mm-3729 === Subject: Re: About Primes ! A fairly simple, if not rigorous, derivation of the approximation pi(n) ~ Li(n) is given at The derivative of Li(x) is 1/ln(x). This implies that the distance between consecutive primes is about ln(n) on average. take out the trash before replying === Subject: what is s in s=vt ? I understand that v is velocity (because velocity starts with a v), that a is acceleration (ditto), that t is time, but what is s ? is it sssdistance ? Richard. === Subject: Re: what is s in s=vt ? Hi. s stands for street KO === Subject: Re: what is s in s=vt ? s is distance Patrick === Subject: Re: what is s in s=vt ? Perhaps the OP wanted to know _why_ s is used to denote distance. My guess -- and it's only that -- is that s was chosen because it's the first letter of the word for distance in another language, such as Latin _spatium_ (for which the English cognate is space, I suppose). David === Subject: Re: what is s in s=vt ? Here's one reference in accord with your guess. Scroll down to Galileo's Discourses on Two New Sciences in 1640. === Subject: Re: what is s in s=vt ? it came up with Strecke as the only German suggestion starting with s. This, it can translate back to distance, but it also suggests stretch, as in the stretch from A to B is 4m. So I suppose stretch will do, but it it probably is from German or Latin. Richard. === Subject: Math Statistics Can anybody help me with these: Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. 2) n = 153, x = 138; 95 percent 2) === Subject: Re: Math Statistics Can you do your own homework? === Subject: Re: Simple Definition interesting... I will consider it thank you. === Subject: Re: Factorization ! 2 isn't really a special case. http://www.btinternet.com/~se16/js/factor.htm just uses Javascipt and trial division to get up to 9007199254740991 =2^53-1 http://www.alpertron.com.ar/ECM.HTM is far better, using Java and Elliptic Curve Method and/or Quadratic Sieve. === Subject: Math teachers move away from methods of kill and drill Despite all the flimflam, scores have been going up. This may be due to the fact that: We have not done a math textbook adoption since I've been here - and that's 11 years, ... -------------------------------------- WC prepares for new style of math instruction Math teachers move away from methods of kill and drill By MICHAEL NEARY, Messenger staff writer WEBSTER CITY - As teacher Sue Stock explained mathematical concepts such as median, mode, clusters and outliers to a group of Webster City sixth-graders, real-world examples were never far behind - even if those examples sometimes leaned toward the bizarre. Stock asked the students to ponder why, for instance, certain people could balance themselves on one leg longer than others. She asked them to consider gymnasts, acrobats, soccer players, adults, students - and their parents. Stock was experimenting with one of several new approaches to mathematics that the Webster City Community School District is considering adopting next year. It's a change that Mindy Mossman, the elementary school principal, says is a long time in coming. We have not done a math textbook adoption since I've been here - and that's 11 years, said Mossman, who added that the district has had strong math test scores over that period. Webster City students, at all tested levels, have scored several percentage points higher than the state average on the math section of the Iowa Tests of Basic Skills, according to the most recent data. Stock said a current practice in the district is to teach a basic computation skill and then follow it up with a more sophisticated conceptual application. The new curriculum, she said, will reverse that order. We (go) from hard to easy instead of easy to hard, she said. Rather than teaching the definition of polygons and then asking them to find examples, Stock explained, she once asked her students to arrive at a definition of polygons themselves by scrutinizing a painting from Pablo Picasso. It's generally meaningful mathematics, said Annette Louk, a math and science school improvement consultant for the Prairie Valley Area Education Agency. Louk said other area school districts - including Storm Lake and Spencer -have adopted similar philosophies in their mathematics instruction. And the Humboldt Community School District uses a mathematics framework called Cognitively Guided Instruction, which also stresses conceptual problem-solving over rote practice. We need a balance, said Linda Williams, the curriculum director for the Webster City Community School District. You need to have computation, but it can't be kill and drill. On Thursday morning, Stock taught from a text called ''Investigations in Number, Data, and Space'' - a text that, like all of the possibilities the district is considering, touches on National Council of Teachers of Mathematics standards. Those standards include problem solving, reason and proof, communication, connections and representation. Stock, a special education teacher, was instructing 12 students in a class she co-teaches with math teacher Jeff Lyons. She said three of the 12 were special education students and noted that their average percentile increase on the Iowa Tests of Basic Skills was 27 points since last year. Students, sitting at a round table during the class, said the lesson was more active than others they'd had. Ashley Meyers, 12, said the students explored more possibilities than they did in other classes. Other students agreed the interaction - along with the active problem- solving - differed from what they'd experienced in other classes. ''In fifth grade we just sat at the desks,'' said Rebecca Gerdes, 11. Gary Ulrick, 12, said he'd noticed the experimental activities in Stock's class to be similar to those at Stratford Elementary School, where he attended classes until February. ''We did a lot of group activities (at Stratford),'' he said. ''We were a little bit ahead of these guys when we came.'' Jamie Elkin, a Webster City fourth-grade teacher, agreed that the district's math classes needed to be more challenging. ''We need to be a lot more rigorous than we are,'' she said, noting that concepts such as fractions, decimals and geometry would soon be introduced in fourth grade. Right now, she said, children don't start seriously studying those concepts until fifth or sixth grade. Like Stock, Elkin said the new curriculum would not only move challenging concepts to earlier grades but also treat them in more complex ways. ''Now we're doing more of the spoon-feeding,'' she said. ''It's 'memorize it,' instead of investigating.'' Elkin said the heavy emphasis on testing now prevalent in schools can discourage teachers from departing from the drill method. But she contended that the new approaches would ultimately demand computation skill - just at a different stage of the learning process. ''We will still teach the algorithm, but after the fact, after the discovery process,'' she said. Contact Michael Neary at (515) 573-2141 or mneary@messengernews.net. === Subject: Re: Math teachers move away from methods of kill and drill I don't know where the kill of kill and drill comes in .... but there's no substitute for drill and practice. Damned if I've gone away from it. Jerry === Subject: Re: Math teachers move away from methods of kill and drill One should not drill on what one can do; this can improve speed, but makes understanding harder. Do you UNDERSTAND mathematics? Being able to compute is not an important part, and I am very strong at computing. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Math teachers move away from methods of kill and drill I disagree only because I've been struggling with trying to teach 5th graders long division. I've noticed that their main problem is that they make mistakes in subtraction.when doing the problem.....and they get very frustrated with the fact that everything that they've done with that problem therefore becomes wrong from the point of that first subtraction mistake.. I need them to understand subtraction......I also need them to be able to do it on demand.. === Subject: Re: Math teachers move away from methods of kill and drill i'm a first year teacher, but a mathematician by training. the one effect i've seen come from not drilling kids is that i have 6th- and 7th-graders who still add and subtract on their hands and have difficulty with simple multiplication/division. i have kids who, when tasked with something like multiplying 17 by 4, will add 17 to itself 4 times. they understand the process, but they expend a ton of energy on the algorithm. right on, new math. === Subject: Re: Math teachers move away from methods of kill and drill At some point, they need to be shown that there are other approaches which are more efficient. It is only AFTER understanding that learning tricks makes sense. But even these tricks should be taught from theory; instead of memorizing addition and multiplication tables, they should produce them, to several bases. They need to understand the distributive laws, and also powers of the base. If you are a mathematician, you are aware that mathematics consists of concepts and the derivations therefrom, not computational methods. If you did not understand the abstract courses as the essence of mathematics, you are not a mathematician. Training is not education, nor education training. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Math teachers move away from methods of kill and drill My son had some amount of drilling in the younger grades. I added to that by having him drill and kill arithmetic facts using a flashcard program. He could do 100 problems in 3 minutes or less with no more than an occasional mistake which was a typo. We did this for 2-3 years. Every time we stopped, if we tried again after a month, he was doing them at a small fraction of the speed, making significant errors (20-30%), and openly counting on his fingers. Meanwhile his rebellion against math homework was growing. I finally realized that no matter how much drill he went through, it wasn't going to stick. A year after he stopped, at whatever age, he would be counting on his fingers, making errors, and needing a calculator to get a useful answer in a reasonable amount of time. Since in fact, drill in basic arithmetic facts DOES stop at some age, added drill was not going to meaningfully change his final mathematical ability. Other than for developing the ability to memorize, memorizing facts and factoids is only significant if those facts get into long term memory and stay there. For some kids, drill helps this. For others, it does not. Some or all of those kids will probably do no better, no matter how you teach them. lojbab === Subject: Polaner Hypothesis Cc: brianbruderlin@verizon.net 1961 Elementary calculus. San Diego State College. My teacher trundles out this business and we talk about it. Please excuse my crude constructions as I am as mathematically naive as I sound. This has stuck in my mind for many years though. Who the hell Polaner is beats me. Polaner Hypothesis: Before you sits a very large! piece of paper . to the left and right of a decimal point extend without limit an infinite sequence of random integers selected from (0 to 9). beneath this sequence is another random expansion extending to the left and right of a decimal point. random is whatever fulfills best definitions of it (im not sure here what to say) and its definition asserts among other things that every finite sequence of numbers must exist within these random expansions. remove the decimal points. Hypothesised: IT IS POSSIBLE to translate one expansion in such a way that ALL integers of one are identical to the expansion beneath. ALL without limit. Polaner asserts that IF IT IS NOT POSSIBLE then there must exist a finite sequence of one expansion which will not correspond in the expressed manner with at least one sequence of the other expansion. As this circumstance violates the very definition of the construction the assertion that the two infinite expansions can be placed into 1 to 1 correspondence in the stated manner is proven by reductio ad absurdem. This is fun stuff for freshmen, which 35 years later Im afraid I still am. But help me be a sophomore. Is it provable that such a translation is not possible? Does this ,as my teacher cheerfully and mystifyingly suggested, imply that all such infinite sequences of integers must be the same but out of joint with one another merely by translation? Michael G. Kramer === 1326 === Subject: Re: solution to various textbook(pdf) Sam === Subject: about maximal ideal What is a maximal ideal in the ring of continuous functions from real line to itself which are periodic with period Pi.And is it unique.81HI am a Chinese,my English is not very well. === Subject: Re: about maximal ideal On Mon, 02 Apr 2007 07:18:38 EDT, liukun in alt.math.undergrad: Let R be the ring of Pi-periodic continuous real-valued functions on the reals. For each c in [0, Pi) let M(c) = {f in R : f(c) = 0}; then M(c) is a maximal ideal of R. Brian === Subject: Re: about maximal ideal On Mon, 2 Apr 2007 09:19:05 -0400, Brian M. Scott And the M(c) are the _only_ maximal ideals in R (although that requires proof, while what you said is more or less obvious). ************************ David C. Ullrich === Subject: Re: Ramanujan circle squaring approximation No. Acctually I wonder how many people of the 6E9 people here on earth knows that's impoissble. I also wonder who many knows that but still tries. === Subject: Re: Ramanujan circle squaring approximation Mmmmkay, just checking ;-) === Subject: Re: Ramanujan circle squaring approximation Actually, the title is Modular equations and approximations to pi No. What he gave was an approximate quadrature of the circle. Given a circle whose radius is _r_, he explains how to obtain, using euclidean tools only, a square whose side if sqrt(p).r, where _p_ is the number that you mentioned. I don't know, but it is reproduced here: Pi: A Source Book Lennart Berggren, Jonathan Borwein, Peter Borwein Springer-Verlag Jose Carlos Santos === Subject: Re: Ramanujan circle squaring approximation === Subject: algebra question I am confused about the difference between solving a linear and nonlinear inequality equation. one question is solve: x+8 x+2 the answer is (-2, oo) it looks like they just took the denominator equal to zero this question looks similar x-1 ----- <=2 x-4 the answer is (-oo,-4)u[9,oo) now is this question i dont know how they got this can someone like type out the steps or in someway help me understand === Subject: Re: algebra question I get the answer (-oo, 4) U [7, oo) === Subject: Re: algebra question I don't think that answer is correct. Check x = 8, or x = 1. One way to solve symbolically is to multiply both sides of the inequality by x - 4, and solve the resulting linear equation. If x - 4 < 0 then the direction of the inequality flips, so there are two separate cases to consider, depending on whether x - 4 is greater than zero or less than zero. A more intuitive method is to sketch the curve and make the obvious deductions. === Subject: Re: algebra question 7 <= x We see that x=6 is not a solution, so do we conclude the solution set is [7,oo]? That's only part of the solution. As already shown in alt.algebra.help (OP, please don't cross-post!) get 0 on one side, rewrite the other side as a single fraction, then utilize the fact that a fraction can change sign only where the numerator and denominator change sign, and that occurs where they =0. In this case, setting the numerator and denominator =0 you get two numbers which divide the number line into three intervals, each of which must be tested with a number within the interval to see if it is a solution (if so, the entire interval is.) Also, the two numbers themselves must be tested to see if it causes the fraction to _equal_ zero. -- Darrell === Subject: Re: algebra question holds only when x - 4 < 0. 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It does not seem the general case where A is a generic C* algebra with unit and B neq A is a sub C*algebra (with the same unit). Can you reduce to this extend in every case? Bye, Valter === Subject: Re: a problem with C*-algebras Content-Length: 2624 Originator: rusin@vesuvius It was an example where B contains the identity of A. So where is the problem? The point is, the result in Dixmier tells you which situation to avoid if you want to construct examples of two different representations of the bigger algebra that restrict to the same representation of the smaller one. If you do that, it is usually possible to come up with lots of examples. Ok, here is yet another example: fix a real q in (0,1). Take A to be the universal C^*-algebra generated by two elements a and b obeying the following relations: a^*a+b^*b=1=aa^*+q^2 bb^*, b^*b=bb^*, ab=qba, ab^*=qb^*a. Take B to be the C^*-subalgebra generated by 1 and b^*b. Take the Hilbert space L_2(N). (N: natural numbers) let e_n be the canonical basis. Let S and T be the operator: S e_n=e_{n+1}, T e_n=q^n e_n. For any zin S^1, gives a reprsentation of A on L_2(N), and these are inequivalent representations. However, their restrictions to B are equal. Thus: In some cases there will exist two such representations, as all these examples tell you. And Dixmier's result tells you that there are situations when two such representations will not exist. By the way, It DOES work --- all you have to do is take B to be the C^*-subalgebra generated by an ideal and the identity element. The above example also answers your other question: And again the answer is: in some cases such a representation will exist (the example above), and in some cases it will not exist (use Dixmier). Hope this answers your question. === Subject: Re: a problem with C*-algebras Content-Length: 1466 Originator: rusin@vesuvius Well, Dixmier result presents just ONE particular case to avoid, which this reason). I remind you that my hypotheses are: (1) B is a sub unital-C*-algebra of a unital C*-algebra A (2) B is different from A. You see that the case treated in Dixmier proposition 2.10.4 is automatically excluded: B fulfilling (1) and (2) cannot be ALSO a two- sided ideal: because due to (1) it would contain the unit of A and thus it would coincide with A, in contradiction with (2). Proof. (trivial) if a in A and b in B then ab and ba in B because B is a two-sided ideal. If 1 in B then a= a1 in B for every a in A. Therefore A=B. Barring your (interesting) examples, it seems to me that my question is still open: If hypotheses (1) and (2) are valid is it *always* possible to find two different representations of A which are equal on B? In other words are there cases of A and B fulfilling (1) and (2) where such two representations do not exits? You cannot use Dixmier proposition because B do not satisfy my hypotheses in that case. I do not think so. Bye, Valter === Subject: Re: a problem with C*-algebras Content-Length: 1000 Originator: rusin@vesuvius Take a unital C^*-algebra A, say the algebra in the earlier example (generated by two elements a and b with those relations). Let J be the ideal generated by the element b. Now take B to be the C^*-subalgebra of A generated by J and the identity of A. If you take any representation pi of B, by that result of Dixmier it would extend uniquely to a representation of A. === Subject: Re: a problem with C*-algebras Content-Length: 478 Originator: rusin@vesuvius OK, thank you very much! Valter === Subject: Re: a problem with C*-algebras Content-Length: 1129 Originator: rusin@vesuvius OK, I know LaTeX commands Yes, I know that result by Dixmier book. However in my case, obviously, it does not work because B contains also the identity and thus B=A. Bye, Valter === Subject: p-adic expansions in Magma Content-Length: 481 Originator: rusin@vesuvius I'm trying to get the p-adic expansion of numbers with Magma (V 2.13) but I can't, even copying and pasting from webpages with examples; I believe my version is too new! [ 127 + O(5^20) ] Instead of [ 2, 0, 0, 1 ]. I've tried with SeriesPrinting but: ^ Runtime error in :=: Invalid attribute 'SeriesPrinting' for this structure Any idea on how to get what I want? D === Subject: Re: p-adic expansions in Magma Content-Length: 1846 Originator: rusin@vesuvius This unfortunately appears to be the case. It seems that series printing was abandoned during the last re-implementation of the p-adic module. For the specific case of p-adics, you could use Intseq(): [ 2, 0, 0, 1 ] It is straightforward to bundle this up in a function for inclusion in your .magmarc (or an intrinsic in your packages, if you prefer): [ 2, 0, 0, 1 ] [ Aside: note that the 'p' that you assigned originally is not the prime as you had hoped. Instead, it will be a generator of the local field over its immediate subfield (this is similar to how other extension fields work). In the case of p-adics, this will turn out to be the value 1. You can recover the prime as an integer using Prime(), or as an element of the p-adic structure using UniformizingElement(). ] I don't think there is any simple way to get the equivalent of the old series printing for local rings. Geoff. ---------------------------------------------------------------------------- - Geoff Bailey (Fred the Wonder Worm) | Programmer by trade -- ftww@maths.usyd.edu.au | Gameplayer by vocation. ---------------------------------------------------------------------------- - === Subject: similar multivariate polynomial matrices Content-Length: 721 Originator: rusin@vesuvius is there a known algorithm to check whether two given multivariate polynomial matrices are similar, with a unimodular similarity matrix? In the case of univariate polynomial matrices computing the Smith normal form of both matrices answers this question, but the Smith form is not available for more than one variable, and simply solving systems of polynomial equations in not feasible. In case there is no such general algorithm, would it help if the given matrices were of block-matrix form with only monomial entries in the two off-diagonal block-matrices and zero entries in the two block- matrices on the diagonal? Any hints or ideas would be very much appreciated. Nils === Subject: Last Call for Papers (extended): The2007 International Conference of Applied and Engineering Mathematics (ICAEM 2007) Content-Length: 10214 Originator: rusin@vesuvius Last CFP (extended): The 2007 International Conference of Applied and Engineering Mathematics (ICAEM 2007) http://www.iaeng.org/WCE2007/ICAEM2007.html Draft Paper Submission Deadline (extended): 22 March, 2007 Camera-Ready papers & Pre-registration Due: 3 April, 2007 WCE 2007: 2-4 July, 2007 The conference ICAEM'07 is held under the World Congress on Engineering 2007. The WCE 2007 is organized by the International Association of Engineers (IAENG), a non-profit international association for the engineers and the computer scientists. Our congress committees have been formed with over two hundred and eighty committee members who are mainly research center heads, faculty deans, department heads, professors, and research scientists from different universities like Cambridge, MIT and Oxford etc. The conference proceedings will be published by IAENG (ISBN: 978-988-98671-5-7) in hardcopy. The full-text congress proceeding will be indexed in major database indexes so that it can be assessed easily. The Technology Research Databases (TRD) of CSA (Cambridge Scientific Abstracts), DBLP and Computer Science Bibliographies have promised to index the print proceeding in advance of its publication. And after the publication of the proceeding, print copies will also be sent to databases like IEE INSPEC, Engineering Index (EI) and ISI Thomson Scientific for indexing. The accepted papers will also be considered for publication in the special issues of the journal Engineering Letters. Some participants may also be invited to submit extended version of their conference papers for considering as book chapters (soon after the conference). The topics of the ICAEM'07 include, but not limited to, the following: Linear algebra and applications: Matrix theory Tensor analysis Combinatorial linear algebra Numerical linear analysis Computational linear algebra Markov chains Iterative methods Large-scale systems Numerical analysis: Numerical methods for ordinary and partial differential equation Computational programming of numerical algorithms Finite Elements Scientific computing Error analysis Stability problems Convergence analysis Non-linear systems Chaos systems Dynamical systems Simulation Differential equations and applications: Ordinary differential equations Partial differential equations Stochastic differential equations Difference equations Integral equations Variation methods Nonlinear systems Perturbation problems Probabilities and statistics: Probability theory Stochastic process Applied statistics Mathematical statistics Estimation Theory Identification Simulation Operations Research and Optimization: Mathematical programming Stochastic modeling Decision theory Game theory Queueing theory Reliability theory Routing theory Transportation problems Financial mathematics Inventory control Scheduling Optimization theory Linear programming Quadratic programming Convex programming Nonlinear programming Stochastic programming Combinatorial programming Discrete Mathematics and Control: Methods of algorithmic analysis Algorithms Combinatorial problems Graph theory Coding Cryptology Signal processing Real time systems Network optimization Control theory Submission: WCE 2007 is now accepting manuscript submissions. Prospective authors are invited to submit their draft paper in full paper (any appropriate style) to WCE{at}iaeng.org by 6 March, 2007. The submitted file can be in MS Word format, PS format, or PDF formats. The first page of the draft paper should include: (1) Title of the paper; (2) Name, affiliation and e-mail address for each author; (3) A maximum of 5 keywords of the paper. Te name of the conference that the paper is being submitted to should also be stated in the email. It is our target that the reviewing process and the result notification for each submitted manuscript can be completed within one month from its submission. The reviewing process is to ensure the quality of the accepted papers in the WCE congress. The conferences have enjoyed high reputation among many research colleagues (for example, see the http://cs.conference-ranking.net/ or http://www.conference-ranking.com/). ICAEM 2007 Conference Committee Dr. Michael Bluck Lecturer, Mechanical Engineering Department, South Kensington Campus, Imperial College London,UK Dr. Pavlos Christodoulides Lecturer, General Studies Department, Higher Technical Institute, Cyprus Prof Bogdan Gabrys Professor, Chair in Computational Intelligence, Computational Intelligence Research Group, School of Design, Engineering & Computing, Bournemouth University, UK Prof. G. Ganesan Dept. of Mathematics, JPN College of Engineering, India Dr Yee Mey Goh Research Officer, Innovative Manufacturing Research Centre, Department of Mechanical Engineering, University of Bath, UK Prof. Rosa Eva Pruneda Gonzalez Associated Professor, Mathematics Department, Civil Engineering School, University of Castilla-La Mancha, Spain Prof. Angel Marin Gracia Professor of Applied Mathematics and Statistical Deparment, de la Universidad Politecnica de Madrid, Spain Dr. Josep R. Herrero Assistant professor, Department of Computer Architecture, Polytechnic University of Catalonia, Spain Dr. Ben James Hicks Senior Research Fellow, Innovative Manufacturing Research Centre, Department of Mechanical Engineering, University of Bath, UK Dr Khalid Hussain Director of Studies in Mechanical & Automotive Engineering, School of Engineering, Design & Technology (EDT4), University of Bradford, UK Prof. Allali Khalid Faculte des Sciences et Techniques de Settat, France Dr. Joshua R Omer Research scientist, Division of Civil Engineering, Faculty of Advanced Technology, University of Glamorgan, United Kingdom Dr. Mourad Oussalah Lecturer, Department of Electronic, Electrical and Computer Engineering, The University of Birmingham, UK Prof. Henry Power (co-chair) Professor and Chair of Computational Fluid Dynamics School of Mechanical, Materials and Manufacturing Engineering, The University of Nottingham, UK Editor of International Journal Engineering Analysis with Boundary Elements Dr. Mihailo Ristic (co-chair) Senior Lecturer, Mechanical Engineering Department, Imperial College London, UK Prof. Sergei Sazhin (co-chair) Professor of Thermal Physics, School of Engineering, Faculty of Science and Engineering, The University of Brighton, UK Professor Riti Singh (co-chair; CENG FIMECHE FRAES FIDGTE) Professor of Gas Turbine Engineering Director of the Gas Turbine Technology Centre Director of the University Technology Centre in Performance Engineering Department of Power & Propulsion, School of Engineering, Cranfield University, United Kingdom Prof Sergei V. Utyuzhnikov (co-chair) Senior Research Fellow, School of Mechanical, Aerospace & Civil Engineering, University of Manchester, UK Dr. Alexander Vikhansky Research scientist, Department of Engineering, Queen Mary University of London, UK Dr Charles H.-T. Wang Reader, School of Engineering & Physical Sciences, University of Aberdeen, King's College, UK Dr Jihong Wang Senior Lecturer, Department of Electrical Engineering and Electronics, University of Liverpool, UK Prof. Jennifer Wen Professor and Director of Research, Faculty of Engineering, Kingston University, UK Dr. Pihua Wen Lecturer, Department of Engineering, Queen Mary, University of London, UK Prof. Alastair Wood Professor of Engineering Mathematics, School of Engineering, Design & Technology, University of Bradford, UK Prof. Valentina Zharkova (co-chair) Professor of Applied Mathematics, School of Informatics, University of Bradford, UK WCE Congress Co-chairs Prof. David WL Hukins, CPhys, FinstP, FIPEM, FRSE Professor of Bio-medical Engineering Head of Department of Mechanical & Manufacturing Engineering, University of Birmingham, UK Prof. Andrew Hunter Professor & Head of Department Head of Vision and AI Research Group, Department of Computing and Informatics, Lincoln University, UK Prof. Alexander M. Korsunsky Professor of Engineering Science Dean, Trinity College Department of Engineering Science, University of Oxford, UK Prof. Leonid Gelman (honorary co-chair) Professor and Chair in Vibro-Acoustic Monitoring, Chairman of COMADIT, British Institute of NDT, Director, Centre of Vibro-Acoustics and Fatigue, Department of Process and Systems Engineering, School of Engineering Cranfield University, UK Dr. Christopher John Hogger (honorary co-chair) Senior Lecturer Department of Computing Imperial College London, UK Prof. Darek J. Ceglarek (ICMEEM honorary co-chair) Professor, International Manufacturing Centre, University of Warwick, UK Professor, Department of Industrial and Systems Engineering The University of Wisconsin-Madison, USA Fellow of CIRP; Associate Editor, IEEE Transactions on Automation Science and Engineering, and ASME Transactions on Manufacturing Science and Engineering Dr. Stephen Payne (ICSBB honorary co-chair) University Lecturer in BioMedical Engineering Dean of Degrees Keble College, Head of Physiological Understanding through Modelling, Monitoring and Analysis Group, Department of Engineering Science, University of Oxford, UK More details about the WCE 2007 can be found at: http://www.iaeng.org/WCE2007/index.html http://www.iaeng.net/WCE2007/index.html http://www.iaeng.com/WCE2007/index.html More details about the International Association of Engineers, and the IAENG International Journal of Computer Science, and the IAENG International Journal of Applied Mathematics can be found at: http://www.iaeng.org/about_IAENG.html http://www.iaeng.org/IJCS/index.html http://www.newswood.org/index.html The official journal web site of Engineering Letters at: http://www.engineeringletters.com Other Engineering Letters web sites at: http://www.engineeringletters.com http://www.engineeringletters.net http://www.engineeringletters.org http://www.engineeringletter.com http://www.engineerletters.com http://www.engineerletter.com ******** It will be highly appreciated if you can circulate these calls for papers to your colleagues. === Subject: A cross-sum of Euclidean norms inequality Content-Length: 442 Originator: rusin@vesuvius Can anyone prove or provide a counter-example to the following inequality? Let p_1,...,p_n and a_1,...,a_n be in R^m, and |.| denote the Euclidean norm. Then, sum[i=1..n] sum[k=1..n] |p_i - a_k| + (1/2) (sum[i=1..n] sum[j=1..n] |p_i - p_j|) I can prove this for m=1 or n<=2, but have been unable to find a proof or counter-example to other cases. === Subject: hypergeometric functions and beta functions Content-Length: 371 Originator: rusin@vesuvius Hi all, I think it is the case that z^{ r + s }/(r+s) F(2r, r+s, 1+r+s, -z) + z^{ s - r} / (r-s) F(2r, r- s, 1+r-s, -1/z) = B( r-s, r+s) where F(.,.,.,.) is Gauss's 2F1 hypergeometric function, B(.,.) is the beta function, r and s are real (may also work if they're complex; I Should this be obvious? Does anyone have a proof? Best, Ian === Subject: Re: hypergeometric functions and beta functions Originator: israel@math.ubc.ca (Robert Israel) Hi Ian. Your result is well known. It's a special case of the following equation : (Gamma[a] * Gamma[b] / Gamma[c]) * Hypergeometric2F1[a, b, c, z] = (Gamma[a] * Gamma[b-a] * (-z)^(-a) / Gamma[c-a]) * Hypergeometric2F1[a, 1+a-c, 1+a-b, 1/z] + (Gamma[b] * Gamma[a-b]* (-z)^(-b) / Gamma[c-b]) * Hypergeometric2F1[b, 1+b-c, 1+b-a, 1/z] where |arg(z)| < Pi. This is a relation between three of the 24 Kummer solutions for the hypergeometric differential equation. In your case, when a = 2r, b = r +s, c = 1+r+s and z replaced with -z, the last hypergeometric function becomes 1, and your result can be easily deduced. The proof can be found in any good book on hypergeometric series. I found this equation (with proof) in L.J. Slater's book. I don't have the complete title with me actually, but I can certainly send it to solutions. Fran.8dois === Subject: Re: hypergeometric functions and beta functions Content-Length: 863 Originator: rusin@vesuvius I did it with Maple 10. Usenet is not conducive to Maple worksheets, so I put it on MaplePrimes, you can view it here: http://www.mapleprimes.com/viewfile/1395 I think you can view it without a login (but you may have to tell your browser to trust the certificate?) -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: hypergeometric functions and beta functions Content-Length: 1035 Originator: rusin@vesuvius Best, Ian === Subject: Chair in Applied Mathematics at UEA Content-Length: 739 Originator: rusin@vesuvius Chair in Applied Mathematics The University of East Anglia Norwich invites applications for a Chair in Applied Mathematics from candidates who are engaged in internationally leading research. Preference may be given to applicants whose research interests complement the existing research strengths of the Applied Mathematics group. The starting date for the appointment is on or before 1 October 2007. A competitive salary package is available for outstanding candidates. The deadline for applications is 15 April 2007. For informal discussions, please contact the Head of School, Professor Tom Ward (+44 (0)1603-592849, email: t.ward@uea.ac.uk). Information about the School of Mathematics is on the web at www.mth.uea.ac.uk === Subject: Competing species Originator: israel@math.ubc.ca (Robert Israel) Hi all, I have a question regarding the partial differential equations. If I have a system of two PDEs such as *du/dt = d^2u/dx^2 + u^2*v *dv/dt = d^2v/dx^2 - u^2*v d: partial derivative in both equations where u and v are two species. How can I know that these species are competing? === Subject: Re: Competing species Originator: israel@math.ubc.ca (Robert Israel) The left sides are describing the change of inviduals in a small space area in a time interval. The second derivative on the right is the divergence of the diffusion current j_u=-du/dx, j_v=--dv/dx generated by nonconstant distribution of indivuals. Since both densities u,v are positive, the u-number at that place is increasing proportional the probability that 2 independent u's and 1 v meet while v is decreasing by the same number. So we conclude, that two u's at the same place occasionally catch an v converting it to a new independent u. -- Roland Franzius === Subject: IICAI-07 Call for papers Originator: israel@math.ubc.ca (Robert Israel) Paper submission deadline is April 2 2007 www.iiconference.org The 3rd Indian International Conference on Artificial Intelligence (IICAI-07) (website: http://www.iiconference.org ) will be held in Pune, INDIA during December 17-19 2007. IICAI-07 is one of the major AI events in the world. This conference focuses on all areas of AI and related fields. We invite paper submissions. Please visit on the conference website for more details. Bhanu Prasad IICAI-07 Chair Department of Computer and Information Sciences Florida A&M University, Tallahassee, FL 32307, USA === Subject: sum of squares of C^infty functions Originator: israel@math.ubc.ca (Robert Israel) I am told that there exists a C^infty real-valued function f on the that f is not the sum of finitely many squares of C^infty functions. fonctions d.8erivables in Bull. Soc. Math. France 133 (2005) 619-639 (in which he proves that such an f can be written g^2+h^2 where g and h are C^m for any given but fixed m). But references in the literature are of the form P. Cohen and/or D. Epstein, private communication, which is not really helpful. Does anyone know how such functions are constructed? I'm not necessarily asking for the details, merely for an idea of how these counterexamples are constructed. (It's already not very to find an example where f is not the square of _one_ C^infty function.) [C^infty = infinitely differentiable, of course] -- David A. Madore ( david.madore@ens.fr, http://www.dma.ens.fr/~madore/ ) Subject: Re: a problem on building a regular expression I agree - the DFA is quite simple. I couldn't remember how to convert that to a regular expression, even though I know it should be possible. Could you point me to a reference? Michael === Subject: Re: a problem on building a regular expression Indeed I do not know a *practical* reference, so I have written my own recipe. Unfortunately, it is in German: http://www.lrz-muenchen.de/services/schulung/unterlagen/regul/regul-13.html# publish4.2.1.0.0.0 It is a part of longer paper. In the cited subpage, there are two DFAs. The first one transforms quite obviously to a regular expression, the steps are explained in points 1 to 3 in the text. I guess I need not explain that. For the DFA in the second sketch, there is no obvious transformation. The trick is then to split a state, C, into an entry-only state, C0, and an exit-only state, C1. Then one considers separately the paths from the initial state, A, to the final state, E, that do not pass by C, then the paths from A to C1 (these are the paths in the original DFA from A to C that do not pass by C), the paths from C0 to E, and the paths from C0 to C1. Then the results are combined into a single regular expression. There is one more example, to wit the decimal numbers divisible by 3, which require more than one such state-split: http://www.lrz-muenchen.de/services/schulung/unterlagen/regul/regul-14.html# publish4.3.2.0.0.0 It has quite some similarity with the problem at hand. -- Helmut Richter === Subject: Re: a problem on building a regular expression looks i made a mistake from start, may be below is right A = 1(01)*(0+%eps) + 0(10)*(1+%eps) L = (A + %eps) (2A)* (2+%eps) === Subject: Re: Hausdorff dimension and differentiable function thank you :) I must have been blind .. === Subject: Re: Wondering if my idea is right concerning Transfinite Sets days. My association with the Department is that of an alumnus. A function f on the cardinals is said to be continuous iff f(lim(kappa_i)) = lim(f(kappa_i)) where the indices i run over some ordered set I (usually an ordinal). I think it might be equivalent to the order topology, but I'm not sure about that. In the case at hand, I = omega, the natural numbers; kappa_0 = 1 kappa_1 = 3 and So lim(kappa_i) = aleph_0, but 2^{lim(kappa_i)} is not equal to lim(2^{kappa_i}), so cardinal exponentiation is not continuous. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: Seeking informed opinion of math website... Mail-To-News-Contact: abuse@dizum.com 1. I always feel reassured when the first thing that I see on a page is the author's degree -- especially so when it's in an area unrelated to the topic. 2. He need to use more quotation marks. What, only two words put in quotes? More quotation marks always make a page better. And Daleks. 3. I'm curious on how he could get a patent on his book or his theory or whatever it is that he says it patented. Or, maybe he's talking about the patents related to the manufacture of DVDs? 4. The connection between prime numbers and whether diamonds produced on the other side of the Universe will have the same properties is intriguing. 5. What do snowflakes and galaxies have in common? I'm guessing that it's six-way symmetry, but it could be that they're both made from quarks. 6. I'm pretty sure that the normal way to announce the results of mathematical research is by selling DVDs. -- Michael F. Stemper If this is our corporate opinion, you will be billed for it. === Subject: Re: pi & 2 pi As far as I know, Kanada et al's 2002 record of: 1,030,700,000,000 hexadecimal digits 1,241,100,000,000 decimal digits [Professor breaks own record -- for thrill of pi] http://seattlepi.nwsource.com/national/98912_pi07.shtml --c === Subject: Re: pi & 2 pi Any more of your Pi-ranting talk, Press, and we'll make you walk the circumference. (-: --c === Subject: Re: total derivative repost (clarified) Looks OK to me. --Lynn === Subject: Phd Position in Adaptive Molecular Dynamics - FRANCE Cc: stephane.redon@inria.fr Title: Adaptive Molecular Dynamics Description: An average human cell contains several billion molecules. Understanding the complexity underlying the mechanisms and interactions of these molecules would have wide-ranging applications. For example, gaining a precise understanding of how proteins fold and interact would have a tremendous impact on drug design. One way to understand molecular mechanisms is through modeling and simulation. Unfortunately, the cost of molecular simulations increases with the size of the simulated systems. This makes it extremely difficult to study large systems (e.g. viruses) over long time periods. To address this problem, researchers typically have to increase the processing power (which might be expensive), or make arbitrary simplifications to the system (which might bias the study). We have recently developed an adaptive simulation method [1,2,4], which rigorously and automatically predicts the most mobile groups in a simulated molecular system, and simulates these groups only. This method allows the user to adapt the simulation to available resources, while guaranteeing that a good approximation is found under the imposed constraints [3,4]. We now want to generalize the adaptive simulation theory. The PhD candidate will develop the mathematics of the generalized adaptive simulation theory, will design and implement the corresponding algorithms, and will validate the work on various biological applications, in collaboration with bio-physicists and biologists at CEA [5]. References (available at http://i3d.inrialpes.fr/~redon) ------------------------------------------------------------------- [1] S. Redon and M. C. Lin. An Efficient, Error-Bounded Approximation Algorithm for Simulating Quasi-Statics of Complex Linkages. Proceedings of ACM Symposium on Solid and Physical Modeling, 2005. [2] S. Redon, N. Galoppo and M. C. Lin. Adaptive Dynamics of Articulated Bodies. ACM Transactions on Graphics, 25(3), 2005. [3] S. Morin and S. Redon. A Force-Feedback Algorithm for Adaptive Articulated-Body Dynamics Simulation. Proceedings of IEEE International Conference on Robotics and Automation, 2007. [5] CEA is a French government-funded technological research organisation (http://www.cea.fr/english portal). Environment --------------- This position is offered at the ñRhone-Alpesî Research Unit of INRIA, located near Grenoble and Lyon. The Unit counts more than 500 people, within 25 research teams and 10 support services. The starting date will be between september 2007 and december 15, 2007. Duration: 3-years position. Salary and benefits : 1529ó NET / month + full health insurance and social benefits included - salary will be upgraded to 1611 ó NET / month the 3rd year. For more information / to apply: ---------------------------------------- - Contact us: stephane.redon@inria.fr Stephane -------------------------------------------------------------------- Stephane Redon INRIA Rhone-Alpes Research Unit Zirst - 655 avenue de l'Europe - Montbonnot 38334 Saint Ismier Cedex - France http://i3d.inrialpes.fr/~redon === Subject: Re:Mueckenheim's Confusion For finite trees, they are, but there is no such thing as a cross section for an infinite tree, at least when using WM's definition of cross section. It is not path lengths that become uncountably in the complete infinite binary tree, but the cardinality of the set of paths, which is easily seen to be the same as the cardinality of the set of all binary sequences, and that Cantor showed to be uncountable. === Subject: Price of XOM Stocks one year from now based on option pricing... According to my model below, which bases future price movements of a stock based on today's an option's TRUE COST, it's volume, the weighted price of XOM should be: $72.54267884 on/around January 2008.Today, the price is $71.32. CALL OPTIONS Expire at close Fri, Jan 18, 2008 XOM= 71.32 Strike Symbol Last Chg Bid Ask Vol Open Int True Cost of Options Market Cap of Options Weighted Price Contribution 35 WXOAG.X 37.3 0 36.5 36.8 19 5,023 -1 18.62 0.312483518 40 WXOAH.X 31.4 0 31.8 32.1 55 3,240 0 4.4 0.084390209 45 WXOAI.X 26.8 0 27.2 27.3 1 2,675 0 0.48 0.01035698 50 WXOAJ.X 23.8 0 22.6 22.8 1 5,047 -2 2.48 0.059456738 55 WXOAK.X 18.5 Up 0.50 18.2 18.4 1 8,289 -2 2.18 0.05749083 60 WXOAL.X 14.4 Down 0.60 14.1 14.3 22 7,352 -3 67.76 1.949413824 68 WXOAU.X 9.7 0 8.8 8.9 11 1,467 -6 64.68 2.093404617 70 WXOAN.X 7.5 Down 0.30 7.2 7.4 11 36,975 -6 67.98 2.281700271 73 WXOAX.X 6.1 Down 0.40 5.9 6.1 144 5,370 -6 878.4 30.53582988 75 WXOAO.X 4.8 Down 0.30 4.8 4.9 4 32,661 -5 19.2 0.690465345 80 WXOAP.X 3 Down 0.10 2.9 3 52 53,700 -3 156 5.984032989 85 WXOAQ.X 1.85 Up 0.05 1.65 1.75 73 13,399 -2 135.05 5.504183549 90 WXOAR.X 0.9 Down 0.10 0.9 1 103 4,285 -1 92.7 4.000383592 95 WXOAS.X 0.5 Down 0.05 0.45 0.55 599 4,240 -1 299.5 13.64268418 2085.55 75.81206396 PUT OPTIONS Expire at close Fri, Jan 18, 2008 Strike Symbol Last Chg Bid Ask Vol Open Int True Cost Market Cap of Costs Weighted Price Contributions 35 WXOMG.X 0.05 0 N/A 0.05 5 4,893 -0.05 0.25 0.002489331 40 WXOMH.X 0.05 0 0.05 0.2 1 8,925 -0.05 0.05 0.00056899 45 WXOMI.X 0.35 0 0.15 0.2 717 7,088 -0.35 250.95 3.212731152 50 WXOMJ.X 0.35 0 0.35 0.4 530 27,146 -0.35 185.5 2.638691323 55 WXOMK.X 0.75 Down 0.05 0.75 0.85 4 31,231 -0.75 3 0.046941679 60 WXOML.X 1.45 Down 0.05 1.45 1.5 28 22,063 -1.45 40.6 0.693029872 65 WXOMM.X 2.45 0 2.55 2.7 1 35,287 -2.45 2.45 0.045305832 68 WXOMU.X 3 0 3.3 3.5 11 8,656 -3 33 0.63371266 70 WXOMN.X 4 Down 0.30 4.2 4.5 55 45,782 -4 220 4.381223329 73 WXOMX.X 5.3 Down 0.01 5.3 5.6 156 8,644 -4 642.72 13.25667141 75 WXOMO.X 6.5 Down 0.10 6.7 6.9 750 21,677 -3 2115 45.12802276 80 WXOMP.X 9.8 Down 0.20 9.9 10.1 3 8,954 -1 3.36 0.076472262 85 WXOMQ.X 13 0 13.8 14 7 2,015 1 -4.76 -0.115106686 90 WXOMR.X 19.2 0 18.6 18.8 22 19 -1 11.44 0.292916074 95 WXOMS.X 24.2 0 23.6 23.8 22 2 -1 11.44 0.309189189 3515 70.60285917 72.54267884 === Subject: Re: Price of XOM Stocks one year from now based on option pricing... your model is dumb the expected price of a Jan08 expiration is below 77 and above 65.50 the true price is the price real people are willing to buy and sell at and not some idiot statical model of your fiat world. Most options are either too expensive or too cheap and therefore expire worthless. === Subject: Re: Price of XOM Stocks one year from now based on option pricing... This just tells people that you don't understand how to work with numbers. === Subject: Re: Price of XOM Stocks one year from now based on option pricing... I am seeing such irrational numbers at my broker site in confirmation emails. What I suspect is software was changed to get an average price of volume and price by lumping some, unknown to me, time period and then dividing by the volume. That would do it so that is my theory - guess. === Subject: Re: Price of XOM Stocks one year from now based on option pricing... Thread-Topic: Price of XOM Stocks one year from now based on option pricing... Thread-Index: AcdikEGsgCcs3s6DEduk1AANk0DKpA== Are you saying that YOU don't carry numbers out to 8(EIGHT) decimal places??? This is VERY important when you're guessing at a price 10 months in the future...... This poor guy would develop carpal tunnel syndrome in his hand and go blind if they took his computer away...... === Subject: Re: Cantor Confusion I thought it was M.9fckenmatics. === Subject: Re: Cantor Confusion Is that a royal we? It is certainly not a plural one. WM mutters his own creed without proofs, whereas the creed that the set of binary infinite sequences is uncountable has a valid proof. There is if one is going to make claims about it as WM keeps doing. It is sufficient to know that 2 + 2 = 5 in order to have all of WM's claims become provable. But unfortunately 2+2 is not 5, and few, if any, of WM's claims are provable. === Subject: Re: Cantor Confusion Then take off your piss colored glasses and take another look. And simultaneoulsy uncountable? In ZF and NBG they are uncountable. WM has yet to produce an axiom system in which they are countable. If it could be counted at all then the set of all infinite binary seqeunces could be counted to, as they are easily seen to be bijectable. And after seeing Cantor's proof of the latter's uncountability, no sane person will accept the countability of either without an explicit surjection from the naturals to one or the other of them. So that anything less that such a surjection in support of WM's claim is mere hot air to anyone else. === Subject: Re: Cantor Confusion The number of branchings of the binary tree is uncountable??? The number of nodes is countable, isn't it? A node is a branching. I is easy, in fact, to show tat there are twice as many branchings than there are paths. If the proof was correct, then we had a contradiction in set theory. But the proof is based upon the assumption that there are infinitely many finite indexes, which is wrong. You will see this yourself, if you consider the Waft Maximum (WM) of the natural numbers, then the WM of the paths of all finite trees, and so on. But for this sake you must think a bit, not only bicker about the meaning of function. === Subject: Re: Cantor Confusion Only in finite trees. In in a complete infinite binary tree, there are as many paths as subsets of N: We can umber the nodes in any infinite path with the members of N. Then for each subset of N, there is a unique path in a complete infinite binary tree having left branchings at those nodes and right branchings at all other nodes. So that the 'number' to paths is the 'number' of members of P(N). Wrong. If the proof is correct, which it is, we have a contradiction of WM's beliefs. Which is quite acceptable to everyone, except possibly WM himself. Not in ZF or NBG. But what is wrong in WM's world is not necessarily wrong in those worlds which do not require his axioms. The WM is, in every sense, irrelevant to mathematics. I have though a bit, and find WM still to be wrong and his WM still to be irrelevant to mathematics, as are his other complaints about how mathematicians do things. In other words, I see no benefit and possible harm, in mathematicians paying any heed to WM. === Subject: Re: Cantor Confusion Then P(N) is countable. I always said that Hessenbergs proof is insufficient. And it is totally clear that the mapping from P(N) into Q cannot be surjective. === Subject: Re: Cantor Confusion If WM can prove that P(N) is countable then he should have no problem also proving that 2+2 = 5. Saying things does not make them make them true, particularly, as in WM's sayings, when they are presented without proof. A proof has been presented here that P(N) is uncountable, at least in ZF and NBG, and no contrary proof has been presented, so that, as far as mathematics is concerned, P(N) is uncountable, at least in ZF and NBG. Except that it is quite easy to describe such a surjection, at least if Q represents the rationals: The rational value correspondintg to a set of naturals will be determined by the number of elements in that set of naturals as follows: Size Corresponding of rational set number 0 0 1 The member of that set 2 The negative of the first member of the set 3 The quotient of first member divided by second member 4 The negative of the number determined by a 3 element set 5 The reciprocal of that from the 3 element set 6 The negative of that from the 5 element set. Other Any rational, as all rationals are already covered. So once again we see that WM has no comprehension of what he is saying. === Subject: Re: Cantor Confusion At each finite position corresponding to a finite tree, the set of infinite paths is empty. But that does not limit the set of infinite paths in an infinite tree to be empty, not to be countable. Then WM must be claiming omega as an index. But it only indexes the length of the paths, not the number of them. WM's limits don't. Non sequitur. If it can only populate finite levels, then WM is denying the existence of an infinite tree. But WM has neither facts nor logic to back his arguments. In ZF and NBG, if there are any finite binary trees at all then there is also a complete infinite binary tree modeled by the set of all finite binary sequences as nodes and the set of infinite binary sequences as paths, and it has uncountably many paths. Since WM does not have any stated system in which he can prove his claims, they fail for lack of proof. === Subject: Re: Cantor Confusion Does an infinite tree possess any level which is *not* enumerated by a finite natural number n, i.e., which has an infinite distance to te tree? === Subject: Re: Cantor Confusion Does the infinite set of naturals need to contain an infinite natural? No more does an infinite set of levels need to contain an infinite level. === Subject: Re: Cantor Confusion Therefore it suffices to show that the number of paths of the tree is countable for every level L(n) which is enumerated by a finite number. This proof shows that all path in the tree are countable- === Subject: Re: Cantor Confusion it only shows that the set of those paths which have a level , i.e., end, are countable. It says nothing at all about the number of endless paths. A path starts at the root node and follows by branching left or right from every node except a leaf node if any. Match each infinite path with the set of naturals containing n if and only if the n'th branching is to the left, and one easily sees that this bijects the set of all such infinite paths with P(N). And since P(N) is uncountable, so is every set bijectable with it. === Subject: Re: Cantor Confusion Nonsense. The cross section does not measure infinite paths which have an end. Every level L(n) measures, by |L(n)|, how many paths are present at level n. Obviously at each level only a finite set of paths is present. As the sequence of levels does never end, there is an unending growth of this number but it remains provably countable. That is the meaning of infinite paths in an infinite tree. === Subject: Re: Cantor Confusion It does not count the set of infinite paths at all, as it is only defined for finite trees. Does that argument work with the set of terminating decimal or terminating binary fractions being extended to infinite? No, because in both cases the infinite cases are uncountable. WM still has not been able to fault card(S) < card(P(S)), which, in view of the validity of its proof is, not surprising. Nor has he been able to fault my bijection between the set of all infinite paths of an infinite binary tree and the uncountable set P(N). === Subject: Re: Cantor Confusion All your problems would be solved if you could find a unique finite charaterist for each subset of N. Otherwise. Let`s suppose the nodes of your graph are enumerated by naturals and for each path on level L(n) there is a unique finite set which contains the numerals of the nodes that the path includes. Now if you take union of all the finite paths (unique finite subset), you surely will get a countable set, BUT the set contains the infinite paths as subsets which aren`t discrete. === Subject: Re: Cantor Confusion What proof? Those paths aren`t discrete. === Subject: Re: Cantor Confusion Please stick to one pseudonym. Otherwise it is tedious to killfile you. I don't want to read you. === Subject: Re: Cantor Confusion I don`t have other pseudonyms, but fine: I will leave you alone with your hallucinations. === Subject: Re: Cantor Confusion Sorry if I have I have insulted you, but I have mixed you up with someone else. My error. To answer your question: If there are single paths in the tree, then their set is subject to the restriction imposed by the cross sections (= numbers of nodes) of the levels. These single paths cannot exist outside of any level. Every level has a countable number of nodes. If there are no single paths in the tree, ... then: where are they? === Subject: Re: Cantor Confusion Nor is their level ennumerated by a finite number. What is true for every finite (natural) number need not be true for things which are not finite (natural) numbers. Otherwise infinite numbers would have to be finite. === Subject: Re: Cantor Confusion I mean, for every finite level L(n) every path is a root for (uncountable) many infinite long paths. So you can`t take an union of all finite paths L(n) and obtain set which contains every path as a discrete subsets. === Subject: Re: Cantor Confusion Forget it. Use only level L(n) and the countable number of nodes C(n) for every n in N to determine the number of paths crossin this level. If you find this insufficient, then tell me what after every n may be imagined. === Subject: Re: Cantor Confusion Nntp-Posting-Host: hera.cwi.nl The paths in finite trees terminate at soe level n. The paths in the infinite tree do not terminate. Do you not see the difference? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor Confusion The levels which have a countable number of nodes do not terminate either. Therefore there is no difference: As long as paths can exist the cardinality of them is restricted to a countable number. An uncountable set of paths cannot exist other than outside of the tree, i.e., outside of mathematics. === Subject: Re: Cantor Confusion For every function, f, from N to {m,w}, there is a corresponding infinite path whose nth branch is to the left child if f(n) = m and to the right child if f(n) = w, and vice versa. Cantor proved that the set of all such function from N to {m,w} is uncountable. If WM wants anyone to believe that the set of all corresponding infinite paths is countable, he will have to find a way to actually count them, i.e., find a concrete surjection from N to the set of all such paths. === Subject: Re: Cantor Confusion It is wrong. A limit is either approached to any positive eps or it isn't a limit. assumption to be contradicted because set theory states that the set of all natural numbers has the cardinal number aleph_0. anyone else. The cross section C(n) = |L(n)| is the number of nodes of the level L(n). I have to show, an I have shown, that every finite Level L(n) is crossed by as many paths as are nodes in this level. This number of nodes is the countable cross section C(n) = |L(n)|. It is sufficient to have proved this in infinity, that is for EVERY level L(n), to show that the number of paths is countable as long as only nodes with finite indexes n contribute to the paths. If you disagree: What part of a path should not be covered by a node of a level L(n) with a finite n? === Subject: Re: Cantor Confusion Nntp-Posting-Host: hera.cwi.nl We are not doing analysis here. But you now state that your definition: is wrong? Strange as you use it. This makes no sense. You provided it as a *definition*. If not, how do you *define* that thing? See page 193 of Hrbacek and Jech where you will find a definition of limit from which you can derive precisely that. I still see no relation between aleph-0 and the paths. But you have not proven that. You have only proven that there are as many paths that *terminate* at nodes at that level as there are nodes at that level. Each node is crossed by more than one path. Actually each node is crossed by uncountably infinite many paths. If you disagree, are there only two paths in the tree because the cross-section C(1) contains only two nodes? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor Confusion What does approached to any positive eps mean when translated into mathematically coherent language? It certainly has no mathematically sensible meaning as is. As it is only your claim which is being contradicted, and not a claim made by mathematics, it is only you who face such contradiction. Which is only the case when n is a natural number. That has long since been agreed upon by everyone. What is not agreed upon, and is not true, is that any part of that carries over to non-naturals. This number of But it is also necessary to show that it holds for non-natural cardinals or ordinals, and this has not been done. The square root of Newton's apple. Not that my answer makes as much sense as your question. === Subject: Re: Cantor Confusion What else is required? Is it? Are there levels which are not indexed by finite numbers n? Are there infinite natural numbers involved. ... Don't worry. Nobody expected that. === Subject: Re: Cantor Confusion For complete infinite binary trees, in which the number of nodes in a path cannot be a natural number, your definition cannot apply. Except that it carefully omits any representation of any complete infinite binary tree. There are paths which cannot be indexed by natural numbers in the sense of having a natural representing the number of nodes in the path. What natural is the index for the last node of a path? How does WM claim to index the last node when there isn't one? === Subject: Re: Cantor Confusion M yproof is valid for every path which is composed of nodes which have a distance from the root node to be expressed by a natural number. Other paths do not express real numbers. What is the difference between the complete infinite binary tree and a tree which covers only all levels L(n) with finite n? i d no want to represent the number of nodes in the path. I only want to represent the distance from the root node. Is that distance infinite for the complete tree? Is there a last node in the complete infinite tree? I claim to index all nodes - each one by a natural number. === Subject: Re: Cantor Confusion WM has yet to produce a valid proof of anything. If a path has a node at every finite distance from the root, as expressed by a natural number, then it has as many nodes as there are naturals. If all such infinite paths are allowed one has a compete infinite binary tree. In a complete infinite binary tree, there are as many paths as subsets of N, as is shown by building a bijection between them. For each infinite path we can pair off the nodes in their natural order starting from the root node with the members of N, with none of either nodes or naturals left out. Then for each subset, S, of of N, there is a unique path in a complete infinite binary tree having left branchings at the nodes corresponding to those naturals in the set, S, and right branchings at nodes whose naturals are not in S. This is clearly a one-to-one correspondence between infinite binary paths and subsets of N. This bijection between the set of all such infinite binary paths and the set of all subsets of N, namely P(N), establishes that they have the same cardinality, and P(N) is known to be uncountable, so the set of paths is also uncountable. If your tree is is limited to countably many paths and mine is not, clearly your tree is missing most of its paths. Where they might be hiding I don't know, or care. But your tree is incomplete until you find where all those missing paths went, or replace them. Some seem people only to want what they cannot have. There are infinitely many finite distances, corresponding to the infinitely many finite naturals, none of which measure the length of a path with no definable length. So WM is asking for a length which he knows cannot exist. No, but WM seems to claim them for his incomplete trees. But then asks for the index of the last one, which falsely presumes what does not exist. === Subject: Big Bertha Thing Battle Big Bertha Thing Battle Cosmic Ray Series Possible Real World System Constructs http://web.onetel.com/~tonylance/battle.html Access page to 600K ZIP file Astrophysics net ring access site Newsgroup Reviews including misc.health.aids 301 files from the first battle of cyberspace. Big Bertha Thing Battle 1. Do not fight. 2. Fight on ground of your own choosing. 3. Make sure you have got an edge. 4. Is it worth dying for? Spam stands for:- Spleen Paucity Artless Monosylabic It includes newbies, spammers, spam busters and even sometimes victim support. The only good spam is branded spam, which is more Sesame Street than Darth Vader. To the victor the spoils, of the 1st Battle of Cyberspace. (20 family jewels and the 301 files.) This has been a serious exercise, under the auspices in conjunction with OUSA Research. Tony Lance Big Bertha Thing farm The preface from An Elementary Treatment of Gyroscopes and Similar Spinning Tops by Crabtree 1909 Classic Cartoon and animated cartoon of Animal Farm by George Orwell Tony Lance judemarie@bigberthathing.co.uk === Subject: Re: Big Bertha Thing testament Thursday, November 13, 1997 01:53:36 PM Message === Subject: Re: spam and tony Guilty as charged, I should not have done any of the following;- 1. defended you in the mods conf. 2. given you an invitation, when the dogs were at the door. 3. offered a refuge for the explosion survivors. 4. ticked off the mods. 5. put your name up in lights on OUSA Astronomy and Astronomy and Space. 6. Put back the release of my software package, just because of the troubles. Personally I would take me out and shoot me, there is no punishment too bad for any mod, who gets even one complaint. We should be above reproach, like Caesers' wife. I will of course remove the words 'Extract to explain the project to Philip Sims' from all further postings. Please accept my appologies for all the bad things I did, before your elevation to mod. I only pick on little people. Tony Lance. === Subject: Find all n Natural so that the remainder of the division (n^3 + 4n + 5) by (n^2 + 1) is (n - 1) i want to try to understand how to solve this using congruence. If (n - 1) is the remainder i know that it is congruent with (n^3 + 4n + 5) like this: (n^3 + 4n + 5) .81§ (n - 1) (mod n^2 + 1) so now if i apply the definition of congruence n^2 + 1 | (n^3 + 4n + 5) - (n - 1) that is n^2 + 1 | n^3 + 3n + 6 and by my thinking that should have a remainder equal to zero? So if i divide i get remainder 2n + 6 = 0 ? But n should be natural!! i dont know how to do this please help me. === Subject: Re: Find all n Natural Hola a todos, gracias por la ayuda! === Subject: Re: Find all n Natural No, 2n + 6 is not supposed to be zero. It is supposed to be divisible by n^2 + 1. So the question is: For which n is 2n+6 divisible by n^2 + 1? === Subject: Re: Find all n Natural n 2n+6 n^2 + 1 divisible? ___________________________________ 0 6 1 yes 1 8 2 yes 2 10 5 yes 3 12 10 no === Subject: Re: Find all n Natural days. My association with the Department is that of an alumnus. Please do not use the subject of the message as part of the message. Include ALL relevant text in the body. So you want to: Find all natural numbers n... May I impose on you to also understand how to use the shift key and type I instead of i? It is particularly important in mathematical contexts, where i has many other meanings. You need to say congruent modulo n^2+1. [.edited non-ASCII characters.] Yes. I have no idea what your thinking is. But if you have a statement that says that an integer k divides an integer m, then this means that when you divide m by k you should have a remainder of 0. So in this case, you are saying that if you divide n^3 + 3n + 6 by n^2 + 1 then the remainder should be 0. Yes. You are dividing them as POLYNOMIALS, but you are actually considering INTEGERS (or should be). This remainder may be larger than n^2+1 as numbers, in which case you still need to do MORE division. For example, if n=2, then n^2+1 = 5, but 2n+6 = 10, so 2n+6 is NOT the remainder. It is the POLYNOMIAL remainder, but not the INTEGER remainder. You are looking at two different problems. Now: IF 0 <= 2n+6 < n^2+1, then this will indeed give the remainder, which means you need 2n+6 = 0, which says you cannot do it with n a natural number. But what if n^2+1 < 2n+6? Then you don't have the remainder yet. Under what conditions will this occur? n^2+1 < 2n + 6 if and only if n^2 - 2n - 5 < 0 The (real) roots of the quadratic are 1 + sqrt(6) and 1-sqrt(6). The product is negative for any value of n that lies strictly between these 2. Since sqrt(6) is approximately 2.44, the possible values of n for which this occurs are -2, -1, 0, 1, and 2; only n=0, 1, and 2 are natural numbers, so those are the only ones we have to check. If n = 0, you are dividing 5 by 1, but your remainder is not n-1=-1, so no go. If n = 1, then you are dividing 10 by 2, and you want the remainder to be n-1=0, which IS the case. If n = 2, then you are dividing 21 by 5, and you want the remainder to be n=1 = 1, which again IS the case. the only correct answer in that range: n = -3. But you did not deal with the cases where 2n+6 was too big. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: Me good boy.Me get hug. No, it's a big French number: million, milliard, billion, billiard, ... baudrillion, baudrilliard. -- | _ | ( ) ASCII Ribbon Campaign | X Against HTML email & news | / www.asciiribbon.org === Subject: Re: Me good boy.Me get hug. I don't remember either, but I can decipher your 733t message coding: Unibrow cyclops with a pointy nose, frown, cleft chin, with a starry wart, half a ZZ Top tatoo and another chin. I think. I thought those smiley-face things were pass.8e, Fernando? -- === Subject: Re: Function question Klueless ain't clueless. Following along If you set n=m=0 you get f(0)=(f(0))^2, so f(0)=0 or 1 As shown above, 1 doesn't work, so f(0)=0 Setting n=m f(n^2)=f(2n)*f(0)+n^2=n^2 Then n^2-m^2=f(n+m)*f(n-m) f(n+m)/(n+m)=(n-m)/f(n-m) As n+m and n-m are independent variables, the terms above must be constant, and can only be +1 or -1. So f(x)=x or f(x)=-x === Subject: Re: Simple Problem with Arc Length Except that inventing substitutions to stupid contrived integrals has nothing to do with mathematics and is more suitable for aspiring fizzisists. Once you understand what the principle behind the curve length integral is, who cares about the tedious computation? Why is turning students into human CASs such a goal? === Subject: Re: Simple Problem with Arc Length Perhaps you're just not a puzzles type. Dirk Vdm === Subject: Gmat Question - have no clue Hi there, Heres the question: A photographer sets the 6 daughters of the Fuches family in two equal rows.In each row, the girls are ordered from the tallest on the lest to the shortest on the right, and no girl on the second row is shorter then the one in front of her.If no girls are of equal height, how many ways are there to set the Fuchs daughters for a photo? Ron === Subject: Re: Gmat Question - have no clue Call the heights 1,2,3,4,5,6 draw some 2x3 grids, and see how many ways you can fill in. With only 6, how long can it take? === Subject: Re: Gmat Question - have no clue 720 grids. Best to take Virgil's suggestion and pare it down a bit before doing the brute force work. === Subject: Re: Gmat Question - have no clue Let 654/321 indicates the 3 tallest in the back row and 3 shortest in front. It is clear that one must have 6**/**1. It is also clear that one must have either 65*/**1 or 6**/5*1 Given 65*/**1 or 6**/5*1 work out the few remaining possibilities. === Subject: Hausdorff's Manifold Hausdorff has squared Hamiltons matrix and solved for the condition where a zero caused the complex plane to be existent! His manifold appears foundationally confusing. A set theorist must affirm. And I believe it a number. And most mathematicians call it a dimension. See, the unit vector was all it created. And a set of a dimension n does not exist until the unit was applied. He only makes the dimension's scale. So over on sci.physics people ask what it means. And you guys got it wrong. It must only be referred to the expert here. === Subject: Re: Hausdorff's Manifold This is very confusing. How about giving a precise reference (paper title, journal, pages, etc.) to where Hausdorff has done what it is you're confused about, and maybe someone will actually try to track down the reference. Or how about posting an exact quote of what Hausdorff actually said -- this will more likely produce results for you. Dave L. Renfro === Subject: Graduate Math Books for sale Hi All, I have a bunch of graduate level math books for sale that I collected during my time as a graduate student at UCB. They are listed on craigslist at http://sfbay.craigslist.org/eby/bks/291293782.html. They are all in excellent condition and I am selling them for about a quarter to a third of what you could buy them for new. If anyone here wants to buy some off of me, send me an e-mail. I am willing to ship them pretty much anywhere you would like. Dan Giaimo === Subject: Re: Locally Compact Subspaces I fail to see why A U B, where B={1} would NOT be locally compact in this case. R.D. === Subject: Re: Locally Compact Subspaces Because no neighborhood of 1 in A U B is compact. Jose Carlos Santos === Subject: Re: Inequality [ ... ] [ ... ] [ ... ] Sorry for the delay! It took me much more time than expexted to write up the proof. But at least my doubts about it have vanished in the meantime. (Although Maple's implicitplot3d function is still playing tricks on me. ;-) ) Let's hope you'll be able to make some sense of it. Please feel free to ask about unclear points! [ ... ] ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Let's do this again, because the version above is - apart from the basic idea behind it - wrong: for which equality in the inequality above is attained. You might not have noticed on reading my first message: THIS is the reason why you can not prove your inequality using Hoelder's inequality the way you do (and I don't mean that you have mixed up m1 and m3 in your application of Hoelder): The statement above says that there is a whole two-dimensional for which equality is obtained. But when you use Hoelder the way you do, (m1^(2/3)*A1^(1/3)+m2^(2/3)*A2^(1/3)+m3^(2/3)*A3^(1/3))^3 with A3 = a(am1m2+bm1m3)(am1m2+cm2m3) A2 = b(am1m2+bm1m3)(bm1m3+cm2m3) A1 = c(am1m2+cm2m3)(bm1m3+cm2m3), then equality in Hoelder is attained iff A1/m1 = A2/m2 = A3/m3, which comes down to a/m3 = b/m2 = c/m1. in your inequality, it turns into the A-G-M inequality for which equality only holds on the *line* m1 = m2 = m3, and not on a whole surface. Ergo: Using Hoelder this way, you've thrown out most cases of equality; and you can't expect the resulting inequality to hold in general any more... BUT NOW - THE PROOF: ;-) [ ... ] [ ... ] REFORMULATION OF THE PROBLEM: ///////////////////////////////////////////////////////////////////// Define a set S as the set of all triples (X,Y,Z) for which the inequality In this terminology, we want to show that the set is a subset of S. ///////////////////////////////////////////////////////////////////// BASICS ABOUT THE SET S: By definition, S is an intersection of closed half-spaces hence a *closed* and *convex* set. Furthermore, if a triple (X,Y,Z) is contained in S, will also be contained in S. Also important is that S is *non-empty*. - The point (9/4,9/4,9/4) lies in S (by Schur's inequality), is included in S. It is also quite easy to see that all of S is included in the polyhedron (e.g. using one of my favourite principles - symmetry-breaking). But don't worry! - This last point will not be used... ;-) FACTS ABOUT THE SURFACE F: Using the implicit function theorem, one sees that the set F is a smooth surface in the positive octant O, A-priori, the surface F might have more than one (path-)connected component. But fortunately, each ray in O of the form meets the surface F exactly once - at t = (X+Y+Z)^2/(4XYZ) - and thus gives rise to a homeomorphism A minor change in the argument even yields that F is homeomorphic to a plane R^2. It also follows that OF consists of two connected components: and an upper part with (X+Y+Z)^2 - (4XYZ) < 0. So far, so good - for the moment... FURTHER FACTS ABOUT THE CLOSED, CONVEX SET S: Denote by Int(S) the set of interior points of S, and by Fr(S) the set of boundary points of S, Fr(S) := S Int(S). Analogously to the surface F Fr(S) is homeomorphic to a plane R^2, and Int(S) is the upper part of OFr(S). - For proof, use the same rays as above and the fact that S contains an infinite cube (e.g. [9/4,+oo) x [9/4,+oo) x [9/4,+oo) ). This means, we only need to determine the boundary Fr(S). THE FORM OF Fr(S): /////////////////////////////////////////////////////////////////////// CLAIM: The boundary Fr(S) is the union of four smooth surfaces with boundaries as follows: /////////////////////////////////////////////////////////////////////// COROLLARY: /////////////////////////////////////////////////////////////////////// /////////////////////////////////////////////////////////////////////// (PROOF: S is the upper part of OFr(S).) COROLLARY: /////////////////////////////////////////////////////////////////////// The surface F = { (X,Y,Z) in O : (X+Y+Z)^2 = 4XYZ } is contained in S. /////////////////////////////////////////////////////////////////////// PROOF: By the CLAIM, the intersection of F For the intersection of F with the other three sectors of O - for example for the intersection with calculate the value of (X+Y-XY) on F: X+Y-XY = X * (X+Y+Z)^2/(4XYZ) + Y * (X+Y+Z)^2/(4XYZ) - X * (X+Y+Z)^2/(4XYZ) * Y * (X+Y+Z)^2/(4XYZ) = ((X+Y)*4XYZ - XY*(X+Y+Z)^2 )* (X+Y+Z)^2/(4XYZ)^2 = -xY * ( X^2+Y^2+Z^2+2XY-2XZ-2YZ ) * (X+Y+Z)^2/(4XYZ)^2 = -xY * (X+Y-Z)^2 * (X+Y+Z)^2/(4XYZ)^2 <= 0. This means that the surface F is contained in the upper of the two parts into which the octant O is divided by the surface { X+Y = XY }. so F is contained in S. This argument also shows that there are no intersection points of the surface F with the surface { X+Y = XY } except along the plane { X+Y=Z }. - This will be used later. Analogously for the remaining two parts of F. Q.E.D. PROOF OF THE FORM OF Fr(S) (Pt. I): HOW TO DISTINGUISH BETWEEN Int(S) AND Fr(S) *MAIN FACT*: Let (X,Y,Z) be an arbitrary point in S, } Then (X,Y,Z) lies in Fr(S) if and only if for which EQUALITY IS ATTAINED in the inequality above. PROOF, PART 1: For (X,Y,Z) in Int(S), if there was a triple (A,B,C) =/= (0,0,0), Then the coefficient A(A+B)(A+C) would be positive for this triple, and thus none of the points (X',Y,Z) with X' < X could be in S, because (A,B,C) would contradict the inequality for these (X',Y,Z). This however contradicts to (X,Y,Z) lying in Int(S). PROOF, PART 2: Given (X,Y,Z) in Fr(S), we have to find a triple (A,B,C) == (0,0,0) which yields equality in the inequality. For each (X,Y,Z) in Fr(S), there holds one of the following alternatives: X = inf { X' : ( X',Y ,Z ) in S } =: X_inf(Y,Z) or Y = inf { Y' : ( X ,Y',Z ) in S } =: Y_inf(Z,X) or Z = inf { Z' : ( X ,Y ,Z') in S } =: Z_inf(X,Y). Otherwise S would contain the points (X_inf(Y,Z),Y,Z), (X,Y_inf(Z,X),Z), (X,Y,Z_inf(X,Y)), their convex combination p := (X_inf(Y,Z)/3+2X/3, Y_inf(Z,X)/3+2Y/3, Z_inf(X,Y)/3+2Z/3), and the whole infinite cube p + [0,+oo)x[0,+oo)x[0,+oo). However, (X,Y,Z) would be an interior point in this infinite cube and hence also an interior point of S. - Contradiction! So assume w.l.o.g. that Z = Z_inf(X,Y). An equivalent definition of Z_inf(X,Y) is as sup { [(A+B+C)^3-X*A(A+B)(A+C)-Y*B(A+B)(B+C)]/[C(A+C)(B+C)] : for which the sequence of quotients [(An+Bn+Cn)^3-X*An(An+Bn)(An+Cn)-Y*Bn(An+Bn)(Bn+Cn)]/[Cn(An+Cn)(Bn+Cn)] Because our inequality is homogeneous (of degree 3) in A,B,C, we can scale the sequence so that it satisfies An+Bn+Cn = 1 for all n. so w.l.o.g. we can assume that the sequence (An,Bn,Cn) i.e. with (A,B,C) =/= (0,0,0). Z_inf = [(A+B+C)^3-X*A(A+B)(A+C)-Y*B(A+B)(B+C)]/[C(A+C)(B+C)], i.e. X*A(A+B)(A+C) + Y*B(A+B)(B+C) + Z_inf*C(A+C)(B+C) = (A+B+C)^3 for A,B,C from this limit point. - Equality is attained. If the limit point satisfies C=0, then in order that the quotient [(An+Bn+Cn)^3-X*An(An+Bn)(An+Cn)-Y*Bn(An+Bn)(Bn+Cn)]/[Cn(An+Cn)(Bn+Cn)] converges to the finite(!) value Z_inf = Z, the numerator has to converge to zero: (An+Bn+Cn)^3-X*An(An+Bn)(An+Cn)-Y*Bn(An+Bn)(Bn+Cn) (A+B+C)^3-X*A(A+B)(A+C)-Y*B(A+B)(B+C) = 0 Fortunately, lim Cn = C = 0 also implies: 0 = Z*C(A+C)(B+C); and equality is again attained, this time for a triple with C=0. End of PROOF of the *MAIN FACT*. APPLICATION OF THE *MAIN FACT* IN THE PROOF OF THE FORM OF Fr(S) (Pt. II): THE FORM OF THE ENDS The MAIN FACT tells us that in order to show that we only have to prove that for every triple (X,Y,Z) and (to each such triple (X,Y,Z)) there are points (A,B,C) =/= (0,0,0) for which equality is attained. (Analogously for the other two ends of Fr(S), PROOF OF THE INEQUALITY: There holds: X * A(A+B)(A+C) + Y * B(A+B)(B+C) + Z * C(A+C)(B+C) Equality is attained if C = 0. The last line has for fixed A,B,C the form X*a+Y*b, the condition (X-1)(Y-1) = 1 [a reformulation of XY = X+Y]: X*a + Y*b = a + b + (X-1)*a + (Y-1)*b Equality is attained if (X-1)*a = (Y-1)*b . In our case: a = A(A+B)(A+C) + C(A+C)(B+C), b = B(A+B)(B+C) + C(A+C)(B+C); and the ratio b/a becomes B^2/A^2 for C = 0. Remains to be proven that I don't know why, but it turns out that if you expand the difference of left-hand side and right-hand side of this inequality into monomials in A,B,C, all coefficients (of these monomials) become positive, thus proving the inequality. In fact the difference can be brought into product form: 4ab - ((A+B+C)^3 - a - b)^2 = 4 * (A(A+B)(A+C) + C(A+C)(B+C)) * (B(A+B)(B+C) + C(A+C)(B+C)) - ((A+B+C)^3 - A(A+B)(A+C) - B(A+B)(B+C) - 2C(A+C)(B+C))^2 = C^2 * (4*B*A+4*A*C+3*C^2+4*B*C) * (A+B+C)^2 with equality if C=0. Thus the inequality is proven; and equality occurs for (A,B,C) = ( 1/sqrt(X-1), 1/sqrt(Y-1), 0 ). Q.E.D. APPLICATION OF THE *MAIN FACT* IN THE PROOF OF THE FORM OF Fr(S) (Pt. III): Fr(S) AS AN ENVELOPING SURFACE AND THE MIDDLE PART OF Fr(S) The MAIN FACT tells us that the boundary Fr(S) of S is *contained* *in the enveloping surface* of the two-parameter family of planes X*A(A+B)(A+C) + Y*B(A+B)(B+C) + Z*C(A+C)(B+C) - (A+B+C)^3 = 0, in other words, of the family X*A(A+B)(1-B) + Y*B(A+B)(1-A) + Z*(1-A-B)(1-B)(1-A) - 1 = 0, A _complication_ is that the domain of the parameters contains boundary points. - Ignore this for the moment: it would be sufficient to set the function X*A(A+B)(1-B) + Y*B(A+B)(1-A) + Z*(1-A-B)(1-B)(1-A) - 1 and its partial derivatives with respect to A, B equal to zero, and solve this system for A,B: X*A(A+B)(1-B) + Y*B(A+B)(1-A) + Z*(1-A-B)(1-B)(1-A) = 1 X*(1-B)(B+2A) + Y*B(1-B-2A) + Z*(1-B)(B-2+2A) = 0 X*A*(1-2B-A) + Y*(1-A)(A+2B) + Z*(1-A)(A-2+2B) = 0 'Solving' in the sense of eliminating the parameters and obtaining an implicit equation for X,Y,Z. Elimination of A yields (according to Maple): ********************************************************************* A = -1/2 (2Z-(Y+X+3Z)*B+(X+Y+Z)B^2) / (-(Z+X)+(X+Y+Z)*B), and two remaining conditions on B: (-4Z-4X+4XZ) +(4Y-8XZ+4X+4Z)*B + (-X^2+2XZ+2YZ+2XY-Y^2-Z^2)*B^2 + 2(X+Y+Z)(-Y+X+Z)*B^3 - (X+Y+Z)^2*B^4 = 0 4XZ(-Y+X+Z) + 2(X+Z)(X^2-2XZ-2YZ-2XY+Y^2+Z^2)*B - (X+Y+Z)(Y^2-8YZ-8XY+10XZ+7Z^2+7X^2)*B^2 + 4(2X+2Z-Y)(X+Y+Z)^2*B^3-3(X+Y+Z)^3*B^4 = 0 ********************************************************************* becomes HOPELESS. - Still, even over the complex numbers, the two polynomial equations in B can only have roots in common if their *resultant* vanishes: Maple gives for the resultant of these to quartic polynomials in B the factorization: -256 * Y^2 * (X-Z)^2 * ((X+Y+Z)^2-4*X*Y*Z) * P(X,Y,Z) * (X+Y+Z)^6 with the symmetric polynomial P(X,Y,Z) = 6 * (X^3*Y^3*Z+X^3*Z^3*Y+X*Z^3*Y^3) + (X^5*Y*Z+X*Z^5*Y+X*Y^5*Z) - 4 * (X^4*Y^2*Z+X^4*Y*Z^2+X^2*Y^4*Z+X*Z^4*Y^2+X^2*Z^4*Y+X*Y^4*Z^2) - 124 * (X^3*Z^2*Y^2+X^2*Z^3*Y^2+X^2*Y^3*Z^2) + 588 * X^2*Z^2*Y^2 + 222 * (X^2*Y^3*Z+X*Y^3*Z^2+X^2*Z^3*Y+X*Z^3*Y^2+X^3*Y^2*Z+X^3*Y*Z^2) + 35 * (X*Y^4*Z+X*Z^4*Y+X^4*Y*Z) + 4 * (Y^4*Z^2+Z^4*Y^2+X^4*Y^2+X^4*Z^2+X^2*Y^4+X^2*Z^4) - (Y^5*Z+Z^5*Y+X^5*Z+X^5*Y+X*Z^5+X*Y^5) - 6 * (X^3*Z^3+X^3*Y^3+Z^3*Y^3) + (Z^5+Y^5+X^5) - 31 * (Z^4*Y+Z^4*X+Y^4*Z+X^4*Z+X^4*Y+X*Y^4) - 98 * (Z^3*Y^2+Z^3*X^2+Y^3*Z^2+X^3*Z^2+X^3*Y^2+X^2*Y^3) - 340 * (Z^3*Y*X+X^3*Y*Z+X*Y^3*Z) - 618 * (X^2*Y*Z^2+X^2*Y^2*Z+X*Y^2*Z^2) + 324 * (X*Z^2*Y+X*Z*Y^2+Y*X^2*Z) + 162 * (Z^2*Y^2+X^2*Y^2+X^2*Z^2) + 108 * (Z^3*Y+Y^3*Z+Z^3*X+X^3*Y+X*Y^3+X^3*Z) + 27 * (Y^4+Z^4+X^4). So one might want to conclude X*A(A+B)(1-B) + Y*B(A+B)(1-A) + Z*(1-A-B)(1-B)(1-A) - 1 = 0, (X+Y+Z)^2 = 4*XYZ or P(X,Y,Z) = 0 or X = Z. There are two points against this simple version: (1.): When A was eliminated, Maple divided by (-(Z+X)+(X+Y+Z)*B) without caring wheter this term might actually become zero: Fortunately, when this term becomes zero - for B = (Z+X)/(X+Y+Z) - one of the equations to be satisfied for the envelope, X*(1-B)(B+2A) + Y*B(1-B-2A) + Z*(1-B)(B-2+2A) = 0 turns into Y*(X-Z)/(X+Y+Z) = 0, so that we arrive at the _already listed_ case X=Z. (2.): When calculating the envelope of X*A(A+B)(1-B) + Y*B(A+B)(1-A) + Z*(1-A-B)(1-B)(1-A) - 1 = 0, we have to take extra care for the boundaries of the parameter region - A = 0, B = 0, resp. A+B = 1: We have to calculate the envelopes of the corresponding one-parameter families - Y*B^2 + Z*(1-B)^2 - 1 = 0, 0 <= B <= 1, X*A^2 + Z*(1-A)^2 - 1 = 0, 0 <= A <= 1, resp. X*A^2 + Y*(1-A)^2 - 1 = 0, 0 <= A <= 1, separately. This is much easier: The envelopes of these families are the cylindrical surfaces ALL IN ALL, we can say, that the boundary Fr(S) of S is contained in the union of the six sets (surfaces) (X+Y+Z)^2 = 4*XYZ Y+Z = YZ X+Z = XZ X+Y = XY P(X,Y,Z) = 0 X = Z. It remains to exclude the two latter alternatives P(X,Y,Z) = 0 and X = Z. EXCLUDING X = Z is easy: Since the set S, and thus also Fr(S), is by definition invariant under arbitrary permutations of the coordinates X,Y,Z, any point in Fr(S) that does not satisfy one of the five first conditions, will not only have to satisfy the sixth condition X = Z, but also X = Y and Y = Z. We already know the *only* point of Fr(S) that lies on the ray X = Y = Z - the point (9/4,9/4,9/4) - and this point satisfies the first condition (X+Y+Z)^2 = 4*XYZ. So the condition X = Z is superfluous. THE LAST STEP: The point (9/4,9/4,9/4) lies in the surface (X+Y+Z)^2 = 4*XYZ, CLAIM: Inside this region, there are no intersection points of the surface (X+Y+Z)^2 = 4*XYZ with either of the other sets Y+Z = YZ, X+Z = XZ, X+Y = XY, P(X,Y,Z) = 0. Once we know this, the part of Fr(S) that connects (9/4,9/4,9/4) to the already known parts of Fr(S) must inevitably lie within the surface (X+Y+Z)^2 = 4*XYZ. intersects Fr(S) in exactly one point; and this point must also be the unique point of intersection of this ray with the surface (X+Y+Z)^2 = 4*XYZ. Thus: Q.E.D. PROOF OF THE CLAIM ABOUT THE INTERSECTION POINTS OF THE SURFACE (X+Y+Z)^2 = 4*XYZ WITH EITHER OF THE OTHER SETS Y+Z = YZ, X+Z = XZ, X+Y = XY, P(X,Y,Z) = 0: We have already established that the sets Y+Z = YZ, X+Z = XZ, X+Y = XY only intersect the surface along the planes Y+Z = X, X+Z = Y, resp. X+Y = Z. Now, for the values of the polynomial P(X,Y,Z) on the surface (X+Y+Z)^2 = 4*XYZ, there holds: P(X,Y,Z) = P(X*(X+Y+Z)^2/(4XYZ),Y*(X+Y+Z)^2/(4XYZ),Z*(X+Y+Z)^2/(4XYZ)) = (X+Y+Z)^12 * (Z^2-2*Y*Z-2*X*Z-2*X*Y+X^2+Y^2)^3 / (16384*(XYZ)^6 ) (Just another 'coincidence'!?) So the only intersection of P(X,Y,Z) = 0 with (X+Y+Z)^2 = 4*XYZ is along the cone X^2 + Y^2 + Z^2 -2XY - 2YZ - 2ZX = 0. This cone is the circular cone with vertex (0,0,0) and axis (1,1,1) which contains the points (1,1,0), (0,1,1), and (1,0,1). The latter are the three points whose positive linear span is the the polyhedral cone so the intersection of P(X,Y,Z) = 0 with (X+Y+Z)^2 = 4*XYZ That's all for the moment. Tell me what you think about it! Thomas Mautsch === Subject: Some pitfalls of dynamic geometry software My homepage at http://mysite.mweb.co.za/residents/profmd/homepage4.html Journal, and the usual math quote & cartoon. Michael de Villiers === Subject: Re: Some pitfalls of dynamic geometry software My homepage also has a (bi-monthly) newsletter which has information on upcoming math & math ed conferences, links to interesting math websites, math quotes & humor, etc. === Subject: Re: The Heliocentric Universe The main point is: Consider two cartesian coordinate systems, one centered with earth and one with th sun. And let both systems have as the principal direction in space the line connecting the center of earth with the center of the sun at the time (spring), when the axis of earth is at an right angle to this line. As a plane of reference take the plane swept out by this line throughout an year. With this - by using right angles - one has three directions, pairwise at an right angle to each other. Now if sun has the coordinates in the earth centered system of ( a, b, c ) , so earth has the coordinates ( - a, - b, - c ) in the sun centered system. Physicists claim, that energy can not be destroyed or created, it just changes form and that is called work. Can You elaborate on this life changing it's appearance? I read things like this in books of Arthur C. Clarke. ...... ...... They are equivalent from a mathematical point of view. Physics teaches that the speed of an object rotating at the distance r from the center of rotation is 2 * pi * r / (time of one revolution). Example for the heliocentric system for the star sirius gives the speed of more than 50 * 300 000 km/ sec, and 365 * 50 * 300 000 km/ sec for a geocentric system ( sirius is rotating in a day minus 4 minutes around the earth - this is what one might think is what he sees, looking at sirius). The calculation of the energy sirius has in these system, i leave to You. So from a physical point of view i would rectrict the use of an egocentric system centered at the relativists head to a smaller radius. With friendly greetings Hero === Subject: Re: The Heliocentric Universe These ideas look to me to be originally those of Pavel Florensky, the Russian polymath. -- LS === Subject: Re: The Heliocentric Universe ... ... I wasn't quite exact. Some improved comments on reference systems. There's no center of universe, so other considerations let us choose in one case the earth as a center of a coordinate system, and in other cases the sun ( or the common center of gravity of sun and earth and the other planets) - and in medicine one chooses the center of a patient. To avoid faults, it is agreed, that if a doctor speaks about the left arm of a patient, it is not the left arm from the point of view of the doctor, but always the patient itself is the reference. More general, are there connections between different parts of the universe, just as with our body? In contrast to a center, there are existing universal directions. In every space-ship is a gyroskop, three axes of rotation, which keep their directions, however the spaceship is moving around. With such a gyroskop, mounted on earth, we also can find out, how long one rotation takes -and it is about 24 hours minus 4 minutes, which matches to the observation of distant stars, for this reason called fixed stars. So there are physical advantages in choosing these directions. But if one chooses the line sun-earth as being the reference, one has a rotation difference of one year between a universal and this system and in this case one get's to the speed for the star sirius of more than 50 * 300 000 km/ sec. Choosing a direction from the center of earth to a point on earth ( like Greenwich for meridian zero) will give a speed of 365 * 50 * 300 000 km/ sec for sirius in such a geocentric system. These arguments shattered before him the crystal sphere as a shell for the universe already in the middle ages ( and probably already in hellenistic times). I read in wiki about him: ,,Among other things he proclaimed that the geometry of imaginary numbers predicted by the theory of relativity for a body moving faster than light is the geometry of the kingdom of God. ,, Just another kind of keeping us fenced in. With friendly greetings Hero === Subject: Re: The Heliocentric Universe He's quite a compelling figure. His real world was shattered by the attack on the Church; he, too, became a martyr. He was connected to the analysts at Moscow State, in particular Egorov, and was a high-powered scientist in his own rite. His cosmology is, in view of modern mathematics, simplistic, but it has the advantage that indeed, it is simple. He put the earth at the center of the universe, and then made arguments regarding where the edge of the observable universe is, given that we seem to be spinning. I like to use it as an example of a subtle notion within relativity: the choice of the origin is arbitrary. Use the one that works best. -- LS === Subject: Re: Integrals don't necessarily preserve strict inequality? I am assuming that the function _f_ is measurable; otherwise, I wouldn't be able to talk about the integral of _f_. But one possible definition of measurable function is: _f_ is measurable if, for every real number _a_, Jose Carlos Santos === Subject: Re: Integrals don't necessarily preserve strict inequality? This is actually a rather subtle question. For Lebesgue integrals, Riemann integrals, using only the machinery of Riemann integration, it's not so easy. Basically, you need to reproduce (part of) the proof that a Riemann-integrable function is continuous a.e. And of course the result is false on an interval [a,a] :-) -- Ron Bruck === Subject: Re: Integrals don't necessarily preserve strict inequality? days. My association with the Department is that of an alumnus. I agree, but the point is that it is equivalent to the question at hand. If the statement is true, then doing the integral of g-f will establish the result. If the statement is false, then it is easy to use any counterexample h to obtain counterexamples f and g that work. I probably agree. Which may be why the answer was marked wrong? -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: Trying to define a set theory. Zuhair, no disrespect but these days set theory is for losers! If you are interested in foundations and that kind of thing, get into category theory and topoi. I highly recommend the following book: Topoi - The Categorical Analysis of Logic by Robert Goldblatt, Dover Books, 2006, ISBN 0-486-45026-0. On the first reading I kind of faded out around page 120, from the plethora of definitions and constructs; but on the second reading I've now reached page 200 and still going strong (just). So it must be good, although it is geared more towards logic, about which I don't care a hoot, than algebraic geometry which is my main motivation for getting into it. John R Ramsden === Subject: Re: Trying to define a set theory. It's a big improvement zuhair. You stated the features of interest; you stated your question as to consistency; and you gave the axioms with theorems justifying the definitions, one by one. I haven't checked out all your proofs and formulas. But at least your presentation is starting to take shape. I see you've made several emendations. So perhaps I'll wait until you've worked on it for a while so that you have it in some kind of semi-finished form altogether again in one post. MoeBlee === Subject: Re: Trying to define a set theory. Ok. Moe the following is the final work. The following is a set theory that I developed latelly. I don't know if it is consistent or not? Hereby I present it to be tested. Introduction: This theory is an extension of first order logic with identity. It contains three types of classes. Types of classes 1) A unique class of all classes that is not a member of any class other than itself. And this class is T. 2) classes other than T, that only T contain them as members, and these classes are called 'proper classes'. 3) classes that are members of a class other than T, and these classes are called 'sets' Structure: All sets are members of one unique proper class V. All proper classes are subclasses of V and are not in V. T contains all classes and itself as members. Defining the theory. Name of the theory is Y . Theory Y is the set of sentences entailed (from first order logic with identity)by these axioms: 2) Universe: E!xAy(yex & ~Ez(~z=x & xez)). 3) Regularity: 4) Schema of Global comprehension: if P is a formula in which x doesn't occure free, then all closures of: are axioms. Theorum 1) V. we get To prove x is unique, let x1 and x2 be defined as for a proof by negation suppose that ~x1=x2 Then there should be a set u: uex1 and ~uex2 Or u: uex2 and ~uex1.(Extensionality). if u: uex1 and ~uex2 , then x2 is not the class of all sets, violating its definition. if u: uex2 and ~uex1 , then x1 is not the class of all sets, violating its definition. Therefore x1=x2. This means the following. Now since x is unique then we proceed to name it. Accordingly V is the proper class of all sets. Theorum 2) ~VeV. Proof: let VeV. then we have Ez(~z=T & Vez) otherwise ~VeV. (The definition of V). so ~V=T, since ~Ez(~z=T & Tez). definition 1. Since VeV then pairing applies to it and we have from pairing B={V}. Now from pairing we have BeV and since ~V=T, then B is a member of a set that is not T therefore ~B=T. And since B has V as a member then ~B=0. Thus regularity applies to B.. But B vioates regularity! A contradiction. Thus ~VeV. Theorum 3) ~T=V Since ~VeV (theorum2) and since VeT (axiom 2). Thus ~V=T (Extensionality). 5) Empty set: ExeVAy(~(yex)). xen))). This axiom simply states that the union class of a set is a set and the union class of a proper class or T is a proper class or T respectivelly. so we have the following theorum. Theorum 5) small binary union Proof: Since a and b are sets then from pairing c={a,b} is a set. and Uc here is the unique x in this theorum. were aeV and beV. were{x}comes from pairing x with itself. were ~a=T. So this axiom simply states that the power class of a set is a set and the power class of a proper class is a proper classs. wo we have the following theorum were ~a=T. were ~a=T. Theorum 7) Large binary union. Proof: follows directly from global comprehension,since y is always a set , because it is a member of a or b which are not T. so we have aUb even when a is a proper class or b is a proper class'. Theorum schema 8) Large separation if Q is a formula in one variable in which x doesn't occure free then all closures of are theorums. Proof: since ~a=T and yea then y is a set in global comprehension yields theorum 8. Theorum 9) Cartesian Product: Proof: since neither a nor b are T then for every u every w such that uea and web ,u and w are sets. ( definition of set ). Therefore substituting it in global comprehension yields theorum 9. axb is the cartesian product of a and b. Another way to prove the existance of axb or every a:~a=T and b:~b=T is to define it in the following manenr. were PP stands for power class of power class. c is unique , this follows from pairing, power, large union , large separation. Theorum schema 10) Relation if P is a formula in one variable in which x doesn't occure free then all closures of are theorums. Proof: it follows from theorum 8 & 9. This theorum schema allows us to define functions, injections, surjections and bijections in the standard way between any two classes weather sets or proper classes provided that any of them is not T. Theorum 11) Non supernumerosity of subclasses. Proof: Let f be the identity function f(x)=x from x to a. Then f is injective from x to a. So it cannot be supernumerous to a ( definition 13). Theorum schema 12) Large replacement. if Q is a formula in two variables in which b is not free,then all closures of are theorums. Proof: since yeV then y is a set, then substituting it in global comprehension we get theorum schema 12. Theorum 13)Non supernumerosity of the range of any function over its domain were dom(f) and range(f) defined in the standard manner. Proof: if y is supernumerous to x , then there exist at least two ordered pairs in f such that for the same uex there are different sets w and k in y. i.e there must exist violating the definition of a function. 10) limitation of size: Thoerum schema 14) small separation if P is a formula in one variable in which x is not free then all Proof: follows from large separation and non supernumerousity of subclasses and axiom 10. Theorum schema 15) small replacement. if P is a formula in two variables in which b is not free then all closures of are theorums. Proof: this follows from large replacement and non supernumerosity of the range of any function over its domain, and axiom 10. Another proof: Since for every yeb there exist x in a and since for every y1eb and y2eb such that ~y1=y2 there exist at least x1ea and x2ea such that ~x1=x2 Then there exit an injection from b to a, Thus ~b is supernumerous to a. Theorum 16) Proper binary union. Proof: This follows from large union and non supernumerousity of subclasses and limitation of size. Theorum 17) global Numeration ( entails global choice present in NBG) Ax(~x=T) Ey( y is an ordinal & y is equinumerous to x). Proof: a theorum in this theory [because if we let substituting P(x) in global comrpehension yields. y is proved to be unique in a similar manner to how uniqueness of V is proved.]. xez))). Now we have ~OeO as a theorum in this theory (Proof similar to ~VeV). Thus O is the proper class, i.e ~Ez(~z=T & Oez) ,of all ordinals that are sets. and O is itself an ordinal( definition 16). Since we have ~Ez(~z=T & Oez) then O is not subnumerous to V ( axiom 10). Now since O subclass_of V, then O is not supernumerous to V(theorum 11). so O is neither super- nor sub- numerous to V. then O is equinumerous with V. Since O is an ordinal, then global numeration is applicable to V. the proof that for every subset of V there exist an ordinal that is equinumerous to is simple, but a little bit long_ It can be proved that every subclass k of O is well ordered and thus there is an ordinal to which k is bijectable. Now for every subclass of V , either it is a proper class in which it would be bijectable to O, or it would be a set in which it would be subnumerous to O, and thus bijectable to a proper subclass k of O, which is bijectable to an ordinal in O and so every set is bijectable to an ordinal in O. Theory definition finished. / The above is the theory with the important defining theorums in it There are a lot of theorums concerning proper classes,but these are to be discussed later. The important results that I think this theory can acheive are 1) Prove ZFC 2) define intersections 3) defining cardinalities of proper classes. As for 1) it is clear that all axioms in ZFC except infinity and pairing and empty are theorums in this theory, while infinity and pairing and empty are axioms here. so ZFC is in this theory. 2)Theorum of Interesection: in this theory. the problem was when y=0 , what would be x. here the answer is x=T. so x=intersection y is solved here. 3) This theory defines relations not only between sets but also between proper classes, and between sets and proper classes. Since we can have equivalence classes of all sets i.e Frege definition of cardinality, but this won't work for the case of proper classes, therefore Von Neuman's method is better, since it defines the cardinality of proper classes which is O. Cardinality is defined here in the following manner. so the intersectional class of all ordinals bijectable to V is O. so card V = O. ( Just to clarify that: Definition of Cardinality ~Ez(z is an ordinal & z is equinumerous to x & zey)). The definition of x is an ordinal is given (see above). The definiton of equinumerousity is also given , it is: Of course Ax(~x=T)E!y(y is an ordinal & y is equinumerous to x & ~Ez(z is an ordinal & z is equinumerous to x & zey)). is a theorum in this theory. It can be easily proved. Now card(V) = O. x is subnumerous to V (axiom 10). we know that O is equinumerous to V. Then there do not exist an ordinal in O that is equinumerous to V. then O is card V.) T in this theory is somewhat detached from the other sets and proper classes, you cannot define subclasses of T that are not subclasses of V, Union T = T. Power is not defined for T. You cannot have separation on T, nor replacement , T is outside the reach of regularity and numeration and thus choice. So the main function of T is intuitive: it has them all including itself as members, besides it might lead to the emergence of theorums that requires the existance ot T, a situation similar to theorum of intersection. Now this theory might be balantently inconsistent, or might be consistent, I don't know, that's why I presented it here , in order to be checked. Zuhair === Subject: Re: Trying to define a set theory. Correction: from a to b). So definiton 12 is: Zuhair === Subject: Re: Trying to define a set theory. I will make some changes so that it look in a better shape, I think I only need some few superficial changes. Zuhair === Subject: Expression simplification I am trying to simplify this expression: -h_bar^2*(-sin(phi) d/d(theta) - cot(theta) cos(phi) d/d(phi)) (-sin(phi) d/d(theta) - cot(theta) cos(phi) d/d(phi)) The result I am getting is: -h_bar^2*(sin(phi) d/d(theta) sin(phi) d/d(theta) - sin(phi) csc^2(theta) cos(phi) d/d(phi) + cot(theta) cos^2(phi) d/d(theta) - csc^2(theta) cos(phi) sin(phi) d/d(phi) - cot^2(theta) cos(phi) sin(phi) d/d(phi) - cot^2(theta) cos(phi) sin(phi) d/d(phi)) I am not sure if this is the correct result. Any input is much appreciated. === Subject: Re: Expression simplification -h_bar^2*(-sin(Phi) * d/dTheta - cot(Theta)*cos(Phi) * d/dPhi) * (- sin(Phi) * d/dTheta - cot(Theta)*cos(Phi)*d/dPhi) It looks like you can re-express this as the square of the divergence of a vector valued function F with F(Theta, Phi) = [ -sin(Phi), -cot(Theta)*cos(Phi) ] yielding -h_bar^2 * (div F)^2 Taking the divergence as the del operator dot F div F = [d/dTheta, d/dPhi] . [ -sin(Phi), -cot(Theta)*cos(Phi) ] = (d/dTheta)(-sin(Phi)) + (d/dPhi)(-cot(Theta)*cos(Phi)) = 0 + cot(Theta)*sin(Phi) = cot(Theta)*sin(phi) Thus giving -h_bar^2 * (div F)^2 = -h_bar^2 * (cot(Theta) * sin(Phi))^2 = -h_bar^2 * cot^2(Theta) * sin^2(Phi) So a simplified result would be -h_bar^2 * cot^2(Theta) * sin^2(Phi) === Subject: Re: Comprehensive Solutions Manual for College Textbooks Please send email to sbooks4sale@hotmail.com if you are interested. === Subject: Re: Comprehensive Solutions Manual for College Textbooks Fast delivery guaranteed within 12 hours of payment. Most of the times under 6 hours. === Subject: Re: Proving the four color theorem Here is something that is bothering me! The sum of the degrees of all the vertices of a graph must be an even number. Therefore, if all the vertices of a cubic planar graph have degree = 3, then a CPG cannot have an odd number of vertices. ---Bill J However, not every cubic planar === Subject: Re: Proving the four color theorem It would probably be more precise to say Cubic means every vertex has at most 3 edges. -- Matthias Hofmann Anvil-Soft, CEO http://www.anvil-soft.com - The Creators of Toilet Tycoon http://www.anvil-soft.de - Die Macher des Klomanagers === Subject: Re: Proving the four color theorem Taking a stab: A Hamiltonian circuit, being closed, already implies n + 1 paths, the origin and endpoint counted twice. An odd degree of vertices, therefore, sums to an even number of paths in a finite graph. To my knowledge, the problem of determining whether every cubic planar graph has a Hamiltonian cycle, is known to be NP-complete. Tom === Subject: Re: Proving the four color theorem ---Bill J === Subject: Re: Proving the four color theorem When Lee R. agreed with you, I realized that you have a valid point. There are two different ways of using a graph to represent a map. 1. The borders become edges and the vertices are where the borders make a T intersection 2. The countries become vertices and the borders become edges. You are thinking in terms of the first definition and I am thinking in terms of the second. But which graph was the op thinking of? ---Bill J === Subject: Re: Proving the four color theorem I was thinking of the second way, which is examplified here: http://en.wikipedia.org/wiki/Four_color_theorem#Formal_statement_in_graph_th eory My assumption is that when you turn a map into a graph this way, a vertice can have at most 3 edges. This seems to prove that no more than 4 colors are needed to color a map in such a way that no two adjacent regions receive the same color. However, the statement I quoted in my original post says something different. -- Matthias Hofmann Anvil-Soft, CEO http://www.anvil-soft.com - The Creators of Toilet Tycoon http://www.anvil-soft.de - Die Macher des Klomanagers === Subject: Re: Proving the four color theorem days. My association with the Department is that of an alumnus. What? Surely you jest. There are certainly maps where there are countries with more than 3 neighbors. Just take two concentric circles, and divide the outside circle into four sections. The graph that represents this map has a vertex in the middle connected to each of the four vertices surrounding it, and hence has more than 3 edges. Yes, I know such a map can be 4 colored. That's not the point. The point is that th eassertion that a verte[x] can have at most 3 edges is just plain wrong. all mutually adjacent in any portion of the map, and so, with respect to those 4 vertices alone, each vertex will have at most 3 edges. But there can be other edges not related to those 4. However, the fact that a map cannot contain 5 mutually adjacent vertices is necessary for a 4-coloring to exist, but not sufficient. You can have maps in which no more than 3 vertices are mutually adjacent, but that need 4 colors anyway. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: Proving the four color theorem No doubt he was thinking that one of the canonical mistakes in erroneous proofs of the four color theorem is to assume, mistakenly, that every vertex has degree 3 somehow implies the graph can be 4-colored. Lee Rudolph === Subject: Re: Proving the four color theorem But where's the mistake in this assumption? It seems obvious to me. -- Matthias Hofmann Anvil-Soft, CEO http://www.anvil-soft.com - The Creators of Toilet Tycoon http://www.anvil-soft.de - Die Macher des Klomanagers === Subject: Re: Proving the four color theorem Huh? A graph where every vertex has degree <=3 can certainly be 4-coloured. I think the canonical mistake is that some vertex has degree <= 3 implies the graph can be 4-coloured. However, in this case the graph where every vertex has degree 3 is the dual of the one you want to 4-colour. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Proving the four color theorem Do you mean vertex coloring or edge coloring? ---Bill J === Subject: Re: Prove this law Husam === Subject: Re: Prove This Law Thnkyou very much rob ,This was important for me. I can not anderstand this statement (My language english is not exelent) Husam === Subject: Re: Prove This Law In your reply, the ASCII graphics (used to display the mathematical formulae) that you quoted were unreadable. Usually this is because either the news reader or news poster (usually the same program) are not set to use a monospace font. A monospace font is necessary to align the characters in ASCII graphics. If you were able to read the formulae in my post, then your reader is set properly, and most likely it is the program you use to post formulae in my post, then your reader is not set properly and should be set to use a monospace font, such as Courier. take out the trash before replying === Subject: Aleph-null Again.. Take the natural number set, N = {1,2,3,4,...}. Cardinality of N is aleph-null. (aleph-null)+1=(aleph-null) and consequently, (aleph)+p=(aleph-null) for any natural number p. n(A+B)=nA+nB for two distinct sets A and B. [nX stands for cardinality of X] For any natural number p, p-n{p}=p-1=q is equal to 0 or is another natural number. {1}+{2}+{3}+{4}+...=N Therefore, nN=n{1}+n{2}+n{3}+n{4}+... Since nN+p=nN for any natural number p, nN=1+2+3+4+... Therefore, 1+2+3+4+...=aleph-null It's my third time writing about transfinite cardinalities and I still haven't got the concept yet. What is wrong with above reasoning (I know there's gotta be something wrong)? === Subject: Re: Aleph-null Again.. Like wit says every term in the language of set theory must be finite. === Subject: Re: Aleph-null Again.. Here the + denotes the usual addition defined for natural numbers. Note that expressions like a_1 + a_2 + a_3 + ... where a_i e IN{0} (for i = 1, 2, 3, ...) are not defined (since the result, if it existed, could not be a natural number). This will be of importance later. See below. I guess, you mean: {1} u {2} u {3} u {4} u ... = N. ~~~~~~~~~~~~~~~~~~~~~~~~~~~ | But actually that's not a defined term in math (set theory). Though some authors DO use it... :-( But there is a correct way to express what you mean/want, it's the /union/ operation: U {n} = N. neN This can also be written the following way: U {{n} : n e IN} = N , or even this way: U {{1}, {2}, {3}, ...} = N. Note that in a set symbol the usage of ... is allowed (if it is used in this way). Nope. Your therefore is wrong. (Note that your claim is not supported by any proof.) Moreover: #N may be defined to be aleph_0, ok. But #{1} + #{2} + #{3} + #{4} + ... or 1 + 1 + 1 + 1 + ... is not defined. (See comment above.) F. === Subject: Re: Aleph-null Again.. I think this is OK. Usually 0 is a natural number. You are not adding any natural number to n{1} + n{2}.....= 1 + 1 ... because all natural numbers are finite. First you are dealing with singleton sets, but cardinality of {3} is not 3, it is |{3}| = 1. === Subject: Re: Aleph-null Again.. My suggestion is that your time would be much better spent trying to obtain a copy of Abraham A. Fraenkel's Abstract Set Theory (2'nd edition, 1961) and going through it. Most college libraries should have a copy, but you should be able to find a used copy on-line. I can't think of a better book for someone really interested in learning cardinal and ordinal numbers without a lot of potentially confusing (to new-comers) overly formal logical symbolism. It's also an excellent reference for who did what, when, and how -- something that most set theory texts either ignore, get wrong, or give the wrong impression by selective omission. The book is very readable and suitable for self-study. Dave L. Renfro === Subject: Re: Aleph-null Again.. Right there, I don't know what that sequence of symbols is supposed to say. What set theory book are you using? MoeBlee === Subject: Re: Aleph-null Again.. There, I meant the union of the subsets of N that contain exactly one element each is equal to N. Sorry about that, I don't know how to write the union symbol on this thing :( === Subject: Re: Aleph-null Again.. Its {1} U {2} U {3} ... But alas the line: Is undefined. I gather you have some form in this newsgroup. You would be better off learning about this subject by studying a book or the web, and posting questions about what you don't understand. Wikipedia has got tons on this (and there are lots of other excellent sites). The real theories are more interesting than any you could ever make up (at least, with your current knowledge). === Subject: Re: Aleph-null Again.. On Sat, 10 Mar 2007 18:24:27 +1100, Peter Webb No, not really. It's properly stated the following way: U {n} = N. neN This can also be written the following way: U {{n} : n e N} = N , or even this way: U {{1}, {2}, {3}, ...} = N. Note that for a /set symbol/ the usage of ... is acceptable (if it is used in this way). F. === Subject: Re: Aleph-null Again.. Peter Whang a .8ecrit : It is the third time people tell you your ideas about addition of cardinalities are, well, lets say ill-defined === Subject: Re: Aleph-null Again.. Could somebody tell me more about it in terms such that I could understand. I know little about the subject and want to know where I got it wrong.. === Subject: Re: Aleph-null Again.. On Sat, 10 Mar 2007 01:00:24 EST, Peter Whang explain where I went wrong? The things you're doing simply don't make any sense at all. ************************ David C. Ullrich === Subject: Re: Aleph-null Again.. Peter Whang a .8ecrit : Why not read a book? And get *definitions* ? === Subject: Re: Do reals really have to be genuine numbers? On Fri, 9 Mar 2007 14:46:12 UTC, Eckard Blumschein [Nit-pick: as someone has pointed out elsewhere, Salviati represented the mature Galileo. The Simplicio who appears in that excerpt is -- unlike the one in the Dialogue -- pretty much the young Galileo who accepted the teachings of the philosopers.] Perhaps you could expand on why his paradox makes any assertion or assumption about the continuum. Simplicio is indeed talking about all the points on a line; but Salviati's demonstration makes no assumption at all about continuity or countability. That is, it has a perfectly sound interpretation without any such assumption. Using the simplest known infinte set, the natural numbers, Salviati shows he concludes that comparing two infinite sets as to which is larger than the other is trickier than Simplicio imagines it to be, which of course is true. He concludes, in fact, that one simply can't compare the magnitudes of two infinite sets, which of course is now known to be false. (Or, if you're anti-Platonic enough: is now false). Win a few, lose a few; but his demonstration shows the weakness in Simplicio's inference, just as it was supposed to do. Naturally he didn't talk specifically about countability, which didn't get into the vocabulary for another 250 years; so I assume the reference above is to the fact that the natural numbers are countable. But it can't be held that he thought that the natural numbers could account for all the points on a line! Hence I don't see any connection here between a countable set and a continuum. -- Dan Drake dd@dandrake.com http://www.dandrake.com/ porlockjr.blogspot.com === Subject: Re: Galileo's Paradox On Fri, 9 Mar 2007 14:50:49 UTC, Eckard Blumschein the question. By methods developed a mere 250 years later, we can make distinctions among infinite sets and make certain comparisons of magnitude. Salviati's broad conclusion that such comparisons are impossible was, as it turned out, over-broad. Mathematicians abandoned it. -- Dan Drake dd@dandrake.com http://www.dandrake.com/ porlockjr.blogspot.com === Subject: Re: Galileo's Paradox Well, you seem to be spouting the same nonsense as before. Where have you been the last couple of months? Someplace nice and quiet, I hope. -- David Marcus === Subject: Re: continuum hypothesis and 0=1 It would help if you included some context in your posts. Whose definition of truth is it that you're finding difficult to grasp? -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: continuum hypothesis and 0=1 I just would like to understand the following excerpt from G?del's incompleteness paper (Section 1) let us suppose that the proposition [R(q);q] were provable; then it would also be true. === Subject: Re: continuum hypothesis and 0=1 Pierre-Yves.Gaillard@iecn.u-nancy.fr says... Would you understand the following statement? If 2+2=4 is provable, then 2+2 = 4. That's a simple implication of the form If A then B where A is a statement about provability, and B is a statement about numbers. In this case, 2+2=4 actually is provable (at least after translating it into whatever language we're working with) and 2+2 actually is equal to 4. Assuming that you understand this, let's try a little trickier case. Do you understand the following statement? If 2+2=5 is provable, then 2+2 = 5. It's a little trickier, because 2+2 is *not* equal to 5. But that's okay, because 2+2=5 is not provable, either. So when Godel says let us suppose that the proposition [R(q);q] were provable; then it would also be true. you can, if you need to, go through Godel's description of how to construct the formula [R(q);q]. Unravel all the definitions, and what you have left is a regular, ordinary formula of arithmetic. It will be a *huge* formula, involving maybe thousands of symbols, and perhaps taking dozens of pages of paper to write it all down. However, although it's complicated, it's just an ordinary mathematical statement, involving arithmetic and logical notions of addition, multiplication, equality, quantification, etc. So you can translate Godel's claim into an ordinary statement of the form let us suppose that [insert your huge formula here, enclosed by quotes] were provable; then [insert your huge formula here, but this time, not enclosed by quotes]. Note that after translating, the word true no longer appears *anywhere* in the translated statement. Of course, if you actually did that, you would probably say: How in the world can anyone understand such a huge complicated statement? Isn't there a more understandable, compact way to say it? The answer is yes, you can greatly simplify it by saying, instead let us suppose that the proposition [R(q);q] were provable; then it would also be true. -- Daryl McCullough Ithaca, NY === Subject: Re: continuum hypothesis and 0=1 Here is a tentative explanation for G?del's let us suppose that the proposition [R(q);q] were provable; then it would also be true. To each statement S G?del attaches a statement Bew(S) whose metamathemathical interpretation is S is provable. He shows that Bew(S) implies S is provable. Then he says S is true to mean S is provable, and S is provable to mean Bew(S) is provable. It seems to me that mainsteam mathematicians use the words provable and true as strict synonyms. At any rate I can see no other meaningful definition of the word true. In particular I'm sorry for not understanding your If 2+2=4 is provable, then 2+2=4. For me 2+2=4 is provable is a metamathematical statement, whereas 2+2=4 is a mathematical statement. I can't see how a metamathematical statement can imply a mathematical statement. But I appreciate your effort. === Subject: Re: continuum hypothesis and 0=1 Pierre-Yves.Gaillard@iecn.u-nancy.fr says... No, S is true does *not* mean S is provable. No, they don't. That's a misunderstanding on your part. Even if it were possible to prove every true statement (which it isn't, according to Godel's theorem), it still is not the case that S is true means the same thing as S is provable. Let's look at the simple case 2+2=4. To say that it is *true* is to say that a certain arithmetic operation, adding two numbers, produces a certain result, 4. To say that it is *provable* is quite different. To say that it is provable is to say There exists a sequence of statements such that each statement is either an axiom or follows from previous statements by the rules of inference and such that the last statement is 2+2=4. Now, you may conjecture that the second statement (2+2=4 is provable) follows from the first (2+2=4 is true) or vice-versa, but they aren't the same statements. Well, that's a flaw in your understanding of mathematics. Yes, that is correct, they are statements about different things. But it is still possible for one type of statement to imply a different type of statement: Joe has 2 apples. Sam has 3 apples. If they combine their apples, they have 5 apples. To reach my conclusion, I had to use some real world facts about how many apples each person has, together with mathematical facts about adding 2 and 3. That's what makes mathematics useful in the real world is that once we've proved that 2+3=5, we can use that fact to draw conclusions about apples, or theorems, or whatever. In Godel's case, the metatheory is an *extension* of the object theory. Godel's proof takes place in a setting in which one can talk about natural numbers and one can *also* talk about formulas and proofs. It's no more mysterious than my being able to talk about apples and also about arithmetic facts. If it were not possible to mix types of sentences this way, then mathematics would be perfectly useless. Maybe that's your conception of mathematics? That it's useless? -- Daryl McCullough Ithaca, NY === Subject: Re: continuum hypothesis and 0=1 He doesn't. Bew(S) implies S is not provable in the formal theory provability in which Bew formalizes, unless S happens to be provable in the theory (for theories meeting the criteria of the incompleteness theorems). Bew(S) implies S is true, and can be recognized as true, when we're dealing with a formal theory which we recognize as sound (for the class of statements S belongs to, e.g. the class of arithmetical statements) such as Peano arithmetic or the system of Principia Mathematica. Bew(S) implies S is not true, on the other hand, when Bew formalizes provability in the theory PA + PA is inconsistent. Truth of statements in the language of arithmetic, in the language of analysis and so forth, is a mathematically defined property, and there is nothing metamathematical or metaphysical about it, any more than there is about any mathematically defined notion. You'll find the definition in any decent textbook on logic. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: continuum hypothesis and 0=1 PRA is more than enough. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: continuum hypothesis and 0=1 Pierre-Yves.Gaillard@iecn.u-nancy.fr a .8ecrit : Why? I am a Bourbaki's fan. But I simply dont think it is a treatise on mathematical logic (nor one on the many other subjects, like probabilities or analytic number theory, I already mentioned) but I can only refer you to the Sorry, I once knew it by rote, but it has been many years. Could you cite? But anyway, I simply dont believe you could be right. Remember the notice : the Treatise addresses to people knowing how to count. And this is why I *everytime* translate in that system. Could you *at last* say what is your point, except arguing for the sake of arguing, with no real agenda? === Subject: Re: continuum hypothesis and 0=1 I just would like to understand G.9adel's incompleteness proof translated in Bourbaki's theory. === Subject: Re: continuum hypothesis and 0=1 Pierre-Yves.Gaillard@iecn.u-nancy.fr a .8ecrit : Is not that your problem, not ours? See, you are the one who want to use for doing formal logic a badly adapted notation. On the other hand, if you would stop playing dumb and just read our answers, it would become very easy : just follow G.9adels definitions and notations, and use P is true for P is provable in Bourbaki's system (like you say you always do); P is true in model M for some complex property of the set (with structure) M, constructed from P, is true, and provable in theory T for there exists a chain of formulas of T such that ..., as we have already explained to you many times. Then G.9adel's results are certainly true, as they are not only provable in the Bourbaki system, but in (much) weaker ones... Their metamathematical meanings, on the other hand... === Subject: Some informal intro to derived categories Yo, I posted on my blog http://sirix.wordpress.com an informal introduction to derived categories. I think it could be useful for someone who already knows some basics of homological algebra and wants to get some view on a subject before starting some real studies. homological algebra? What I basically refer to is that HA is very unexciting to learn and I'm curious how other people cope with that. -- [ sirix ]------------sirix-at-univ-szczecin-pl------------- http://sirix.wordpress.com (my maths&physics-oriented blog) === Subject: Re: Some informal intro to derived categories Originator: jdolan@math.UUCP (James Dolan) |Yo, |I posted on my blog http://sirix.wordpress.com an informal |introduction to derived categories. I think it could be useful for |someone who already knows some basics of homological algebra and |wants to get some view on a subject before starting some real |studies. | |homological algebra? What I basically refer to is that HA is very |unexciting to learn and I'm curious how other people cope with |that. i learned homological algebra by trusting my instincts that told me that something that seemed so incredibly boring couldn't possibly be worth learning; instead i learned other intrinically more exciting subjects such as homotopy theory and category theory which revealed the true motivating ideas behind homological algebra so clearly as to make it largely unnecessary to learn homological algebra itself. (see also: and: .) -- jdolan@math.ucr.edu === Subject: Re: Some informal intro to derived categories This one showed me how strongly I don't get at all homological algebra and category theory. And actually I don't have an idea what to do to get them. Let me try to ask some meaningful question, though. Apart from studying structures for itself, mathematicians tend to invent new definitions to solve some concrete problems (say, Fermat's Last Theorem or Poincare Hypothesis), in the very end. So, what is derived category invented for? More precisely, why Grothendieck came up with this idea AND why people are still interested in it? -- [ sirix ]------------sirix-at-univ-szczecin-pl------------- http://sirix.wordpress.com (my maths&physics-oriented blog) === Subject: compactification Let X be a countably compact subspace of the Stone-Cech compactification BN of the discrete space N of natural numbers. Let a be a point of X and Y = X {a}. Show that the subspace Y is also countably compact. === Subject: Re: automorphism of quadratic field K need not be an ordered field in general, so it's not clear what you mean by assuming a < b when a,b are from K. Is K necessarily a real quadratic field? === Subject: Re: automorphism of quadratic field You can start by simplifying the problem: Define c = b-a. Now, since s is an automorphism, s(b-a) = s(b) - s(a), so the problem now is: If c is a positive element of Q(sqrt(d)), when is s(c) positive? If c = a + b sqrt(d), then s(c) = a - b sqrt(d). So, if b is negative then s(c) is necessarily positive, but if b is positive, it seems you have basically just saying that s(c) is positive. So I don't think you can find anything better than: If the sqrt(d)- part of b is smaller than the sqrt(d)-part of a then it must be true. Avital. === Subject: Re: Innumeracy: Where to start? I don't know about yours, but my local Borders has an excellent collection of low-cost math review books. (The Barnes and Noble does not seem to have the same selection.) They go down to the lowest level, usually called basic mathematics. Look at them and pick out ones that cover the material you need in a manner you would like. You can also hire a tutor to help you out, if you feel you need one. Keep up the good work! -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: Innumeracy: Where to start? [...] Eli Maor has written at least three books on the history of mathematics; I have his book Trigonometrical Delights and I relly like it. Maor has a gift for explaining concepts without much assumed background. Also, he's good at telling stories. The other book I have from him is all about the history of the number e and John Napier's invention of logarithms; also very interesting, in part because it traces the developments of ideas about logarithms from Napier through Briggs and after that, I don't know. Another of Maor's books on the history of mathematics is To Infinity and Beyond: A Cultural History of the Infinite. it seems interesting. There is an on-line review of _To Infinity and Beyond_ at the page with link < http://mathdl.maa.org/convergence/1/?pa=content&sa=viewDocument&nodeId=1029 === Subject: Re: Innumeracy: Where to start? boundary=----=_NextPart_000_0093_01C763D6.F1AB3930 --------------------------------------------------------------------- It is unclear to me how your very interesting books about the history of maths will help someone who wants to improve their understanding of fundamental mathematical concepts. I know that the British Prof Ian Stewart is very highly regarded as a maths writer. I have found this by him: Concepts of Modern Mathematics If you look at the above titles on Amazon then you can read the reviews of this and other related books - and make your own choice. Nick === Subject: Re: Innumeracy: Where to start? I teach probability and statistics in the oil drilling industry. Most students just what to understand math well enouph to do their job well, which I assume is similar to your interest. To get by better in the course you are in I suggest the paperback book the cartoon guide to statistics by Larry Gonnik and Walcott Smith. It is $20 at Barnes and Noble. It is a good book and may bring what you are hearing in class down where the rubber meets the road; where variance and standard deviation are described in terms of hours you spend watching TV or weights of students in your class. I may be biased (another part of the study of statistics) but I think you are starting in th right place. Statistics and probability allow you to forcast and if you can forcast with knowledge of the uncertainty involved you can make better decisions. Most people do math to solve problems. Most problems have to do with life decisions about the future. Life problems about the past and present are accounting. Maybe that is your next class. === Subject: Re: How to solve the probability density of z=ax+w? Hui Zhou === Subject: Re: How to solve the probability density of z=ax+w? E[X | Z] = (1 - c) EX + c Z, where c = Var(X) / {2 [Var(X) + Var(W)]}. Hints towards the derivation follow. E[X | Z] = E[E[X | Z, A] | Z] X and Z are conditionally independent given A = 0. X and Z are conditionally jointly normal given A = 1. Apply the bilinearity of covariance and the formula E[U | V] = EU + Cov(U,V) / Var(V) (V - EV) for jointly normal U and V. Others in sci.math have given you the density. If this is homework, be sure to acknowledge sources of assistance. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: How to solve the probability density of z=ax+w? I just don't know how do you get this Pr(x and z) ? -- Hui === Subject: Re: How to solve the probability density of z=ax+w? If a = 0, then z = w. Now, x and z are independent. So, the density is N(mu_x, sigma_x, x)*N(mu_w, sigma_w, z). If a = 1, then z = x + w. There are a couple of ways to handle this. You could note that (z,x) is bivariate normal and calculate the mean and variance of this vector. Or, you can notice that given x, z is normal with mean mu_w + x and variance the same as w. So, the density is N(mu_x, sigma_x, x)*N(mu_w + x, sigma_w, z). -- David Marcus === Subject: Re: How to solve the probability density of z=ax+w? === Subject: *** The most profound and the most fragrant words *** The most profound and some of the most morally fragrant words by the True jewish rabbis of the neturei karta ... please visit www.nkusa.org for more information When the fear of the Almighty is Absent All Becomes Permitted including adulteries, 911 controlled demolitions and mass murders ..... ....................................................... When the fear of the Almighty is Absent All Becomes Permitted A Neturei Karta Statement on the Parade and Demonstration in Favor of Moral Abominations in Jerusalem November 9, 2006 The Glory of Mercy of the Almighty fills the universe. He is the center of existence and of the lives of all men. We are summoned to serve Him and obey His law in a spirit of joy and gratitude. If this is done, all will be well. It will be well for us as individuals and as members of groups and nations and peoples. The rejection of Divine Law and its good and just norms can only yield tragedy. Today in the Holy Land in the holy city of Jerusalem just such a sorrowful tragedy is being played out. A march is planned that will advance the absurd and evil notion that the most basic elements of Divine moral law, accepted by all nations down through the ages, are a source, not of peace and blessing but of oppression, G-d forbid. The public acceptance and assent to perversion as a legitimate way of life is cruel and barbaric. It wars against the very decency that lies at the root of life's spiritual and material bounty, purity and happiness. Thousands of Jews and many others have taken to the streets of Jerusalem to protest this planned desecration. But at root the desecration is a symptom of far greater evils. It is a symptom of all the materialist rejections of the Divine that have plagued our people over the last two centuries. This rejection has spawned the evil of Zionism with its rejection of the Divine plan of exile and redemption and its cruelty and arrogance towards other nations. Without the light and true kindness of Torah there will be, Heaven forbid, rebellion and cruelty and, yes, perversion. With humility and compassion in our hearts we humbly pray that those Jews and all people ensnared by these evils should repent and realize that they are the victims of pernicious propaganda. When returning to Torah and the Divine plan they will find that which they seek, truth and beauty, family and empathy and then and, only then, will they be able to be truly gay, in the real sense of the word. May the Almighty grant that it be so, speedily in our days. === Subject: Re: *** The most profound and the most fragrant words *** Amen to everything including the 9/11 remarks. Let us all pray that the evil forces behind the 9/11 attack are exposed and brought to justice. Some folks talk about a news item that mentions some Israelis were dancing with joy as the towers fell. I firmly believe that the original unadulterated Torah, even the one written by Ezra (may peace be upon him)can bring peace, tranquility, and chastity to the world. Muhammad === Subject: Re: *** The most profound and the most fragrant words *** ************************************************** ...and so have we seen along the centuries, brother: peace, love, justice ALL over the place anywhere in the world where religion (ANY religion) rules....oh yeah! Tonio === Subject: attn: Aatu Koskensilta (Godel's incompleteness Thm.) For FOL theory of Peano arithmetic, I've been thinking about what is provable in a system (PA+G) but not in PA, where G is a Godel sentence for FOL PA. First off, it seems to me that the G obtained by following through on the methods exposed in, e.g. Hofstadter's ``Godel, Escher Bach would tend to be very long since one has to render in arithmetic predicates such as : the sequence of sentences with Godel numbers s1, ... s_k is a proof in PA of the sentence with Godel number t. As for consequences of PA+G that are unprovable in PA, clearly some have many characters (such as G itself). I've tried to determine how short a theorem of PA+G that is a non-theorem of PA can be. Although I've thought about this for a number of hours, and used Google to search for anything related to this, I've made zero progress. The only thing I've found which is somewhat related to this is that some researchers went looking for short hard results in FOL PA and they observed, by submitting problems they couldn't solve to number theorists, that statements equivalent to Diophantine equations occurred often as short hard results (or maybe also as problems as yet unsolved). David Bernier === Subject: Re: attn: Aatu Koskensilta (Godel's incompleteness Thm.) That's a non-trivial task. Off hand, I don't recall any interesting result on that question. But see Friedman's posts on FOM on completeness of PA for simple sentences and schemas, e.g. http://cs.nyu.edu/pipermail/fom/2006-October/011050.html Of course, it's also good to bear in mind that the length of the formalisation of some mathematical statement in the language of arithmetic needs have nothing to do with the complexity of the statement as ordinarily understood. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: sequential === Subject: Congruence Theorem Hy I dont understand this theorem in respect with an example of congruence relation. Theorem: The necessary and sufficient condition for 2 integers a and b to be congruent modulo m, is that they give the same remainder when divided by m. Problematic example: 14 .81§ -8 (11) excuse me please if what I am asking too simple or stupid, but -8 isnt divisible (with a remainder, like 14) by 11, so how can I apply the theorem? === Subject: Re: Congruence Theorem === Subject: Re: Congruence Theorem If we write a = q m + r, then we always have to pick the quotient q so that the remainder r satisfies 0 <= r < m. If a = -8, then we have to take q = -1 in order to make r = 3. So, -8 = -1 * 11 + 3, 3 = 0 * 11 + 3, 14 = 1 * 11 + 3. -- David Marcus === Subject: Is this function integrable? Let X={(r,theta): 0