> I need an automated method to fit a curve of the form y = > c*x*(1+a*e^(b*x)) to test data. This is a magnetic saturation curve > for an electric machine. The exponential term is the saturation > function, and is negligible at low values of x. We have several > points at low x values. At high values of x, the exponential term > dominates. We usually have a few points out to where the exponential > term is 0.5 or more, but not always. The data at low x are often a > bit noisy. When we fit by hand, we draw a straight line approximating an > asymptote to the lower part of the curve, then we subtract that line > from the data to get approximately the exponential part of the > function (the saturation factor). By taking the logarithm of both > sides for the dataabove a certain value of x, we can get a pretty good > fit on the upper part of the curve. Combining the two functions gives > the full approximation. The problem being, it requires a bit of human judgement to pick out > the spot where the linear part ends and the exponential begins. If I > get it wrong, either the linear portion will be skewed by having part > of the exponential term mixed in, or else it won't fit the transition > if not enough data is used for the exponential part (particularly if > it is not large). The approximation only needs to fit within 1% or so of the real > values (which don't actually fit the function exactly). Accuracy near > the transition region is most important, although this form of > equation was selected because it gives reasonable results when > extrapolated to higher values than we have data for. We're doing this in LabView, so we don't have access to numerical > libraries for nonlinear curve fitting and such. Would prefer a quick > and dirty solution, though we might be able to port in some Fortran > code. I don't know how much procedural code is possible. A table of sample data follows: Voc(PU) Igf(A) > 0.196 5.00 > 0.304 7.85 > 0.399 10.43 > 0.616 16.20 > 0.797 21.54 > 0.906 25.89 > 0.942 29.01 > 1.014 32.75 > 1.051 37.25 > 1.105 44.80 > 1.196 61.18 > 1.304 105.61 This is approximately fit with: a = 0.000232 > b = 7.005 > c = 25.74 > My fitted values, and approx. std. errors using a Fortran program are: a = 0.000250(0.000070) b = 6.95 (0.20) c = 25.69 (0.49) I see that someone else agrees with these values. -- Alan Miller http://users.bigpond.net.au/amiller Retired Statistician Allen > -- > Allen Windhorn (507) 345-2782 FAX (507) 345-2805 > Kato Engineering (Though I do not speak for Kato) > P.O. Box 8447, N. Mankato, MN 56002 > Allen.Windhorn@LSUSA.com ==== > I need an automated method to fit a curve of the form y = > c*x*(1+a*e^(b*x)) to test data. This is a magnetic saturation curve > for an electric machine. The exponential term is the saturation > function, and is negligible at low values of x. We have several > points at low x values. At high values of x, the exponential term > dominates. We usually have a few points out to where the exponential > term is 0.5 or more, but not always. The data at low x are often a > bit noisy. When we fit by hand, we draw a straight line approximating an > asymptote to the lower part of the curve, then we subtract that line > from the data to get approximately the exponential part of the > function (the saturation factor). By taking the logarithm of both > sides for the dataabove a certain value of x, we can get a pretty good > fit on the upper part of the curve. Combining the two functions gives > the full approximation. The problem being, it requires a bit of human judgement to pick out > the spot where the linear part ends and the exponential begins. If I > get it wrong, either the linear portion will be skewed by having part > of the exponential term mixed in, or else it won't fit the transition > if not enough data is used for the exponential part (particularly if > it is not large). The approximation only needs to fit within 1% or so of the real > values (which don't actually fit the function exactly). Accuracy near > the transition region is most important, although this form of > equation was selected because it gives reasonable results when > extrapolated to higher values than we have data for. We're doing this in LabView, so we don't have access to numerical > libraries for nonlinear curve fitting and such. Would prefer a quick > and dirty solution, though we might be able to port in some Fortran > code. I don't know how much procedural code is possible. A table of sample data follows: Voc(PU) Igf(A) > 0.196 5.00 > 0.304 7.85 > 0.399 10.43 > 0.616 16.20 > 0.797 21.54 > 0.906 25.89 > 0.942 29.01 > 1.014 32.75 > 1.051 37.25 > 1.105 44.80 > 1.196 61.18 > 1.304 105.61 This is approximately fit with: a = 0.000232 > b = 7.005 > c = 25.74 > Allen > -- > Allen Windhorn (507) 345-2782 FAX (507) 345-2805 > Kato Engineering (Though I do not speak for Kato) > P.O. Box 8447, N. Mankato, MN 56002 > Allen.Windhorn@LSUSA.com You need a nonlinear optimizing routine minimizing the penalty function Sum( ( y - Yest) ^2 ) where Yest is your formula, At the first clance your problem is well behaving so a simple algorithm can do it. There are plenty of them. I can give my favorite ( F77 ) if needed. ==== See my address at www.estlab.com Arto Huttunen ==== >> I need an automated method to fit a curve of the form y = >> c*x*(1+a*e^(b*x)) to test data. This is a magnetic saturation curve >> for an electric machine. The exponential term is the saturation >> function, and is negligible at low values of x. We have several >> points at low x values. At high values of x, the exponential term >> dominates. We usually have a few points out to where the exponential >> term is 0.5 or more, but not always. The data at low x are often a >> bit noisy. >> >> When we fit by hand, we draw a straight line approximating an >> asymptote to the lower part of the curve, then we subtract that line >> from the data to get approximately the exponential part of the >> function (the saturation factor). By taking the logarithm of both >> sides for the dataabove a certain value of x, we can get a pretty good >> fit on the upper part of the curve. Combining the two functions gives >> the full approximation. >> >> The problem being, it requires a bit of human judgement to pick out >> the spot where the linear part ends and the exponential begins. If I >> get it wrong, either the linear portion will be skewed by having part >> of the exponential term mixed in, or else it won't fit the transition >> if not enough data is used for the exponential part (particularly if >> it is not large). >> >> The approximation only needs to fit within 1% or so of the real >> values (which don't actually fit the function exactly). Accuracy near >> the transition region is most important, although this form of >> equation was selected because it gives reasonable results when >> extrapolated to higher values than we have data for. >> >> We're doing this in LabView, so we don't have access to numerical >> libraries for nonlinear curve fitting and such. Would prefer a quick >> and dirty solution, though we might be able to port in some Fortran >> code. I don't know how much procedural code is possible. >> >> A table of sample data follows: >> >> Voc(PU) Igf(A) >> 0.196 5.00 >> 0.304 7.85 >> 0.399 10.43 >> 0.616 16.20 >> 0.797 21.54 >> 0.906 25.89 >> 0.942 29.01 >> 1.014 32.75 >> 1.051 37.25 >> 1.105 44.80 >> 1.196 61.18 >> 1.304 105.61 >> >> This is approximately fit with: >> >> a = 0.000232 >> b = 7.005 >> c = 25.74 >> >> >> Allen Allen: I believe the Levenberg-Marquardt method of curve-fitting can be used to >fit your data. A quick search on Google found many freely available >programs, some with source code. Also, here is what the Matlab v6.5 curve fit GUI predicts for your data: General model: > f(x) = c*x + a*c*exp(b*x) should be: c*x + a*c*x*exp(b*x) Then you get: a = .0002504 b = 6.946 c = 25.685 >Coefficients (with 95% confidence bounds): > a = 0.0001179 (4.303e-005, 0.0001928) > b = 7.735 (7.286, 8.183) > c = 25.52 (24.39, 26.65) Goodness of fit: > SSE: 3.104 > R-square: 0.9996 > Adjusted R-square: 0.9996 > RMSE: 0.5873 Good luck, >OUP ==== Dear Group, I`m trying to calculate a 2D fourier transform (2DFFT) of a 2d exponential decay. Upon comparisions with a lorentzian form ( i compare the normalized values of the real and the imaginary parts seperately) i see that with my best efforts i can achieve an accuracy of only 10^(-3). Even this is possible only if go to a totally unacceptable time step (In order to get a reasonable resolution in the frequency domain i would need to do a 8192x8192 FFT calculation). I`m unable to locate where the error is coming from. I don`t think my problem is related to zero padding as i can choose a decay const which ensure that the value goes to zero within the interval. I would be really grateful to any comments and pointers as i`m really stuck. (i need an accuracy of at least 10^(-5) to proceed). Ravi ==== What is the data you are trying to analyse ???? > Dear Group, > I`m trying to calculate a 2D fourier transform (2DFFT) of > a 2d exponential decay. Upon comparisions with a lorentzian form ( i > compare the normalized values of the real and the imaginary parts > seperately) i see that with my best efforts i can achieve an accuracy > of only 10^(-3). Even this is possible only if go to a totally > unacceptable time step (In order to get a reasonable resolution in the > frequency domain i would need to do a 8192x8192 FFT calculation). > I`m unable to locate where the error is coming from. I > don`t think my problem is related to zero padding as i can choose a > decay const which ensure that the value goes to zero within the > interval. I would be really grateful to any comments and pointers as > i`m really stuck. (i need an accuracy of at least 10^(-5) to proceed). > > Ravi ==== > I want to know how we can find the coefficients of this serie? > _L_ > > C =/_ (a_n)^2 * f_n(t) > n=0 > C = is a constant > a_n = series coefficients (we want to find) > f_n = constitue an orthogonal basis function ======== CASE 1: a_0,...,a_L are real numbers and the set (1) {f_0(t),f_1(t),...,f_L(t)} , (L >= 1), is a polynomial system which is orthogonal. Therefore f_0(t)=c_{0,0}=/=0 and for 1== 0 there is only one (real) solution, namely a_0=sqrt(C/c_{0,0}) and a_1=a_2=...=a_L= 0 . Other situations,casers, must be considered (e.g. complex coefficients,...). ==== >>In X a normal space, there is a continuous function f: X -> [0,1] such >>that f(x) =0 on A and f(x) = 1 on B AND 0 < f(x) <1 OTHERWISE, if and >>only if A and B are disjoint closed G-delta sets in X. >>Does anybody know where I can find a proof of this theorem? I don't >> I have not seen this theorem before, but I find it trivial. >> All that one needs to use about G-delta sets is that they are >> countable intersections of open sets, the definition. >Well ok, would you detail a quicker proof in a few lines? I prefer to toss out another hint. An equivalent form of Urysohn's Lemma is that in a normal space, if A is closed, and U is an open superset of A, there is a continuous function which is 0 on A and l outside of U. -- 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 ==== UBLAS supports sparse matrix stuff fairly efficiently I understand. I have used it for small dense matrix work, for which it is not bad, but its maintainer seems more keen on the sparse support. http://www.boost.org/libs/numeric/ublas/doc/overview.htm It supports most data types, exploits lazy evaluation through some heavy template use etc. Some work has been done on LAPACK interfaces. ==== I am new to the world of stats and probabilty was wondering if anyone could explain a wee bit about probabilty density I am trying to write a piece of software quite like Granular Synthesis Demo 1 in http://www.dcs.gla.ac.uk/~jhw/audioclouds/ . I was wondering if anyone could explain the difference between a probabilty distribution and a probabilty density? And if you have a probabilty distribution can a density be obtained from it and if so how? Also the main bit i am stuck on is how a density can be conditioned (in the example given it is conditioned by the x and y positions of a mouse) how would you go about doing such a thing Any help would be greatly appericated thanks Gregory ==== I was wondering if anyone > could explain the difference between a probabilty distribution and a > probabilty density? for a real random variable? > And if you have a probabilty distribution can a density be obtained > from it and if so how? Differentiate. If r.v. X has distribtion function F that means F(x) = Prob(X<=x), and the probability density (if it exists, that is if F is absolutely continuous) is f(x) = F'(x). I agree with the other answer. Coding something you do not understand well is likely to lead to inferior (or incorrect) code. ==== >I am new to the world of stats and probabilty was wondering if anyone >could explain a wee bit about probabilty density >I am trying to write a piece of software quite like Granular Synthesis >Demo 1 in >http://www.dcs.gla.ac.uk/~jhw/audioclouds/ . I was wondering if anyone >could explain the difference between a probabilty distribution and a >probabilty density? >And if you have a probabilty distribution can a density be obtained >from it and if so how? >Also the main bit i am stuck on is how a density can be conditioned >(in the example given it is conditioned by the x and y positions of a >mouse) how would you go about doing such a thing I would strong recommend getting hold of a good book on probability and reading up BEFORE coding. It is nearly 40 years since I started to learn it, and nearly 35 since I taught it, so I am not a good person to give you references! The books I used would still be OK, but are probably hard to get :-) Nick Maclaren. ==== I have been trying to figure out why there is a /constrained/ quadratic progamming step in SQP methods. I don't understand why the QP isn't formulated without constraints, instead using Langrange multipliers since it would be much easier to solve. If you can tell me the answer I'd be very grateful! I'm not sure if its a stupid question or not, so I've described my thinking below: I currently understand that SQP is an extension of quasi-Newton methods to constrained optimisation. A quasi-Newton method is used to maintain a positive definite approximation to the Hessian of the objective function. Each iteration involves finding the optimum step direction by solving the QP based on the second order approximation of the objective using this approximate Hessian with first order approximations to the constraint functions. The solution of this constrained QP yields the step direction for both the parameters and the Lagrange multipliers for the full NLP and a line search yields the best scaling of the step vector. What I don't understand is this: why not formulate the QP as an unconstrained problem using an approximated Hessian to the Lagrangian of the objective function. Since the quasi-Newton method yeilds a positive definite approximate Hessian this unconstrained QP will be trivially solved with a Linear system. BFGS can give an inverse to the approximate Hessian quite easily using a Choleski factorization which can also be easily updated. This would mean that the Lagrange multipliers would have to be updated outside of the QP solution, -but this would be fairly straghtforward with an augmented Lagrangian method based on some penalty function? I'm guessing that either (i) it would be hard/impossible to find the multipliers outside of the QP step for some reason, or (ii) it would be possible but it is more efficient to find as part of a constrained QP. Very grateful for any help! Best, Josh. ==== How to integrate sinc(x) (without Fourier transformation) ?? 1018 ==== Good evenind, there's an interesting fact.. i'd like discuss with you: Terminology: differentiable = C^{infty}; cap = symbol of intersection between sets, otimes = symbol of tensorial product; x in S means that x belongs to S. I know the following standard result: ************************************************* Let X be a topological space (or a differentiable manyfold) and U = ( U_i )_{i in I} an open covering of X. forall i in I, consider the trivial vector bundle U_{i} x R^{p} and suppose that for each (i,j) in I^{2} such that U_{i} cap U_{j} not equal emptyset is assigned a continuous (or differentiable if it makes sense) application f_{ij}:U_{i} cap U_{j} --> GL(R^{p}). In the disjoint union of such trivial bundles, let's glue a point (x,v_{i}) in U_{i} x R^{p} with (y,v_{ j}) in U_{ j} x R^{p} if and only if x = y in U_{i}cap U_{j} and v_{j} = f_{ij}(x)[v_{i}]. If, forall (i,j,k) in I^{3} and x in U_{i}cap U_{j}cap U_{k} we have f_{ij}(x) o f_{jk}(x) = f_{ik}(x) ==> the quotient for such a relation defines a vector bundle over X. ************************************************* Now, M_{i} will denote a differentiable manyfold; If M is a differentiable manyfold then V(M) will denote the additive category of C^{infty} real vector bundles over M. Let M = M_{1} x ... x M_{k} be the product manyfold of M_{i}. If xi^{i} in V(M_{i}), then xi^{1} otimes ... otimes xi^{k} is a bundle over M whose fiber at x = (x_{1},...,x_{k}) is xi^{1}_{x_{1}}otimes ... otimes xi^{k}_{x_{k}}. If f ^{i} in C^{infty}(M_{i}, xi^{i}), we define f^{1}otimes ... otimes f^{k}, a function on M, by (f^{1}otimes...otimes f^{k})(x) = = f^{1}(x_{1})otimes...otimes f^{k}(x_{k}). The bundle structure of xi^{1}otimes ... otimes xi^{k} is characterized by the property that f^{1}otimes ... otimes f^{k} in C^{infty}(M, xi^{1} otimes ... otimes xi^{k}) wherever f ^ {i} in C^{infty}(M_{i}, xi^{i}). I don't understand the reason for which ``the bundle structure of xi^{1} otimes... otimes xi^{k} is characterized by the property that f ^{1} otimes... otimes f^{k} in C^{ infty}(M, xi^{1}otimes ... otimes xi^{k}) wherever f^{i} in C^{infty}(M_{i}, xi^{i}) ''. May you to help me, please? Tern ternnret@yahoo.it ==== I've heard that there is no complex analogue of the Darboux Theorem. Its a local question so suppose I have a smooth, closed, non-degenerate, complex valued two-form omega on complex n-space M. When is it possible to find coordinates z_i, w_j (smooth maps from M to C) such that omega = dz_1.dw_1 + .... + dz_n.dw_n ? Are there simple counterexamples? Also, in the holomorphic case (omega holomorphic), I think the theorem may actually hold. Any refferences? Please forward any postings to orenb@hans.math.upenn.edu ---Oren Ben-Bassat http://www.math.upenn.edu/~orenb ==== > I've heard that there is no complex analogue of the Darboux Theorem. > Its a local question so suppose I have a smooth, closed, > non-degenerate, complex valued two-form omega on complex n-space M. > When is it possible to find coordinates z_i, w_j (smooth maps from M > to C) such that > omega = dz_1.dw_1 + .... + dz_n.dw_n ? Are there simple > counterexamples? I don't know too much about the subject, but it seems to me that for dz_i to be *defined*, z_i needs to be differentiable with respect to each component axis of M (= C^n). Generalizing, this probably leads to omega being holomorphic, through the same arguments as are used in the theory of functions of one complex variable. OTOH, given the coordinates z_i, one can construct a two-form by summing the basis two-forms, sum_ij alpha_ij * dz_i * dz_j, for *any* complex-valued functions alpha_ij. If the alpha's aren't differentiable, the sum probably won't be. Dale ==== On 7 Jan 04 23:39:13 -0500 (EST), Oren Ben-Bassat >I've heard that there is no complex analogue of the Darboux Theorem. >Its a local question so suppose I have a smooth, closed, >non-degenerate, complex valued two-form omega on complex n-space M. >When is it possible to find coordinates z_i, w_j (smooth maps from M >to C) such that >omega = dz_1.dw_1 + .... + dz_n.dw_n ? Are there simple >counterexamples? Also, in the holomorphic case (omega holomorphic), I think the theorem >may actually hold. I can't figure out what a complex Darboux Theorem would even _state_, much less whether it's true. Maybe we're talking about different results: The Darboux Theorem I have in mind is the fact that if f : (a,b) -> R is differentiable then the derivative f' satisfies the intermediate value property. (What _is_ the intermediate value property in the complex case?) >Any refferences? Please forward any postings to orenb@hans.math.upenn.edu ---Oren Ben-Bassat >http://www.math.upenn.edu/~orenb ************************ David C. Ullrich ==== Let f be a continuous complex function defined on the unit circle such that |f| is increasing and f(1)=1. Let g be a continuous real function defined on the unit circle such that g is increasing and g(1)=1. def; A(n):=Sum[f(exp(2*k*Pi*i/n)),{k,0,n-1}] , B(n):=Sum[f(exp(2*k*Pi*i/n))*g(exp(2*k*Pi*i/n)),{k,0,n-1}] , C(n):=Sum[f(exp(2*k*Pi*i/n))*(g(exp(2*k*Pi*i/n)))^2,{k,0,n-1}]. PROBLEM: Is |B(n)-A(n)|<=|C(n)-A(n)| & |B(n)-C(n)|<=|C(n)-A(n)| for infinite number of n ??? ==== >Let f be a continuous complex function defined on the unit circle such >that |f| is increasing and f(1)=1. Let g be a continuous real function >defined on the unit circle such that g is increasing and g(1)=1. def; What does increasing mean for functions defined on a circle? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== Here are this week's titles in the mathematics arXiv, available at: http://front.math.ucdavis.edu/ http://front.math.ucdavis.edu/submissions This week in the mathematics arXiv may be freely redistributed with attribution and without modification. Titles in the mathematics arXiv (15 Dec - 19 Dec) ------------------------------------------------- AC: Commutative Algebra ----------------------- math.AC/0312368 Joe Buhler, Zinovy Reichstein: Symmetric functions and the phase problem in crystallography math.AC/0312333 Mordechai Katzman: The support of top graded local cohomology modules math.AC/0312302 Martin Lorenz: On the Cohen-Macaulay property of multiplicative invariants AG: Algebraic Geometry ---------------------- math.AG/0312362 space of semistable vector bundles over smooth curves math.AG/0312351 Giuseppe Borrelli: On the classification of surfaces of general type with non-birational bicanonical map and Du Val Double planes math.AG/0312349 Charles Cadman: Using stacks to impose tangency conditions on curves math.AG/0312343 Spencer Bloch, H'el`ene Esnault: Local Fourier transforms and rigidity for D-modules math.AG/0312324 Shihoko Ishii: The arc space of a toric variety math.AG/0312301 R. M. Miro-Roig, K. Ranestad: Intersection of ACM-curves in P^3 math.AG/0312285 T. Shaska: Curves of genus 2 with (n, n)-decomposable jacobians math.AG/0312284 T. Shaska: Determining the automorphism group of a hyperelliptic curve math.AG/0312280 Fabrizio Coiai, Yogish I. Holla: Extension of semistable principal bundles in positive characteristic math.AG/0312278 Trond Stolen Gustavsen: On the cotangent cohomology of rational surface singularities with almost reduced fundamental cycle math.AG/0312273 Nikita A. Karpenko: Holes in I^n math.AG/0312271 D. Kaledin: Sommese Vanishing for non-compact manifolds math.AG/0312264 Carla Dionis: Stabilizers for nondegenerate matrices of boundary format and Steiner bundles math.AG/0312262 Geometry math.AG/0312260 Adrian Langer: Semistable principal G-bundles in positive characteristic math.AG/0312259 Fr'ed'eric Bihan: Asymptotiques de nombres de Betti d'hypersurfaces projectives r'eelles AP: Analysis of PDEs -------------------- math.AP/0312367 Christiaan C. Stolk: A pseudodifferential equation with damping for one-way wave propagation in inhomogeneous acoustic media math.AP/0312319 Michael Goldberg, Wilhelm Schlag: Scaling invariant smoothing estimates for the schroedinger flow in dimension three math.AP/0312281 Kim Dang Phung: Nearly a polynomial decay rate for the wave equation math.AP/0312276 M. L. Frankel, V. Roytburd: Finite-Dimensional Attractor for a Nonequilibrium Stefan Problem with Heat Losses AT: Algebraic Topology ---------------------- math.AT/0312277 Martin Markl, Steve Shnider: Associahedra, cellular W-construction and products of $A_infty$-algebras CA: Classical Analysis and ODEs ------------------------------- math.CA/0312353 F. Pakovich, N. Roytvarf, Y. Yomdin: Cauchy Type Integrals of Algebraic Functions math.CA/0312321 Julius Borcea: Hyperbolicity-preserving linear operators and Gaa rding-convexity of twisted root maps math.CA/0312320 D.V. Gorbachev, A.S. Manoshina: Relation between Tur'an extremum problem and van der Corput sets math.CA/0312317 P. Chladek, L. Klapka: The Cauchy atlas on the manifold of all complete ODE solutions math.CA/0312312 N. M. Atakishiyev, A. U. Klimyk: A set of orthogonal polynomials, dual to alternative q-Charlier polynomials math.CA/0312310 Hjalmar Rosengren: An elementary approach to 6j-symbols (classical, quantum, rational, trigonometric, and elliptic) math.CA/0312307 Luca Brandolini, Allan Greenleaf, Giancarlo Travaglini: $L^p - L^{p'}$ estimates for overdetermined Radon transforms math.CA/0312248 Jos'e Manuel Marco, Javier Parcet: Taylor series for the Askey-Wilson operator and classical summation formulas CO: Combinatorics ----------------- math.CO/0312358 Masao Ishikawa, Masato Wakayama: Applications of Minor Summation Formula III, Plucker Relations, Lattice Paths and Pfaffian Identities math.CO/0312350 Nicholas A. Loehr, Gregory S. Warrington, Herbert S. Wilf: The combinatorics of a three-line circulant determinant math.CO/0312297 David Speyer, Lauren K. Williams: The tropical totally positive Grassmannian math.CO/0312287 Tuerker Biyikoglu, Josef Leydold: Faber-Krahn Type Inequalities for Trees math.CO/0312282 Jason Burns: Bijective Proofs for Enumerative Properties of Ferrers Graphs math.CO/0312272 Animesh Datta: Turbo Codes over the Real Field math.CO/0312250 Ezra Miller: Alternating formulas for K-theoretic quiver polynomials math.CO/0312247 Jos'e Manuel Marco, Javier Parcet: A new approach to the theory of classical hypergeometric polynomials CT: Category Theory ------------------- math.CT/0312339 Volodymyr Lyubashenko, Oleksandr Manzyuk: Free ${A}_infty$-categories CV: Complex Variables --------------------- math.CV/0312304 Julien Duval: Une sextique hyperbolique dans P^3(C) math.CV/0312258 Mikhail Sodin, Boris Tsirelson: Random complex zeroes, III. Decay of the hole probability DG: Differential Geometry ------------------------- math.DG/0312364 D.E.Blair, J.Davidov, O.Mushkarov: Hyperbolic Twistor Spaces math.DG/0312325 A. M. Ionescu, G. E. Vilcu: A note on para-quaternion manifolds hep-th/0312154 Paolo Aschieri, Luigi Cantini, Branislav Jurco: Nonabelian Bundle Gerbes, their Differential Geometry and Gauge Theory math.DG/0312318 Jeff Viaclovsky, Gang Tian: Moduli spaces of critical Riemannian metrics in dimension four gr-qc/0312047 Hubert L. Bray, Piotr T. Chrusciel: The Penrose Inequality nlin.SI/0312030 B.G.Konopelchenko, W.K.Schief: Conformal geometry of the (discrete) Schwarzian Davey-Stewartson II hierarchy math.DG/0312251 Haibao Duan: A Problem of Hsiang-Palais-Terng on isoparametric submanifolds math.DG/0312249 C. Dunn, P. Gilkey, R. Ivanova, S. Nikcevic: The Spectral Geometry of the DS: Dynamical Systems --------------------- math.DS/0312361 B. Demir, V. Dzhafarov, S. Kocak, M. Ureyen: Restrictions of Harmonic Functions on the Sierpinski Gasket to Segments math.DS/0312356 F'ethi Grabsi, James Montaldi, Juan-Pablo Ortega: Bifurcation and forced symmetry breaking in Hamiltonian systems math.DS/0312346 Alex Furman: Outer automorphism groups of some ergodic equivalence relations math.DS/0312328 JR Chazottes, F. Durand: Local rates of Poincare recurrence for rotations and weak mixing math.DS/0312323 Claire Moura: On the multiplicity of the hyperelliptic integrals math.DS/0312306 Volodymyr Nekrashevych: Iterated Monodromy Groups math.DS/0312291 J.-R. Chazottes, R. Leplaideur: Birkhoff averages of Poincare cycles for Axiom-A diffeomorphisms math.DS/0312279 Wolf Jung: Homeomorphisms of the Mandelbrot Set FA: Functional Analysis ----------------------- math.FA/0312365 Gelu Popescu: Entropy and multivariable interpolation math.FA/0312341 Kamthorn Chailuek, Wicharn Lewkeeratiyutkul: A Pointwise Bound for a Holomorphic Function which is Square-Integrable with Respect to an Exponential Density Function GM: General Mathematics ----------------------- math.GM/0312360 Paola Cattabriga: Beyond Uncountable math.GM/0312309 Craig Alan Feinstein: The Collatz 3n+1 Conjecture is Unprovable GN: General Topology -------------------- math.GN/0312288 R. N. Gumerov: On finite-sheeted covering mappings onto solenoids GR: Group Theory ---------------- math.GR/0312352 Robert W. Baddeley, Cheryl E. Praeger, Csaba Schneider: Innately transitive subgroups of wreath products in product action math.GR/0312342 Rafael Villarroel-Flores: Group Orders That Imply Existence Of Nontrivial Normal p-Subgroups math.GR/0312331 Sean Cleary, Jennifer Taback: Metric properties of the lamplighter group as an automata group math.GR/0312257 Michael Mueger: On the center of a compact group GT: Geometric Topology ---------------------- math.GT/0312363 Boris A. Springborn: Variational principles for circle patterns math.GT/0312347 Blake Mellor: Intersection Graphs for String Links math.GT/0312337 Alexis Virelizier: Kirby elements and quantum invariants math.GT/0312329 Gennaro Amendola, Bruno Martelli: Non-orientable 3-manifolds of complexity up to 7 math.GT/0312322 P. B. Kronheimer, T. S. Mrowka: Dehn surgery, the fundamental group and SU(2) math.GT/0312311 Tahl Nowik: Framings and Projective Framings for 3-Manifolds math.GT/0312295 Greg Friedman: All frame-spun knots are slice math.GT/0312269 A. V. Karasev: On two problems in extension theory math.GT/0312266 Brendan Owens, Saso Strle: A characterisation of the Z^(n-1) + 3Z lattice and applications to rational homology spheres ------------------------ math.HO/0312293 Josh Isralowitz: Classical Lebesgue Integration Theorems for the Riemann Integral KT: K-Theory and Homology ------------------------- math.KT/0312305 Ralf Meyer: The cyclic homology and K-theory of certain adelic crossed products math.KT/0312292 Ralf Meyer, Ryszard Nest: The Baum-Connes Conjecture via Localization of Categories LO: Logic --------- math.LO/0312308 Arnold W. Miller: The gamma - Borel conjecture MG: Metric Geometry ------------------- math.MG/0312268 Alexander Barvinok, Grigoriy Blekherman: Convex Geometry of Orbits math.MG/0312253 Ezra Miller, Igor Pak: Metric combinatorics of convex polyhedra: cut loci and nonoverlapping unfoldings MP: Mathematical Physics ------------------------ math-ph/0312047 Thomas H. Otway: Anomalies of the tides math-ph/0312046 Simon P. Eveson, Christopher J. Fewster, Rainer Verch: Quantum inequalities in quantum mechanics math-ph/0312045 M. Hartmann, G. Mahler, O. Hess: Gaussian quantum fluctuations in math-ph/0312044 Anna Jencova: Geodesic distances on density matrices hep-th/0312186 Giovanni Arcioni, Claudio Dappiaggi: Holography in asymptotically flat space-times and the BMS group math-ph/0312043 Romuald A. Janik, Waldemar Wieczorek: Multiplying unitary random matrices - universality and spectral properties hep-th/0312168 S.L. Lukyanov, E.S. Vitchev, A.B. Zamolodchikov: Integrable Model of Boundary Interaction: The Paperclip hep-th/0312158 Petr P. Kulish, Anton M. Zeitlin: Group Theoretical Structure and Inverse Scattering Method for super-KdV Equation hep-th/0312116 Euler-Heisenberg Effective Actions cond-mat/0312361 Roberto Franzosi, Marco Pettini: A Theorem on the origin of Phase Transitions math-ph/0312042 Alberto De Sole, Victor Kac: Freely generated vertex algebras and non-linear Lie conformal algebras math-ph/0312041 Marek Biskup, Christian Borgs, Jennifer T. Chayes, Roman Kotecky: Partition function zeros at first-order phase transitions: Pirogov-Sinai theory math-ph/0312040 A. H. El Kinani, M. Daoud: Generalized coherent and intelligent states for exact solvable quantum systems math-ph/0312039 Shamgar Gurevich, Ronny Hadani: On Berry-Hannay Equivariant Quantization of the Torus math-ph/0312038 Anna Mikhailova, Boris Pavlov, Lev Prokhorov: Modelling of Quantum Networks hep-th/0312129 Ion I. Cotaescu, Mihai Visinescu: The induced representation of the isometry group of the Euclidean Taub-NUT space and new spherical harmonics quant-ph/0312096 C. Quesne, K.A. Penson, V.M. Tkachuk: Geometrical and physical properties of maths-type q-deformed coherent states for $0 1$ math-ph/0312037 Kouichi Takemura: On the Inozemtsev model hep-th/0312112 Amitabha Lahiri: Parallel transport on non-Abelian flux tubes cond-mat/0303425 W. K. Theumann: Mean-field dynamics of sequence processing neural networks with finite connectivity NA: Numerical Analysis ---------------------- math.NA/0312316 Nathanial P. Brown: Quasidiagonality and the finite section method math.NA/0312315 Nathanial P. Brown: AF embeddings and the numerical computation of spectra in irrational rotation algebras math.NA/0312296 Sergei V. Shabanov: Electromagnetic pulse propagation in passive media by path integral methods NT: Number Theory ----------------- math.NT/0312366 Enric Nart, Christophe Ritzenthaler: Non hyperelliptic curves of genus three over finite fields of characteristic two math.NT/0312359 Robin de Jong: On the Arakelov theory of elliptic curves math.NT/0312357 Robin de Jong: Arakelov invariants of Riemann surfaces math.NT/0312303 Aleksandar Ivi'c: On certain sums over ordinates of zeta zeros math.NT/0312255 Aleksandar Ivi'c: A note on the Laplace transform of the square in the circle problem OA: Operator Algebras --------------------- math.OA/0312348 Slawomir Klimek: Quantum Dynamical Systems with Quasi--Discrete Spectrum math.OA/0312300 Javier Parcet, Gilles Pisier: Non-commutative Khintchine type inequalities associated with free groups math.OA/0312286 Eberhard Kirchberg, Mikael Rordam: Purely infinite C*-algebras: ideal-preserving zero homotopies math.OA/0312283 Astrid an Huef, S. Kaliszewski, Iain Raeburn: Extending representations of subgroups and the duality of induction and restriction math.OA/0312275 Javier Parcet: Multi-indexed p-orthogonal sums in non-commutative Lebesgue spaces math.OA/0312261 Severino T. Melo: Norm closure of classical pseudodifferential operators does not contain Hormander's class math.OA/0312246 Javier Parcet: B-convex operator spaces math.OA/0312245 Jos'e Garc'ia-Cuerva, Javier Parcet: Quantized orthonormal systems: A non-commutative Kwapie'n theorem PR: Probability Theory ---------------------- math.PR/0312344 Noam Berger, Christopher Hoffman, Vladas Sidoravicius: Nonuniqueness for specs in $ell^{2+epsilon}$ math.PR/0312340 Mark Jerrum: On the approximation of one Markov chain by another math.PR/0312335 Carl Graham: Out of equilibrium functional central limit theorems for a large network where customers join the shortest of several queues math.PR/0312334 Carl Graham: A functional central limit theorem in equilibrium for a large network in which customers join the shortest of several queues math.PR/0312326 Roderich Tumulka: Some Jump Processes in Quantum Field Theory math.PR/0312314 Michael Barnsley, John E. Hutchinson, {O}rjan Stenflo: V-variable fractals and superfractals cond-mat/0312262 G.Oshanin, R.Voituriez: Random walk generated by random permutations of {1,2,3, ..., n+1} math.PR/0312298 Mikhail Menshikov, Dimitri Petritis, Serguei Popov: Bindweeds or random walks in random environments on multiplexed trees and their asympotics math.PR/0312294 Hans-Otto Georgii, Roderich Tumulka: Global Existence of Bell's Time-Inhomogeneous Jump Process for Lattice Quantum Field Theory math.PR/0312256 Balint Toth, Benedek Valko: Perturbation of singular equilibria of hyperbolic two-component systems: a universal hydrodynamic limit QA: Quantum Algebra ------------------- math.QA/0312336 David Hernandez: Representations of Quantum Affinizations and Fusion Product math.QA/0312330 Alexis Virelizier: Graded quantum groups math.QA/0312327 Masahiro Kasatani: Zeros of Symmetric Laurent Polynomials of Type $(BC)_n$ and Self-dual Koornwinder-Macdonald Polynomials Specialized at $t^{k+1}q^{r-1}=1$ math.QA/0312313 Markus Rosellen: On OPE-Algebras hep-th/0312159 Petr P. Kulish, Anton M. Zeitlin: Integrable Structure of Superconformal Field Theory and Quantum super-KdV Theory math.QA/0312289 Nicola Ciccoli, Fabio Gavarini: A quantum duality principle for subgroups and homogeneous spaces RA: Rings and Algebras ---------------------- math.RA/0312263 Peter Jorgensen: The Gorenstein projective modules are precovering RT: Representation Theory ------------------------- math.RT/0312338 G. Lusztig: Character sheaves on disconnected groups, IV math.RT/0312332 Melnikov Anna: The combinatorics of orbital varieties closures of nilpotent order 2 in sl(n) math.RT/0312299 Volker Heiermann: Une remarque sur le degre formel d'une serie discrete d'un groupe lineaire general p-adique math.RT/0312290 Melnikov Anna: Description of B-orbit closures of order 2 in upper-triangular matrices math.RT/0312270 Sergei Kerov, Grigori Olshanski, Anatoly Vershik: Harmonic analysis on the infinite symmetric group math.RT/0312244 Jos'e Garc'ia-Cuerva, Jos'e Manuel Marco, Javier Parcet: Sharp Fourier type and cotype with respect to compact semisimple Lie groups SG: Symplectic Geometry ----------------------- math.SG/0312355 E. Meinrenken: Witten's formulas for intersection pairings on moduli spaces of flat $G$-bundles math.SG/0312354 Paolo Lisca: On Symplectic Fillings of Lens Spaces math.SG/0312345 Lisa Jeffrey, Joon-Hyeok Song: A new proof of formulas for intersection numbers in q-Hamiltonian reduced spaces math.SG/0312274 Alan Weinstein: The Maslov Gerbe math.SG/0312265 Lev Buhovski: Homology of Lagrangian Submanifolds in Cotangent Bundles math.SG/0312252 Bertram Kostant: Minimal coadjoint orbits and symplectic induction SP: Spectral Theory ------------------- math.SP/0312267 Fritz Gesztesy, Konstantin A. Makarov: (Modified) Fredholm Determinants for math.SP/0312254 Kwang C. Shin: Trace Formulas for Non-Self-Adjoint Periodic Schrodinger Operators and some Applications -- / Greg Kuperberg (UC Davis) / / Visit the Math ArXiv Front at http://front.math.ucdavis.edu/ / * All the math that's fit to e-print * ==== The powerset topology for a powerset P(S) is defined as the topology produced by subbase sets of the form { A in P(S) | a in A }, { A in P(S) | a not in A } A base for this topology are the sets of the form { X | A subset X subset SB } for A,B finite subset S. A local base for U in P(S) are the sets of the form { X | A subset X subset SB } A finite subset U, B finite subset SU We have the following theorems: The powerset topology for P(S) is homeomorphic to the product topology of {0,1}^S, the space of characteristic functions. {0,1} has discrete topology. Thus P(S) is zero-dimensional compact Hausdorff. P(S) is 1st countable iff S countable iff P(S) is 2nd countable -- set theory limits In set theory, the limit of a sequence Aj of subsets of P(S) is defined as lim Aj = liminf Aj = limsup Aj provided liminf Aj = limsup Aj, where liminf Aj = /{ /{ Aj | j >= n } | n in N } limsup Aj = /{ /{ Aj | j >= n } | n in N } / /; cap cup; intersection union Where are such limits used? Do they show up in modern analysis? -- limit equivalence The powerset topology has the desired property A = set-lim Aj iff A = topology-lim Aj Hence when S is countable, the powerset topology can be described by set theory limits. When S is uncountable extend the definition of set theory limits to nets. Again is theorem A = set-net-lim Aj iff A = topology-net-lim Aj so the powerset topology can be described by set-net limits -- Questions Has any use of the powerset topology been made? Have these notions been extended or refined? Does this topic show up in modern analysis? Who else has consider the powerset topology? Comments most welcome as well as additions, refutations or requests for proofs. -- latticed ordered topological spaces As P(S) is the protype for complete complemented atomic distributive lattices or complete atomic Boolean algebras, the powerset topology can be extended to those spaces. Upon using 'subset' for the order <=, of P(S), the powerset topology presents a base of convex sets and the T_2-ordered property, that <= is closed subset P(S)xP(S). Thus it is an appropriately well mannered topology for a partially ordered set. ---- ==== For 1,2,3... omega, a sequence aj with limit a in S is an element f = (a1,a2,... a) = (aj,a) of S^(omega+1) with f(j) = aj for j < omega and f(omega) = a A topology for S assures (the limits of) A subset S^(omega+1) when for all (a1,a2,... a) in A, aj -> a in S The indiscrete topology is the smallest topology for S assuring A. If T is bunch of topologies that assure A, then sup T assures A. if (aj,a) in A, a in U some subbase set of sup T, then some tau in T with U in tau; aj -> a in tau aj eventually in U now use theorem aj -> a iff for all subbase sets U containing a, aj eventually in U Thus there's a largest topology for S assuring A. The sup of all topologies for S that assure A. -- coinduction Consider f:omega+1 -> S, f = (aj,a) For (aj,a) in A with aj -> a in S, let U be open nhood a. Then for f = (aj,a) in A, f^-1(U) is open. case a in U: omega in f^-1(U); aj eventually in U cofinite many j a in S We may ask when a topology for S models A, is it the largest topology that assures A ? This is may not be so. For example let S be uncountable and A all the eventually constant sequences with limit the eventual constant. There are two topologies for uncountable S that model this A. The discrete topology and the cocountable topology. -- open conjecture The powerset topology for a powerset P(S) is defined as the topology produced by subbase sets of the form { A in P(S) | a in A }, { A in P(S) | a not in A } A base for this topology are the sets of the form { X | A subset X subset SB } for A,B finite subset S. The powerset topology is know to be homeomorphic to the product topology {0,1}^S ({0,1} discrete), ie the space of characterist functions for the sets of P(S) In set theory, the limit of a sequence Aj of subsets of P(S) is defined as lim Aj = liminf Aj = limsup Aj provided liminf Aj = limsup Aj, where liminf Aj = /{ /{ Aj | j >= n } | n in N } limsup Aj = /{ /{ Aj | j >= n } | n in N } / /; cap cup; intersection union Does anyone know of applications for these 'set-limits' To continue, it is know for sequences Aj subset P(S) set-limit Aj = A iff topology-limit Aj = A ie, the power set topology models the sequences with set-limits of P(S) Now a topology for P(S) can be coinduced as above by all f in P(S)^(omega+1) for which f = (Aj,A) with set-limit Aj = A Let L be the collection of sequences with set-limits in P(S). Thus this coinduced topology for P(S) is the largest assuring L. Now it would seem that the powerset topology, could be homeomorphic to the coinduced topology. One may surmise this as the powerset topology models L. However as the example above showed, modeling may not imply being the largest that assures. So the conjecture: is the coinduced topology, the largest assuring L, homeomorphic to the powerset topology that models L? One way to establish this homeomorphism is to show the powerset topology is the largest assuring L. Any suggestions? ---- ==== > I would like an open statement by all 3 authors, in english, > as to whether they think they have proven P!=NP. based on g.sterten's > publicize their work without anyone endorsing it yet, they seem to be playing > coy or evasive or something-- one of the few sins in mathematical research. What evidence do you have that they went to the press first? The media can be very aggressive in following leads, and strange as it may seem, nowadays solving a famous math problem can be a hot story. An open statement would be nice, but I doubt we'll get one anytime soon. After all, their paper is supposed to be published in the near future. They may appear evasive to you, but I'm sure they are busy and have better things to do than enlighten sci.math.research readers. ==== Assume: f(z)=c^z, (c, z in C, not both 0) f^(n)(z)=[f(z), n=1, f(f^(n-1))(z), if n>1] The sequence {f^(n)(1), n in N} is chaotic for c=e^{t/e^t}, t=e^{alpha*Pi*i}, alpha irrational, while it converges (it's a m-cycle, coalescing into a 1-cycle eventually) if t=e^{2*k*Pi*i/n}, k in {0,1,2,...n-1}, some n in N. The process z->c^z always posesses the fixed point: c_0=e^{-LW(-Log(c))} (where LW is the prinipal branch of the LambertW function), however, in both cases above, |f '(c_0)|=1. Does anyone have any idea how one can go about proving convergence or divergence in the two cases above without resorting to the theory of iteration by Fatou and Julia? -- Ioannis Galidakis http://users.forthnet.gr/ath/jgal/ ------------------------------------------ Eventually, _everything_ is understandable ==== Given any z_0 in the punctured unit disk, one can find find f in Sigma such that f(z_0)=0. E.g. f(z)=1/z-1/z_0. Maybe you have omitted the condition a(0)=0? If so, then you can find the answer to your question in the 2nd edition of Goluzin's book. Theorem 3 in section 3 of Chapter IV describes the range of all possible values of f(z), where z is fixed and f varies over all functions in Sigma with a(0)=0. This range turns out to be a closed disk, and it should be easy to find (at least numerically) for what values of z this disk contains 0. Sincerely, Leonid Kovalev > Define Sigma to be the set of all univalent functions of the form: f(z) = 1/z + a(0) + a(1)*z + a(2)*z^2 + a(3)*z^3 + ... That is, f is one-to-one and complex analytic in the disk |z|<1 > except for a simple pole at the origin. If N is the supremum of all r such that an arbitrary f in Sigma > never vanishes on the disk |z|<=r, then Goluzin [1] proved that 0.86 < N <= sqrt(3)/2 = 0.8660254037... Has anyone seen a sharpening of this theorem anywhere? Goodman [2] referred to N as the radius of zeroness (although > he didn't mention Goluzin's particular problem). Help or > references would be gratefully received. For more background, > please visit http://pauillac.inria.fr/algo/bsolve/ and open the file entitled Radii in geometric function theory. Steve Finch > 1. G. M. Goluzin, Zur Theorie der schlichten konformen Abbildungen > (in Russian), Mat. Sbornik 42 (1935) 169--190. 2. A. W. Goodman, Univalent Functions, v. 2, Mariner, 1983, pp. 85 > & 110. _________________________________________________________________ > Tired of slow downloads? Compare online deals from your local high-speed > providers now. https://broadband.msn.com == What I like about adding to Rick Decker's example is that it makes it easy. In case you missed it, Rick Decker is a professor at Hamilton College (do a web search to find out about his school) who posted a *quadratic* in an attempt to refute some of my conclusions. Well, I found his example grabbed me, for various reasons, and finally found that I could modify it slightly to suit my purposes, and now confident in your understanding--even as undergrads--of independence between variables, I feel I can finally show you how I have been defending mathematics against people working to undermine it! Here's the quadratic that Decker gave: (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30xy + 2y^2) where the a's are roots of a^2 - (x - 1)a + 7(x^2 + x). And my modification was just to add in the variable y, to get (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) where the a's are roots of a^2 - (x - y)a + 7(x^2 + xy). Notice that x and y are *independent* variables, as I haven't related them to each other in any way. That second polynomial a^2 - (x - y)a + 7(x^2 + xy) is the polynomial being analyzed by considering (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2). You find it from 7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7y(x - y)(5) + 7^2 y^2 where what's on the right side is just a regrouping of terms from what's on the left, as you can find out by simplifying the right side. Notice that with (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) it *looks* like you have 7y and 7y as terms independent of x on the left side, but that contradicts with 7(2y^2) on the right! So what gives? Well the problem is that x is in the way, and a simple technique in analysis is to clear out one variable by setting it to 0. Now letting x=0, gives a(a + y) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, you have a_2(x,y) = b_2(x,y) - y, so (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). And now everything matches up correctly. Obviously, now the coefficients of the terms that have y but don't have x as a factor are revealed, and not surprisingly, 7y(2y) matches up with what you see on the right side with 7(2y^2). So if x does not equal y then it follows that (5a_1(x,y)/7 + y)(5b_2(x,y) + 2y) = 25x^2 + 30xy + 2y^2 but if x-y=0, then you have a^2 + 7(x^2 + xy), defining the a's, and you can directly calculate that a = +/- sqrt(-14)y, and picking a_1(y,y) = sqrt(-14)y, you have (5 sqrt(-14)y + 7y)(5 sqrt(-14)y + 7y) = 7(25y^2 + 30 y^2 + 2y^2). Now considering when x does not equal y it's easy enough to see that regardless of anyone's feelings on the matter, it's not possible for the terms *independent* of x, to actually have a dependency on x. So, with (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2) if (5b_2(x,y) + 2y) had any factors in common with 7, then those factors would have to be a factor of 2y, as that term is both visible and independent of x. So you can see my conclusion about what happens when you divide both sides by 7, when x does not equal y, follows from some basic algebra. What's fascinating is that you can also consider what happens when y=0 to separate out further, so letting y=0, I have a^2 - xa + 7x^2 = 0, so a = (1+/-sqrt(-27))x/2, and introducing c_1(x,y) and c_2(x,y) I then have (5(1-sqrt(-27))x/2 + c_1(x,y) + 7y)(5(1+sqrt(-27))x/2 + c_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). What's fascinating here is that completely broken is *any* idea that there's a variable factor in common with 7 that divides differently dependent on the value of x, as you have the lead coefficients 1-sqrt(-27))/2 and 1+sqrt(-27))/2 which are, of course, constant. What I like here is that quadratics are simpler than cubics to show the result, the example I modified was presented first by Rick Decker, a professor at Hamilton College, and my conclusion follows from independence between independent variables. That is to doubt me here you need to doubt that concept in mathematics itself--the concept of independence between variables. Of course, the problem for the old view on algebraic integers, is that if you supposed that you were in the ring of algebraic integers (notice I made no mention of a ring before now) then you run into a problem specific to that ring, where dividing both sides by 7 pushes you out of the ring of algebraic integers. It might seem esoteric and distant as a problem, but my discovery of that problem is actually a fascinating story, and now you can see how mathematicians *worldwide* answer the question: What do you do when your paradigm is forced to shift by a discoverer who thought outside of the box? Want more? Then read my other threads on sci.math or check out my blog archives: James Harris ==== > What I like about adding to Rick Decker's example is that it makes it > easy. In case you missed it, Rick Decker is a professor at Hamilton College > (do a web search to find out about his school) who posted a > *quadratic* in an attempt to refute some of my conclusions. Your conclusions have already been conclusively refuted by others. Rick's example appeared to be an attempt to simplify the arguments so that even YOU could understand your own errors. Apparently it is still not simple enough... -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com ==== [snip latest repetitive and flawed argument eschewing proofs in favor of leaping to conclusions] > Want more? Prefer less. Or none at all. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com ==== > What I like about adding to Rick Decker's example is that it makes it > easy. It seems that JSH finds it easier to make silly mistakes in quadratics that in cubics. ==== > It looks like there was more to adding in that y as a variable than > was realized as I saw the usual suspects replying--yet again--trying > to attack the mathematics, attacking algebra, as if they could still > convince all of you. What was being attacked was not mathematics, it was, as usual, JSH's non-mathematics. However, the other shoe fell as I set y=0 giving a^2 - xa + 7x^2 = 0, so a = (1+/-sqrt(-27))x/2, and introducing c_1(x,y) and c_2(x,y) I then have (5(1-sqrt(-27))x/2 + c_1(x,y) + 7y)(5(1+sqrt(-27))x/2 + c_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2) from my modification of Rick Decker's example. Now I guess that Decker is a good enough mathematician to recognize > that now it's completely over, but I wonder about the others. It was over a long time ago, but JSH doesn't realize yet that it went gainst him. Will they *dare* step out now despite the mathematics clearly showing > that I've been right all along? JSH has been consistently wrong, metaphorically at the top of his lungs, for 8 years now. It would be a shame to spoil such a consistent record. Fortunately, he hasn't. ==== [snip] > However, the other shoe fell... Sorry, that wasn't the other shoe. The other shoe will fall when you post your next Oops!. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com ==== Could someone please tell me in which branch of mathematics the following expression belongs, as well as the ffollowing problem : n [SIGMA] = (1^2) + (2^2) + ... (n^2) i=1 Prove that this expression is equal to the following expression n(n+1)(2n+1) / 6 (SIGMA is of course the mathematical symbol for summation) Lucas ==== : >Could someone please tell me in which branch of mathematics the following >expression belongs, as well as the ffollowing problem : n >[SIGMA] = (1^2) + (2^2) + ... (n^2) > i=1 It's algebra. Specifically, it's a simple series. >Prove that this expression is equal to the following expression >n(n+1)(2n+1) / 6 Look in the index of your textbook for induction or mathematical induction. Not only is that the method to be used, you'll probably find this problem used as an example. You might also look under series. (Some calculus textbooks also prove this, at the beginning of the section on definite integrals.) -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com My god, you can be insensitive sometimes. Sometimes? I must be slipping. -- Mrs Winterbourne (1996) ==== > Could someone please tell me in which branch of mathematics the following > expression belongs, as well as the ffollowing problem : n > [SIGMA] = (1^2) + (2^2) + ... (n^2) > i=1 Prove that this expression is equal to the following expression > n(n+1)(2n+1) / 6 (SIGMA is of course the mathematical symbol for summation) > Lucas The Calculus of Finite Differences is one area. Bill ==== if (5b_2(x,y) + 2y) had any factors in common with 7, then those > factors would have to be a factor of 2y, as that term is both visible > and independent of x. > No. The factors have to be factors of 2y only if they are factors of (5b_2(x,y) + 2y) for all x. We are dealing with a case where the GCD(7,(5b_2(x,y) + 2y)) varies with both x and y. Trivial example. Let b_2(x,y) = 0 if x=0, (7x/5 + y) if x not equal to 0. Then (5b_2(x,y) + 2y) = 2y if x=0, 7(x+y) if x not equal to 0. So (5b_2(x,y) + 2y) is divisible by 7 for all x not equal to 0. However, in general, 2y is not divisible by 7. -William Hughes ==== >What I like about adding to Rick Decker's example is that it makes it >easy. In case you missed it, Rick Decker is a professor at Hamilton College >(do a web search to find out about his school) who posted a >*quadratic* in an attempt to refute some of my conclusions. Well, I found his example grabbed me, for various reasons, and finally >found that I could modify it slightly to suit my purposes, and now >confident in your understanding--even as undergrads--of independence >between variables, I feel I can finally show you how I have been >defending mathematics against people working to undermine it! > ... See below. >Here's the quadratic that Decker gave: (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30xy + 2y^2) where the a's are roots of a^2 - (x - 1)a + 7(x^2 + x). And my modification was just to add in the variable y, to get (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) where the a's are roots of a^2 - (x - y)a + 7(x^2 + xy). Notice that x and y are *independent* variables, as I haven't related >them to each other in any way. That second polynomial a^2 - (x - y)a + 7(x^2 + xy) is the polynomial being analyzed by considering (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2). You find it from 7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7y(x - y)(5) + 7^2 y^2 where what's on the right side is just a regrouping of terms from >what's on the left, as you can find out by simplifying the right side. Notice that with (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) it *looks* like you have 7y and 7y as terms independent of x on the >left side, but that contradicts with 7(2y^2) on the right! So what gives? Well the problem is that x is in the way, and a simple technique in >analysis is to clear out one variable by setting it to 0. Now letting x=0, gives a(a + y) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, you have a_2(x,y) = b_2(x,y) - y, so (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). And now everything matches up correctly. > The previous expression was correct also. You are factoring your polynomial as if 5, not x or y, were the polynomial 7*((x^2 + x*y)*5^2 + (x - y)*y*5 + 7*y^2) = (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) makes the most sense. But keep reading ... >Obviously, now the coefficients of the terms that have y but don't >have x as a factor are revealed, and not surprisingly, 7y(2y) matches >up with what you see on the right side with 7(2y^2). So if x does not equal y then it follows that (5a_1(x,y)/7 + y)(5b_2(x,y) + 2y) = 25x^2 + 30xy + 2y^2 but if x-y=0, then you have a^2 + 7(x^2 + xy), defining the a's, and you can directly calculate that a = +/- sqrt(14)y, and picking >a_1(y,y) = sqrt(14)y, you have > This should be a = +/- sqrt(-14)y. But continue ... >(5 sqrt(14)y + 7y)(5 sqrt(14)y + 7y) = 7(25y^2 + 30 y^2 + 2y^2). Now considering when x does not equal y it's easy enough to see that >regardless of anyone's feelings on the matter, it's not possible for >the terms *independent* of x, to actually have a dependency on x. So, with (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2) if (5b_2(x,y) + 2y) had any factors in common with 7, then those >factors would have to be a factor of 2y, as that term is both visible >and independent of x. So you can see my conclusion about what happens when you divide both >sides by 7, when x does not equal y, follows from some basic algebra. What's fascinating is that you can also consider what happens when y=0 >to separate out further, so letting y=0, I have a^2 - xa + 7x^2 = 0, so a = (1+/-sqrt(-27))x/2, [*****] > Note this for reference below ... >and introducing c_1(x,y) and c_2(x,y) I then have (5(1-sqrt(-27))x/2 + c_1(x,y) + 7y)(5(1+sqrt(-27))x/2 + c_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). > Clearly here c_1(x, y) = 0 and c_2(x, y) = 5y. But keep going ... >What's fascinating here is that completely broken is *any* idea that >there's a variable factor in common with 7 that divides differently >dependent on the value of x, as you have the lead coefficients 1-sqrt(-27))/2 and 1+sqrt(-27))/2 which are, of course, constant. > OK, in [*****] above, you present two roots of the polynomial a^2 - xa + 7x^2, namely r1*x = (1 + sqrt(-27))*x/2, and r2*x = (1 - sqrt(-27))*x/2. The product of the roots of a quadratic must equal the constant term. Therefore (r1*x)*(r2*x) = 7*x^2, or, when x <> 0, r1*r2 = 7. Thus in this example, as it happens, the x's cancel out. And so, from what you have claimed, it must be true that r1 = 7 or r2 = 7. If one of them is 7, the other one must be coprime to 7. Now: does r1 = (1 + sqrt(-27))/2 look like 7 to you ? How about r2 = (1 - sqrt(-27))/2 ??? Which one of those is 7 ? Nora B. >What I like here is that quadratics are simpler than cubics to show >the result, the example I modified was presented first by Rick Decker, >a professor at Hamilton College, and my conclusion follows from >independence between independent variables. That is to doubt me here you need to doubt that concept in mathematics >itself--the concept of independence between variables. Of course, the problem for the old view on algebraic integers, is that >if you supposed that you were in the ring of algebraic integers >(notice I made no mention of a ring before now) then you run into a >problem specific to that ring, where dividing both sides by 7 pushes >you out of the ring of algebraic integers. It might seem esoteric and distant as a problem, but my discovery of >that problem is actually a fascinating story, and now you can see how >mathematicians *worldwide* answer the question: What do you do when your paradigm is forced to shift by a discoverer >who thought outside of the box? Want more? Then read my other threads on sci.math or check out my blog archives: James Harris ==== > >What I like about adding to Rick Decker's example is that it makes it >easy. > [snip ...] > >and introducing c_1(x,y) and c_2(x,y) I then have (5(1-sqrt(-27))x/2 + c_1(x,y) + 7y)(5(1+sqrt(-27))x/2 + c_2(x,y) + > 2y) = 7(25x^2 + 30xy + 2y^2). > Clearly here c_1(x, y) = 0 and c_2(x, y) = 5y. But keep going ... > Here I misinterpreted Harris's intentions regarding c_1(x,y) and c_2(x,y). The preceding statements were wrong with the consequence that my argument below is also incorrect. So far as I can tell Harris is not claiming that r1 or r2 is divisible by 7. It is still true however as explained at length in other threads that what Harris does claim is false. Nora B. >What's fascinating here is that completely broken is *any* idea that >there's a variable factor in common with 7 that divides differently >dependent on the value of x, as you have the lead coefficients 1-sqrt(-27))/2 and 1+sqrt(-27))/2 which are, of course, constant. > OK, in [*****] above, you present two roots of the > polynomial a^2 - xa + 7x^2, namely r1*x = (1 + sqrt(-27))*x/2, and r2*x = (1 - sqrt(-27))*x/2. The product of the roots of a quadratic must equal > the constant term. Therefore (r1*x)*(r2*x) = 7*x^2, or, when x <> 0, r1*r2 = 7. Thus in this example, as it happens, the x's cancel out. And so, from what you have claimed, it must be true that r1 = 7 > or r2 = 7. If one of them is 7, the other one must be coprime to 7. Now: does r1 = (1 + sqrt(-27))/2 look like 7 to you ? How about r2 = (1 - sqrt(-27))/2 ??? Which one of those is 7 ? Nora B. >What I like here is that quadratics are simpler than cubics to show >the result, the example I modified was presented first by Rick > Decker, >a professor at Hamilton College, and my conclusion follows from >independence between independent variables. That is to doubt me here you need to doubt that concept in > mathematics >itself--the concept of independence between variables. Of course, the problem for the old view on algebraic integers, is > that >if you supposed that you were in the ring of algebraic integers >(notice I made no mention of a ring before now) then you run into a >problem specific to that ring, where dividing both sides by 7 pushes >you out of the ring of algebraic integers. It might seem esoteric and distant as a problem, but my discovery of >that problem is actually a fascinating story, and now you can see how >mathematicians *worldwide* answer the question: What do you do when your paradigm is forced to shift by a discoverer >who thought outside of the box? Want more? Then read my other threads on sci.math or check out my blog archives: James Harris ==== > Now: does r1 = (1 + sqrt(-27))/2 look like 7 to you ? How about r2 = (1 - sqrt(-27))/2 ??? Which one of those is 7 ? On balance, I'd say r2 looks more like 7. Gib ==== >[...] And so, from what you have claimed, it must be true that r1 = 7 >or r2 = 7. If one of them is 7, the other one must be coprime to 7. Now: does r1 = (1 + sqrt(-27))/2 look like 7 to you ? How about r2 = (1 - sqrt(-27))/2 ??? Which one of those is 7 ? This is a trick question, right? My advice to James is not to answer, because actually neither one is equal to 7. Probably he should drop this thread at this point. > Nora B. >[...] >>It might seem esoteric and distant as a problem, but my discovery of >>that problem is actually a fascinating story, and now you can see how >>mathematicians *worldwide* answer the question: >>What do you do when your paradigm is forced to shift by a discoverer >>who thought outside of the box? >>[...] Makes you wonder if there's any way to get back inside that box once we've wandered out of it - things make a lot more sense inside... >>James Harris >> ************************ David C. Ullrich ==== > A while back I found these algebraic tools for finding out about roots > of a polynomial using another polynomial, but the problem is that > apparently (from the arguing) mathematicians had never found my > technique before, and not having had it, they'd come up with some > false assumptions about roots of polynomials. So not surprisingly, worldwide, mathematicians when contacted about my > work have tried to run away from it. I've tried to explain it before, > but it is just complicated enough that it seems to shoot past people, > but recently a Rick Decker, a professor at Hamilton College came up > with a *quadratic* example to try and refute my position, but I found > that by modifying his example, I could use it to explain, knowing that > quadratic are simple enough that there's less room for confusion. What does Professor Decker have to say concerning your use of his example? Or have you not checked back with him? David Ames Consider (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) where the a's are roots of > > a^2 - (x - y)a + 7(x^2 + xy). Notice that x and y are independent variables, as I haven't related > them to each other in any way. That second polynomial a^2 - (x - y)a + 7(x^2 + xy) is the polynomial being analyzed by considering (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2). You find it from 7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7y(x - y)(5) + 7^2 y^2 where what's on the right side is just a regrouping of terms from > what's on the left, as you can find out by simplifying the right side. Now letting x=0, gives a(a + y) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, you have a_2(x,y) = b_2(x,y) - y, so (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). Obviously, now the coefficients of the terms that have y but don't > have x as a factor are revealed, and not surprisingly, 7y(2y) matches > up with what you see on the right side with 7(2y^2). That's the simple result which mathematicians can't handle. So instead posters fighting the mathematical reality have made claims > that here would mean that that 7 divides as some kind of factor of x, > which is to say that depending on what value x has, 7 divides through > what's on the left hand side in different ways. Mathematicians, you see, believe that if the quadratic a^2 - (x - y)a + 7(x^2 + xy) doesn't have *integer* roots, then *both* roots would have to share > factors with 7. If you ask them why you'll probably get a lot of complicated > explanations in return, and they're very defensive on the issue from > what I've gathered, as well as being wrong. Here, for physicists it's easy to see that their wrong, as x and y are > INDEPENDENT of each other, so there's no way with (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). that (5b_2(x,y) + 2y) can in general have any factor of 7. It's just not mathematically possible, as if it did that factor would > show up multiplied times y, along with 2. Notice you *do* see that > factor showing up with (5a_1(x,y) + 7y). Importantly, there is a special case here when x=y, as then the > quadratic becomes a^2 - (x - y)a + 7(x^2 + xy) = a^2 + 14y^2. At that value the equation belongs to a different family as the > solution then is identical to that from a^2 + 7(x^2 + xy) when x=y. So why would mathematicians argue against such a simple result? Pride? Fear? Want more advanced polynomial factorization? Then check out my blog archives: James Harris ==== factoring by inspection, I like that characterization. so, when are you going to learn some numbertheory, Inspector Jimi? > Divide through by 7: 7*b1^2 - (x - y)*b1 + (x^2 + x*y) = 0. Now, in general this polynomial in b1 is irreducible, non-monic, > and primitive. These properties are all true for most > choices of x and y. For example, x = 5 and y = 3; x = 1, y = 0; > x = 1, y = -3; etc., etc.. This means (by a basic theorem in algebraic number theory) > that b1 in general cannot be an algebraic integer. This contradicts the statement that a1 is divisible by 7, > and therefore it also contradicts Harris's statement that > one of the roots must have no factors in common with 7. Now: is that a complicated explanation? Is it defensive? > Is it *wrong* ? > Here, for physicists it's easy to see that their wrong, as x and y are > INDEPENDENT of each other, so there's no way with (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). that (5b_2(x,y) + 2y) can in general have any factor of 7. It's just not mathematically possible, as if it did that factor would > show up multiplied times y, along with 2. Notice you *do* see that > factor showing up with (5a_1(x,y) + 7y). > Just proving what we have said repeatedly: the only way you have > ever learned to factor is by inspection. Good old high-school math. > Importantly, there is a special case here when x=y, as then the > quadratic becomes a^2 - (x - y)a + 7(x^2 + xy) = a^2 + 14y^2. At that value the equation belongs to a different family as the > solution then is identical to that from a^2 + 7(x^2 + xy), when x=y. > This is not a special case. This is just an example of what > was described above, where the polynomial is irreducible. --Give the Gift of Trickier Dick Cheeny -- out of office, at last! http://www.benfranklinbooks.com/ http://www.wlym.com/pages/music.html http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac http://www.wlym.com/PDF-68-76/CAM7606.pdf ==== factoring by inspection, I like that characterization. so, when are you going to learn some numbertheory, Inspector Jimi? > Divide through by 7: 7*b1^2 - (x - y)*b1 + (x^2 + x*y) = 0. Now, in general this polynomial in b1 is irreducible, non-monic, > and primitive. These properties are all true for most > choices of x and y. For example, x = 5 and y = 3; x = 1, y = 0; > x = 1, y = -3; etc., etc.. This means (by a basic theorem in algebraic number theory) > that b1 in general cannot be an algebraic integer. This contradicts the statement that a1 is divisible by 7, > and therefore it also contradicts Harris's statement that > one of the roots must have no factors in common with 7. Now: is that a complicated explanation? Is it defensive? > Is it *wrong* ? > Here, for physicists it's easy to see that their wrong, as x and y are > INDEPENDENT of each other, so there's no way with (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). that (5b_2(x,y) + 2y) can in general have any factor of 7. It's just not mathematically possible, as if it did that factor would > show up multiplied times y, along with 2. Notice you *do* see that > factor showing up with (5a_1(x,y) + 7y). > Just proving what we have said repeatedly: the only way you have > ever learned to factor is by inspection. Good old high-school math. > Importantly, there is a special case here when x=y, as then the > quadratic becomes a^2 - (x - y)a + 7(x^2 + xy) = a^2 + 14y^2. At that value the equation belongs to a different family as the > solution then is identical to that from a^2 + 7(x^2 + xy), when x=y. > This is not a special case. This is just an example of what > was described above, where the polynomial is irreducible. --Give the Gift of Trickier Dick Cheeny -- out of office, at last! http://www.benfranklinbooks.com/ http://www.wlym.com/pages/music.html http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac http://www.wlym.com/PDF-68-76/CAM7606.pdf ==== What I like about adding to Rick Decker's example is that it makes it easy. In case you missed it, Rick Decker is a professor at Hamilton College (do a web search to find out about his school) who posted a *quadratic* in an attempt to refute some of my conclusions. Well, I found his example grabbed me, for various reasons, and finally found that I could modify it slightly to suit my purposes, and now confident in your understanding--even as undergrads--of independence between variables, I feel I can finally show you how I have been defending mathematics against people working to undermine it! Here's the quadratic that Decker gave: (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30xy + 2y^2) where the a's are roots of a^2 - (x - 1)a + 7(x^2 + x). And my modification was just to add in the variable y, to get (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) where the a's are roots of a^2 - (x - y)a + 7(x^2 + xy). Notice that x and y are *independent* variables, as I haven't related them to each other in any way. That second polynomial a^2 - (x - y)a + 7(x^2 + xy) is the polynomial being analyzed by considering (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2). You find it from 7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7y(x - y)(5) + 7^2 y^2 where what's on the right side is just a regrouping of terms from what's on the left, as you can find out by simplifying the right side. Notice that with (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) it *looks* like you have 7y and 7y as terms independent of x on the left side, but that contradicts with 7(2y^2) on the right! So what gives? Well the problem is that x is in the way, and a simple technique in analysis is to clear out one variable by setting it to 0. Now letting x=0, gives a(a + y) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, you have a_2(x,y) = b_2(x,y) - y, so (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). And now everything matches up correctly. Obviously, now the coefficients of the terms that have y but don't have x as a factor are revealed, and not surprisingly, 7y(2y) matches up with what you see on the right side with 7(2y^2). So if x does not equal y then it follows that (5a_1(x,y)/7 + y)(5b_2(x,y) + 2y) = 25x^2 + 30xy + 2y^2 but if x-y=0, then you have a^2 + 7(x^2 + xy), defining the a's, and you can directly calculate that a = +/- sqrt(-14)y, and picking a_1(y,y) = sqrt(-14)y, you have (5 sqrt(-14)y + 7y)(5 sqrt(-14)y + 7y) = 7(25y^2 + 30 y^2 + 2y^2). Now considering when x does not equal y it's easy enough to see that regardless of anyone's feelings on the matter, it's not possible for the terms *independent* of x, to actually have a dependency on x. So, with (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2) if (5b_2(x,y) + 2y) had any factors in common with 7, then those factors would have to be a factor of 2y, as that term is both visible and independent of x. So you can see my conclusion about what happens when you divide both sides by 7, when x does not equal y, follows from some basic algebra. What's fascinating is that you can also consider what happens when y=0 to separate out further, so letting y=0, I have a^2 - xa + 7x^2 = 0, so a = (1+/-sqrt(-27))x/2, and introducing c_1(x,y) and c_2(x,y) I then have (5(1-sqrt(-27))x/2 + 5c_1(x,y) + 7y)(5(1+sqrt(-27))x/2 + 5c_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). What's fascinating here is that completely broken is *any* idea that there's a variable factor in common with 7 that divides differently dependent on the value of x, as you have the lead coefficients 1-sqrt(-27))/2 and 1+sqrt(-27))/2 which are, of course, constant. What I like here is that quadratics are simpler than cubics to show the result, the example I modified was presented first by Rick Decker, a professor at Hamilton College, and my conclusion follows from independence between independent variables. That is to doubt me here you need to doubt that concept in mathematics itself--the concept of independence between variables. Of course, the problem for the old view on algebraic integers, is that if you supposed that you were in the ring of algebraic integers (notice I made no mention of a ring before now) then you run into a problem specific to that ring, where dividing both sides by 7 pushes you out of the ring of algebraic integers. It might seem esoteric and distant as a problem, but my discovery of that problem is actually a fascinating story, and now you can see how mathematicians *worldwide* answer the question: What do you do when your paradigm is forced to shift by a discoverer who thought outside of the box? Want more? Then read my other threads on sci.math or check out my blog archives: James Harris ==== > What I like about adding to Rick Decker's example is that it makes it > easy. In case you missed it, Rick Decker is a professor at Hamilton College > (do a web search to find out about his school) who posted a > *quadratic* in an attempt to refute some of my conclusions. Well, I found his example grabbed me, for various reasons, and finally > found that I could modify it slightly to suit my purposes, and now > confident in your understanding--even as undergrads--of independence > between variables, I feel I can finally show you how I have been > defending mathematics against people working to undermine it! Here's the quadratic that Decker gave: (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30xy + 2y^2) where the a's are roots of > > a^2 - (x - 1)a + 7(x^2 + x). And my modification was just to add in the variable y, to get (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) where the a's are roots of > > a^2 - (x - y)a + 7(x^2 + xy). Notice that x and y are *independent* variables, as I haven't related > them to each other in any way. That second polynomial a^2 - (x - y)a + 7(x^2 + xy) is the polynomial being analyzed by considering (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2). You find it from 7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7y(x - y)(5) + 7^2 y^2 where what's on the right side is just a regrouping of terms from > what's on the left, as you can find out by simplifying the right side. Notice that with (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) it *looks* like you have 7y and 7y as terms independent of x on the > left side, but that contradicts with 7(2y^2) on the right! So what gives? Well the problem is that x is in the way, and a simple technique in > analysis is to clear out one variable by setting it to 0. Now letting x=0, gives a(a + y) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, you have a_2(x,y) = b_2(x,y) - y, so (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). And now everything matches up correctly. Obviously, now the coefficients of the terms that have y but don't > have x as a factor are revealed, and not surprisingly, 7y(2y) matches > up with what you see on the right side with 7(2y^2). So if x does not equal y then it follows that (5a_1(x,y)/7 + y)(5b_2(x,y) + 2y) = 25x^2 + 30xy + 2y^2 but if x-y=0, then you have a^2 + 7(x^2 + xy), defining the a's, and you can directly calculate that a = +/- sqrt(-14)y, and picking > a_1(y,y) = sqrt(-14)y, you have (5 sqrt(-14)y + 7y)(5 sqrt(-14)y + 7y) = 7(25y^2 + 30 y^2 + 2y^2). Now considering when x does not equal y it's easy enough to see that > regardless of anyone's feelings on the matter, it's not possible for > the terms *independent* of x, to actually have a dependency on x. So, with (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2) if (5b_2(x,y) + 2y) had any factors in common with 7, then those > factors would have to be a factor of 2y, as that term is both visible > and independent of x. So you can see my conclusion about what happens when you divide both > sides by 7, when x does not equal y, follows from some basic algebra. What's fascinating is that you can also consider what happens when y=0 > to separate out further, so letting y=0, I have a^2 - xa + 7x^2 = 0, so a = (1+/-sqrt(-27))x/2, and introducing c_1(x,y) and c_2(x,y) I then have (5(1-sqrt(-27))x/2 + 5c_1(x,y) + 7y)(5(1+sqrt(-27))x/2 + 5c_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). What's fascinating here is that completely broken is *any* idea that > there's a variable factor in common with 7 that divides differently > dependent on the value of x, as you have the lead coefficients 1-sqrt(-27))/2 and 1+sqrt(-27))/2 which are, of course, constant. > There are serious flaws with your argument as have been explained previously. For example: 1. If x = y, a_1(x) = sqrt(-14)*x, and this is neither coprime to 7 nor divisible by 7. 2. If x = 1 and y = 0, a_1(x) = (1 + sqrt(-27))/2, which is neither coprime to 7 nor divisible by 7. 3. If x = 2 and y = 1, a_1(x) = (1 + sqrt(-167))/2, which Keith Ramsay has shown by an explicit calculation is neither coprime to 7 nor divisible by 7. 4. There are more general arguments which you evidently do not understand which show the same as 1-3 for most choices of x and y. 5. There are other ways to factor 7 out of the Decker polynomial so that the coefficients of all the factors are algebraic integers, which is what you wanted in the first place. By other ways, I mean, other than as 7*1. These have been specified by both Dik Winter and me. But anyway ... Adding in the functions c_1(x,y) and c_2(x,y) does essentially nothing for your argument. It is still the case that you are claiming that if Decker's polynomial is factored as (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) you conclude that a_1(x,y) must be divisible by 7 and therefore a_2(x,y) must be coprime to 7. There is another slightly subtle flaw with your argument. You know that a_1(x,y) and a_2(x,y) are roots of a^2 - (x - y) + 7(x^2 + xy). There are two roots. Perhaps you choose as the 'negative' root, a_1(x,y) = ((x - y) - sqrt((x - y)^2 - 28*(x^2 + xy)))/2 and the other as the 'positive' root, a_2(x,y) = ((x - y) + sqrt((x - y)^2 - 28*(x^2 + xy)))/2. The choice is arbitrary. Your subsequent argument makes no use of whether the negative or positive root is chosen. They are, however, different *numbers*. You conclude that one of them (a_1(x,y), the 'negative' root) is divisible by 7, and the other (a_2(x,y), the 'positive' root) is coprime to 7. If you had chosen them oppositely, you would have come to the opposite conclusion: i.e., you would conclude that the 'negative' root is coprime to 7, while the 'positive' root is divisible by 7. They cannot both be true. This is not a matter of ambiguity about square roots. You can make the square root function well-defined for all positive arguments by saying that sqrt(z) is positive for all z > 0. In any case, however you define the square root, the two numbers here are *different*, and you argument makes no distinction. The arbitrariness here is simply a matter of labelling. You arbitrarily pick one of the roots to express in the form b_2(x,y) = a_2(x,y) + y and go from there. Switch the labels, you get the opposite conclusion. This tells you something important. It says your argument has to be wrong, even without having to look at the guts of it. The right conclusion thus has to be that both a_1(x,y) and a_2(x,y) are not coprime to 7 and both are not divisible by 7. This is actually a common way that mathematicians use to check that they have not made a mistake: if they have made an arbitrary choice somewhere along the way, would the arrive at a conclusion that they know to be false if another choice had been made? The same problem actually occurs with your argument about your cubic polynomial, except there you are arbitrarily choosing one of three roots a_1, a_2, and a_3 instead of one of two. You conclude that two of the three must be divisible by 7, and the other coprime to 7. Permute them and choose a different one to be replaced by b_3, etc., and you come to a different conclusion. Nora B. > What I like here is that quadratics are simpler than cubics to show > the result, the example I modified was presented first by Rick Decker, > a professor at Hamilton College, and my conclusion follows from > independence between independent variables. That is to doubt me here you need to doubt that concept in mathematics > itself--the concept of independence between variables. Of course, the problem for the old view on algebraic integers, is that > if you supposed that you were in the ring of algebraic integers > (notice I made no mention of a ring before now) then you run into a > problem specific to that ring, where dividing both sides by 7 pushes > you out of the ring of algebraic integers. It might seem esoteric and distant as a problem, but my discovery of > that problem is actually a fascinating story, and now you can see how > mathematicians *worldwide* answer the question: What do you do when your paradigm is forced to shift by a discoverer > who thought outside of the box? Want more? Then read my other threads on sci.math or check out my blog archives: James Harris ==== thank you, but how could I possible have missed it?... I mean, I was away from the Jimi Harris Experience for weeks, and that's teh first thing that I saw, a few days ago -- and repeatedly. nice marketing effort, I suppose, though, if it's just to keep the newbies on their toes. I mean, there's bound to be a Homeland Security issue in this exercise! but, yes; stick to the quadratic case, if it helps to lessen the ration of obfuscation in your work (the Ten-year Programme, Decade Two .-) > In case you missed it, Rick Decker is a professor at Hamilton College > (do a web search to find out about his school) who posted a > *quadratic* in an attempt to refute some of my conclusions. > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30xy + 2y^2) where the a's are roots of > > a^2 - (x - 1)a + 7(x^2 + x). And my modification was just to add in the variable y, to get (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) where the a's are roots of > > a^2 - (x - y)a + 7(x^2 + xy). Notice that x and y are *independent* variables, as I haven't related > them to each other in any way. > That is to doubt me here you need to doubt that concept in mathematics > itself--the concept of independence between variables. Of course, the problem for the old view on algebraic integers, is that > if you supposed that you were in the ring of algebraic integers > (notice I made no mention of a ring before now) then you run into a > problem specific to that ring, where dividing both sides by 7 pushes > you out of the ring of algebraic integers. > Want more? --Give the Gift of Trickier Dick Cheeny -- out of office, at last! http://www.benfranklinbooks.com/ http://www.wlym.com/pages/music.html http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac http://www.wlym.com/PDF-68-76/CAM7606.pdf ==== (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2) if (5b_2(x,y) + 2y) had any factors in common with 7, then those > factors would have to be a factor of 2y, as that term is both visible > and independent of x. Repeating this in a large number of threads will not make it true. This only holds if the factors are constant. In the present case the factors are not constant. -William Hughes ==== > What I like about adding to Rick Decker's example is that it makes it > easy. In case you missed it, Rick Decker is a professor at Hamilton College > (do a web search to find out about his school) who posted a > *quadratic* in an attempt to refute some of my conclusions. Well, I found his example grabbed me, for various reasons, and finally > found that I could modify it slightly to suit my purposes, and now > confident in your understanding--even as undergrads--of independence > between variables, I feel I can finally show you how I have been > defending mathematics against people working to undermine it! Well I made a mistake, as I added in that y variable wrong. Oh well. James Harris ==== [snip] > Well I made a mistake, as I added in that y variable wrong. Oh well. THAT'S the other shoe! -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com ==== >>What I like about adding to Rick Decker's example is that it makes it >>easy. >>In case you missed it, Rick Decker is a professor at Hamilton College >>(do a web search to find out about his school) who posted a >>*quadratic* in an attempt to refute some of my conclusions. >>Well, I found his example grabbed me, for various reasons, and finally >>found that I could modify it slightly to suit my purposes, and now >>confident in your understanding--even as undergrads--of independence >>between variables, I feel I can finally show you how I have been >>defending mathematics against people working to undermine it! > Well I made a mistake, as I added in that y variable wrong. Oh well. If you didn't trumpet your discoveries so loudly and abuse those who help you so crudely you'd get some sympathy. Instead you get scorn. Gib ==== [snip zillionth failed argument] James, your posting history makes it clear that mathematics is simply too difficult for you. Every time you leap triumphantly to a false conclusion, instead of acknowledging your error you announce that you have discovered some breakthrough new principle. That simply isn't the behavior of a rational person. Perhaps you should switch to some activity better suited to your capabilities, such as basket weaving, for example. (But do wear eye protectors!) -- == Using residue method to evaluate the following integral, LoopIntegrate[(3z+2)/(z^4+1)] along circle |z|=3... I used Cauchy Residue Theorem to compute the four residues at: sqrt(1/2)+sqrt(1/2)*i sqrt(1/2)-sqrt(1/2)*i -sqrt(1/2)+sqrt(1/2)*i -sqrt(1/2)-sqrt(1/2)*i Then add them up, multiply with 2*pi*i...and get the final result... The procedure was horrible... but the final answer is simple: ZERO! I wonder if there is any simpler way to sense out the answer should be ZERO without tedious residue computation? Please give me some help! -Walala ==== > Dear all, > > Using residue method to evaluate the following integral, > > LoopIntegrate[(3z+2)/(z^4+1)] along circle |z|=3... > > I used Cauchy Residue Theorem to compute the four residues at: > > sqrt(1/2)+sqrt(1/2)*i > sqrt(1/2)-sqrt(1/2)*i > -sqrt(1/2)+sqrt(1/2)*i > -sqrt(1/2)-sqrt(1/2)*i > > Then add them up, multiply with 2*pi*i...and get the final result... > > The procedure was horrible... but the final answer is simple: ZERO! > > I wonder if there is any simpler way to sense out the answer should > be ZERO without tedious residue computation? Yes. Instead of integrating about |z| = 3, integrate about |z| = R and let R -> infinity. Alternaively, use substitution z = 1/w in the orignal integral. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== I have a similar question regarding the bound in the Waring problem. For a fixed positive integer, k, what is the best bound discovered so far for n(k) where n(k) is such that every positive integer can be expressed as a sum of <= n(k) kth powers? Three Pearls of Number Theory constructs a bound for this problem, too. However, the construction is not particularly explicit, so I will have to do some more work before I can derive the Three Pearls bounds, even for small k. I'm sure that this bound is astronomical, too. Paul Epstein ==== Jamis Harres > Dear mathematicians, old friend JSH; the fact is you're all 100 miles far from the truth: I > know James Harris, and he is the most brilliant mathematician ever. > To tell the truth... well... I'm not sure James will appreciate... but > you can check the following link, where you'll find: 1¡) The social status of JSH. Now you can meet the myth ... http://www.jal.cc.il.us/math/faculty/jim.html :) They say you're not really a star until people start doing parodies of you. Well, maybe JSH has become a star at last. LH ==== Wanted: James Harris, amateur mathematician (discoverer), future porn star. Alive: 1$ Dead: 10$ Those prizes may change without prior notice, depending on the evolution of the porn industry. If you are interested in buying James Harris, any offers should be sent to jamis_harres@yahoo.com. Jamis Harres ==== This is the last message of a thread that has been only visible on The Math Forum, see http://mathforum.org/discuss/sci.math/t/521337 > I'm stuck on this problem that was posed on another >> message board. The problem is to find which positive >> integers n can be the result of a/b + b/c + c/a, where >> a, b, and c are positive integers. >Marcus I don't know why this question hasn't shown up on Usenet or >Google, but that may be why you've gotten so few answers. I >think the problem is well-known (though not to me). The thread seems to be invisible on Usenet and Google. Anyway, an approach is to assume gcd(a,b,c)=1, though a, b, >and c may not be pairwise relatively prime. Then it turns >out that we can find integers t,u,v that are pairwise >relatively prime, and for which a=t^2 u, b=u^2 v, c=v^2 t. >The same (t,u,v) also yields a triple a=t^2 v, b=v^2 u, >c=u^2 t, which is different if the t,u,v are different >(which can only occur with (1,1,1) and (1,1,2)). Anyway, n=(t^3+u^3+v^3)/(tuv). My program tries >u=1,2,...,50 (though it could easily go further on a better >Lisp). For each u it tries all 1 <= t <= u with >gcd(t,u)=1. For each t and u it tries all divisors >v|t^3+u^3 with v >= u. The result (sorted by n) is: n t u v > 3 1 1 1 > 5 1 1 2 > 6 1 2 3 > 9 2 3 7 > 10 5 7 18 > 13 9 13 38 > 14 2 7 13 > 17 5 18 37 > 18 13 42 95 > 19 1 5 9 > 21 2 13 21 > 26 9 38 91 > 29 27 43 182 > 30 2 21 31 [...] I had cross-posted the problem to The SeqFan Mailing List http://www.ccr.jussieu.fr/gmpib/seqfan/seqfan.html There were various contributions (e.g. by Hans Havermann, the terms given above. There is already a corresponding sequence in the On-Line Encyclopedia of Integer Sequences: http://www.research.att.com/projects/OEIS?Anum=A072716 This sequence will probably get an update during the next few days, with a conjectured continuation ...26,29,30,38,41,51,53,54,57,66,... A yet unsolved problem seems to be how far up you have to go with t,u,v to be sure that there are no further terms {t,u,v,n}={133, 2502, 4607, 74} Dan Hugo Pfoertner ==== > I'm stuck on this problem that was posed on another > message board. The problem is to find which positive > integers n can be the result of a/b + b/c + c/a, where > a, b, and c are positive integers. >>Anyway, an approach is to assume gcd(a,b,c)=1, though a, b, >>and c may not be pairwise relatively prime. Then it turns >>out that we can find integers t,u,v that are pairwise >>relatively prime, and for which a=t^2 u, b=u^2 v, c=v^2 t. >>Anyway, n=(t^3+u^3+v^3)/(tuv). Perhaps I'm simply retreading ground covered in another conversation about this problem but... That equation describes (for each n) an elliptic curve. You want a (positive) integer solution on the curve. Equivalently, you can set, say, v=1 and look for rational solutions to x^3+y^3+1 = n xy and then scale back to integers. You can write the elliptic curve in canonical form if you like; a transformation which works is x = -1/2*(X*n^2-2*Y)/n/(3*X-27*n+n^4) y = -1/2*(X*n^2+2*Y)/n/(3*X-27*n+n^4) giving the form Y^2 = X^3 + 1/4*n*(n^3-108)*X^2 - 9*n^2*(n-3)*(n^2+3*n+9)*X -n^3*(n^2+3*n+9)^2*(n-3)^2 We can then check for the existence of rational points. There is always torsion of order 3 corresponding to solutions with {t,u,v}={1,-1,0}. The curve is singular when n=3. It has rank 0 but additional torsion when n=5. It has rank 0 (no nontrivial rational points) when n=1,2,4,7,8,11,12,... It has rank 1 (infinitely many nontrivial solutions) when n=6,9,10,13,14, ... These are in agreement with the table posted, which shows solutions: >> n t u v >> 3 1 1 1 >> 5 1 1 2 >> 6 1 2 3 >> 9 2 3 7 >> 10 5 7 18 >> 13 9 13 38 >> 14 2 7 13 >> 17 5 18 37 This skips over are the cases n=15 and n=16, which also have rank 1. For example, for n=15, we have the generator (X,Y)=(11160, 1674000), which gives x = 1/3, y = -7/3, z = 1 (and then t=1,u=-7,v=3). You wanted _positive_ t,u,v, which means we need to find a multiple of the generator [*] which makes x>0 and y>0. I believe there are no such points. (All multiples of a point on the unbounded portion of an elliptic curve in normal form will also be in the unbounded portion.) The same thing happens for n=16,20,31,35,36,40,44,... The case n=41 is the first with rank 2, which doesn't mean very much in this setting except that there will be more than the usual number of solutions (t,u,v). (The next rank-2 case is n=69.) The behaviour for all the values of n < 67 mentioned in the previous post follows the pattern of one of the preceding paragraphs, except that I was unable to verify that there were no solutions when n=62 and n=64 (I suppose the ranks are one but my software found no generator for the curves in the little time I allotted, so I couldn't check to see whether the generator lay on the bounded part of the curve or not.) dave [*] Torsion doesn't help; adding one of the elements of order 3 just permutes t,u,v cyclically. And taking negatives on the elliptic curve corresponds to performing a permutation of odd sign on these three. ====