leave it. Anyway, here comes. Consider the following iterates: x := (x/(1-2.x))^2 . And give us the closed formula(s) for (x) as a function of (k), where k denotes the k'th iterand. Where are the invariant points ? And are they stable or unstable? > But there is no program needed to get the surprise. A program that iterates x := 1/(x-1) and outputs x that always seems to converge to a number that is close to (1-sqrt(5))/2 comes as a surprise, at first sight, if you don't have any theory. Why would you try to deny? Han de Bruijn === Subject: Re: Testable Predictions by HdB ... > Consider the following iterates: x := (x/(1-2.x))^2 . > And give us the closed formula(s) for (x) as a function of (k), where k > denotes the k'th iterand. Where are the invariant points ? And are they > stable or unstable? Off-hand I would say that the fixed points are 0, 1/4 and 1. Of these only 0 is stable. I do not know whether there is a closed formula. -- 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: Testable Predictions by HdB > ... > > Consider the following iterates: x := (x/(1-2.x))^2 . > > > > And give us the closed formula(s) for (x) as a function of (k), where k > > denotes the k'th iterand. Where are the invariant points ? And are they > > stable or unstable? > Off-hand I would say that the fixed points are 0, 1/4 and 1. Of these > only 0 is stable. These are fixed points, indeed. But there are (infinitely) many more, in the domain 1/4 < x < oo . In that domain, they come with n-cycles of any magnitude (n). Outside this domain there is only one, 0 as you said, and that one is stable, indeed. Examples: (3/2+sqrt(5)/2) and (3/2-sqrt(5)/2) come with a 2-cycle. > I do not know whether there is a closed formula. The I do not know is relevant four our debate. What about theorems, indeed, that we do not know? And yet we have that computer program .. Han de Bruijn === Subject: Re: Testable Predictions by HdB > They do not realise that different floating point systems have > different properties, so what may be refuted by one system, will not > be refuted by another system. Moreover, to refute a theorem in a > floating point system you need to know how it works exactly, and > that gives you the ability to construct examples that contradict a > theorem in reals when applied to floating point systems. With random > searching you will almost never get a wrong result. > Would you please elaborate on random searching. I don't know what it > is supposed to mean in this context. Random searching means writing a program and using random inputs to check whether the theorem fails or not. That is just how you use your program iterate. ( About triangle inequality.) > Random search > will in general *not* show failure in a particular f-p system, and > constructed counter-examples in some f-p system are *not* in general > counter-examples in other f-p systems. (And, of course, compiler > optimisation can also play a role.) > Again, I don't even have a clue what random search means here. Random search is contrasted here with the input of *all* possible inputs. If you leave some out it is pretty likely that you leave out precisely those cases where it will fail, unless you base what you leave out on the f-p system characteristics and much more knowledge about how everything works. As an example, on Intel machines in general operations are performed in extended (80 bit) precision and are rounded to double precision when storing the result. This means that the result is not necessarily the same as when the operation was performed in double precision initially. It is however unlikely that you will find with a program cases where there is a difference, unless you have knowledge about where to search for such cases. ... > A quote: > So what we see is that initially convergence is very good (and proven), > when we come too close to the correct answer the rounding errors make it > difficult to get ultimate convergence. I think you will find this > behaviour in many (if not most) iterative algorithms. > So the conclusion is that we should not come too close to the correct > answer, by which you mean: the answer as provided by the theory. That's > quite an important observation, I think. And furthermore, it's according > to my (rather crude) experience. It is indeed true in general, but there are exceptions to the rule. Newton-Raphson for the square root is one example. That is an example where a next iterate is *never* much worse than a previous iterate. On the other hand, the problem shows up when you try to do numerical differentiation. (As I have stated, numerically differentiation is much more difficult than integration, on paper it is the other way around.) ... > OK. Apart from terminology. I am distinguishing invariant numbers and > stable numbers. In my paper, the invariant numbers are (sqrt(5)+1)/2 > and (sqrt(5)-1)/2. The first one is unstable while the second is stable. > You claim that there could be an infinity of those points. And who am I > to deny that you are right. But then we have the above quote from that > IEEE thread. Could we conclude then that our invariant points are _all_ > quite closely nested around the real solutions? Yes, that is something we can conclude. On binary machines the iteration of 1/(x-1) is quite stable when x is outside the range [1/2, 2]. But inside that range the relative error of the result can be quite large. -- 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: Testable Predictions by HdB > > Would you please elaborate on random searching. I don't know what it > > is supposed to mean in this context. > Random searching means writing a program and using random inputs to check > whether the theorem fails or not. That is just how you use your program > iterate. Right. I could also write a loop around it which generates random input. > > So the conclusion is that we should not come too close to the correct > > answer, by which you mean: the answer as provided by the theory. That's > > quite an important observation, I think. And furthermore, it's according > > to my (rather crude) experience. > It is indeed true in general, but there are exceptions to the rule. > Newton-Raphson for the square root is one example. That is an example > where a next iterate is *never* much worse than a previous iterate. On > the other hand, the problem shows up when you try to do numerical > differentiation. (As I have stated, numerically differentiation is > much more difficult than integration, on paper it is the other way > around.) Well, not really much more difficult, but rather very different. See: http://huizen.dto.tudelft.nl/deBruijn/programs/shannon.htm Your comments are quite welcome. > Yes, that is something we can conclude. On binary machines the iteration > of 1/(x-1) is quite stable when x is outside the range [1/2, 2]. But > inside that range the relative error of the result can be quite large. That's why I'd prefer to stay outside that range [1/2,2]. Han de Bruijn === Subject: Re: Testable Predictions by HdB ... > It is indeed true in general, but there are exceptions to the rule. > Newton-Raphson for the square root is one example. That is an example > where a next iterate is *never* much worse than a previous iterate. On > the other hand, the problem shows up when you try to do numerical > differentiation. (As I have stated, numerically differentiation is > much more difficult than integration, on paper it is the other way > around.) > Well, not really much more difficult, but rather very different. See: > http://huizen.dto.tudelft.nl/deBruijn/programs/shannon.htm > Your comments are quite welcome. That is something quite different. Assume a (given) function f(x) in an algorithmic way. Getting the derivative in good precision is bothersome, getting the integral in good precision is peanuts. -- 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: Testable Predictions by HdB > > Well, not really much more difficult, but rather very different. See: > > http://huizen.dto.tudelft.nl/deBruijn/programs/shannon.htm > > Your comments are quite welcome. > That is something quite different. Assume a (given) function f(x) in > an algorithmic way. Getting the derivative in good precision is > bothersome, getting the integral in good precision is peanuts. Use the algorithmic way to obtain a reasonable sampling of the function. Then interpolate with Gaussians where the spread = 2*sampling. Then do the differentiations. Then: what's the problem? Han de Bruijn === Subject: Re: Testable Predictions by HdB > > Well, not really much more difficult, but rather very different. See: > > http://huizen.dto.tudelft.nl/deBruijn/programs/shannon.htm > > Your comments are quite welcome. > > That is something quite different. Assume a (given) function f(x) in > an algorithmic way. Getting the derivative in good precision is > bothersome, getting the integral in good precision is peanuts. > Use the algorithmic way to obtain a reasonable sampling of the function. > Then interpolate with Gaussians where the spread = 2*sampling. Then do > the differentiations. Then: what's the problem? Do you know how it is done the standard way? Did you do an error analysis of your algorithm? The problem with numerical differentiation is that when you make the stepsize too small the numerical stability suffers due to rounding errors. This kind of problems do not occur with numerical integration. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: a curve fitting Hi folks. I've been thinking that it'd be very nice, since I'm a sysadmin, to have some software that'll graph disk usage over time, and use some form of curve fitting to predict when a disk might fill up based on past utilization changes. Does anyone have any thoughts on what might be a good curve fitting algorithm for this sort of purpose? About all I'm familiar with are least squares and splines, and I don't really know when to use them. === Subject: Re: a curve fitting > I've been thinking that it'd be very nice, since I'm a sysadmin, to have > some software that'll graph disk usage over time, and use some form of > curve fitting to predict when a disk might fill up based on past > utilization changes. > Does anyone have any thoughts on what might be a good curve fitting > algorithm for this sort of purpose? About all I'm familiar I have some lightly developed KBH Code for a least squares parabola that can make a newsgroup post: {KBH Code} Procedure Wrt; Begin WriteLn(p:14:4, q:14:4, r:14:4, s:14:4); WriteLn(pp:14:4, qq:14:4, rr:14:4, ss:14:4); WriteLn(ppp:14:4, qqq:14:4, rrr:14:4, sss:14:4); WriteLn; End; Procedure Noso; Begin WriteLn(' [ No solution ]'); flg:= 1; End; Procedure Frstsec; Begin pppp:= p; qqqq:= q; rrrr:= r; ssss:= s; p:= pp; q:= qq; r:= rr; s:= ss; pp:= pppp; qq:= qqqq; rr:= rrrr; ss:= ssss; End; Procedure Secthrd; Begin pppp:= ppp; qqqq:= qqq; rrrr:= rrr; ssss:= sss; ppp:= pp; qqq:= qq; rrr:= rr; sss:= ss; pp:= pppp; qq:= qqqq; rr:= rrrr; ss:= ssss; End; begin {KBH Code} {y = a + b + cx^2 Least squares parabola} WriteLn; WriteLn(' [ Enter q to end input ]'); WriteLn; flg:= 0; n:= 0; For i:= 1 To 50 Do Begin Write(' Input x coordinate: '); ReadLn(sa); Val(sa, x[i], cod); If (cod <> 0) Then Break; Write(' Input y coordinate: '); ReadLn(sa); Val(sa, y[i], cod); If (cod <> 0) Then Break; n:= i; End; p:= n; q:= 0; r:= 0; rr:= 0; rrr:= 0; s:= 0; ss:= 0; sss:= 0; {Begin Least Squares Parabola} For i:= 1 To n Do Begin q:= q + x[i]; ra:= Sqr(x[i]); r:= r + ra; rr:= rr + (x[i] * ra); rrr:= Sqr(ra); s:= s + y[i]; ss:= ss + (x[i] * y[i]); sss:= sss + (ra * y[i]); End; pp:= q; qq:= r; ppp:= r; qqq:= rr; {End Least Squares Parabola} Wrt; {Begin 3x3 Gaussian Elimination} If (p = 0) And (pp = 0) And (ppp = 0) Then Noso; If (flg = 0) Then Begin If (p = 0) And (pp <> 0) Then Frstsec; If (p = 0) And (pp = 0) Then Begin Secthrd; Frstsec; End; If (pp <> 0) Then Begin clc:= -(pp / p); pp:= pp + (p * clc); qq:= qq + (q * clc); rr:= rr + (r * clc); ss:= ss + (s * clc); Wrt; End; If (ppp <> 0) Then Begin clc:= -(ppp / p); ppp:= ppp + (p * clc); qqq:= qqq + (q * clc); rrr:= rrr + (r * clc); sss:= sss + (s * clc); Wrt; End; End; If (qq = 0) And (qqq = 0) Then Noso; If (flg = 0) Then Begin If (qq = 0) Then Secthrd; If (qqq <> 0) Then Begin clc:= -(qqq / qq); qqq:= qqq + (qq * clc); rrr:= rrr + (rr * clc); sss:= sss + (ss * clc); Wrt; End; End; If (rrr = 0) Then Noso; If (flg = 0) Then Begin sss:= sss / rrr; rrr:= rrr / rrr; Wrt; {End 3x3 Gaussian Elimination} {Begin substitution} c:= sss; If (qq <> 0) Then b:= (ss - (rr * c)) / qq Else b:= 0; a:= (s - (q * b) - (r * c)) / p; WriteLn(a:14:8); WriteLn(b:14:8); WriteLn(c:14:8); {End substitution} WriteLn; For i:= 1 To n Do Begin yy:= a + (b * x[i]) + (c * Sqr(x[i])); WriteLn(x[i]:14:2, yy:14:2); End; End; ReadLn; end. === Subject: Re: a curve fitting > I've been thinking that it'd be very nice, since I'm a sysadmin, to have > some software that'll graph disk usage over time, and use some form of > curve fitting to predict when a disk might fill up based on past > utilization changes. > Does anyone have any thoughts on what might be a good curve fitting > algorithm for this sort of purpose? About all I'm familiar with are least > squares and splines, and I don't really know when to use them. Dan, I am the author of a curve fitting/nonlinear regression program called NLREG (http://www.nlreg.com). I think your idea is reasonable. I would being by plotting your data and looking at the nature of the trend line. It may turn out that a simple linear regression would work, but you may have a more complicated function like an asymptotic curve or a sigmoid (S-shaped) curve. Once you have a general idea of the shape of the curve, you can then begin trying various functions to see how well they fit. If you can send me your data file via e-mail, I'll be happy to look at a plot of it and run some trial function fits through NLREG. Or if you want to do it yourself, you can download a free demonstration copy of NLREG from http://www.nlreg.com -- Phil Sherrod (phil.sherrod 'at' sandh.com) http://www.dtreg.com (decision tree and SVM predictive modeling) http://www.nlreg.com (nonlinear regression) === Subject: Re: a curve fitting > Hi folks. > I've been thinking that it'd be very nice, since I'm a sysadmin, to have > some software that'll graph disk usage over time, and use some form of > curve fitting to predict when a disk might fill up based on past > utilization changes. > Does anyone have any thoughts on what might be a good curve fitting > algorithm for this sort of purpose? About all I'm familiar with are least > squares and splines, and I don't really know when to use them. I think that it's a bad idea to apply curve-fitting techniques when you do not understand, broadly speaking, why the data is the way it is. These methods can't magically work out the reasons why disk space usage changes over time if you don't already roughly know this yourself. I would look at the data and try to understand the patterns (e.g. how does disk usage depend on the number of users and the work they are doing?). Once you've figured out roughly why usage is changing THEN you can apply some curve-fitting to extrapolate and predict. === Subject: Re: math disput btwn friends (dimension of vector spaces) > My friends and I are taking an intro linear > algebra class, and were discussing dimensions > of vector spaces. > Our teacher gave us some examples, such as: > 1) Over the field of complex numbers, > the vector space of complex numbers has > dimension 1. > 2) Over the field of real numbers, the vector > space of complex numbers has dimension 2. > So, my friend said that we could generalize > that if a vector space, V, is over the complex > numbers, then it will have dimension n. > And if we change the vector space,V, to be over > the real numbers then its dimension will change > to 2n. > Is he right??? And if so, why hasn't my teacher > made that connection? Can someone explain this, > if it is true? One can generalize this result: let V be a vector space of dimension n over a field L and let K be a subfield of L such that L viewed as a vector space over K has dimension m. Then via the inclusion map K-->L one can view V as a vector space over K and its dimension is mn. This fact is particularly useful in the algebra of fields: here not only K and L but also V is assumed to be a field. A simple but at first glance surprising consequence is this: let L be a field containing the field K as a subfield and assume that the dimension of L as a K-vector space is a prime number. Then there exists no proper subfield between K and L. H === Subject: Re: math disput btwn friends (dimension of vector spaces) > But I was not right about more abstract V^n where > the vectors are not themselves tuples of complex > numbers. When the hole gets deep enough, you have to stop digging. In general, vectors are not tuples of numbers, though (given a basis) they are specified by tuples of numbers. (Where have I seen this before?) An n-dimensional complex vector space may be specified by a bunch of basis vectors e_1 to e_n, and then a general vector is given by z_1 e+1 + ... +z_n e_n where the z_i are complex numbers. Now, you can also write such a vector as (a_1 + ib_1)e_1 + ... + (a_n + ib_n)e_n where the a_i and b_i are all real. Then you can write *that* as (a_1e_1 + .. +a_n e_n) + (b_1 ie_1 + ... +b_n ie_n) which is an element of a 2n real dimensional vector space with basis e_1...e_n, ie_1...ie_n since those vectors are linearly independent over the reals. So yes, if you have an n-dimensional vector space over C, you can make a 2n-dimensional vector space over R from it. OK, I've glossed over how natural this construction is: it isn't a natural construction, you need more structure before you have a natural identification of an n-dimensional vector space over C with a 2n-dimensional vector space over R. But I hope this gets the general idea across. === Subject: Re: math disput btwn friends (dimension of vector spaces) ... >So yes, if you have an n-dimensional vector space >over C, you can make a 2n-dimensional vector space >over R from it. OK, I've glossed over how >natural this construction is: it isn't a natural >construction, you need more structure before >you have a natural identification of an n-dimensional >vector space over C with a 2n-dimensional vector >space over R. What? There is, I thought, one-count-them-one embedding of R into C as a subfield. That is surely quite enough to give any vectorspace over C (finite dimensional or not) a natural structure as a vectorspace of R; if the first vectorspace is n-dimensional over C, then the second vectorspace(-structure on the exact same set) is 2n- dimensional over R. Perhaps you meant to say that you need more structure before you have a natural identification of a 2n-dimensional vector space over R with an n-dimensional vector space over C? *That* I will buy (with the exception of n = 0). Lee Rudolph === Subject: Re: math disput btwn friends (dimension of vector spaces) >What? There is, I thought, one-count-them-one embedding >of R into C as a subfield. No, there are infinitely many, unless you assume continuity or a specific automorphism of C of order 2 (a.k.a. complex conjugation). You can't identify R as a subfield of C using only field operations. Pertti's ghost is smiling smugly. dave === Subject: Re: math disput btwn friends (dimension of vector spaces) >>What? There is, I thought, one-count-them-one embedding >>of R into C as a subfield. >No, there are infinitely many, unless you assume continuity >or a specific automorphism of C of order 2 (a.k.a. complex >conjugation). You can't identify R as a subfield of C >using only field operations. heh-heh. You math guys like to make things sooo complicated. Where I come from we don't need no steenking embeddings, R _is_ a subfield of C, and hence a complex vector space of dimension n _is_ (ok, is, if you want to be picky) a real vector space of dimension 2n. >Pertti's ghost is smiling smugly. RIP. You know you're really supposed to send Rudolph a letter with the offending statement underlined in red. (I don't recall what the next step is if he fails to acknowledge the error of his ways...) >dave ************************ David C. Ullrich === Subject: Re: math disput btwn friends (dimension of vector spaces) >>Pertti's ghost is smiling smugly. > RIP. You know you're really supposed to send Rudolph > a letter with the offending statement underlined in > red. (I don't recall what the next step is if he fails > to acknowledge the error of his ways...) Didn't Pertti used to maintain a web page of people's mistakes? === Subject: Re: math disput btwn friends (dimension of vector spaces) >>What? There is, I thought, one-count-them-one embedding >>of R into C as a subfield. >No, there are infinitely many, unless you assume continuity >or a specific automorphism of C of order 2 (a.k.a. complex >conjugation). You can't identify R as a subfield of C >using only field operations. >Pertti's ghost is smiling smugly. I did worry about that briefly (well, not the ghost part), but I confess that I don't see how I'd be able to use a wild embedding of R into C to give an unnatural R-vectorspace structure on a set already equipped with a given C-vectorspace structure. ... I guess the trick, if there is one, is hidden in the word natural. We might as well assume our original C-vectorspace V has dimension 1 (over C). We want to pick out *one*, among all the possible R-vectorspace structures on the underlying set of V, which is natural with respect to the given C-vectorspace structure. Well, that natural must include in its meaning (mustn't it?) that the construction respects *the embedding of R in C*. And, indeed, *given such an embedding*, I claim that the forgetful construction (which forgets how to multiply elements of V by scalars that aren't in the embedded copy of R, but otherwise preserves the given structure of V) is indeed natural. My error was, therefore, including the extraneous (and false) comment about a unique embedding of R into C as a subfield. Is that right now? I have to go back to grading.... Lee Rudolph === Subject: Re: math disput btwn friends (dimension of vector spaces) <4042048.1128015026228.JavaMail.jakarta@nitrogen.mathforum.org> <3q2t3pFd25v3U1@individual.net> So yes, if you have an n-dimensional vector space >over C, you can make a 2n-dimensional vector space >over R from it. OK, I've glossed over how >natural this construction is: it isn't a natural >construction, you need more structure before >you have a natural identification of an n-dimensional >vector space over C with a 2n-dimensional vector >space over R. > What? There is, I thought, one-count-them-one embedding > of R into C as a subfield. That is surely quite enough > to give any vectorspace over C (finite dimensional or not) > a natural structure as a vectorspace of R; if the first > vectorspace is n-dimensional over C, then the second > vectorspace(-structure on the exact same set) is 2n- > dimensional over R. Damn, *that's* what I wanted to say in my first posting; instead of recanting it all I should have simply corrected my stupid non-embedding and the is not an exception. > Perhaps you meant to say that you need more structure > before you have a natural identification of a 2n-dimensional > vector space over R with an n-dimensional vector space > over C? *That* I will buy (with the exception of n = 0). Heh, I think this is what I was groping toward in my failed attempt at damage control. Let's see if I get it right this time: when the set in question *is* R^2n, then there *is* a natural identification, though strictly speaking you have to map your set to (R^2)^n to get the needed structure. For other sets the mapping might not be so natural. {And yes I'm tired of digging. Here's hoping this is not just another shovelful.) === Subject: Re: math disput btwn friends (dimension of vector spaces) > ... > Perhaps you meant to say that you need more structure > before you have a natural identification of a 2n-dimensional > vector space over R with an n-dimensional vector space > over C? Yes: I was confusing myself about extracting the real space from the complex one and putting a complex structure on the real one to get a complex one. === Subject: Klein Bottle homeomorphism I am trying to show that the Klein bottle (the square with opposite edges identified, where two opposite arrows are in the same direction, and two opposite arrows in the opposite direction) is homeomorphic to an annulus whose antipodal points on the outer circle are identified, and whose antipodal points on the inner circle are identified. I have thought about it a bit, and I'd like to use the following strategy : Let S be a square, A be an annulus, and let Klein bottle be S / ~ , where ~ was defined above, and let the second space above be A / ~ where ~ was defined above as well. I'd like to construct a commutative diagram where S ---> S / ~, A ----> A / ~ are canonical, and then I will define maps f : S ----> A and F : S / ~ -----> A / ~. I decided that the most natural thing for f would be just exponentiation in the complex plane. exp takes a square to an annulus (where the height and length of the square is 2pi). You can probably see the ensuing problem. If I identify opposite edges on the square in the complex plane as I explained above (one pair using the same arrows, the other pair using opposite arrows), I run into problems. What happens after I exponentiate i s that I end up identifying an point on the outer circle of the annulus, with the corresponding point on the inner circle. This map also ends up identifying purely real points on the annulus in a funny way. I don't really want any hints, I guess I would just like to know if I am close. I feel like my strategy here is very natural, and the map f : S ----> A is also very natural. But I feel like my map f may not be correct and I should probably find a different map. James