mm-3 === Subject:Subject: 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 Millerhttp://users.bigpond.net.au/amillerRetired 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 === Subject: 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.comYou need a nonlinear optimizing routine minimizing the penalty functionSum( ( y - Yest) ^2 ) where Yest is your formula,At the first clance your problem is well behaving so a simple algorithmcan do it.There are plenty of them. I can give my favorite ( F77 ) if needed. === Subject: See my address at www.estlab.comArto Huttunen === Subject:>> 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>> AllenAllen: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: 0.5873Good luck,>OUP === Subject: Dear Group, I`m trying to calculate a 2D fourier transform (2DFFT) ofa 2d exponential decay. Upon comparisions with a lorentzian form ( icompare the normalized values of the real and the imaginary partsseperately) i see that with my best efforts i can achieve an accuracyof only 10^(-3). Even this is possible only if go to a totallyunacceptable time step (In order to get a reasonable resolution in thefrequency domain i would need to do a 8192x8192 FFT calculation). I`m unable to locate where the error is coming from. Idon`t think my problem is related to zero padding as i can choose adecay const which ensure that the value goes to zero within theinterval. I would be really grateful to any comments and pointers asi`m really stuck. (i need an accuracy of at least 10^(-5) to proceed).Ravi === Subject: 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 === Subject: > 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 === Subject: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,...). === Subject: >>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 ofUrysohn's Lemma is that in a normal space, if A is closed,and U is an open superset of A, there is a continuous functionwhich is 0 on A and l outside of U.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University === Subject: UBLAS supports sparse matrix stuff fairly efficiently I understand.I have used it for small dense matrix work, for which it is not bad, butits maintainer seems more keen on the sparse support.http://www.boost.org/libs/numeric/ublas/doc/ overview.htmIt supports most data types, exploits lazy evaluation through some heavytemplate use etc. Some work has been done on LAPACK interfaces. === Subject: I am new to the world of stats and probabilty was wondering if anyonecould explain a wee bit about probabilty densityI am trying to write a piece of software quite like Granular SynthesisDemo 1 inhttp://www.dcs.gla.ac.uk/~jhw/audioclouds/ . I was wondering if anyonecould explain the difference between a probabilty distribution and aprobabilty density?And if you have a probabilty distribution can a density be obtainedfrom 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 amouse) how would you go about doing such a thingAny help would be greatly appericatedthanks Gregory === Subject: 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 meansF(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 understandwell is likely to lead to inferior (or incorrect) code. === Subject: >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 thingI would strong recommend getting hold of a good book on probabilityand reading up BEFORE coding. It is nearly 40 years since I startedto learn it, and nearly 35 since I taught it, so I am not a goodperson to give you references! The books I used would still be OK,but are probably hard to get :-)Nick Maclaren. === Subject: I have been trying to figure out why there is a /constrained/ quadraticprogamming step in SQP methods. I don't understand why the QP isn'tformulated without constraints, instead using Langrange multiplierssince it would be much easier to solve.If you can tell me the answer I'd be very grateful! I'm not sure if itsa stupid question or not, so I've described my thinking below:I currently understand that SQP is an extension of quasi-Newton methodsto constrained optimisation. A quasi-Newton method is used to maintain apositive definite approximation to the Hessian of the objectivefunction. Each iteration involves finding the optimum step direction bysolving the QP based on the second order approximation of the objectiveusing this approximate Hessian with first order approximations to theconstraint functions. The solution of this constrained QP yields thestep direction for both the parameters and the Lagrange multipliers forthe 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 anunconstrained problem using an approximated Hessian to the Lagrangian ofthe objective function. Since the quasi-Newton method yeilds a positivedefinite approximate Hessian this unconstrained QP will be triviallysolved with a Linear system. BFGS can give an inverse to the approximateHessian quite easily using a Choleski factorization which can also beeasily updated. This would mean that the Lagrange multipliers would haveto be updated outside of the QP solution, -but this would be fairlystraghtforward with an augmented Lagrangian method based on some penaltyfunction?I'm guessing that either (i) it would be hard/impossible to find themultipliers outside of the QP step for some reason, or (ii) it would bepossible but it is more efficient to find as part of a constrained QP.Very grateful for any help!Best,Josh. === Subject: How to integrate sinc(x) (without Fourier transformation) ??1018 === Subject: Good evenind, there's an interesting fact.. i'd like discuss with you:Terminology: differentiable = C^{infty}; cap = symbol of intersectionbetween 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 emptysetis assigned a continuous (or differentiable if it makes sense) applicationf_{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 arelation 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 additivecategory 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 Mwhose 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 thatf^{1}otimes ... otimes f^{k}in C^{infty}(M, xi^{1} otimes ... otimes xi^{k}) whereverf ^ {i} in C^{infty}(M_{i}, xi^{i}).I don't understand the reason for which ``the bundle structure ofxi^{1} otimes... otimes xi^{k} is characterized by the property thatf ^{1} otimes... otimes f^{k}in C^{ infty}(M, xi^{1}otimes ... otimes xi^{k}) whereverf^{i} in C^{infty}(M_{i}, xi^{i}) ''.May you to help me, please?Ternternnret@yahoo.it === Subject: 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 Mto C) such that omega = dz_1.dw_1 + .... + dz_n.dw_n ? Are there simplecounterexamples?Also, in the holomorphic case (omega holomorphic), I think the theoremmay actually hold. Any refferences?Please forward any postings to orenb@hans.math.upenn.edu---Oren Ben-Bassathttp://www.math.upenn.edu/~orenb === Subject: > 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 fordz_i to be *defined*, z_i needs to be differentiable with respect toeach component axis of M (= C^n). Generalizing, this probably leadsto omega being holomorphic, through the same arguments as are used inthe theory of functions of one complex variable.OTOH, given the coordinates z_i, one can construct a two-form bysumming the basis two-forms, sum_ij alpha_ij * dz_i * dz_j, for *any*complex-valued functions alpha_ij. If the alpha's aren'tdifferentiable, the sum probably won't be.Dale === Subject: 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 iff : (a,b) -> R is differentiable then the derivative f' satisfies theintermediate value property. (What _is_ the intermediate valueproperty 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 === Subject: Let f be a continuous complex function defined on the unit circle suchthat |f| is increasing and f(1)=1. Let g be a continuous real functiondefined 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 ??? === Subject: >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.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Here are this week's titles in the mathematics arXiv, available at: http://front.math.ucdavis.edu/ http://front.math.ucdavis.edu/submissionsThis week in the mathematics arXiv may be freely redistributedwith 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 crystallographymath.AC/0312333 Mordechai Katzman: The support of top graded local cohomology modulesmath.AC/0312302 Martin Lorenz: On the Cohen-Macaulay property of multiplicative invariantsAG: Algebraic Geometry--math.AG/0312362 space of semistable vector bundles over smooth curvesmath.AG/0312351 Giuseppe Borrelli: On the classification of surfaces of general type with non-birational bicanonical map and Du Val Double planesmath.AG/0312349 Charles Cadman: Using stacks to impose tangency conditions on curvesmath.AG/0312343 Spencer Bloch, H'el`ene Esnault: Local Fourier transforms and rigidity for D-modulesmath.AG/0312324 Shihoko Ishii: The arc space of a toric varietymath.AG/0312301 R. M. Miro-Roig, K. Ranestad: Intersection of ACM-curves in P^3math.AG/0312285 T. Shaska: Curves of genus 2 with (n, n)-decomposable jacobiansmath.AG/0312284 T. Shaska: Determining the automorphism group of a hyperelliptic curvemath.AG/0312280 Fabrizio Coiai, Yogish I. Holla: Extension of semistable principal bundles in positive characteristicmath.AG/0312278 Trond Stolen Gustavsen: On the cotangent cohomology of rational surface singularities with almost reduced fundamental cyclemath.AG/0312273 Nikita A. Karpenko: Holes in I^nmath.AG/0312271 D. Kaledin: Sommese Vanishing for non-compact manifoldsmath.AG/0312264 Carla Dionis: Stabilizers for nondegenerate matrices of boundary format and Steiner bundlesmath.AG/0312262 Geometrymath.AG/0312260 Adrian Langer: Semistable principal G-bundles in positive characteristicmath.AG/0312259 Fr'ed'eric Bihan: Asymptotiques de nombres de Betti d'hypersurfaces projectives r'eellesAP: Analysis of PDEsmath.AP/0312367 Christiaan C. Stolk: A pseudodifferential equation with damping for one-way wave propagation in inhomogeneous acoustic mediamath.AP/0312319 Michael Goldberg, Wilhelm Schlag: Scaling invariant smoothing estimates for the schroedinger flow in dimension threemath.AP/0312281 Kim Dang Phung: Nearly a polynomial decay rate for the wave equationmath.AP/0312276 M. L. Frankel, V. Roytburd: Finite-Dimensional Attractor for a Nonequilibrium Stefan Problem with Heat LossesAT: Algebraic Topology--math.AT/0312277 Martin Markl, Steve Shnider: Associahedra, cellular W-construction and products of $A_infty$-algebrasCA: Classical Analysis and ODEs---math.CA/0312353 F. Pakovich, N. Roytvarf, Y. Yomdin: Cauchy Type Integrals of Algebraic Functionsmath.CA/0312321 Julius Borcea: Hyperbolicity-preserving linear operators and Gaa rding-convexity of twisted root mapsmath.CA/0312320 D.V. Gorbachev, A.S. Manoshina: Relation between Tur'an extremum problem and van der Corput setsmath.CA/0312317 P. Chladek, L. Klapka: The Cauchy atlas on the manifold of all complete ODE solutionsmath.CA/0312312 N. M. Atakishiyev, A. U. Klimyk: A set of orthogonal polynomials, dual to alternative q-Charlier polynomialsmath.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 transformsmath.CA/0312248 Jos'e Manuel Marco, Javier Parcet: Taylor series for the Askey-Wilson operator and classical summation formulasCO: Combinatorics-math.CO/0312358 Masao Ishikawa, Masato Wakayama: Applications of Minor Summation Formula III, Plucker Relations, Lattice Paths and Pfaffian Identitiesmath.CO/0312350 Nicholas A. Loehr, Gregory S. Warrington, Herbert S. Wilf: The combinatorics of a three-line circulant determinantmath.CO/0312297 David Speyer, Lauren K. Williams: The tropical totally positive Grassmannianmath.CO/0312287 Tuerker Biyikoglu, Josef Leydold: Faber-Krahn Type Inequalities for Treesmath.CO/0312282 Jason Burns: Bijective Proofs for Enumerative Properties of Ferrers Graphsmath.CO/0312272 Animesh Datta: Turbo Codes over the Real Fieldmath.CO/0312250 Ezra Miller: Alternating formulas for K-theoretic quiver polynomialsmath.CO/0312247 Jos'e Manuel Marco, Javier Parcet: A new approach to the theory of classical hypergeometric polynomialsCT: Category Theory---math.CT/0312339 Volodymyr Lyubashenko, Oleksandr Manzyuk: Free ${A}_infty$-categoriesCV: 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 probabilityDG: Differential Geometry-math.DG/0312364 D.E.Blair, J.Davidov, O.Mushkarov: Hyperbolic Twistor Spacesmath.DG/0312325 A. M. Ionescu, G. E. Vilcu: A note on para-quaternion manifoldshep-th/0312154 Paolo Aschieri, Luigi Cantini, Branislav Jurco: Nonabelian Bundle Gerbes, their Differential Geometry and Gauge Theorymath.DG/0312318 Jeff Viaclovsky, Gang Tian: Moduli spaces of critical Riemannian metrics in dimension fourgr-qc/0312047 Hubert L. Bray, Piotr T. Chrusciel: The Penrose Inequalitynlin.SI/0312030 B.G.Konopelchenko, W.K.Schief: Conformal geometry of the (discrete) Schwarzian Davey-Stewartson II hierarchymath.DG/0312251 Haibao Duan: A Problem of Hsiang-Palais-Terng on isoparametric submanifoldsmath.DG/0312249 C. Dunn, P. Gilkey, R. Ivanova, S. Nikcevic: The Spectral Geometry of theDS: Dynamical Systems-math.DS/0312361 B. Demir, V. Dzhafarov, S. Kocak, M. Ureyen: Restrictions of Harmonic Functions on the Sierpinski Gasket to Segmentsmath.DS/0312356 F'ethi Grabsi, James Montaldi, Juan-Pablo Ortega: Bifurcation and forced symmetry breaking in Hamiltonian systemsmath.DS/0312346 Alex Furman: Outer automorphism groups of some ergodic equivalence relationsmath.DS/0312328 JR Chazottes, F. Durand: Local rates of Poincare recurrence for rotations and weak mixingmath.DS/0312323 Claire Moura: On the multiplicity of the hyperelliptic integralsmath.DS/0312306 Volodymyr Nekrashevych: Iterated Monodromy Groupsmath.DS/0312291 J.-R. Chazottes, R. Leplaideur: Birkhoff averages of Poincare cycles for Axiom-A diffeomorphismsmath.DS/0312279 Wolf Jung: Homeomorphisms of the Mandelbrot SetFA: Functional Analysis---math.FA/0312365 Gelu Popescu: Entropy and multivariable interpolationmath.FA/0312341 Kamthorn Chailuek, Wicharn Lewkeeratiyutkul: A Pointwise Bound for a Holomorphic Function which is Square-Integrable with Respect to an Exponential Density FunctionGM: General Mathematics---math.GM/0312360 Paola Cattabriga: Beyond Uncountablemath.GM/0312309 Craig Alan Feinstein: The Collatz 3n+1 Conjecture is UnprovableGN: General Topologymath.GN/0312288 R. N. Gumerov: On finite-sheeted covering mappings onto solenoidsGR: Group Theorymath.GR/0312352 Robert W. Baddeley, Cheryl E. Praeger, Csaba Schneider: Innately transitive subgroups of wreath products in product actionmath.GR/0312342 Rafael Villarroel-Flores: Group Orders That Imply Existence Of Nontrivial Normal p-Subgroupsmath.GR/0312331 Sean Cleary, Jennifer Taback: Metric properties of the lamplighter group as an automata groupmath.GR/0312257 Michael Mueger: On the center of a compact groupGT: Geometric Topology--math.GT/0312363 Boris A. Springborn: Variational principles for circle patternsmath.GT/0312347 Blake Mellor: Intersection Graphs for String Linksmath.GT/0312337 Alexis Virelizier: Kirby elements and quantum invariantsmath.GT/0312329 Gennaro Amendola, Bruno Martelli: Non-orientable 3-manifolds of complexity up to 7math.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-Manifoldsmath.GT/0312295 Greg Friedman: All frame-spun knots are slicemath.GT/0312269 A. V. Karasev: On two problems in extension theorymath.GT/0312266 Brendan Owens, Saso Strle: A characterisation of the Z^(n-1) + 3Z lattice and applications to rational homology spheresmath.HO/0312293 Josh Isralowitz: Classical Lebesgue Integration Theorems for the Riemann IntegralKT: K-Theory and Homology-math.KT/0312305 Ralf Meyer: The cyclic homology and K-theory of certain adelic crossed productsmath.KT/0312292 Ralf Meyer, Ryszard Nest: The Baum-Connes Conjecture via Localization of CategoriesLO: Logic-math.LO/0312308 Arnold W. Miller: The gamma - Borel conjectureMG: Metric Geometry---math.MG/0312268 Alexander Barvinok, Grigoriy Blekherman: Convex Geometry of Orbitsmath.MG/0312253 Ezra Miller, Igor Pak: Metric combinatorics of convex polyhedra: cut loci and nonoverlapping unfoldingsMP: Mathematical Physicsmath-ph/0312047 Thomas H. Otway: Anomalies of the tidesmath-ph/0312046 Simon P. Eveson, Christopher J. Fewster, Rainer Verch: Quantum inequalities in quantum mechanicsmath-ph/0312045 M. Hartmann, G. Mahler, O. Hess: Gaussian quantum fluctuations inmath-ph/0312044 Anna Jencova: Geodesic distances on density matriceshep-th/0312186 Giovanni Arcioni, Claudio Dappiaggi: Holography in asymptotically flat space-times and the BMS groupmath-ph/0312043 Romuald A. Janik, Waldemar Wieczorek: Multiplying unitary random matrices - universality and spectral propertieshep-th/0312168 S.L. Lukyanov, E.S. Vitchev, A.B. Zamolodchikov: Integrable Model of Boundary Interaction: The Papercliphep-th/0312158 Petr P. Kulish, Anton M. Zeitlin: Group Theoretical Structure and Inverse Scattering Method for super-KdV Equationhep-th/0312116 Euler-Heisenberg Effective Actionscond-mat/0312361 Roberto Franzosi, Marco Pettini: A Theorem on the origin of Phase Transitionsmath-ph/0312042 Alberto De Sole, Victor Kac: Freely generated vertex algebras and non-linear Lie conformal algebrasmath-ph/0312041 Marek Biskup, Christian Borgs, Jennifer T. Chayes, Roman Kotecky: Partition function zeros at first-order phase transitions: Pirogov-Sinai theorymath-ph/0312040 A. H. El Kinani, M. Daoud: Generalized coherent and intelligent states for exact solvable quantum systemsmath-ph/0312039 Shamgar Gurevich, Ronny Hadani: On Berry-Hannay Equivariant Quantization of the Torusmath-ph/0312038 Anna Mikhailova, Boris Pavlov, Lev Prokhorov: Modelling of Quantum Networkshep-th/0312129 Ion I. Cotaescu, Mihai Visinescu: The induced representation of the isometry group of the Euclidean Taub-NUT space and new spherical harmonicsquant-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 modelhep-th/0312112 Amitabha Lahiri: Parallel transport on non-Abelian flux tubescond-mat/0303425 W. K. Theumann: Mean-field dynamics of sequence processing neural networks with finite connectivityNA: Numerical Analysis--math.NA/0312316 Nathanial P. Brown: Quasidiagonality and the finite section methodmath.NA/0312315 Nathanial P. Brown: AF embeddings and the numerical computation of spectra in irrational rotation algebrasmath.NA/0312296 Sergei V. Shabanov: Electromagnetic pulse propagation in passive media by path integral methodsNT: Number Theory-math.NT/0312366 Enric Nart, Christophe Ritzenthaler: Non hyperelliptic curves of genus three over finite fields of characteristic twomath.NT/0312359 Robin de Jong: On the Arakelov theory of elliptic curvesmath.NT/0312357 Robin de Jong: Arakelov invariants of Riemann surfacesmath.NT/0312303 Aleksandar Ivi'c: On certain sums over ordinates of zeta zerosmath.NT/0312255 Aleksandar Ivi'c: A note on the Laplace transform of the square in the circle problemOA: Operator Algebras-math.OA/0312348 Slawomir Klimek: Quantum Dynamical Systems with Quasi--Discrete Spectrummath.OA/0312300 Javier Parcet, Gilles Pisier: Non-commutative Khintchine type inequalities associated with free groupsmath.OA/0312286 Eberhard Kirchberg, Mikael Rordam: Purely infinite C*-algebras: ideal-preserving zero homotopiesmath.OA/0312283 Astrid an Huef, S. Kaliszewski, Iain Raeburn: Extending representations of subgroups and the duality of induction and restrictionmath.OA/0312275 Javier Parcet: Multi-indexed p-orthogonal sums in non-commutative Lebesgue spacesmath.OA/0312261 Severino T. Melo: Norm closure of classical pseudodifferential operators does not contain Hormander's classmath.OA/0312246 Javier Parcet: B-convex operator spacesmath.OA/0312245 Jos'e Garc'ia-Cuerva, Javier Parcet: Quantized orthonormal systems: A non-commutative Kwapie'n theoremPR: 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 anothermath.PR/0312335 Carl Graham: Out of equilibrium functional central limit theorems for a large network where customers join the shortest of several queuesmath.PR/0312334 Carl Graham: A functional central limit theorem in equilibrium for a large network in which customers join the shortest of several queuesmath.PR/0312326 Roderich Tumulka: Some Jump Processes in Quantum Field Theorymath.PR/0312314 Michael Barnsley, John E. Hutchinson, {O}rjan Stenflo: V-variable fractals and superfractalscond-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 asympoticsmath.PR/0312294 Hans-Otto Georgii, Roderich Tumulka: Global Existence of Bell's Time-Inhomogeneous Jump Process for Lattice Quantum Field Theorymath.PR/0312256 Balint Toth, Benedek Valko: Perturbation of singular equilibria of hyperbolic two-component systems: a universal hydrodynamic limitQA: Quantum Algebra---math.QA/0312336 David Hernandez: Representations of Quantum Affinizations and Fusion Productmath.QA/0312330 Alexis Virelizier: Graded quantum groupsmath.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-Algebrashep-th/0312159 Petr P. Kulish, Anton M. Zeitlin: Integrable Structure of Superconformal Field Theory and Quantum super-KdV Theorymath.QA/0312289 Nicola Ciccoli, Fabio Gavarini: A quantum duality principle for subgroups and homogeneous spacesRA: Rings and Algebras--math.RA/0312263 Peter Jorgensen: The Gorenstein projective modules are precoveringRT: Representation Theory-math.RT/0312338 G. Lusztig: Character sheaves on disconnected groups, IVmath.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-adiquemath.RT/0312290 Melnikov Anna: Description of B-orbit closures of order 2 in upper-triangular matricesmath.RT/0312270 Sergei Kerov, Grigori Olshanski, Anatoly Vershik: Harmonic analysis on the infinite symmetric groupmath.RT/0312244 Jos'e Garc'ia-Cuerva, Jos'e Manuel Marco, Javier Parcet: Sharp Fourier type and cotype with respect to compact semisimple Lie groupsSG: Symplectic Geometry---math.SG/0312355 E. Meinrenken: Witten's formulas for intersection pairings on moduli spaces of flat $G$-bundlesmath.SG/0312354 Paolo Lisca: On Symplectic Fillings of Lens Spacesmath.SG/0312345 Lisa Jeffrey, Joon-Hyeok Song: A new proof of formulas for intersection numbers in q-Hamiltonian reduced spacesmath.SG/0312274 Alan Weinstein: The Maslov Gerbemath.SG/0312265 Lev Buhovski: Homology of Lagrangian Submanifolds in Cotangent Bundlesmath.SG/0312252 Bertram Kostant: Minimal coadjoint orbits and symplectic inductionSP: Spectral Theory---math.SP/0312267 Fritz Gesztesy, Konstantin A. Makarov: (Modified) Fredholm Determinants formath.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 * === Subject: The powerset topology for a powerset P(S) is definedas 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 SUWe have the following theorems:The powerset topology for P(S) is homeomorphic to the producttopology 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 limitsIn set theory, the limit of a sequence Ajof subsets of P(S) is defined as lim Aj = liminf Aj = limsup Ajprovided liminf Aj = limsup Aj, where liminf Aj = /{ /{ Aj | j >= n } | n in N } limsup Aj = /{ /{ Aj | j >= n } | n in N } / /; cap cup; intersection unionWhere are such limits used? Do they show up in modern analysis?-- limit equivalenceThe powerset topology has the desired property A = set-lim Aj iff A = topology-lim AjHence when S is countable, the powerset topology can be describedby set theory limits. When S is uncountable extend the definitionof set theory limits to nets. Again is theorem A = set-net-lim Aj iff A = topology-net-lim Ajso the powerset topology can be described by set-net limits-- QuestionsHas 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 spacesAs P(S) is the protype for complete complemented atomic distributivelattices or complete atomic Boolean algebras, the powerset topology can beextended to those spaces.Upon using 'subset' for the order <=, of P(S), the powerset topologypresents 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 manneredtopology for a partially ordered set. === Subject: 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) = aA topology for S assures (the limits of) A subset S^(omega+1) when for all (a1,a2,... a) in A, aj -> a in SThe 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 Unow use theorem aj -> a iff for all subbase sets U containing a, aj eventually in UThus there's a largest topology for S assuring A.The sup of all topologies for S that assure A.-- coinductionConsider 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 SWe 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 allthe 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 conjectureThe powerset topology for a powerset P(S) is definedas 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 Ajof subsets of P(S) is defined as lim Aj = liminf Aj = limsup Ajprovided liminf Aj = limsup Aj, where liminf Aj = /{ /{ Aj | j >= n } | n in N } limsup Aj = /{ /{ Aj | j >= n } | n in N } / /; cap cup; intersection unionDoes 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 = Aie, 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 = ALet 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 homeomorphicto 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 powersettopology is the largest assuring L. Any suggestions? === Subject: > 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 mediacan 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 anytimesoon. After all, their paper is supposed to be published in the nearfuture. They may appear evasive to you, but I'm sure they are busy andhave better things to do than enlighten sci.math.research readers. === Subject: 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 bothcases above, |f '(c_0)|=1.Does anyone have any idea how one can go about proving convergence ordivergence in the two cases above without resorting to the theory ofiteration by Fatou and Julia?--Ioannis Galidakishttp://users.forthnet.gr/ath/jgal/--- ---Eventually, _everything_ is understandable === Subject: Given any z_0 in the punctured unit disk, one can find find f in Sigmasuch that f(z_0)=0. E.g. f(z)=1/z-1/z_0. Maybe you have omitted thecondition a(0)=0?If so, then you can find the answer to your question in the 2nd editionof Goluzin's book. Theorem 3 in section 3 of Chapter IV describes therange of all possible values of f(z), where z is fixed and f varies overall functions in Sigma with a(0)=0. This range turns out to be a closeddisk, and it should be easy to find (at least numerically) for whatvalues 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 iteasy.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 finallyfound that I could modify it slightly to suit my purposes, and nowconfident in your understanding--even as undergrads--of independencebetween variables, I feel I can finally show you how I have beendefending 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 relatedthem to each other in any way.That second polynomiala^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 from7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7y(x - y)(5) + 7^2 y^2where what's on the right side is just a regrouping of terms fromwhat'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 theleft 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 inanalysis 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 havea_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'thave x as a factor are revealed, and not surprisingly, 7y(2y) matchesup 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^2but if x-y=0, then you havea^2 + 7(x^2 + xy), defining the a's, and you can directly calculate that a = +/- sqrt(-14)y, and pickinga_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 thatregardless of anyone's feelings on the matter, it's not possible forthe 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 thosefactors would have to be a factor of 2y, as that term is both visibleand independent of x.So you can see my conclusion about what happens when you divide bothsides 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=0to separate out further, so letting y=0, I havea^2 - xa + 7x^2 = 0,soa = (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 thatthere's a variable factor in common with 7 that divides differentlydependent on the value of x, as you have the lead coefficients1-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 showthe result, the example I modified was presented first by Rick Decker,a professor at Hamilton College, and my conclusion follows fromindependence between independent variables.That is to doubt me here you need to doubt that concept in mathematicsitself--the concept of independence between variables.Of course, the problem for the old view on algebraic integers, is thatif 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 aproblem specific to that ring, where dividing both sides by 7 pushesyou out of the ring of algebraic integers.It might seem esoteric and distant as a problem, but my discovery ofthat problem is actually a fascinating story, and now you can see howmathematicians *worldwide* answer the question:What do you do when your paradigm is forced to shift by a discovererwho thought outside of the box?Want more?Then read my other threads on sci.math or check out my blog archives:James Harris === Subject: > 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'sexample appeared to be an attempt to simplify the arguments so that evenYOU could understand your own errors. Apparently it is still not simpleenough...--There are two things you must never attempt to prove: the unprovable -- andthe obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Subject: [snip latest repetitive and flawed argument eschewing proofs in favor ofleaping to conclusions]> Want more?Prefer less. Or none at all.--There are two things you must never attempt to prove: the unprovable -- andthe obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Subject: > 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. === Subject: > 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 givinga^2 - xa + 7x^2 = 0,soa = (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. === Subject: [snip]> However, the other shoe fell...Sorry, that wasn't the other shoe. The other shoe will fall when you postyour 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 === Subject: Could someone please tell me in which branch of mathematics the followingexpression belongs, as well as the ffollowing problem : n[SIGMA] = (1^2) + (2^2) + ... (n^2) i=1Prove that this expression is equal to the following expressionn(n+1)(2n+1) / 6(SIGMA is of course the mathematical symbol for summation)Lucas === Subject: 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=1It's algebra. Specifically, it's a simple series.>Prove that this expression is equal to the following expression>n(n+1)(2n+1) / 6Look 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.comMy god, you can be insensitive sometimes.Sometimes? I must be slipping. -- Mrs Winterbourne (1996) === Subject: > 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)> LucasThe Calculus of Finite Differences is one area.Bill === Subject: 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 casewhere 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 === Subject: >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, andfinally>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 ofa^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 ofa^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 polynomiala^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 from7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7y(x - y)(5) + 7^2 y^2where 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 rightside.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, givesa(a + y) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, you havea_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 factoringyour 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^2but if x-y=0, then you havea^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 wheny=0>to separate out further, so letting y=0, I havea^2 - xa + 7x^2 = 0,soa = (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 coefficients1-sqrt(-27))/2 and 1+sqrt(-27))/2which 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 equalthe 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 = 7or 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 RickDecker,>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 inmathematics>itself--the concept of independence between variables.Of course, the problem for the old view on algebraic integers, isthat>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 === Subject: >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. Sofar as I can tell Harris is not claiming that r1 or r2 is divisibleby 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 coefficients1-sqrt(-27))/2 and 1+sqrt(-27))/2which 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 === Subject: > 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 === Subject: >[...] 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, becauseactually neither one is equal to 7. Probably he shoulddrop 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 boxonce we've wandered out of it - things make a lot more senseinside...>>James Harris>>**David C. Ullrich === Subject: > 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 AmesConsider(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 polynomiala^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 from7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7y(x - y)(5) + 7^2 y^2where 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 havea_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 quadratica^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 becomesa^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 froma^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 === Subject: 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 becomesa^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 froma^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.htmlhttp://www.rand.org/publications/randreview/issues/ rr.12.00/http://members.tripod.com/~american_almanachttp:// www.wlym.com/PDF-68-76/CAM7606.pdf === Subject: 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 becomesa^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 froma^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.htmlhttp://www.rand.org/publications/randreview/issues/ rr.12.00/http://members.tripod.com/~american_almanachttp:// www.wlym.com/PDF-68-76/CAM7606.pdf === Subject: What I like about adding to Rick Decker's example is that it makes iteasy.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 finallyfound that I could modify it slightly to suit my purposes, and nowconfident in your understanding--even as undergrads--of independencebetween variables, I feel I can finally show you how I have beendefending 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 relatedthem to each other in any way.That second polynomiala^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 from7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7y(x - y)(5) + 7^2 y^2where what's on the right side is just a regrouping of terms fromwhat'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 theleft 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 inanalysis 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 havea_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'thave x as a factor are revealed, and not surprisingly, 7y(2y) matchesup 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^2but if x-y=0, then you havea^2 + 7(x^2 + xy), defining the a's, and you can directly calculate that a = +/- sqrt(-14)y, and pickinga_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 thatregardless of anyone's feelings on the matter, it's not possible forthe 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 thosefactors would have to be a factor of 2y, as that term is both visibleand independent of x.So you can see my conclusion about what happens when you divide bothsides 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=0to separate out further, so letting y=0, I havea^2 - xa + 7x^2 = 0,soa = (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 thatthere's a variable factor in common with 7 that divides differentlydependent on the value of x, as you have the lead coefficients1-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 showthe result, the example I modified was presented first by Rick Decker,a professor at Hamilton College, and my conclusion follows fromindependence between independent variables.That is to doubt me here you need to doubt that concept in mathematicsitself--the concept of independence between variables.Of course, the problem for the old view on algebraic integers, is thatif 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 aproblem specific to that ring, where dividing both sides by 7 pushesyou out of the ring of algebraic integers.It might seem esoteric and distant as a problem, but my discovery ofthat problem is actually a fascinating story, and now you can see howmathematicians *worldwide* answer the question:What do you do when your paradigm is forced to shift by a discovererwho thought outside of the box?Want more?Then read my other threads on sci.math or check out my blog archives:James Harris === Subject: > 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 polynomiala^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 from7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7y(x - y)(5) + 7^2 y^2where 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 havea_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^2but if x-y=0, then you havea^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 havea^2 - xa + 7x^2 = 0,soa = (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 coefficients1-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 essentiallynothing for your argument. It is still the case that you areclaiming 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 thereforea_2(x,y) must be coprime to 7. There is another slightly subtle flaw with your argument. You knowthat 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)))/2and 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 nouse of whether the negative or positive root is chosen. They are, however, different *numbers*. You conclude thatone 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 tothe opposite conclusion: i.e., you would conclude that the'negative' root is coprime to 7, while the 'positive' rootis divisible by 7. They cannot both be true. This is not a matter of ambiguity about square roots. Youcan make the square root function well-defined for all positivearguments by saying that sqrt(z) is positive for all z > 0.In any case, however you define the square root, the two numbershere 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) + yand 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 ofit. The right conclusion thus has to be that both a_1(x,y) anda_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 madean arbitrary choice somewhere along the way, would the arriveat a conclusion that they know to be false if another choicehad been made? The same problem actually occurs with your argument aboutyour cubic polynomial, except there you are arbitrarily choosingone 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 === Subject: thank you, but how could I possible have missed it?... I mean,I was away from the Jimi Harris Experience for weeks, andthat's teh first thing that I saw, a few days ago -- andrepeatedly. nice marketing effort, I suppose, though, ifit's just to keep the newbies on their toes. I mean,there's bound to be a Homeland Security issuein this exercise! but, yes; stick to the quadratic case, ifit helps to lessen the ration of obfuscationin 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.htmlhttp://www.rand.org/publications/randreview/issues/ rr.12.00/http://members.tripod.com/~american_almanachttp:// www.wlym.com/PDF-68-76/CAM7606.pdf === Subject: (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 casethe factors are not constant. -William Hughes === Subject: > 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 === Subject: [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 === Subject: >>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 === Subject: [snip zillionth failed argument]James, your posting history makes it clear that mathematics is simply toodifficult for you. Every time you leap triumphantly to a false conclusion,instead of acknowledging your error you announce that you have discoveredsome breakthrough new principle. That simply isn't the behavior of arational person.Perhaps you should switch to some activity better suited to yourcapabilities, such as basket weaving, for example. (But do wear eyeprotectors!)--==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 )*iThen 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 shouldbe ZERO without tedious residue computation?Please give me some help!-Walala === Subject: > 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| = Rand 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 === Subject: I have a similar question regarding the bound in the Waring problem. For a fixed positive integer, k, what is the best bound discovered sofar for n(k) where n(k) is such that every positive integer can beexpressed 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 Iwill have to do some more work before I can derive the Three Pearlsbounds, even for small k.I'm sure that this bound is astronomical, too.Paul Epstein === Subject: 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 ofyou. Well, maybe JSH has become a star at last.LH === Subject: Wanted: James Harris, amateur mathematician (discoverer), future pornstar.Alive: 1$Dead: 10$ Those prizes may change without prior notice, depending on theevolution of the porn industry.If you are interested in buying James Harris, any offers should besent to jamis_harres@yahoo.com.Jamis Harres === Subject: This is the last message of a thread that has been only visibleon The Math Forum, seehttp://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.>MarcusI 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 toThe SeqFan Mailing Listhttp://www.ccr.jussieu.fr/gmpib/seqfan/seqfan.htmlThere were various contributions (e.g. by Hans Havermann,the terms given above.There is already a corresponding sequence in theOn-Line Encyclopedia of Integer Sequences:http://www.research.att.com/projects/OEIS?Anum= A072716This sequence will probably get an update during the nextfew 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 togo with t,u,v to be sure that there are no further terms{t,u,v,n}={133, 2502, 4607, 74}DanHugo Pfoertner === Subject: > 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 xyand 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)^2We can then check for the existence of rational points. There isalways 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 37This 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 multipleof the generator [*] which makes x>0 and y>0. I believe thereare no such points. (All multiples of a point on the unbounded portionof 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 muchin this setting except that there will be more than the usualnumber 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 postfollows the pattern of one of the preceding paragraphs, except thatI 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 whetherthe generator lay on the bounded part of the curve or not.)dave[*] Torsion doesn't help; adding one of the elements of order 3 justpermutes t,u,v cyclically. And taking negatives on the elliptic curvecorresponds to performing a permutation of odd sign on these three. === Subject: