mm-473 === Subject: Re: Quintic equation and Galois Theory > Since this group is not solvable, general Galois theory shows > the x_i do not have a rational expression in terms of radicals > involving the a_j. > This is certainly a good perspective but I have to say I also like the > first interpretation. It's one thing to say that there's no formula for > the roots involving radicals, which is what this general result shows. > But there are occasions in mathematics in which we know that there is > no _general_ result of a certain type, while at the same time we know > that there are _specific_ results which can be found with some > cleverness.Indeed. Consider the roots of a^2 + (1 - x)a + 7(x^2 + x)with integer x (which came up not too long ago in a lengthy, eh,discussion). For x != 0 and x != 1 the quadratic is irreducible,so both roots are not coprime to 7 (by Galois). Stronger, the commonfactor can be expressed in radicals, but there is no general formula,far from it. So showing that there is no general solution does notshow that there is also no solution for each case.However, it is a bit different. Galois theory shows that *whenever*the group is not solvable, there are no expressions in radicals. Withoutexceptions. It not only shows that a generic formula does not exist, butalso that no specific formulas do exist. http://www.cwi.nl/~dik/ === Subject: Re: Courant vs. Spivak vs. ApostolApostol is dry?... well, I didn't actuallu go through it, butI like the fact that he *begins* with integrals,a la Leibniz, using his ... forgot the word. so, at least,look through the first chapter! > per se, either Apostol or Spivak might be preferable. I find Apostol--Give Earth a Trickier Dick Cheeny -- out of office, after GIGA yearshttp://www.benfranklinbooks.com/http://members.tripod.com /~american_almanachttp://www.wlym.com/pdf/iclc/ howthenation.PDF === Subject: 3x3 differential homogenous linear systemSuppose (x_1,y_1,z_1) and (x_2,y_2,z_2) are given linear independentsolutions of a 3x3 homogenous linear differential system. Is there a methodto obtain a third linearly independent solution in terms of the two givensolutions and the coefficient functions? === Subject: Re: 3x3 differential homogenous linear system> Suppose (x_1,y_1,z_1) and (x_2,y_2,z_2) are given linear independent> solutions of a 3x3 homogenous linear differential system. Is there a method> to obtain a third linearly independent solution in terms of the two given> solutions and the coefficient functions?you can try constructing a solution of the formy = a1(t)y1 + a2(t)y2,where the ai(t)'s are to be determined. === Subject: Re: 3x3 differential homogenous linear system> Suppose (x_1,y_1,z_1) and (x_2,y_2,z_2) are given linear independent> solutions of a 3x3 homogenous linear differential system. Is there a method> to obtain a third linearly independent solution in terms of the two given> solutions and the coefficient functions?Yes, vector cross product. === Subject: Re: 3x3 differential homogenous linear system> Suppose (x_1,y_1,z_1) and (x_2,y_2,z_2) are given linear independent> solutions of a 3x3 homogenous linear differential system. Is there a method> to obtain a third linearly independent solution in terms of the two given> solutions and the coefficient functions?>Yes, vector cross product.But the vector cross product of two solutions is not a solution of thedifferential equations system in general. === Subject: Re: Conway's comment on large numbers> I was reading a paper by Conway and Doyle that proves that division by> three is possible (without the axiom of choice).> http://math.dartmouth.edu/~doyle/docs/three/three.pdf> In the last paragraph the authors explain their doubts about the ZF axioms, saying,> Indeed, we're somewhat doubtful whether large natural numbers (like> 80^5000, or even 2^200) exist in any very real sense...> I don't understand this at all. What do very large numbers have to do> with the consistency of ZFC? Nothing, directly. > Is this a joke?Well, Conway has a sense of humor, and this is humorous but it neednot be taken as a joke; the somewhat serious interpretation is as aquestion about the provenance of numbers that are larger than thethe number of distinct things in the observable physical universe --are they really ontologically different from actual infinities?OTOH this was posted in mid-April; allowing a couple of weeks betweenthe time it was first presented or circulated in MS and the date the.pdf was created, it's plausible that it first came into being inearly April ... === Subject: Re: Conway's comment on large numbers> I was reading a paper by Conway and Doyle that proves that division by> three is possible (without the axiom of choice).> http://math.dartmouth.edu/~doyle/docs/three/three.pdf> In the last paragraph the authors explain their doubts about the ZF axioms, > saying,> Indeed, we're somewhat doubtful whether large natural numbers (like> 80^5000, or even 2^200) exist in any very real sense...> I don't understand this at all. What do very large numbers have to do> with the consistency of ZFC? Is this a joke?I had a professor once who used to joke that you can't tell the difference between 42 marbles and 43 marbles. In other words if someone dumped a bag of marbles on the table, you couldn't really tell the difference. I always understood that to be an ironic comment that even though we have this theory of the natural numbers all the way up to infinity, in practice we can only recognize a very small handful of numbers, up to about 6 or 7. I was reminded of this during the 2000 Presidential election, when it turned out that it's actually impossible, in practice, to do something as simple as counting all the votes. In principle you can count, but in practice you can't. === Subject: =?ISO-8859-1?Q?Unterteilung_einer_Kugeloberfl?= =?ISO-8859-1?Q?=E4che_durch_=E4quidistante_Punkte?=Mein Problem ist folgendes: Ich arbeite mit einem 3D Helmoltz-Spulensystem, das ich zu programmieren versuche.Das generierte Magnetfeld wird als Vektor betrachtet. Durch dieAnordnung von 6 Spulen, soll das Magnetfeld durch ver.8andern desStromes durch Spulen rotiert werden. Dieses Testfeld (Magnetfeld)kann also als Kugel betrachtet werden. Nun mu ich verschiedenTestpunkte auf der Kugel fest legen. In meinem Programm m.9achte ich denBenutzer fragen, aus wievielen Punkten der Test bestehen soll. Wiekann ich dann die Kugel in diese Anzahl von Punkten unterscheiden, soda diese gleichm.8aig .9fber die gesamte Oberfl.8ache verteilt sind?Der einfachste Weg wird wohl .9fber Sph.8arische Koordinaten f.9fhren. Ich bin f.9fr jede Anregung dankbar und hoffe, da ich mich halbwegsverst.8andlich ausgedr.9fckt habe. === Subject: Re: =?ISO-8859-1?Q?Unterteilung_einer_Kugeloberfl?= Keywords: sphere surface triangulation|> Mein Problem ist folgendes: Ich arbeite mit einem 3D Helmoltz-Spulen|> system, das ich zu programmieren versuche.|> Das generierte Magnetfeld wird als Vektor betrachtet. Durch die|> Anordnung von 6 Spulen, soll das Magnetfeld durch ver.8andern des|> Stromes durch Spulen rotiert werden. Dieses Testfeld (Magnetfeld)|> kann also als Kugel betrachtet werden. Nun mu ich verschieden|> Testpunkte auf der Kugel fest legen. In meinem Programm m.9achte ich den|> Benutzer fragen, aus wievielen Punkten der Test bestehen soll. Wie|> kann ich dann die Kugel in diese Anzahl von Punkten unterscheiden, so|> da diese gleichm.8aig .9fber die gesamte Oberfl.8ache verteilt sind?|> Der einfachste Weg wird wohl .9fber Sph.8arische Koordinaten f.9fhren. |> Ich bin f.9fr jede Anregung dankbar und hoffe, da ich mich halbwegs|> verst.8andlich ausgedr.9fckt habe.|> ...http://www.mathematik.uni-dortmund.de/lsx/research/projects /fliege/nodes/nodes.htmlhttp://www.research.att.com/~njas/ icosahedral.codes/index.html === Subject: Re: =?ISO-8859-1?Q?Unterteilung_einer_Kugeloberfl?= =?ISO-8859-1?Q?=E4che_durch_=E4quidistante_Punkte?=> Mein Problem ist folgendes: Ich arbeite mit einem 3D Helmoltz-Spulen> system, das ich zu programmieren versuche.> Das generierte Magnetfeld wird als Vektor betrachtet. Durch die> Anordnung von 6 Spulen, soll das Magnetfeld durch ver.8andern des> Stromes durch Spulen rotiert werden. Dieses Testfeld (Magnetfeld)> kann also als Kugel betrachtet werden. Nun mu ich verschieden> Testpunkte auf der Kugel fest legen. In meinem Programm m.9achte ich den> Benutzer fragen, aus wievielen Punkten der Test bestehen soll. Wie> kann ich dann die Kugel in diese Anzahl von Punkten unterscheiden, so> da diese gleichm.8aig .9fber die gesamte Oberfl.8ache verteilt sind?> Der einfachste Weg wird wohl .9fber Sph.8arische Koordinaten f.9fhren. > Ich bin f.9fr jede Anregung dankbar und hoffe, da ich mich halbwegs> verst.8andlich ausgedr.9fckt habe. In general the larger the number of test points from sphericalsymmetry, the better it is. But it may be limited from practicalconsiderations: Are the coils equally spaced,like vertices of anoctohedron? What size is the sphere radius in comparision to thedistance between the coils?If you choose a dodecahedron,there are 20 vertices,12 faces. If anicosahedron is chosen, 20 faces,12 vertices.You can consider sphericalgeodesic dome design after Buckminster Fuller,where each triangle sideof an icosahedron is sub-divided into p=2,3,4 etc. parts.If you choose an icosi-dodecahedron, there are 30 vertices, 32 facesto mount the test points. Please seehttp://mathworld.wolfram.com/Icosidodecahedron.html. === Subject: Re: Unterteilung einer =?iso-8859-1?Q?Kugeloberfl=E4che?= durch =?iso-8859-1?Q?=E4quidistante?= Punkte> Mein Problem ist folgendes: Ich arbeite mit einem 3D Helmoltz-Spulen> system, das ich zu programmieren versuche.> Das generierte Magnetfeld wird als Vektor betrachtet. Durch die> Anordnung von 6 Spulen, soll das Magnetfeld durch ver.8andern des> Stromes durch Spulen rotiert werden. Dieses Testfeld (Magnetfeld)> kann also als Kugel betrachtet werden. Nun mu ich verschieden> Testpunkte auf der Kugel fest legen. In meinem Programm m.9achte ich den> Benutzer fragen, aus wievielen Punkten der Test bestehen soll. Wie> kann ich dann die Kugel in diese Anzahl von Punkten unterscheiden, so> da diese gleichm.8aig .9fber die gesamte Oberfl.8ache verteilt sind?> Der einfachste Weg wird wohl .9fber Sph.8arische Koordinaten f.9fhren.> Ich bin f.9fr jede Anregung dankbar und hoffe, da ich mich halbwegs> verst.8andlich ausgedr.9fckt habe.This is a tough global optimization problem. I have seen it called theFekete problem. === Subject: Re: =?ISO-8859-1?Q?Unterteilung_einer_?= =?ISO-8859-1?Q?Kugeloberfl=E4che_dur?= =?ISO-8859-1?Q?ch_=E4quidistante_Pun?= =?ISO-8859-1?Q?kte?=This should have been posted to the German newsgroupThomas [X-post and Fup to de.sci.mathematik]>Mein Problem ist folgendes: Ich arbeite mit einem 3D Helmoltz-Spulen>system, das ich zu programmieren versuche.>Das generierte Magnetfeld wird als Vektor betrachtet. Durch die>Anordnung von 6 Spulen, soll das Magnetfeld durch ver.8andern des>Stromes durch Spulen rotiert werden. Dieses Testfeld (Magnetfeld)>kann also als Kugel betrachtet werden. Nun mu ich verschieden>Testpunkte auf der Kugel fest legen. In meinem Programm m.9achte ich den>Benutzer fragen, aus wievielen Punkten der Test bestehen soll. Wie>kann ich dann die Kugel in diese Anzahl von Punkten unterscheiden, so>da diese gleichm.8aig .9fber die gesamte Oberfl.8ache verteilt sind?>Der einfachste Weg wird wohl .9fber Sph.8arische Koordinaten f.9fhren. >Ich bin f.9fr jede Anregung dankbar und hoffe, da ich mich halbwegs>verst.8andlich ausgedr.9fckt habe.>Mfg,>Florian. === Subject: Re: =?ISO-8859-1?Q?Unterteilung_einer__Kugeloberf?= > This should have been posted to the German newsgroup> Thomas [X-post and Fup to de.sci.mathematik]>Mein Problem ist folgendes: Ich arbeite mit einem 3D Helmoltz-Spulen>system, das ich zu programmieren versuche.>Das generierte Magnetfeld wird als Vektor betrachtet. Durch die>Anordnung von 6 Spulen, soll das Magnetfeld durch ver.8andern des>Stromes durch Spulen rotiert werden. Dieses Testfeld (Magnetfeld)>kann also als Kugel betrachtet werden. Nun mu ich verschieden>Testpunkte auf der Kugel fest legen. In meinem Programm m.9achte ich den>Benutzer fragen, aus wievielen Punkten der Test bestehen soll. Wie>kann ich dann die Kugel in diese Anzahl von Punkten unterscheiden, so>da diese gleichm.8aig .9fber die gesamte Oberfl.8ache verteilt sind?>Der einfachste Weg wird wohl .9fber Sph.8arische Koordinaten f.9fhren. >Ich bin f.9fr jede Anregung dankbar und hoffe, da ich mich halbwegs>verst.8andlich ausgedr.9fckt habe.> This should have been posted to the German newsgroup> Thomas [X-post and Fup to de.sci.mathematik]>Mein Problem ist folgendes: Ich arbeite mit einem 3D Helmoltz-Spulen>system, das ich zu programmieren versuche.>Das generierte Magnetfeld wird als Vektor betrachtet. Durch die>Anordnung von 6 Spulen, soll das Magnetfeld durch ver.8andern des>Stromes durch Spulen rotiert werden. Dieses Testfeld (Magnetfeld)>kann also als Kugel betrachtet werden. Nun mu ich verschieden>Testpunkte auf der Kugel fest legen. In meinem Programm m.9achte ich den>Benutzer fragen, aus wievielen Punkten der Test bestehen soll. Wie>kann ich dann die Kugel in diese Anzahl von Punkten unterscheiden, so>da diese gleichm.8aig .9fber die gesamte Oberfl.8ache verteilt sind?>Der einfachste Weg wird wohl .9fber Sph.8arische Koordinaten f.9fhren. >Ich bin f.9fr jede Anregung dankbar und hoffe, da ich mich halbwegs>verst.8andlich ausgedr.9fckt habe.S.g.H. Florian, Ich setze es moeglich auf Englisch ueber, undhoeffentlich Sie koennen die Antworten wieder auf Deutsch selbersetzen.I try giving a free translation of Florian's message in German,hopinghe can get responses back into German.---------Interdivision of Spherical surface by equidistant points.I am programming a 3D Helmholtz coil system. The magnetic fieldgenerated is a vector. By an ordered arrangement of six coils andvarying currents running through them, the magnetic field can beoriented in any direction. The test field is considerd as sphere andlocations for magnetic field (measurement) can be considered locatedon a spherical surface. I should determine the locations of severaltest points on this sphere. I would like to ask you,how many testpoints should be included or would be adequate so that they are asmuch uniformly separated/located/distributed apart as possible overthe entire spherical surface ? The simplest way is of course to usespherical coordinates.Florian.------------ === Subject: Re: =?ISO-8859-1?Q?Fragmentation_of_a_sphere_in_equidistant_points _/_Untert?= =?ISO-8859-1?Q?eilung_einer__Kugeloberf_l=E4c?= =?ISO-8859-1?Q?he_durch_=E4quidistante_Punkte?=Hello!I am sorry, but I thought that I posted the message to the Germangroup. Anyway, thank you for the translation!> This should have been posted to the German newsgroup> Thomas [X-post and Fup to de.sci.mathematik]> This should have been posted to the German newsgroup> Thomas [X-post and Fup to de.sci.mathematik]> S.g.H. Florian, Ich setze es moeglich auf Englisch ueber, und> hoeffentlich Sie koennen die Antworten wieder auf Deutsch selber> setzen.> I try giving a free translation of Florian's message in German,hoping> he can get responses back into German.> ---------> Interdivision of Spherical surface by equidistant points.> I am programming a 3D Helmholtz coil system. The magnetic field> generated is a vector. By an ordered arrangement of six coils and> varying currents running through them, the magnetic field can be> oriented in any direction. The test field is considerd as sphere and> locations for magnetic field (measurement) can be considered located> on a spherical surface. ! The user is asked the amount of test points on the sphere. Now, Ishould determine the locations of these> test points on this sphere. How do I find these the vectors, so that the points on the sphere are as> much uniformly separated/located/distributed apart as possible over> the entire spherical surface ? The simplest way I guess, is to use> spherical coordinates and then convert them to cartesians (for my Power supplies).> ------------ === Subject: ask for help on computing the probabilitySuppose T1,T2...Tn are i.i.d. nonnegative random variable with pdf f(t).For two fixed positive value a and b, I have to compute the followingPr(T1+T2+...Tn < a, T1Suppose T1,T2...Tn are i.i.d. nonnegative random variable with pdf f(t).>For two fixed positive value a and b, I have to compute the following>Pr(T1+T2+...Tn < a, T1 P(sum(T_i) < a | max(T_i) < b) P{max(T_i) < b}> Or perhaps it is easy somehow to compute the n-fold convolution of the> conditional distribution?It seems for me that the conditional probabilityP(sum(T_i) < a | max(T_i) < b) === Subject: Re: ask for help on computing the probability> P(sum(T_i) < a | max(T_i) < b) P{max(T_i) < b}> Or perhaps it is easy somehow to compute the n-fold convolution of the> conditional distribution?It depends on what one means by easy, and this dependson the precise distribution. If m(z) = E(exp(z*T_k)|T_k < b},then P(sum(T_k) > a | max(T_k) < b) is given by(1/2*pi*i)*int_{c-i*infty}^{c+i*infty) m(z)^n*exp(-za)/z dz,where c > 0. Frequently a good approximation can be madeby steepest descent.>It seems for me that the conditional probability>P(sum(T_i) < a | max(T_i) < b)If you want closed form, these are generally unavailable.The easiest case is that of the uniform; it is just theprobability that sum(U_k) < a/b, where the U_k are uniform (0,1), and this is in a closed form given inthe 19th century. Probably the next easiest is that ofa gamma distribution, where the distribution of the T'sgiven the sum is independent of the sum, but this is still a little nasty, although studied.This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Question about textI have always been interested in this kind of foundational mathematics stuff but never got a chance to study it (except godel's incompleteness theorem). What is a good book to read on philosophical / foundational issues in mathematics?-----------------------------------------Remove obviously faked punctuation fromburningspear79 a.t ly-cos peri-od c.omand ignore the rest of this sigmayuresh79@angelfire.com----------------------------------- ------> [added sci.logic]>I'm currently studying first-order logic, but I have two questions>that doesn't seem to be answered in the book I'm using. I hope someone>can spare a moment of their time and briefly explain it :)>a) The book starts by defining what are first-order expressions, what>is a model, what does it means that a model satisfies a first-order>expression, what does it mean for an expression to be valid etc. Then>it goes on to define how to go from some valid expresions to other>valid expressions and proofs that certain sets of expression are>valid. This is used to create a proof system with the properties that>a) any expression provable in the system is valid and (amazingly) b)>every valid expression has a proof in the system.>Now my question is, in all these proofs of completeness and soundness>etc. all kinds of axioms (such as the induction axiom, the axiom of>choice etc.) are implicitly used along with techniques such as>indirect proofs. Now I'm wondering what is the justification for this?> That is indeed a very good question. Using generally-accepted forms of > mathematical reasoning, one proves various things about a logical system, > which then provides a foundation for those very same generally-accepted forms > of mathematical reasoning. There are various criticisms of such > Foundationalism in the literature.>I mean, first-order logic should be able to support many kinds of>sets of axioms and the axiom of induction, for instance, certainly>doesn't have to be among them. And we carefully prove that things like>indirect proofs works inside of the system. However when proving these>things we are - outside of the system - using all kinds of arbitrary>axioms and inference rules (maybe even the rules we are proving)! So>my question is what framework it is that these proofs take place in>and where does all these axioms come from and how can we use them>right away? And isn't there some kind of logical problem in using>these rules when studying something that is even more fundamental?>b) Exactly how does one define a model? I understand how one can>create a vocabulary of first-order expressions for the natural>numbers (it's just a set of sequences of certain symbols). Also, I>see how we can select some of these expressions and call them axioms>and use them inside of the proof system we have constructed. But>exactly how does one specify a model for the natural numbers? I>mean, doesn't that in itself require some kind of axioms?> Yes. Models of FOL are usually represented as sets, and set theory has some > very nice FOL axiomatizations, but it certainly appears to be begging the > question to be using a set theory to define truth in a logical system that > will then be used to axiomatize the set theory.>I mean, it must be an infinite construction.> It's even rather worse than that. Tarski showed that NO 2-valued logical > system on its own can consistently define its own truth function (i.e. > model-theoretic validity): if there is a truth function then you can construct > a liar proposition L that asserts L is false, which is clearly inconsistent.> Tarski's solution to this is to have an infinite chain of meta-language (to > define validity in the base language), meta-meta-language (to define validity > in the meta-language), etc.> Such an infinite skyscraper of logical systems appears to be a pretty dubious > foundation, n'est-ce pas?>So it seems that the specification>of a model will have to take place in some other kind of logical>system which confuses me a bit because it seems first-order logic>should be very fundamental?! The book I'm using seems to say something>like: Well, 0 should correspond to 0, 1 should correspond to 1 the>successor function is simply the function n+1 etc. But how do we know>what those objects we are using in the definitions are?!>Surfing === Subject: Re: Question about text> I have always been interested in this kind of foundational mathematics > stuff but never got a chance to study it (except godel's incompleteness > theorem). What is a good book to read on philosophical / foundational > issues in mathematics?Some standard texts:pre-Godel development of foundations.The Undecidable, Martin Davis ISBN 0-911216-01-4 covers post-Godeldevelopments concerning incompleteness in Logic and the Theory ofComputation.Classic Set Theory, Derek Goldrei ISBN 0-412-60610-0 is a good,well-organized exposition of the Set Theory side of foundations with afair number of specific results and exercisesDover reprints a number of classic texts such as the above.http://www.cs.nyu.edu/mailman/listinfo/fom/ is a lively discussion ofcurrent day thinking on foundational issues. It tends to be a bitconservative (pro status quo), but search for my name to find a fewchallenges to the establishment.And if you're interested in a rigorous formulation of these ideas,hold on to your seat and check outhttp://www.arxiv.org/html/cs.lo/0003071Charlie VolkstorfCambridge, MA === Subject: convexity of an implicit functionLet f and g be two functions defined on [0,1]^n with values in R. Allpartial derivates of both f and g are strictly positive at any pointin [0,1]^n.Let u be a function defined on R with values in R.u(T)=g(x) where f(x)=T and for any x' in [0,1]^n we have g(x') > g(x).Can we conclude anything about the convexity of u(T)?Sorin === Subject: Puthoff-Sarfatti Debate on Zero Point Energybcc (Hal's original list + a few more)HP: Jack, with regard to my statements that you misunderstand/misrepresent the PV program, concerning which you've asked for clarification, I'm going to spell out here just what I mean in detail.JS: Good, finally. Do you want me to put this in Super Cosmos? I have been delaying because I do not want to misrepresent your physics ideas. It was never my intention to be unfair. I have no motive for that. If I thought PV could work I would use it.HP: My reason is this. I know that if you truly understood what the PV effort is designed to do, you would laud it rather than disparage it, appreciate it for what it is designed to do, rather than condemn it. Since I give you credit for being bright, I'm choosing to go on the assumption that you simply misunderstand what the PV effort is designed to do.JS: That is an accurate assumption.JP: My best approach is to use an EM analogy, for it's precise. We have Maxwell's equations which represent how charges and currents generate fields, and how fields affect charge motions via the Lorentz force equation. Straightforward. JS: OK, agreed.HP: Under certain conditions, however, when dealing with the design and analysis of simple circuits, we take an approximate approach and use Kirchoff's equations to solve for currents, voltages, etc., given circuit parameters like resistance, inductance and capacitance. Agreed this is an approximation (though a very useful one) because it simplifies our effort, e.g., we do not have to calculate radiation from the interconnecting wires at bends and junctions, etc.JS: OK, agreed.HP: Now would it be appropriate at this point to say Maxwell's equations ain't broke, so who needs Kirchoff's equations to fix it?! JS: No, of course not.HP: Kirchoff's equations are not even wrong!JS: Let me make it clear. PV is better than Cargo Cult pseudo-science. PV is real physics because it is testable. It's better than not even wrong. It's simply wrong, which is fine. A lot of initially plausible theories have been proved wrong. There may even be some flaw in my theory, which will make it wrong.HP: Kirchoff's equations are CRAP! (Sound Familiar? :-) )JS: Now wait a minute. The big difference is that Kirchoff's model, as currently formulated, agrees with all experiments in its proper domain. In contrast your own assistant Michael Ibison has published a very excellent paper showing that your PV does not agree with now well established facts of precison cosmology as well as the binary pulsar gravity radiation data whose agreement with Einstein's GR got a Nobel prize for the people who did the analysis. The fact is that problems easy to solve in GR are hard in PV such as the rotating source problem with the Kerr solution in GR. You have had ample time to solve that problem and good evidence for gravi-magnetism frame drag already exists even without Gravity B space probe. Now this is quite apart from philosophical issues in the informal language of your PV model such as your Tables I & II which to me suggest you are violating the locality of the equivalence principle and also your use of the isotropic radial coordinate and your lack of clarity on the role of general coordinate symmetrye.g. gu'v'(P) = Xu'^u(P)Xv'^v(P)guv(P)For two LNIF observers Alice and Bob almost colliding with each other at collision point P each on different nearly intersecting time-like non-geodesics. It is interesting to point out that without electrical forces and the Pauli exclusion principle, timelike non-geodesics would not be possible. You could not detect light rays either. You could not you could not have gravity in Einstein's sense at all! I mean from the operational point of view. So obviously, gravity is NOT a fundamental force but is emergent out of electromagnetism like non-renormalizable Fermi theory of beta decay is emergent out of U(1)SU(2) renormalizable gauge theory with Higgs vacuum coherence. You could not even detect timelike geodesic deviation for local tidal curvature if e -> 0 and h -> 0. That is, you need e, h, c before you can get G. That much is obvious.HP: No, not unless someone were saying that Maxwell's equations were wrong, or incomplete, or some such, which is not the case here.JS: There is a mathematical limiting case Maxwell's EM -> Kirchoff's circuit laws. Where is the mathematicsGR -> PVIn the simplest SSS case?HP: Similarly, the PV approach to GR or representation of GR is a program put forward in the vein of Kirchoff's engineering equations as a (useful) approximation to Einstein's GR. JS: You have not proved that. GR uses tensors, you do not seem to. What parameters of GR go to some limit, presumably, zero so that PV emerges with, in the SSS caseK = e^2GM/c^2r ?Simply demonstrate that with the mathematics.HP: Its purpose is to calculate certain simple cases without the full tensor math, just as Kirchoff's equations for electric circuits permit facile calculation without the full Maxwellian vector math. JS: My point still stands. The analogy GR -> PV with Maxwell -> Kirchoff does not stand until you can prove it mathematically. You must show in the SSS case thatgoo(Einstein) = (1 - 2GM/c^2r) -> goo = e^-2GM/c^2rr > 2GM/c^2You cannot do that. Indeed, you do just the opposite! You havegoo(Einstein) = (1 - 2GM/c^2r) <- goo = e^-2GM/c^2ri.e. you have GR is an approximation to PV! So your logic is wrong. You have contradicted yourself. It's not that PV is a representation of GR, but, rather GR is a representation of PV! At least that's the way I learned philosophy of physics under Max Black at Cornell in late 1950's. I wish Zielinski and other theoretical physicists would chime in on this.HP: As with Kirchoff's equations, certain issues become clear that can otherwise be obscured in a full-fledged Maxwellian analysis. JS: Indeed, but again you have it backwards. The simple fact is that problems easy in GR are hard in PV, e.g. rotating source problem. Note Schwarzschild, Kerr & Kerr-Newman are all vacuum solutions toRuv = 0 nonexotic vacuumIn the case of Schwarzschild, the exterior solution for r > 2GM/c^2 is fitted to an interior solution where, excluding black holes with event horizons, absent in PVGuv = Tuv/(String tension) = (Metric Elasticity)TuvMetric elasticity = 8piG/c^4 ~ 10^-33 cm per 10^19 Gevi.e. very stiff, very inelastic - unless ...HP: Similarly with regard to PV vis a vis GR. Does PV replace GR? No, nor was it meant to. JS: I have shown a very obvious confusion in your logic, in your meta-analysis of your theory.You have in SSS case PV -> GR when GM/c^2r -> 0So your analogy above with Maxwell -> Kirchoff is wrong. The real analogy, in the sense of category theory, functors and morphisms and all that fancy stuff, would beKirchoff -> Maxwellwhich is obviously wrong.HP: Should we trash PV because it does not (in its present form) handle rotating systems, any more than we should trash Kirchoff's equation because it does not handle radiation? No, of course not. JS: Yes, for reasons I gave above. Also, you will probably never solve the rotating system problem. You have had plenty of time to solve it. You have not been able to. Ibison is a good mathematician and he cannot solve it. Also flying saucers rotate some of the time, or parts of them do, so you cannot have a good metric engineering model with PV for that reason alone!HP: Should we ridicule PV because its few lines of scalar calculation for GR radiation give only 2/3 the more complex GR tensor-derived value, any more than we should ridicule Kirchoff's equations because its circuit-equivalent analysis of microwave circuits gives only approximate answers compared to a full-fledged Maxwellian analysis? No, not at all.JS: We need not ridicule it, but we should certainly reject it, which is why the top people in the field pay no attention to it really. GR works well where it is supposed to work well. PV does not. PV is falsified by the facts that GR agrees with. HP: The PV approach is simply a handy guide to certain 'metric engineering' calculations, no more, no less. PV is not being promoted as a replacement for Einstein's equations (any more than Kirchoff's equations are meant to replace Maxwell's equations).JS: Handy? How is it handy? Until you can solve the Kuv field from a spinning top you cannot say it is handy. Shady would be more accurate! :-) Also, you cannot claim it is metric engineering until you can show that saucers work according to PV given that we both accept the saucer evidence as given by Eric Davis in the NIDS sanctioned MUFON 2001 paper. You must get plausible numbers and you do not get it from Levi-Civita effect using Newton's G. GR has that same effect of course with same problem unless G* > G on small scale and you never mention that in writing. I do of course.HP: In addition, there seems to be a misunderstanding in your characterization of PV to the effect that it cannot handle vacuum energy; Ibison's PV analysis of cosmological issues shows that vacuum energy is incorporated into PV analysis in essentially the same way as in standard GR.JS: Perhaps. Can you point out the exact pages and equations in his latest paper where that is done? In Einstein's GR there is a / term in the action that yieldsGuv + /guv = 8pi(G/c^4)Tuvwhere are the analogous equations in PV? Even if you can do that my objection above still applies. Furthermore, you cannot explain WHY / and under what conditions? I can and do. Second, you cannot derive guv from QED. I can and do. Third, there is no PV in your PV. I mean I see no connection to QED or to the ZPE aspect of QED in your actual PV math.HP: Finally, to address another issue, application of the PV approach to determining whether, e.g., strong electric or magnetic fields can be used to significantly affect the spacetime metric (Levi-Civita Effect) was soundly ridiculed by you as being a wasted effort, because the of the PV enterprise, however, (1) it is useful to demo the fact that the PV metric engineering approach permits one to generate in a few lines what to engineers is more difficult to produce by standard GR tensor techniques, (2) and it is important to generate a catalog of all effects, regardless of their strength/significance, so that further effort is not spent on dead-end avenues, while promising avenues are taken up for further evaluation.JS: It is important that you point out the effect you get is too weak. Something crucial is missing. The fact that the saucers are out there doing what they do shows you are barking up the wrong tree. You are evading what you believe are the UFO facts. You have no plausible model here for metric engineering that I can see. You act as if we have a lot of time. We do not. The Right Stuff is needed now. Also, I see no mathematical model from you showing how even if you get ZPE over-unity how that changes K in your PV. I can show you how to get it in PV with my theory very easy BTW. You haveK ~ 1 + (String Tension)(Electric Energy Density)(Area)In my theory that is simply, if I replace GR with PVK ~ 1 + /zpf(Area)where for your Levi-Civita effect/zpf ~ (Chiao Impedance Match)|Vacuum Coherence||Superconductor Coherence|(Volume Factor)cosine(2pi(Electric Flux)/(Quantum of Electric Flux)|/zpf > 0 is zero point induced strong short-range anti-gravity field (dark energy w = -1 exotic vacuum)/zpf < 0 is zero point induced strong short-range gravity field (dark matter w = -1 exotic vacuum)And you configure the /zpf field like TraceK* in Alcubierre's weightless warp drive acceleration field zero g-force, zero time dilation globally faster-than-light G-Engine warp drive solution. K* is not your K of course.HP: In summary, the scalar PV 'metric engineering' approach is to the full-fledged tensor GR approach to spacetime metric calculations what Kirchoff's circuit engineering approach is to the full-fledged vector Maxwellian approach to EM calculations, and, analogously, just as useful when appropriately applied in an engineering setting.JS: No that is wrong. You made an error there as I showed above. === Subject: FYI: Power Units of Measurement converterI noticed a lot of posts regarding power measurement.http://www.xdweb.net/~dibblego/utilities/power/ index.htmlTony Morris(BInfTech, Cert 3 I.T.)Software EngineerSun Certified Programmer for the Java 2 Platform (1.4)Sun Certified Developer for the Java 2 Platform === Subject: A Limit QuestionWhat is 1 / -- lim | n! | 3nn->oo | ---- | | (3n)! | /or in tex formatlim_{n rightarrow infty} frac{ n! }{ 3n! } ^ frac{1}{3n}I put it in Maple and it says it's 1/3.How does that happen? === Subject: Re: A Limit Question: SORRY CORRECTION> What is> 1> / --> lim | n! | 3n> n->oo | ---- |> | (3n)! |> /> or in tex format> lim_{n rightarrow infty} frac{ n! }{ 3n! } ^ frac{1}{3n}> I put it in Maple and it says it's 1/3.> How does that happen?Oops, sorry.I missed n!^3 in the ascii. 1 / 3 -- lim | n! | 3nn->oo | ---- | | (3n)! | /and in texlim_{n rightarrow infty} frac{ n!^3 }{ 3n! } ^ frac{1}{3n}The original problem is to find the radius of convergence of the seriesoo / 3 -- | n! | 3n | ---- | z/ | (3n)! |__ /n=0and I used the Cauchy-Haddamard thrm to find the radius which gives radius as 1------------------- 1 / 3 -- lim | n! | 3nn->oo | ---- | | (3n)! | / === Subject: Re: A Limit Question: SORRY CORRECTION>The original problem is to find the radius of convergence of the series>oo / 3 >-- | n! | 3n> | ---- | z>/ | (3n)! |>__ />n=0>and I used the Cauchy-Haddamard thrm to find the radius which gives >radius as You can estimate (n!)^3 / (3n!) by observing thatthe trinomial expansion (x + y + z)^(3n)has (3n + 1)*(3n + 2)/2 terms x^i * y^j * z^k where0 <= i, j, k <= 3n and i + j + k = 3n.The coefficients sum to 3^(3n) (by setting x = y = z = 1)and the largest is (3n)! / (n!)^3. Therefore (3n)! 3^(3n) 3^(3n) >= ------ >= ------------------- (n!)^3 (3n + 1)(3n + 2)/2Now take reciprocals and the (3n)-th root.John Adams served two terms as Vice President and one as President, but lostreelection. Later his son became President despite losing the popular vote.That son lost his reelection attempt badly. Now history is repeating itself.pmontgom@cwi.nl Microsoft Research and CWI Home: San Rafael, California === Subject: Re: A Limit Question: SORRY CORRECTION> What is> 1> / --> lim | n! | 3n> n->oo | ---- |> | (3n)! |> /> or in tex format> lim_{n rightarrow infty} frac{ n! }{ 3n! } ^ frac{1}{3n}> I put it in Maple and it says it's 1/3.> How does that happen?> Oops, sorry.> I missed n!^3 in the ascii.> 1> / 3 --> lim | n! | 3n> n->oo | ---- |> | (3n)! |> /> and in tex> lim_{n rightarrow infty} frac{ n!^3 }{ 3n! } ^ frac{1}{3n}> The original problem is to find the radius of convergence of the series> oo / 3 > -- | n! | 3n> | ---- | z> / | (3n)! |> __ /> n=0> and I used the Cauchy-Haddamard thrm to find the radius which gives > radius as> 1> -------------------> 1> / 3 --> lim | n! | 3n> n->oo | ---- |> | (3n)! |> /Then use the ratio test, not the Cauchy-Hadamard theorem:|((n + 1)!^3 z^{3(n + 1)} / (3(n + 1))!) / (n!^3 z^{3n}/ (3n)!) | = = (n + 1)^3/((3n + 3)(3n + 2)(3n + 1)) |z|^3and the limit of this sequence is |z|^3/27 = (|z|/3)^3. So, the seriesconverges absolutley when |z| < 3 and diverges when |z| >3; therefore,the radius of convergence is 3.Jose Carlos Santos === Subject: Re: A Limit Question> What is> 1> / --> lim | n! | 3n> n->oo | ---- |> | (3n)! |> /> or in tex format> lim_{n rightarrow infty} frac{ n! }{ 3n! } ^ frac{1}{3n}> I put it in Maple and it says it's 1/3.> How does that happen?Here's the maple command I used.limit ( ( (n!)^3/(3*n)! )^(1/(3*n)), n=infinity) === Subject: Re: A Limit Question> What is> 1> / --> lim | n! | 3n> n->oo | ---- |> | (3n)! |> /> or in tex format> lim_{n rightarrow infty} frac{ n! }{ 3n! } ^ frac{1}{3n}> Here's the maple command I used.> limit ( ( (n!)^3/(3*n)! )^(1/(3*n)), n=infinity)Much nicer and more readable than TeX but not surpassed by your vividascii rendition however Maple command isn't the same formulaUse Sterlings approximation for n! = (n/e)^n sqr(2n.pi) === Subject: Re: A Limit Question : OOPS SORRY CORRECTION !!!!>What is> 1> / --> lim | n! | 3n>n->oo | ---- |> | (3n)! |> />or in tex format>lim_{n rightarrow infty} frac{ n! }{ 3n! } ^ frac{1}{3n}>Here's the maple command I used.>limit ( ( (n!)^3/(3*n)! )^(1/(3*n)), n=infinity)> Much nicer and more readable than TeX but not surpassed by your vivid> ascii rendition however Maple command isn't the same formula> Use Sterlings approximation for n! = (n/e)^n sqr(2n.pi)Oops, sorry.I missed n!^3 in the ascii. 1 / 3 -- lim | n! | 3nn->oo | ---- | | (3n)! | /and in texlim_{n rightarrow infty} frac{ n!^3 }{ 3n! } ^ frac{1}{3n}The original problem is to find the radius of convergence of the seriesoo / 3 -- | n! | 3n | ---- | z/ | (3n)! |__ /n=0 === Subject: Re: sqrt(2)^sqrt(2) rational?> I remember something about a strange theorem but since it involves> square roots it's hard to search using Google. It's something like this:> Either sqrt(2)^sqrt(2) or sqrt(2)^sqrt(2)^sqrt(2) is rational. Is there> also a proof you can't say which of them it is?> Anyone knows where to get more information?This is probably the famous theorem:there exist positive, irrational numbers a and b such that a^b is rational.the proof rests on the fact that you don't need to know if sqrt(2)^sqrt(2)is rational or not to prove the theorem, since in both case, you canchoose a and b for proving the theorem.American Mathematical Monthly, vol. 108, january 2001) nous devons agir comme si la chose qui peut-.90tre ne sera pas devait.90tre (Kant, M.8etaphysique des moeurs, doctrine du droit, II conclusion) Thomas Baruchel I wonder finite field = finite integral domain ?Yes, a standard result, in many textbooks. === Subject: Re: finite field = finite integral domain ?* fuzzykyh@nate.com> I wonder finite field = finite integral domain ?Homework?I guess you want to prove that a finite integral domain is a field.Hint 1: Let a in G and a != 0. Examine the sequence a,a^2,a^3,... http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 85 24 92 === Subject: Re: Help> Some Help with the following would be appreciated (mixture of first> year Analysis and Algebra):> A = #64830; 2 .961 #64831;> #64830; 3 4 #64831;Huh? === Subject: Re: If xyz = 1, prove that (xy)^2 + (yz)^2 + (zx)^2 - 2x - 2y - 2z - 3 >= 0.>Narasimham G.L. escribi.97 en el mensaje> Can you solve this problem?> Hint: if the problem would be to to prove ( without OP typo) that> (xy)^2 + (yz)^2 + (zx)^2 - 2x - 2y - 2z + 3 >= 0.,> see what happens to other squares and (x+y+z) together.>What with x = y = z = 3/4?>3x^4 - 6x + 3 has a minimum at 2^(2/3)/2, that is (12 - 9*2^(2/3))/4>~= -0.57165>Ignacio Larrosa Ca.96estro>A Coru.96a (Espa.96a)>ilarrosaQUITARMAYUSCULAS@mundo-r.com It suffices to consider the case where x, y, z are all positive.Use the arithmetic-geometric mean inequality: t1^3 + t2^3 + t3^3 >= 3*t1*t2*t3when t1, t2, t3 are nonnegative (actually when t1 + t2 + t3 >= 0). |^/| Maple 9 (SUN SPARC SOLARIS) MAPLE / All rights reserved. Maple is a trademark of <____ ____> Waterloo Maple Inc. | Type ? for help.> assume(x > 0);> assume(y > 0);> assume(z > 0);> t1 := (1 + x)/x^(2/3); 1 + x~ t1 := ------ 2/3 x~> t2 := (1 + y)/y^(2/3); 1 + y~ t2 := ------ 2/3 y~> t3 := (1 + z)/z^(2/3); 1 + z~ t3 := ------ 2/3 z~> u := t1^3 + t2^3 + t3^3 - 3*t1*t2*t3 /(x^(1/3)*y^(1/3)*z^(1/3)); 3 3 3 (1 + x~) (1 + y~) (1 + z~) 3 (1 + x~) (1 + y~) (1 + z~) u := --------- + --------- + --------- - ---------------------------- 2 2 2 x~ y~ z~ x~ y~ z~> simplify(u, {x*y*z = 1}); 2 2 2 2 2 2 x~ y~ + x~ z~ - 2 x~ + y~ z~ - 2 y~ - 2 z~ + 3> ;quit;bytes used=759012, alloc=720764, time=0.19John Adams served two terms as Vice President and one as President, but lostreelection. Later his son became President despite losing the popular vote.That son lost his reelection attempt badly. Now history is repeating itself.pmontgom@cwi.nl Microsoft Research and CWI Home: San Rafael, California === Subject: Bounds on multiplicity of an eigenvalueOriginator: fab@soda.csua.berkeley.edu (Fabio Rojas)What are some basic results on bounds for the multiplicity of an eigenvalue?I want to show that a not particularly complicated matrix has an eigenvalueof 1 (it does) and the multiplcity of the eigenvalue is 1.Is there way other than to explicitly compute the characteristic poly and analyze it? Fabio Rojas === Subject: Re: Bounds on multiplicity of an eigenvalue> What are some basic results on bounds for the multiplicity of an eigenvalue?> I want to show that a not particularly complicated matrix has an eigenvalue> of 1 (it does) and the multiplcity of the eigenvalue is 1.> Is there way other than to explicitly compute the characteristic> poly and analyze it?> Fabio RojasThe eigenspace related to the eigenvalue 1 is the kernel of the linearmapping with matrix A-I_n. Apply the rank-nullity theorem to this to getmultiplicity of 1 = n - rank(A-I_n).The rank can be computed by reducing the matrix into a row echelon formand counting the number of non-zero rows. That's all there is to it.Jyrki Lahtonen, Turku, Finland === Subject: Re: Bounds on multiplicity of an eigenvalue> What are some basic results on bounds for the multiplicity of an eigenvalue?> I want to show that a not particularly complicated matrix has an eigenvalue> of 1 (it does) and the multiplcity of the eigenvalue is 1.> Is there way other than to explicitly compute the characteristic > poly and analyze it? I don't know, but is it so complicated to calculate the characteristicpolynomial P(x) and to see that P'(1) is different from 0?Jose Carlos Santos === Subject: Re: Bounds on multiplicity of an eigenvalue> What are some basic results on bounds for the multiplicity of an eigenvalue?> I want to show that a not particularly complicated matrix has an eigenvalue> of 1 (it does) and the multiplcity of the eigenvalue is 1.> Is there way other than to explicitly compute the characteristic > poly and analyze it? > Fabio Rojas Bound -> Dimension :)Can't you diagonize the matrix? === Subject: Re: Bounds on multiplicity of an eigenvalue> What are some basic results on bounds for the multiplicity of an eigenvalue?> I want to show that a not particularly complicated matrix has an eigenvalue> of 1 (it does) and the multiplcity of the eigenvalue is 1.> Is there way other than to explicitly compute the characteristic > poly and analyze it? > Fabio Rojas > Bound -> Dimension :)> Can't you diagonize the matrix?Or given matrix A, determine that the dimension of the kernel of(A-Id) is one. Here, Id is the usual identity map. In general, givena potential eigenvalue x, the multiplicity of x is the dimension ofthe kernel of the linear map (A - x*Id). For all but a finite numberof values of x (for finite dimension matrices that is!), the kernel istrivial (consists solely of the zero vector) and hence has dimensionzero.Karl Hallowellkhallow@hotmail.com === Subject: Re: Characteristic polynomials of cyclic graph>Is there a simple way to compute the characteristic polynomial of>the adjacency matrix of a cycle with 1's along the diagonal?> 1 1 0 0 ... 1> 1 1 1 0 ... 0> 0 1 1 1 ... 0> 0 0 1 1 1.. 0> 0 0 0 1.....0> ........> 1 0 0 0 ..1 1 >I'm trying to calculate the multiplicity of the eigenvalues of this >creature, and I'm not getting a nice simple form for the matrix. Trying>to see if I made a mistake...If the matrix is n x n, for k = 0 to n-1 we have eigenvector v(k) with v(k)_j = exp(2 pi i k j/n) for eigenvalue 1 + 2 cos(2 pi k/n). Since k and n-k correspond to the same eigenvalue, all eigenvalues except 3 and (if n is even) -1 have multiplicity 2. Note that the Chebyshev polynomial T_n(t) has T_n(cos(2 pi k/n)) = 1.The characteristic polynomial is 2 (T_n((t-1)/2)-1). === Subject: Re: bounded function === > Subject: bounded function>Define f(0,0)=0 and f(x,y)=(x^3)/(x^2+y^2) if (x,y)!=(0,0).>1) Prove that D_1(f) and D_2(f) are bounded functions in R^2.> f_x(0,0) = 1, f_y(0,0) = 0 and for (x,y) /= (0,0):> f_x(x,y) = 3x^2 / (x^2 + y^2) - 2x^4 / (x^2 + y^2)^2> = x^2 (x^2 + 3y^2) / (x^2 + y^2)^2> a discontinuity at (0,0)> <= 3x^2 (x^2 + y^2) / (x^2 + y^2)^2> <= 3x^2 / (x^2 + y^2) <= 3> So you decided to do his homework for him.I decided to review my skills.> f_y(x,y) = 2x^3 y / (x^2 + y^2) unbounded along x = y> f_y(x,x) = 2x^4 / 2x^2 = x^2> False.f_y(x,y) = -2x^3 y / (x^2 + y^2)^2>2) Let u be any unit vector in R^2. Show that the directional>derivative D_u(f)(0,0) exists, and that its absolute value is>at most 1.> x = x0 + u.cos t; y = y0 + u.sin t; (x0,y0) = (0,0)> What does u.cos t mean?It's about a line along the direction vector.> f_u(x,y) = (cos t)f_x(x,y) + (sin t)f_y(x,y)> = (cos t)^3 (1 + 2(sin t)^2) + 2u^2 (sin t)^2 (cos t)^3> = (cos t)^3 (1 + 2(1 - (cos t)^2) at u = 0> = (cos t)^3 (3 - 2(cos t)^2)> = 3(cos t)^3 - 2(cos t)^5> -9(cos t)^2 sin t + 10(cos t)^4 sin t = 0> sin t = 0 is one extremum which gives +-1> cos t = 0 another extremum which gives 0> -9 + 10(cos t)^2 = 0 is the last critical point> cos t = sqr 9/10 which gives> +- (9/10)^(3/2) (3 - 2*9/10) = +- (9/10)^(3/2) (12/10)> = +- (108/100) sqr 9/10 = +- (324/1000) sqr 10 = 1.025> At t = pi/4, cos t = (1/2) sqr 2> (1/4) (sqr 2)(3 - 1) = (1/2) sqr 2 = 0.707> I have no idea what you're doing here.> Just let u = (a,b), compute> [f(ta,tb) - f(0,0)]/t, and let t -> 0.Using the chain rule to do the same with (a,b) = (cos t, sin t)with my u being your t.D_t f(ta,tb) = a.f_x(ta,tb) + b.f_y(ta,tb) = aa^2 (a^2 + 3b^2) / (a^2 + b^2)^2 - 2a^3 b / (a^2 + b^2)^2 = a^3 as (a,b) is unit vector> [f(ta,tb) - f(0,0)]/t, and let t -> 0.= (at)^3 / t((at)^2 + (bt)^2)= a^3 / (a^2 + b^2) = a^3 as u is unit vector. === Subject: Re: Frequentist probability confusion <405EC787.2070703@univie.ac.at As to the original question about natural numbers. If one requires that> probability measures be countably additive, then there is on probability> measure defined on the integers that I would want to label uniform.> Countably additive means that if {A_n} is a countable collection of> disjoint (measurable) sets, then the probability of the union of the A_n> is equal to the sum of the probabilities of each of the A_n. I certainly> want my probability measures to be countably additive, but some authors> argue that it is enough to be finitely additive. >It seems to me that no countable probability space for which each>element has probability zero can have a countably additive measure. >By a countably additive measure we mean that we have the useful>theorem that a countable union of sets of measure zero has measure>zero. In the hypothesized uniform measure over the integers each>singleton set has measure zero. The integers are a countable union of>singleton sets. So a countably additive measure must have a measure>of zero for the entire space and it fails to be a probability space.> Right; the problem would be solved if our number system had genuine> infinitesimals. Then each integer could be assigned a probability> 1/aleph_0 and everything would work out ok. Such number systems> exist and there is nothing inherent in probability theory which> ties it to real numbers except its current formulation. Actually measure theory can be applied to just about any set. Complexnumbers, vector spaces, groups, rings, topological spaces, and soforth are commonly used. I'd be very much surprised if non-standardanalysis had never been applied.> We could> even imagine a formal definition of probability distributions as> equivalence classes of algorithms for producing numbers from random> seeds. Then we could honestly claim to be talking about processes> generating numbers when we do probability.> If one allows finitely> additive probability measures, then one can defined a probability measure> over the natural numbers that some might want to label uniform. However,> the construction of this measure uses the the axiom of choice (or at least> some large portion of the axiom of choice).> Do you have a reference for this?> R.Finitely Additive Measures on Groups and Ringswith M. Pasteka, R. Tichy and R. Winkler On arbitrary topological groups a natural finitely additive measurecan be defined via compactifications. It is closely related toHartman's concept of uniform distribution on non-compact groups (cf.S. Hartman, Remarks on equidistribution on non-compact groups, Comp.Math. 16 (1964) 66-71). Applications to several situations arepossible.Some results of M. Pasteka and other authors on uniform distributionwith respect to translation invariant finitely additive probabilitymeasures on Dedekind domains are transfered to more generalsituations. Furthermore it is shown that the range of a polynomial ofdegree >1 on a ring of algebraic integers has measure 0.---There seems to be a fair amount of literature under finitely additivemeasure.I also tried non-Kolmogorov probability and found some Russiansrefuting the non-locality result of Bell's theorem using the p-adicsto produce sets with negative probabilities. Humph. === Subject: Re: Frequentist probability confusion As to the original question about natural numbers. If one requires that> probability measures be countably additive, then there is on probability> measure defined on the integers that I would want to label uniform.> Countably additive means that if {A_n} is a countable collection of> disjoint (measurable) sets, then the probability of the union of the A_n> is equal to the sum of the probabilities of each of the A_n. I certainly> want my probability measures to be countably additive, but some authors> argue that it is enough to be finitely additive. >It seems to me that no countable probability space for which each>element has probability zero can have a countably additive measure. >By a countably additive measure we mean that we have the useful>theorem that a countable union of sets of measure zero has measure>zero. In the hypothesized uniform measure over the integers each>singleton set has measure zero. The integers are a countable union of>singleton sets. So a countably additive measure must have a measure>of zero for the entire space and it fails to be a probability space.>Right; the problem would be solved if our number system had genuine>infinitesimals. Then each integer could be assigned a probability>1/aleph_0 and everything would work out ok.Wrong. In non-standard analysis, there existinfinitesimals, but all of them are much smaller thananything looking like that. All non-standard positiveintegers have at least as many smaller integers as thereare ordinary real numbers. But they behave like finiteintegers within the model. Such number systems>exist and there is nothing inherent in probability theory which>ties it to real numbers except its current formulation.Non-standard models of the real numbers are much moredifferent from the usual ones than you seem to think,and are at the same time more similar. We could>even imagine a formal definition of probability distributions as>equivalence classes of algorithms for producing numbers from random>seeds. Then we could honestly claim to be talking about processes>generating numbers when we do probability.This is already the case in probability as we have itnow, but with random seeds being real numbers uniformbetween 0 and 1. > If one allows finitely> additive probability measures, then one can defined a probability measure> over the natural numbers that some might want to label uniform. However,> the construction of this measure uses the the axiom of choice (or at least> some large portion of the axiom of choice).>Do you have a reference for this?One does not need much of the axiom of choice, but someis needed. If one only wants some of the sets to bemeasurable, nothing is needed; consider the field ofsets which are periodic from some point on, and giveit the limiting frequency. But what are you going todo with it? This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: what does close mean in topology?defined from another toplogical space S (which is a subuset of R^3) :two elements a,b in T are close if f(a) and f(b) are close in S and I don't see how this fits with the notion of neighbourhoods in theHausdorff axioms.Probably I'm jsut missing a very siple point. But how can a topology be defined from this notion of closeness? === Subject: Re: what does close mean in topology?> defined from another toplogical space S (which is a subuset of R^3) :> two elements a,b in T are close if f(a) and f(b) are close in S and > help a lot.Best ragards,Jose Carlos Santos === Subject: Re: what does close mean in topology?> help a lot.Thomas C. Hales: A Proof of the Kepler Conjecturehttp://www.math.pitt.edu/~thales/kepler04/ fullkepler.pdfpage 71, first paragraph Put a topology...I think all notations used are defined on pages 69/70. Lambda is a discrete subset of R^3, whose elements are called vertices. The Q-System is a set of tetrahedra defined in section 4. === Subject: Re: what does close mean in topology?> Thomas C. Hales: A Proof of the Kepler Conjecture> http://www.math.pitt.edu/~thales/kepler04/fullkepler.pdf> page 71, first paragraph Put a topology...> I think all notations used are defined on pages 69/70. Lambda is a > discrete subset of R^3, whose elements are called vertices. The > Q-System is a set of tetrahedra defined in section 4.I am not sure about this, but it would seem that the topology isdefined as follows: consider some pair (f,t). For each set W ={V_i : i in the domain of f}, where each V_i is a neighbourhood off(i), consider the set of all pairs (f',t) such that f and f' havethe same domain and, furthermore, each f'(i) belongs to V_i. CallW' to this set. Then the topology is the only topology such thata set V is a neighbourhood of a pair (f,t) iff V contains some W'.I hope that this makes sense to you.Jose Carlos Santos === Subject: Re: what does close mean in topology?Now I understand it as follows:V(f,t) = { (f',t') | t' = t / dom f' = f / forall i, f'(i) in V(f(i)) }where V(f,t) is the set of neighbourhoods of (f,t)and V(f'(i)) the set of neighbourhoods of f'(i)thank you for your quick response. === Subject: Re: what does close mean in topology?> defined from another toplogical space S (which is a subuset of R^3) :> two elements a,b in T are close if f(a) and f(b) are close in S and> Waylay my notion that you're reading math trash.What's the rest of the definition and what's f?> I don't see how this fits with the notion of neighbourhoods in the> Hausdorff axioms.> Probably I'm jsut missing a very siple point. But how can a topology be> defined from this notion of closeness?Weird. a,b are close in a metric space such as R^3 when d(a,b) < e forsome small e and the topology might be bunches of close points such asB(a,r) for all a and small r or B(a,1/n) for all a in R^3, n in N ?? === Subject: Re: integral problemJames> If I know that sum_(1)^(infty)(sin(nk))^2/((n^2)k)=(Pi-k)/2 how> can I show that int_(0)^(infty)(sin(x)/x)^2dx=Pi/2 letting k->0.We havesum_{n=1}^(infty) (sin(nk))^2 / ((n^2)k) =sum_{n=1}^(infty) k (sin(nk))^2 / ((nk)^2)which is a Riemann sum.LH === Subject: Re: VonNeumannGametheory of StockMarket; Crossover today 20 free shares BMY vs. SBCI am rather saddened and disappointed with myself for last friday BMYbriefly touched 25.05 and SBC was down to about 24.30. Today BMY was about24.48and SBC was about 24.30. You see, BMY made a crossover since my lasttransactions when I sold about 3,000 shares of SBC and bought 3,060 sharesof BMY with the proceeds. Last friday I could have utilized the crossoverand sold 3,000 BMY and with the proceeds gotten 3,060 SBC having made 60free shares of BMY and then 60 free shares of SBC.What kept me back on friday was pure greed in that I wanted to see at leasta full $1. spread.And this raises the issue of what in the VonNeumann gametheory of theOptimal Strategy for playing StockMarket does one act and execute upon aCrossover. Is there some math that hints of a spread minimum? Should Iexecute on a Crossover when it has reached at least a minimum of $1.spread?Archimedes Plutoniumwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxieswww.archimedesplutonium.comwww.iw.net/~a_plutonium === Subject: osculating circle and curvature> f, g, f', g', x and y are known, calculate f'' and g''. Hint ;) f''(t)= d f'(t)/dt , g''(t)= d g'(t)/dt, it is like findingaccelerations when velocities are given. You need not proceed to theevolute.You need to appreciate properties of curves or invariants beforetrying to manipulate them locally.For a parabola, {f = 2 a t,g = a t^2}. To manipulate g'' =2a you canchange the constant a which is its focal length property.Read further on to find what is a constant,what is a (time) variableand what are constants that enter as boundary conditions inintegration. Read about intrinsic/natural equations.E.g., if s=ArcLength, CornuSpiral has R*s = a^2, an Involute has R^2= a*s as intrinsic equations, where you can see on integration thenatural constant a to which it tends to, or, is a starting geometricalconstant property/ parameter.But R is continuously manipulated, orsimply put, bent. For cycloids, radius of generating circle is theconstant property in its intrinsic equation. Practically try bending atwistie to whatever R you want to,to get a physical feel. Later onsimulations could be useful.:).. If you want to manipulate Earth's acceleration around the Sun(and influence the seasons), the only way is to add or remove massfrom the Sun. r''=h^2/r**3-GM/r**2 . G is a universal constant, M ismass of Sun, and h is a constant angular momentum to be evaluatedfrom boundary conditions during a related integration.Narasimham === Subject: Re: Commutative rings aren't physical spaces (This Week's Finds 205) > This is a rather presumptuous post! Mr. Pratt presumes to know how > things should be in physics. Everyone knows that the mathematics > used in physics is just a model for describing the real world. Anyone > who says that the real world is or is not some mathematical > structure doesn't really get it. Take for example supposedly curved > space as in general relativity. Whitehead showed you could do > relativity in flat space. It's all a device for thinking about things > and shouldn't be confused with reality.Hopefully you will not consider me presumptuous if I am humbly tryingto add two mistakable but urgent cries for help which I did not yetmanage to get understandable posted in sci.physics.research.Dealing with similar questions I am, however, not plying theorybut I felt driven by serious deficits in understanding of auditoryfunction. Not every model fits well. Lighthill's passive travelingwave was definitely not the correct feet in Cinderella's shoe.My suspicion against adequacy of complex frequency analysis startedwhen I compared the spectrogram with my own auditory perception.Getting aware of many paradoxes and dilemmas I looked for reasons.In the end I am blaming ancient notions for what I consider anongoing mistake affecting much more than just the spectrogram.It was Ren.8e Descartes whose xyz-coordinates might have suggesteda scale of time extending from minus infinite to plus infinite. WhileDescartes strived for as little arbitrariness as possible, the missingreference points gives rise to unnatural choices. Wouldn't it bereasonable to use the only available alternative, i.e. the current border between past and future, instead? Language does so. I tried it too, and it works well.As a consequence, I had to separate spectral analysis from complexcalculus. Elapsed time is always positive as is distance and would bereasonably defined frequency and wave number. I was told, excessive useof complex quantities is like a bubble to be carefully protected againstperforation. It's up. Never mind. I was anyway unhappy with unjustifiedredundant symmetries and striking non-causality. Recently I found thatPCT means Hermitian symmetry, the same as in electrical engineering.Nonetheless I would like to abstain from pertaining speculations.Complex Fourier analysis relies on Heaviside's trick: The lack of futureinformation is disguised by zero padding the empty half of IR with evenand odd functions canceling each other. There is a catch in it: With thetraditional notion of time, Heaviside's function is only valid for a moment. So it works for deterministic prediction but analysis of processes with direct link to reality requires permanent relocation of the temporal window because elapsed time cannot be shifted at will. Please test my natural spectrogram: no arbitrary window, no trade-off. This is my first cry for help.Do not worry about lacking mathematical basics for real-valued frequencyanalysis. Fourier cosine transform in *IR+ instead of complex-valuedFourier transform in IR is proven as DCT for MP3 coding. Subsequent complex calculus also works. It took me 18 month of demanding discussion until a physicist apologized for wrongly blaming me wrong.Let's return to the relationship between mathematical model and reality:Hermitian symmetry of complex models usually stems from transformationinto the redundant fictitious complex domain. So it disappears with the return to reality where perfect symmetry tends to be an exception, provided the pertaining variable has not just been introduced by speculation. The same is true for the neglect of either clockwise or anticlockwise rotating phasors.The higher sophistication of complex models does not necessarilyconvey more details of reality. It carries the burden of mere arbitrariness beginning with the reference point of time and eventually including the choice of position, size, and shape of temporal or spectral windows.On the other hand, the two components of the complex model canreflect some relationship. For instance, they can be interpreted asrelated to particular elements of a circuit. Bilateral quantities yieldcomplex-valued models, unilateral quantities yield real-valued models.While axis of a cylinder is bilateral, implying propagating real andevanescent imaginary wave numbers. Its radius is unilateral.So the modal expansion of a cylinder is a mixed one.So called singularity functions got much simpler and reached a logicalstructure in *IR+. Several notorious problems vanished. Newly definedstep and ramp function allow integration Inversion of integration doesno longer cause loss of information. There are just a few minor deviations from mathematical tenets:I am sure, t=0 is not just irrelevant because it does no longer mark aninitial value. Any single value out of a continuum is not worth to benoticed separately, since it has infinitely times less weight than theensemble adjacent to it. Accordingly, I see it as nonsense to excludezero but not to exclude zero plus or minus an infinitely small value.In other words, I cannot see any justification for the distinctionbetween open and closed intervals along a continuous quantity.Please consider my incapacity to abandon above opinion as my secondcry for help.I was already pointed to non-standard-analysis. Admittedly, I am not yetsure whether or not it will answer my questions.Respectfully,Eckard Blumschein === Subject: Re: Peter Robson Garden - May 8th 1975New email address Thomas? Did hotmail shut you down for newsgroup abuse?Welcome to another set of killfiles - again.> <<Let vector F(vector R) = (vector A)f((vector R)(vector B)), where vector>A, vector B are constant vector fields. Prove that curl vector F is>orthogonal to both vector A and vector B.This seems like homework, so I will give some hints.First, I assume that by f((vector R)(vector B)) you mean f of the dotproduct of R and B.Remember the formula for the curl of the product of a scalar field anda vector field: curl(f(x.B) A) = grad(f(x.B)) x A + f(x.B) curl(A)What is the curl of a constant vector field?Remember the definition the gradient: grad(f(x.B)) = < D f(x.B),D f(x.B),D f(x.B) 1 2 3Where D_k is the partial with respect to x_k. Remember that D_k treatsall components other than x_k as constant, then compute D f(x.B) = D f(x B + x B + x B ) 1 1 1 1 2 2 3 3 D f(x.B) = D f(x B + x B + x B ) 2 2 1 1 2 2 3 3 D f(x.B) = D f(x B + x B + x B ) 3 3 1 1 2 2 3 3Remember that BxA is orthogonal to both A and B.These are hopefully enough hints. If not, post what you have done. === Subject: Re: homology> Show that H_1(X,A) is not isonorphic to ~H_1(X/A) if X = [0,1] and A isthe> sequence 1, 1/2, 1/3, ... together with its limit 0.> Here is what I have done so far:> Let Y be the the space obtained by collapsing all the points in A /in X toa> single point, Y =X/A this space must be the shrinking wedge of circles.> Let C_n be the circle of Y with radius 1/n> Considering the retractions r_n: Y --> C_n collapsing alle C_i's exceptC_n.> Each of these retractions r_n induces a surjection R_n: H_1(Y) --H_1(C_n).> Since the reduced holomogy H_1 of a circle is Z,> I get that R_n: H_1(Y) --> Z> Now the product of all these R_n's is a homomorphism R: H_1(Y) --> the> infinite direct product Z x Z x ...> This product is as far as I can see uncountable.This is discussed in Hatcher book Algebraic Topology.H1(X/A) is the abelianisation of Pi1(X/A) and is uncoutably generated, butit is all that can be proven simply.> Now I have some questions:> Is it possible to determine H_(Y) this way?> Is it possible to show that R is surjective? And if so...how?It's a difficult problem. But no need to solve it to answer your question.Tou have just to show that ~H1(A,A) can be countably generated.> Now I look at the reduced holomogy H_1(X,A)> I know that H_1(X) = 0 and H_0(X) = 0 so I have the short exact sequence:> 0 --> H_1(X,A) --> H_0(A) --> 0> This mean that H_1(X,A) is isomorphic to H_0(A)> I know that A is a countable set of points, so my guess is that H_0(A) is> the countably infinite direct sum of Z.Yes H_n(X) is always the direct sum of H-n( path components of X)> But I dont know if this is correct or how to show it.It's done in Hatcher's Book> Can anybody help me with this or tell me about the relation between the> uncountable direct product and the countable infinite direct sum.First has no countable basis. That all we have to know so far. === Subject: Help Needed Understanding ArticleHello all,I am interested in the Theory of Computation and also the use offormal methods of proof. So naturally I became very excited when Iacquired a copy of the paper, A Mechanical Proof Of The Unsolvabilityof The Halting Problem athttp://citeseer.ist.psu.edu/boyer82mechanical.html (click on PDF inthe upper right hand corner.)However, I am having a hard time understanding how they actually reachthe conclusion that the Halting Problem is unsolvable. I know aboutTuring's original proof in 1937 and don't have any problem with that. I just don't see that the above paper proves it using their method.Is there anyone here who believes that this paper does prove that theHalting Problem is undecidable, and if so, could you explain the logicand reasoning that is used? Are there axioms and rules of inference? What do they state, in plain English?Charlie VolkstorfCambridge, MA === Subject: Re: Help Needed Understanding Article> Hello all,> I am interested in the Theory of Computation and also the use of> formal methods of proof. So naturally I became very excited when I> acquired a copy of the paper, A Mechanical Proof Of The Unsolvability> of The Halting Problem at> http://citeseer.ist.psu.edu/boyer82mechanical.html (click on PDF in> the upper right hand corner.)> However, I am having a hard time understanding how they actually reach> the conclusion that the Halting Problem is unsolvable. I know about> Turing's original proof in 1937 and don't have any problem with that. > I just don't see that the above paper proves it using their method.> Is there anyone here who believes that this paper does prove that the> Halting Problem is undecidable, and if so, could you explain the logic> and reasoning that is used? Are there axioms and rules of inference? > What do they state, in plain English?> Charlie Volkstorf> Cambridge, MAThe authors inverted the LISP complier and applied the relationproving it is solvable to appear the solution using symbols. A keyconcept is the letter or byte as the symbol or the machine state. Theauthors have done nothing but confuse everybody.Now adays the existence of the LISP self compilation proves thehalting problem solved. === Subject: Re: Help Needed Understanding Article: Hello all,: I am interested in the Theory of Computation and also the use of: formal methods of proof. So naturally I became very excited when I: acquired a copy of the paper, A Mechanical Proof Of The Unsolvability: of The Halting Problem at: http://citeseer.ist.psu.edu/boyer82mechanical.html (click on PDF in: the upper right hand corner.): However, I am having a hard time understanding how they actually reach: the conclusion that the Halting Problem is unsolvable. I know about: Turing's original proof in 1937 and don't have any problem with that. : I just don't see that the above paper proves it using their method.: Is there anyone here who believes that this paper does prove that the: Halting Problem is undecidable, and if so, could you explain the logic: and reasoning that is used? Are there axioms and rules of inference? : What do they state, in plain English?: Charlie Volkstorf: Cambridge, MAPages 19 and 20 briefly describe the basics of their logic system,and refer you to references 1 and 2 for more details. The theoremprover uses rewrite rules, which are presumably based on LISP.Stephen === Subject: Re: Help Needed Understanding Article>I am interested in the Theory of Computation and also the use of>formal methods of proof. So naturally I became very excited when I>acquired a copy of the paper, A Mechanical Proof Of The Unsolvability>of The Halting Problem at>http://citeseer.ist.psu.edu/boyer82mechanical.html (click on PDF in>the upper right hand corner.)>However, I am having a hard time understanding how they actually reach>the conclusion that the Halting Problem is unsolvable. It's a research paper, not Unsolvalbility for Dummies. In particularit assumed prerequisites. (i) Do you know Lisp fairly well?(ii) Are you familiar with the logical system they're using?(Note the sentence in the Summary that reads The unsolvabilitytheorem is proved in a constructive logic like those of Skolem[7]and Goodstein[4]. Are you familiar with [7] and/or [4]?)>I know about>Turing's original proof in 1937 and don't have any problem with that. >I just don't see that the above paper proves it using their method.>Is there anyone here who believes that this paper does prove that the>Halting Problem is undecidable, and if so, could you explain the logic>and reasoning that is used? Are there axioms and rules of inference? >What do they state, in plain English? === Subject: Re: Help Needed Understanding Article>I acquired a copy of the paper, A Mechanical Proof Of The>Unsolvability of The Halting Problem at>http://citeseer.ist.psu.edu/boyer82mechanical.html (click on PDF in>the upper right hand corner.)>However, I am having a hard time understanding how they actually reach>the conclusion that the Halting Problem is unsolvable. > It's a research paper, not Unsolvalbility for Dummies. In particular> it assumed prerequisites. (i) Do you know Lisp fairly well?> (ii) Are you familiar with the logical system they're using?> (Note the sentence in the Summary that reads The unsolvability> theorem is proved in a constructive logic like those of Skolem[7]> and Goodstein[4]. Are you familiar with [7] and/or [4]?)Of course. Now, every proof can be expressed self-contained, can'tit? In fact, they can all be explained at a high level with as muchor as little detail as we wish to include, would you agree?So how about a little bit of self-contained explanation as to what inthat paper convinces you that the Halting Problem is unsolvable? Axioms, Rules and Theorems? Ok - what do the axioms and theoremsmean (I can supply the rules myself)?Charlie Volkstorf> ************************> === Subject: Re: Help Needed Understanding Article>I acquired a copy of the paper, A Mechanical Proof Of The>Unsolvability of The Halting Problem at>http://citeseer.ist.psu.edu/boyer82mechanical.html (click on PDF in>the upper right hand corner.)>However, I am having a hard time understanding how they actually reach>the conclusion that the Halting Problem is unsolvable. > It's a research paper, not Unsolvalbility for Dummies. In particular> it assumed prerequisites. (i) Do you know Lisp fairly well?> (ii) Are you familiar with the logical system they're using?> (Note the sentence in the Summary that reads The unsolvability> theorem is proved in a constructive logic like those of Skolem[7]> and Goodstein[4]. Are you familiar with [7] and/or [4]?)>Of course. Of course? If you're familiar with the system of logic used inthe paper then why do you keep asking people to explain itto you?>Now, every proof can be expressed self-contained, can't>it? In fact, they can all be explained at a high level with as much>or as little detail as we wish to include, would you agree?>So how about a little bit of self-contained explanation as to what in>that paper convinces you that the Halting Problem is unsolvable? When did I say that the paper convinced me that the haltingproblem was unsolvable?I personally _don't_ know any Lisp, and I'm _not_ familiar withthe constructive logic they say they're using. Hence I'mnot qualified to judge the correctness of the paper.Which is why I have not made any statements about itscorrectness.I do otoh have an advantage over some people: I understandthat there are things that I don't understand, and I don'tgo around saying that a paper I don't understand simplecannot contain what it claims to contain. Or that I'm the only one who's ever done something, when there exist papers I don't understand that claim to do the same thing. If I'ddone something, thought I was the only one who'd doneit, and was told about a paper where it had been donepreviously I'd make certain to understand that paper_before_ informing the world that the authors were lying.(No, you _don't_ understand the contents of the paper.This is evident from the way you continually ask peopleto explain it to you. You're reminding people of JamesHarris, explaining why Wiles' proof of Fermat's lasttheorem is wrong although he doesn't even know whata ring is.)>Axioms, Rules and Theorems? Ok - what do the axioms and theorems>mean (I can supply the rules myself)?>Charlie Volkstorf> ************************> === Subject: Re: Sorry>If xyz = 1, prove that (xy)^2 + (yz)^2 + (zx)^2 - 2x - 2y - 2z + 3 >= 0.Writing z = 1/(xy), we want to show that 0 2 2 1 1 2 <= x y + --- + --- - 2x - 2y - --- + 3 x^2 y^2 x y 2 2 1 1 2 = x y - 2x - 2y + 3 + ( - - - ) x y 1 1 2 = f(x,y) + ( - - - ) x ySetting the partials of f(x,y) to 0, we get that its minimum occurs at(x,y) = (1,1). Since f(1,1) = 0, the inequality holds. === Subject: Re: Sorry>If xyz = 1, prove that (xy)^2 + (yz)^2 + (zx)^2 - 2x - 2y - 2z + 3 >= 0.>Writing z = 1/(xy), we want to show that> 0> 2 2 1 1 2> <= x y + --- + --- - 2x - 2y - --- + 3> x^2 y^2 x y> 2 2 1 1 2> = x y - 2x - 2y + 3 + ( - - - )> x y> 1 1 2> = f(x,y) + ( - - - )> x y>Setting the partials of f(x,y) to 0, we get that its minimum occurs at>(x,y) = (1,1). Since f(1,1) = 0, the inequality holds.>Rob Johnson take out the trash before replying The partials of f do vanish at (1, 1), but this is a saddlepoint, not a minimum. For example, x = 0.9 and y = 1.1 give f = 0.9801 - 1.8 - 2.2 + 3 = -.0199 < 0Easier for hand computation is x = 0 and y = 2, giving -1.John Adams served two terms as Vice President and one as President, but lostreelection. Later his son became President despite losing the popular vote.That son lost his reelection attempt badly. Now history is repeating itself.pmontgom@cwi.nl Microsoft Research and CWI Home: San Rafael, California === Subject: name ?hello.....sir~i saw the Fleury's Algorithm.i want to know his perfect name.perhaps, i think that he is the French.if possible, let me introduce web-site for his life.thank you very much. === Subject: Re: name ?>hello.....sir~>i saw the Fleury's Algorithm.>i want to know his perfect name.>perhaps, i think that he is the French.>if possible, let me introduce web-site for his life.>thank you very much.It could be a math guy named Henri Fleury.Here, it says he invented a solitaire game in 1890http://www.recreomath.qc.ca/dict_tricolore_paradoxal.htm and another onehttp://www.recreomath.qc.ca/dict_cameleon.htmwhich look like they have something to do with paths. There is also one document published under that name in 1865 here:http://gallica.bnf.fr/Th.8eorie .8el.8ementaire des convergents des fonctions d'une seulevariable...etc...by Henri Fleury,chief of instruction in Marseille, diploma inmathematical sciences. === Subject: ArcsA path p:[0,1] -> S from a to b is an arc when1) p([0,1]) is homeomorphic to [0,1]2) p embeds [0,1] into S ?---- === Subject: Anti-compactA space is anti-compact when every compact subset is finite.Discrete spaces are anticompact, Hausdorff.Uncountable cocountable spaces are anticompact, T1 and not Hausdorff.Are all anticompact, Hausdorff spaces discrete?Assume S anti-compact and let aj -> a.Then { aj | j in N } / {a} is compact and the only convergent sequences are finite.In addition if S is T1, then S is kc (compact closed) space. as compact sets are finite and finite sets are closed.Now let S be anti-compact, 1st countable T1.Assume S isn't discrete.Thus some infinite subset K sans isolated points.Take k in K and let U_j be a local base for k.As k isn't isolated, every U_j is infinite and further by T1we may demand U_j be strictly descending local base for k.Pick aj from U_j U_(j+1). Thus aj -> k but whoa, nowI've an infinite converging sequence and infinite compact set.Hence anti-compact, T1 or Hausdorff, spaces aren't 1st countable.Same arguement fails for nets, because a convergent net with it'sconverging point may not be compact.Is N^R anti-compact Hausdorff not discrete example? No, becauseN^N is a 1st countable infinite T1 non-discrete subspace.---- === Subject: need help on Normalizing dataI have a problem on working with infinite power series in Matlabi am using this seriesfn=summation(an*x^n)where x=50 and n=1,2,...........when x>182, the term x^n is going to infinity....This may be due tothe value is out of matlab limit (its a very big number)but I want to use more number of terms in my apllication (may bearound 350 terms)is it possible to overcome this problem by normalizing x??If yes, How can I do this in matlab?Please suggest me how to deal with this problem?is there any method to scaling the x^n term in the series? === Subject: Re: need help on Normalizing data> I have a problem on working with infinite power series in Matlab> i am using this series> fn=summation(an*x^n)> where x=50 and n=1,2,...........> when x>182, the term x^n is going to infinity....This may be due to> the value is out of matlab limit (its a very big number)> but I want to use more number of terms in my apllication (may be> around 350 terms)> is it possible to overcome this problem by normalizing x??> If yes, How can I do this in matlab?> Please suggest me how to deal with this problem?> is there any method to scaling the x^n term in the series?> RaviYou might calculate sum(an*x^(n-c)) for some constant c (as long as x isn't0, in which case you'll have your answer) then multiply that by x^c. If c issomewhere around 350/2 = 175, you'll not run into the explosion, but youmight have the opposite problem when x is small, though in that case you canuse your original calculation. === Subject: Re: need help on Normalizing data> You might calculate sum(an*x^(n-c)) for some constant c (as long as x isn't> 0, in which case you'll have your answer) then multiply that by x^c. If c is> somewhere around 350/2 = 175, you'll not run into the explosion, but you> might have the opposite problem when x is small, though in that case you can> use your original calculation.I am getting right answer from this method for upto x=30, but for x>30i am facing the same overflow problem...and coefficients (an) are of alternative sign and a(2n+1)=0.........Ineed to try for higher value of xis there any other way to overcome this problem? === Subject: Re: need help on Normalizing data> You might calculate sum(an*x^(n-c)) for some constant c (as long as xisn't> 0, in which case you'll have your answer) then multiply that by x^c. Ifc is> somewhere around 350/2 = 175, you'll not run into the explosion, but you> might have the opposite problem when x is small, though in that case youcan> use your original calculation.> I am getting right answer from this method for upto x=30, but for x>30> i am facing the same overflow problem...> and coefficients (an) are of alternative sign and a(2n+1)=0.........I> need to try for higher value of x> is there any other way to overcome this problem?The only other thing I can think of is that perhaps the coefficients aresmall enough so that you can calculate an*x, then iteratively multiply thatby x until you get to an*x^n. If you want to do this without iteration, youmight calculate exp(ln an + n ln x), which will hopefully keep you in thebounds of the program.To take advantage of this in Matlab, I should note, you might have to usefloating point operations, since, if Matlab is like Maple, it'll try tocompute it exactly in fraction form, which might lead you to anotheroverflow. === Subject: MeanderHi!I'm looking for any info about meander curve.Barbara === Subject: Re: Meander> I'm looking for any info about meander curve.A Google search for meander curve gave me more than 10 000answers.Jose Carlos Santos === Subject: Hybrid Conjoint AnalysisHi I'm a research professional based in India. I'm facing a researchproblem, which requires conjoint analysis. The study involves 27hybrid conjoint as the most pertinent conjoint design. Sawtoothpublished a per title Adaptive Choice Based Conjoint (ACBC).I would like to adiminister Choice Based Conjoint (CBC)involving 10attributes to each individual. However, these 10 attributes would bethe ones considered moost pertinent/important by him. Similarly, iwould like to have those attribute levels which are relevant to him. Iwas hoping to use Self explicatory ratings to obtain this information.So, a hybrid conjoint involving ACA and CBC would have been perfect.Is there any software which I could use. Sawtooth doesn't have acommercial version yet for ACBC.Can someone please help me out? Finally, the entire conjoint exercisehas to be web-based. There is no alternative for that. We also wantthe two stages to happen immediately after each other.Awaiting reply === Subject: Re: Resistance to Change>I am referring to the sarcasm which suggests that my paper is the>opposite of brilliant without any justification. Once again sarcasm>is used to make a point without backing it up with any kind of a>technical analysis - just as I originally pointed out.> And once again, the fact that you have to manufacture imaginary> playmates in order to make it look like someone agrees with you> makes all this simply hilarious, regardless of what you're referring> to above.> Oh, I forgot, the fact that the two IP addresses are the same> must be due to someone stealing yours when you were at> Kinko's. A truly amazing coincidence, the fact that the _one_> person on usenet who speaks up in support is the same> as the person who was at Kinko's when you were there> that day.>I guess there's a lot of amazing things about my paper.*Your paper? Huh - based on the content of your posts here(in particular on what you seem to think is interesting and/orimportant) I would not have guessed you'd ever publishedanything.Where did it appear?>You also forget that I live in Harvard Square (where I do my research)>and there's a lot of smart people around here. That would probably>explain it.>Or maybe he read the post as I sent it?>Actually, I think one guy once agreed with me. *sigh* It's lonely at>Kinko's as me.)> Guffaw. Chortle. Do you _really_ think anyone buys that?> If you have any interest in maintaining any sort of > credibility you'd... come to think of it admitting you> made the guy up won't help your credibility either.> Chortle.>Wait a minute - maybe all the Kinko'ses have the same IP address? >Then he could have been anywhere. Then maybe even you did it to make>me look bad!>* The 1st paper to ever axiomatize 3 branches of Mathematics with one>set of Rules of Inference! (better than Euclid) === Subject: Re: Resistance to Change> Your paper? Huh - based on the content of your posts here> (in particular on what you seem to think is interesting and/or> important) I would not have guessed you'd ever published> anything.What did I think was interesting or important that isn't?> Where did it appear?http://www.arxiv.org/html/cs.lo/0003071All this discussion and you hadn't even read the paper??? :(Charlie Volkstorf> ************************> === Subject: Re: Resistance to Change> Your paper? Huh - based on the content of your posts here> (in particular on what you seem to think is interesting and/or> important) I would not have guessed you'd ever published> anything.>What did I think was interesting or important that isn't?> Where did it appear?>http://www.arxiv.org/html/cs.lo/0003071Uh, right. So it hasn't actually been published in a refereedjournal? >All this discussion and you hadn't even read the paper??? :(I haven't _said_ anything about this paper that I haven't read...>Charlie Volkstorf> ************************> === Subject: Re: Resistance to Change> It's lonely at the top.> At the top of idiocity?> F.In a sense, yes. And what do you find idiotic?that it proved some significant theorem, but was unable to provide anylogic or reasoning as to how they reached that conclusion? Would youfind that idiotic?Children believe whatever you tell them - Santa Claus, the ToothFairy. For children, being innocent and naive is quite charming.Charlie VolkstorfCambridge, MAPS Read The Wizard of Oz. It's very insightful. === Subject: Re: Resistance to Change> (http://citeseer.ist.psu.edu/boyer82mechanical.html), or is it> open-source Nqthm proof checker) is available at:> Now, Messrs. Boyer and Moore have their own notion of a proof. It is> the exhibition of a collection of LISP programsMore precisely, it is the exhibition of a sequence of expressions in aside-effect-free, Lisp-like language. Some expressions are definitionsand others are theorems.> and the assertion that they may be input into another program that> then declares them to be acceptable.I wouldn't consider that part of the proof. Nqthm just verifies thatthat the definitions in the proof are admissible and that the theoremsfollow from either the previous expressions or the axioms via theinference rules of Boyer & Moore's Computational Logic (sketched inAppendix A of boyer82mechanical and described precisely in _AComputational Logic Handbook_.)> But is that really a compelling argument to you that the> Halting Problem is unsolvable? Could you explain how that follows?> What is the logic and reasoning by which we reach that conclusion?Since the proof weighs in at just 5.2 kbyte (minus comments and extrawhitespace) it's easy to verify that it supplies a model of universalcomputation and an expression asserting the unsolvability of its haltingproblem. One can then check that each lemma follows from the axiomsand/or previous definitions or lemmas via the rules of inference. As abonus, a widely-used open-source proof-checker confirms this.> And is there a reason that the myriad of related theorems don't also> flow from their approach?But many other theorems have flowed from their approach, including: * the invertibility of the RSA public key encryption algorithm, * the Church-Rosser theorem of the lambda calculus, * Goedel's first incompleteness theorem, andFor many more, see: More precisely, it is the exhibition of a sequence of expressions in a> side-effect-free, Lisp-like language. Some expressions are definitions> and others are theorems.How many intermediate theorems are there and what do they mean, inordinary English? For example, here is an explanation of the first 5intermediate theorems in my axiomatic system's reductio ad absurdumproof of the unsolvability of the self-applicability problem (I willlist all of them if you wish, but I take it that my point is welltaken):Theorem V-HALT(I,I)The Self-applicability Problem is unsolvable. Proof1. HALT(I,I) Given2. ~HALT(I,I) NOT 13. TRUE(x) Axiom 1 We can list the Universal Set.4. TRUE(x)^~HALT(x,x) DO 3,2 I=x5. ~HALT(x,x) DEF-7 4 Property of Universal Set1 = The self-halting predicate is recursive.2 = The non-self-halting predicate is recursive.3 = The universal set (the natural numbers) is r.e.4 = The intersection of the universal set and the non-self-halting setis r.e.5 = The non-self-halting set is r.e.etc.What are the intermediate theorems to which you refer (that convinceyou that the Halting Problem is unsolvable)?> it supplies a model of universal> computation and an expression asserting the unsolvability of its halting> problem.What are the input and output (if any) of this program and what doesit compute internally?> each lemma follows from the axioms> and/or previous definitions or lemmas via the rules of inferenceAnd what are the axioms and lemmas used, in plain English (as wediscussed above)? How many steps are there in this proof?> And is there a reason that the myriad of related theorems don't also> flow from their approach?> But many other theorems have flowed from their approach.I was specifically referrring to variations of the Halting Problemflowing from this proof, such as those that occur in my axiomaticsystem's proof (sample above and in my previous post.) However, theanswers to the above questions should explain what its byproducts are.Charlie VolkstorfCambridge, MA === Subject: Re: Resistance to Change> How many intermediate theorems are there and what do they mean, in> ordinary English?Your questions seem to be answered directly in the paper. Earlier yousince? If so, how recently?Anyway, a quick glance at the proof reveals that it uses 10 intermediatelemmas and 14 definitions. In addition to the descriptions in thepaper, each definition (DEFN) and lemma (PROVE-LEMMA) statement in theinput file is preceded by an explanatory comment in in ordinary English(e.g. This function returns T or F according to whether X occurs in thebody of some defn in LST.).> What are the input and output (if any) of this program and what does> it compute internally?According to section 2, The LISP Interpreter: The programming language used in our statement of the halting problem is a version of Pure LISP [5]. We present our version by defining the logical function EVAL, which takes four arguments: 1. an S-expression to be evaluated, 2. a variable alist* assigning values to variable symbols, 3. a function alist assigning definitions to nonprimitive function symbols, and 4. a natural number, indicating the maximum depth of function calls. EVAL returns either the value of the S-expression in the given environment or else it returns the object (BTM). * An alist is a list of pairs.The internal workings of the function are described in section 2.1,Formal Description of EVAL. See also section 2.2, An EnglishParaphrase of EVAL, and section 2.3, Examples of EVAL. === Subject: Re: Resistance to Change> Now, Messrs. Boyer and Moore have their own notion of a proof. It is> the exhibition of a collection of LISP programs>More precisely, it is the exhibition of a sequence of expressions in a>side-effect-free, Lisp-like language. Some expressions are definitions>and others are theorems.> and the assertion that they may be input into another program that> then declares them to be acceptable.>I wouldn't consider that part of the proof. Nqthm just verifies that>that the definitions in the proof are admissible and that the theorems>follow from either the previous expressions or the axioms via the>inference rules of Boyer & Moore's Computational Logic (sketched in>Appendix A of boyer82mechanical and described precisely in _A>Computational Logic Handbook_.)> But is that really a compelling argument to you that the> Halting Problem is unsolvable? Could you explain how that follows?> What is the logic and reasoning by which we reach that conclusion?>Since the proof weighs in at just 5.2 kbyte (minus comments and extra>whitespace) it's easy to verify that it supplies a model of universal>computation and an expression asserting the unsolvability of its halting>problem. One can then check that each lemma follows from the axioms>and/or previous definitions or lemmas via the rules of inference. As a>bonus, a widely-used open-source proof-checker confirms this.> And is there a reason that the myriad of related theorems don't also> flow from their approach?>But many other theorems have flowed from their approach, including:> * the invertibility of the RSA public key encryption algorithm,> * the Church-Rosser theorem of the lambda calculus,> * Goedel's first incompleteness theorem, andThat's interesting. It's not clear to me that inserting actual content into a thread likethis is appropriate. But I suppose it's ok if the content isinteresting enough...>For many more, see:> More precisely, it is the exhibition of a sequence of expressions in a>side-effect-free, Lisp-like language. Some expressions are definitions>and others are theorems.>Nqthm verifies that>that the definitions in the proof are admissible and that the theorems>follow from either the previous expressions or the axioms via the>inference rules of Boyer & Moore's Computational Logic (sketched in>Appendix A of boyer82mechanical and described precisely in _A>Computational Logic Handbook_.)>Since the proof weighs in at just 5.2 kbyte (minus comments and extra>whitespace) it's easy to verify that it supplies a model of universal>computation and an expression asserting the unsolvability of its halting>problem. One can then check that each lemma follows from the axioms>and/or previous definitions or lemmas via the rules of inference. As a>bonus, a widely-used open-source proof-checker confirms this.>Many other theorems have flowed from their approach, including:> * the invertibility of the RSA public key encryption algorithm,> * the Church-Rosser theorem of the lambda calculus,> * Goedel's first incompleteness theorem, and> That's interesting. > It's not clear to me that inserting actual content into a thread like> this is appropriate. But I suppose it's ok if the content is> interesting enough...Oh, so what in the above convinces you that the Halting Problem isunsolvable? (And don't worry - I will find your answer to thisquestion EXTREMELY interesting.)Charlie VolkstorfCambridge, MA>For many more, see:> ************************> === Subject: Re: Is Information really matter/energy, (randomness and entropy)NO LONGER WRT: Is Information really matter/energy, (randomness andentropy)I was going to say something else, then I read what Uncle Al had tosay...I *may* have met my match, that much is clear. So instead I thoughtI'd challenge this uncle Al fellow to a duel. And not owning eitherswords or weapons, I decided that the duel should be in the domain ofanti-randomness.As anyone knows (whose been paying attention, non-randomness can beexpressed aseither compression or predictions.) These days I don't bother tocompress, I predict.I know who will be the next US president, and a whole lot of otherthings that I have cajoled out of the depths of my computer. But thatstuff is far off. For today, (well, tomorrow) I will do somethingmore definite.So if you accept my challenge Al, my first 'shot' will be to predict apost the OPEN, HIGH, LOW, and CLOSE'ing prices for some major USstocks.And what can you do?PS: You'll need to post your actual name and email address if youaccept my challenge. I want to know who you are. === Subject: Re: Is Information really matter/energy, (randomness and entropy)> As anyone knows (whose been paying attention, non-randomness can be> expressed as> either compression or predictions.) These days I don't bother to> compress, I predict.> I know who will be the next US president, and a whole lot of other> things that I have cajoled out of the depths of my computer. But that> stuff is far off. For today, (well, tomorrow) I will do something> more definite.Jaysesskryst Gilbert, somehow I was under the impression that you areonly one sorry deluded wretch of a crank. Now you are trying for afull featured big time lunatic. Makes one feel really sorry for you. > PS: You'll need to post your actual name and email address if you> accept my challenge. I want to know who you are.bother? Predict it or uncompress it... === Subject: Re: Is Information really matter/energy, (randomness and entropy)Also maybe, is there a Shannon brachistochrone and or tautochrone is polymerasechain reaction travelling salesman problem workup.Dream on nacelles' wind, Caught in tepid Tome, an urge, Gin in from the cold. === Subject: Re: Is Information really matter/energy, (randomness and entropy)Or in the same sense, is there a least entropy dpendent on time of solution ofa traveling saleman problem, shortest receipt route to cover target sales.Important post.Dream on nacelles' wind, Caught in tepid Tome, an urge, Gin in from the cold. === Subject: Shopping basket special offerI need to create an equation that calculates the following info for the contents of a shopping basket:1) Number of items that should be charged at the normal price.2) Number of items that should be free....when we know the following information:1) The total number of items in the shopping basket.2) The number of items that the shopper has to buy at normal price to be eligible for some free items.3) The number of free items that the shopper can get for each iteration of the special offer.e.g. - dealing with apples:Offer = buy 3 apples get 2 apples free (i.e. total 5 apples).Number of apples in shopping basket = 10Number of apples shopper has to buy = 6Number of apples shopper gets free = 4 === Subject: Re: Shopping basket special offer> I need to create an equation that calculates the following info for > the contents of a shopping basket:> 1) Number of items that should be charged at the normal price.> 2) Number of items that should be free.> ...when we know the following information:> 1) The total number of items in the shopping basket.> 2) The number of items that the shopper has to buy at normal price > to be eligible for some free items.> 3) The number of free items that the shopper can get for each > iteration of the special offer.> e.g. - dealing with apples:> Offer = buy 3 apples get 2 apples free (i.e. total 5 apples).> Number of apples in shopping basket = 10> Number of apples shopper has to buy = 6> Number of apples shopper gets free = 4Well, we can get a ratio of # of free items per normal items, we'll call f/n.And we'll call the total t, and we'll call the amount of normal priced items N. It naturally follows that the amount of free items, F, is t-N. But also, F=(f/n)*N.t - N = (f/n)*Nt = (f/n)*N - Nt = N*(f/n - 1)N = t / (f/n - 1)And again, F = t - N. === ===Subject: Re: Antidiagonal, Infinity> Ok, I took this at face value and decided to start over, reread what I> had read, and try to figure out where I was going wrong. I think I> understand a bit more fully the nature of the binary arguments special> circumstances now.> The number constructed using th diagonal is designed in general to> produce a number not on the list, in binary because of the duel> representation issue the diagonal isn't neccesarily the best way to> construct a number not on the list, so an alternative contruction must> be done. Most of my previous arguements were only to demonstrate that> another number exists that isn't on the list, and attempts to find> another construction which would work every time a duel representation> number appeared, and I convinced myself I could always reproduce such> a number. I could simply swap a couple well chosen rows each time, or> construct a specific number not on the list by some other means.> For example,in order to construct a diagonal number with a particular> diagonalization which appears already on the list we need to consider> any two elements on the list. Each element intersects the diagonal> number at some digit, we'll call them the p and qth digit of these two> numbers respectively. the pth digit is either 0 or 1, and the qth> digit is either 0 or 1, and the diagonal number is the opposite. Now> consider a third real number r. We'll construct r as follows, r will> agree with the qth and pth digit of those numbers, and agree at every> other digit with the diagonal number. Since this number r can only be> intersected by the diagonal at the pth or qth digit, all that's left> to show is that p and q exist which are different from r and cannot be> anywhere else on the list. So consider the real number l and m. l> agrees with the diagonal line at every position except p, and m agrees> at every position except q. I now have three numbers, and only two> spots on the diagonal to put them, since they must intersect at p and> q. One of these three numbers is not on the list. Does this work to> demonstrate the uncountability of the reals?Not good enough.Take the following listing:. 1 0 0 0 0 (all other columns are zeroes). 0 0 1 0 0. 0 1 0 0 0. 0 1 1 0 0. 0 0 0 0 0 (all other rows are zeroes)Take p = 2 and q = 3The diagonal is: .10000... = 1/2The anti-diagonal is .01111... = 1/2 On the list at position 1r is .0001111... = 1/8 On the list at position 2l is .0011111... = 1/4 On the list at position 3m is .0101111... = 3/8 On the list at position 4Your construction fails. The anti-diagonal and each of the threeproposed modifications thereof all appear on the original list.Yes, yes, it is easy to find a real number that is not on this list.3/4, for instance.And yes, for this particular list, you could choose p and q differentlyand come up with a real number not on the original list. But yourclaim seems to have been that any p and q would work.Your task is to come up with a mechanical rule (or at least an existenceproof) that will, given an arbitrary countable list of real numbers,generate a real number not on that list.Such rules exist. But, so far, you haven't given one. === Subject: Re: Antidiagonal, InfinityIn considering that in bases other than binary, that differentantidiagonals or rather everywhere-non-diagonals can be generated,I think a key point to consider is the reversion of the argument inthat in any base the antidiagonal can by the simple rule that 0->1,1->2, etcetera be generated and that it may well dually represent anelement of the list.That is to say, for any 1-1 mapping between the finite set of eachpossible value of an integral modulus for a given base b, the set {0,..., b-1}, and itself, to generate the antidiagonal, the antidiagonalmay well have dual representation as a list element, and it fact inconsideration of the set-antidiagonal of the complete set, it must,for any rule. The set-antidiagonal of the set of infinite binarysequences either does not exist, or there is dual representation of atleast a pair of those elements. Constraint to a contrived rule forthe list may lead to an unconstructible antidiagonal.What do you think about the leading zeros? Any expansion can have theradix shifted and infinitely many zeros prefixed to it, changing therepresentation but not value.The antidiagonal argument is not complete until you explain how toapply it to binary, because those sequences in binary do representeach element of the unit interval of reals, and dual representationdoes allow the construction of the antidiagonal. I agree that you candefine a rule that will generate a set different than each element ofa set, except for the set of all sets, but not that it will be anelement of that set.The argument against the antidiagonal argument by the sole binaryantidiagonal is not complete without explaining how, in bases otherthan binary, antidiagonals by arbitrary rules can generate non-dualrepresentations. For the set-wise application of that rule, it mayfail. Many or most rules to generate antidiagonals do have thecapacity to generate dual representations.Please note that it's dual, not duel, representation. I thinktransfinite set theory can, and perhaps should, be totally ignored inprimary and secondary education. Middle school students should belearning about algebra and trigonometry, and introductory complexanalysis. They are to encounter the infinite in counting numbers inlower grades, and later as an introduction to the integral calculus. That is biased from my own education and not as an educator.I want to address something about set theory and paradoxes: noparadoxes are allowed, and a class is a set.I say that a function can exist between the naturals and the unitinterval of reals, and that the natural/unit Equivalency Function issuch a function. === Subject: Re: Antidiagonal, Infinity> But that was precisely the starting point for this whole discussion. You> began by asking why it would ever be necessary to use an alternative> construction. You seemed to be arguing that such constructions were not> necessary, but every time you tried to justify this claim, you found it> necessary to call on some alternative construction in order to prove it.Yes exactly so, and I think what I was getting caught up on was that Ididn't realize any list was equivilent to a countable subset of reals.The equality wasn't clear to me till I saw the list presented for theduel representation form in binary. At that point I failed to make theconnection to any infinite countable subset of the reals though, andit took me some time to figure out that the list defined the subset.> It's not sufficient to consider just two elements on the list. You need> to consider all of the elements on the list. The point is not that,> given any two numbers on the list, it is possible to construct a third> number that differs from both of them. What we need is a reliable way (a> single rule) that will produce, for every list, a number not already on> the list.Ah I was only attempting such a contruction in the case where thediagonal produces a number on the list. I use the properties of such aconstruction to create an element not on the list. Let me show youwhat I have in mind, and you can perhaps let me know if I'm stillwrong.Firstly we note, that any such construction must rely on duelrepresentation, because the diagonal number differs at every place.Now, since any duel representation number must terminate in all 1s orall 0s in binary after some number of digits say, n then for all n+mfor m > 0 are all 0s or 1s. If the list is 0s for positions greaterthen n, then the diagonal is all 1s for positions greater then n. Howdo we know this?Thinking about what a duel representation means, and the nature ofwhat's we're calling the same number here, we clearly see that itrelies on our ability to show that a number of the form 1/(2^n) isequivilent to the limit of the series created by the sumation of1/(2^(k)) from k = n+1 to infinite. Now if we take the numberterminating in all 1s, we notice that by changing any 1 to a 0, it'sequivilent to subtracting 1/(2^j) where j > k. Subtracting from bothsides of the equality and observing that 1/(2^n) does not equal1/(2^n) - 1/(2^j) makes it clear a position n exists. This isn't trueof reals in general, but the duel representation type of numbers itdoes.Now this is where I introduce the two numbers I have in mind, but in asomewhat different setting, since I polished up the idea a bit. Itoccured to me I actually was thinking of a set of numbers, which areall the numbers of the following form. Consider the subset of thereals P, such that all the numbers that disagree with the diagonalnumber at position p, and agree with it everywhere else for any p. Thediagonal number must intersect these numbers at position p, because ifthey intersected them anywhere else, wouldn't produce the diagonalnumber in question.Now consider another real number, well call it C. We'll say that thisnumber agrees with the diagonal number until position n, and theninstead has infinitely many 0s and 1s after position n. I'm going toattempt to show, that this second number is on the list, or thenumbers of the other kind are not on the list. So consider the firstposition where C can intersect the list, which is position n+1. If n+1agrees with the diagonal, then C cannot be in position n+1. ThereforeC must disagree at some position n+m with the diagonal to intersectit, and we're assured that such points exist because C has infinite 0sand infinite 1s, so it must disagree with the diagonal number sincethat number is a duel representation.Choose the smallest m such the n+m = p where p disagrees with thediagonal number. Now we know that C is a different number then any ofthe numbers in P, because C has infinite 0s and 1s while all thenumbers in P have a finite number of either 0s or 1s. Also now, noteat position p, an element of P must intersect the diagonal at p.Therefore C cannot intersect the diagonal at that position.diagonal, and we notice that a distinct element of P for this newposition, say p + q, must intersect the diagonal at that position,therefore C is not at position p + q on the list. In fact, C cannot beat any position on the list because each element of P must also belisted, and for each potential place to list C, an element of Pprevents us from listing it at that point, since we cannot list anyelement of P anywhere else then position p.Basically I believe every number of the form C represent an unlistedreal number, or if C is listed, then the coresponding element of Pthat must intersect the diagonal wherever C is listed isn't listed. Byplacing one on the list, you've pushed the other one off it. Whatmakes this possible is that our original diagonal produced a duelrepresentation, this duel representation gives us a finite number of1s or 0s, while C has infinite of both. I'm really hoping this ismaking sense. === Subject: Re: Antidiagonal, Infinity> It's not sufficient to consider just two elements on the list. You need> to consider all of the elements on the list. The point is not that,> given any two numbers on the list, it is possible to construct a third> number that differs from both of them. What we need is a reliable way (a> single rule) that will produce, for every list, a number not already on> the list.> Ah I was only attempting such a contruction in the case where the> diagonal produces a number on the list. I use the properties of such a> construction to create an element not on the list. Let me show you> what I have in mind, and you can perhaps let me know if I'm still> wrong.> Firstly we note, that any such construction must rely on duel> representation, because the diagonal number differs at every place.> Now, since any duel representation number must terminate in all 1s or> all 0s in binary after some number of digits say, n then for all n+m> for m > 0 are all 0s or 1s. If the list is 0s for positions greater> then n, then the diagonal is all 1s for positions greater then n. How> do we know this?> Thinking about what a duel representation means, and the nature of> what's we're calling the same number here, we clearly see that it> relies on our ability to show that a number of the form 1/(2^n) is> equivilent to the limit of the series created by the sumation of> 1/(2^(k)) from k = n+1 to infinite. Now if we take the number> terminating in all 1s, we notice that by changing any 1 to a 0, it's> equivilent to subtracting 1/(2^j) where j > k. Subtracting from both> sides of the equality and observing that 1/(2^n) does not equal> 1/(2^n) - 1/(2^j) makes it clear a position n exists. This isn't true> of reals in general, but the duel representation type of numbers it> does.A position n exists? Did you intend to say that a position n existshaving the property ____________, where ______________ is some propertythat you forgot to state? Sorry, but I can't follow what you are gettingat here. It's not sufficient to consider just two elements on the list. You need> to consider all of the elements on the list. The point is not that,> given any two numbers on the list, it is possible to construct a third> number that differs from both of them. What we need is a reliable way (a> single rule) that will produce, for every list, a number not already on> the list.> A position n exists? Did you intend to say that a position n exists> having the property ____________, where ______________ is some property> that you forgot to state? Sorry, but I can't follow what you are getting> at here.Ok n is defined as the position where the diagonal number begins torepeat either 0s or 1s indefinitely. I couldn't just assume such aposition existed, so I had to show why it did, in the case of duelrepresentation numbers. I used the definition of a duel representationto show this was true.> Why does it have to be so convoluted? We already have three perfectly> good ways to avoid the problem, each of which can be concisely stated:> 1. Chuck binary and use decimal, avoiding 0's and 9's.> 2. Group binary digits in pairs and treat them as base 4,> avoiding 0's and 3's.> 3. Add all the dual-representation numbers to the list> (twice each) before diagonalizing, which is possible> because there are only countably many of them.Well the first two suggestions do not deal with the binary casedirectly, I would assume anyone who generated such a list would like adirect counter example rather then simply producing the arguement theygave before. The formal proof of the claim is obviously fine, it's thesocial proof I'm after.I'm never going to convince a 12 year old that it's ok to add the duelrepresentaion number to the list before diagonalizing, they're goingto argue that it changes the cadinality of the list for that mattereither. If I'm presented with a list, and they say the diagonalnumber is already on the list then I can produce a number not on thelist using the property they used to generate the list directly. Thestarting point is exactly the same for both people, so it's mucheasier to convince them in case the regular base 10 case doesn't.Moreover, I find my explaination more obvious then these othersolutions, call it personal bias if nothing else. === Subject: Re: Antidiagonal, Infinity> Why does it have to be so convoluted? We already have three perfectly> good ways to avoid the problem, each of which can be concisely stated:> 1. Chuck binary and use decimal, avoiding 0's and 9's.> 2. Group binary digits in pairs and treat them as base 4,> avoiding 0's and 3's.> 3. Add all the dual-representation numbers to the list> (twice each) before diagonalizing, which is possible> because there are only countably many of them.> Well the first two suggestions do not deal with the binary case> directly, I would assume anyone who generated such a list would like a> direct counter example rather then simply producing the arguement they> gave before. The formal proof of the claim is obviously fine, it's the> social proof I'm after.There are no cases to consider here. The hypothesis says we are givena mapping f: N -> R. That means the elements of the list are realnumbers, not digit strings. We may choose to represent the numbers inthat list in any way we find convenient.There seems to be a widespread misconception that the diagonal argumentis not complete until we have explained how to apply it in everyconceivable base. That's tantamount to suggesting that the argument isnot complete until we have translated it into every known language,including Sanskrit and Klingon. Just one correct proof is quitesufficient. In fact, if your 12-year-olds have trouble grasping thispoint, I submit that it is more important that the diagonal proof itself,since it gets to the heart of the matter of what constitutes a validproof.> I'm never going to convince a 12 year old that it's ok to add the duel> representaion number to the list before diagonalizing, they're going> to argue that it changes the cadinality of the list for that matter> either. If I'm presented with a list, and they say the diagonal> number is already on the list then I can produce a number not on the> list using the property they used to generate the list directly. The> starting point is exactly the same for both people, so it's much> easier to convince them in case the regular base 10 case doesn't.> Moreover, I find my explaination more obvious then these other> solutions, call it personal bias if nothing else.Given a list of reals, you can rearrange it so that the original numbersoccupy all the odd-numbered positions and the dual-representation numbersoccupy all the even positions.If you really think your five-paragraph explanation is moreunderstandable to 12-year-olds than any of my three one-sentencedescriptions, then have at it.Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. let A :{a <= x <=b} and B:{c<=x<=d}, with a show that if f is uniformly continuous on each of A and B,> then it is uniformly continuous on the set D consisting of> both A and B.Take e > 0. Choose some d_1 such that if x,y belong to [a,b] and|x - y| < d_1, then |f(x) - f(y)| < e. Choose some d_2 such that if x,ybelong to [c,d] and |x - y| < d_2, then |f(x) - f(y)| < e. Now, taked no greater than d_1, d_2 or c - b. Then |x - y| < d => (x, y are in[a,b]) or (x, y are in [c,d]). In both cases, |f(x) - f(y)| < e.> i think......that> if A = [0,1] , B = [3,4], f is not continuous on AUB.If f is continuous in [01] and in [3,4], then it is continuous in theirreunion.Jose Carlos Santos === Subject: Re: Yes, these are homework questions!point of recognition - and when I had it at 'ac+a=b' the mere idea ofchanging it to 'a(c+1)=b' was a less-than-tiny speck in the distance from myfeeble-little-mind.I see that many responses were attempting to edge me in that direction - butwithout Robert J. Kolker's reply, I never would have got it - unfortunately,because it's so obvious now... can anyone recommend a book on algebra - forevery algebra problem I have resolved, there are 10 new ones to hate me.(and if I was any good at algebra, I may have been able to make an algebrajoke, instead of saying a boring ol' 10)And question two, I wasn't sure of what I was doing, and was making mistakeswhen testing my answer; so from your answers, I've tried again knowing I wasgoing about it the right way, and got it to work out - so I'm happy asLarry's King's long day. MRe === Subject: Re: Yes, these are homework questions!X-NFilter: 1.2.0>I was going to ask my teacher, but ran out of time... and as for working>these out myself, well there's only so may times I can shake my fist at>them, I'm lost.>Anyone care to help me out, please?>Express a in terms of b and c> a=(b-a)/c> -- Don't have a clue where to begin.Multiply both sides by the appropriate term(s) to remove alldenominator(s). Collect terms involving a the left side and terms notinvolving a on the right. Factor the left side to the form a*(...).Divide both sides by the parenthetical term to get a in terms of b andc.>Given a factor, find k> x^3-x^2+kx+n=0> -- This isn't the actual question - I'm just looking for a kick in the>right direction for this one - I've tried dividing by the factor and also>replacing x by the root, but don't know which (if any) is the right answer!If you have a factor (call it q), then you can divide the originalpolynomial by (x-q) exactly. Perform the division and select a kwhich eliminates the remainder.< === Subject: Re: Yes, these are homework questions!>I was going to ask my teacher, but ran out of time... and as for working>these out myself, well there's only so may times I can shake my fist at>them, I'm lost.>Anyone care to help me out, please?>Express a in terms of b and c> a=(b-a)/c> -- Don't have a clue where to begin.> Multiply both sides by the appropriate term(s) to remove all> denominator(s). Collect terms involving a the left side and terms not> involving a on the right. Factor the left side to the form a*(...).> Divide both sides by the parenthetical term to get a in terms of b and> c.>Given a factor, find k> x^3-x^2+kx+n=0> -- This isn't the actual question - I'm just looking for a kick inthe>right direction for this one - I've tried dividing by the factor and also>replacing x by the root, but don't know which (if any) is the rightanswer!> If you have a factor (call it q), then you can divide the original> polynomial by (x-q) exactly. Perform the division and select a k> which eliminates the remainder.If (s)he can't do the first question, I doubt very much (s)he will know howto do polynomial long division ... === Subject: Re: Yes, these are homework questions!Given a factor, find k> x^3-x^2+kx+n=0> -- This isn't the actual question - I'm just looking for a kick in> the>right direction for this one - I've tried dividing by the factor and also>replacing x by the root, but don't know which (if any) is the right> answer!> If you have a factor (call it q), then you can divide the original> polynomial by (x-q) exactly. Perform the division and select a k> which eliminates the remainder.> If (s)he can't do the first question, I doubt very much (s)he will know how> to do polynomial long division ...Polynomial division is just division in base x. === Subject: Re: Yes, these are homework questions!>Variable a represents a tablevalue.>First determine the weight of each variable.>b>c and b for b=(a*c)+a>cfor c=1 b has to be as big as twice to a:>(c*a)+a=b>or coming back to a=(b-a)/c:>(c*a)=b-a>In any table c will be the multiplier to tablevalue a where b will adept by>+a upgoing the table (from c=1 b=a+a, c=2 b=a+a+a etc.).Well, that certainly clears everything up....-- Lynn === Subject: Theory behind algebraWhere can I find books that rigorously proves all of the facts,theorems, and properties one is taught in High School algebra?Specifically, I am looking for proofs of the theorem of partialfraction decomposition, But I would like a book of the abovedescription anyway for other purposes. === Subject: Re: Theory behind algebra>Where can I find books that rigorously proves all of the facts,>theorems, and properties one is taught in High School algebra?>Specifically, I am looking for proofs of the theorem of partial>fraction decomposition, But I would like a book of the above>description anyway for other purposes.You want an old book on what used to be called college algebra. I have one such book on my shelf:Lehmann, CH. College Algebra. New York: John Wiley & Sons, 1962.This one explains the things you want in a clear manner. I don't know if you'll be able to find it. What you should do is go to a library (perhaps at a university) and look around for similar books.Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Theory behind algebra> Where can I find books that rigorously proves all of the facts,> theorems, and properties one is taught in High School algebra?> Specifically, I am looking for proofs of the theorem of partial> fraction decomposition, But I would like a book of the above> description anyway for other purposes.Algebra (2 volumes), by van der Waerden, satisfies both requirements.Jose Carlos Santos === Subject: limit~hi....teacher.....iflim f(x) = 00 ,x->00can we say that f(x) exists limit at x=00 ??in addition,----------------theorem)suppose f is continuous on [a,00).iflim f(x) exists ,x->00f is uniformly continuous on [a,00)----------------it's right theorem ??i am not sure to limit existence.if f(x) = x^2 or f(x) = x , is that theroem satisfy ??let me advice ~please. thank you. === Subject: Periodic functionA function like cos x^2 produces a cosine wave that compresses itself in both directions away from x=0. What function would produce regularly periodic compressions and rarefactions?Tim === Subject: Re: Periodic function>A function like cos x^2 produces a cosine wave that compresses itself in >both directions away from x=0. What function would produce regularly >periodic compressions and rarefactions?>Tim Try cos(2*pi*cos(kx)). Choose k to your liking. === Subject: Re: Geometric interpretation of the trace of a matrix? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i3KFPLE02351;Sorry,My last post was wrong in the sense that the perimeter is notmultiplied by the trace when you transform any figure. That certainlyhappens with the determinant.But still, there must be a geometric interpretation for the invarianceof the coeficients of the characteristic equation, and thesecoeficients are lines, areas, volumes and so on. Why is it so hard toidentify them?If you start with unitary eigenvectors of the transformation, as sides of a paralelepiped, the trace is certainly what I said in the lastpost, i.e., the sum of the edges of the image paralelepiped divided by2^{n-1}. === Subject: Re: Geometric interpretation of the trace of a matrix?>If you start with unitary eigenvectors of the transformation, as sides >of a paralelepiped, the trace is certainly what I said in the last>post, i.e., the sum of the edges of the image paralelepiped divided by>2^{n-1}.Consider a rotation transformation in R^2 (with two exceptions): no real eigenvectors at all, and any trace in the open interval from-2 to 2. How do you fit that example into some interpretation ofyour claim?Lee Rudolph === Subject: Re: Definition of one-component> According to the Wolfram website, scalar is A one-component quantity> which is invariant under rotations of the coordinate system.Sounds like a physics definition. Or a 19th century defintion.Or both.Nowadays, mathematicians would not include anything about rotationsin their definition of scalar.> What is meant by one-component? It sounds like a contradictory term.> A component is part of a whole. If there are no parts, then you have> no components. === Subject: Re: Source for Russell quote?> : A logical theory may be tested by its capacity for dealing with> : puzzles, and it is a wholesome plan, in thinking about logic, to stock> : the mind with as many puzzles as possible, since these serve much the> : same purpose as is served by experiments in physical science.> : > : Source:> : Bertrand Russell, On Denoting, 1905> : http://cscs.umich.edu/~crshalizi/Russell/denoting/> One stock of experiments being used in my subfield is at> http://www.tptp.orgThe TPTP Problem Library for Automated Theorem Proving? Oh hell,that's nothing. Check out these lists of sites:http://www.cs.kun.nl/~freek/digimath/index.htmlhttp:// www.cl.cam.ac.uk/users/jrh/atp/index.htmlhttp://www.ora.on.ca/ biblio/biblio-prover-a-b.html As you can see, there are literally hundreds of sites that talk aboutautomated (often computerized) theorem-proving. Systems to performpropositional calculus deductions have been around since probably the1960's, with an early big push being the development of Resolution. And even more.Hey Frege, I just got a good idea. Why don't you look through theseand see if you can find a formal deduction of the unsolvability of theHalting Problem (preferably predating my 3/9/00 paper)? You've got 28days left! Heck, make it a year - you've got lots of work to do. (Just make a progress report after 30 days.) But wait a minute, Idon't remember your agreeing to my challenge. Tell you what, agreewithin the next two days and it's on. No excuses, now. Do we have adeal? : )Charlie VolkstorfCambridge, MAPS If you ask nice and polite, I'll tell you about 3 that actuallyclaim to have automated the Halting Problem proof. One of them is theCheck it out - maybe it has more than the on-line version.) You'd bedoing me a great favor by helping me with my research. Tell me aboutevery one that says that they proved the Halting Problem, so I canAnybody know of any automated proof of the unsolvability of theHalting Problem? Please post it here! === Subject: Re: freedom and silence. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i3KFk2J05912;> It is better to keep your mouth closed and let people think you are a> fool than to open it and remove all doubt.>If you don't know, ask. You will be a fool for the moment, but a wise man>for the rest of your life. ----Senecabut you must know enough to ask an intellegent question. H.J.S.the word 'bored' should be removed from the English Language James Balleu the === Subject: Re: Ability to read> 1. If you produce a formal proof of the unsolvability of the Halting> Problem within 30 days (or thereafter), I will admit that someone else> has done it.Another contender is Hougaard's _Computability in HOL_ (1994):http://citeseer.ist.psu.edu/ hougaard94computability.htmlAccording to the abstract: This paper describes the implementation of a formal model for computability theory in the logical system HOL. [...] I will define the concepts of computable functions, recursive sets, and decidable or partially decidable predicates, and show how they relate to each other. [...] The report will then evolve in two directions: The first subject is the reduction of recursive sets, leading to the unsolvability of the halting problem. The other is two general results of computability theory: The s-m-n theorem and Kleene's version of the 2nd recursion theorem.Section 1 gives an informal overview of the syntax, axioms, andinference rules of HOL (another widely-used open-source proof checker).HOL's 8 inference rules (ASSUME, REFL, BETA_CONV, SUBST, ABS, INST_TYPE,DISCH, and MP) and the axioms for its core theories are described indetail in section 2 of By the standard definition an integral domain D is called a Dedekind> domain if it has the following properties:> (i) D is a Noetherian ring.> (ii) D is integrally closed, that is if K is the fraction field of D> and x in K satisfies an equation> x^n + a1x^(n-1) + ... + an where a1, ..., an are in D than x is in D.> (iii) Every prime ideal of D is maximal.> This properties are used to prove that every ideal of D can be> uniquely written as a product of maximal (prime) ideals.> But I don't see why condition (iii) is essential. Perhaps someone> could clarify that.I do not know which kind of proof you have in mind. At least if oneconsiders this to be a special case of the general primary decompositionin Noetherian rings, the reasons why (iii) is essential are clearlyvisible as follows.First recall the general primary decomposition. Let R be a (commutative,unitary) Noetherian ring. An ideal I != R of R is called *prime* if forall elements x, y in R the fact that x*y in I implies x in I or y in Y.Or, in other terms, I is prime iff the quotient ring R/I has nozerodivisors, i.e. is an integral domain. I is called *primary* if forall x, y in R the fact that x*y in I implies x in I or y in rad(I),where rad(I) := {y in R | y^n in I for some n} is the *radical* of I.Or, in other terms, I is primary iff in the quotient ring R/I anyzerodivisor is nilpotent. One has the following basic facts - Any prime ideal is primary - For any primary ideal Q, the radical P := rad(Q) is prime, called the *prime associated to Q*The *primary decomposition* says that any ideal I has a representation(1) I = Q_1 cap ... cap Q_sas an intersection of finitely many primary ideals which is irredundantin the sense that the rad(Q_i) are all distinct and no Q_i is containedin the intersection of the others. The *associated primes* P_i :=rad(Q_i) are then unique as are those Q_i with *minimal* P_i. These Q_iare called *isolated*, the other ones *embedded*, as are the associatedprimes. For a minimal prime P_i the corresponding primary Q_i is called the P_i-primary component.The proof of existence is deceptively easy: Call an ideal*irreducible* if it cannot be represented as a nontrivial intersectionof two ideals. It is then almost immediate that any ideal is theintersection of finitely many irreducible ones. The crucial fact then isthat an irreducible ideal is primary.One of the nice things about the primary decomposition is itsinteresting interplay with localization. Let S be a multiplicativelyclosed subset of R containing 1. Let R_S be the localization of R at S,i.e. the ring of fractions x/s with denominators in S. Every ideal Igenerates an ideal IR_S of R_S, called the *extension* I^e of I. Anyideal J of R_S generates an ideal J^c, the inverse image of J under thecanonical map R --> R_S and called the *contraction* of J. Given anideal I of R, the contraction of the extension of I, I^ec, is an idealof R containing I and called the *saturation* of I w.r.t S. It clearlyconsists of all those x in R such that there is an s in S with sx in I.In particular, if I meets S, some element of I becomes a unit in R_S andso I^ec = R, then. On the other hand, if I = Q is primary and Q does notmeet S, suppose sx in Q with s in S; then s^n in S for any n, so s^n notin Q for all n, whence x in Q, so that Q^ec = Q in this case. So thesaturation of a primary ideal w.r.t a multiplicatively closed setcontaining 1 reproduces the ideal.This is as much as can be said in the general case. In particular - in general, I is *not* a product of primary ideals - in general, a primary ideal is *not* a prime power - in general, a prime power need *not* be primary.These facts, however, become true in case R is a Dedekind domain D thusleading to Dedekind's result that any ideal is a product of primepowers, and it is, in fact, just the condition (iii) which plays acrucial role in esablishing these facts as follows.a) If I and J are two ideals in a ring R, we always have (I + J)(I cap J) subseteq IJ subseteq I cap JIf I + J = (1), in which case I and J are said to be *coprime*, we havefrom this I cap J = IJi.e. the intersection can be replaced by the product in this case. Byinduction, if I_1, ..., I_s are pairwise coprime, we have I_1 cap I_2 cap ... cap I_s = I_1I_2 ... I_sNow let D be a Dedekind domain, I an ideal of D. Because of (iii), thereare no embedded primaries, so (1) is the unique primary decomposition ofI into isolated primaries. In particular, the associated primes arepairwise distinct, so by (iii) again, are pairwise coprime. Now, for inot= j,rad(Q_i + Q_j) = rad(rad(Q_i) + rad(Q_j)) = rad(P_i + P_j) = (1),since the P_i are pairwise coprime, so Q_i + Q_j = (1), whence the Q_iare pairwise coprime. There follows:(A) In a Dedekind domain, any ideal is a unique up to order product of primary ideals with distinct associated primesb) Let Q be a primary ideal of D, P = rad(Q) its associated prime. Thenthe localization D_P w.r.t S = D P is a local Dedekind domain withmaximal ideal m = PD_P. In particular, D_P is integrally closed. SinceD_P is local, this implies D_P is a PID and so any nonzero ideal is apower m^n of m [PROOF: It suffices to show that m is principal. Now m != m^2, otherwise, by Nakayama, m = 0, and D_P would be a field. Let x be an element of m m^2. Then 1/x is not in D_P, since otherwise x would be a unit. Since D_P is integrally closed in its quotient field, 1/x cannot be integral and so multiplication by 1/x cannot map m to itself. Since, on the other hand, (1/x)m is contained in D_P, we have (1/x)m = D_P, whence m = D_P x, i.e. x generates m]. Therefore, the extension of Q in D_P satisfies Q^e = m^n = (P^n)^efor some n, whence Q = Q^ec =(P^n)^ecNow, if P is a prime in a general noetherian ring, the ideal (P^n)^ec is called the *n-th symbolic power of P* and denoted P^(n). It is the P-primary component of P^n, in particular, it is primary, and P^(n) contains P^n. Moreover, P^n = P^(n) iff P^n is primary. But, in our case R = D, there are no embedded primes because of (iii), whence P^n is primary, and Q = P^(n) = P^nand there results(B) In a Dedekind domain, any primary ideal is a prime power.(A) and (B) together imply that any ideal in D can be uniquely written as a product of prime ideals, and the given line of argumentation shows, I hope, that (and how) (iii) is essential in more than one way.Boudewijn === Subject: Logistic Curve Goodness of fitI am analysing some data and have a simple linear equation and alogistic/sigmoid equation. To measure the Goodness of Fit for thelinear equation I am using simple X:Y R^2 but am unsure of the bestway to compare this goodness of fit to that of the logistic equation.Is a direct comparison of the Sum of Squares accurate or is there abetter way.Any advice would be appreciated.