mm-396 Subject: Re: Galios Field: want to understand conjugacy class?> The eight element of GF(8) can be arranged into conjugacy classes with> respect to GF(2) in the following classes:> {0},> {a^0=1},> {a, a^2, a^4},> {a^3, a^6, a^5}> where a is the root of the primitive polynomial x^3+x+1;> I don't understand where is a^7, and also why {a^3, a^6, and a^5} are in a> conjugacy class?Well, a^7 is equal to a^0.Secondly, the fourth conjugacy class consist of elements that are inverse tothe third conjugacy class, i.e. a^(-1) = a^(7-1) = a^6, etc.> -Walala===Subject: Re: naive geometry questionsX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith said:>1) Are there surfaces that can be bent to each other (the way a flat>piece of paper is bent to a cylinder) but not to anything flat?What do you mean by surface and what do you mean by bent to eachother? Are you limiting your question to surfaces in Euclidean3-space?>2) Besides planes and spheres, is there any other surface S such>that a piece of S can be moved around adlibitum while each of its>points remains in contact with S?If I understand your question, the cylinder and the torus areexamples. Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to spamtrap@library.lspace.org===Subject: Re: Naive Q: Set theory, logic - which comes first? But what we say is>consistent with their being sets, functions in lamda calculus, or>something else. We need to exclude such interpretations,Why? Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to spamtrap@library.lspace.org===Subject: Re: The need of geodesics in physics <2202379a.0310310905.20a78472@posting.google.com <3fa63eb8$6$fuzhry+tra$mr2ice@news.patriot.net The equations of motion for r(t), v(t) are still the same> r'(t) = v(t), v'(t) = -kr(t)/|r(t)|^3No, they are replaced by Schrodinger's Equation, and variables in thepotential are replaced with the corresponding operators. A physicalstate may be a superposition of distinct eigenstates, which doesn'tcorrespond to any solution of the classical equations. Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to spamtrap@library.lspace.org===Subject: Re: The need of geodesics in physics We should make sure we're talking about the >same thing. It is *invariant* energy which can>only vary discretely and NOT continuously.No. Energy is quantized only in a bound system. In an unbound system,e.g., an atom that can be ionized, you need to deal with continuousspectra. Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to spamtrap@library.lspace.org===Subject: Re: question about book on calculus in 7d>Is there a book on calculus in 7 dimensions?I doubt it.>Can it answer why calculus works best in 7 dimensions?Who says it does? Are you talking about the existence of exotic spheres that are smooth and homeomorphic but not diffeomorphic to S^7? How does that become works best?Department of Mathematics http://www.math.ubc.ca/~israel ===Subject: Re: I can't stand it anymore> What gave you the idea that I was upset about the term Asian?> My only point was that IQ tests to determine people's intelligence are> bs.> Ah, of course. It would be far better indeed to administer> intelligence tests rather than IQ tests to determine people's> intelligence, right?> Really, you sound like someone who is disappointed by their> own results on such a test.I knew that someone would come up with that stupid opinion. It justhappens to be you. Sorry about that Don't you realize that us Asians don't need the tests invented bynon-Asians to find out whether we are intellgent or not?It was only at the turn of 2oth century that the Western World started to reliaze that Asians are not inferior to them as they havebeen told by the 19th century European racists who believed that theywere superior to all other humans.As someone was throwing away a bunch of old books, piling up outisdethe lab I was doing my research worl in grad school, I saw just such abook written by a European scientists. Anyway,before that time, did the opinion of the West affected the Asians? Ever read the book soverign Individual. The author stated thatwhite people hate Asians but that didn't affect .... He was justtrying to say how the Asians knew who they were and didn't act basedon others' opinion.We certainly don't need these IQ tests to find out about ourintelligent level.===Subject: Re: Galios Field: Conjugacy classes and minimal polynomials> Dear all,> suppose a is an element of order 65535 in GF(65536), how to find the> conjugates of a with respect to GF(2), GF(4), GF(16), and GF(256)...> Can anybody show me how to do that?> -WalalalSorry, could you clarify the question? The multiplicative group ofGF(65536) is abelian so all the conjugacy classes have just oneelement.===Subject: Re: A NOTy problem.>I thought there had to be a way to avoid backtracking ie a method of>doing the grid WITHOUT using an eraser. I imagined solving a Boolean>must arrive at the answer using reductio absurdum ie it can't be this>answer or that one so it must be whatever is left. I do a lot of these for fun. A grid helps to record the initial facts,but almost always I end up abandoning the grid halfway through andthen writing down and eliminating trial solutions. And you eliminatethem with what you call reductio ad absurdum (really just arriving ata contradiction from some starting hypothesis; therefore thehypothesis is false). - Randy===Subject: Re: Key core error argument, stepped outgood grief, Nora B. Baron;it's clearly a Leftwing Proof, andyou shouldn't draw conclusions about itwith your Rightwing methods -- sheesh! > So here we have these 7 steps about which much arguing has > occurred in the last few days, but there isn't any conclusion. > At least it is not stated here. What have we been arguing > about? What truth are we supposed to be acknowledging?> What is JSH now claiming to have proved ? When is a proof? http://www.maa.org/devlin/devlin_06_03.html > http://mathforprofit.blogspot.com--les ducs d'Enron!http://www.wlym.com/covers/7101contents.png===Subject: Re: I can't stand it anymore>I have been biting my tongue about the IQ test but I can't any more. >How reliable is a test that use the term Asian to represent the most>diverse of ethnic and cultural groups?> IQ tests are not particularly reliable, but what does this have to do> with it? What test is using that term? And how is any of this relevant> to these newsgroups?>If you want to know more about the diversity, ask me and I will give>you tons of specific examples.> ...> Your examples may be interesting, but what do they have to do with > sci.math?Sci.math was one of the group listed in the thread where I saw thediscussion of IQ test. I had no idea who were from which group in thatthread and so decided to posts in all 3 groups. The only science groupI posted before was sci.chem group.I don't browse sci.chem group regularly either. So when I broswe theposts once in a while and see some supposedy intelligent people fromthese 3 groups talking about IQ tests as a ver important tests (withsome seeing it as the ultimate test to determine human intelligence),I get very disappointed.===Subject: Re: The need of geodesics in physics permission for an emailed response.> No, they are replaced by Schrodinger's Equation, and variables in the> potential are replaced with the corresponding operators. A physical> state may be a superposition of distinct eigenstates, which doesn't> correspond to any solution of the classical equations.Almost right. A physical state *is* a superposition of distincteigenstates.Remember, there is no privileged basis. What is not a superpositionin one basis will be in another basis.And, for any physical state, there is some basis for which it is aneigenstate and not a superposition at all. But, that eigenstate mightnot correspond to any solution of the classical equations.===Subject: Re: Continuum> yours truly can promptly word it into a hypothesis :> there exist an _infinite_ (technically wrong term, i know) amount of> non-interequivalent axioms, each of them completing ZF; cardinality of this> _set_ of axioms is strictly greater then card(N) and strictly lower then> card(Z).> Not of much use without a proof, which took inventing another level of> abstraction, but who cares ?> yours,> amateurThe proof of this one should be quite interesting, considering that card(N)=card(Z)===Subject: Re: Math dependency logic REVISEDAin't it touchin' the lengths to which Logikoi will go to bailone another out? See Camaraderie of the Experts======================================================= ==============You concentrated on mathematics because it is predictable, becausethere is always a right answer you can check in the back of the book,because you like following very precise rules, because it allowsyou to escape from everyday life into a world that has nothing todo with everyday life, and because mathematics does not requirethe creativity that you completely lack.Keith Devlin, Commencement address to the mathematics graduating classof UC Berkeley, May 23, 1997A performance system is designed to work in a defined task domain,accepting particular goals and seeking to reach them by some kind ofhighly selective search. The system must be told what goal is to bereached and must be given a description of the structure andcharacteristics of the task domain in which it is to operate: itsproblem space. [...] In contrast, a learning system, is capable ofacquiring a problem space, in whole or part, by interacting with theexternal environment and without being instructed about it directly. A performance system is designed to work in a defined task domain,accepting particular goals and seeking to reach them by some kind ofhighly selective search. The system must be told what goal is to bereached and must be given a description of the structure andcharacteristics of the task domain in which it is to operate: itsproblem space. [...] In contrast, a learning system, is capable ofacquiring a problem space, in whole or part, by interacting with theexternal environment and without being instructed about it directly.Herbert >What about functions not defined at zero, such as f(x) = 1/x?>Just an addition: With ln(b)-ln(a) = ln(b/a) the number of parameters >reduces from two to one in this case, too.> That only considers one out of infinitely many cases of functions > with antiderivatives not defined at zero, so it hardley solves the > problem.I don't see any practical problem. Aren't there infinitely many admissible values of 1/x in close vicinity of x=0? The real world including application of physics in technology, medizine, management, etc. does perhaps not need the only forbidden wightless value at exactly x=0.On the other hand, the function 1/x is perhaps more important than the infinitely many functions worrying you.===Subject: Re: More symmetry between derivative and antiderivative?>While dx/dt at x has only one parameter x,>the integral from x1 to x2 has two parameters x1 and x2.>Iff x2 was always equal to zero, then derivative and antiderivative >would be more similar to each other.>Mathematicians might feel this idea somewhat cheeky.>However, it has a physical background.>Eckard> Congratulations. You just discovered the fundamental theorom of> calculus! I'd submit a paper if I were you.> Really - why all the debating over this? Flip open any calc book and> read through the appendicies.> And, regarding the integeral of 1/x, it's solved via numerical> analysis. Again, check the same appendicies.> I suppose I should be prepared for the onslaught of pets now...Der Ubergeck sounds German. I mention that because I didn't reveal any reasonable argument in your reply. So I would like to judge: armer Irrer. Maybe, I am wrong and you are able to justify your attack more in detail.===Subject: Re: Galios Field: Conjugacy classes and minimal polynomials> Dear all,> suppose a is an element of order 65535 in GF(65536), how to find the> conjugates of a with respect to GF(2), GF(4), GF(16), and GF(256)...> Can anybody show me how to do that?> -Walalal1) It is spelled 'Galois'2) This kind of basic questions are surely explained in your textbookand/or lecture notes. Ask your teacher for help. He/she is paid to answerquestions like this (and won't mind answering them). 3) Most likely you did an example in class in, e.g. GF(16) or GF(8).Use the same ideas!Good luck!Jyrki Lahtonen, Turku, Finland===Subject: Re: A NOTy problem.This post not CC'd by email>I do a lot of these for fun. A grid helps to record the initial facts,>but almost always I end up abandoning the grid halfway through and>then writing down and eliminating trial solutions. And you eliminate>them with what you call reductio ad absurdum (really just arriving at>a contradiction from some starting hypothesis; therefore the>hypothesis is false).> - RandyG'day G'day Randy, Is there is a better name for this process. ... arriving at a contradiction from some starting hypothesis;therefore concluding the hypothesis is false ... I wasn't implying the process was structurally the same as the oneFermat used which had a name that included a phrase like infinitedescent. Are there two different processes, reductio ad absurdum andreductio adsurdum; the former involving an infinite series and theother a finite one? Best wishes and thank you. Quentin Grady ^ ^ /New Zed, >#,#< [ / / ... and the blind dog was leading. http://homepages.paradise.net.nz/quentin===Subject: Re: Continuum> yours truly can promptly word it into a hypothesis :> there exist an _infinite_ (technically wrong term, i know) amount of> non-interequivalent axioms, each of them completing ZF; cardinality of this> _set_ of axioms is strictly greater then card(N) and strictly lower then> card(Z).> Not of much use without a proof, which took inventing another level of> abstraction, but who cares ?> yours,> amateur> The proof of this one should be quite interesting, considering that > card(N)=card(Z)If N and mean , as usual, the set of naturals and the set of integers, respectively, such a proof would indeed be interesting.Possibly something that Ross Findlayson or James Harris might assent to.===Subject: Re: A NOTy problem.This post not CC'd by email>Note that you know exactly how many cases there will be before>you start. Thus, you know whether it will be feasible to consider>them all. You will also know that you have not left any cases out.>When exactly three out of four lie, you know there are 4 cases.>When exactly two out of four lie, you know there are 6 cases.>When exactly one out of four lie, you know there are 4 cases.>When no one lies, you know there is exactly one case.>You can do similar calculations when there are 5 or 6 individuals>in the puzzle.G'day G'day William, I figured there were four cases since any one of the four could bethe one who told the truth ... until proved otherwise. general situations. Best wishes, Quentin Grady ^ ^ /New Zed, >#,#< [ / / ... and the blind dog was leading. http://homepages.paradise.net.nz/quentin===Subject: Re: A Question for James Harris>In seven years you have not convinced even ONE of them about ANYTHING.On the contrary, James has convinced me of several things about him. Justnot intentionally on his part, I'm sure.===Subject: Re: Hard tensor question (kst).> at 01:04 PM, dynamics@vianet.on.ca (Ken S. Tucker) said:>1) Suppose K= |g^uv|, where g^uv is the >contravariant metric tensor, and K is>it's determinant and is a relative tensor. >And then form a covariant tensor K_uv, >by K_uv = K*g_uv where g_uv is the>covariant metric,>That doesn't give you a covariant tensor.> Shmuel (Seymour J.) Metz, SysProg and JOATRight, set a determinant J = |&x'^a/&x^b| thenK' = J^2 K , K=|g^uv| .Sometimes the determinant J is called a Jacobian,and (according to my references) K is a *relativetensor* with weight = -2.To the immediate RHS above outer multiply with the metric g_uv to get, K*g_uv. As I understand, the indices forming the relativelydefined tensor K are meaningless after the definition,but I have some uncertainty.So now I should write out the transformation equationof the tensor K_uv = K*g_uv, as follows,K'_ab = J^2*(&x^u/&x'^a)*(&x^v/&x'^b)*K_uvand the determinant of this is,|K'_ab| = |J^2*(&x^u/&x'^a)*(&x^v/&x'^b)*K_uv|.This is where it gets complicated (for me), can i move the det bars | to reform the above to create,|K'_ab| = J^2* | &x^u/&x'^a|*|&x^v/&x'^b|*|K_uv|.|K'_ab| = |J^2|*|J^-2|*|K_uv|.where ( J^-1 = |&x^u/&x'^a| ).If this is so, then |K'_ab| = |K_uv| and these are invariants,if the reasoning is rational.You may ask, why am I doing this?I'm trying to find a short circuit to the obtainan orthogonal metric without employing theRiemann-Christoffel Curvature tensor B^a_bcd =0to prove orthogonality. The relative tensor K_uv is actually a 4th rank(counting weight) like the Riemann-Christoffel Curvature tensor. Any mathematician who has read this far will see some of the difficulties (and possible benefits)of exploring this research.It is very evident the Kronecker delta (d^u_v) serves as an orthogonal metric, it has a mixed form. How can the invariance of this mixedform be preserved if expressed as a 2nd rankcovariant metric?If this is possible, then a displacement in any orthogonal CS isdX^2 = K_uv dx^u dx^v andX^2 = K_uv x^u x^vbecause K_uv describes orthogonal CSs.IOW's one can easily see the compatiblity of the above two equations if the basis was the cartesian i,j,k unit vectors.At this point nonorthogonality would be fun.Lets introduce universal base vectors definedas,u_u = e_u + A_uwhere e*A=0 and form the metrics K_uv = e_u*e_v andg_uv = u_u * u_v = K_uv + A_u*B_vTo conclude g_uv is NOT a relative tensorhowever K_uv is, but when summed withnonorthogonal components A_u*B_v therelativity of K_uv vanishes.Anybody know why?Ken S. Tucker===Subject: Re: More symmetry between derivative and antiderivative?Just one more comment: Riemann, not Reimann----------------->Maybe, mathematicians believe they are independent.> Do you claim that physics is the only use for mathematical theories? > There are applications of mathematics outside of physics which are > not in any way dependent on the theories of physics. In that sense, > if no other, mathematics IS independent of physics.I am an electrical engineer, and was born in Berlin where much industry was based on application of mathematics via physics. I live in Magdeburg, a city where Siemens started his carreer after Ohm trained him in mathematics and where nowadays neuroscience plays an increasing meant application of mathematics.> Actuarial sciences and economics and other non-physics areas fund > more math than physics does. Perhaps mathematicians should dump > physics and only go where the big money is?History tells us that mathematics never dictated the development of practical needs but vice versa some branches of mathematics grew after new tools came up. Mathematics is not free to decide going somewhere. It has to do so on demand.>I will try and give the gist of an utterance by Oliver Heaviside: Mathematicians said, this series does not converge. On that >condition it will be of use. He hated unjustified rigor. >Ironically, his revenge was made as rigorous as possible.> The whole mathematical theory of asymptotic series deals usefully > and rigorously with series that diverge. So what is your point?The context. >>I am aware of some facts concerning Riemann type >> integration. Perhaps nobody defined integrals exclusively for open >> domains while the world excessively integrates from minus infinite to >> plus infinite. Is this correct?> Were you, as an old engineer or otherwise aware that Reimann > integrals are defined only on closed intervals? Yes, that's what I tried to express in my shaky English.However, isn't (-inf,inf) an open domain too?> To integrate over an open domain, including the set of all reals, > requires an extension of the Reimann definition though some sort of > limiting process. Why not?> I am not sure what you mean by excessively in this context.Almost like a gospel. In other words, like self suggesting confirmation that those who are reluctant to swallow negative frequencies and anticipatory symmetry are stupid ignorant heretics. > The realities of mathematics are quite independent of>>your view of physics.>I was told, every paying costumer is the king at least in America. >Doesn't physics pay for mathematics? > No! See above! Well, substitute physics by its children: information technology, genomics, etc.> Doesn't physics provide tailor-made >models of nature?> Yes, but it is not the exclusive provider of models of interest to > mathematicians. Very often it has been mathematics developed without > physical models that has been later adapted to the needs of physics. Of course.> Group theory springs to mind as one example.> Thus one can produce a strong argument that mathematics does not > need physics as much as physics needs mathematics.> You seem to want the tail (physics) to wag the dog (mathematics).Perhaps, you will agree that applications like physics apply mathematics rather than the other way round. So they are mighty customers.Concerning basic understanding of the world, I do not see mathematics the dog (reality).===Subject: Re: Who contributed most to mathematics?|I think France has made more contributions to mathematics than any|other nation.The French have made many important contributions. Remember,though, that the mathematical results you've heard of are a verysmall and not very representative sampling of all mathematicalresults. Only a small minority of the mathematics that existstoday was done before the 20th century (although the importanceof the old results is greater on average, since more of it wasfundamental or pioneering work). Nowadays a huge volumeof work is generated. So comparing one century to anotherwill always be a little odd.It's also a little odd to be comparing country-to-country. TheU.S. does well in many of these comparisons by being big.A chunk of the rest of the western world of similar size alsowould look good if taken together. I don't know what a percapita comparison would look like.Keith Ramsay===Subject: Re: JSH: It beginsjstevh@msn.com (James Harris) pushed briefly to the front of the queue(snip)^ The process has already begun.Thank God for that. Now do us non-professionals in here a favour andplease shut the fu**k up until the process produces unequivocaloutput.AndyHell! - don't worry about old raving Dave Ullrich ...Basically he's a sociopath who can't see a red ragwithout regarding it as a personal insult. Taylor, sci.math===Subject: Re: Equation Object converting?> Probably can do so with a quick macro.> To what version of Word are you converting?Please, give me more details how to do with a quick macro.===Subject: Re: Galios Field: Conjugacy classes and minimal polynomials> Sorry, could you clarify the question? The multiplicative group of> GF(65536) is abelian so all the conjugacy classes have just one> element.The OP undoubtedly meant field conjugates, i.e. images of a givenelement under field automorphisms fixing a chosen subfield elementwise.(e.g. the conjugates of the complex number 2+3i with respect to the realsare 2+3i and 2-3i).To do the exercise the OP needs to:A) describe all the automorphisms of GF(65536) (done in class)B) determine which of these keep all the elements of the chosen subfield fixed (the nature of the answer to the question A is such that this is quite straightforward given that the answer to question A for all the subfield is also done in class)C) apply all the remaining (after step B) automorphisms to the chosen element, whose conjugates we want to study.All of these are straightforward. Particularly as an example with GF(16)instead of GF(65536) was undoubtedly done in class (or was contained in thetextbook as an example).Cheers,Jyrki Lahtonen, Turku, Finland===Subject: Re: JSH: Attacking a proof is attacking yourselves> Now I've found out that even though I've stepped out my proof and> explained in detail there are posters who just keep trotting out the> same things despite getting refuted.> Maybe it's because _you_ keep trotting out the same thing despite being > shown the errors in PAINFUL DETAIL!> That's not true. > http://www.mentalhealth.com/dis/p20-ps02.htmlYou mean this one: Narcissistic Personality Disorder, which Mr. Harris suffers from by his own admission. http://www.mentalhealth.com/dis/p20-pe07.htmlI'll quote the Diagnostic Criteria from this web page; that is a precise description of Mr. Harris. What is going on in this newsgroup is probably the worst thing that could happen to him; everything is just designed to reinforce his illness. One could feel sorry for him, except that his behavior doesn't allow this.Diagnostic CriteriaA pervasive pattern of grandiosity (in fantasy or behavior), need for admiration, and lack of empathy, beginning by early adulthood and present in a variety of contexts, as indicated by five (or more) of the following: 1. has a grandiose sense of self-importance (e.g., exaggerates achievements and talents, expects to be recognized as superior without commensurate achievements) 2. is preoccupied with fantasies of unlimited success, power, brilliance, beauty, or ideal love 3. believes that he or she is special and unique and can only be understood by, or should associate with, other special or high-status people (or institutions) 4. requires excessive admiration 5. has a sense of entitlement, i.e., unreasonable expectations of especially favorable treatment or automatic compliance with his or her expectations 6. is interpersonally exploitative, i.e., takes advantage of others to achieve his or her own ends 7. lacks empathy: is unwilling to recognize or identify with the feelings and needs of others 8. is often envious of others or believes that others are envious of him or her 9. shows arrogant, haughty behaviors or attitudes===Subject: Re: JSH: It begins> Now that I have a proof that's simple enough to explain to kids, I can> move back to email.> Yup, you're leaving other mathematicians open to a major sandbagging> as I go back to a process that's netted me Andrew Granville, Barry> Mazur, and other notables over the years.> And you're leaving them open and unready.> I learned a while back that few professional mathematicians seem to> read sci.math, so it's not like what's been discussed here is likely> to get to them.> So I have a wide open field of mathematicians all over the world> contactable by email with proven techniques for getting their> attention.> The process has already begun.Did you ever try therapy?===Subject: Re: A NOTy problem. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA65nrQ05064;1. 's and Bob's statements are contradictory to each other.2. Hence, one of them tells true.3. Only one from four tells true. Hence, Dave and Chris lie.4. Chris lies, so Bob say true. So did it.The main conclusion is the 1-st.===Subject: Re: Stone's Representation Theorem(s?): help!Student...> Then, some properties get demonstrated, which use in parallel h and psi,and> precisely:> 1) For every A in B(X), h[X] / psi(A) = h[A]> 2) h is a continuous function> 3) For every A in B(X), the closure in U(B(X)) of h[A] is psi (A).> 4) in particular, the closure of h[X] in U(B(X)) is U(B(X)) itself, and> therefore h[X] is dense in U(B(X)).> So we have demonstrated that h is continuous, injective, and also that its> image is dense in U(B(X)).> Here ends the part of the theorem that I have....There is a little about it in Halmos Measure Theory, in the exercises forSection 40. I think it is also in Bourbaki Commutative Algebra. Stone'spapers are in Transactions of the American Mathematical Society 40(1936)37-111, and 41(1937) 375-481.===Subject: Re: More symmetry between derivative and antiderivative?Thank you for the intriguing details.Perhaps, you are not related to the famous P. Rubin (www.physics.orst.edu/~rubin/)The very old measures you mentioned are perhaps limited to the zero of time at now. I wonder how you can be sure about measures 5000 years ago. This age is about the same as ascribed to the sky disc recently found near Nebra, showing sun, moon and stars.>Antiderivative is the same like integral. Therefore one could expect >some similarity.>I blame Rn Descartes for the missing fix point of our coordinates.> Antiderivative is not the same as integral; it gives a> means of calculating integrals.> Integral is FAR older than derivative. In fact, integrals> with respect to discrete measures were evaluated> approximately 5000 years ago, and all measures (and most> integrals) are limits of these.===Subject: Re: JSH: It begins>Now that I have a proof that's simple enough to explain to kids, I can>move back to email.>Yup, you're leaving other mathematicians open to a major sandbagging>as I go back to a process that's netted me Andrew Granville, Barry>Mazur, and other notables over the years.>And you're leaving them open and unready.>I learned a while back that few professional mathematicians seem to>read sci.math, so it's not like what's been discussed here is likely>to get to them.>So I have a wide open field of mathematicians all over the world>contactable by email with proven techniques for getting their>attention.>The process has already begun.> Did you ever try therapy?Forget therapy. He needs strong psycho-active drugs.Gib===Subject: Re: Riemann's Zeta Function by H. M. Edwards>No Real Mathematician thinks of the logarithm having a basis:-)> Oops I meant base :-(No Real Mathematician thinks of the logarithm having a base:-)Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Partridge, _Bouncing Back_ (14 times)===Subject: Re: I NEED HELP BADLY (sorry, maths not psych)Expires: 28 days>Meanwhile, why don't you explain to me how this model applies to an>attractive electric field. What are the ping-pong balls and where are>they coming from?>This is called theoretical research Randy.>The aim is to try to match a hypothetical physical process with the prediction>of an equally hypothetical maths theory, namely relativity.It's just a theory? Are we a creationist now?>Of course the theory is just a theory. The point is, it>makes predictions about such things like the maximum speed>You have an alternate theory.>are accelerated by receiving momentum from little objects>moving at c.>Fine. If this is your explanation for the limiting speed, please>tell me what corresponds to the little ping-pong balls in the>case of an attractive force, and where they are coming from.dimension too, just like space and time.>You have now alluded to theoretical research without>a theory, differential equations without an equation,>and in general spun a lot of words while refusing to answer>the question.I can see this is far too difficult for you Randy.>Could it be that you know that your model is not a modelCan you or your favorite theory explain action-at-a-distance?>Or do you really think nobody noticed that you avoided answering any>questions about your model?>Paul Anderson understood perfectly. Is he sarter than you?.>If Paul Anderson has given you an answer about your model>which you can crib and offer as your answers, by all means>feel free to do so.>Do you really think nobody has noticed that you've>avoided answering any questions about your model?It is merely a straightforward and interesting mechanical problem, involving abasic differential equation. Something like D^2+KD=CSurely you have SOME scientific ability Randy.>Explain to me how this model applies to an attractive >electric field.>What are the ping-pong balls and where are they coming from?Don't worry Randy, It is a hypothetical.But OK, consider a small frictionless wheeled vehicle sitting on the desk infrom of you. You have a gun that fires perfectly elastic ping-pong balls atvelocity v, at the rate of one per nanosecond. You fire them at the vehicle.Plot the 'distance versus time' graph for the vehicle.Why do you automatically criticize something that you don't even understand? Doyou want to prove that supporters of relativity are the REAL crackpots?> - RandyHenri Wilson.See why relativity is wrong:http://www.users.bigpond.com/HeWn/index.htm===Subject: Re: I NEED HELP BADLY (sorry, maths not psych)Expires: 28 days>OK, so you have something which is neither a model for>a rocket NOR an EM field. Of what use is that supposed to>be, exactly?>I want to drive a boat across a river by hosing it with water from the bank.>I want to know how fast it will travel.>I want to know exactly how it reaches the maximum speed.>By the way the river water has no viscosity.>For this you need to specify three things:> - Water stream velocityThe jet velocity is Vo wrt the nozzle.> - Water flowm/sec> - Boat massM>Hopefully, the computation itself should be obvious by applying the>concepts of impulse and momentum.You have it in a nutshell Harry.Mdv/dt=d(mv)/dtand dm/dt is f(Vo-v) where v is the velocity of the boat.there are three separate situations. ! inelastic, 2) elastic, 3) where thewater from the jet is taken onboard and adds to the boat's mass.> Harry C.Henri Wilson.See why relativity is wrong:http://www.users.bigpond.com/HeWn/index.htm===Subject: Re: JSH: My victories get lost:) good one!===Subject: Re: someone told me ...> Well, if normality is unknown for pi, what are some> numbers that are proven normal? Every one of them would> satisfy Ben's requirement.Champernowne's number0.123456789101112131415161718192021222324....is normal in base 10.Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Partridge, _Bouncing Back_ (14 times)===Subject: Re: any good lecture notes on algebraic coding, finite field, minimal polynomial with application to error control coding?===>Subject: any good lecture notes on algebraic coding, finite field,> minimal polynomial with application to error control coding?>Dear all,>I am craming for a homeworkyou're craming for a homework... what does that even mean?adam===Subject: Re: It begins> Now that I have a proof that's simple enough to explain to kids, I can> move back to email.And proof of *what* is that? FLT?===Subject: Re: Key core error argument, stepped out <3c65f87.0311031501.63257981@posting.google.com EwTYfhf*u~,Eu,tf6 $HN*MY&)u0G=N' x<%)/s=GZ_BD2Qz9m=S%4v^I+>T|'1{w70ZY=ih,=)kMY_}?{%)x0)];K~@ J6m5.EN?>ZhXh;Y V|',x(js'Jfq02joVpj|#x linux)[...]> - William HughesAny reason you put quotation marks around your own name? Just tryingto beat James to the punch?However, you presuppose that certain numbers *are* prime ideals,... when in fact ... they are not... (Maybe I should look up 'primeideals' but the effort doesn't seem to be worth it. I assume someposter will get excited ... if I messed up.) --James Harris===Subject: Re: Cantor's diagonal number: why do we just ASSUME such a number exists?X-ID: Xc+-I0ZXwemf9lTK6ztNWp04we03fNE+0tFUcF6fJYPPn9k+Z2H96s> Your correspondent,> Nathan Deeth> Age 11> I'm interested in how you have remained at age 11 for the past 5> years? One merely needs to look up Nathan the Great (or Nathaniel> Deeth) and Age 11 on Google Groups to see that you have been> claiming this age for some time now. You were even accused of this> back then: http://mathforum.org/discuss/sci.math/a/m/219236/219315> I can only guess that you are speaking about the equivelent age of> your thinking and spelling.> Jonathan HoyleNo, but this just proves the incountability of N:You start at 11 and not even ever get to 12... Gottfried Helms===Subject: Re: Math dependency logic REVISED <2qtiqv4uegckp1j5ssg4ntcs2hsbr0goib@4ax.com EwTYfhf*u~,Eu,tf6 $HN*MY&)u0G=N' x<%)/s=GZ_BD2Qz9m=S%4v^I+>T|'1{w70ZY=ih,=)kMY_}?{%)x0)];K~@ J6m5.EN?>ZhXh;Y V|',x(js'Jfq02joVpj|#x linux)> Ain't it touchin' the lengths to which Logikoi will go to bail> one another out? See Camaraderie of the ExpertsI did go to that link. Here's how it begins.Lonely, are you?Anyway, the lengths the Logikoi will go to bail each other out of*what*? Was there something threatening in this thread? Oooh, werewe almost exposed as charlatans and frauds? Dear oh dear, what withmy mortgage[1] and all, we can't have that.There's no bailing out going on here. You said something stupid andirrelevant to the question at hand (namely, whether anyone disputesthat from true premises and correct inferences, true conclusionsfollow). I chose, all by my lonesome and with no concern forUllrich's well-being, to point out that you're being stupid. Hardlythe first time.Footnotes: [1] Not really, but James always talks about how mathematicians (notthat I am necessarily one) are worried about the effect his work willhave on their mortgage payments. I guess John's work is similarlythreatening. Once it's accepted, then one's mortgage payment may nolonger be self-identical.What I've learned is that [mathematicians are] the gatekeepers, andseem to have almost absolute power when it comes to mathematics. -- James Harris, on All I Really Ever Needed to Know I Learned in /Ghostbusters/.===Subject: Re: Your Attention Please, Sci.* NGX-ID: TDGWUBZroenDVcX4yc-YGeSoq3sxKoN535CSjF7cDSwBsH+m7eZUUl> Abhi:>This is very serious attempt indeed. My life is at stake.> I hope you have a will.>I request>you to discuss and understand this very seriously and then execute>this action device to bring before this world.>I just want to give justice to this invention.> Then burning at the stake is probably not out of the question.>I will begin my attempt in new post titled, Geometry And Newtonian>Mechanics of Action Device> I understand alt.test has a very large readership.No, this only means, that the OP just has forgotten to mention> Einstien, Hawkins, and Feynmann were all wellGottfried Helms===Subject: PM TO INAUGURATE 91st INDIAN SCIENCE CONGRESSX-Url: http://www.mantra.com/jai http://www.flex.com/~jaiX-Warning: Not for commercial use. Views expressed by others not necessarily poster's.PM to inaugurate 91st Indian Science Congress UNIKolkata, Nov 6 (UNI) - Prime Minister Atal BihariVajpayee will inaugurate the 91st Indian Science Congress(ISC) in Chandigarh on January 3. The Congress, to be hosted jointly by Punjab Universityand Institute of Microbial Technology (IMTECH), wouldfocus on Science & Society in the 21st Century: Quest forExcellence, said ISC president Prof. Ashish Datta. Datta said the Congress would comprise of plenary talks,panel discussions, public fora and a special programmefor school children - Science for School Children. also be organised in the Science expo. The aim of thisevent is to showcase the products and technology alongwith research work being carried out in various publicand private sector organisations. Concurrent to the Expo, Genesis, a symposiumrepresented by important personalities from industry andscientific fields from the country and abroad would alsobe held, said Datta. The congress will bring together over 5000 delegates fromhome and abroad, over 1000 participants from theGovernment organisations, PSUs, research institutions,corporates, and NGOs in India. http://www.hinduonnet.com/holnus/00206130012.htmJai Maharajhttp://www.mantra.com/jaiOm ShantiShubhanu Nama Samvatsare Dakshinaya Jeevan Ritau Tula Mase Shukl Pakshe Buddh Vasara YuktayamUttaraprostapad-Revati Nakshatr Harshan-Vajr Yog Balav-Kaulav Karan Dvadashi-Trayodashi Yam TithauHindu Holocaust Museumhttp://www.mantra.com/holocaustHindu life, principles, spirituality and philosophyhttp://www.hindu.orghttp://www.hindunet.orgThe truth about Islam and Muslimshttp://www.flex.com/~jai/satyamevajayate o Not for commercial use. Solely to be fairly used for theeducational purposes of research and open discussion. The contents ofthis post may not have been authored by, and do not necessarily representthe opinion of the poster. The contents are protected by copyright lawand the exemption for fair use of copyrighted works. o If you send private e-mail to me, it will likely not be read,considered or answered if it does not contain your full legal name,current e-mail and postal addresses, and live-voice telephone number. o Posted for information and discussion. Views expressed by othersare not necessarily those of the poster.===Subject: Re: Probability of a Run>As far as I can see, if q is the probabilty of a loss and P(n,m) is>the probability of at least 1 run of at least length n in m trials>then>P(n,m) = 0 if m < n,> = q^n.(1 + (1-q).( (m-n) - sum(i=n..m-n-1, P(n,i) ) ).>Having read the material in this thread, I assume I've made a simple>mistake or that a similar formula is given in Feller but not>considered a practical method of calculation in 1968.I'm afraid I can't evaluate your formula without a detailedexplanation of the logic behind it, but the change in the computingenvironment is precisely the reason I am seeking a new solution atthis time. Nobody uses Weddle's rule (for numerical integration) anymore, but it was a Godsend in the days of pencil and paper arithmetic.A train of logic used by both Burnside and Uspensky, and which mighthave originated with DeMoive, leads to the following differenceequationu(m + 1) = u(m) + (1 - u(m - n)) (1 - p) p^nwhereu(m) = Pr that a run of n or more has happened by the m'th trial.The initial values areu(n) = p^n and u(m - n) = 0 if m < 2nBurnside and Uspensky develop solutions to the difference equation,find them unwieldy, and go on to derive approximations. Today, anykid with a TI-83 or a home computer can calculate particular solutionsnumerically. Here is the logic:--------What is the probability that an event shall occur at least n times ina row in m trials?u(m) = Pr that a run of n or more has happened by the m'th trialv(m + 1) = Pr that a run of n or more has been completed for the firsttime at the (m + 1)th trialthenu(m + 1) = u(m) + v(m + 1)In order that the (m + 1)th trial may complete the first run thefollowing conditions must be satisfied:1. There must not have been a run of n or more up to and including the(m - n)th trial. Pr = 1 - u(m - n)2. It must not happen at the (m - n + 1)th trial. Pr = 1 - p3. It must happen at each trial from the (m - n + 2)th to the (m +1)th. Pr = p^nso thatu(m + 1) = u(m) + (1 - u(m - n)) (1 - p) p^n===Subject: Re: {Group theory} Number of units?>I have asked a question like this before and the answer was that no>set could be big enough.>Here is a slightly different question.>Is it possible for a set to exist which contains all the units of the>various isomorphically distinct groups that exist?>In other words, a set S such that:> --> for any group g with unit g_e, then either g_e is in S, or>there> exists some group h, isomorphic to g and with unit h_e,>such> that h_e is in S> --> if g,h are isomorphic but unequal groups with units g_e, h_e,> then either g_e is in S or h_e are in S, but not both>(unless> they're equal)This condition seems self-contradictory. Suppose that g,h,k are isomorphicgroups with distinct identities g_e,h_e,k_e. Then exactly one element ofeach of the sets {g_e,h_e}, {g_e,k_e} and {h_e,k_e} must be in S, whichis impossible.> --> if x is in S then x is the unit of some group g and for any>group> h that is isomorphic to g, with unit h_e, either h_e = x or> h_e is not in SIf you omit your second condition, then the answer is yes - just takeany set S with |S| = 1.But if, as I suspect, you want S to contain distinct identity elements fornon-isomorphic groups then, as another poster pointed out, the answer is no,because there is at least one group of any given cardinality.Derek Holt.===Subject: Re: I NEED HELP BADLY (sorry, maths not psych)>> accelerated beyond the operating speed of the accelerating fields, ie at 'c'.>Explain that phrase, please.>As you know, the accelerating electic field in an accelerator>So what do you mean by:the operating speed of the accelerationg field?> You don't know what happens in close proximity to the accelerating charge. Back> radiation?When you can't answer, ask a question, eh?Yes, we know very well what happens to the accelerating charge.No mysteries here.You didn't answer.I suppose your words are supposed to mean something.My simple question is:Can you please explain what:the operating speed of the acceleration fieldmeans?Or are you babbling words void of meaning?Paul===Subject: Re: Who contributed most to mathematics?> Any historian - not just a good one - would know that Switzerland > produced Euler and the whole Bernoulli family. And where are the Czechs?And the es?===Subject: Re: NOVA strings and branes> For the record, I do not think that the intelligence agencies of the the > major powers, US in particular, have the ET UFO anti-gravity technology > on the shelf and are hiding it. In other words I disagree with Nick > Cook's book Hunt for the Zero Point and with Stephen Greer's thesis in > the Disclosure Project. Why do you think that such a thing would NOT be kept secret? And ifit were secret, how would you find out about it? Come on now, thinkabout this. Suppose, an alien vehicle crashes in New Mexico somewhere.The army goes and grabs the wreckage. Now you as a government personin authority have to decide what to do.I think the response is TOTALLY obvious to any student of government.It has absolutely NOTHING to do with the usual arguements of publicpanic and the like and EVERYTHING to do with advantage such technology would give one country over the others. It's a power trip at LEASTon the order of the Atomic bomb. Was the atomic bomb kept secret(at least until they blew a few up)? You bet. Well, except from the countries with spies in the project. The same goes for UFOsand more!And it's even BIGGER than that. Let's consider zero point energy.If there is indeed zero point energy that can be easily tapped,then this spells doom for oil/gas/coal and all other fuel industries.It's a MAJOR loss and upheaval economically speaking. If nothingelse officials would deem it prudent to get their ducks in a rowBEFORE releasing any information on how to tap this energy. It wouldbe totally irresponsible to not keep the lid on such information and just let nature take it's course crashing the economy and causing pandemonium! Hence we can readily conclude that your position that the governmentisn't and wouldn't keep such things under wraps truly makes nosense. One might argue that crashed technology was never retreivedbut how could one know? You can't. It's secret! Therefore anydenials have to be considered as possibly bogus.Either the technology is real and being kept secret. In which case you'll never prove it, because even if you had hard proof,you'd just disappear if you tried to show it. Or there is nothingthere so there is nothing to prove. > It is quite obvious that no human physicist > today understands how such stuff would work. We are only now beginning > to get to that level of understanding of how metric engineering would > work by manipulating dark energy. It may be possible that USG et-al > has retrieved damaged craft and alien creatures from the Universe Next > Door perhaps only a millimeter away across the extra bosonic space > dimensions like Robert Bigelow's NIDS people reported on his Utah > Ranch, or that may have happened at Roswell in 1947 etc. Even if that > were true, and I am not saying it is true, the fact is that none of the > intelligence people who might have that stuff have the slightest > understanding how that hypothetical and/or alleged alien technology > actually works. It's The Sorceror's Apprentice.Think so? Back-engineering is quite a separate thing from a trueunderstanding of physics. And I see no way you can be sure thatno human physicist might understand some of this stuff. Youhave no idea who or what is involved with black projects.You are making arguments here that seem to make sense on the surface, but the problem is they don't make any sense at thepolitical level. And it is at the political level where governmentsoperate. I'm sure you know that.===Subject: Re: I NEED HELP BADLY (sorry, maths not psych)> Let's be serious for once.> In the elastic case, the water (or stream of ping pong balls) ends up moving> backwards at Vo-2v, so the change in momentum per second is 2m(Vo-v)Note that this is based on the assumption that the mass ofa ping-pong ball is very much smaller than M.This is of course an approximation.>>v = Vo*(1 - exp(-2*m*t/M))> Incidentally, particularly in the case of the elastic 'ping-pong ball drive',> momentum is bced but does the (kinetic) energy equation match?Not quite, due to the approximation mentioned above.Paul===Subject: Re: Intersection of concentric circles - divide by zero??> Please: Clean up your formula by introducing some temporary variables to> replace common sub expressions. It is simply too hard to read a formula that> spans more than one line. For instance, the sub expressions ((R1 + R2) ^> 2) - (d ^ 2), (d ^ 2) - ((R2 - R1) ^ 2)), and (R1 ^ 2) - (R2 ^ 2) appear> frequently. Consider replacing them everywhere with variables, say, A, B,> and C. This will simplify your equations considerably.> Incidentally, what is 'd' in the above equation?While breaking up your expressions you might keep numerators and denominators separate. Then you can test the value of denominators before dividing by them, and so detect errors. Similarly you could check the sqrt arguments. What happens to your method when the circles do not intersect?Gib===Subject: Re: White Noise Dilemma boundary=------------040605070702000205090804----------------- ---------------------------------------------------->Here is a question that has had me in a quandary for several>years now.>Let X(t) be a zero-mean IID process with variance of >sigma^2. Now we know that since this function is>IID, it has a white PSD and therefore its autocorrelation>function at lag 0, R_XX(0), should be b*delta(tau), where>delta(tau) is the usual Dirac delta function and b is some>constant.>However, we also know that the autocorrelation function R_XX(tau)>is defined to be E[X(t)*X(t+tau)], and therefore that R_XX(0)>is E[X^2] (where we have dropped the dependence on t since the>process is IID). Therefore in this sense R_XX(0) = E[X^2] = sigma^2,>which is not infinite.>How do you resolve this discrepancy?As I understand it, a strictly white process must have a uniform power spectrum density, say 1 Watt per MHz.But this will apply for every 1 MHz interval of spectrum....So in total : 1+1+1+1 + ..... = infinity (?!) Watts ; the details:MHz Watts-----------------------------------------------------0->1 11->2 12->3 1.....600->601 1...1000000->1000001 1...and so onto infinity.....At the Web page : http://cnx.rice.edu/content/m11105/latest/_BEGIN_QUOTE_ > ``so the White Noise Process is unrealizable in practice, because of its infinite bandwidth. However, it is very useful as a conceptual entity and as an approximation to 'nearly white' processes which have finite bandwidth, but which are 'white' over all frequencies of practical interest. For 'nearly white' processes, r X X ? r X X ? is a narrow pulse of non-zero width, and SX? SX ? is flat from zero up to some relatively high cutoff frequency and then decays to zero above that. <<<<<<<<_END_QUOTEWhat I'm wondering about is if some mathematical sense can be made of thisinfinite total power of the white process X.... , or in other words:Is there a rigorous mathematical definition , treatment , idea or theoryfor the intuitive idea of a White Noise Process?I don't know, but I'm curious.David Bernier ===Subject: Re: I NEED HELP BADLY (sorry, maths not psych)>> In the elastic case, the water (or stream of ping pong balls) ends up moving>> backwards at Vo-2v, so the change in momentum per second is 2m(Vo-v)>v = Vo*(1 - exp(-2*m*t/M))Note:According to Newton the momentump = m*Vo*(1 - exp(-2*m*t/M)) decreases with time.Newton predicts that there is a speed limit because the ping pongball will cease to transfer momentum to the mass when its speed approaches Vo.[..]>There is no reason for making this more complicated than it is, Henry.>The mass M will in all cases end up moving with the speed Vo.>Or rather, v will approach Vo asymptotically.>But you knew this, didn't you?>And this have absolutely nothing to do with what happens>Which you also knew, didn't you?> Not at all Paul.> I have to compare the curve given by the above solution with that predicted by> SR for a charge accelerated in a constant field.> Can you tell me what that might be?Sure I can.But first, I will tell you what Newtonian mechanics predicts.Let the mass m with the charge q be in a static electric field Eo.Let v be the normalized speed, that is the speed is v*c.Let the mass accelerate from standstill.What is v(t)?Solution according to Newton:----------------------The force on the the mass is constant Fo = q*EoNewton predicts that the momentum p = Fo*tincreases linearly with time.Fo*t = m*v*cv = t/T where T = m*c/q*EoNewton predicts that the speed will increase lineray with time,and will pass the speed of light at the time T = m*c/q*Eo.There is no speed limit because the electric field neverceases to transfer momentum to the mass.Solution according to SR:--------------------The force on the the mass is constant Fo = q*EoSR predicts that the momentum p = Fo*tincreases linearly with time.Fo*t = m*v*c/sqrt(1 - v^2)v = (t/T)/sqrt(1+(t/T)^2) where T = m*c/q*EoSR predicts that the speed will approach the speed oflight asymptotically. v = 2^-0.5 = 0.707 at the time T = m*c/q*Eo.c is the speed limit despite the fact that the electric field neverceases to transfer momentum to the mass.Of course two curves which both are starting at zero andare approaching a limit asymptotically will show a superficialresemblance.But it makes no sense to compare the Newtonian predictionfor a scenario to the SR prediction for a completely differentscenario, and conclude that because the predictions showa slight resemblance there must be a causal connectionbetween the predictions.What makes very much sense, though, is to compareWhich I did above.They are very different.And we both know which of them are in accorce withexperimental evidence.Don't we?Paul===Subject: Re: Stone's Representation Theorem(s?): help!>We start from X, topological space, and build the Boole algebra of its>clopens, B(X). Then we build the set of the ultrafilters of this Boole>algebra, U(B(X)). Then we define a function h: X -> U(B(X)) defined like>this:>for every x belonging to X, h(x) is {A in B(X) | x is in A}.>It's shown quite easily that this subset of B(X) is indeed an ultrafilter,>and therefore is indeed an element of U(B(X)). We structure U(B(X)) as a>topological space using the topology induced from the old psi-function,>which makes U(B(X)) compact and totally disconnected (it was proved in the>First Theorem).>We then suppose that X is totally disconnected (this hypothesis is always>taken as granted , from now on), and from that we deduce that h is>injective.>Then, some properties get demonstrated, which use in parallel h and psi, and>precisely:>1) For every A in B(X), h[X] / psi(A) = h[A]>2) h is a continuous function>3) For every A in B(X), the closure in U(B(X)) of h[A] is psi (A).>4) in particular, the closure of h[X] in U(B(X)) is U(B(X)) itself, and>therefore h[X] is dense in U(B(X)).>So we have demonstrated that h is continuous, injective, and also that its>image is dense in U(B(X)).>Here ends the part of the theorem that I have.>I have the feeling that we're very near to demonstrate that h is also>surjective (maybe, it could be enough to demonstrate that h[X] is closed in>U(B(X)), therefore its closure, U(B(X)), is h[X] itself, but I don't know if>it's possible to prove it), and that it has an inverse function that is>continuous (maybe it could be shown that h is open, or that it's closed, I>don't know...), to prove that h is a homeomorphism and finish the theorem.>Can anybody either help me to fill the blanks or give me an advice of the>right text to look at in my university's library (we have tons of books and>publications in english, here) to complete the proof? A big big thank you to>anyone who'll want to help me.If you only asume that X is totally disconnected that this is as far as you can carry the theorem.You can ask for two more properties of h:1. it is an embedding, i.e., X is homeomorphic to h[X]. For this it is necessary and sufficient to assumethat X is zero-dimensional (the clopen sets form a base for the topology). Your formula 1) will be instrumentalin proving this.2. it is a homeomorphism between X and U(B(X)) for this it is necessary and sufficient to assume that X iscompact Hausdorff and zero-dimensional. Compactness is used to find a point common to all membersof a given ultrafilterKPPS See http://aw.twi.tudelft.nl/~hart/teaching.html for some lecture note of mine on the subject.E-MAIL: K.P.Hart@EWI.TUDelft.NL PAPER: Faculty EWIPHONE: +31-15-2784572 TU DelftFAX: +31-15-2786178 Postbus 5031URL: http://aw.twi.tudelft.nl/~hart 2600 GA Delft the Netherlands===Subject: Re: I NEED HELP BADLY (sorry, maths not psych)>> Let's be serious for once.> Consider an object being accelerated by a idealistic jet of water or a>> continuous 'stream of elastic ping pong balls'. What is its subsequent velocity>> pattern?> accelerated beyond the operating speed of the accelerating fields, ie at 'c'. I>> hope it might also produce a relationship that is equivalent to mass>> 'appearing' to increase with velocity by gamma.)>>Please define what you mean by the operating speed of an>accelerating field. I've been in physics research for over 20>years, and I've never heard that term.>Do you mean phase velocity, group velocity, how fast the>operators turn the knobs, what?I think you forgot to answer the question, Henry.Why is that?Paul===Subject: Re: cardinality of monotone set-functions>Federico grava .88 la saucisse et au marteau:>> I suspect the solution to the following is known:>> Let X be a finite set. A set-function on X is a>> function f: 2^X rightarrow 2^X. f is monotone>> if A subset B implies f(A) subset f(B).>> What is the cardinality of the set of monotone>> set-functions on X ?> If f is monotone, then for each x in X the set> G(x) = {A in 2^X: x in f(A)} is monotone, i.e. if A is in G(x) and> A subset B then B is in G(x).Another small point: I thought a collection U of subsets of X such thatA in U and A subset B ==> B in U was called an upset or filterAre they also called monotone sets? Of course they are the supportsof monotone functions f: 2^X ->{0,1}. --Edwin> Conversely, if for each x in X the set G(x) is monotone, then> f is monotone, i.e. if A subset B then B is in G(x) for all x in f(A),> and so f(A) subset f(B).> And the family {G(x): x in X} determines> f, since f(A) = {x: A is in G(x)}. So the cardinality of the set of> monotone set-functions is |X|^N(|X|) where N(|X|) is the cardinality of> Oops! That should be N(|X|)^|X|.> the set of monotone subsets G of 2^X, or equivalently monotone Boolean> functions. That is sequence A000372 from the On-Line Encyclopedia> of Integer Sequences,> (look up> sequence number A000372). There is no known formula for it. See themany> references there.>> (Actually, I'm looking for the cardinality of>> the set functions such that f(A)subset A for all A.)> That's a different question. Are you also requiring f to be> monotone?>I'd say that you've got C(a,x) sets or card x in A. For each of those>sets, your function can take 2^x values. Therefore, you must have>Sum(x=0,a) 2^x C(a,x) = 3^a.>Am I right ?> Yes, for the second question (if f is not required to be monotone).> Oops again! You want> Product_{A in 2^X} 2^|A| = 2^(Sum_{A in 2^X} |A|) = 2^(|X| 2^(|X|-1))> Department of Mathematics http://www.math.ubc.ca/~israel> University of British Columbia> Vancouver, BC, Canada V6T 1Z2===Subject: Association of authorsI'd like to know whether there exists an association devoted to theprotection of the copyrights of scientific papers written by freelancescientists.Thank you very muchMichel===Subject: What is meant by Subrepresentation? Hi there! In a book I encountered the term subrepresentation, whichwas meant in the way of restricting the representation on a subspace. If Ais a C*-algebra and r a representation of A on a Hilbert space H, and H' aclosed subspace of H, what is then r restricted to H' ? I could imaginethe following: If P is the orthogonal projector onto H', then thesubrepresentation r' is representing A as operators on H', given by :r'(a) := P r(a) P(Is this a representation at all?) Benjamin===Subject: Re: Uncle Al is Sadistic .>> 3 Africans I met in Computer science department in the last 4 years>> were way above average in thier programming skills in the midst of>> Chinese and Indian grad students who are the overwhelming majority in>> that department. One was so good that he even got a parttime teachign>> position to teach an undergrad programming course, while he's still a>> student himself.>You will always find bright smart people in just about any naturally >occuring group of humans. Race is nothing. Culture is everything.>She said Africa. I would have asked what country? Ethiopia>used to send their brightest to the States for grad study. It>would be interesting which countries are starting to get their>act together.>/BAH>Subtract a hundred and four for e-mail.>None of them Ethiopians.Answer the ing question. Which countries?/BAHSubtract a hundred and four for e-mail.===Subject: Re: Naive Q: Set theory, logic - which comes first?>But what we say is>consistent with their being sets, functions in lamda calculus, or>something else. We need to exclude such interpretations,> Why?For some purposes it is neither necessary nor desirable to excludesuch interpretations, but if we do not want natural numbers to getidentified with what is really a representation of them, then we needexclude such interpretations.Mattias===Subject: Re: Naive Q: Set theory, logic - which comes first?>>Two questions whose answers I am seeking are:>>1) What exactly might we mean by representation?>>2) How do we define a mathematical object if we do not want to>>identify it with some representation of it?>> This is done quite often. The Peano Postulates CHARACTERIZE>> the positive (or non-negative) integers. There are many other>> ways to characterize them. >So we say that 0 is a natural number, that the successor operation>takes us from a natural number to another natural number, and then>list properties that natural numbers satisfy. But what we say is>consistent with their being sets, functions in lamda calculus, or>something else. We need to exclude such interpretations, but how?> Why? We only characterize what it means for a particular > object to be considered as a model for the integers. It> is a mistake to attempt to say what an integer is.By protesting when I said that a natural number is a sequence ofsymbols in which only symbol the symbol '|' occurs, you said somethingabout what a natural number is.A mathematical object is uniquely determined if we know what counts asa model of it, so instead of working with a mathematical object wecould work with the property of being a model of it, but I think thatwould be inellegant. Analogy: in geometry, instead of talking aboutpoints, we could talk about the property a line may have of passingthrough that point, but this is inellegant.Mattias===Subject: [Probability] Convergence in probability metrizable?I am given a probability space (W,F,P) and a seperable metric space E.I consider the set M(E) of all maps f:W->E which are Borel measurable.I consider on M(E) the topology generated by the base {B(f,a,e): f inM(E), a,e>0} where:B(f,a,e)={g in M(E): P(d(f,g)>=a)f with respect to thistopology is equivalent to the convergence in probability, i.e.:For all a>0, P(d(fn,f)>=a)->0 as n->+ooI was trying to find a metric for this topology and am starting tosuspect it may not be metrizable. Any hint is appreciated.Noel.===Subject: Re: Sets before logicX-ID: ZK0iLqZdQeLu7ud4OBNPlEQhbPUpVwoATxizwxqwxOvTx+zhH1krw-> At a bare minimum you owe us a simple small axiom set that> is consistent with the existence of a universal set [...]You might take this one (due to W. V. O. Quine): Ax. 1 EyAx(x e y <-> phi(x))* // Comprehension Ax. 2 AxAy(Az(z e x <-> z e y) -> x = y) // Extensionality * If phi(x) is /stratified/, and y does not occur free in phi(x).Note that = is not a logical primitive in this system, but x = y is an abbreviation for the formula Az(x e z <-> y e z).F.P.S.For the notion of stratification see M. Randall Holmes, Identity ofindiscernibles in Quine's New Foundations and related theories.(http://math.boisestate.edu/~holmes/holmes/define_ equality.ps)===Subject: Re: Usenet Posting Guide?>I prefer 'tin' but remember 'rn' (and 'readnews') fondly and could use it>if necessary. Graphical interfaces do have their uses; for instance, the>X Window System lets me open multiple xterm windows to run command-line>clients like tin (or rn).> What benefit do you get by having two newsreader windows open at once?He did not refer to several newsreaders at once but rather toseveral command-line clients at once===Subject: Re: How to define a function to be smooth?>Hey all>When we say a function f(t) is smooth, does this mean that>f has infinite differentials with respect to t?> That's _often_ what it means - sometimes it means less than that.I thought a function was smooth at a point a if there was aneighborhood of a in which f could be represented as a Taylor seriesabout a. So, I had a wrong definition.Artur===Subject: Re: Key core error argument, stepped out>>7. But to divide 7 from those constant terms requires dividing>through two of the factors, so>(5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22>from reverse use of the distributive property, which gives a constant>term coprime to 7, as required.>>>Let's take a simpler example and see how your argument works.>>Let>> Q(x) = 7(25x^2 + 30x + 2) [1]>> = 7(x^2 + x)(5^2) + 7(x - 1)(5) + 7^2>> That's [...] > [...] the truth.> Even considering the source, that response is worth keeping. I'm archiving> this puppy. Sounds to me like James has no intention of responding to> my post, which is a pity since he might have learned something. I guess> I underestimated his obdurate attachment to this particular argument.Readers notice how Rick Decker is playing you for fools as I have a*stepped* out math proof.Now then, if you've gone through the trying and difficult process offinding a spectacular proof, and then face people who decide to ignoreyour hard work and effort, to trot out their own pet examples, to tryand distract people from your proof, how would you take it?How would Rick Decker take it with his own work? > I'm left wondering how particularly abhorrent type compares to stupid> f**k. Not that I'm all that competitive, but I'd like to know where> I stand compared to Dave Ullrich on the Harris Abuse of Critics scale.> I figure I'm way below David and Arturo, somewhat below Dik and Nora--> perhaps in the neighborhood of Will. No matter, I'm evidently in good> company and proud to be a card-carrying member of the Evil Math Cabal.Notice how Rick Decker played you for an emotional reaction, while thereal issue here, the reason I'm pissed, is that I have a PROOF andwhen you have a proof, mathematicians aren't supposed to try and usetactics to convince others that what is true is not.> Rick> p.s. Cabal members, don't forget our next meeting. Be sure to bring> your black robes: the rites look so much more dignified when we're> all properly clad.Yet if Rick Decker's work were facing the same treatment, would he beso cavalier?What if he knew he had a paper that could win him the Field's Medal,do you think he'd be so casual about others trying to find ways aroundproper process to discredit him?The reality is I have a stepped out proof that posters like RickDecker can't refute because it's a proof, so they try differenttactics to make you believe something false, while they also work toconvince you that they're decent people.But a Rick Decker is math slime, willing to attack the basis of thediscipline: mathematical proof.Do you, if you are a mathematician, really want to move to a newsystem, a Rick Decker system?James Harris===Subject: Re: Math dependency logic REVISED>Ain't it touchin' the lengths to which Logikoi will go to bail>one another out? Giggle. What's fascinating is the consistency with which you replywith non-sequiturs when it becomes apparent even to you thatyou've been making an idiot of yourself.As here. Let's give this statement a label:[*] If one starts from something true and applies correct reasoning then one's conclusions must be true.James says people are denying [*]. I point out this is ridiculous,people are not denying [*], they're pointing out errors in hisreasoning. He says I'm lying, and in support of that repeatshis reasoning, overlooking the fact that whether his reasoningis correct has _no_ relevance whatever to whether I was lyingin claiming that nobody has disputed [*].We'd expect that from him, he's never been real good atkeeping things straight. But then, motivated by nothingbut the fact that James is a hero in your eyes becauseeveryone despises him, you agree that clearly I'm lying.(Well, I don't know that that was your sole motivation,but it's hard to imagine what else it could have been,since if you were instead considering what had actuallybeen said you would never have said anything soridiculous.) I point out that you're being stupid, andask for an example _showing_ I was lying.Then, incredibly, you post a totally irrelevant example,showing (or at least purporting to show) that I saidsomething false about the correctness of a bit ofreasoning of yourse. People point out that the quotehas no relevance whatever.You decide you'd better change the subject, becauseyou _finally_ realize that what I'd said was clearly trueso there's no way you're going to be able to show itwas a lie. So you reply with comments like above.Congratulations.>See Camaraderie of the Experts>====================================================== ===============You concentrated on mathematics because it is predictable, because>there is always a right answer you can check in the back of the book,>because you like following very precise rules, because it allows>you to escape from everyday life into a world that has nothing to>do with everyday life, and because mathematics does not require>the creativity that you completely lack.>Keith Devlin, Commencement address to the mathematics graduating class>of UC Berkeley, May 23, 1997A performance system is designed to work in a defined task domain,>accepting particular goals and seeking to reach them by some kind of>highly selective search. The system must be told what goal is to be>reached and must be given a description of the structure and>characteristics of the task domain in which it is to operate: its>problem space. [...] In contrast, a learning system, is capable of>acquiring a problem space, in whole or part, by interacting with the>external environment and without being instructed about it directly. >A performance system is designed to work in a defined task domain,>accepting particular goals and seeking to reach them by some kind of>highly selective search. The system must be told what goal is to be>reached and must be given a description of the structure and>characteristics of the task domain in which it is to operate: its>problem space. [...] In contrast, a learning system, is capable of>acquiring a problem space, in whole or part, by interacting with the>external environment and without being instructed about it directly.>Herbert Ain't it touchin' the lengths to which Logikoi will go to bail>one another out? See Camaraderie of the Experts>I did go to that link. Here's how it begins.>Lonely, are you?If not for the way _he_ tends to speculate on people's personal liveswhen he can't refute their arguments I'd say this wasn't very nice,pointing out what a pathetic character he must be, forced to talk tohimself like this in public where everyone can see.>Anyway, the lengths the Logikoi will go to bail each other out of>*what*? Was there something threatening in this thread? I know _I've_ been terrified of the possible consequences. I meanof course everything he's said here has been nonsense, butregardless, what if someone at OSU found out that there wassomeone on the internet saying bad things about me?(Giggle. Worse yet: You may not have noticed, but he oftenquotes me saying wild things like everything is equal toitself. What if someone at OSU found out I was promulgatingthat sort of heresy? I can just picture it, once that post-tenurereview he mentioned elsewhere is implemented: There's acommittee meeting in Whitehurst. Ullrich says everything isequal to itself? Off with his head.)>Oooh, were>we almost exposed as charlatans and frauds? Dear oh dear, what with>my mortgage[1] and all, we can't have that.>There's no bailing out going on here. You said something stupid and>irrelevant to the question at hand (namely, whether anyone disputes>that from true premises and correct inferences, true conclusions>follow). I chose, all by my lonesome and with no concern for>Ullrich's well-being, to point out that you're being stupid. Hardly>the first time.>Footnotes: >[1] Not really, but James always talks about how mathematicians (not>that I am necessarily one) are worried about the effect his work will>have on their mortgage payments. I guess John's work is similarly>threatening. Once it's accepted, then one's mortgage payment may no>longer be self-identical.===Subject: Re: Naive Q: Set theory, logic - which comes first?I was too quick with posting.> [.snip.]>Natural numbers pose a special problem since they may be viewed in two>different ways. > Actually, they can be defined in many, in many, many different ways,> and viewed in many, many, many different ways.> It seems to me that there are essentially two ways of viewing them.>We may view them either as something defined by>induction, or as the smallest algebraic structure which is closed>under the operations 0, 1 , +, and * (theese being subject to the>usual laws).> As far as I can tell, {0,1} satisfies the hypothesis you give, through> suitable definitions. Presumably, you mean the smallest subset of> R/Q/Z which contains 0, 1 and is closed under + and *...> Yes, or natural numbers as an algebraic structure characterized> elsehow (for example, adding an axiom saying that there are infinitely> many elements would do the trick).Adding the infinite set of axioms {0/=1, 0/=2, 1/=2, 0/=3, 1/=3, 2/=3,...} would give what we want. Requiring there to be infinitely manyelements would not.> But there are plenty of other ways to characterise the natural> numbers; they are the smallest non-finite ordinal; they are the first> singular cardinal; > Here you view the natural numbers as an algebraic structure. The> number 1 is simultaneously a natural number, an integer, a real> number, an ordinal number, a cardinal number, and who knows what else.> This is because the natural numbers form a substructure of the various> other structures.This is of course not true with the usual set representations oftheese things, but I cannot see how it could be false for the thingsthemselves.As for the idea of identifying an ordinal number with the set of allordinal numbers less than it, I do not see any more reason for makingthat identification than for identifying a real number with the set ofall real (or rational) numbers less than it.> Suppose now we define natural numbers in the same way that we define> finite lists or binary trees. I do not see any way of viewing the> latter two structures as algebraic structures, and therefore I think> it is fair to say that natural numbers defined this way are not an> algebraic structure. Maybe I am wrong, though.It is of course possible to turn finite lists or binary trees intoalgebraic structures, but I doubt there is a uniform (that is, itshould do the same thing to all the things it is applied to) way ofturning inductively defined things into algebraic structures whichwhen applied to natural numbers will give the structure (N, 0, 1, +,*, <) or something similar.If when one introduces natural numbers one says that there is a firstobject and a successor for each object, then it is possible tointerpret the first object either as 0 or as 1. For some purposes itdoes not matter. This alone is a reason for thinking that naturalnumbers as an inductively defined structure is something distinct fromnatural numbers as an algebraic structure.Mattias===Subject: Re: Key Core Error Argument> Define three functions w1(x), w2(x), w3(x), such that w1(x).w2(x).w3(x) = 49> for all x, and w1(0) = w2(0) = 7, w3(0) = 1. Now> (5 a1(x)/w1(x)+7/w1(x))(5 a2(x)/w2(x)+7/w2(x))(5 b3(x)/w3(x)+22/w3(x)) => 300125 x^3 - 10375 x^2 - 360 x + 22> So why is that *not* a possibility? If I understand your terminology> correctly (you still have *not* defined the concept constant term> as you use it), the constant terms of the three factors are 1, 1 and 22.> Let's say you have f(x)/g(x) in the ring of algebraic integers. That> is, both f(x), and g(x) give algebraic integers for algebraic integer> x. Then necessarily, you have *another* algebraic integer function> I'll call h(x).> Understand?> So, h(x) = f(x)/g(x)?Given algebraic integer functions f(x), and g(x), where f(x)/g(x) isan algebraic integer as well then obviously you have yet anotheralgebraic integer function, which I've called h(x) for this disussion. It's not even a big leap to realize that Dik winter, and strange toquestion.Your reply should have been, yeah, that's right.> Now then, if you have h(x), then you can set x=0 to get the constant> term.> Ok. So you define the constant term of a function as the value of the> function when the argument is 0. True? That is what I expected.That's the definition! That's why it's constant!!!If you have f(x) = x + 1, the constant term is distinctive in that asx changes it dos not.Why present it as if it's some weird thing that has to be figured out?I want other readers to see the foot-dragging and in general childishbehavior that I'm facing from posters.So you have this poster Dik Winter acting as if the simplest thingsare difficult or weird, which is a tactic. And his post was overlongso I've shortened it. > 1. f(x) = (5 a1(x) + 7), g(x) = w1(x), h(x) = f(x)/g(x).> h(0) = f(0)/g(0) = 7/7 = 1> 2. f(x) = (5 a2(x) + 7), g(x) = w2(x), h(x) = f(x)/g(x).> h(0) = f(0)/g(0) = 7/7 = 1> 3. f(x) = (5 b3(x) + 22), g(x) = w3(x), h(x) = f(x)/g(x).> h(0) = f(0)/g(0) = 22/1 = 22> So the constant terms of the functions work out as 1, 1 and 22.> This is even independent of whether the h's are algebraic integers> or not.Here the poster Dik Winter has again behaved childishly by how he'spicked his f(x), and it's probably deliberate based on what he saysfurther down.That is, he chose f(x) = 5 a_1(x) + 7 rather than f(x) = 5a_1(x) or f(x) = 7probably because he knows that he would promptly lose the argument ifhe did either of the latter.I said:> Otherwise you'd have a backdoor to making numbers like 7, that are> constants, into variables, which would be mathematically inconsistent.> So if you have 7/w_1(x), and it gives algebraic integers for any> algebraic integer x, then you have some function h(x) = 7/w_1(x),> which STILL has to have a constant term of 1, found by checking at> x=0.> Note that I maybe do not have that 7/w1(x) is an algebraic integer, I> do have that (5 a1(x)/w1(x) + 7/w1(x)) is an algebraic integer, and asHere he tries his latest trick which is to NOW claim that 7/w1(x) isnot necessarily an algebraic integer.However, remember that 49 is being divided off, so w_1 w_2 w_3 = 49.Think about why the poster tried to float such a puzzling assertionpast you.James Harris===Subject: Re: Key core error argument, stepped out>>7. But to divide 7 from those constant terms requires dividing>through two of the factors, so>(5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22>from reverse use of the distributive property, which gives a constant>term coprime to 7, as required.>>>Let's take a simpler example and see how your argument works.>>Let>> Q(x) = 7(25x^2 + 30x + 2) [1]>> = 7(x^2 + x)(5^2) + 7(x - 1)(5) + 7^2>That's [...] >[...] the truth.>Even considering the source, that response is worth keeping. I'm archiving>this puppy. Sounds to me like James has no intention of responding to>my post, which is a pity since he might have learned something. I guess>I underestimated his obdurate attachment to this particular argument.>I'm left wondering how particularly abhorrent type compares to stupid>f**k. Not that I'm all that competitive, but I'd like to know where>I stand compared to Dave Ullrich on the Harris Abuse of Critics scale.>I figure I'm way below David and Arturo, That's correct. A particularly abhorrent type has a _lot_ of work todo before becoming a ing piece of dog. (I'm not sure on thefiner points of the rating, but I _think_ that I even have that stupid Magidin beat.)>somewhat below Dik and Nora-->perhaps in the neighborhood of Will. No matter, I'm evidently in good>company and proud to be a card-carrying member of the Evil Math Cabal.>Rick>p.s. Cabal members, don't forget our next meeting. Be sure to bring>your black robes: the rites look so much more dignified when we're>all properly clad.===Subject: Re: Key Core Error Argument> If I understand your terminology> correctly (you still have *not* defined the concept constant term> as you use it), the constant terms of the three factors are 1, 1 and 22.> So despite all your efforts Dik Winter, you end up with the same> result: constant terms for the factors that are 1, 1, and 22.> Boy that's telling him!> -William HughesActually it is, as Dik Winter is trying to find a way that 7, 7 and 22become 1, 1 and 22 based on a *varying* x.That is, he's trying to make the change in *constants* dependent on avariable.I want readers to understand that his behavior is crank, while I guessmany of you may sympathize with his strong desire for me to be wrong,remember, it's not about people as the math didn't just decide tochange.What you should be sympathetic to, is the truth.James Harris===Subject: Algebraic number theory booksCan anyone suggest a good book(s) on algebraic number theory?===Subject: Re: Key Core Error Argument> Start with> g(x) = ( a(x) + f )> where we know nothing about a(x). Why?You *wish* to believe that the varying function drives thefactorization I use, when I can prove that it does not, but you wantme and other readers to go off on some wild goose chase because you'reirrational?If you think there's a problem in the proof, point at a step.A proof begins with a truth and proceeds by logical steps to aconclusion which then MUST be true.You need go no further than the presented proof, where you do knowabout the a's.Readers, notice that posters seem to want to change the rules!!!Suddenly the presented proof is not enough, as they expect people totake off on tangents to listen to them prattle, when they can't findan error in the proof.Do you want such a rule change in effect with your work?What if you present a proof, but because people don't like it, theydecide to talk about something else and claim it invalidates yourproof?James Harris===Subject: Re: JSH: It begins>Now that I have a proof that's simple enough to explain to kids, I can>move back to email.We're back to the simple enough to explain to kids, eh?Have you figured out yet why kids can understand it but professionalmathematicians can't?>Yup, you're leaving other mathematicians open to a major sandbagging>as I go back to a process that's netted me Andrew Granville, Barry>Mazur, and other notables over the years.>And you're leaving them open and unready.>I learned a while back that few professional mathematicians seem to>read sci.math, so it's not like what's been discussed here is likely>to get to them.>So I have a wide open field of mathematicians all over the world>contactable by email with proven techniques for getting their>attention.Right. Like the people who send me those emails abouthow I can increase the size of parts of my body - theyuse the same proven technique.>The process has already begun.>James Harris===Subject: Re: A Question for James Harris> James:> Something's been bothering me. The world population> is now approximately 6.3 ion. In seven years you> have not convinced even ONE of them about ANYTHING.> The thing that is a source of puzzlement to me is that> in view of this obvious fact, is why have you not just> once sat down and said to yourself, Hmmm. Maybe.> Just possibly. Something's wrong here.That's a rather easy falsehood to correct.Like, seehttp://www.megasociety.net/NoesisHighlights.htmland a version of the paper mentioned there is still under peer reviewat the Southwest Journal, Pure and Applied Math.Or read up on my prime counting function, which finds the count ofprime numbers by integrating a partial difference equation.I've got lots of research that is being taken seriously enough to keepme from totally losing heart, and you people need to remember: Usenetis not all there is.It seems to me that many of you are strangely deluded about your realrole and influence, as you seem to believe that just *saying* thingsthat are wrong on sci.math makes a huge difference in the real world.But my analysis has led me to wonder how many professionalmathematicians bother to read sci.math at all.James Harrishttp://mathforprofit.blogspot.com/===Subject: Re: is there a general way to approach series?>i feel that everytime i look at a new series problem i am starting>from the beginning and struggle to finish it. is there a general>approach to simple series?>for example, why does the divergence of Sum[an] imply the divergence>of Sum[an/(1+an)].It doesn't. (You're also given that an >= 0 in the exercise, right?_Then_ it does. Hint: either an tends to 0 or an does not tend to 0.)>it seems to me that self study from rudin's book is nearly impossible.Possibly you need to read it much more slowly and carefullly.Or possibly you're not ready for a book that makes the sort of demands on the reader that Rudin does - you could trysomething like Ross Elementary Analysis: the Theory of Calculusinstead.>thanks===Subject: Normailzers of Sylow subgroupsLet G be a group and say it has k > 1 distinct Sylow p-subgroups P1,...Pk.Can we say anything in general about how large the intersection of theirrespective normalizes in G are? Can Pi and Pj have the same normailzer??Also, I have a question about acting on Sylow subgroups by conjugation.Suppose G is a simple group with k > 1 Sylow p-subgroups and m > 1 Sylowq-subgroups where p and q are distinct primes. Now acting on the Sylow p andq subgroups by conjugation we get and injective homomorphism to a transitivesubgroup of Sp and Sq respectively (It is injective since G is simple).What's confusing me is that G is now isomorphic to a transitive subgroup ofSp and Sq which I find strange since I can't see how 2 transitive subgroupsof of Different Symmetric groups can be isomorphic? For example I wasworking on a problem where G was isomorphic to A5 and also (under adifferent action) isomorphic to a transitive subgroup of A6 (Also underanother action was isomorphic to a transitive subroup of A10!). Is thispossible?===Subject: Re: Key core error argument, stepped out>>>That's [...] >>[...] the truth.>Even considering the source, that response is worth keeping. I'm archiving>this puppy. Sounds to me like James has no intention of responding to>my post, which is a pity since he might have learned something. I guess>I underestimated his obdurate attachment to this particular argument.I'm left wondering how particularly abhorrent type compares to stupid>f**k. Not that I'm all that competitive, but I'd like to know where>I stand compared to Dave Ullrich on the Harris Abuse of Critics scale.>I figure I'm way below David and Arturo, somewhat below Dik and Nora-->perhaps in the neighborhood of Will. No matter, I'm evidently in good>company and proud to be a card-carrying member of the Evil Math Cabal.p.s. Cabal members, don't forget our next meeting. Be sure to bring>your black robes: the rites look so much more dignified when we're>all properly clad. You might take this one (due to W. V. O. Quine) W. V. Quine, New Foundations for Mathematical Logic, The American Monthly 44 (1937), 70-80.F.===Subject: Re: A 3rd Grade Word Problem---HELP>My 3rd grade son brought this word problem home the other day and he was>given the answer (127). His job, for extra credit, was to figure out how to>get 127. I'm no genius but not a dope either. I couldn't figure out how to>get 127. Nobody in the whole neighborhood could figure out how to get 127.>Is this something of a trick question or is there something in the wording>that I am missing? Any Help????>The Problem:>You know a very good story. On Sunday you tell the story to a friend. On>Monday you tell it to two new people. (So far, a total of three people have>heard the story). Each day after Monday, you double the number of new>people you tell the story to. What will be the total number of people that>will have heard your story after you tell it on Thursday?>I get 31 reading it in what is to me the obvious>way. Looks to me like there's a cheap trick, some way>to read the words in a non-obvious way.Could it be asking for the number of times the story was heard?The old people never left the room and had to hear the sameold story over and over./BAHSubtract a hundred and four for e-mail.===Subject: Re: Sets before logicX-ID: VPw9DgZcweFWemVyhmClHrX-FXTwvixlS3zdJ3-m2rZ5vRSl2F4soT> OBVIOUSLY -- JEEZUS! -- THAT is going to depend on whether> YOUR UNDERLYING SET THEORY WAS DEFINED in Classical> FOL xor in Intutionistic FOL! You simply canNOT say, Presume SOME set theory. [...] You really are going to> have to define your own set theory. [...]Indeed, the reader may consult W. V. Quine, Set-theoretic foundations for logic, Journ. of Symb. Logic 1 (1936), 45-57.F.===Subject: Re: [Probability] Convergence in probability metrizable?>I am given a probability space (W,F,P) and a seperable metric space E.>I consider the set M(E) of all maps f:W->E which are Borel measurable.>I consider on M(E) the topology generated by the base {B(f,a,e): f in>M(E), a,e>0} where:>B(f,a,e)={g in M(E): P(d(f,g)>=a)(fn) in M(E) and f in M(E), the convergence fn->f with respect to this>topology is equivalent to the convergence in probability, i.e.:>For all a>0, P(d(fn,f)>=a)->0 as n->+oo>I was trying to find a metric for this topology and am starting to>suspect it may not be metrizable. Any hint is appreciated.Haven't considered the details of your situation. But if wewere talking about ordinary convergence in measure ona finite measure space:Suppose rho : [0, infinity] -> [0, infinity] has the propertythat rho(t) <= t for all t and also rho(t) <= 1. Let d(f,g) = int rho(|f-g|).Then it's not hard to show that f_n -> f in measure if andonly if d(f_n,f) -> 0, and you just have to find an rho asabove such that d is a metric.>Noel.===Subject: Re: Key core error argument, stepped out>>7. But to divide 7 from those constant terms requires dividing>through two of the factors, so>(5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22>from reverse use of the distributive property, which gives a constant>term coprime to 7, as required.>>>Let's take a simpler example and see how your argument works.>>Let>> Q(x) = 7(25x^2 + 30x + 2) [1]>> = 7(x^2 + x)(5^2) + 7(x - 1)(5) + 7^2>That's [...] >[...] the truth.>Even considering the source, that response is worth keeping. I'm archiving>this puppy. Sounds to me like James has no intention of responding to>my post, which is a pity since he might have learned something. I guess>I underestimated his obdurate attachment to this particular argument.>Readers notice how Rick Decker is playing you for fools as I have a>*stepped* out math proof.Right. Saying it's stepped out means that you've numbered thevarious steps. And the fact that you've numbered the steps meansthat it can't possibly be wrong.>Now then, if you've gone through the trying and difficult process of>finding a spectacular proof, and then face people who decide to ignore>your hard work and effort, to trot out their own pet examples, to try>and distract people from your proof, how would you take it?>How would Rick Decker take it with his own work?I don't think that he'd call people ing pieces of dog.We'll never know of course, because it's never going tohappen: When he thinks he's proved something he'sright, so people don't insist that he's wrong (exceptwhen he's wrong, in which case he follows when peopleexplain why he's wrong and says Oh, I was wrong.) >I'm left wondering how particularly abhorrent type compares to stupid>f**k. Not that I'm all that competitive, but I'd like to know where>I stand compared to Dave Ullrich on the Harris Abuse of Critics scale.>I figure I'm way below David and Arturo, somewhat below Dik and Nora-->perhaps in the neighborhood of Will. No matter, I'm evidently in good>company and proud to be a card-carrying member of the Evil Math Cabal.>Notice how Rick Decker played you for an emotional reaction, while the>real issue here, the reason I'm pissed, is that I have a PROOF and>when you have a proof, mathematicians aren't supposed to try and use>tactics to convince others that what is true is not.>Rick>p.s. Cabal members, don't forget our next meeting. Be sure to bring>your black robes: the rites look so much more dignified when we're>all properly clad.>Yet if Rick Decker's work were facing the same treatment, would he be>so cavalier?>What if he knew he had a paper that could win him the Field's Medal,Actually I don't think anyone's ever won a Field's Medal for onepaper. Regardless, the FM committee doesn't pay attention to sci.math;all you have to do is (giggle) _publish_ this paper.>do you think he'd be so casual about others trying to find ways around>proper process to discredit him?>The reality is I have a stepped out proof that posters like Rick>Decker can't refute because it's a proof, so they try different>tactics to make you believe something false, while they also work to>convince you that they're decent people.>But a Rick Decker is math slime, willing to attack the basis of the>discipline: mathematical proof.>Do you, if you are a mathematician, really want to move to a new>system, a Rick Decker system?>James Harris===Subject: Re: How to define a function to be smooth?>Hey all>When we say a function f(t) is smooth, does this mean that>f has infinite differentials with respect to t?>That's _often_ what it means - sometimes it means less than that.I thought a function was smooth at a point a if there was a>neighborhood of a in which f could be represented as a Taylor series>about a. That's certainly a _type_ of smoothness, and I suppose there couldbe a context in which people define smooth to mean exactly that.But it's not the usual definition - the usual terminology for functions with this property is analytic, or sometimesreal-analytic>So, I had a wrong definition.>Artur===Subject: Re: {Group theory} Number of units?> The set { 6 } has the requisite property: because for every group,> there is another group isomorphic to it in which 6 is the unit.Ah yes, I ought to have seen that. And yes, by unit i meant identityelement, my apologies for the confusion.This trivial solution is not very satisfying. What I am really tryingto get at is something like the set{0,<0,0>(vector),<0,0,0>(vector),...,0 matrix, function f(x)=0, 0 quaternion, 0-degree angle, etc.}But I cannot figure out a good way of defining this, as your {6}illustrates above. :( Can you think of anything?===Subject: Re: A Question for James Harris>James:>Something's been bothering me. The world population>is now approximately 6.3 ion. In seven years you>have not convinced even ONE of them about ANYTHING.>The thing that is a source of puzzlement to me is that>in view of this obvious fact, is why have you not just>once sat down and said to yourself, Hmmm. Maybe.>Just possibly. Something's wrong here.>That's a rather easy falsehood to correct.>Like, see>http://www.megasociety.net/NoesisHighlights.html>and a version of the paper mentioned there is still under peer review>at the Southwest Journal, Pure and Applied Math.>Or read up on my prime counting function, which finds the count of>prime numbers by integrating a partial difference equation.>I've got lots of research that is being taken seriously enough to keep>me from totally losing heart, and you people need to remember: Usenet>is not all there is.>It seems to me that many of you are strangely deluded about your real>role and influence, as you seem to believe that just *saying* things>that are wrong on sci.math makes a huge difference in the real world.>But my analysis has led me to wonder how many professional>mathematicians bother to read sci.math at all.A very small percentage - wondering about this is stupid, becauseit's been explained to you many times.The fact that so few mathematicians read sci.math is why you havebetter luck with the famous mathematicians you harass via email,and the editors of the journals you send your stuff to. Giggle.>James Harris>http://mathforprofit.blogspot.com/===Subject: Re: Sets before logicX-ID: V3RHorZaQeyXYpZIjJmz53VNtzilmjhYCXBI8TTeB0W1wq6Dn5bCkQ> The mere fact that you have alleged the existence of a > universal set AUTOMATICALLY takes you AWAY from most > of the world's known set theories!Indeed, but luckily not from ALL (of them), see: Bibliography: Set Theory with a Universal Set http://math.boisestate.edu/~holmes/holmes/setbiblio.htmlQuote: IntroductionThis is a comprehensive bibliography on axiomatic set theories which have auniversal set. (Zermelo-Fraenkel set theory, the most widely studied set theory,does not have a universal set.) This field presently includes three main areasof study: New Foundations, a set theory devised by W. van Orman Quine, thepositive set theory of Helen Skala, and model-based extensions ofZermelo-Fraenkel set theory, initiated by Alonzo Church.Some books: Forster, T.E. [1995] Set Theory with a Universal Set, second edition. Clarendon Press, Oxford. Holmes, M. R. [1998] Elementary set theory with a universal set. volume 10 of the Cahiers du Centre de logique, Academia, Louvain-la-Neuve (Belgium), 241 pages, ISBN 2-87209-488-1. See: http://www.lofs.ucl.ac.be/cnrl/Cahiers/Cahier10.htmlOnline document: Church's Set Theory with a Universal Set by Forster 'Church's Set Theory with a Universal Set' is a revision and expansion of the last chapter of my book on set theory with a universal set, and supercedes it. It was written for the Alonzo Church festschrift. The version here is to be preferred to the version in print, as I remove typos and mathematical errors from it as they come to my notice. http://www.dpmms.cam.ac.uk/~tf/church2001.psF.===Subject: Re: Uncle Al is Sadistic .>Actually, there are a lot of studies that make the connection pretty>obvious. I don't have the book here, though, and it's been years since I>read it; I can't remember the details. If you're curious you can either>find a copy of _The Bell Curve_ yourself, or I can get back to you on>Thursday.> I have read -TBC- throughly and twice. What it was really about was an > internal brain drain in the United States, where a few professions > seemed to be sopping up all the Brights.Ahh. Sorry. It's been a few years since I read it, and I must have primarily remembered the stuff that interested me -- extrapolation of future effects. > The one or two chapters on different intellectual capacities among > various ethnic groups caused Political Correct Fanatical Liberals to go > into near earth orbit.It was amusing.> The underlying lesson I took away from the book is that one cannot put > in what God left out. As my grandmother (peace be upon her) once taught > me, you can't make a Gucci Bag from a pig's tuchas.:)- Laurel * * * http://amberdine.com===Subject: Re: Key Core Error Argument> If I understand your terminology> correctly (you still have *not* defined the concept constant term> as you use it), the constant terms of the three factors are 1, 1and 22.> So despite all your efforts Dik Winter, you end up with the same> result: constant terms for the factors that are 1, 1, and 22.> Boy that's telling him!> -William Hughes> Actually it is, as Dik Winter is trying to find a way that 7, 7 and 22> become 1, 1 and 22 based on a *varying* x.> That is, he's trying to make the change in *constants* dependent on a> variable.> I want readers to understand that his behavior is crank, while I guess> many of you may sympathize with his strong desire for me to be wrong,> remember, it's not about people as the math didn't just decide to> change.You are wrong, that's the point. It is clear to me that you do notcompletely understand algebra and I strongly believe you attack argumentsthat you don't understand. Example: I showed you an example a few weeks agothat was basic algebra 1 and you attacked it, which means to me you don'thave a good grasp on algebra. You don't care about the math, do you? Youjust care about fame, which you probably won't get without a correctargument.David Moran> What you should be sympathetic to, is the truth.> James Harris===Subject: Re: Cantor's diagonal number: why do we just ASSUME such a number exists?in part:>Although some parts of Cantor have truth, I>believe that Cantor is completely bereft of truth. (Actually, that>last sentance was inconsistent and needs correction, but we can still>adhere to it without loss of generality.)Ah, you're not just an interested young fellow. You're a deliberatetroll.>I'm not good at math,>but my theory is conceptually right, so all I need is for someone to>express it in terms of equations.How can you *tell* if something is conceptually right?>We must remember that the current>axiomatic models are only theories, nothing more. Although ZFC can be>used to accurately predict the outcome of typing things into a>calculator, it hardly explains *why* the calculator gives the answers>it gives. Like Einstein, I have a special insight into these matters.> My work is on the cutting edge of a paradigm shift.Oh, wow. Perhaps you may yet achieve your goal, and be recognized asone of the great comedians of USENET, like that Anaximander Uraniumfellow.John Savardhttp://home.ecn.ab.ca/~jsavard/index.html===Subject: Re: Do finite strings over a countable alphabet form a countable set?>Given a countable set of letters A={L_1,L_2,L_3,...}, consider the set of>finite strings S(A) which can be formed using letters in A. Is S(A) a>countable set?Well, of course. Example: let A={1, 2, 3, 4, 5, 6, 7, 8, 9}; in thatcase, every member of S(A) corresponds to an element of N.The generalization to any other A, with however many elements in it,is trivial.If this is a homework problem, you obviously need to abandon Cantorand take up divinity.John Savardhttp://home.ecn.ab.ca/~jsavard/index.html===Subject: Re: Do finite strings over a countable alphabet form a countable set?>How could you have a rational discussion with ZZBunker? :-)But what's wrong with giving someone who posts a homework problem thewrong answer he deserves?John Savardhttp://home.ecn.ab.ca/~jsavard/index.html===Subject: Re: [Probability] Convergence in probability metrizable?> I am given a probability space (W,F,P) and a seperable metric space E.> I consider the set M(E) of all maps f:W->E which are Borel measurable.> I consider on M(E) the topology generated by the base {B(f,a,e): f in> M(E), a,e>0} where:> B(f,a,e)={g in M(E): P(d(f,g)>=a) (fn) in M(E) and f in M(E), the convergence fn->f with respect to this> topology is equivalent to the convergence in probability, i.e.:> For all a>0, P(d(fn,f)>=a)->0 as n->+oo> I was trying to find a metric for this topology and am starting to> suspect it may not be metrizable. Any hint is appreciated.Here's one: d_{M(E)}(f,g) = int_W min[d(f(w),g(w)), 1] P(dw).A.===Subject: Re: Who contributed most to mathematics?>I think France has made more contributions to mathematics than any>other nation. There is no Nobel prize for math but the French out>rank the U.S there too or are very close.There sure have been a lot of important French mathematicians;Fourier, Galois, d'Alembert...but the Greeks started it.However, I really suspect that the Germans have to take top honors.And thanks to just *two* guys, even though there were other Germanmathematicians.Euler and Gauss.John Savardhttp://home.ecn.ab.ca/~jsavard/index.html===Subject: Re: Making Star Trek Real>sarfatti@pacbell.net says...>Note some minor typographical corrections to the macro-quantum >geometrodynamical equations for the metric engineering of Star Gate Time >Travel Machines and Weightless FTL Warp Drives that underlie the physics >of UFO super-technology that I was put to work on in 1954 at the age of >14 in a USG Project headed by Eugene Mc Dermott a co-founder of Texas >Instruments and a high level Defense Intelligence Honcho since WWII.>Can you say that in one breath?He must have watched The Time Tunnel as a child, and had nightmaresimagining he was James Darren.John Savardhttp://home.ecn.ab.ca/~jsavard/index.html===Subject: Re: A 3rd Grade Word Problem---HELP===>Subject: Re: A 3rd Grade Word Problem---HELP>Message-id: > My 3rd grade son brought this word problem home the other day and he was>> given the answer (127). His job, for extra credit, was to figure out >how to>> get 127. I'm no genius but not a dope either. I couldn't figure out how >to>> get 127. Nobody in the whole neighborhood could figure out how to get >127.>> Is this something of a trick question or is there something in the >wording>> that I am missing? Any Help????>> The Problem:> You know a very good story. On Sunday you tell the story to a friend. >On>> Monday you tell it to two new people. (So far, a total of three people >have>> heard the story). Each day after Monday, you double the number of new>> people you tell the story to. What will be the total number of people >that>> will have heard your story after you tell it on Thursday?>I get 31 reading it in what is to me the obvious>way. Looks to me like there's a cheap trick, some way>to read the words in a non-obvious way.>Could it be asking for the number of times the story was heard?>The old people never left the room and had to hear the same>old story over and over.57>/BAH>Subtract a hundred and four for e-mail.===Subject: Re: Bible 1, Darwin 0! And we are talking pure archeology?part:>The aim of the book is not to prove things that the Bible said, I just>put a catchy title.I remember looking at it at the library. I thought its aim was toprove things the Bhagavad-Gita said.So that we know of the great dispensation offered to us unfortunatedenizens of the Age of Kali, that we can improve our karma simply bychanting a simple hymn of praise to our teachers and to Lord Krsna.John Savardhttp://home.ecn.ab.ca/~jsavard/index.html===Subject: Re: What is meant by Subrepresentation?> Hi there! In a book I encountered the term subrepresentation, which> was meant in the way of restricting the representation on a subspace. If A> is a C*-algebra and r a representation of A on a Hilbert space H, and H' a> closed subspace of H, what is then r restricted to H' ?My guess is that your representation r is such that for every v in H' andevery a in A, a.v belongs again to H'. Therefore r induces a representationof A in H' thus defined: if a in A and v in H', then the action of a on v issimply equal to a.v; this makes sense, since a.v belongs to H'.I hope that this helps.Jose Carlos Santos===Subject: Re: I can't stand it anymoreamanda replied:>>What gave you the idea that I was upset about the term Asian?>> My only point was that IQ tests to determine people's intelligence are>bs.>Ah, of course. It would be far better indeed to administer>intelligence tests rather than IQ tests to determine people's>intelligence, right?>Really, you sound like someone who is disappointed by their>own results on such a test.> I knew that someone would come up with that stupid opinion. It just> happens to be you. Sorry about that> Don't you realize that us Asians don't need the tests invented by> non-Asians to find out whether we are intellgent or not?> It was only at the turn of 2oth century that the Western World > started to reliaze that Asians are not inferior to them It's interesting that in the Asian countries, the inferiority ofnon-Asians is still widely accepted and believed. There is no attemptfor or desire for reciprocity.> as they have been told by the 19th century European racistsIs there an Asian country which is not racist? Just look at the nicethings high Japanese govt officials have said about Americans.> who believed that they were superior to all other humans.As do all Asian races currently. But somehow you don't seem to seeanything wrong with this. This entire post would be ironic were itnot so hypocritical. Why don't you also complain about Asian racism?Probably because you support it. It's clear that you will profitfrom any pro-Asian racism. It's almost certain that you have.===Subject: Re: I can't stand it anymore> What gave you the idea that I was upset about the term Asian?> My only point was that IQ tests to determine people's intelligenceare> bs.> Ah, of course. It would be far better indeed to administer> intelligence tests rather than IQ tests to determine people's> intelligence, right?> Really, you sound like someone who is disappointed by their> own results on such a test.> I knew that someone would come up with that stupid opinion. It just> happens to be you. Sorry about thatOh, boo hoo. It was an observation, not an opinion. Do youunderstand the difference? Should we test you on that?> Don't you realize that us Asians don't need the tests invented by> non-Asians to find out whether we are intellgent or not?So? You have your own tests then? Fine. Get on with it.> It was only at the turn of 2oth century that the Western World> started to reliaze that Asians are not inferior to them as they have> been told by the 19th century European racists who believed that they> were superior to all other humans.> As someone was throwing away a bunch of old books, piling up outisde> the lab I was doing my research worl in grad school, I saw just such a> book written by a European scientists. Anyway,> before that time, did the opinion of the West affected the Asians?> Ever read the book soverign Individual. The author stated thatwhite people hate Asians but that didn't affect .... He was just> trying to say how the Asians knew who they were and didn't act based> on others' opinion.> We certainly don't need these IQ tests to find out about our> intelligent level.So what IQ tests would you apply? Stop whining and get on with it.===Subject: Re: I can't stand it anymore>I have been biting my tongue about the IQ test but I can't any more.>How reliable is a test that use the term Asian to represent the most>diverse of ethnic and cultural groups?> IQ tests are not particularly reliable, but what does this have to do> with it? What test is using that term? And how is any of this relevant> to these newsgroups?>If you want to know more about the diversity, ask me and I will give>you tons of specific examples.> ...> Your examples may be interesting, but what do they have to do with> sci.math?> Sci.math was one of the group listed in the thread where I saw the> discussion of IQ test. I had no idea who were from which group in that> thread and so decided to posts in all 3 groups. The only science group> I posted before was sci.chem group.> I don't browse sci.chem group regularly either. So when I broswe the> posts once in a while and see some supposedy intelligent people from> these 3 groups talking about IQ tests as a ver important tests (with> some seeing it as the ultimate test to determine human intelligence),> I get very disappointed.So let's see; in your opinion, talking about IQ tests isan indication of low intelligence or some social backwardness?===Subject: Re: Greek Alphebetchristina@norman31.freeserve.co.uk (arsenal) quoted, in part:> In my studies on triginometry, and calculus, I am comming across many>> symbols which appear to be the Greek Alphebet, as there was a chart of>> the Alphebet in the front of the book. I am wondering if someone could>> tell me what they all mean, or point me to a site where I could find>> out.>You need this rule: The symbols mean whatever the author says they mean.>Take each one in the context where it is discussed. Somewhere else, the>same symbol is very likely to mean something different.>Even the venerable PI does not necessarily mean the constant 3.1415...,>although that is probably the meaning in the context of the subjects you>mentioned.That's true for mathematics. Although there are other conventionaluses. For example, epsilon often means a small number. And epsilon subzero has one meaning in Physics (a constant involved in the relationbetween electricity and magnetism) and another common one inMathematics (an order-class larger than omega to any power).But in other contexts, the letters of the Greek alphabet are fraughtwith meaning!Thus, alpha comes from aleph, which means Ox. And gamma comesfrom gimel... so one can look at a book on the history of thealphabet. And the 22 letters of the Hebrew alphabet are alleged tohave meaning on other levels too.But then to study mathematics is to go in one direction, and to studyTarot cards is to go in a different direction...John Savardhttp://home.ecn.ab.ca/~jsavard/index.html===Subject: Re: Maple question> Does anyone know of a way to extract coefficients of certain expressions,> like if I have z = A*f(x) + B*g(x), how can I get A (and B) without copying> it from the output? I want something like A = get_coeff(z, f(x))> Try the op(.) commandThis may work, depending on what f(x),g(x) you actually have:subs({g(x)=0,f(x)=1},z);===Subject: Re: JSH: It begins> Now that I have a proof that's simple enough to explain to kids, I can> move back to email.> Yup, you're leaving other mathematicians open to a major sandbagging> as I go back to a process that's netted me Andrew Granville, Barry> Mazur, and other notables over the years.> And you're leaving them open and unready.> I learned a while back that few professional mathematicians seem to> read sci.math, so it's not like what's been discussed here is likely> to get to them.> So I have a wide open field of mathematicians all over the world> contactable by email with proven techniques for getting their> attention.> The process has already begun.> James HarrisDoes this mean you are *finally* ready to stop degrading the SNR ofsci.math? Are you really leaving sci.math this time -- or is this justanother one of your tricks to lull its readers into a false sense ofrelief?A fool and his proof are soon refuted.Democracy: The triumph of popularity over principle.http://www.crbond.com===Subject: Re: c-number>Can someone please tell me what the hell a c-number is?a trajectory t |-> (x1(t),x2(t),x3(t)) = r(t), with velocityv(t) = r'(t) = (x1'(t), x2'(t), x3'(t)), everything commutes: xi(t) xj(t) = xj(t) xi(t) xi(t) xj'(t) = xj'(t) xi(t) xi'(t) xj'(t) = xj'(t) xi'(t)All the quantities in Classical Physics commute and are calledc-numbers.In Quantum Physics, for the same system, you still have: xi(t) xj(t) = xj(t) xi(t) xi'(t) xj'(t) = xj'(t) xi'(t)as well as: xi(t) xj'(t) = xj'(t) xi(t), for i, j differentbut now xj(t) xj'(t) = xj'(t) xj(t) + i h/(4 pi m)where h is Planck's constant.The corresponding quantities in Quantum Physics don't generallycommute and are called q-numbers.In the classical limit (as h -> 0) the q-numbers approach theircorresponding c-numbers of classical physics.The inverse problem of finding a quantum system whose classicallimit is a given classical system is called quantizing theclassical system.In a general system with N degrees of freedom q(t) = (q1(t),q2(t),...,qN(t))satisfying equations of motion of the form qi'(t) = vi(t), vi'(t) = ai(q(t),v(t)); i=1,2,...,Nif you assume likewise that qi(t) qj(t) = qj(t) qi(t)but that A(t) A'(t) != A'(t) A(t)for any linear combination A(t) = (c1(t) q1(t) + .. + CN(t) qN(t))then the requirement that this be compatible with the equations ofmotion means that the matrix given by: Wij = (qi(t) vj(t) - vj(t) qi(t))/ (h/(2 pi))(which is symmetric, as seen by taking the derivative of the q-qcomutators) has a classical limit given by: W -> M^{-1} Mij = @^2L/@vi @vj (@ = ASCII partial derivative operator)for some Lagrangian L(q,v) such that (1) ai(q,v) = @L/@qi - @^2L/@vi@t - sum (vj @^2L/@vi@qj) (2) (A(q,v)B(q,v) - B(q,v)A(q,v)) / (h/(2 pi)) -> {A(q,v),B(q,v)} as h -> 0; {,} = corresponding Poisson Bracketand (3) the matrix S of velocity commutators Sij = (vi(t)vj(t) - vj(t)vi(t))/ (h/(2 pi)) has the limit S -> M^{-1} s M^{-1}, as h -> 0 with sij = @^2L/@qi@vj - @^2L/@qj@vi = {vi,vj}.So, under these conditions (q-q's commute, 2nd order equations ofmotion, q-v's don't commute), a quantum system can only be thequantization of a classical system that arises from a Lagrangian,which (because M = @^2L/@v^2 is non-singular) will also have aHamiltonian.The q numbers arise from quantizing a classical system given by aHamiltonian.===Subject: Re: A NOTy problem. Adjunct Assistant Professor at the University of Montana.>This post not CC'd by email>>G'day G'day Folks, >> I have been puzzling over what ought to be a simple logic problem>>but am at a loss to come up with a method. >These sort of things can be solved by brute force, by assigning all>possible values of truth or false to each and seeing which yield>contradictory information. On the other hand, sometimes there are>shortcuts based on noticing certain things.>>A crime is committed by one of the following; , Bob, Chris or>>Dave. >>Each makes a statement to the police but three of them lie. >> says, I didn't do it. >>Bob says, is lying. >>Chris says,Bob is lying. >>Dave says, Bob did it. >There's two easy observations here. Note that Dave and Allan's>statements are compatible.> Yes. That is neat. Dave and are talking about who did the>crime. Bob and Chris are talking about who is lying. This forms>natural groupings to test for contradiction. >There's two easy observations here. Note that Dave and Allan's>statements are compatible.>Compatible? I can see that the statements are not inconsistent. That was the point.>If Dave is telling the truth, then so is . >Since only one person is telling the truth, Dave must be lying.>You lost me there. What if I had a let's pick on day? You lost me there. What the heck does it matter if you had that?Dave says Bob did it. says I did not do it, and inparticular, is saying Either Bob, Chris, or Dave did it. IfDave is telling the truth, and Bob did it, then is ALSO tellingthe truth, since he did not do it.So: If Dave is telling the truth, then so is .Since there can be at MOST one person telling the truth, it isimpossible for Dave to be telling the truth (if he were telling thetruth, there would necessarily be at least TWO people telling the truth).>If is telling the truth can we say that so it Dave. No, you cannot say that. Because if is telling the truth, and hedid not do it, that does NOT imply that Bob did it (what Dave issaying). It means that EITHER Bob, Chris, or Dave did it. It could bethat Dave is lying.===========Subject: Re: Key Core Error Argument... > Let's say you have f(x)/g(x) in the ring of algebraic integers. That > is, both f(x), and g(x) give algebraic integers for algebraic integer > x. Then necessarily, you have *another* algebraic integer function > I'll call h(x). > Understand? > So, h(x) = f(x)/g(x)? > Given algebraic integer functions f(x), and g(x), where f(x)/g(x) is > an algebraic integer as well then obviously you have yet another > algebraic integer function, which I've called h(x) for this disussion. > It's not even a big leap to realize that Dik winter, and strange to > question. > Your reply should have been, yeah, that's right.Nope. When you say then necessarily you have *another* algebraic integerfunction*, it is not clear what you mean. Obfuscation is it. > Now then, if you have h(x), then you can set x=0 to get the constant > term. > Ok. So you define the constant term of a function as the value of the > function when the argument is 0. True? That is what I expected. > That's the definition! That's why it's constant!!!Well, you never have stated so, although I have asked a few times already. > If you have f(x) = x + 1, the constant term is distinctive in that as > x changes it dos not.In this case f(x) is a polynomial, so there is a standard definition ofconstant term. When f(x) is *not* a polynomial, there is *no* standarddefinition of constant term. Understand? That is why I ask. You usenon-standard terminology, and I simply ask what you mean. The terminologyis distinctive in that as x changes it does not is just obfuscation.Take sqrt(x + 4). When x changes, 4 does not change. Is 4 the constantterm? No, 2 is. When f(x) = 2x. Is 2 the constant term (it does notchange when x does change), no it is not. > Why present it as if it's some weird thing that has to be figured out?Because there is no standard definition, and you explain it only inobfuscated terminology. > I want other readers to see the foot-dragging and in general childish > behavior that I'm facing from posters. > So you have this poster Dik Winter acting as if the simplest things > are difficult or weird, which is a tactic. And his post was overlong > so I've shortened it.If you use non-standard terminology in mathematical text, it is best to*define* your terms rather than relying on others to understand what youmean. > 1. f(x) = (5 a1(x) + 7), g(x) = w1(x), h(x) = f(x)/g(x). > h(0) = f(0)/g(0) = 7/7 = 1 > 2. f(x) = (5 a2(x) + 7), g(x) = w2(x), h(x) = f(x)/g(x). > h(0) = f(0)/g(0) = 7/7 = 1 > 3. f(x) = (5 b3(x) + 22), g(x) = w3(x), h(x) = f(x)/g(x). > h(0) = f(0)/g(0) = 22/1 = 22 > So the constant terms of the functions work out as 1, 1 and 22. > This is even independent of whether the h's are algebraic integers > or not. > Here the poster Dik Winter has again behaved childishly by how he's > picked his f(x), and it's probably deliberate based on what he says > further down. > That is, he chose > f(x) = 5 a_1(x) + 7 > rather than > f(x) = 5a_1(x) or f(x) = 7 > probably because he knows that he would promptly lose the argument if > he did either of the latter.But what would be the reason to pick either 5a_1(x) or 7? > I said: > Otherwise you'd have a backdoor to making numbers like 7, that are > constants, into variables, which would be mathematically inconsistent. > > So if you have 7/w_1(x), and it gives algebraic integers for any > algebraic integer x, then you have some function h(x) = 7/w_1(x), > which STILL has to have a constant term of 1, found by checking at > x=0. > Note that I maybe do not have that 7/w1(x) is an algebraic integer, I > do have that (5 a1(x)/w1(x) + 7/w1(x)) is an algebraic integer, and as > Here he tries his latest trick which is to NOW claim that 7/w1(x) is > not necessarily an algebraic integer. > However, remember that 49 is being divided off, so w_1 w_2 w_3 = 49. > Think about why the poster tried to float such a puzzling assertion > past you.Yup. w_1.w_2.w_3 = 49. Can you give a good reason why 7/w_1(x) *should*be an algebraic integer? The only thing you are doing is dividing aproduct of three factors by 49. It is only those three that need bealgebraic integers when divided by a w(x), not the parts that make upthe factors. Or give a good reason why the parts that make up thefactors must be divisible by w(x) in the algebraic integers.dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/===Subject: Re: some coin tossing> If I toss a fair coin n times, what is the probability that the outcomeof> my experiment will contain m consecutive heads (m If the exact probability is difficult to compute, can we have a goodupper> bound on the probability?> A trivial upperbound is (n-m+1)(0.5^m)> I'll do counting, you can divide by 2^n at the end to get a probability.> Condition on the beginning point of the run. Let X denote an ARBITRARY> outcome of the toss. Then an enumeration of all the possibilities which> yield a run is:> Pattern Count> 1 2 3 4 .. m m+1 m+2 m+3 ... n The choice comes from choosing X's> H H H H .. H X X X ... X 2^(n-m-1)Shouldn't this be 2^(n-m)?> T H H H .. H H X X ... X 2^(n-m-2) <-- This is (2^1-1)2^(n-m-2)> X T H H .. H H H X ... X (2^2-1) 2^(n-m-3)So now we are doing an overcount. For example let n=5, m=2.The case HHTHH will be counted twice!> and the pattern which has i X's followed by a T and m H's followed by X's> to the end has count> (2^i-1)*2^(n-m-1-i)What you mean is 2^(i-1)*2^(n-m-1-i). Right?> If you do the sums the answer should be> 2^(n-m-1) + sum_{i=1}^{n-m-1} (2^i-1)*2^(n-m-1-i)> or> (n-m)*2^(n-m-1) - sum_{i=1}^{n-m-1} 2^(n-m-1-i)> or> (n-m)*2^(n-m-1) - (1+2+2^2+..+2^(n-m-2)) = (n-m-1)*2^(n-m-1)> or> (n-m-1)*2^(-m). SerdarSo with this overcount I now have an upper bound of (n-m+2)*(0.5^(m+1)),which is better than the pervious upper bound, but still not quite good. Anyother suggestions?===Subject: Re: Algebraic number theory books Adjunct Assistant Professor at the University of Montana.>Can anyone suggest a good book(s) on algebraic number theory?I'm partial to iel Marcus's _Number Fields_; Ireland and Rosen's _AClassical Introduction to Modern Number Theory_ is also a good firststep.===========Subject: Re: Key Core Error Argument... > Actually it is, as Dik Winter is trying to find a way that 7, 7 and 22 > become 1, 1 and 22 based on a *varying* x. > That is, he's trying to make the change in *constants* dependent on a > variable.Eh? > I want readers to understand that his behavior is crank, while I guess > many of you may sympathize with his strong desire for me to be wrong, > remember, it's not about people as the math didn't just decide to > change. > What you should be sympathetic to, is the truth.You claim that having P(x) = g1(x).g2(x).g3(x) with g1(0) = g2(0) = 7and g3(0) = 22; the *only* way to divide P(x) by 49 is by dividing g1(x)and g2(x) by 7 and g3(x) by 1. Because now g1(0)/7 = 1, g2(0)/7 = 1 andg3(0)/1 = 22.I claim there are other ways to do that. Have w1(x), w2(x), w3(x), suchthat w1(0) = w2(0) = 7, w3(0) = 1, w1(x).w2(x).w3(x) = 49. Because nowg1(0)/w1(0) = 1, g2(0)/w2(0) = 1 and g3(0)/w3(0) = 22.Where do I change constants?dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/===Subject: Re: how to identify data from different distributions?>distribution. Is it possible to use some learning or statistical>algorithms to separate the two data set out?>For example,>X=(x1,x2,...,xn) ~ Gaussia>Y=(y1,y2,...,yn) ~ Uniform>Given (x1, y1, x2,x3,y2,x4,y3,...), can you separate out X and Y?Given two distributions r1(x), r2(x), you'll sort a point X intothe first if r1(X) > r2(X), and into the second if r1(X) < r2(X).It's a tie of r1(X) = r2(X) and the set { X: r1(X) = r2(X) }represents the boundary betweey the two zones.===Subject: Re: Uncle Al is Sadistic .> The genetic inheritance of mental efficiency is NOT obvious.It is.> It is very hard to isolate genetics from upbringingOnly if you suffer amnesia of your life from 0-2, so as not toknow first-hand precisely what came from where, when and how -- a ratherextraordinary deficit, to say the least; but certainly one I don't share.It's not a question that needs to be surmised any more than the questiondoes of what I ate yesterday for breakfast. I was there when it happened.===Subject: Re: Normailzers of Sylow subgroups>Let G be a group and say it has k > 1 distinct Sylow p-subgroups P1,...Pk.>Can we say anything in general about how large the intersection of their>respective normalizes in G are? Can Pi and Pj have the same normailzer??You mean `normalizer'. Distinct Pi and Pj cannot have the same normalizer, because that would mean that Pi normalized Pj, and thenPi and Pj would generate a p-group larger than P, which is impossible.In general, you can say that the intersection of distinct normalizersdoes not contain a Sylow p-subgroup of G, so its order cannot be divisibleby the largest power of p dividing |G|.>Also, I have a question about acting on Sylow subgroups by conjugation.>Suppose G is a simple group with k > 1 Sylow p-subgroups and m > 1 Sylow>q-subgroups where p and q are distinct primes. Now acting on the Sylow p and>q subgroups by conjugation we get and injective homomorphism to a transitive>subgroup of Sp and Sq respectively (It is injective since G is simple).I think you mean Sk and Sm, not Sp and Sq.>What's confusing me is that G is now isomorphic to a transitive subgroup of>Sp and Sq which I find strange since I can't see how 2 transitive subgroups>of of Different Symmetric groups can be isomorphic?They certainly can be! By Cayley's Theorem, every group G is isomorphic toa transitive group of degree |G|, as well as any smaller degree transitiveactions that it may have.>For example I was>working on a problem where G was isomorphic to A5 and also (under a>different action) isomorphic to a transitive subgroup of A6 (Also under>another action was isomorphic to a transitive subroup of A10!). Is this>possible?A5 is isomorphic to transitive groups of degrees 5, 6, 10, 12, 15, 20,30 and 60.Derek Holt.===Subject: Re: cardinality of monotone set-functionsContent-Length: 301Originator: rusin@vesuviusMy previous post on the Dedekind problem is not (obviously) relevantto the original problem of monotone functions 2^X ---> 2^X. TheDedekind problem deals with monontone function 2^X ----> {0,1}.Clark for pointing this out.Victor===Subject: associative one-way functionX-Mimeole: Produced By Microsoft MimeOLE V6.00.2800.1165X-Msmail-Priority: NormalDoes anyone know if there is any potential one-way functionwhich is associative other than modular exponentiation?That is, given a function f(x,y) (which is easy to compute),- given z and x, it is difficult to find y such that z = f(x,y)- f(f(x, y1), y2) = f(f(x, y2), y1)===Subject: Re: some coin tossingfigure out what exactly he (Bryce Carlson) was saying. But I still don'tunderstand how he computes those p1, p2, etc ... It seems that he iscontinuing some discussion here. Can you please send me the link where hedoes the analysis for 2 consecutive heads?> If I toss a fair coin n times, what is the probability that the outcomeof> my experiment will contain m consecutive heads (m This is NOT a binomial distribution> If the exact probability is difficult to compute, can we have a goodupper> bound on the probability?> Here's a guy who's looked at probabilities of successive> heads and found a connection with Fibonacci numbers:> http://www.bjmath.com/bjmath/probable/flips.htm> A trivial upperbound is (n-m+1)(0.5^m)> - Randy===Subject: Re: Why is math so difficult for some people?> Many of the so-called elegant proofs completely [sic] hide the concepts.Elegance is characterized by the propensity toward dispensing withextraneous inessentials.> Proving the Central Limit Theorem for sums of> independent identical distributions is certainly the most> elegant, but it [sic] hides virtually everything; it has no> probability in it at all.Exactly.===Subject: Re: A 3rd Grade Word Problem---HELP>My 3rd grade son brought this word problem home the other day and he was>given the answer (127). His job, for extra credit, was to figure out how to>get 127. I'm no genius but not a dope either. I couldn't figure out how to>get 127. Nobody in the whole neighborhood could figure out how to get 127.>Is this something of a trick question or is there something in the wording>that I am missing? Any Help????>The Problem:>You know a very good story. On Sunday you tell the story to a friend. On>Monday you tell it to two new people. (So far, a total of three people have>heard the story). Each day after Monday, you double the number of new>people you tell the story to. What will be the total number of people that>will have heard your story after you tell it on Thursday? complete solution sent to the e-mail address you so courteouslyprovided.===Subject: Re: I NEED HELP BADLY (sorry, maths not psych)>Meanwhile, why don't you explain to me how this model applies to an>>attractive electric field. What are the ping-pong balls and where are>>they coming from?> This is called theoretical research Randy.>> The aim is to try to match a hypothetical physical process with the prediction>> of an equally hypothetical maths theory, namely relativity.It's just a theory? Are we a creationist now?>Of course the theory is just a theory. The point is, it>makes predictions about such things like the maximum speed>You have an alternate theory.>are accelerated by receiving momentum from little objects>moving at c.>Fine. If this is your explanation for the limiting speed, please>tell me what corresponds to the little ping-pong balls in the>case of an attractive force, and where they are coming from.> dimension too, just like space and time.>You have now alluded to theoretical research without>a theory, differential equations without an equation,>and in general spun a lot of words while refusing to answer>the question.> I can see this is far too difficult for you Randy.How could it be? You haven't yet provided the thisOK, we're up to three statements that there exists somemysterious content that, were it to be revealed, wouldsay what you mean by the ping-pong ball model. But ifyou were to reveal the content, it would be too difficultfor me.Why don't you reveal it and see?>>Or do you really think nobody noticed that you avoided answering any>>questions about your model?Do you really think nobody has noticed that you've>avoided answering any questions about your model?> It is merely a straightforward and interesting mechanical problem, involving a> basic differential equation. Something like D^2+KD=CSomething like?So what IS it?For that matter, do you have any idea what the symbols inthe DE that you did write mean?> Surely you have SOME scientific ability Randy.some momentum carrier analogous to ping-pong balls movingat c (there, I've given you even more words to plagiarize).So now it's something like an equation with undefinedDo you really think nobody has noticed that you'veavoided answering any questions about your model?>Explain to me how this model applies to an attractive >electric field.>What are the ping-pong balls and where are they coming from?> Don't worry Randy, It is a hypothetical.Yes? And what's the hypothesis?Do you really think nobody has noticed that you'veavoided answering any questions about your model?Explain to me how this model applies to an attractive electric field.What are the ping-pong balls and where are they coming from?> But OK, consider a small frictionless wheeled vehicle sitting on the desk in> from of you. You have a gun that fires perfectly elastic ping-pong balls at> velocity v, at the rate of one per nanosecond. You fire them at the vehicle.> Plot the 'distance versus time' graph for the vehicle.> Why do you automatically criticize something that you don't even understand? I understand that perfectly well. If you try to propelsomething forward by transferring momentum from constantvelocity momentum carriers, there is a limit. This willmodel lots of things, including your vehicle (the carriersare actual ping-pong balls), a balloon (air molecules),a solar sail (photons), a boat pushed by a water jet.Nobody is denying that all of these things have theHowever, there are lots of things that this model doesnot apply to. YOU have made the claim that it appliesas opposed to real ping-pong balls.Explain to me how this model applies to an attractive electric field.What are the ping-pong balls and where are they coming from?Do you really think nobody has noticed that you'veavoided answering any questions about your model? - Randy===Subject: Re: Who contributed most to mathematics?Don't forget about the Lichtensteiniens!Lurch> I am a patriotic American who does not cheat on his taxes,does not> drive drunk, does not carry a gun or knife, obeys traffic laws,> obeys the rights of others, is courteous to other people,> How about having the courtesy not to post this dreck to sci.math?> -- > Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)===Subject: Re: Uncle Al is Sadistic .>The genetic inheritance of mental efficiency is NOT obvious.> It is.>It is very hard to isolate genetics from upbringing> Only if you suffer amnesia of your life from 0-2, so as not to> know first-hand precisely what came from where, when and how -- a rather> extraordinary deficit, to say the least; but certainly one I don't share.> It's not a question that needs to be surmised any more than the question> does of what I ate yesterday for breakfast. I was there when it happened.Be shocked to recognize then how many people haveamnesia from age 0 through 5.===Subject: Re: A 3rd Grade Word Problem---HELP> My 3rd grade son brought this word problem home the other day and he was> given the answer (127). His job, for extra credit, was to figure out how to> get 127. I'm no genius but not a dope either. I couldn't figure out how to> get 127. Nobody in the whole neighborhood could figure out how to get 127.> Is this something of a trick question or is there something in the wording> that I am missing? Any Help????> The Problem:> You know a very good story. On Sunday you tell the story to a friend. On> Monday you tell it to two new people. (So far, a total of three people have> heard the story). Each day after Monday, you double the number of new> people you tell the story to. What will be the total number of people that> will have heard your story after you tell it on Thursday?My guess:1+2+4+8+16+32+64=127and the last word Thursday should have been Saturday===Subject: Re: Who contributed most to mathematics? grava .88 la saucisse et au marteau:> Euler and Gauss.Wasn't Euler from Switzerland? They also have Hilbert and Cantor who didsome interesting stuff :)Nicolas===Subject: Re: Probability of a Run>As far as I can see, if q is the probabilty of a loss and P(n,m) is>the probability of at least 1 run of at least length n in m trials>then>P(n,m) = 0 if m < n,> = q^n.(1 + (1-q).( (m-n) - sum(i=n..m-n-1, P(n,i) ) ).>The logic is simple. Either the first n trials are losses or the first>trial is a success and then we have n losses or we have no runs of n>or more in i trials followed by a success and then n losses (for i =>1.. m-n-1). I think this enumerates all the possibile combinations>without duplication.I apologize for having missed your logic at the bottom of the messageat first reading. I shall take some time to study it in detail. Inthe interim I have been working with Burnside's and Uspensky'srecurrence equation, which is difficult to solve algebraically becauseit is of high order. It may be written asu(m) = u(m - 1) + (1 - u(m - n - 1)) (1 - q) q^nwhich would be better suited to computing u(m) from the top down.function run(m, n, q);/* Returns the probability of a run of at least n *//* consecutive failures in m independent trials where *//* the probability of failure at any one trial is q */integer m, n; float q; if m < n then return 0; else if m = n then return q^n; else return run(m - 1, n, q) + (1 - run(m - n - 1, n, q)) * (1 -q) * q^n; end if;end run;This is merely an outline in PL/I. If implemented, one must use aprogramming language that supports recursive function calls. One alsoshould write so that q^n and 1 - q do not have to be re-computed witheach call.This makes it look simple but forces the computer to compute in aninefficient manner. For fast execution it would be better to take moretime and trouble and program the calculation from the bottom up. Sincethe logic of the recursive function is so simple it is easy toconvince oneself that it is correct. Also it has the mellowness of 300years of aging. Sample outputs could be used to verify a moreefficient but logically more complex program or a formula such asyours.===Subject: Re: JSH: It begins> Now that I have a proof that's simple enough to explain to kids, I can> move back to email.> Yup, you're leaving other mathematicians open to a major sandbagging> as I go back to a process that's netted me Andrew Granville, Barry> Mazur, and other notables over the years.Hey James, I have a challenge that's right up youralley:http://www.nanowrimo.org/index.php?s=250000 words by midnight, Nov 30. And there's no requirementthat they be GOOD words either. I'll bet you could do thatjust with your FLT posts. - Randy===Subject: Re: JSH: It begins> jstevh@msn.com (James Harris) pushed briefly to the front of the queue> (snip)> ^ The process has already begun.> Thank God for that. Now do us non-professionals in here a favour and> please shut the fu**k up until the process produces unequivocal> output.> AndyThat's not how it works. I may be posting more on sci.math goingforward.The issue is, how do I break through the recalcitrance of a group ofpeople desperate to hide from the truth?It may be necessary to pump up the volume.James Harrishttp://mathforprofit.blogspot.com/===Subject: Re: Riemann's Zeta Function by H. M. EdwardsIt seems to me that proving the functional equation for the Riemann Zetafunction is easier without using complex analysis at all. The proof thatI referenced earlier basically does it as follows: Let Psi(t) = sum from n=1 to infinity of exp(-pi n^2 t). Let R(s) = Integral from t=0 to infinity of Psi(t) t^{s/2} dt/t Let Q(s) = 1/s + Integral from t=0 to 1 of Psi(t) t^{s/2} dt/tThen you can prove (using Fourier transforms) that R(s) = Q(s) + Q(1-s)It immediately follows that R(s) = R(1-s). On the other hand, integratingthe power series for Psi(t) term by term gives R(s) = sum from n=1 to infinity of pi^{-s/2} n^{-s} Gamma(s/2) = pi^{-s/2) Zeta(s) Gamma(s/2)So from R(s) = R(1-s), you get pi^{-s/2) Zeta(s) Gamma(s/2) = pi^{-(1-s)/2) Zeta(1-s) Gamma((1-s)/2)Of course, now that I think about it, this proof is a littlesuspicious, because it uses a definition of the Zeta functionthat is only valid for s > 1 (the power series representation),while the result necessarily involves Zeta(s) for s < 1.===Subject: Re: A NOTy problem.> This post not CC'd by email>I do a lot of these for fun. A grid helps to record the initial facts,>but almost always I end up abandoning the grid halfway through and>then writing down and eliminating trial solutions. And you eliminate>them with what you call reductio ad absurdum (really just arriving at>a contradiction from some starting hypothesis; therefore the>hypothesis is false).> - Randy> G'day G'day Randy, > Is there is a better name for this process. > ... arriving at a contradiction from some starting hypothesis;> therefore concluding the hypothesis is false ... To me, it's proof by contradiction. I'm no logician, butwhen I read reductio ad absurdum I interpret it literallyfrom my schoolboy latin as reduction to the absurd. Andto me that says you take an extreme example of somethingand show how you get an impossible conclusion. Mostproofs by contradiction don't involve an extremeexampleHowever, your usage may be more correct than mine.> I wasn't implying the process was structurally the same as the one> Fermat used which had a name that included a phrase like infinite> descent.Yes, I think that was something like mathematicalinduction, but in a descending direction. http://www.mathpages.com/home/kmath288.htm> Are there two different processes, reductio ad absurdum and> reductio adsurdum;I'd assumed that reductio absurdum was merely a typofor reductio ad absurdum> the former involving an infinite series and the> other a finite one? No, no, no.Here's a page that suggests reductio ad absurdum isjust another name for proof by contradiction as I useit.http://www.msu.edu/~wolfalli/logic-handout.htm - Randy===Subject: Re: A NOTy problem.This post not CC'd by email>1. 's and Bob's statements are contradictory to each other.G'day G'day , Does are contradictory mean can't both be true. Yesterday Ithought I knew what a contradiction was. Today I'm not so sure. says, I didn't do it Bob says, did it Would seem to be a direct contradiction. While I am struggling forthe terminology I sense a certain symmetry in the contradiction. says, I didn't do it Bob says, lies. I can see the above pair of statements can't both be true. There also seems to an asymmetry about it. It seems to me they areincompatible statements ... My apologies for struggling over such simple things. Best wishes, >2. Hence, one of them tells true.>3. Only one from four tells true. Hence, Dave and Chris lie.>4. Chris lies, so Bob say true. So did it.>The main conclusion is the 1-st.Quentin Grady ^ ^ /New Zed, >#,#< [ / / ... and the blind dog was leading. http://homepages.paradise.net.nz/quentin===Subject: Re: Usenet Posting Guide?Originator: rusin@cs.niu.edu (David Rusin)>I prefer 'tin' but remember 'rn' (and 'readnews') fondly and could use it>if necessary. Graphical interfaces do have their uses; for instance, the>X Window System lets me open multiple xterm windows to run command-line>clients like tin (or rn).>What benefit do you get by having two newsreader windows open at once?I get a kick out of trying to beat myself to the punch!dave===Subject: Re: Key Core Error Argument> The following proof steps through a rather basic argument which is key> in proving an over one hundred year old error resulting from> previously unexpected consequences resulting from the definition of> the ring of algebraic integers.> Note that ultimately the proof relies on 22 NOT having 7 as a factor,> and constant terms like 7 and 22, being constant, and not variables> dependent on x, which may seem like odd things to emphasize, but I've> faced posters who've gotten away with challenging those truths because> people seem unaware that's what they're doing. Let's just cut to the heart of it. Below I am going to askone real simple question. One of the factors of your polynomial P(x) is 5*b3 + 22. Here I think we all agree that b3 is a function of x, and that b3(0) = 0. We all agree that 7 is coprime to 22. We all know that you can have two numbers, r and s for example,each of which is coprime to 7, but their sum is not coprime to 7:for example, r + s = 3 + 4. Now consider 5*b3(1) + 22. Again, 22 is coprime to 7. What you need is that the whole expression is coprime to 7. You don't know what b3(1) is. It can be computed, but it is a mess. Again, we do know the constant term is 22, and 22 is coprimeto 7. It is constant, and no one is saying here that it changesin any way. Now let r = 5*b3(1) and s = 22. Thus s is coprime to 7. We don't know about r. So now tell us why r + s is coprime to 7. Nora B.===Subject: Re: Usenet Posting Guide?>I prefer 'tin' but remember 'rn' (and 'readnews') fondly and could use it>if necessary. Graphical interfaces do have their uses; for instance, the>X Window System lets me open multiple xterm windows to run command-line>clients like tin (or rn).>What benefit do you get by having two newsreader windows open at once?It's fun trying to be able to respond faster than myself!dave===Subject: Re: Math dependency logic> Ultimately my argument relies on numbers like 7 being NUMBERS, not> variables dependent on x, and on the distributive property.> 7 is a number, but so is x. Neither can change (in a given context);> each refers to exactly one number at a time.> What does time have to do with it?> How do you define time with mathematical argument Bushnell, BSG?> Good grief.Time in that sentence refers to scope, it's a repetition of the in> a given context paretheses. This is English usage. I refer you to > In any given context, x refers to exactly one number (that is, x> refers to the number x), never more than one. It certainly doesn't> change. It doesn't vary> For example, in the sentence x = 2, x refers to 2. In the> sentence x - 2 = 0, the symbol x again refers to 2. In both> sentences, x refers to exactly one number, and doesn't vary or> change at all.I've clipped the rest as I get the gist of your idea, which is bogus.Let f(x) = x^2, so x is referring to a *set* of values, which is howalgebra gets its power.If every time a variable were used it had to have a value, like x=2,then there wouldn't really be a need for algebra, now would there?It's easy to use English as well to give the idea, so there's lesslikelihood that you'll stay confused Bushnell, BSG.A human being has a brain. That's a set characteristic which while referring to a single humanbeing, is understood to carry over to many more human beings.Here the variable is a human beingYou might say f(human being) = has a brain.Now your idea would be like saying that Bob, one particular humanbeing, has a brain. Which gives the idea that *every* single humanbeing needs to be listed out to say they have a brain! So nextthere's Susie, and oh yeah, don't forget non-English people either,like Singh. While you are going checking every human being for abrain, the rest of the world can just accept that a variable meansthat you get wide coverage.Sets are quite powerful and important in algebra, and it might helpyou a lot if you try to understand them Bushnell, BSG.James Harris===Subject: Re: Riemann's Zeta Function by H. M. Edwards> No Real Mathematician thinks of the logarithm having a basis:-)> Oops I meant base :-(> No Real Mathematician thinks of the logarithm having a base:-)Hmm OK. What do you call 'e' then? :-)===Subject: Re: JSH: It begins> jstevh@msn.com (James Harris) pushed briefly to the front of the queue> (snip)> ^ The process has already begun.> Thank God for that. Now do us non-professionals in here a favour and> please shut the fu**k up until the process produces unequivocal> output.> Andy> That's not how it works.Correction: That's not how *you* work.> I may be posting more on sci.math going> forward.Which means: littering the internet with more crap intended to proclaim your own superiority over peoplewho are better than you.> The issue is, how do I break through the recalcitrance of a group of> people desperate to hide from the truth?That is not and never was the issue. The issue is whether your proof and the supporting arguments aretrue or false. You will succeed in breaking through whenever you post something that bears up under closescrutiny.> It may be necessary to pump up the volume.Ha! The chickens come home to roost. Your answer to every refutation and counter-example, is to pump upthe volume. Is 'volume' the tool of truth?> James Harris> http://mathforprofit.blogspot.com/There are two things you must never attempt to prove: the unprovable -- and the obvious.Democracy: The triumph of popularity over principle.http://www.crbond.com===Subject: Re: Key core error argument, stepped outIn sci.math, Arturo Magidin:> [.snip.]>> Using Dot's proof that the values of a polynomial with algebraic>> integer coefficients are always divisible by an integer if and only if>> each coefficient is a multiple of that integer (in the algebraic>> integers) gives you that the statement is correct for polynomials in>> A[x]. I had not noticed Dot's theorem. The x^2 + x example seems to belie>it: for any integer x, x^2 + x is divisible by 2, but the coefficients >are not multiples of 2, and both of the coefficients are algebraic>integers. > But the values of x^2+x are not multiples of 2 for every algebraic> integer value of x: if x=i, then you i-1, which is not a multiple of> 2. y = (i-1)/2y' = (-i-1)/2Equation satisfied: y^2 + y + 1/2 = 0.: y is not an algebraic integer and (i-1) is not divisible by 2.Interesting. I'm now wondering what assumption(s) break(s) downbetween integers and algebraic integers, as x^2 + x is a multipleof 2 for any integer x, but not for any algebraic integer.[rest snipped]#191, ewill3@earthlink.netIt's still legal to go .sigless.===Subject: Re: Campbell's theorem for filtered point processes> I am looking for an english reference for the Campbell theorem for> stationary filtered point processes (not only Poisson). What Campbell theorem do you have in mind?In any event, a likely place to look is An Introduction to the Theory of Point Processes by Daley & Vere-Jones.> I have only a reference in> Koenig/Schmidt: Einfuehrung in Punktprozesse,> have a reference which is easily accessible everywhere.A.===Subject: Re: Key core error argument, stepped out Adjunct Assistant Professor at the University of Montana.>In sci.math, Arturo Magidin>:> [.snip.]> Using Dot's proof that the values of a polynomial with algebraic> integer coefficients are always divisible by an integer if and only if> each coefficient is a multiple of that integer (in the algebraic> integers) gives you that the statement is correct for polynomials in> A[x]. >> I had not noticed Dot's theorem. The x^2 + x example seems to belie>>it: for any integer x, x^2 + x is divisible by 2, but the coefficients >>are not multiples of 2, and both of the coefficients are algebraic>>integers. >But the values of x^2+x are not multiples of 2 for every algebraic>integer value of x: if x=i, then you i-1, which is not a multiple of>2. >y = (i-1)/2>y' = (-i-1)/2>Equation satisfied: y^2 + y + 1/2 = 0>.: y is not an algebraic integer and (i-1) is not divisible by 2.Easier: 1-i is a FACTOR of 2: 2 = (1-i)(1+i). Moreover neither 1+i nor1-i are units: for the minimal polynomial of either is x^2 - 2x +2. If a|b and b|a, then a and b would be associates, and that wouldmean that 1+i is a unit, which is false.>Interesting. I'm now wondering what assumption(s) break(s) down>between integers and algebraic integers, as x^2 + x is a multiple>of 2 for any integer x, but not for any algebraic integer.Easy: 2 is not a prime, and the index of (2) in the ring is largerthan 2. If q is a prime, then either q divides a or q is coprime toa. In the case of 2, moreover, since [Z:2Z] = 2, the sum of twoelements which are coprime is divisibly by 2. You use this to showthat all values of x^2+x are multiples of 2: if they are both in (2),then clearly so is their sum; if x is not in (2), then neither is x^2(since Z/2Z is a field of 2 elements), and their sum is therefore in(2). ============================================================== ========Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan======================================================== =================Subject: Re: cardinality of monotone set-functions> If f is monotone, then for each x in X the set> G(x) = {A in 2^X: x in f(A)} is monotone, i.e. if A is in G(x) and> A subset B then B is in G(x). > Another small point: I thought a collection U of subsets of X such that> A in U and A subset B ==> B in U was called an upset or filter> Are they also called monotone sets? Of course they are the supports> of monotone functions f: 2^X ->{0,1}. --EdwinUpset, yes. Filter AFAIK also requires being closed under intersections.I didn't mean to imply that monotone was standard terminology there; infact I was in a bit of a hurry and didn't bother to search for the correctword, but thought monotone would be reasonably understandable in thiscontext. Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2===Subject: Re: A Question for James Harris>But my analysis has led me to wonder how many professional>mathematicians bother to read sci.math at all.>James HarrisThere is a searchable on-line database of mathematicians,_The Mathematics Genealogy Project_, and the homepage is at http://genealogy.math.ndsu.nodak.edu/David Bernier===Subject: Re: Who contributed most to mathematics?> Wasn't Euler from Switzerland?Also claimed by Russia.===Subject: Which side of 2D line is Point OnAny help with the following would be appreciated.I have a set of points in the plane and want to sort them intocyclical order (say anti-clockwise) about another given point P0.The naive way would to have a function F1(point PN) which takes oneargument - a point PN. The function makes the vector from P0 to PN andreturns the angle this vector makes with the positive x axis.The algorithm would have a list of cyclically ordered points about P0which is initially just one point (any point from the set of points).Then for all other points (PNi) in the set call F1(PNi) to determinewhere in the ordered list PNi belongs (by looping though all points inthe list until I come to the first one which makes a greater anglewith the positive x axis than PNi). If there is no fuch point, thenPNi belongs on the end of the list.The result is a list of points cyclically ordered about P0.I've implemented this, and it works fine.However, F1 can be greatly improved by not calculating an actualangle. F1 could simply determine whether a given point lies to the'right' of each line defined by P0 and the next point from the list,if it does then it should be inserted here. If it is not to the rightof any then it should be inserted on the end of the list.So F1 should be a function which take three points Po,P1,P2 andreturns whether p2 is on the left or right or on the infinite linedefined in the sense Po to P1.I am using this function to determin this:(P1.x - Po.x)*(P2.y - Po.y) - (P2.x - Po.x)*(P1.y - Po.y)It returns < 0.0, > 0.0 or 0.0 depending on the side P2 is from theline defined in the sense Po p1Well, this works if P2 is entirely outside of the point set but not ifP2 is interior to the set.Is this the correct function, will it always return >0.0 if P2 is onthe left of Po P1. If so I should expect it to work if P2 is interiorto the point set..===Subject: Re: Math dependency logic permission for an emailed response.> Let f(x) = x^2, so x is referring to a *set* of values, which is how> algebra gets its power.No. If f is the function R->R such that f(x) = x^2, then the *range*of the function is a set of values. (R, in this case.) The range ofthe function is a property of the function, and not a property of x,which is a symbol.> If every time a variable were used it had to have a value, like x=2,> then there wouldn't really be a need for algebra, now would there?The power of variables is quantification, but the whole point ofquantification is to make the variables go away. > It's easy to use English as well to give the idea, so there's less> likelihood that you'll stay confused Bushnell, BSG.The problem is that English is ambiguous.> A human being has a brain. This is an ambiguous sentence. It could mean that every human beinghas a brain, or it could mean that there exists at least one humanbeing that has a brain. In neither case is there any specific humanbeing under consideration, and there is no real variable, only anapparent variable:for all x, if x is a human being, then x has a brainthere exists an x such that x is a human being and x is has a brainThose are two possible meanings for your sentence, but the variablex is only apparent, not real. > Sets are quite powerful and important in algebra, and it might help> you a lot if you try to understand them Bushnell, BSG.I do understand them you arrogant twit.===Subject: CGAL 3.0 Released, Computational Geometry Algorithms LibraryWe are pleased to announce the release 3.0 of CGAL, the Computational GeometryAlgorithms Library. Version 3.0 differs from version 2.4 in licensing, in theplatforms that are supported and in functionality.The license has been changed to either the LGPL (GNU Lesser General PublicLicense v2.1) or the QPL (Q Public License v1.0) depending on each package.So CGAL remains free of use for you, if your usage meets the criteria ofthese licenses, otherwise, a commercial license has to be purchased fromGeometry Factory (www.geometryfactory.com).Major changes in this release include the following:o Apollonius graph: the dual of the Voronoi diagram of a set of circles under the Euclidean metric. The implementation is dynamic.o Min_sphere_of_spheres_d: Algorithms to compute the smallest enclosing sphere of a given set of spheres in d-dimensional space.o Spatial Searching: Provides exact and approximate distance browsing in a set of points in d-dimensional space (such as nearest neighbor searching).o Largest_empty_iso_rectangle_2: Given a set of points P in the plane, computes the largest empty iso-rectangle that are inside a given iso-rectangle bounding box, and that do not contain any point of P.o Interval_skip_list: A data strucure for finding all intervals in R that contain a value, and for stabbing queries, that is for answering the question whether a given value is contained in an interval or not.o Existing packages have been improved in various area: 2D and 3D triangulations, Planar Maps, Arrangements...o The CORE library (http://www.cs.nyu.edu/exact/core/) for exact computations is now distributed as part of CGAL as well.o We support the latest versions of the C++ compilers from GNU, Microsoft, Intel, Sun, SGI.o All demos are now using the portable Qt window toolkit.See http://www.cgal.org/releases_frame.html for a complete list of changes.The CGAL project is a collaborative effort to develop a robust,easy-to-use, and efficient C++ software library of geometric datastructures and algorithms. The CGAL library contains:o Basic geometric primitives such as points, vectors, lines, predicates for testing things such as relative positions of points, and operations such as intersections and distance calculation.o A collection of standard data structures and geometric algorithms, such as convex hull, (Delaunay, Regular, Constrained) triangulation, Voronoi diagrams, planar map, arrangements, polyhedron, smallest enclosing sphere, multidimensional query structures...o Interfaces to other packages, e.g. for visualization, and I/O, and other support facilities.For further information and for downloading the library and itsdocumentation, please visit the CGAL web page: http://www.cgal.org/===Subject: Re: naive geometry questions| at 08:00 PM, Elaine Jackson said:|| >1) Are there surfaces that can be bent to each other (the way a flat| >piece of paper is bent to a cylinder) but not to anything flat?|| What do you mean by surface and what do you mean by bent to each| other? Are you limiting your question to surfaces in Euclidean| 3-space?YES|| >2) Besides planes and spheres, is there any other surface S such| >that a piece of S can be moved around adlibitum while each of its| >points remains in contact with S?|| If I understand your question, the cylinder and the torus are| examples.|rotate it as well.| --| Shmuel (Seymour J.) Metz, SysProg and JOAT|| Unsolicited bulk E-mail will be subject to legal action. I reserve| the right to publicly post or ridicule any abusive E-mail.|| Reply to domain Patriot dot net user shmuel+news to contact me. Do| not reply to spamtrap@library.lspace.org|===Subject: Number theory question about sums of primesI have no idea whether this is a trivial or extremely difficult numbertheory problem. Amazingly enough it came up in real life. Sortof--I'm writing a computer book and while demonstrating a certain kindthis question. So here it is: If P(n) is nth prime number, does there exist an n such that (P(1) + P(2) + ... + P(n-1)) < P(n)This came up because I ended up with a loop something like this andcouldn't figure out whether was infinite or not: while (true) sum = 5 then (sum + p) p = 5 then next_prime(1 + p) if (p > sum) returnwhere next_prime returns the next prime number greater or equal to itsargument. (And the `x = foo then bar' means it has the value 'foo' onthe first iteration and on subsequent iterations gets its value fromevaluating `bar'.)It seems likely that this is an infinite loop since we keep adding tosum, moving it farther and farther out of reach. However, I know thatthere can be arbitrarily large gaps between prime numbers[1] soperhaps its possible for a really big gap to come along at the righttime and let p get ahead of sum and terminate the loop.Like I said, I have no idea if this is a trivial or unsolvableproblem. I haven't been that much into math since my freshman vectorcalc/linear algebra course convinced me I didn't really want to be amath major.-Peter[1] For any n greater than 1 consider the sequence of n numbers n!+2, n!+3, ... n!+nThe first is divisible by 2, the second by 3, up to n!+n is divisibleby n. Thus we have a sequence of n-1 composite numbers. Since n > 1,there is a prime number p < n!+2 and the next prime has to be greaterthan n!+n, thus the gap is > n-1. To get a larger gap, pick a largern.Peter Seibel peter@javamonkey.com Lisp is the red pill. -- John Fraser, comp.lang.lisp===Subject: Re: cardinality of monotone set-functionsContent-Length: 622Originator: rusin@vesuvius> Let X be a finite set. A set-function on X is a> function f: 2^X rightarrow 2^X. f is monotone> if A subset B implies f(A) subset f(B).> What is the cardinality of the set of monotone> set-functions on X ?> (Actually, I'm looking for the cardinality of> the set functions such that f(A)subset A for all A.)thanks for the answers. Thought I had a strategy forfinding the cardinality of monotone set-functions. Looks a lot less promising now...One clarification about f(A)subset A: I meant thecardinality of the monotone set-valued f such thatf(A)subset A.Federico===Subject: Re: A Question for James HarrisX-ID: Xo5B0BZbreGEF8NBEM0Thh1Bjq-Qlj4pDLQL8IsBzukoKmiprPkCZA> But my analysis has led me to wonder how many professional> mathematicians bother to read sci.math at all.I don't think that it`s surprising that most mathematicians avoid newsgroupswhere people like James Harris post the same over and over again.reverse my forename for mail!===Subject: One homomorphism questionHi The commutator subgroup G' of G is the subgroup generated by allelements of the form a^{-1}b^{-1}ab, with a in G, b in G.If H subset G is any subgroup, show that if G' subset Hthen H is a normal subgroup in G and G/H is abelianHere is my proof first try to prove H is normal to GFor any h1 in H for any g in Gso gh1g^{-1} = gh1g^{-1}h1^{-1} h1because G' is a subgroup of H soh2=gh1g^{-1}h1so gh1g^{-1} = gh1g^{-1}h1^{-1} h1 =h2h1 in Hso H is a normal subgroup of GBut I don't know how to prove G/H is abelianLet Ha is one element of G/H, let Hb is another element of G/HHaHb ...Thank you very much!===Subject: Re: Number theory question about sums of primesPeter Seibel grava .88 la saucisse et au marteau:> If P(n) is nth prime number, does there exist an n such that> (P(1) + P(2) + ... + P(n-1)) < P(n)For each n, there exists a prime number between n and 2n (strictly).That's Bertrand's postulate.Therefore, P(n-1) > [P(n)-1]/2P(n-1) >= [P(n)+1]/2 > P(n)/2P(n-2) >= [P(n-1)+1]/2 >= [P(n)+3]/4 > P(n)/4Recursively, you find:P(1) + ... + P(n-1) > P(n)[1/2 + 1/4 + ... + 2^(1-n)] = P(n)(1-2^(1-n))Or P(n) < 2^(n-1), so P(n)2^(1-n) <1Therefore P(1) + ... + P(n-1) >= P(n)There is probably a simpler proof, but that's the first one I thought ofNicolas===Subject: Re: Sets before logic permission for an emailed response.X-Zippy-Says: Just imagine you're entering a state-of-the-art CAR WASH!!> Random thoughts on creating a theory of sets prior to a theory of> propositions and quantifiers:> Let's start with the empty set, 0, and logical identity, =, then we can> define T, for true, by> T =def 0 = 0> Let's define ordered pair a la Kuratowski, then we can define> conjunction byKuratowksy defines as {a,{a,b}}. But how do you make sense ofthat latter notation at this stage of the presentation? Note that inZF, you can only make sense of it because of the Axiom of Pairing; youcan only verify that it satisfies the ordered pair axiom because ofthe Axiom of Extensionality. Those axioms both seem to require theapparatus of first order logic to be formulated. So how do youpropose to make sense of {a,{a,b}} without quantifiers?It's not that this seems like a waste of time; maybe it's interesting.But what I'd want to see is a formal system (in the way that we canwrite formal systems for first order logic), in which you show me howderivations in your set theory are going to proceed.===Subject: Re: Math dependency logic permission for an emailed response.> Let f(x) = x^2, so x is referring to a *set* of values, which is how> algebra gets its power.> No. If f is the function R->R such that f(x) = x^2, then the *range*> of the function is a set of values. (R, in this case.) The range of> the function is a property of the function, and not a property of x,> which is a symbol.Erp, typo. I meant the *domain* of the function. (Of course, therange is also a set, but in the context, it is the domain of thefunction that is under discussion.)===Subject: help me, please!I'm not very good in math!A question:Is there a fast way to determine X and Y both integer such that:X^2 - Y^2 = AThank you very much!Marco===Subject: Re: help me, please! permission for an emailed response.X-Zippy-Says: Are you mentally here at Pizza Hut??> Is there a fast way to determine X and Y both integer such that:> X^2 - Y^2 = AYes.===Subject: sum of random variablesLet X1 and X2 be two known random variables.Let X1 denote the distribution of the number of channels allocated in interval T1. Let X2 denote the distribution of the number of channels allocated in interval T2.Let X be the distribution of the number of channels allocatedin steady state. May I state that X = X1 + X2? If not, whatassumptions should I have in order to state this? Is there anyformal proof for that?Carlo.===Subject: Re: help me, please! Bushnell, BSG ha scritto nel messaggio> Is there a fast way to determine X and Y both integer such that:> X^2 - Y^2 = A> Yes.???? Are you joking?What is the answer?Thank you===Subject: Re: Jaynes' book on probability - thoughts?> This url will connect you with a thread of discussion on Sci.stat.math> ng which was an interesting and lengthy debate in part between Hardy> and me of one aspect Bayesian statistics. Based on your comment about> Hardy's review--I got to Amazon to read his review for myself as fast> as I could. > http://home.rochester.rr.com/jbxroads/interests/sci.stat.math/ > John Bailey> http://home.rochester.rr.com/jbxroads/mailto.html===Subject: Re: Key Core Error Argument> Start with> g(x) = ( a(x) + f )> where we know nothing about a(x). > Why?Because I defined a(x) to be an arbitrary function. The pointbeing that we need to know something more about a(x) to reachconclusions about divisibility. The rest of the post demonstratedwhat we need to know about a(x), and why we cannot gainthis knowledge from a non-polynomial factorization.> You *wish* to believe that the varying function drives the> factorization I use, when I can prove that it does not, but you want> me and other readers to go off on some wild goose chase because you're> irrational?> If you think there's a problem in the proof, point at a step.I did. Others have. You ignore.I suspect you stopped reading my post after the first coupleof lines.> A proof begins with a truth and proceeds by logical steps to a> conclusion which then MUST be true.> You need go no further than the presented proof, where you do know> about the a's.The a's you use are non-polynomial parts of non-polynomial factorizations.Your conclusions about divisibility are not supportable. - William Hughes===Subject: Re: Probability of a RunThis is to illustrate my approach to the problem with a specificexample. Let us say you toss a coin 4 times. What is the probabilitythat the longest run of tails will be 2? For a single throw theprobability is p, and q=1-p. The value to be found is maxprob(4).Streakprob(3,2,1) is the probability of having the situation 3 throws,maximum run of 2 and current streak 1. Here goes:Part 1:Maxprob(4,2)=streakprob3,1,1).p+streakprob(3,2,2).q+ streakprob(3,2,0)+streakprob(3,2,1)Streakprob(3,1,1)= streakprob(2,0,0).p+streakprob(2,1,0).pStreakprob(3,2,2)= streakprob(2,1,1).p+streakprob(2,2,1).pStreakprob(3,2,0)= maxprob(2,2).qStreakprob(3,2,1)=streakprob(2,2,0).pStreakprob( 2,0,0)=maxprob(1,0).qStreakprob(2,1,0)=maxprob(1,1). qStreakprob(2,1,1)=streakprob(1,0,0).p+streakprob(1,1,0). pStreakprob(2,2,1)=0Maxprob(2,2)=p^2Streakprob(2,2,0)=0Maxprob (1,0)=qMaxprob(1,1)=pStreakprob(1,0,0)=qStreakprob(1,1,0)= 0Part 2:Streakprob(2,1,1)=q.p+0.p=pqStreakprob(2,1,1)=q.p+0.p= pqStreakprob(2,1,0)=p.q=pqStreakprob(2,0,0)=q.q=q^2Streakprob( 3,2,1)=0.p=0Streakprob(3,2,0)=p^2.qStreakprob(3,2,2)=pq.p+0.p= p^2.qStreakprob(3,1,1)=q^2.p+pq.p=(p^2)q+p(q^2)Maxprob(4,2)=[ q^2.p+pq.p=(p^2)q+p(q^2)].p+p^2.q.q+p^q.+0I have written a program (function) for Visual Basic for Excel basedon this. However, I found that this forward and backward approachtakes up too much memory. For maxprob(100,5) my program finds all thevalues from maxprob(0,0) to maxprob(100,5) and streakprob(0,0,0) tostreakprob(100,5,5). Here is the code:Function maxprob(num_throws, maxval, prob)Dim mprob(200, 6)Dim streakprob(200, 6, 6)throw = num_throwsMax = maxvalp = probq = 1 - pmprob(0, 0) = 1streakprob(0, 0, 0) = 1For i = 0 To throwIf i > 0 Thenmprob(i, 0) = q ^ istreakprob(i, 0, 0) = q ^ iEnd IfFor j = 0 To MaxIf i = j And j > 0 Thenmprob(i, j) = p ^ jEnd IfIf i > j And j > 0 Thenstreakprob(i, j, 0) = mprob(i - 1, j) * qEnd IfIf i < j Thenmprob(i, j) = 0End IfIf i > j And j > 0 Thenmprob(i, j) = streakprob(i - 1, j - 1, j - 1) * p + streakprob(i - 1,j, j) * qm = 0While m < jmprob(i, j) = mprob(i, j) + streakprob(i - 1, j, m)m = m + 1WendEnd IfFor k = 0 To MaxIf i < j Or j < k Thenstreakprob(i, j, k) = 0End IfIf i = j And j = k And k > 0 Thenstreakprob(i, j, k) = p ^ iEnd IfIf i = j And j > 0 And k < i Thenstreakprob(i, j, k) = 0End IfIf i > j And j = k And k > 0 Thenstreakprob(i, j, k) = streakprob(i - 1, j - 1, j - 1) * p +streakprob(i - 1, j, j - 1) * pEnd IfIf i > j And j > k And k > 0 Thenstreakprob(i, j, k) = streakprob(i - 1, j, k - 1) * pEnd IfNext kNext jNext imaxprob = mprob(throw, Max)End Function===Subject: Re: Key Core Error Argument> If I understand your terminology> correctly (you still have *not* defined the concept constant term> as you use it), the constant terms of the three factors are 1, 1 and 22.> So despite all your efforts Dik Winter, you end up with the same> result: constant terms for the factors that are 1, 1, and 22.> Boy that's telling him!> -William Hughes> Actually it is, as Dik Winter is trying to find a way that 7, 7 and 22> become 1, 1 and 22 based on a *varying* x.> That is, he's trying to make the change in dependent on a> variable.No actually, he pointed out that it is possible to have more thanone factorization with the same constant terms (as this is equivalentto having more than one function f with f(0)=c, this is notsurprising). - William Hughes===Subject: Re: naive geometry questionsIt *sounds* extremely cool but, being spatially challenged, I can't follow ityet. I'm still trying to figure out what the surfaces look like. Is thisright?...HELICOID:plots[cylinderplot]([r*cos(theta),r*sin( theta),5*theta],theta=0..2*Pi,r=0..5,coords=cylindrical); CATENOID: plots[implicitplot3d]((y^2+z^2)^(1/2)=(1/2)*(exp(x)+exp(-x) ),x=-2...2,y=-2..2,z=-2..2);| > Here are two geometry questions from a curious layperson.|| I think these are wonderful questions; such a curious layperson is| welcome into my classes any time! Let met take them in opposite order.|| > 2) Besides planes and spheres, is there any other surface S such that| > a piece of S can be moved around adlibitum while each of its points| > remains in contact with S?|| What is meant by a surface is not always clear, but let me| assume that S is what would technically be called a smooth| 2-dimensional submanifold of R^3. For piece of S I will take an| open neighborhood around a point P.|| If you zoom in very near to P, you will see that this piece of S| looks flat, that is, there is a well-defined tangent plane to S at| the point P, which approximates the shape of S very well| near P. (The existence of such a plane is the definition of| differentiability or smoothness.) For convenience, spin S around| in space so that this plane is horizontal.|| Now zoom out a little bit -- not too far! -- and you will see that the| region around P isn't really flat. It will have hills and valleys| near P, looking very much like the graph of a function f defined| on the horizontal plane. (That's the Implicit Function Theorem at work.)| If you see several hills and complicated valleys, zoom in a bit more.| In a small enough region around P, you can approximate the shape| with the graph of a _quadratic polynomial_ Q instead of f itself.| (That's Taylor's theorem -- the right Q can be computed if you| know the partial derivatives of f at P.) With a bit of high school| geometry you can rotate coordinates on the horizontal plane so that| Q is just a x^2 + b y^2 + c for some constants a, b, c.| (This c is just the height of the point P off the horizontal plane,| so if this plane -- which I have never yet specified! -- is just the| tangent plane mentioned earlier, then c = 0.)|| The constants a and b are pivotal. They measure how markedly the| graph of Q (and so of f, and so of S itself) twists up or down| as you travel along the axes. Moreover, these two directions are the| extremes: along any other line, Q changes more slowly (and if ab<0,| there are even lines of travel where Q stays constant).|| The importance of a and b is that these are invariants of the| shape of S at P. Although they can be computed from various| coordinates or other formulas, they are clearly geometric in nature| and uniquely specify what is known as the _curvature_ of S at P.| (Be careful of that word; it's also used for other quantities which| summarize a and b , e.g. reporting only their sum or product.)|| So your request comes down to this: you want to know what are the| surfaces which have the same curvature (tensor) at every point.| You have provided two important examples: on planes, we have a=b=0 at| each point; on spheres we also have a=b (but nonzero, and constant)| at each point. It turns out there are no other surfaces of this| particular type: if at the point P we have a=b, then P is said| to be an _umbilic point_ of S . If every point is an umbilic point| (with, a priori, possibly different values of a=a(P) at each point)| then it turns out that a is indeed constant, and the surface is| part of a plane or sphere.|| You're forgetting another obvious example: on a cylinder we have| a=0 at each point P, and a nonzero constant value for b.| (Unlike the previous case, here it IS now possible for every point| to be parabolic in this sense but to have b vary with P,| giving generalized cylinders. But they don't have the property| you asked for.)|| Any other examples of surfaces with the property you seek would have| at each point a pair of distinct, nonzero values of a and b| which remain constant on the surface. In particular, their product| would remain constant, which means your surface would be a| surface of constant Gaussian curvature. Their sum is also constant,| making the surface a constant mean curvature space. These are| interesting categories of surfaces, each containing some beautiful| examples. (Soap bubbles stretching across a wire frame form CMC surfaces.)| But it turns out that you can't find any more examples having both| Gaussian and mean curvatures be constant -- not among surfaces| which are immersed in R^3.||| > 1) Are there surfaces that can be bent to each other (the way a flat| > piece of paper is bent to a cylinder) but not to anything flat?|| You're asking for a one-parameter family of local isometries.| I don't know if you intended flat in the technical sense but you've| got it right: flat means Gaussian curvature is zero, i.e. at| each point P we have a=0 in the discussion above.|| (The Moebius strip is flat in this sense too, but maybe isn't what| you had in mind as being flat. You can certainly bend a Moebius| strip into many configurations.)|| Here's a nontrivial example, using what are called _minimal surfaces_.| (These are surfaces of constant mean curvature equal to zero, that is,| at each point we have a = -b.)|| One such surface is the helicoid, formed by taking a double helix| wrapping around a vertical line and then joining points in one helix| with a line segment stretching horizontally to the other helix. They| sell such things as yard ornaments, packaged with all the horizontal| line segments lying in a plane, but the helicoid is not really flat:| the true surface would have some fabric stretching from one horizontal| line to the next, with more fabric near the helices, so that when| you try to lay the thing flat there will be ruffles of fabric which| are very noticeable far from the central vertical line.|| The other surface is the catenoid, formed by rotating a catenary,| e.g. the graph of y=(1/2)( exp(x)+exp(-x) ), around the horizontal axis.| This is really a different surface from the helicoid. But you can| make one from exactly one full turn of the helicoid by forming each| of the helices into a circle; the central vertical line of the helicoid| then also forms a circle (smaller than the other two circles). The| horizontal parallels have been bent into copies of the catenary, and| the fabric in the ruffles smoothes out to give the surface of the| catenoid.|| You can perform this transformation in a such a way that at every| intermediate moment the fabric is fully stretched out, that is,| this is a continuous bending|| Cool, huh?|| dave===Subject: The Elegant Multiverse 1PaulThe MTW argument against Yilmaz is simply that one cannot make alocalclassical vacuum pure gravity stress-energy density tensor from thetorsion-free non tensor connection without violating the equivalenceprinciple.But we have to be very careful about what we mean by the equivalenceprincipleThere is the so-called EEP; there is Einstein's originalequivalence principle; whichcan be viewed as an interpretation of the EEP; and there is thealternative interpretationof the EEP in terms of *inertial compensation* which rejectsEinstein's fundamentalidentification of gravitational and inertial fields.What you need to do is spell out carefully all of these differentversions in formal terms and in clearoperational terms to see if there is really any scientific differenceof any importance.Of course. That's what I'm working on.I'm looking for a tame mathematician who really knows differentialgeometry,since I can't seem to find the theorem I need in the standard sources.The key is decomposing the metric gradients (with respect to thecoordinates)into inertial (transformational) and gravitational (deformational)components.Define this distinction mathematically. Look at Hagen Kleinert's freebooks online.I already have. The gradients of the g_uv at a spacetime point with respectto the x_v are each the sum of two distinct contributions, as a matter ofpure mathematics. These map separately to the inertial and permanentgravitational contributions to the local connection field.In that case please show me the exact math here. I have the books. Right now I do not understandwhat you mean until I see the actual math. Presumably the permanent term is some kind of tidal termthat vanishes if the local curvature tensor vanishes. This will be a very simple matter to decide if youshow me the math. Ordinary words not good enough here. They are too imprecise and ambiguous - goodfor poetry not for physics.The torsion-free connection can probably be split into the homogeneous tensor part + the inhomogeneous part.Indeed I remember that and can quickly look it up. So if that is all you mean, then that is standard Einstein.What then is the point here?I gather deformational means Ruvwl curvature tensor =/= 0 at P ?It means the contribution to the connection field (i.e. metric gradients) that resultsfrom the deformation of the spacetime manifold, where Ruvwl =/= 0.Transformational is when Ruvwl = 0?It's what you have left when you set Ruvwl = 0. It is obviously stillpresent even when Ruvwl =/= 0 in almost all frames.Fine.This is really a matter of pure differential geometry, since it holdsgenerally in theabstract, without any specific reference to spacetime manifolds.I do not understandwhat you mean by inertial compensation above.Just think of how inertial and electrical forces interact and how thisis describedmathematically in *standard GR*. The EM field is not geometrodynamic,while thepure inertial field (no gravitational sources) is.EM field is geometrodynamical in Kaluza-Klein with one extra spacedimension.That's not *standard GR*, JackI think the fictitious or inertial g-force i.e. the metrictorsion-free connection for parallel transport in the original 1915theory is LOCALLY equivalent to agravity forceEinstein's principle was that a uniform gravitational field is*fundamentallyindistinguishable* from an inertial field. That is all in thehistorical record.Depends what one means by fundamentallyIt is quite clear what Einstein meant. He was quite forthcoming on this.I think it's important to recognize that this is the cornerstone of Einstein'sconcept of general relativity. The meaning of Einstein's principle is quiteunambiguous when placed in context.Fine, but nothing wrong with what Einstein said above except that physically we never havea global uniform gravity field. That's only an idealization.Tidal differences in the g-force between P and P + dP means curvaturei.e. real gravity as opposed to simulated gravity.Yes, exactly. That is what is meant by a permanent or intrinsicgravitational field.Fine, so what's the problem?It turns out (and I didn't know this when I first got onto it) that the position Iam advancing here is essentially Eddington's as set out in 1924. At that time heviewed Einstein equivalence as heuristic and tentative in character, and not asa fundamental principle.I say he was right on the money.This is the view that was later echoed by Bergmann, Synge, and others.EEP as technically formulated by Wheeler is fine -- but it is consistent with eitherinterpretation (Einstein's or Eddington's).So it seems your point is one of the history of relativity on how Einstein's ideas evolved.I do not see however how this will lead to new physics today unless you mean thataddition torsion local gauge field and possibly local gauge fields from ALL of theconformal group will modify EEP which is only based upon the 4-parametertranslation group of the 15 parameter Conformal Group for Minkowski space-time.I refer to the paper by Utiyama I think in mid-60's also Kibble's later paper inwhich the general relativity principle is rederived as a kind of local Yang Millsgauge principle in the curved space-time base space of the fiber bundle where thefiber is the U(1)SU(2)SU(3) or its GUT internal symmetry generalization. Roughly, the gauge forces arein the principle bundle with unstable globally flat space-time as a coset space mod the internal symmetrysubgroup of the total group and the lepto-quarks are the associated vector bundle. One thendoes the same trick again inside the coset space whose symmetry group is the Conformal Group.Einstein's. It isinconsistent with Eddington's as stated in Mathematical Theory of RelativityMTW seem to use both, which strikes me as an example of Bohrian complementarity-- AKA speaking with forked tongue -- in the context of spacetime physics.That is a very bold hypothesis, as Feynman pointed out on manyoccasions.But it is at the core of Einstein's concept of general relativity,as Einsteinhimself consistently emphasized (specifically to von Laue).Of course, this was later formulated as a local cancellation of thegravitationalforce field in a free-fall frame, together with our ability to rendertidal effectsnegligible by going to a sufficiently small neighborhood. This makesit lookon the surface like the Einstein principle is still valid -- hence themisleadingacronym EEPForget what Einstein allegedly thought in his early days.Jack, this was integral and indispensable to Einstein's idea of general relativity.So, if you forget about this -- and also Pauli's entire 1921 exposition of GR --you then have general relativity *without actual general relativity*, as I said earlyon.I mean, this is a historical point about the initial immature conception of relativity that gotfurther refined into the modern ideas launched in MTW.Looks like I was right.But if you are willing to drop this and hew to a consistent non-Einsteinian interpretationof EEP, then fine.But then, the MTW argument against localized vacuum gravitational stress-energyfalls apart (though of course that does not in itself mean there are no other valid objections).The only way you can attack MTE there is to say that their initial assumption that the localizedvacuum tensor is quadratic in the connection is a Straw Man, i.e. not complete. That there isanother form based only on the nonlinear products of the curvature tensor but that wouldchange the field equation from (neglecting zero point /zpfguv)Guv = -(8piG/c^4)TuvtoGuv + Nonlinear products of 4th rank curvature tensor = -(8piG/c^4)TuvThe extra terms are a power series in Lp^2 = hG/c^3they may be there, but are too small on the macro-scale.Is that what you mean?I cannot understand really what you mean until I see formal statements in math that pin down the ambiguities of ordinary language.This is why I feel uneasy about Bohr's reliance on ordinary language in the quantum reality problem.Which is what I originally argued: the bare EEP does not establish that thegravitational energy density must vanish at a point, or that its components dependupon the frame of reference in a non-covariant manner. It only requires that thenet measurable combined gravitational and inertial forces vanish at a point in somecoordinate system.To get the true gravitational stress-energy, we simply ignore the inertial contribution-- that's all.Eddington himself explicitly states that in GR the inertial field has a stress-energythat is frame-dependent and represented by a non-tensor. But this is a *fictitious*stress-energy, unless you believe in Einstein equivalence *as originally stated* --which it now appears that you don't.He wasperhaps a bit confused whilst the theory was immature.The point is that it's coherent conceptually today in the form MTW have.I cannot agree. See above. This is a *real interpretive dichotomy* at the most fundamentallevel.Now you could say somethinglike this about Schroedinger, who really thought he had awave theory in the classical sense -- but later had to give this up for a number of reasons.Einstein never explicitly abandoned equivalence and true general relativity. Not even inhis Autobiographical Notes, which is just about his last word on the subject.You cannot render tidal effects negligible in the approach to asingularity until effective Planck scale is reached.That scale may not be 10^19 Gev after all. It may be 1 Gev?That may well be the case, but this is not part of canonical GR. No doubt a deepertheory, if it ever finally comes, will answer such questions.Ed Witten thinks that's M-theory. It's now on NOVA! I am beginning to read Witten. I am getting a little uneasyabout all this duality stuff in string theory because it is beginning to remind me of mydebate with Hal Puthoff on his use of isotropic r vs. curvature r in the SSS model.This debate is described in my Dec 2002 book Space-Time and Beyond IIMy debate with Puthoff was that he was misinterpreting the physics not realizingthat the expansion of space inside a critical radius was an illusion of not realizingwhat the coordinate patch was in the curved differential manifold. Matt Vissermakes this clear in Lorentzian Wormholes in contest of Kruskal coordinates.There is a kind of duality there withr' = r*^2/rbut r' and r are both OUTSIDE the horizon. Puthoff and Ibison make the conceptual error that r' is INSIDE the horizon!I am wondering if Witten, Greene & Co are making a similar conceptual error in that really they ignore the physics INSIDELp*^2 which G.E. Volovik in Universe in a Helium Drop says is essential for solving the smallness of the CosmologicalConstant problem. It's like in condensed matter physics not going inside the unit cell of the crystal lattice. If you do that youwill get a zero point energy for phonons that is much too large! This is Volovik's insight.String theory is like 2D field theory because the string sweeps out a world sheet in ordinary space-time rather than the 1D world line. This is a great thing because it essentially does away with the infinities of Cosmological Constant problem, which worries Witten and rightly so.There are analogies of string theory with lattice spin Ising models which have a duality relating a T theory to a 1/T theory in a dual lattice.Witten introduces the famous alpha' as basically a new string fundamental constant. He does not seem to connect it to the Salam gravity I worked on in 1973.Using his h = c = 1 units. Witten's alpha' is essentially an effective field Planck area. If we believe Susskind's world hologram we havealpha' = Lp*^2 = Lp^4/3L^1/3at scale LIf we choose L = c/Ho in a kind of Mach Principle of large scale influence on small scaleHo = HubbleLp* ~ 1 fermiThe problem with this is the time dependence in H(t) where Ho is the present epoch value.Note, in Witten's conventions, energy is 1/LengthThereforealpha' is the universal Regge slope, which for hadronic resonance hasalpha' ~ (1 fermi)^2Quantized spin of resonance = alpha'(Energy of resonant peak)^2 + interceptalpha' is reciprocal string tension since Energy/Length ~ 1/Area in the h = c = 1 convention used by Witten & Co.In my theory/zpf = (alpha')^-1[(alpha')3/2|Vacuum Coherence|^2 - 1}Einstein's gravity emerges fromguv = (Minkowski)uv + (alpha'){Goldstone Phase of Vacuum Coherence)(,u,v)}anholonomic torsion field potential = (alpha'){Goldstone Phase of Vacuum Coherence)[,u,v]The O(2) internal symmetry of this complex numbered scalar Vacuum Coherence local field implies 1D string topological defects in 4D spacetime asa rigorous theorem in the text books.More generalized models of O(N) hypercomplex numbered MACRO-QUANTUM VACUUM More is different (P.W. Anderson) coherence order parameters will give the brane world topological defect generalizations of these strings that should also have torsion on the microscale.BTW, how does your theory explain mechanical inertia?Exactly as in Wheeler's book Geometrodynamics with Mass without mass except now I have G* = 10^40 G(Newton) at the fermi scale.The characteristic rest mass of the spatially extended lepto quark of Charge without charge rotating with Spin without spin ism = e^2/zpf^1/2 = e^2/Lp* = e^2(alpha')^-1/2because Vacuum Coherence -> inside the Type II superconductor quantized flux vortex vibrating string exotic vacuum dark matter core with ends pinned to the 3Dim brane world we are stuck on like Ed Abbott's Flatlanders if the basic Witten & Co ideas are on right track.These 1 Dim strings vibrate in the extra boson space dimensions and the extra fermion supersymmetry matrix space-dimensions forming the Calabi-Yau space generalization of the Kaluza-Klein hyperspace of the 1920's.But it only takes one physical effect that is not locally negligiblein this senseto blow out Einstein equivalence.Einstein equivalence is only a classical weak field approximation oryour correspondence principle in the informal sense for the zerotorsion limit.I'm not sure exactly what you mean here. Einstein thought it held withoutrestriction.Looks like you don't agree.I am not sure without doing a careful look. In any case, of course there have beensubtle modifications or refinements of Einstein's early thinking. These are smallcorrections so to speak. Only minor clarifications of his great insights. There isnothing revolutionary in the distinctions you propose here that I am able tounderstand at least.I am not sure what happens when there is torsion-spincoupling. It can break down for a variety of reasons.Exactly -- you're not sure. The fact is it's anybody's guess and is ultimatelyan empirical question.Not quite - there is a lot of work in the published literature - it's only my personal lack of knowledge here I allude to.Yet Einstein thought he was completely sure, and said so. That's why Feynmanmade fun of him in Lectures on GravitationZ.Again this is very minor compared to what is happening now and what I am talking about above in relation toWitten & Co.I do not have time to absorb all your points below. Later.In order to save his concept of strict equivalence, Einstein wasforced toassume that Riemannian curvature -- which is absolute -- has no directphysical significance from the standpoint of local measurements.This is actually not true, as Feynman has pointed out. It only holds,strictlyspeaking, at a *point* in spacetime. But Riemann curvature is stilldefined, andneed not vanish, at any point in a GR gravity field.That is why I have argued that Einstein equivalence has now beentacitly restrictedto a (weak) *correspondence principle*, which does not in itself implyanything about thefundamental physical indistinguishability of gravitational andinertial fields.The problem with MTW is that they try to have it both ways -- leadingto confusion.They call everything curvature. Wheeler claims in Wheeler andCiufolino that Einsteins'theory was that gravity is spacetime curvature -- which is nonsenseif you interpret thisas *Riemann* curvature.But if you call even a purely linear distortion of the Lorentz metricspacetime curvature,then you might as well say that acceleration of one's frame ofreference curves spacetimeeven in Newtonian mechanics.Try it. Write ds^2 = g_uv dx^u dx^v in a purely Newtonian context andsee what happenswhen you accelerate the frame (non-linear coordinate transformation).Whether or not that gravity force is real dependson whether or not there is a local tidal curvature differentialin the g-force as described by the geodesic deviation equation betweentwo neighboring timelike geodesic LIF free float observers.I am not so interested in what is real, and what is not, as I am inthe hypothesisof fundamental physical indistinguishability -- which I argue issimply false.The effects of deformation and transformation are certainlydistinguishable in differentialgeometry -- but this basic mathematical distinction is papered over incanonical GR so thatwe can pretend that Einstein equivalence is still valid. Themathematical distinction betweendeformational and transformational effects on the metric componentsg_uv (appearanceof non-vanishing gradients) is de-emphasized .As far as I can see, this is done pedagogically in order to sustainthe myth of generalrelativity (to be sharply distinguished from mere general covariance).Stephen Weinberg devotes the better part of a chapter to thisdistinction, after callinghimself a hereticThis curvature tensor tidal differential force cannot be eliminatedlocally.Yes. And of course everyone knows this. Riemann curvature -- which inGR distinguishesreal gravitational fields from pure inertial fields -- is absolute.All I mean effectively by EEP is the set of following formalstatements.1. There exist local tetrads ea^u(P) at point P such thatnab = ea^u(P)eb^v(P)guv(P)2. The local symmetric torsion-free connection from first derivativesof the metric guv(P) can be set to zero locally at P.This is one formulation. As far as I can see the use of tetrads doesnot add to the physicalcontent of EEP, which is that a sufficiently small neighborhood ofspacetime can be *modeled*as a flat Lorentz manifold *with respect to a certain class ofphysical measurements*.However, this does not mean that inside GR spacetime is actuallyflat within *any*neighborhood, no matter how small, in the presence of permanentgravitational fields.The point is that canonical GR uses the same metric g_uv to representboth inertial andgravitational phenomena. This is not a mathematical necessity, asshown by various bi-metricapproaches.The historical origins of this fusion of gravitational and inertialmetrics lies with Einstein'sbold hypothesis of strict gravitational-inertial equivalence, and hisclosely related conceptof general relativityWhat I am trying to do is to relax Einsteinian *strict* equivalencewhile retaining a singlemetric -- which means the theory can be physically re-interpreted,breaking the conceptuallogjam, with minimal disturbance to the existing mathematicalformalism.If this allows the definition of a localized gravitational fieldstress-energy density representedby a Diff(4) world-tensor, it could have significant bottom-linephysical consequences.This defines the distinction between LIFs on timelike geodesics andLNIF's on timelike non-geodesics both intersecting at same P.Sure. Again, I have no problem with this. The problem for me is, Whatexactly does this meanphysically in relation to Einstein equivalence and Einsteiniangeneral relativity?Note timelike non-geodesics could not exist without electrical charge.This is a contingent fact that I do not regard as fundamental to theinterpretation of GR. Anyrepulsive forces would do.Neither presumably could light hence no light cones hence nothing -not even null geodesics. This suggests in itself that gravity is anemergent collective effect primarily from electromagnetism andHeisenberg quantum uncertainty assuggested by Sakharov's metric elasticity.OK.Part of the confusion is that curvature = field strength in themodern fiber bundle picture of the spin 1 gauge forces.That word curvature again. What does this mean? What kind ofcurvature?The connection for parallel transport in the internal dimensionsbeyond space-time (or extra space dimensions of Kaluza-Klein et-al) istheYang-Mills potential like the EM 4-vector potential Au(P) in theU(1) EM simplest case.OK, this is how you draw your correspondence between thephenomenologicalg-field and the underlying spin-1 gauge field?On the other hand it is the metric, specifically goo(P) ~ (1 +2U(Newton)/c^2) in weak field slow motion static spherically symmetricvacuum case outside Mfrom source mass Mleading to notion of gravity force as connection from Newton's 2nd lawF = -m GradU(Newton)F is locally eliminitable on the LIF timelike geodesics in curvedspace-time.Tidal inhomogeneous differentials F(P+dP) - F(P) are not.Which appears to say that the metric deformations of GR have a*physical origin*in the properties of the underlying spin-1 field?It seems to me that however you slice this, the implication is thatthe deformable metricof GR is a reflection of the physical properties of the quantumvacuum, as opposed tosimply characterizing an abstract Riemannian geometry of purespacetime (vacuum asvoid).If so, then this reduces the combined gravitational-inertial metric ofGR to the status of amere predictive mathematical model -- much as we now view Ptolemaicastronomy.EEP, as I understand it, merely requires that the netgravitational-inertial forceswhile tidal effects canalways be locally neglected.This last statement needs clarification. It is strictly not true. Itisonly a weak field approximationthat the tidal effects are of order (L/r)^2 where r is the local scaleof radius of curvature.I am packing all this into the convenient term locally, suitablydefined. Soof course I agree.As Feynman said, this is only *strictly* valid at a spacetime point.MTW makethis approximate nature of the remark very clear when I read them. Youhave misinterpreted themas making a much stronger statement and I think this is the root ofthepseudo-problem you pose.No, Jack.You are absolutely right and I agree with you about the clarity oftheir minimalformulation of EEP. However, I am right that they go beyond this whenapplyingthis EEP in certain -- though not all -- contexts. They quietly addadditionallayers of interpretation in those contexts. That is how they try tohave it bothways.You have to keep your eye on that pea. Remember, MTW was written by acommittee!I agree Wheeler is very clear in his 1992 formulation of EEP, and onthe absolutecharacter of *Riemann* curvature. The dispute is about what thesethings meanphysically in regard to Einstein equivalence.If Einstein was a bit confused on this in the early days, I don'tknow,it is of no real consequenceexcept in the history of the evolution of the idea.This is completely inaccurate. Einstein was quite emphatic that strictequivalencewas at the core of his theory of general relativity. No strictequivalence, no generalrelativity.I have not been able to find any explicit abandonment of this boldidea *anywhere*in Einstein's writings.Again, the MTW and Wheeler formulations of EEP are technically OK.The problemI have is how they further interpret EEP as they have presented itin *certain* contexts.So I think you may be barking up the wrong tree here. This is a rathersubtle but veryimportant point.This can be consistently interpreted in at least two different ways:(1) as the consequenceof a (local) *fundamental physical identity* of gravitational andinertial fields (Einstein, Pauli);or (2) as merely the result of inertial compensation of theseparatelyexisting gravitationalfield (Eddington et al.) in conjunction with the natural attenuationof tidal effects withinarbitrarily small neighborhoods of spacetime.Either way , the theory remains generally covariant, and a weakcorrespondencerelationship with SR holds -- locally measurable effects that locallydiscriminate betweenthe predictions of SR (flat spacetime) from those of GR (curvedspacetime) can beregarded as practically negligible in a sufficiently smallneighborhood.Yes, in the weak field limit, but not at a black hole, which Puthoffwrongly claims does not exist.Which is immaterial to the *correspondence* relation between GR andSR, sinceblack holes are clearly beyond the domain of SR and there is noempiricalsuccess of the predecessor theory to explain.In any case, Einstein didn't believe in them.I would argue that a weak correspondence relation is all that onecan reasonablydemand of the re-interpreted theory. There is in reality no flatspacetime *anywhere*.Fine, that's how I read MTW anyway. Also in my theory, globally flatspace-time is intrinsicallydynamically unstable from quantum electrodynamics.That's how MTW reads in some places, but not in others.Also, it is important to understand that none of this depends on theempirical correctness ofYilmaz's particular theory vis a vis GR.I find Yilmaz's idea completely obscure intuitively.It's just a different theory. I think it's at least internallyconsistent. For example, the alternative approach is taken in anumber of bimetric theories that agree empirically with GR. Theseemploy a backgroundmetric that accounts for inertial effects in conjunction with aseparate (though commensurable)gravitational metric that accounts for physical gravitation.Very ugly. Less with more.Unless it leads to conceptual errors of the kind I have beeninvestigating.Although I agree it would be nice to keep a single unified g_uv, whichiswhat I am trying to do under an alternative interpretation.My point is that there is no *mathematical necessity* for this.In the alternative interpretation, I believe the way is open toconsistently defining a truegravitational field stress-energy density, which is fullylocalizable,and an inertially-compensated*net effective* stress-energy density which corresponds to the forcefield that is actuallymeasured in an accelerated frame of reference.Show me the math otherwise the words have no meaning to me.I'm working on it.I have to show that when the purely inertial component of thegravitational stress-energy density is removed, one is left with a true Diff(4) tensor.This means that Mach's principle is effectively abandoned -- alongwith his rather loony brandof strict empiricism (remember, Mach flatly rejected the atomictheory).Mach's idea reenters in the Holographic World whereLp* = Lp^2/3(c/H)^1/3 = 1 fermi nowBut this may have a fatal flaw since H is cosmic time dependent in theFRW regime.OK.The connection for parallel transport locally vanishes inLIF's on timelike geodesics.Of course it does. But we can (at least locally) distinguish betweenthe inertial andthe gravitational *components* of the canonical connection field(distinct contributionsto the net metric gradients). That comes right out of thedifferentialgeometry (also, see e.g.Weinberg's Gravitation and Cosmology).I think I said that above.You did?Does this mean you now agree with my proposal for a local decompositionof the Christoffel connection?I don't know what you really mean until I see some math.===Subject: Re: help me, please! permission for an emailed response.X-Tom-Swiftie: Who drank the last beer? Tom asked, hopping mad Bushnell, BSG ha scritto nel messaggio> Is there a fast way to determine X and Y both integer such that:> X^2 - Y^2 = A> Yes.> ???? Are you joking?> What is the answer?The answer is yes. As for what is the fast way, I'm afraid youhave to do your own homework.===Subject: Re: help me, please!In schrieb MeC :> Is there a fast way to determine X and Y both integer such that:> X^2 - Y^2 = AWrite X^2 - Y^2 as a product (X-Y)*(X+Y). Then both factors have to be divisors of A ...===Subject: Re: Cantor's diagonal number: why do we just ASSUME such a number exists?> in part:>Although some parts of Cantor have truth, I>believe that Cantor is completely bereft of truth. (Actually, that>last sentance was inconsistent and needs correction, but we can still>adhere to it without loss of generality.)> Ah, you're not just an interested young fellow. You're a deliberate> troll.No, just someone aiming to hit every point on the Baezcrackpot index, and succeeding. In order. Other peoplewith sharper eyes than mine have already caught this.http://math.ucr.edu/home/baez/crackpot.htmlI think the Baez index was originally written forscoring sci.physics crackpots, and doesn't lenditself well to math crackpots. So the fact that Deethsucceeded in hitting every point is somewhat ofan achievement.>I'm not good at math,>but my theory is conceptually right, so all I need is for someone to>express it in terms of equations.> How can you *tell* if something is conceptually right?15. 10 points for each statement along the lines of I'm not good at math, but my theory is conceptually right, so all I need is for someone to express it in terms of equationsThat's one of the ones that works better for revolutionaryphysics theories proving Einstien wuz wrong. Hard to fit in a mathematical claim. So Deeth just gave it to usverbatim.Deeth has set a high mark, but I think he can be beaten.Note that many items can be scored multiple times. - Randy===Subject: Re: help me, please! Bushnell, BSG ha scritto nel messaggio Bushnell, BSG ha scritto nel messaggio> Is there a fast way to determine X and Y both integer such that:X^2 - Y^2 = A> Yes.> ???? Are you joking?> What is the answer?> The answer is yes. As for what is the fast way, I'm afraid you> have to do your own homework.I repeat: I'm not very good in math. I'was not able to find the answer (Ifit exists). Before asking a question in a group I always try to find ananswer...Is it so difficult to give a concrete answer? In any case don't worry... Ifyou don't want to help me I'll wait for someone else help.Thank you anyway.Marco===Subject: Re: One homomorphism questionHello> Hi The commutator subgroup G' of G is the subgroup generated by all> elements of the form a^{-1}b^{-1}ab, with a in G, b in G.> If H subset G is any subgroup, show that if G' subset H> then H is a normal subgroup in G and G/H is abelian1/ Take g in G and h in H, then u = ghg^(-1)h^(-1) is in G', and in H. Thenuh = ghg^(-1) is in H, hence H is normal.2/ Take a and b in G. We have: aHbH=abH and bHaH=baH (because H is normal).Take v in abH: v = abh with h in H.a^(-1)b^(-1)ab is in H, a^(-1)b^(-1)abh is in H, hence:abh is in baH. We proved abH C baH.Similarly you can prove that baH C abH, hence the result.Julien Santini,France.===Subject: Re: Algebraic number theory books>Can anyone suggest a good book(s) on algebraic number theory?> I'm partial to iel Marcus's _Number Fields_; Ireland and Rosen's _A> Classical Introduction to Modern Number Theory_ is also a good first> step.> ============================================================== ========It's not denial. I'm just very selective about> what I accept as reality.> --- Calvin (Calvin and Hobbes)> ============================================================== ========> Arturo Magidin> magidin@math.berkeley.eduI warmly recommend Marcus as well. In addition I have found Problemsin Algebraic Number Theory, By Esmonde and Murty, quite entertainingand useful. And of course, one should always study the masters -Hecke's classic is just that.There's one book that has for long intrigued me, but I never gotaround to actually buying it (it's expensive!) or spending serioustime with it in a library - this is the fairly recent book by Neukirch(trans. by Schappacher). Can anyone comment on this one?===Subject: Re: One homomorphism question Adjunct Assistant Professor at the University of Montana.>Hi The commutator subgroup G' of G is the subgroup generated by all>elements of the form a^{-1}b^{-1}ab, with a in G, b in G.>If H subset G is any subgroup, show that if G' subset H>then H is a normal subgroup in G and G/H is abelian>Here is my proof >first try to prove H is normal to G>For any h1 in H for any g in G>so gh1g^{-1} = gh1g^{-1}h1^{-1} h1>because G' is a subgroup of H so>h2=gh1g^{-1}h1>so gh1g^{-1} = gh1g^{-1}h1^{-1} h1 =h2h1 in H>so H is a normal subgroup of GYes.>But I don't know how to prove G/H is abelian>Let Ha is one element of G/H, let Hb is another element of G/H>HaHb HINT: The reason they are called commutators is that for any twoelements a and b,ab = ba*(a^{-1}b^{-1}ab).So, if (a^{-1}b^{-1}ab) is in H, what can you say about Hab and Hba?And what can you say about HaHb and HbHa?===========Subject: Re: The need of geodesics in physics>That's a bit confusing, one can set the Lorentz >Force to generally vanish ie. >f_u = q*F_uv *U^v =0>and in effect decribe EM geodesics, then put>q*F_uv in the metric forming a non-symmetrical >metric.>>The problem with that is that both sides of that equation are tensors.>>If they vanish in one frame, then they must vanish in all frames.>>Geodesics are based on affine connections, which are not tensors, so>>they have non-homogeneous transformation properties. >Ah, but the geodesic equation is the absolute derivative>of the 4 velocity U^u = dx^u/ds, ie>DU^u/ds = U^u;v * U^v =0.>> True, but how does that relate to the LF?>Well multiply by mass (scalar invariant) then, >m*DU^u/ds =0 >but QT requires m to change discretely, and >since dm >0 requires a continuous change in m, >dm=0, therefore set P^u = m*U^u (the 4->momentum) and find >DP^u/ds =0 (P^u = m*U^u, dm=0).>>I don't know where you got that notion from. QT says nothing about m>>changing discretely. In fact, the rules of QT maintain continuous>>change of mechanical variables (of course under the restrictions of>>the operator algebra). What is discrete in QT are the quantized>>values of the so-called stationary states, but these tend to be the>>exception rather than the rule. Continuous fluctuation tends to occur>>everywhere in the quantum world.>We should make sure we're talking about the >same thing. It is *invariant* energy which can>only vary discretely and NOT continuously.>A simple atom emits and absorbs energy>by photons. >Is true generally, and applies to LF, so >f^u=DP^u/ds =0.>>Moreover,>>geodesic acceleration should be independent of mass, which applies to>>forces such as centrifugal, coriolis, and of course, gravitation.>True for acceleration of point masses, (but not true>for relatively large masses, that's why acceleration>is a better term than force in GR, otherwise you'll>need to include the term k*t_uv >0 we figuring G_uv).>>Acceleration due to Lorentz force will be inversely dependent on mass,>>making it a no go.>motion must respect Quantum Theory, Lorentz force does>not, so the stated objection evaporates.>>But why are you making that assumption? I can make that statement>>certainly, but I'm excluding a large variety of systems by doing it.>>So I don't think it is very realistic.>Consider, for example f_0 = q*F_0i*U^i,>which in vectors is approximately >f_0 = q*E_i (dot) V_i.>If q moves in direction of E_i then it's>potential energy will vary continuously and>this is prohibited by QT, therefore >E_i (dot) V_i =0 and thus f_0 =0.>Simply stated, charge q's V_i is always >perpendicular to E_i. >>But potential still varies continuously in QT and its changes are very>>easily tracked in the position representation. >We're talking about Lorentz force not potential.>>I don't know how you came to that conclusion.>Fair enough. Just to make clear definitions, define >invariant energy P by P*P = P_u*P^u ok?>An invariant force would be f = dP/ds, and >dP = f*ds ok?>The Lorentz force f_u *dx^u = f*ds, ok?>f_u*dx^u = q*F_uv *(dx^v/ds)* dx^u ok?>Since dx^u*dx^v is symmetrical and F_uv is >antisymmetrical F_uv*dx^u*dx^v =0 when>summed ok?>Then f*ds = 0 = dP ok?>That's it! >To summarize: The Lorentz Force Equation>predicts dP=0. But we know P is not a constant,>therefore P varies discretely, ie by quanta. >Yeah, I think I see what you're trying to say. The quantity f ds is>actually invariant work. Right (and thanks). Perhaps I might check out the following with you to ascertain co-understanding,($ = integral)f = dP/ds =0 ok?$ f ds = constant (= invariant energy) ok?>And your result threw me for a few minutes>since magnetic fields do no work, but electric fields do. Then I>realized that it could be broken down as f_i * d x^i + f _0 * d x^0>= 0. The first term is the work done by the electric field and the>second term is non-zero so they will bce out. Note that f_0 is>essentially E dot v and will be related to power. Ah, but f_0 is an enigma, lets study it together.1) If f_0 is non-zero then E dot v is non-zero. If E dot v is non-zero then *spiral* orbits become possible, for example, the charge q can move slowly into the electric field E. (this is the old classical idea ofhow electrons move in the nuclear field, and was replaced by Plancks Quantum hypothesis (imho)).So a non-zero f_0 violates QT.2)You pointed out E dot v is power, well consider electrical current moving threw a resistor and thisproduces power by the product Volts x Current and in turn produces heat. The production of heatis by quanta, ie. infared photons, hence power isquantized. This can be solved and explained by,$ f_0 ds = q*$ F_0i dx^i = constant.For simplicity, work the RHS and get q*E*x = q*Q/r = energy.=constant =quanta.>I'm still not sure where you're going with this, but I think IYou're very welcome, thank YOU.I'm sorry my understanding is not simple, blame QT,I (we) are working in the system. I think we agreef=0, so the discussion is on the LF components f_0and f_i being zero or not. (I'm arguing they are zero).Ken S. Tucker===Subject: Re: help me, please! Bushnell, BSG ha scritto nel messaggio>>Is there a fast way to determine X and Y both integer such that:>>X^2 - Y^2 = A>Yes.> ???? Are you joking?> What is the answer?> Thank youYes, he's kidding, but only by reading your request literally, andignoring the implied request for him to do your homework for you.There are two implied messages in Bushnell's note: 1. This is a simple exercise in algebra (i.e., factor the left side of your equation and do what comes naturally), suitable for homework in an elementary math course. 2. If this is indeed a request to have someone do your homework problem, then you deserve to be teased, at least gently. 3. If this is *not* a homework problem, please state that much, and I'm sure someone can step forward and provide the few little steps that solve the problem.Dale.P.S. I nearly forgot: 4. No one expects the Spanish Inquisition!===Subject: [JSH] Re: A Question for James Harris charset=iso-8859-1> Like, see> http://www.megasociety.net/NoesisHighlights.htmlSee alsohttp://www.google.com/groups?selm= qm7jmvo01dsmu499gv3enerr13n16jfo16@4ax.com===Subject: Re: some coin tossing> figure out what exactly he (Bryce Carlson) was saying. But I still don't> understand how he computes those p1, p2, etc ... It seems that he is> continuing some discussion here. Can you please send me the link where he> does the analysis for 2 consecutive heads?No.Seriously, that one turned up in a google search, butdoesn't seem linked to the main structure of www.bjmath.com,talking about.However, this problem does get discussed here and therearound the web. Here's an entry at Ask Dr. Mathhttp://mathforum.org/library/drmath/view/52217. htmlSomewhere I've also seen it developed as a recursiveexpression, but when I've tried to work that out myselfI get lost in the counting. One recursion goes somethinglike:I can get m (or more) successive heads in N tossesby either: (a) getting m or more in the first (N-1) or (b) Having the first (N-1) *not* contain a run of m, but ending with (m-1) heads in a row, followed by heads in the last toss.I *think* that means P(N,m) = P(N,m-1) + (1 - P(N, m-1))*(1/2)^mWith P(N,m) = probability of getting m or more headsin a row in N tosses, andP(N,m) = 0 if N < m.The Mathworld entry on Fibonacci numbers mentionscoin tossing:http://mathworld.wolfram.com/FibonacciNumber.htmlThe probability of not getting two heads in a row in n tosses of a coin is F_(n+2)/2^n (Honsberger 1985, pp. 120-122).[Look right below the picture of dominoes on acheckerboard.]More relevantly, there's an entry on n-step Fibonaccinumbers, which generalizes the above result:http://mathworld.wolfram.com/ Fibonaccin-StepNumber.htmlThe probability that no runs of k consecutive tailswill occur in n coin tosses is given by F^k_(n+2)/2^n,where F^k is a Fibonacci k-step number.forth, which makes it a little confusing to read]The normal Fibonacci numbers are the sum of the twoprevious terms. A k-step number is the sum of thek previous terms.So you could actually work out your exact probabilityby calculating the appropriate k-step Fibonacci numberrecursively. You'd want a multi-precision numberpackage of some kind, like GMP. - Randy===Subject: Re: Key core error argument, stepped out> Readers notice how Rick Decker is playing you for fools as I have a> *stepped* out math proof.In fact, you have dozens of posts with your proof steppedout. And each one is followed by a number of posts showingthe error you made around step 6 (or possibly 7, I can'trecall).> Now then, if you've gone through the trying and difficult process of> finding a spectacular proof, and then face people who decide to ignore> your hard work and effort, to trot out their own pet examples, to try> and distract people from your proof, how would you take it?Hardly ignored. People go step by step through your steppedout proof to show in great stepped-out detail why step 6(or 7) is wrong. There's a fair amount of work involved in composing those arguments.Nor can a post which includes your proof and analyzes yoursteps in detail be considered to be distracting. Any morethan if I took a calculus text and spent a page analyzingwhat took one line in the text could be considered to beignoring or distracting from the text. - Randy===Subject: Re: Greek Alphebet charset=iso-8859-7 John Savard > But in other contexts, the letters of the Greek alphabet are fraught> with meaning!> Thus, alpha comes from aleph, which means OxReferences please? The _name_ alpha might be a degenerate form of theHebrew aleph, but the letter itself started AS the symbol for an ox, inGreek. It actually comes from the upside down capital alpha, which was theactual symbol for the head of an ox. (Just picture an upside down A).Who says it's not the other way around and aleph is not a degenerate formof alpha?> And gamma comes> from gimel... so one can look at a book on the history of the> alphabet.I am not convinced that easily, particularly given the long history of bothHebrews and Greeks. If you have some online linguistic references, i'dappreciate them. And since I am not a linguist myself, I am posting this tosoc.culture.greek as well, so that people who are more familiar with itsetymology can drop in.> And the 22 letters of the Hebrew alphabet are alleged to> have meaning on other levels too.> But then to study mathematics is to go in one direction, and to study> Tarot cards is to go in a different direction...> John Savard> http://home.ecn.ab.ca/~jsavard/index.htmlhttp:// users.forthnet.gr/ath/jgal/----------------------------------- -------Eventually, _everything_ is understandable===Subject: Re: Key Core Error Argument> ...> Actually it is, as Dik Winter is trying to find a way that 7, 7 and 22> become 1, 1 and 22 based on a *varying* x.> That is, he's trying to make the change in *constants* dependent on a> variable.> Eh?> I want readers to understand that his behavior is crank, while I guess> many of you may sympathize with his strong desire for me to be wrong,> remember, it's not about people as the math didn't just decide to> change.> What you should be sympathetic to, is the truth.> You claim that having P(x) = g1(x).g2(x).g3(x) with g1(0) = g2(0) = 7> and g3(0) = 22; the *only* way to divide P(x) by 49 is by dividing g1(x)> and g2(x) by 7 and g3(x) by 1. Because now g1(0)/7 = 1, g2(0)/7 = 1 and> g3(0)/1 = 22.> I claim there are other ways to do that. Have w1(x), w2(x), w3(x), such> that w1(0) = w2(0) = 7, w3(0) = 1, w1(x).w2(x).w3(x) = 49. Because now> g1(0)/w1(0) = 1, g2(0)/w2(0) = 1 and g3(0)/w3(0) = 22.> Where do I change constants?The basic algebra is that if you have 7 and divide it by *something*and get 1, then you divided by 7.And no tricks will change that fact, and it's simply crank behavior totry and act like there's some complicated way you can divide 7 to get1, without actually just dividing it by 7.I noted that the trick of using ratios of functions to try and hidethings just doesn't work, yet the crank Dik Winter is still backtrying to find a way to divide 7 as a *variable* rather than justdivide it by 7 to get 1.I think it satisfying to *finally* be able to show that Dik Winter andthe posters like him are indeed the ones who are cranks, as here allyou need to be able to realize is that dividing 7 by *something* toget 1 means you divided it by 7.That's it. If you're still doubting, you're a crank as well.James Harris===Subject: Re: A Question for James Harris> That's a rather easy falsehood to correct.Unlike the many falsehoods that JSH promulgates?JSH deals with corrections to his own lies in one of two ways:(1) ignoring the correction entirely,(2) repeating the lie without answering the points raised by the correction and asserting that this somehow refutes the correction.On unfortunately rare occasions, when the correction to one of his lies becomes too obvious for even him to ignore, he will admit to a typo.===Subject: Re: Probability of a Run> I am trying to calculate the probability that a gambler with capital C> and who uses a Martingale betting strategy will be wiped out in m> turns at the game. This would happen with a run of n consecutive> losses and I want to calculate the probability of this.> The textbooks treat this problem at an advanced level, invoking> generating functions or difference equations, but I wonder if a> solution satisfactory for the purpose could be arrived at in a simpler> way. All I need is the probability that there would be AT LEAST one> run of length AT LEAST n.> If in n independent trials the probability of a loss at a single trial> is q, the probability of all losses in n trials would be q^n. A string> of n losses could begin at any trial from the first up to the m - n +> 1 th, so would the probability be (m - n + 1) * q^n?There's a related question in sci.math right now aboutthe probability of k heads in a row in a series of ncoin tosses. According to the Mathworld entry formulti-step Fibonacci numbers,http://mathworld.wolfram.com/ Fibonaccin-StepNumber.htmlthe probability of NO runs of k heads or more is givenby F^k_(n+2)/2^n, where F^k_n represents the n-thk-step Fibonacci number.I suspect whatever argument was used to generate thatresult can be easily modified for the general case ofprobability of loss = q, but Mathworld gives no referencefor this interesting result. - Randy===Subject: Re: JSH: It begins> Now that I have a proof that's simple enough to explain to kids, I can> move back to email.> Yup, you're leaving other mathematicians open to a major sandbagging> as I go back to a process that's netted me Andrew Granville, Barry> Mazur, and other notables over the years.> And you're leaving them open and unready.> I learned a while back that few professional mathematicians seem to> read sci.math, so it's not like what's been discussed here is likely> to get to them.> So I have a wide open field of mathematicians all over the world> contactable by email with proven techniques for getting their> attention.> The process has already begun.> James Harris> Does this mean you are *finally* ready to stop degrading the SNR of> sci.math? Are you really leaving sci.math this time -- or is this just> another one of your tricks to lull its readers into a false sense of> relief?> A fool and his proof are soon refuted.> Democracy: The triumph of popularity over principle.What's with cranks like yourself C. Bond that you keep making thingsup?I didn't say anything about leaving the newsgroup.I'm simply pointing out that now I can use email effectively onceagain, and sci.math mathematicians are leaving their colleagues openand vulnerable.I have a world of mathematicians to email.As for my posting, it may increase dramatically if I see that pumpingup the volume is effective.So there may be many MORE posts from me on sci.math, not fewer.James Harris===Subject: Re: Uncle Al is Sadistic .> Ahem. Is that the correlation between SAT scores and grades FOR THE> college tend to start with remedial this and that before they are> ready for the main major courses, whereas people with the higher> scores tend to dive right into the meaty college stuff with no> pre-prep. So you're comparing apples and oranges-- performance in two> completely different grades of academic difficulty.> Example: I personally had pretty high SAT scores. And I had some AP> credit, so I figured I'd dive right into calculus 111, chem 111, and> physics 111, tech writing, upper division history, and so on. What a> shock. It was like 55 F water. I didn't sink, but I had to> study/paddle like hell for the first time in my student life, and I'm> sure a few of my A- grades would have been A's if I'd taken remedial> algebra and English, conceptual physics and the like. Then the> correlation between my SAT, ACT and grades would have been perfect,> instead of just good.> No kidding. I AP'd 31 credits in calculus, chemistry, and biology> thereby entering as a sophomore for all the real classes. Dr.> Gerasimos Karabatsos gets up in front of first term organic lecture,> says there are 1200 people in the course and room for 400 in lab> starting next term. 800 will either drop the course or flunk out> solid. Then he began the first lecture. About 380 made it. Look to> the left of you, look to the right of you... That was when education> worked.> Going from high school calculus to third term college stuff was a> nightmare. I managed 176 of 180 credits needed in three years, then> spent my senior year mostly doing independent research. I'd already> had seven terms and a summer with the prof, so it was a chance to> really rip into it. Classes are only foreplay. I didn't hit this so called wall until graduate school at Stanford. Ace'd AP calculus and physics in high school (wasn't allowed into APenglish, more on this later). Ace'd all the technical coursesthroughout my undergraduate years. Graduated Summa Cum Laude Decidedto go to graduate school at Stanford and was dropped in with the creamof the crop. Boom!!! But the most ironic part was I didn't feel Iwas being better educated, it was just harder to get a good grade. Itstruck me that Stanford was more of a filter than an educator. By thistime I was also getting bored with it all and eager to make somemoney. So I got my masters and split with an MSEE. Thoughout myentire college career, I never was motivated by the end ofprofessional developement. I was motivated by the desire to learn newthings. I always viewed it as an educational playground. As soon asit ceased to serve this function I was outta of there! :) I was justlucky that my desires and abilities matched up well with a lucrativefield. Throughout all this time the only courses I ever had trouble withwere non-objective courses such as humanities, or subjects whoseanswers relied on memorized facts rather than deductability. The most ironic thing is now, 20 years later, although I'm stillfascinated by the latest developements in physics, science and math,etc. I'm now interested in some of the non-objective subjects Ioriginally hated, such as history and politics.-Bruce===Subject: Re: It begins> Now that I have a proof that's simple enough to explain to kids, I can> move back to email.> And proof of *what* is that? FLT?No, FTL (faster than light travel).===Subject: Re: More symmetry between derivative and antiderivative?>What about functions not defined at zero, such as f(x) = 1/x?>Just an addition: With ln(b)-ln(a) = ln(b/a) the number of parameters >>reduces from two to one in this case, too.>> That only considers one out of infinitely many cases of functions > with antiderivatives not defined at zero, so it hardley solves the > problem.> I don't see any practical problem. It is the theoretical problems involved in extending *every* antiderivative to be defined at zero that I see. I see no need for doing so, and a lot of reasons for not doing so. What do you do about antiderivatives for, say, f(x) = 1/sqrt(x^2-1) near zero?> Aren't there infinitely many admissible values of 1/x in close > vicinity of x=0? The real world including application of physics > in technology, medizine, management, etc. does perhaps not need > the only forbidden wightless value at exactly x=0.If mathematics were satisfied with such slovenly thinking, then it would never have been able to develop most of those mathematical tools that are so useful outside of (formal) mathematics.> On the other hand, the function 1/x is perhaps more important than the > infinitely many functions worrying you.===Subject: Number theory problem, or somethingThis is a homework problem from a course I'm not taking.Prove that none of the numbers 1, 11, 111, 1111, 11111 and so on, aresquares of natural numbers greater than 1.I can't seem to do it completely. Here's my best attempt so far: nConsider the number Sum 10^i, where n is positive and even. i=0 n n-1It is greater than 10^n, because Sum 10^i - 10^n = Sum 10^i > 0. i=0 i=0It is also smaller than 11^n, because 2 n n-1Sum 10^i = 111 < 121 = 11^2, and Sum 10^i = 10Sum 10^i + 1 < i=0 i=0 i=0 n-1 n-1 n-110Sum 10^i + Sum 10^i = 11Sum 10^i, and because of this it is i=0 i=0 i=0majorated by a geometric sequence whose multiplicand (or what isthe correct term?) is 11, and as it already started out smaller than11^n, and is growing slower, it will also remain smaller than 11^n.Because it is greater than 10^n and smaller than 11^n, the only wayfor it to be a square of a natural is for it to be a square of x^(n/2)where x is some natural greater than 10 but smaller than 11. There areno such naturals, so therefore the sum cannot be a square of anatural.But this only applies when n is positive and even. How to do it inthe case n=0 and for odd n?-- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland ---------- http://www.helsinki.fi/~palaste --------------------- rules! --------/He said: 'I'm not Elvis'. Who else but Elvis could have said that? - ALF===Subject: Re: Number theory problem, or somethingJoona I Palaste scribbled the following:> But this only applies when n is positive and even. How to do it in> the case n=0 and for odd n?D'oh! I mean of course how to do it for odd n?. The case n=0 istrivial. The sum for n=0 is 1, and 1 is not the square of any naturalgreater than 1./-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland ---------- http://www.helsinki.fi/~palaste --------------------- rules! --------/As a boy, I often dreamed of being a baseball, but now we must go forward, notbackward, upward, not forward, and always whirling, whirling towards freedom! - Kang===Subject: Re: Number theory problem, or something Adjunct Assistant Professor at the University of Montana.>This is a homework problem from a course I'm not taking.>Prove that none of the numbers 1, 11, 111, 1111, 11111 and so on, are>squares of natural numbers greater than 1.>I can't seem to do it completely. Here's my best attempt so far:> n>Consider the number Sum 10^i, where n is positive and even.> i=0 n n-1>It is greater than 10^n, because Sum 10^i - 10^n = Sum 10^i > 0.> i=0 i=0>It is also smaller than 11^n, because> 2 n n-1>Sum 10^i = 111 < 121 = 11^2, and Sum 10^i = 10Sum 10^i + 1 <> i=0 i=0 i=0> n-1 n-1 n-1>10Sum 10^i + Sum 10^i = 11Sum 10^i, and because of this it is> i=0 i=0 i=0>majorated by a geometric sequence whose multiplicand (or what is>the correct term?) is 11, and as it already started out smaller than>11^n, and is growing slower, it will also remain smaller than 11^n.>Because it is greater than 10^n and smaller than 11^n, the only way>for it to be a square of a natural is for it to be a square of x^(n/2)>where x is some natural greater than 10 but smaller than 11. There are>no such naturals, so therefore the sum cannot be a square of a>natural.Ehr, no, that's not true.Say n=4. You are claiming that there are no squares strictly between10^4 = 10,000 and 11^4 = 14,641, because any such squarehave to be asquare of x^2 where x is some natural greater than 10 but smallerthan 11. Unfortunately, 110^2 = 12,100, showing that your argument isincorrect at this step.What you COULD say, for n even, is that it must be a square x^2, withx strictly between 10^{n/2} and 11^{n/2}. But that is not nearlyenough to get your claimed conclusion.===========Subject: Re: help me, please!> I'm not very good in math!> A question:> Is there a fast way to determine X and Y both integer such that:> X^2 - Y^2 = A> Thank you very much!> MarcoIf A = BC, then let X = ( B + C )/2 and Y + (C - B )/2. Will this alwaysgive you X and Y in the integers? Why does this work? This will not helpunless you understand how I got X and Y. Now go and study.===Subject: Number theory congruenceHi group,I'm a little confused over some homework problems which dont appear to haveany supporting examples/theorems (to me anyway).For integer n > 0, use congruence theory to establish the divisibilitystatement:7 | 5^(2n) + 3 * 2^(5n-2)among other theorems, i have the one that saysif ca congruent cb (mod n) and gcd(c, n) = 1,then a congruent b (mod n)...which implies factoring something out, which i dont see any way to do.any help is appreciatedIra===Subject: Re: Product of Reals>> Is it possible to determine what the uncountable product of all real>> numbers in the interval [0.5, 1.5] is? Intuitively it seems like 1,>> but is this concept studied in general anywhere (e.g. the concept of>> uncountably infinite products or sums of real numbers)?>> Lookup the term summable family. This is a concept which allows definition>> of convergence for any numbers a_i with i in I where I is an arbitrary>> index set. By applying exp-log-formalism (transform your product into a>> sum) you get the analogue theory for products.>> A necessary condition for such a family to converge is that its support (the>> largest set J subset I such that a_j <> 0 forall j in J) is countable ->> this means that your product cannot converge.>> Because equally well than your reasoning I could argue that it converges to>> any other real number <> 1. Cf. Riemann's rearrangement theorem.> This is something that I have wondered about. Occasionally in> previous threads the idea of summing an uncountable set of numbers has> come up. It is usually commented this is not possible unless all but> a countable number are zero. However my reaction has always been: is> there any definition for infinite sums of any sort other than the> common series.> Suppose {a_i} is a family of real numbers, where the subscripts i range> over some (possibly uncountable) index set I. If S is a finite subset of> I, we can consider the finite sum A_S = sum_{s in S} a_s.> We say the family {a_i} for i in I is summable (and the sum = L) if, for> every epsilon > 0, there exists a finite subset S of I (depending on> epsilon) such that | A_S' - L | < epsilon for every finite set S' such> that S subseteq S' subseteq I.> This can be stated more concisely by saying that the net of finite sums> converges to L.> I wondered about summing the values of a function f from a set to R or> C. I wondered what sort of structure the set would have to have and> what class of functions could be handled. Even when the set is> countable, the answer is not obvious since the sequence can affect the> result. I mentioned in a recent thread, the distinction between> absolutely and conditionally convergent series.> In the definition above there is no structure imposed on the index set I.> It does not even have an order.> It does turn out that with this definition, a summable family must> necessarily be zero except for an countable number of terms. If> uncountably many terms are nonzero, then for some n we must have > | a_i | > 1/n for uncountably many i, implying that the sum diverges.> This is the first reference that I have noticed to a generalisation of> infinite sums. Unfortunately I could not find much on Summable> families on the net. A search of the groups revealed little but this> thread. A search of the web found more but many seemed to mention it> only in passing or were in formats that I could not read. Mathworld> did not seem to have anything.> Can you point me somewhere or give a bit more detail?> For non-countable sets, could measure theory be considered a> generalisation of these sums?> Yes. In fact, counting measure is a particular type of measure, and thenet of finite sums approach amounts to integration with respect to> counting measure.It took me a couple of minutes to realise why the uncountable sums didnot work. Is this right? If the sum converges then clearly thenumber of terms with absolute value greater than 1 must be finite. Similarly the number of terms with absolute value between 1 and 1/2 isalso finite, ditto 1/2 and 1/3, 1/3 and 1/4 so the non-zero terms arethe union of a countable set of finite sets and hence are countable. Simple, once you know.If the structure of N (natural numbers) is ignored and this definitionof sum is used then I expect that only the series which are absolutelyconvergent in the common sense will be convergent in this sense.Considering the countability limitation, is this generalised notion ofconvergence useful and commonly used?J===Subject: Re: Uncle Al is Sadistic . charset=Windows-1252> 3 Africans I met in Computer science department in the last 4 years> were way above average in thier programming skills in the midst of> Chinese and Indian grad students who are the overwhelming majority in> that department. One was so good that he even got a parttime teachign> position to teach an undergrad programming course, while he's still a> student himself.>You will always find bright smart people in just about any naturally >>occuring group of humans. Race is nothing. Culture is everything.> She said Africa. I would have asked what country? Ethiopia>> used to send their brightest to the States for grad study. It>> would be interesting which countries are starting to get their>> act together.>None of them Ethiopians.> Answer the ing question. Which countries?Yo, listen, sweet yenta BAH, shouldn't that rather/better be:Answer the geographic question?A ing question is pointing elsewhere, besideyour Freudian implication of in countriesAre you a wilting lezzy, coming out of the closet, long last?ahahahaha........ahahahahanson===Subject: Re: Who contributed most to mathematics?>Wasn't Euler from Switzerland?Born and raised there, but I don't think he spent much time there afterthe age of 20.>Also claimed by Russia.... where he did much of his work. And which country gets the credit for the many other mathematicians whose careers were mainly outside their countries of origin? Many of the top American mathematicians of the past century were born elsewhere.Department of Mathematics http://www.math.ubc.ca/~israel ===Subject: Re: Number theory problem, or something Adjunct Assistant Professor at the University of Montana.>This is a homework problem from a course I'm not taking.>Prove that none of the numbers 1, 11, 111, 1111, 11111 and so on, are>squares of natural numbers greater than 1. [.snip incorrect argument.]Say that you have a rep-unit (number all of whose digits are equal to1) which is a perfect square, and it has n digits, n>1. Call it r(n).If a square is 1 mod 10, then it must be the square of a number whichis 1 or -1 mod 10; so the number must end in 1 or in 9; that is, r(n)= (10x+1)^2 or (10x+9)^2, x>0.(10x+1)^2 = 100x^2 + 20x + 1 = 100x^2 + 10(2x) + 1. The last digit of [(10x+1)^2-1]/10is supposed to be 1 as well. But[(10xx+1)^2-1]/10 = 10x^2 + 2x, so the last digit is the last digit of2x, which clearly cannot equal 1. Therefore, we cannot have it asquare of the form (10x+1)^2.So we must have (10x+9)^2 = r(n). But(10x+9)^2 = 100x^2 + 180x + 81. Again, the last digit of (r(n)-1)/10is equal to 1; but(100x^2+180x+80)/10 = 10x^2 + 18x + 8. The last digit is equal to theremainder of dividing 8 + 8x by 10. But again, this cannot equal1. Therefore, it cannot be of the form (10x+9)^2 either.===========Subject: Re: someone told me ...>Well, if normality is unknown for pi, what are some>numbers that are proven normal? Every one of them would>satisfy Ben's requirement.>Champernowne's number>0.123456789101112131415161718192021222324....>is normal in base 10.... and has the additional convenience that it's easy to find anyfinite sequence of digits you might want to look for.Department of Mathematics http://www.math.ubc.ca/~israel ===Subject: Re: Number theory problem, or somethingX-ID: bomGX6Zc8e0fgWksNwhZJvoR1PxWXRkv6xLb-5tPJaKhTPa8Jdw-EC> Because it is greater than 10^n and smaller than 11^n, the only way> for it to be a square of a natural is for it to be a square of x^(n/2)> where x is some natural greater than 10 but smaller than 11. There are> no such naturals, so therefore the sum cannot be a square of a> natural.No. 110^2 lies between 10^4 and 11^4, but is a square.Hint: consider the number 11, 111, 111,... modulus 4.What possible values modulus 4 can square numbers have?reverse my forename for mail!===Subject: Re: Uncle Al is Sadistic .> The most ironic thing is now, 20 years later, although I'm still> fascinated by the latest developements in physics, science and math,> etc. I'm now interested in some of the non-objective subjects I> originally hated, such as history and politics.Congratulations!===Subject: Re: Number theory congruence Adjunct Assistant Professor at the University of Montana.>Hi group,>I'm a little confused over some homework problems which dont appear to have>any supporting examples/theorems (to me anyway).>For integer n > 0, use congruence theory to establish the divisibility>statement:>7 | 5^(2n) + 3 * 2^(5n-2)>among other theorems, i have the one that says>if ca congruent cb (mod n) and gcd(c, n) = 1,>then a congruent b (mod n)>...which implies factoring something out, which i dont see any way to do.>any help is appreciatedConsider 5^(2n) modulo 7. 5^2 = 4 (mod 7), 5^3 = -1 (mod 7), 5^4 = 2(mod 7), 5^5 = 3 (mod 7), 5^6 = 1 mod 7, 5^7 = 5 (mod 7).So the summand 5^(2n) cycles through the residue classes 4, 2, 1, 4,2, 1, etc; if n=1 (mod 3), then you get 4; if n=2 (mod 3), you get 2;if n=0 (mod 3) you get 1.Now consider the powers of 2 modulo 7. You have2^1 = 2 (mod 7)2^2 = 4 (mod 7)2^3 = 1 (mod 7)2^4 = 2 (mod 7)So 2^j = 1 (mod 7) when j=0 (mod 3); 2^j = 2 (mod 7) when j=1 (mod 3),and 2^j = 4 (mod 7) when j=2 (mod 3).What does that mean for 2^(5n-2)? Well, we need to consider 5n-2 (mod3). This is equal to 2n+1 (mod 3).So for n=0 (mod 3) you get 1; for n=1 (mod 3) you get 0, and for n=2(mod 3) you get 2 (mod 3). So: if n=0 (mod 3), then 5^(2n) = 1 (mod 7) and 2^(5n-2) = 2 (mod 7). if n=1 (mod 3), then 5^(2n) = 4 (mod 7) and 2^(5n-2) = 1 (mod 7). if n=2 (mod 3), then 5^(2n) = 2 (mod 7) and 2^(5n-2) = 4 (mod 7).You want 5^(2n) + 3 * 2^(5n-2). So:If n=0 (mod 3), we get 1+3(2) = 0 (mod 7).If n=1 (mod 3), we get 4+3(1) = 0 (mod 7).If n=2 (mod 3), we get 2+3(4) = 0 (mod 7).These three cases cover all possibilities, so you are done.===========Subject: Re: someone told me ... grava .88 la saucisse et aumarteau:> Champernowne's number> 0.123456789101112131415161718192021222324....> is normal in base 10.Actually, it's normal in all bases (I mean, the number obtained by thesame construction process).Nicolas===Subject: Re: Number theory problem, or something> This is a homework problem from a course I'm not taking.> Prove that none of the numbers 1, 11, 111, 1111, 11111 and so on, are> squares of natural numbers greater than 1.As Arturo Magidin already pointed out, your attempt didn't makeany sense. You didn't narrow your numbers between two consequtivesquares. Below I give you a few hints to get you started(you may want to always hide the following hint at first:),Good luck!Hint 1: What do you know about the squares of odd integers?Hint 2: Have you tried to simply look at the remainders ofthese numbers modulo a certain single digit number?Hint 3: The number n in hint 2 has the nice property that in orderto know the remainder of an integer m modulo n, one only needs toknow the two least significant digits of m.Hint 4: Your technique could be easily made to work in the caseof an even number of ones. Simply make the observation thatif 11....11 is a square, then so is 99....99 ! This latternumber is easy to squeeze between the squares of two consecutiveintegers.Cheers,Jyrki Lahtonen, Turku, Finland===Subject: Re: Maple question>Does anyone know of a way to extract coefficients of certain expressions, like if I>have z = A*f(x) + B*g(x), how can I get A (and B) without copying it from the output? I>want something like A = get_coeff(z, f(x))coeff(z, f(x));Department of Mathematics http://www.math.ubc.ca/~israel ===Subject: Re: Number theory congruenceIra escribi.97 en el mensaje> Hi group,> I'm a little confused over some homework problems which dont appear tohave> any supporting examples/theorems (to me anyway).> For integer n > 0, use congruence theory to establish the divisibility> statement:> 7 | 5^(2n) + 3 * 2^(5n-2)> among other theorems, i have the one that says> if ca congruent cb (mod n) and gcd(c, n) = 1,> then a congruent b (mod n)> ...which implies factoring something out, which i dont see any way to do.> any help is appreciated> IraStudy 5^n (mod 7) and 2^n (mod 7), and then 5^(2n) and 3*2(5n-2) (mod 7),and add it ...===Subject: Re: Number theory congruence> Show 7 | 5^(2n) + 3 2^(5n-2) for integer n > 0Hint: 5^2 == 2^5 and 3 == - 2^2 (mod 7)===Subject: Four Color GraphsEvery complete 4-partite graph is four-colorable. The Mathworlddescriptions of Complete Graph, Complete k-Partite Graph,k-Partite Graph are essentially appropriate to the followingdiscussion.The complete graph Kn is of course n-partite. In Mathworld, thecomplete k-partite graph (Ck) is denoted Kp,q, ... ,r; where p+q+ ...+r = n. Here, a slightly different nomenclature is adopted; ie, Ck =(P1,P2,P3, ... Pk).To minimize confusion, let Pi represent partition 'i' and pi representthe nummer of vertices in partition 'i'. So we can also write that Ck = {p1, p2, p3, ... , pk}Then, C4 = {p1, p2, p3, p4}. The number of edges (Ec4) in C4 is, Ec4 = p1*(p2+p3+p4) + P2*(p3+p4) + p3*p4 For example, let p1 = p2 = p3 = p4 = n/4; then Ec4 = n/4*(n/4+n/4+n/4) + n/4*(n/4+n/4) + n/4*n/4 =[6*(n/4*n/4)} Ec4 = .375*(n^2)Let n = 16, then pi = 4; and Ec4 = .375*(16*16) = 96. To check Ec4 = 4*(4+4+4) + 4*(4+4) + 4*4 = 48 + 32 + 16 = 96.If we let Ec4 = INT(0.375*(n^2)), then the formula works for allvalues of n when p1 ~ p2 ~ p3 ~ p4. For example; for n = 14, C4 = {4,4, 3, 3}. Without wrting out the calculations in detail; Ec4 = 4*10 + 4*6 + 9 = 40 + 24 + 9 = 73 Ec4 = INT(.375*(14^14)) = INT(.375*(196)) = 73At the other end of the spectrum, let p1= (n-3), p2 = p3 = p4 = 1. Then Ec4 = (n-3)*(1+1+1) + 1*(1+1) + 1*1 = 3*n -6 We propose that 3*n-6 <= Ec4 <= 0.375*(n^2) for all n and all possiblepartitionings thereof.===Subject: Re: Key Core Error Argument>[snip umpteenth repeat of previous flawed argument]>I thought you were going to just post your proof elsewhere and stop>>cluttering up this newsgroup. You've given a URL, now go away.> Such a remarkably silly posting.> What's the definition of insanity again?> Do the same thing over and over, and expecting different results... such > as acceptance of your proof.> But you see there's talk and there's substance. My proof is> substance, and in fact, there is no error in it.> Except in step 6.> If you disagree, then point out an error in what follows.> I have. You never responded.Oh yeah, I've looked over your stuff and it's wrong.Hey, I just assumed that Twentyman was a pseudonym, and some posterpointed out that it's an actual name, which I verified, so sorry.Well given that you're giving your real name and I assumed you didn't,I'll rethink the possibility that you're serious.Now then, what you have said so far isn't correct, so why don't you gothink about it, and if you're really, really sure that you've foundsome problem with my argument, go ahead and repost and I'll try tocheck it out, and maybe comment.James Harris===Subject: Re: Number theory problem, or somethingTobias Fritz scribbled the following:>Because it is greater than 10^n and smaller than 11^n, the only way>for it to be a square of a natural is for it to be a square of x^(n/2)>where x is some natural greater than 10 but smaller than 11. There are>no such naturals, so therefore the sum cannot be a square of a>natural.> No. 110^2 lies between 10^4 and 11^4, but is a square.> Hint: consider the number 11, 111, 111,... modulus 4.> What possible values modulus 4 can square numbers have?Um. Naturals can have values 0, 1, 2 or 3 modulus 4. Squares of evennaturals can only have value 0 modulus 4. Squares of odd naturals aresquares of (even naturals + 1), so if n is an even natural, the squareof (n+1) is equal to n^2 + 2n + 1. Both n^2 and 2n have value 0modulus 4, so the square has value 1 modulus 4.But 11 has value 3 modulus 4. Adding 100 doesn't help, because 100has value 0 modulus 4. So has 100, so has 10000, and so on.I think I understand this now. It is *very* much easier than what I/-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland ---------- http://www.helsinki.fi/~palaste --------------------- rules! --------/C++ looks like line noise. - Fred L. Baube III===Subject: Aut(Petersengraph)Hello everyone.How can one show directly, without using any theorems about automorphisms ofKn and L(Kn), that Aut (L(K5)) isomorph to S5? Or how can one prove thesetheorems?sasha.mal===Subject: Re: I NEED HELP BADLY (sorry, maths not psych)Expires: 28 days>Explain to me how this model applies to an attractive >electric field.>What are the ping-pong balls and where are they coming from?>But OK, consider a small frictionless wheeled vehicle sitting on the desk in>from of you. You have a gun that fires perfectly elastic ping-pong balls at>velocity v, at the rate of one per nanosecond. You fire them at the vehicle.>Plot the 'distance versus time' graph for the vehicle.>Why do you automatically criticize something that you don't even understand? >I understand that perfectly well. If you try to propel>something forward by transferring momentum from constant>velocity momentum carriers, there is a limit. This will>model lots of things, including your vehicle (the carriers>are actual ping-pong balls), a balloon (air molecules),>a solar sail (photons), a boat pushed by a water jet.>Nobody is denying that all of these things have the>However, there are lots of things that this model does>not apply to. YOU have made the claim that it appliesNo I haven't.I merely stated that I wanted to compare the x/t curves for these models withDo I now qualify as a super-crackpot?>as opposed to real ping-pong balls.>Explain to me how this model applies to an attractive >electric field.>What are the ping-pong balls and where are they coming from?>Do you really think nobody has noticed that you've>avoided answering any questions about your model?By now, everyone will have noticed that you have no idea at all how scienceworks. Henri Wilson.See why relativity is wrong:http://www.users.bigpond.com/HeWn/index.htm===Subject: Re: I NEED HELP BADLY (sorry, maths not psych)Expires: 28 days>> accelerated beyond the operating speed of the accelerating fields, ie at 'c'.>Explain that phrase, please.>As you know, the accelerating electic field in an accelerator>So what do you mean by:the operating speed of the accelerationg field?>You don't know what happens in close proximity to the accelerating charge. Back>radiation?>When you can't answer, ask a question, eh?>Yes, we know very well what happens to the accelerating charge.>No mysteries here.You think you kow the maths. You don't know the 'physics'. >You didn't answer.>I suppose your words are supposed to mean something.>My simple question is:>Can you please explain what:the operating speed of the acceleration field>means?No I cannot explain. That is why I am looking for physical ideas tat mighthelp.>Or are you babbling words void of meaning?Please read 'The Real Einstein' again.HaHaHaHa!!!!>PaulHenri Wilson.See why relativity is wrong:http://www.users.bigpond.com/HeWn/index.htm===Subject: Re: {Group theory} Number of units? permission for an emailed response.X-Zippy-Says: Thousands of days of civilians ... have produced a... feeling for the aesthetic modules --> But I cannot figure out a good way of defining this, as your {6}> illustrates above. :( Can you think of anything?If you want it so that non-isomorphic groups have different identities,then you can have a class of such, but it's a proper class, not a set,because there is a group of every cardinality.===Subject: Re: naive geometry questions>It *sounds* extremely cool but, being spatially challenged, I can't follow it>yet. I'm still trying to figure out what the surfaces look like. Is this>right?...>HELICOID:>plots[cylinderplot]([r*cos(theta),r* sin(theta),5*theta],theta=0..2*Pi,r=0..5,>coords=cylindrical); >CATENOID: plots[implicitplot3d]((y^2+z^2)^(1/2)=(1/2)*(>exp(x)+exp(-x) ),x=-2...2,y=-2..2,z=-2..2);> ...Google for helicoid catenoid to find web sites showing this inaction. For example:http://www.cogsci.indiana.edu/farg/harry/mat/ helicoid.htmIf you want details, this classical result is discussed in many textson differential geometry (e.g., O'Neill, Elementary DIfferentialGeometry, or DoCarmo, Differential Geometry of Curves and Surfaces,which has illustrations of the deformation).John ell===Subject: Re: I NEED HELP BADLY (sorry, maths not psych)Expires: 28 days> Let's be serious for once.> In the elastic case, the water (or stream of ping pong balls) ends up moving> backwards at Vo-2v, so the change in momentum per second is 2m(Vo-v)>Note that this is based on the assumption that the mass of>a ping-pong ball is very much smaller than M.>This is of course an approximation.I don't want an approximation. the relative masses shouldn't make anydifference.>>v = Vo*(1 - exp(-2*m*t/M))>Incidentally, particularly in the case of the elastic 'ping-pong ball drive',>momentum is bced but does the (kinetic) energy equation match?>Not quite, due to the approximation mentioned above.Why are you only producing an approximation?The (classical) equation is Md2x/dt2=2m(Vo-dx/dt) , is it not?That is MD^2+2mD-mVo=0, where m is the mass per second leaving the source and Mis the mass of the body..The roots are -K(+/-)sqrt(K^2-4KVo), where K=m/MSo a solution is: x=(e^-mt)[e^root()-e^root()] +AIsn't this tanh(t)?I'm a bit rusty on this stuff.>PaulHenri Wilson.See why relativity is wrong:http://www.users.bigpond.com/HeWn/index.htm===Subject: Re: Number theory problem, or something> Um. Naturals can have values 0, 1, 2 or 3 modulus 4. Squares of even> naturals can only have value 0 modulus 4. Squares of odd naturals are> squares of (even naturals + 1), so if n is an even natural, the square> of (n+1) is equal to n^2 + 2n + 1. Both n^2 and 2n have value 0> modulus 4, so the square has value 1 modulus 4.2n isn't 0 mod 4. It is 0 or 2 mod 4./David===Subject: Re: I can't stand it anymore>I have been biting my tongue about the IQ test but I can't any more.>How reliable is a test that use the term Asian to represent the most>diverse of ethnic and cultural groups?> IQ tests are not particularly reliable, but what does this have to do> with it? What test is using that term? And how is any of this relevant> to these newsgroups?>If you want to know more about the diversity, ask me and I will give>you tons of specific examples.> ...> Your examples may be interesting, but what do they have to do with> sci.math?> Sci.math was one of the group listed in the thread where I saw the> discussion of IQ test. I had no idea who were from which group in that> thread and so decided to posts in all 3 groups. The only science group> I posted before was sci.chem group.> I don't browse sci.chem group regularly either. So when I broswe the> posts once in a while and see some supposedy intelligent people from> these 3 groups talking about IQ tests as a ver important tests (with> some seeing it as the ultimate test to determine human intelligence),> I get very disappointed.> So let's see; in your opinion, talking about IQ tests is> an indication of low intelligence or some social backwardness?Unbelievably childish response.===Subject: Re: I NEED HELP BADLY (sorry, maths not psych)Expires: 28 days>> In the elastic case, the water (or stream of ping pong balls) ends up moving>> backwards at Vo-2v, so the change in momentum per second is 2m(Vo-v)>v = Vo*(1 - exp(-2*m*t/M))>Note:>According to Newton the momentum>p = m*Vo*(1 - exp(-2*m*t/M)) decreases with time.>Newton predicts that there is a speed limit because the ping pong>ball will cease to transfer momentum to the mass when its speed approaches Vo.Is that your approximte solution? See my other message.>[..]>There is no reason for making this more complicated than it is, Henry.>The mass M will in all cases end up moving with the speed Vo.>Or rather, v will approach Vo asymptotically.>But you knew this, didn't you?>And this have absolutely nothing to do with what happens>Which you also knew, didn't you?>Not at all Paul.>I have to compare the curve given by the above solution with that predicted by>SR for a charge accelerated in a constant field.>Can you tell me what that might be?>Sure I can.>But first, I will tell you what Newtonian mechanics predicts.>Let the mass m with the charge q be in a static electric field Eo.>Let v be the normalized speed, that is the speed is v*c.>Let the mass accelerate from standstill.>What is v(t)?>Solution according to Newton:>---------------------->The force on the the mass is constant Fo = q*Eo>Newton predicts that the momentum p = Fo*t>increases linearly with time.>Fo*t = m*v*c>v = t/T where T = m*c/q*Eo>Newton predicts that the speed will increase lineray with time,>and will pass the speed of light at the time T = m*c/q*Eo.>There is no speed limit because the electric field never>ceases to transfer momentum to the mass.Well we know that is certainly wrong.Why it is wrong remains to be explained 'physically'.>Solution according to SR:>-------------------->The force on the the mass is constant Fo = q*Eo>SR predicts that the momentum p = Fo*t>increases linearly with time.>Fo*t = m*v*c/sqrt(1 - v^2)>v = (t/T)/sqrt(1+(t/T)^2) where T = m*c/q*Eo>SR predicts that the speed will approach the speed of>light asymptotically. v = 2^-0.5 = 0.707 at the time T = m*c/q*Eo.>c is the speed limit despite the fact that the electric field never>ceases to transfer momentum to the mass.>Of course two curves which both are starting at zero and>are approaching a limit asymptotically will show a superficial>resemblance.>But it makes no sense to compare the Newtonian prediction>for a scenario to the SR prediction for a completely different>scenario, and conclude that because the predictions show>a slight resemblance there must be a causal connection>between the predictions.Except that one is a physical explanation and the other a purely mathematicalone that sheds no light on anything.>What makes very much sense, though, is to compare>Which I did above.>They are very different.You didn't take into account any of the limiting factors. You must know this is plain silly.>And we both know which of them are in accorce with>experimental evidence.>Don't we?Yes Paul . Read all about the evidence in 'The Real Einstein' .>PaulHenri Wilson.See why relativity is wrong:http://www.users.bigpond.com/HeWn/index.htm===Subject: Re: I NEED HELP BADLY (sorry, maths not psych)Expires: 28 days>> Let's be serious for once.>> Consider an object being accelerated by a idealistic jet of water or a>> continuous 'stream of elastic ping pong balls'. What is its subsequent velocity>> pattern?>> accelerated beyond the operating speed of the accelerating fields, ie at 'c'. I>> hope it might also produce a relationship that is equivalent to mass>> 'appearing' to increase with velocity by gamma.)>Please define what you mean by the operating speed of an>accelerating field. I've been in physics research for over 20>years, and I've never heard that term.>Do you mean phase velocity, group velocity, how fast the>operators turn the knobs, what?>I think you forgot to answer the question, Henry.>Why is that?yes, for some strange reason I 'forget' to answer anything EjP asks.>PaulHenri Wilson.See why relativity is wrong:http://www.users.bigpond.com/HeWn/index.htm===Subject: Re: Greek Alphebet> Thus, alpha comes from aleph, which means Ox. And gamma comes> from gimel... so one can look at a book on the history of the> alphabet. And the 22 letters of the Hebrew alphabet are alleged to> have meaning on other levels too.Sounds like a case of putting two and two together to make twenty-two : )===Subject: Re: JSH: My victories get lost> :) good one!> LHConfused by the [original] subject line. Plenty of null sets in theuniverse, no need to lose one.===Subject: Re: JSH: It begins> As for my posting, it may increase dramatically if I see that pumping> up the volume is effective.> So there may be many MORE posts from me on sci.math, not fewer.> James HarrisSpoken like a sullen, petulant juvenile. Pumping up the volume has nothing to do withmathematical correctness. It is behavior more appropriate for an ape than a man. Did youlearn that shouting louder was an algebraic method for proving something?There are two things you must never attempt to prove: the unprovable -- and the obvious.Democracy: The triumph of popularity over principle.http://www.crbond.com===Subject: prime primitive rootIs it true that every odd prime has a prime primitive root?How does the proof go or where can I find one?===Subject: Re: Number theory congruence> Hi group,> I'm a little confused over some homework problems which dont appear to have> any supporting examples/theorems (to me anyway).> For integer n > 0, use congruence theory to establish the divisibility> statement:> 7 | 5^(2n) + 3 * 2^(5n-2)> among other theorems, i have the one that says> if ca congruent cb (mod n) and gcd(c, n) = 1,> then a congruent b (mod n)> ...which implies factoring something out, which i dont see any way to do.> any help is appreciated> IraIf nothing else seems to work, you can do it by direct evaluation for enough cases from n = 1 , n = 2, ..., to prove the pattern. To simplify some of your calculations, note that 7 | (k^7 - k) for all integers k, or, equivalently, 7 | (k^6 - 1) for every integer k coprime to 7.In mod notation, k^7 == k (mod 7), for all integers k, ork^6 == 1 (mod 7) for integers k such that k =/= 0 (mod 7).===Subject: Re: Product of Reals> It took me a couple of minutes to realise why the uncountable sums did> not work. Is this right? If the sum converges then clearly the> number of terms with absolute value greater than 1 must be finite. > Similarly the number of terms with absolute value between 1 and 1/2 is> also finite, ditto 1/2 and 1/3, 1/3 and 1/4 so the non-zero terms are> the union of a countable set of finite sets and hence are countable. > Simple, once you know.Yes, that's right.> If the structure of N (natural numbers) is ignored and this definition> of sum is used then I expect that only the series which are absolutely> convergent in the common sense will be convergent in this sense.Exactly.> Considering the countability limitation, is this generalised notion of> convergence useful and commonly used?Measure theory also depends heavily on countable additivity, and yet thishas not proved to be much of a handicap.Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. This is a homework problem from a course I'm not taking.> Prove that none of the numbers 1, 11, 111, 1111, 11111 and so on, are> squares of natural numbers greater than 1.> I can't seem to do it completely. Here's my best attempt so far:Hint: The square of an odd number is congruent to 1 mod 4.===Subject: Re: Number theory problem, or somethingDavid Rasmussen escribi.97 en el mensaje> Um. Naturals can have values 0, 1, 2 or 3 modulus 4. Squares of even> naturals can only have value 0 modulus 4. Squares of odd naturals are> squares of (even naturals + 1), so if n is an even natural, the square> of (n+1) is equal to n^2 + 2n + 1. Both n^2 and 2n have value 0> modulus 4, so the square has value 1 modulus 4.> 2n isn't 0 mod 4. It is 0 or 2 mod 4.Joona said: ... so if n is an even natural ...===Subject: Re: I can't stand it anymore