mm-3899 === Subject: : looking for a coauthor Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I am Prof. of Economics. Developed some results in application of Godel's theorem. Look for a coauthor to finalize the formalization. Anatoly Kandel === Subject: : Re: looking for a coauthor Originator: tchow@lebesgue.mit.edu.mit.edu (Timothy Chow) Originator: bergv@math.uiuc.edu (Maarten Bergvelt) It would be helpful if you could post an abstract of your results. This would help identify the area of expertise that your potential coauthor needs to have. Merely saying Goedel's theorem is too broad and vague. -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === Subject: : Re: looking for a coauthor Originator: bergv@math.uiuc.edu (Maarten Bergvelt) The following is honestly an attempt to be helpful: You might first try posting the results on sci.logic or somewhere to get opinions on whether they're valid. If things are not already formalized that doesn't sound good - most informal applications of Godel's theorem, especially to fields other than mathematical logic, are somewhat bogus. === Subject: : Re: A question on differential eqution Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Or what amounts to the same thing: If we want A, B, C such that y_1 and y_2 both satisfy A y'' + B y' + C y = 0 that says that the vector (A, B, C) should be orthogonal to (y_j'', y_j', y_j) for j = 1, 2. So we take (A, B, C) to be the formal cross product) (A, B, C) = (y_1'', y_1', y_1) x (y_2'', y_2', y_2). === Subject: : existence of fractional iteration Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Let S be the class of strictly increasing continuous functions on the there is at most one set of fractional iterates f^t of S with f^t(x)/x However for the *existence* of f^t(x) there is always some additional property (for example first derivative of f) assumed. Is this due to the inability to prove without those additional properties, or are there counterexamples? === Subject: : Re: existence of fractional iteration Originator: bergv@math.uiuc.edu (Maarten Bergvelt) f^[r](x) r in R is not always uniquely defined. --See f1^[1/2](x) with f1(x) = 4x +3 verified by f1(x) = 2x +1 and -2x -3 . -- f^[2](x) = x, here any involutive function will work. A well known general case is f(x) = phi^-1(k +phi(x)), k a constant , defined for fractional iterates. Example: f(x) =cx may be written :c^(1 +ln(x)/ln(c)). Derivatives for f is not a nice condition to legitimate fractionnal iteration or f^[r](x) Alain === Subject: : Principal ring Originator: bergv@math.uiuc.edu (Maarten Bergvelt) let p be a prime number and u=exp(2iPI/p). Z refers to integers. My question: is Z[u] a PID ? What about the cases p=7,11,13,17,19 ? === Subject: : Re: Principal ring Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Yes for all these. This is due to Montgomery and Uchida. Indeed all cyclotomic fields of class number one are known due to Masley and Odlyzko: they are Z[exp(2pi i/m)] for m = 1, 3, 4, 5, 7, 8, 9, 11, 12, 13, 15, 16, 17, 19, 20, 21, 24, 25, 27, 28, 32, 33, 35, 36, 40, 44, 45, 48, 60 and 84. This is dealt with in detail in Chapter 11 of Washington's book on cyclotomic fields. Victor Meldrew === Subject: : Re: Principal ring Originator: bergv@math.uiuc.edu (Maarten Bergvelt) The answer is yes. You are talking about class numbers of cyclotomic fields, these keywords give you lots of references, e.g., Washington's Introduction to Cyclotomic Fields. o. === Subject: : structure constants of commutative matrix algebras Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I'm studing the properties of structure constants of matrix algebras. For some particular (commutative) algebras that I am looking at the structure constants themselves define an algebra with the same structure constants...this is a somewhat unexpected self referencing property. Here's a concrete example to show what I mean : let C1,...,Cn be the matrices corresponding to the classes of a finite group in its regular (right or left) representation. Then C_i C_j = Sum_k a_{ijk} C_k and C_i C_j = C_j C_i Now define Bi as the matrix with ij-th entry a_{ijk} then you have B_i B_j = Sum_k a_{ijk} B_k and B_i B_j = B_j B_i I don't know if this result is known (or maybe even follows trivially from something else). Any references or keywords that might lead to references on this are appreciated. R.N. PS. Below are such matrices for symmetric group of order 6 C_1= [[1,0,0,0,0,0], [0,1,0,0,0,0], [0,0,1,0,0,0], [0,0,0,1,0,0], [0,0,0,0,1,0], [0,0,0,0,0,1]] C_2= [[0,1,0,1,0,1], [1,0,1,0,1,0], [0,1,0,1,0,1], [1,0,1,0,1,0], [0,1,0,1,0,1], [1,0,1,0,1,0]] C_3= [[0,0,1,0,1,0], [0,0,0,1,0,1], [1,0,0,0,1,0], [0,1,0,0,0,1], [1,0,1,0,0,0], [0,1,0,1,0,0]] B_1= [[1,0,0], [0,1,0], [0,0,1]] B_2= [[0,1,0], [3,0,3], [0,2,0]] B_3= [[0,0,1], [0,2,0], [2,0,1]] (a=[B1,B2,B3]) === Subject: : multiplicative number theory over finite field Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Please give me a pointer for the following known(I presume) result; In Hardy and Wright's Introduction to Number Theory one can find the asymptotic formula for the number of integers which has k prime factors and is less than X as X goes to infinity. (Here k is fixed.) An analogue should be as follows; p = a prime number F = finite field with p elements F[X] = the polynomial ring M(N) = the number of monic polynomials od degree N in F[X] which has k(fixed integer) irreducible factors Then M(N) is asymptotic to (log N)^{k-1} / (k-1)! N as N goes to infinity. with best wishes, S. Hahn KAIST, South Korea === Subject: : Re: multiplicative number theory over finite field Originator: tchow@lebesgue.mit.edu.mit.edu (Timothy Chow) Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I think this is in Knopfmacher's book Abstract Analytic Number Theory. If not, that should at least get you pointed in the right direction. -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === Subject: : Re:Minimum d'une fonction convexe Originator: bergv@math.uiuc.edu (Maarten Bergvelt) On peut citer deux reponses,bien connues: La premiere.Dans la direction du resultat mentionne par le premier message, on rappelle la formulation suivante ( peut-etre plus pratique): Theoreme. Soient (E,I.I) un espace de Banach reflexif ,A subset E un convexe ferme,non vide et f:A ----]-infty,infty ] une fonction convexe,semi-continue- -inferieuremet,f non identitiquement egal .88 infty et telle que: lim f(x)=infty Alors f admet un minimum sur A. Comme exemple simple f(x)=Ix-aI.(a in E) Pour la demonstration de ce resultat,voir H.Brezis-Analyse fonctionnelle(Masson).P46 La deuxieme.Le resultat suivant generalise un resulat dans la droite reelle: Theoreme. Soit E un espace de Banach quelconque.f est une application convexe de E dans IR.On suppose qu'il existe un z dans E tel que delta_{+}f(z)=0. (delta_{+}f(z).x=lim [f(z+hx)-f(z)]/h ,avec z,x dans E) Alors , f admet un minimum en z. pour la definition de delta_{+}f(z),voir R.H.Martin.Nonlinear operators and differential equations in Banach spaces.p.37.Ensuite,on montre aisement l'inegalite: Enfin,remarquons que, si f est strictement convexe,alors,f admet un minimum unique. E.Hanebaly(Rabat) http://www.analysefonctionnelle.blogspot.com === Subject: : question about Fourier transforms Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Perhaps I should say first that I am no analyst! I have what is probably a very naive question about Schwartz space. If G is a locally compact abelian group and G-hat its Pontryagin dual then my understanding is that Fourier transform gives an isometric isomorphism between L^2(G) and L^2(G-hat)---a very general duality result. However I am wondering whether one can formulate duality results of this nature for certain spaces of continuous functions rather than L^2 spaces. As an example, let S be Schwartz space on R^n, i.e. the rapidly decreasing C^infinity functions on G:=R^n. In this case G-hat is also isomorphic to R^n and Fourier transform induces an isomorphism of vector spaces between S and S. Another example: if G is locally compact, abelian, and totally disconnected, and moreover G is the union of compact subgroups, then I think that one can let S_G be the locally constant functions on G with compact support; similarly one can let S_{G-hat} be the locally constant functions on the dual group G-hat with compact support: if my understanding is correct then Fourier transform now induces an isomorphism of S_G with S_{G-hat} (and will send convolution to pointwise multiplication, as in the real case). I am naively wondering whether there is a notion of Schwartz space for a general locally compact abelian group G, and this Schwartz space should be a (biggish) *subspace* of the continuous functions on G (so this rules out L^2(G), at least for most G), and whether one can prove that Fourier transform induces a bijection between Schwartz space for G and for G-hat. I'm sorry that this question is so vague! For example one logically valid answer would be one could define Schwartz space to be zero for all G---but I would rather have some general natural-looking definition that specialised to the cases above where I think I know what Schwartz space should be. Perhaps a test case would be G=T, the circle group, so G-hat is the integers. Now Fourier transform sends a continuous function on T to an element of ell^2(Z) but I don't really know what the image of the continuous functions in ell^2(Z) is. I am wondering whether one can isolate a specific biggish subspace of continuous complex-valued functions on T such that one can say precisely what the image of this subspace is in ell^2(Z). Kevin Buzzard === Subject: : Re: question about Fourier transforms Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I'm pretty sure nobody else knows a nice characterization of this either. I have no idea what the answer to the general question is; I doubt that a satisfactory answer exists in that generality. But the answer for the circle is well known - it's analogous to the answer for R^n. If f is in C^infinity(T) then the sequence f^(n) is rapidly decreasing, in the sense that for any k there exists c with |f^(n)| <= c/(1+|n|)^k for all n. Conversely, if c_n is a rapidly decreasing sequence then there exists f in C^infinity(T) with c_n = f^(n). The proofs in both directions are quite simple. === Subject: : Re: looking for a coauthor Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Ä.8c.87.98 Ä.87.9a.8e.8a ï.8f.87.94you for your message. I agree that many (if not most) applications of Godel's theorem outside logic might look bogus for a pure mathematician. But, like beautiful, a judgment regarding what is and is not bogus depends on the beholder's personal perception. In my industry, the applications are centered around the following idea: under such-and-such conditions, decision makers work with noncomputable functions and, hence, cannot be sure that the results of their models produce consistent expected values of control variables. [Times].8fis this idea worth of attention? Because it introduces .9b.94.8b.8c.98.99.87.8e.94[ OHat].9d .8e¨.8c¨ .8a.8c.8b.8e.97.8e.95.94 .97.8e.99.9b.87.99.8e.95[IHa t].97in which only nonnormal distributions make sense. By implication, decision makers should develop some rather special mechanisms of diagnostics what they are dealing with. Specifics of these mechanisms are the subject of major concern and interest. work with a recursive decision problem whose time horizon is, say, 10-15 years. Dynamics of 15-20 control variables is described by multi-valued (interval-valued) recursive functions. In other words, the control variables take values in some intervals. The number of such functions in uncountably infinite and, hence, some of these functions are not algorithmically (recursively) solvable. Decision makers cannot decide whether or not expected values of certain control variables are consistent. In Economic analysis, multi-valued functions are usually referred to as $,1r|(Bcorrespondences.$,1r}(B Undecidability of such correspondences has been used in studies of expectations of equilibrium prices. The authors built their proofs on theorems by Gold (Information & Control, 1967, 10: 447-474) & Blum & Blum (Information & Control, 1975, 28: 125-155). These theorems are covered in Osherson et al (Systems that Learn, 1986, MIT). ï.95 .99.8fbest of my knowledge, multi-valued functions were never applied to solving the class of decision problems that I deal with. The minimum task is to apply the already developed formalisms to this class. å.8c.97.99 '.94.87.99.95.93 === Subject: : =?iso-8859-1?q?2=BDD_convex_hull/pseudo-traveling_salesman_problem?= Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Hi - I'd appreciate any suggestions or references on that might help solve the following problem. In reality, the problem I'm trying to solve is quite abstract, and that's I thought it best to recast it as an everyday the following problem: Given: two irregularly-shaped confines situated on different parts of a gently undulating terrain. Each confine is defined by a series of stakes driven into the ground at various distances, and a wire passed round the stakes forming a closed region. For example, two square- shaped regions (confines) on different parts a sphere (terrain). Task: select a: - starting location or stake on either confine border, and a - sequence of stakes on both confine borders, ending with the starting stake, such that visiting the selected stakes in the dictated order results in the quickest coverage of both confines i.e. distance traveled would be minimum. (If both confines were on the same plane, this problem would reduce to the familiar 2D convex hull problem.) (If it matters, the terrain is represented as a discrete surface.) - Olumide === Subject: : Re: More properties of mutated quaternions -- Correction Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Had to quickly revise this. The multiplicaiotn table should read: 1*ek = ek*1 = ek for k = 1, 2, 3 e1*e1 = e2*e2 = -1 e3*e3 = +1 <------- e1*e2 = e2*e1 = e3 e1*e3 = e3*e1 = -e2 e2*e3 = e3*e2 = -e1 --OL === Subject: : Re: More properties of mutated quaternions Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Not to burst your bubble, but your mutaated quaternions do not satisfy associativity: (e1*e2)*e3 = -e3*e3 = +1 e1*(e2*e3) = e1*e1 = -1 This complicates multiplication of three or more elements. Quaternions with zero divisors are not new. Try this multiplication table: 1*ek = ek*1 = ek for k = 1, 2, 3 ek*ek = -1 for k = 1, 2, 3 e1*e2 = e2*e1 = e3 e1*e3 = e3*e1 = -e2 e2*e3 = e3*e2 = -e1 Multiplication is commutative and associative. Zero divisors include 1 (+/-) e3 and e1 (+/-) e2, and there exists an idempotent pair (1 (+/-) e3)/2 along with the trivial pair of 0 and 1. --OL === Subject: : JSH: Strange will to be wrong I think one of the biggest puzzles for me as I wait and hope that the latest very simple explanation of a problem in number theory might finally get some traction is this strange will to be wrong that many of you clearly have. One would think that some of you actually would like to use correct mathematics, and not waste time and effort with mathematical ideas proven to be wrong, and that at least some of you would actually care about learning powerful mathematical techniques that DO WORK versus wasting time on flawed ones that have been proven not to work. But I guess most of you tell yourselves that's not what you're doing so you can do it, which is how this situation has continued for so many years. And then you must feel pride at learning useless crap that doesn't work, but maybe impresses people because so many people are a part of the error? So this time I updated a technique that I first thought up late last year as now I start with the factorization 2975x^2 - 15x + 2 = 2(f(x) + 1)*(g(x) + 1) where the change was to leverage two prime numbers so that I can step through the derivation with 7 and 17 versus just 7 as that didn't work with you as I got some of the same people on sci.math who would come over to other newsgroups mind you as well, tossing up specious objections which sounded intelligent, I guess, and that crap worked like it has worked so many times before. By using two prime numbers though I remove the ability of those people to make seemingly intelligent objections but people remember I have not only explained and explained as I've come up with different ways to try and get mathematicians to behave like mathematicians and accept mathematical proof, I've even had the result published in a peer reviewed mathematical journal!!! Usenet crushed the process as sci.math'ers managed to convince the editors to pull my paper, and later the entire journal died. A an entire mathematical journal died here yet still the error has stayed in place along with the denial in this will to be wrong. That takes a will to be wrong for that to happen and years to go by while I labor to try and find some way to explain that doesn't allow you to tell yourselves lies about the math so you can keep doing wrong mathematics. Wrong mathematics is EASIER. There is a thrill in feeling you are correct because you are using flawed ideas that allow you to convince yourself that you have proven something you haven't. But now we have a situation where a lot of people around the world are choosing easy, doing wrong mathematics, on a huge scale. So yes, I can understand if say Andrew Wiles wouldn't want to accept this result as it takes away his research, so no, he did not prove Fermat's Last Theorem. And Ribet might want to hide from this because it takes away his, and I'm not saying either of them are doing so, but I'm giving them as dramatic examples to explain the why of the very human behavior here. But wrong is wrong. You people can keep doing wrong math to prove things all you want and you are not doing anything of value no matter how many of you get together to live in the error. Wrong mathematical ideas are easier because they let you prove anything, like the classic examples that boil down to divide by zero errors. So yeah, do the wrong math, and use the ring of algebraic integers wrong, without understanding its quirks and real mathematical properties, and you can think you proved Fermat's Last Theorem when you didn't. Wrong math is easier, but it's still wrong. My hope is that some of you start appreciating mathematics. Because of the last few years Usenet has been doing the opposite, using group processes to make this situation much harder in a fight to be wrong. And in a fight to keep bringing other people into the error as pity the poor students, who could have been learning the truth years ago, but instead are currently wasting time and mental energy on bogus mathematical ideas that are appealing because they are wrong, and in mathematics, wrong is easier. It is so, so much harder to be right in mathematics. James Harris === Subject: : JSH: Galois Theory, so what's wrong? I am sure there will be a lot of confusion about the significance of the result that I have showing a problem with use of the ring of algebraic integers. It is such a huge problem in number theory that it's hard to grasp the full impact, but I can maybe help at least with Galois Theory. The result shows that Galois Theory tells you nothing more about non- rationals than it does about rationals. That is the succinct way to explain the impact there, and why I say it does not say Galois Theory is wrong, exactly, but it greatly limits its usefulness to number theorists as to taking it away for the most part as a meaningful tool. And I want to emphasize that mathematically it just never was. People just can make mistakes, and as time goes on those mistakes can be found and the truth learned. It is a process that has gone on for as long as there have been people. We live. We learn. James Harris === Subject: : Re: JSH: Galois Theory, so what's wrong? Nice to see you back. Sorry you are recycling the same old material. === Subject: : Re: JSH: Galois Theory, so what's wrong? Since it's such a huge problem in number theory, you should have no difficulty coming up with an example of an *actual theorem* that you believe you have shown to be false. So please do. Really? Then what theorem of Galois theory concerning non-rationals is false? In that case, presumably there is some theorem that number theorists claim to have proved using Galois theory which is incorrect. What is it? You certainly don't seem to learn. -Rotwang === Subject: : Re: Pseudo-Education Software Fails The Test I think any generalization about technology is not useful. If the product being used has been tested with accepted methods AND if it addresses instructional purposes important to the course of which it is a part AND if it is properly implemented as part of a program AND if the teachers understand how to use it AND if the students use it as intended, then the outcomes may or may not reveal to you something about learning styles. But these ANDs rarely meet up with one another in the known PS universe. It's because no teacher in a school can expect to be given the time, money, equipment or clout to implement anything, regardless of what one is told he is supposed to accomplish. The time, money, equipment and clout go to people who don't know you, don't know what you're doing or what you need, but who experience some sense of importance and accomplishment when they drop shrinkwrapped packages on some lowly, much-hassled teacher's wobbly schoolroom desk. TSK ---------------------------------- May those who damn us be damned. alhuriyehNOBOTS@NOBOTSyahoo.com ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: : Is there any way to factor this x^3 - 5x^2 - 8x + 4 ??? I need to factor [ x^3 - 5x^2 - 8x + 4 ] in order to solve a recurrence for an algorithms class, but I seem to be stuck. The closest I've been able to get is (x - 1)(x - 2)(x - 2), which yields [ x^3 - 5x^2 + 8x - 4 ] , but as you can see, the signs on the 3rd and last term are off. The particular method for solving the recurrence strictly calls for factoring of this polynomial, but I'm starting to think this polynomial can't be factored. Anyway, before I say this to my professor, I want to make sure there is no other way to factor this.... Ed === Subject: : Re: Is there any way to factor this x^3 - 5x^2 - 8x + 4 ??? No nice roots. There are a number of online polynomial root finders. Here's one: === Subject: : Re: Is there any way to factor this x^3 - 5x^2 - 8x + 4 ??? Hint: If the roots are r1, r2, and r3, what is their product? What does this say about possible rational roots? Hint: If the polynomial is reducible then it must have at least one linear factor and correspondingly at least one rational root. Hint: Since the polynomial is monic, if it has a rational root, then one can conclude something further about that root. Hint: If you find one root, (say) r, then you can divide by (x-r) to get a quadratic. I assume you know how to factor a quadratic. === Subject: : Re: Is there any way to factor this x^3 - 5x^2 - 8x + 4 ??? Must it be factored into three monomials with integer coefficients? === Subject: : Re: Is there any way to factor this x^3 - 5x^2 - 8x + 4 ??? Hmm... I belive so because I need to plug-in both the roots and the coefficients into the general solution for the recurrence I'm trying to solve (which is of the form [ a_n = C_1r_1^n + C_2r_2^n + C_3r_3^n ] (note I used the underscore to indicate a subscript). === Subject: : Re: How do you solve this simple equation? really need a refresher. === Subject: : Re: SOLVE IT I am getting a different answer: 100! * (-7)/12 -Vishvas Vasuki