mm-2 problem:How to solve this integer matrix equation?[ a1^2*u1, a1*a2*u2, a1*a2*u3, a2^2*u4][ a1*a3*u1, a1*a4*u2, a3*a2*u3, a2*a4*u4][ a1*a3*u1, a3*a2*u2, a1*a4*u3, a2*a4*u4][ a3^2*u1, a3*a4*u2, a3*a4*u3, a4^2*u4]=[ 16 16 -16 -4][ 4 -16 -4 4][ 4 4 16 4][ 1 -4 4 -4]a1, a2, a3, a4, u1, u2, u3, u4 should be 2s positive integer power, suchas -2, +1, etc.I considered to factor the right hand side(the constant matrix) ?st:a1^2*u1 =16=4*4*1a1*a2*u2=16=4*4*1a1*a2*u3=-16=-4*4*1a2^2*u4=-4=-2*2* 1The try to allocate/assign these 2s factors to the variables a1, a2, a3,a4, u1, u2, u3, u4...Since there are more equations than unknowns, sometimes there arecon?g assignment to the variables, so only approximated assignment canbe made to minimize the mean square error between the left hand side andright hand side in the above equation...It is doable by exhaustive search, for example, since there are 16 equationsand 8 unknowns, I can do 8-level iteration like the following:for a1=[-4, -2, -1, 1, 2, 4] do for a2=[-4, -2, -1, 1, 2, 4] do for a3=[-4, -2, -1, 1, 2, 4] do for a4=[-4, -2, -1, 1, 2, 4] do for u1=[-4, -2, -1, 1, 2, 4] do for u2=[-4, -2, -1, 1, 2, 4] do for u3=[-4, -2, -1, 1, 2, 4] do for u4=[-4, -2, -1, 1, 2, 4] do compare MSE between LHSvsRHS for all cases keep the best one end end end end end end end endThe only problem is that one the matrix gets larger, with hundreds ofunknowns, the above method cannot be used any more...Can anybody give me more ideas and pointers?-Walala =Let H(n) = sum{k=1 to n} 1/k, the n_th harmonic number.Prove, for each m = positive integer,sum{k=1 to m-1} H(k) k! (m-k)!is congruent to(1/2) m! H(?/2))) (-1)^m (mod (m+1)) .(I do not know if the result is trivial {or am certain it iscorrect}.)By the way, for those interested in harmonic-number math puzzles(limits, this time, instead of congruences), here is a link to arecent sci.math math-puzzle no one has yet answered.(I am including the link because I am cross-posting this torec.puzzles, as well as to lr=&ie=UTF-8&group=sci.math&safe=off&selm=b4be2fdf P. 50, of Contemporary Abstract Algebra, Theorem 2.1 Uniqueness of theIdentity states In a group G, there is only one identity element.1. ae=a for all a in G, and2. ea=a for all a in G.The choice of a=e in (1) and a=e in (2) yields ee=e and ee=e. Thus, e ande are both equal to ee and so are equal to eachother. QEDI do not understand the proof. I dont see how, after 1. and 2., he makes thejump to say the choice of a=e in (1) and a=e in (2) yields ee=e and ee=e. I actually have a different proof that seems ?e but it really bothers methat I cant follow where he went with this proof. I dont see how he could(or why he would want to!) claim that a=e from (1) or a=e from (2). Why didnt he just start off with the identities themselves, and demonstratethat they were equal to eachother? => On P. 50, of Contemporary Abstract Algebra, Theorem 2.1 Uniqueness of the> Identity states In a group G, there is only one identity element.> 1. ae=a for all a in G, and> 2. ea=a for all a in G.The choice of a=e in (1) and a=e in (2) yields ee=e and ee=e. Thus, e> and e are both equal to ee and so are equal to eachother. QED> I do not understand the proof. I dont see how, after 1. and 2., he makes> the jump to say the choice of a=e in (1)This is called instantiation, that is the process of gettinga particular proposition by substituting in a general proposition.For another example,consider the proposition(i) for all odd integers n, n^2 is odd.By choosing say n = 7 in (i) we get a consequence:(ii) 7^2 = 49 is oddor by choosing n = 9999 we get(iii) 9999^2 = 99980001 is odd.So here we substitute e for a in 1. (OK as e is an element of G)and obtain e e = e.> Why didnt he just start off with the identities themselves, and> demonstrate that they were equal to eachother?What?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) => On P. 50, of Contemporary Abstract Algebra, Theorem 2.1 Uniqueness of the> Identity states In a group G, there is only one identity element.> 1. ae=a for all a in G, and> 2. ea=a for all a in G.> The choice of a=e in (1) and a=e in (2) yields ee=e and ee=e. Thus, e and> e are both equal to ee and so are equal to eachother. QEDThis doesnt immediately answer your question, but theres apparently aee=e and ee=e, which leads to the conclusion that e = e.> I do not understand the proof. I dont see how, after 1. and 2., he makes the> jump to say the choice of a=e in (1) and a=e in (2) yields ee=e and ee=e.> I actually have a different proof that seems ?e but it really bothers me> that I cant follow where he went with this proof. I dont see how he could> (or why he would want to!) claim that a=e from (1) or a=e from (2). He didnt claim that a had any particular value; he said a particularstatement holds for every a in G.Let P(a) be the property that ae=a. Then statement (1) says P(a) holdsfor every a in G. In particular, e is a member of G and therefore P(e)holds. That means ee = e. Applying similar reasoning in (2) we getee = e.> Why didnt he just start off with the identities themselves, and demonstrate> that they were equal to eachother?You mean, e = ee = e? The ?st equality holds because e is anidentity, and the second equality holds because e is an identity. Thatsbasically what he did, except that (1) and (2) spell out the reasoning abit more fully.-- Dave SeamanJudge Yohns mistakes revealed in Mumia Abu-Jamal ruling. =>On P. 50, of Contemporary Abstract Algebra, Theorem 2.1 Uniqueness of the>Identity states In a group G, there is only one identity element.>1. ae=a for all a in G, and>2. ea=a for all a in G.>The choice of a=e in (1) and a=e in (2) yields ee=e and ee=e. Thus, e and>e are both equal to ee and so are equal to eachother. QED>I do not understand the proof. I dont see how, after 1. and 2., he makes the>jump to say the choice of a=e in (1) and a=e in (2) yields ee=e and ee=e.Since ae=a for all a in G, then it is true that ae=a for any speci?example of a as an element of G. Since e is an element of G, then itfollows that ee=e. Since ea=a for all a in G, then it true that ea=afor any speci? example of a as an element of G. Since e is an elementof G, then it follows that ee=e. I cannot see what the trouble is. A statement which is known to be true for all elements of G must be true for any speci? element. That is all that is needed to conclude that ee=eand that ee=e.> I actually have a different proof that seems ?e but it really bothers me>that I cant follow where he went with this proof. I dont see how he could>(or why he would want to!) claim that a=e from (1) or a=e from (2). He did not make such a claim. He substituted a=e into (1). He has theright to do this since (1) holds for every possible value that a can takefrom G, and e is such a possible value. He substituted a=e into (2). Hehas the right to do this since (2) holds for every possible value that acan take from G, and e is such a possible value.>Why didnt he just start off with the identities themselves, and demonstrate>that they were equal to eachother?So how would you recommend that he does this without using the quanti?r rules in the manner that he did above?David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills. => Leroy Quet>> ....>> And A(r) is, if I am right, ...>oo>> --- (1+1/2+1/3+...+1/k)>> ------------------->> / k^(r+1)>> --->> k=1> >= sum{k=1 to inf} (1+1/2+1/3+...+1/k)/ k^(1+r),> >which is an Euler sum.>> [ http://mathworld.wolfram.com/EulerSum.html ]>> I wonder naively>> if there is some kind of zeta-function-like re?n formula for the>> Euler-sum analytical continuation which relates> >E(r) and E(2-r),> >where E(r) is>> sum{k=1 to inf} (1+1/2+1/3+...+1/k)/ k^r.> >(I know of no study regarding analytical continuating Euler sums at>> all, nor anything beyond the consideration of such at r = integers >=>> 2.)> I looks to me as if these Euler sums just havent been studied very much.> You might try this question on the pros over at sci.math.research. There> might be some vast literature to which most of us are quite oblivious :)> LHI would post to sci.math.research, but I feel a little queezy aboutdoing this because there have been SO MANY(!) math-questions I haveposted to (our lower -level) sci.math over the years which might beapt for sci.math.research posting also; and I feel like I am slightingthose other questions somehow...;) In any case, I feel comfortable staying here at this relativelylow-pressure math newsgroup...(I have posted in my other reply to this thread a *little* regardingthe zeros of E(r), by the way. => I will copy/paste the last half of my post made at:> prev=> and then bottom-post more.> ....>And A(r) is, if I am right, ...> oo>--- (1+1/2+1/3+...+1/k)> ------------------->/ k^(r+1)>--->k=1>= sum{k=1 to inf} (1+1/2+1/3+...+1/k)/ k^(1+r),>which is an Euler sum.>[ http://mathworld.wolfram.com/EulerSum.html ]>(For example,>A(1) = 2*zeta(3).)>(Right?)>So, since q(r,m) = >sum{k|m,k<=sqrt(m)} k^r /m^r>(note inequalitys direction here, as opposed to in q()s de?ition),>I wonder naively>if there is some kind of zeta-function-like re?n formula for the>Euler-sum analytical continuation which relates>E(r) and E(2-r),>where E(r) is >sum{k=1 to inf} (1+1/2+1/3+...+1/k)/ k^r.>(I know of no study regarding analytical continuating Euler sums at>all, nor anything beyond the consideration of such at r = integers >=>2.)I must ask again, is there anything commonly known about the> analytical continuation of Euler-sums, such as where its zeros (if> any) are??> > Is there any kind of zeta-like continuation-formula (analogous to that> which relates zeta(z) with zeta(1-z)) for Euler sums?Are there any known closed forms, in terms of zeta-functions or> whatever, for any E(r)s where r is *not* a positive progress regarding ?ding the zeros of E(r).;) If the sums converge,E(r) = zeta(r+1) *F(r), where F(r) = sum{k=1 to oo} G(k)/k^r, and whereG(k) = sum{1<=j<=k, GCD(j,k)=1} 1/j.So, if by analytic continuation, r is such that:zeta(r+1) = 0 and F(r) is ?ite,or F(r) = 0 and r is not 0 (because zeta(r+1) is then a pole of zeta),then E(r) = 0 by analytic continuation.(And *perhaps* E(r) is still zero if F has a pole =what the probe observed was not beyond the heliopause;this could be the tenperature of sub-galactic space; eh -- unless,you believe in a vacuum; right? anyway, if Universe really is acceleratingaway from us, ever-faster, then superluminal metric engineeringwill surely be needed. just dont count on Jack, who never botherswith his readers, to dyscuss his up-and-coming breakthrough!... besides,his great-great-great-granchildren are coming after him -- and*their* great-great-greats are going to get them, two,just to be sure! ah, so; notice how he uses a bevy of Astronomers and Spooks Royal,like Rees Flipbook Universe. > The ordinary matter in the Attractive Stuff is only ~ 0.04 which is > same order as the error bars on these numbers. > Not only that, the technological potential of manipulating the exotic > vacua in metric engineering for fast space and also time travel is > compelling. Herman Bondi saw some of this coming 45 years ago when he > was Chief Scientist of the British Ministry of Defence. I heard him > lecture on this at Cornell back then.Its invisible, it permeates every pore of space and it is remorselessly> driving the galaxies apart. The discovery of dark energy nearly six> years ago was a bombshell dropped on the world of cosmology - and voted> international team of astronomers is controversially claiming that it> could all be a big mistake. The dark energy could be a huge cosmic> mirage, says Subir Sarkar of the University of Oxford. It may not> exist at all. > Because dark energy is a property of space, as space expands, the amount> of dark energy grows and, with it, its anti-gravity effect.> Consequently, the gravity that is sucking in matter to make new galaxy> clusters is eventually overwhelmed by dark energy, preventing any more> galaxy clusters from being born. If the Universe contains dark energy,> therefore, the number of galaxy clusters in a given volume should have> levelled off at some time, not continued increasing until the present> time, says Sarkar.Astronomers such as Cambridges Sir Martin Rees think the behaviour of> that our> lack of understanding of the complexities may be leading Blanchard and> his colleagues to an erroneous conclusion. Taken at face value, however,> the XMM results imply that there is no dark energy. > NO! What these people do not realize is that dark matter and dark > energy ARE BOTH THE made the original supernova> observations that led to the announcement of dark energy stand by their> result. According to Alexei Filippenko of the University of California> at Berkeley, leader of a second team at Berkeley beside Perlmutters,> the supernova observations are really pretty convincing now.Bizarrely, even if the supernova observations are wrong, it may not> matter. According to Rees, this is because the dark energy result is now> not dependent on a single observation but on a large, interlocked set of> observations. One of the most important is an observation made last year> by Nasas Wilkinson Microwave Anisotropy Probe (WMAP).The probe observed the cosmic background radiation, the relic heat of> the Big Bang ?eball. Tiny variations in the temperature of this> afterglow across the sky turned out to be compatible with the Universe> having the critical density, with 30 per cent of its mass in the form of> matter - both visible and dark - and 70 per cent in the form of dark> energy. In other words, they con?med the existence of dark energy.--Give the Gift of Trickier Dick Cheeny -- out of of?e, at last!http://www.benfranklinbooks.com/http://www.wlym.com/pages /music.htmlhttp://www.rand.org/publications/randreview/issues/ rr.12.00/http://members.tripod.com/~american_almanachttp:// =bccNo closure yet. ;-)I am trying to get some closure on this debate of at least 2 years thathas been interesting in that it has deepened my heuristic understandingof Einsteins magni?ent achievement and of the cheap attempts toreplace it as in Hal Puthoffs PV approach to metric engineering(exotic UFO warp drive). I support a lot of Hals scienti? work BTW,but not this particular item.PZ: Lets be clear that I also view Einsteins theory as a magni?ent achievement, evenif the position I have been arguing here is eventually vindicated.It is certainly a beautiful theory. A little too beautiful, perhaps.JS: Not possible IMHO.PZ: The question in my POV is, from the standpoint of gravitational *physics*, is it theproper point of departure for heuristic development of a viable theory of quantumgravity? And, does classic GR, with its Einsteinian equivalence principle, faithfullyre?he true nature of physical gravitation?JS: In my opinion it does. There is no problem with Einsteins original idea of the equivalence principle i.e.The non-tensor g-force is locally equivalent to an inertial force. This is always true, independent of the local tensor curvature at the same point.Also, the g-force can be locally eliminated by switching the test with it at the same point.Paul Zielinski is concerned with distinguishing arti?ial gravity withreal gravity.PZ: In relation to a dialectical examination of the true character of Einsteins equivalencehypothesis and its implications for the energy-momentum content of the permanentgravitational ?ld.JS: I think I have solved all that - at least to my satisfaction. Einsteins ideas have survived your critique IMHO.The latter, would only happen when the local 4th rankcurvature tensor did not vanish. Paul seems to think there is alogical contradiction in the foundations of Einsteins thinking that isglossed over in MTW (1973) Gravitation. I do not think Paul iscorrect here.PZ: I have been arguing that there is an interpretive inconsistency in the MTW treatment.JS: I fail to see there is any problem here once it is clear that the total experimental arrangements (Niels Bohr) for measuring gravity in contrast, for measuring local tensor curvature at 4th rank level with ZERO g-force) are incompatible or complementary in Bohrs sense, there is no longer any conceptual problem at all here! It is curious that Bohrs quantum measurement ideas work as well for General Relativity as they do for Quantum Theory. This may be a clue for the correct theory of quantum gravity. Note my claim that Einsteins GR is a MACRO-QUANTUM theory based upon Vacuum Coherence missing in most approaches although not in Chaplines and Voloviks independent approaches.PZ: I am saying that the MTW committee apply both a modern interpretation, in whichthe question of non-vanishing Riemann curvature is taken seriously as a mark o?equivalence between inertial and permanent gravitational ?lds, and also the classicEinsteinian interpretation, in which it is not -- all depending on the context.JS: That distinction has absolutely no impact on Einsteins local equivalence principle as I de?e it above.mish-mash.JS: I do not think so at all.The relevant form of Einsteins equivalence principle is that:weight or g-force proportional to the 3rd-rank non-tensora time-like geodesic (neutralize the charge) and the g-force vanishes.In this sense, gravity is locally equivalent to an accelerating framedependent inertial force.PZ: Yes, this is the local translational aspect.This is also true in Newtonian theory -- but the Newtonian *interpretation* is of coursevery different.The point here is that the GR inertial its physicalenergy content.JS: Your last sentence seems to garble things.The TOTAL g-force IS an inertial force not just This may be the root of your confusion?The LNIF connection ?ld is of the formg-force = Indexed coeffcient (Third rank tensor + inhomogeneous term) with summation convention, i.e. big sum of termsApply the local tetrad transformation to get(g-force) in LIF = 0The gravity force is the TOTAL SUM of the tensor part and the inhomogeneous part.It is an error to think that the inhomogeneous part is the inertial force.g-force = inertial force is local EEPThe 4th rank curvature tensor is irrelevant at this level.In math terms(^u|vw) = tensor + non-tensoru,v,w are LNIF indicesa,b,c from LNIF to LIFi.e. eu^aeb^vec^w(^u|vw) = (^a|bc) = 0The you take partial derivatives of the connection to get curvature you cannot maintain this clean separation.You have missed this mathematical fact IMHO.You cannot consistently think of the tensor part of the connection ?ld as the real gravity with the non-tensor part as the fake gravity.PZ: A good model for this is the rest-frame behavior of a charged a *pure* electrostatic ?ld, as treated in standard GR, which is there interpreted as*inertial compensation*: the net apparent translational forces on the this frame, but the *electrostatic ?ld* does not.rest frame of the charge here is an LNIF on a timelike non-geodesic with a net g-force, or weight, on the charge from the local EM ?ld that is equal and opposite to the electrostatic reaction force from the charge on the EM ?ld (or equivalently on the distant charges making the EM ?ld). That is, the total momentum of charge and EM ?ld is conserved. There is no free ?eightlessness in the LNIF rest frame of the charge.Off the top of my head (I could be making a mistake?):P = p - (e/c)A = 0 by de?ition in the LNIF rest frame.withDP/ds = Dp/ds - (e/c)DA/ds = 0in the rest LNIFwhere D/ds is a covariant proper time derivative?Your example is wrong IMHO. A tiny strain gauge on the charge in the electrostatic ?ld shows a weight. The non-geodesic time-like motion depends on e/m. In no way could the electrostatic force on a charge be confused with a gravity force. There is no universality. Your example is equivalent to us standing on the surface of the Earth. We have no net charge, but there are induced electric and magnetic multi-pole forces on neutral bodies that have a similar effect as in your less complex example.line that break down in quantum gravity, also lack of a torsion-spincoupling.PZ: Yes. Here we are at the least ignoring local rotational effects.without an electric charge in an external EM ?ld. Neglect weak andstrong charges for simplicity for now.PZ: OK.JS: Note that I is independent on the contingency that the 4th rankcurvature tensor vanishes or does not vanish on the world line of theTherefore, the decomposition suggested by ZielinskiPZ: ?!JS: guv = guv(arti?ial) + guv(real)makes no physical sense at all IMHO.PZ: I agree this makes no physical sense. However, I NEVER SUGGESTED ANYSUCH THING.PZ: I am NOT proposing a linear decomposition of the uni?d g_uv of this sort; I amproposing a linear decomposition of the GRADIENTS g_uv, w of the metriccomponents g_uv with respect to the coordinates x_u ONLY.JS: OK even for the gradients the distinction is no good at the next curvature level. BTW I seem to rememberPZ: Neither do bimetric theories entail any such decomposition: they de?e two separatemetrics, a deformable physical metric g_uv, and a transformable kinematicalmetric gamma_uv over a ?ckground manifold.JS: Angels dancing on the head of a pin IMHO.PZL So yourguv = guv(arti?ial) + guv(real)is a straw man of your own invention.Jack, you will at least have to clear up this basic confusion if you want real closureof the debate!JS: I handled the gradients above.II. Measurement of the local 4th rank curvature tensor, outside theregime of quantum gravity, uses the time-like geodesic motion of a pairEinsteins geodesic deviation equation for stretch-strain relativetidal accelerations.PZ: This a merely an assertion. In fact it is, strictly speaking, false. The correct versionis:... Measurement of the local 4th rank curvature tensor, outside the regime ofquantum gravity, CAN USE the time-like geodesic motion of a pair of neighboringThis is a very important detail. This is explained at length in Ohanian & Ruf?i,Gravitation and Spacetime, Sec 1.9.JS: I do not have immediate access to 1.9. What is the relevant quote? Any valid alternative cannot contradict the method with a pair of test completely independent of the measurement of g-force on a single testPZ: Yes.JS: Indeed, the two kinds of measurement are incompatible in the sense ofBohrs principle of complementarity.PZ: OK, I have to agree that you seem to be injecting Bohrian complementarity into thisdiscussion of GR.This is a very serious issue IMO. In my POV it is a license for *semantical incoherence*.It reduces all discussion of interpretation of a formal-empirical framework in terms ofconcrete models to a matter of subjective preference (personal belief -- W. Heisenberg).It encourages equivocacy and pedagogical caprice.This attitude represents a troglodyte theory of science that is no longer taken seriously amongstlack of understandingof the epistemologically essential role that concrete physical models -- and the semantic integrityof such models -- play in scienti? explanation, empirical prediction, and heuristic development.It is what is fundamentally wrong with so-called modern physics, IMHO.Ironically, the born-again realist Einstein later himself emphatically rejected this pseudo-positivist approach. And as you know, Jack, the late-Einsteinian neo-realist torch was latertaken up by David Bohm.So Jack, where do you really stand on all of this? Are you now ?ally unmasked as aCopenhagenist?JS: Not really, only on Tuesdays, Thursdays and Sundays. ;-)I am a pragmatist. If the shoe ?s, wear it.III. What about the nonlocality of energy and momenta of the puregravity ?ld? This has to do with the issue of gravity waves. You needto split the near ?ld from the far ?ld. You need to de?e a global?tegral over a space-like slice for total Pu (also Muv forangular momentum), for example, using an effective non-tensor forONLY the far ?ld dynamical degrees of freedom!PZ: The point is that moving matter exchanges energy-momentum with the ?ld. Itcan do this in the standard theory by gravitational radiation.JS: What moving matter? Take a gravity wave detector in the Hubble ?r example.PZ: A gravitational pulse propagates through the vacuum in a lawful manner.JS: Wait! An EM wave propagates relative to a ?ed space-time geometry in Special Relativity.A gravity wave IS wiggly space-time geometry. So what is the reference frame for the wiggles?You need to split the degrees of freedom of gravity into propagating and non-propagating pieces.PZ: It carriesenergy momentum. Yet, according to the canonical theory, while it is in ? wecannot say exactly where the energy-momentum carried by the pulse is. It is,strictly speaking, everywhere and nowhere.At the same time, we are told that the pulse propagates at two separate speeds;one (the propagation speed of the curvature component) is objective, while the other(the propagation speed of the gradient component) is frame-dependent.My take on all this is that modern physicists are so punch-drunk with Alice-in-Wonderland quantum mechanics that they are now ready to swallow just aboutanything. You could almost convince them that the moon is made of green cheese if some Copenhagenist techno-priest says so.What energy-momentum conservation principles apply to the matter-?ld system?There is no sensible answer to this basic physical question in orthodox GR. This isa real problem: Einstein himself saw this, and so did Schrodinger, Bauer, and Laue,among others, in the early days of GR.JS: What do you mean? In orthodox GRTuv^;v = 0Philosophy ofVacuum (The Mass of the Classical Vacuum). Even Penrose understands that thisis a serious problem in orthodox GR.PZ: Now we have Lee Smolin chasing his tail trying to ?d quasi-local conservedquantities in GR. All such efforts are doomed IMO for the reasons I have been arguing:there is a fundamental problem with the Einstein equivalence hypothesis.The solution of this problem, I am arguing, must go to the roots of Einsteinsrelativistic thinking. It will require a radical deconstruction of the early Einsteinianphilosophy -- of the depth undertaken, ironically, by none other than Einstein himself,starting in the early 1920s.WILL THE REAL ALBERT EINSTEIN PLEASE STAND UP?JS: Note that the elimination of dynamical degrees of freedom of theelectromagnetic ?ld by Wheeler and Feynman, extended to the gravity?ld by Hoyle and Narlikar, only applies to the far ?ld on the lightcone not to the near ?ld off the light cone!PZ: OK.JS: What about Freuds identity as in Yilmazs theory? It is not relevantbecause it attempts to replace curved space-time covariant divergencesby global ?ce -time ordinary divergences.PZ: The Freud theorems (Freud decomposition and Freud identity) are purelymathematical propositions that are true for all geometrodynamic theories ofthe GR type. These do not depend on speci? versions of the RHS of the?ld equations.JS: Agreed. Nothing I said contradicts that.unphysical use ofcovariant divergences, and this is part of the problem.JS: The Question is: What is The Question? WheelerIt is useful intreating gravity waves in asymptotic ?ace-times, but it is notmore than that. Yilmaz is mistaken to try to elevate it as a basis fora new theory.PZ: But Yilmaz doesnt do that, Jack. He simply uses it as a mathematical *point ofdeparture* for an alternative development and interpretation of GR. You havesimply not understood his argument.JS: A delicate distinction. ;-)PZ: IMO this is no different from Bohms use of a trivial decomposition of theSchrodinger wave function as a *point of departure* for his neo-realist interpretationof QM.JS: No comparison IMHO, but that is another long story.PZ: Or do you now have a problem with that? Was Bohm mistaken to try to elevateit as a basis for a new theory? In your current opinion?JS: Loaded question based on a false comparison IMHO.There is a local stress-energy density covariant tensor for the puregravity ?ld, it is simplytuv(Gravity) = [(Wittens String Tension)/(QED dimensionlesscoupling)]Guv(Einstein)PZ: That is not phenomenological GR! There is no string tension in GR! There is noQED dimensionless coupling in GR!JS: You are wrong. It is there, but hidden because many theoretical physicists today are really elegant tailors more into the gloss of fashion than substance. They are too quick to take G = h = c = k = 1 and thereby miss some deep connection a bit under the surface of phenomenological GR. I am offering a constructive theory like Lorentz and Fitzgerald but for GR in the sense of Sakharov and PW Andersons More is different. I am offering a kinetic theory to Einsteins 1915 thermodynamics. Einstein always was interested in doing that.PZ: You are arguing all over the shop!JS: I am asking you to scratch the surface. There is a deeper insight of Einsteins GR as an emergent MACRO-QUANTUM coherent vacuum theory.PZ: In any case, if there are such things in a deeper theory, and your statement aboveJS: tuv(Gravity) = [(Wittens String Tension)/(QED dimensionlesscoupling)]Guv(Einstein)PZ: is valid, then by correspondence this tensor tuv(Gravity) should show up at themacro-level -- which supports my argument.How do you answer this?JS: Exotic vacuum dark energy + dark matter add up to Omega ~ 0.96 with our kind of matter a mere Omega ~ 0.04.This is IMHO a crucial fact and test of my Vacuum Coherence theory of Einsteins gravity. I do not think the recent Blanchard argument that dark energy is a mirage will hold up. I agree with Martin Rees on that.Guv(Einstein) = Ruv - (1/2)RguvIn the non-exotic vacuum Guv = 0 exactly.PZ: Yes, because R_uv = 0 and R = 0. But we still have the Weyl components of Ruvwlto play with.JS: Note that if you try to writetuv(Gravity) = tuv(Gravity)near-?ld +tuv(Gravity)far-?ldthe two terms on the RHS are not covariant tensors separately.PZ: In orthodox GR.JS: This may have been part of the confusion.PZ: What confusion? You are unwittingly supporting the Yilmaz thesis by correspondence!Perhaps this is the confusion?This has nothing to do with near-?ld vs. far-?ld. There are NO less-than-global *exactly*frame-independent vacuum energy-momentum densities *anywhere* in orthodox GR.JS: Straw Man. The generally covariant vacuum stress-energy density tensor istuv(vacuum gravity) = [(String Tension)/(QED coupling)]Guv(Einstein)--> ZERO in non-exotic vacuum, i.e. no dark energy and no dark matter in the small region around P.In exotic vacua with positive zero point energy pressure (attractivedark matter Omega ~ 0.3) and negative zero point energy pressure(repulsive dark energy Omega ~ 0.7)tuv(Gravity) = - (Witten String Tension)/(QED dimensionlesscoupling)/zpfguv/zpf = (Loop Gravity Area Quantum)^-1[(Loop Gravity AreaQuantum)^3/2|Vacuum Coherence|^2 - 1]Witten String Tension = hc/(Loop Gravity Area Quantum)The discrete stringy link edge in a spin network is dual to the LoopGravity Area Quantum) ~ Bekenstein BIT.guv = Einsteins Special Relativity Metric + (Loop Gravity AreaQuantum)(Phase of Vacuum Coherence)(,u,v)where passive general coordinate transformations at a ?ed event Pemerge from local gauge transformations on the Goldstone Phase of theVacuum Coherence ?ld.PZ: Ah hah! So this is indeed closely analogous to the Feynman spin-2 case, wherehe gets macro-covariance from the micro- gauge invariance of the underlyingquantum ?ld.JS: No! because my decomposition is NON-PERTURBATIVE. The second term on RHS is NOT SMALL compared to ?st.PZ: This does NOT give you Einstein equivalence in the correspondence model!Thats the point I have been trying to get over to you.Read Feynman!JS: False comparison IMHO.PZ: The viability of Einstein equivalence, the uni?d g_uv, and the entire Riemannianmanifold model, is all simply contingent on the symmetries of the underlying quantum?ld -- in your case a virtual BEC. But the same Feynmanian arguments apply.This was all laid out by Feynman in his Lectures on Gravitation. He thought that thesuccess of the Einstein geometric model was the result of a mere mathematicalcoincidence.JS: You only get GR by summing an in?ite number of Feynman diagrams of certain topological class. This is non-perturbative like in BCS theory.PZ: So you are in fact unwittingly supporting my Venutian (Feynman 1963) position bycorrespondence!Or so it seems to me.JS: ,u is ordinary partial derivative.PSI = |Vacuum Coherence|e^i(Goldstone Phase of Vacuum Coherence)Curvature comes from stringy vortex core topological defects in theU(1) order parameter PSI as shown by Hagen Kleinert.More speci?ally curvature from disclination defects and torsion fromdislocation defects in 4D world crystal lattice, whose discretesymmetry groups inside the unit cell on scale (Loop Gravity AreaQuantum) are different vacuum phases.This theory can be extended to hyperspace with supersymmetry matrixdimensions of M-theory in which gauge force charges are reduced toKaluza-Klein geometrodynamics.PZ: This all looks very interesting. The only problem I have with it so far -- within thelimits of my current comprehension -- is your stubborn insistence on Einsteinequivalence at the macro-level, which does not IMHO seem to be supported byany reasonable correspondence model that is based on your virtual BEC/Goldstonephase theory of gravity.JS: Sure it is. One can get the tetrads and hence vanishing connection ?ld in LIFs.Indeed Hagen Kleinert works all that out in detail.PZ: Neither can it be.Z. =Sorry for using inappropriate terminology.I noticed that non-existence of the most slowly diverging sequence could beeasily demonstrated using reduction to absurdityThat is:Assume there be the sequence pn which most slowly diverges.Then we make log(pn).lim (log(pn)/pn) ----->zero (n--->in?ite)So log(pn) is a slower sequence than pn.This is contradictory to the previous assumption.So there is not the most slowly diverging sequence.But I have another quetion about the following value related to the abovesequence while I thought about the above.Let fn(x) be fn(x) = log(log(log(....log(x))..) n times log nesting.What is the value of the following suppose an be a diverging sequence?lim fn(an) =? (n--->in?ite)Is it 0?or another value?Does it depend on the sequence an?G.C. > I have the following question:>Does the most slowly diverging seriesexist?>Let an and bn (n:integer) be series of real numbers.>If lim an and bn --> in?ite (n-->in?ite) and lim an/bn --> 0>(n-->in?ite)>then lets call an diverges slower than bn and designate it withan<Is there any lower limit of these series?>I think such lower limit doesnt exist intuitively but is it right?>G.C> > So you are talking about sequences, not series, strictly speaking. -- > G. A. Edgarhttp://www.math.ohio-state.edu/~edgar/ =Both of these sum results are true ifp (p=prime) divides m(and if I am right).Let phi(x,n) = the number of positive integers which are <= x and arecoprime with n.Let phi(m) = phi(m,m) = the Euler phi function.So,sum{k>=1} phi(m/p^k,m)= phi(m)/(p-1).(Again, true if p|m; And the sum is ?ite because m/p^k < 1 for highenough ks.)And, related,Let q(p,m) = the sum of the digits of m as represented in base-p.Let mu() be the Mobius (Moebius) function.So,then, if p|m,sum{k|m} q(p,k) mu(m/k) = 0.Right?...Must be trivial, but maybe =I must admit I am a bit baf?y . would repost an old post ofmine with no new material..Was this reposting intentional?(Is . even a real person, or is . a malfunctioning server orsomething else more or less inspired more by the poetic half of my brain,> >rather than the mathematical side.> >A polygon with 4 sides: quadrilateral> >A polygon with 3 sides: triangle> >With 2 sides: line segment? (twice drawn)> >With 1 side (but closed, as polygons of 3 or more sides): ???????> >Of course, such a polygon does not exist in the REAL world.> >But what about in an abstract sense?> >Of course, of course....> >What would it be called? A monogon? A monoangle?> >It wouldnt be a line segment (as the bigon/biangle), because a line> >segment has 2 vertexes, its endpoints.> >Calling a line segment a bigon/biangle is a stretch itself, though I> >might have seen this before.> >And, last question, if a monogon/monoangle is simply a point, how is> >this to be distinguished from a 0-sided polygon?> >Hmmm....> >(All in fun...)> >Leroy Quet> > =Squares of 0.999... tend toward 0, not toward 1 .9^2 = .81 .9^3 = .729 .9^4 = .6561 .99^2 = .9801 .99^3 = .970299 .99^4 = .96059601 .999^2 = .998001 .999^3 = .997002999 .999^4 = .996005996001 .9999^2 = .99980001 .9999^3 = .999700029999 .9999^4 = .9996000599960001 .99999^2 = .9999800001 .99999^3 = .999970000299999 .99999^4 = .99996000059999600001 .999999^2 = .999998000001 .999999^3 = .999997000002999999 .999999^4 = .999996000005999996000001 .9999999^2 = .99999980000001 .9999999^3 = .999999700000029999999 .9999999^4 = .9999996000000599999960000001 .99999999^2 = .9999999800000001 .99999999^3 = .999999970000000299999999 .99999999^4 = .99999998000000059999999600000001 .999999999^2 = .999999998000000001 .999999999^3 = .999999997000000002999999999 .999999999^4 = .999999996000000005999999996000000001 .9999999999^2 = .99999999980000000001 .9999999999^3 = .999999999700000000029999999999 .9999999999^4 = .9999999996000000000599999999960000000001 .99999999999^2 = .9999999999800000000001 .99999999999^3 = .999999999970000000000299999999999 .99999999999^4 = .99999999996000000000059999999999600000000001 .999999999999^2 = .999999999998000000000001 .999999999999^3 = .999999999997000000000002999999999999.999999999999^4 = .999999999996000000000005999999999996000000000001[snip rest for brevity]Prove squares of 0.999... tend toward 1or stop claiming that 0.999... = 1 Garry Denke, GeologistDenoco Inc. of Texas =>Squares of 0.999... tend toward 0, not toward 1NONSENSE! LIM(n->+oo) { (1- 1/(10^n)) * [1 - 1/(10^n))} = LIM(n->+oo)[1 - (1/10^n)) * LIM(n->+oo)[ 1 - (1/10^n)) = 1*1 .9^2 = .81> .9^3 = .729> .9^4 = .6561> > .99^2 = .9801> .99^3 = .970299> .99^4 = .96059601 .999^2 = .998001> .999^3 = .997002999> .999^4 = .996005996001> > .9999^2 = .99980001> .9999^3 = .999700029999> .9999^4 = .9996000599960001 .99999^2 = .9999800001> .99999^3 = .999970000299999> .99999^4 = .99996000059999600001 .999999^2 = .999998000001> .999999^3 = .999997000002999999> .999999^4 = .999996000005999996000001 .9999999^2 = .99999980000001> .9999999^3 = .999999700000029999999> .9999999^4 = .9999996000000599999960000001> > .99999999^2 = .9999999800000001> .99999999^3 = .999999970000000299999999> .99999999^4 = .99999998000000059999999600000001 .999999999^2 = .999999998000000001> .999999999^3 = .999999997000000002999999999> .999999999^4 = .999999996000000005999999996000000001 .9999999999^2 = .99999999980000000001> .9999999999^3 = .999999999700000000029999999999> .9999999999^4 = .9999999996000000000599999999960000000001 .99999999999^2 = .9999999999800000000001> .99999999999^3 = .999999999970000000000299999999999> .99999999999^4 = .99999999996000000000059999999999600000000001 .999999999999^2 = .999999999998000000000001> .999999999999^3 = .999999999997000000000002999999999999>.999999999999^4 = .999999999996000000000005999999999996000000000001[snip rest for brevity]Prove squares of 0.999... tend toward 1>or stop claiming that 0.999... = 1 > Garry Denke, Geologist>Denoco Inc. of Texas =>Squares of 0.999... tend toward 0, not toward 1NONSENSE!> LIM(n->+oo) { (1- 1/(10^n)) * [1 - 1/(10^n))}> = LIM(n->+oo)[1 - (1/10^n)) * LIM(n->+oo)[ 1 - (1/10^n))> = 1*1So what youre saying is the sequence .9^2, .9^3, .9^4 ....9^2 = .81.9^3 = .729.9^4 = .6561etc the sequence .99^2, .99^3, .99^4 ....99^2 = .9801.99^3 = .970299.99^4 = .96059601etcthe sequence .999^2, .999^3, .999^4 ... .999^2 = .998001.999^3 = .997002999.999^4 = .996005996001etc tend to 1 making the sequence .999...^2, .999...^3, .999...^4 tend to 1.Is that what you are saying?Garry Denke, GeologistDenoco Inc. of Texas =>>Squares of 0.999... tend toward 0, not toward 1>NONSENSE!> LIM(n->+oo) { (1- 1/(10^n)) * [1 - 1/(10^n))}> = LIM(n->+oo)[1 - (1/10^n)) * LIM(n->+oo)[ 1 - (1/10^n))> = 1*1So what youre saying is the sequence .9^2, .9^3, .9^4 ....9^2 = .81> .9^3 = .729> .9^4 = .6561> etc> > the sequence .99^2, .99^3, .99^4 ....99^2 = .9801> .99^3 = .970299> .99^4 = .96059601> etcthe sequence .999^2, .999^3, .999^4 ... .999^2 = .998001> .999^3 = .997002999> .999^4 = .996005996001> etc> > tend to 1 making the sequence .999...^2, .999...^3, .999...^4 tend to 1.Is that what you are saying?Garry Denke, Geologist> Denoco Inc. of TexasFor each ?ed n, the sequences f(m) = (1 - 1/10^m)^n go to 1 as m -> oo.That is quite different from the behavior of the sequencesg(n) = (1 - 1/10^m)^n), for ?ed m, which got to 0 as n -> oo. garrydenke@1webhighway.com (Garry Denke)0 where n is the number of nines behindthe dot.So the number 0.999... can be found by looking at:lim{n->in?ity}{f(n)} = lim{n->in?ity}{1 - (1/(10^n))} =lim{n->in?ity}{1} - lim{n->in?ity}{1/(10^n)}Since the limit of the fraction-part goes to zero, the whole thingbecomes equal to 1. In other words 0.999... is equal to 1.BTW Im sure there are better ways (a probably more mathematicalcorrect ways) to explain this, but perhaps this helps. => Squares of 0.999... tend toward 0, not toward 1 .9^2 = .81> .9^3 = .729> .9^4 = .6561 .99^2 = .9801> .99^3 = .970299> .99^4 = .96059601 .999^2 = .998001> .999^3 = .997002999> .999^4 = .996005996001 [snip rest for brevity] Prove squares of 0.999... tend toward 1> or stop claiming that 0.999... = 1I think you mean *Powers* of 0.999... tend toward 0, not toward 1.The problem is that your 0.999... notation doesnt have a well de?edmeaning.Something like the following is easier to work with:f(p, n) = [1 - 1/(10^p)]^n, with p, n > 0As examples: f(1, 2) = [1 - 1/(10^1)]^2 = 0.9^2 = 0.81 f(1, 3) = [1 - 1/(10^1)]^2 = 0.9^3 = 0.729 f(2, 2) = [1 - 1/(10^2)]^2 = 0.99^2 = 0.9801 f(2, 3) = [1 - 1/(10^2)]^3 = 0.99^3 = 0.970299Using your empirical results, one can extrapolate f(p, n) as p or n ->in?ity:For ?ite p, as n -> in?ity, f(p, n) -> 0For ?ite n, as p -> in?ity, f(p, n) -> 1But what happens when both p and n -> in?ity?The most straightforward analysis gives:For p and n -> in?ity, f(p, n) = [1 - 1/(10^p)]^n -> [1 - 0]^n = [1]^n = 1The longstanding dispute over the limit of 0.999... is the limit of f(p,1) as p -> in?ity, which is of course 1. The faq goes into greater detailhttp://www.faqs.org/faqs/sci-math-faq/specialnumbers/ 0.999eq1/.Russell Harper, Harmless DrudgeGetting Ready for Work in Ontario> Garry Denke, Geologist> Denoco Inc. of Texas => Prove squares of 0.999... tend toward 1> or stop claiming that 0.999... = 1 > Garry Denke, Geologist> Denoco Inc. of TexasProve that 0.999... is not a single number, or stop claiming that 0.999... < 1.And while your at it, quit wasting bandwith repeatedly posting all those meaningless computations. =So, with(5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2)if (5b_2(x,y) + 2y) had any factors in common with 7, then those> factors would have to be a factor of 2y, as that term is both visible> and independent of x.>No, yet again. Just for something new, lets take x = -3, y = 1. Then a_1 = -2 + sqrt(-38) a_2 = -2 - sqrt(-38)so b_2 = a_2 + 1 = -1 - sqrt(-38)hence (5a_1 + 7)(5b_2 + 2) = (5(-2 + sqrt(-38)) + 7)(5(-1 - sqrt(-38)) + 2) = (-3 + 5sqrt(-38))(-3 - 5sqrt(-38)) = 9 + 25*38 = (7)(137) = 7(25(-3)^2 + 30(-3) + 2) = P(-3)as expected.However, in the algebraic integers its easy to show that 1. The ?st term, 5a_1 + 7, is NOT divisible by 7.and harder to show--but true--that 2. BOTH terms have factors in common with 7.In simple terms, the visible factors 7 and 2 have NOTHING to dowith the divisibility properties.As Ive said all along.As hard as it might be, you should really give up on looking atany constant terms. As has been pointed out, its a blind alley. > So you can see my conclusion about what happens when you divide both> sides by 7, when x does not equal y, follows from some basic algebra.Whats fascinating is that you can also consider what happens when y=0> to separate out further, so letting y=0, I havea^2 - xa + 7x^2 = 0,soa = (1+/-sqrt(-27))x/2,and introducing c_1(x,y) and c_2(x,y) I then have> Why spoil an interesting example by introducing spuriousfunctions? Just do as you did above and write (5a_1 + 7y)(5(a_2 + y) + 2y) = 7(25x^2 + 30xy + 2y^2)which, since y = 0, gives us (5a_1)(5a_2) = 7(25x^2)so a_1*a_2 = 7x^2but (1 - sqrt(-27)x/2)(1 + sqrt(-27)x/2) = (1 + 27)x^2/4 = 7x^2,exactly as expected. The divisibility properties areleft to the reader, but depend on x. Rick => It looks like there was more to adding in that y as a variable than> was realized as I saw the usual suspects replying--yet again--trying> to attack the mathematics, attacking algebra, as if they could still> convince all of you.However, the other shoe fell as I set y=0 givinga^2 - xa + 7x^2 = 0,soa = (1+/-sqrt(-27))x/2,and introducing c_1(x,y) and c_2(x,y) I then have(5(1-sqrt(-27))x/2 + c_1(x,y) + 7y)(5(1+sqrt(-27))x/2 + c_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2)from my modi?ation of Rick Deckers example.> Thats an interesting modi?ation, and I address it in myresponse to your post 2 hours before.Youll need to show us what you intend by introducing the cs,though, before I can give a complete response.Rick => It looks like there was more to adding in that y as a variable than>was realized as I saw the usual suspects replying--yet again--trying>to attack the mathematics, attacking algebra, as if they could still>convince all of you.>However, the other shoe fell as I set y=0 giving>a^2 - xa + 7x^2 = 0,>so>a = (1+/-sqrt(-27))x/2,>and introducing c_1(x,y) and c_2(x,y) I then have>(5(1-sqrt(-27))x/2 + c_1(x,y) + 7y)(5(1+sqrt(-27))x/2 + c_2(x,y) + 2y)> = 7(25x^2 + 30xy + 2y^2)>from my modi?ation of Rick Deckers example.> Thats an interesting modi?ation, and I address it in myresponse to your post 2 hours before.Youll need to show us what you intend by introducing the cs,> though, before I can give a complete response.While I could have left the cs as above Ive decided to simplifytheir de?ition slightly by using a slightly different expression:(5(1-sqrt(-27))x/2 + 5c_1(x,y) +7y)(5(1+sqrt(-27))x/2 + 5c_2(x,y) +2y) = 7(25x^2 + 30xy + 2y^2) Its fascinating. The factorization looks similar to one withpolynomial factors but there are the terms 5c_1(x,y), and 5c_2(x,y)where x and y are somehow entangled, but where clearly x and y in theterms inside of those functions cant have rational or even algebraicinteger degree.James Harris =it isnt really fascinating, if its algebra; at least,you admit that they attack [your] math, althoughyou seem to pretend that theyre attacking the use of algebra per se. obviously, there is method to your quaintness, butIm not interested in it, sinceyou dont even use the whole complex domain,while trying to invoke the ghost of Gauss and his FTA. seriously,I cant beleive that you actually believe what youre typing,so that it must be your Grand Course in Marketing. oh, well; dome day, youll have some result,mathematically speaking, if thats really what you like --the math will ?ally win you over! >> to attack the mathematics, attacking algebra, as if they could still>> convince all of you. > While I could have left the cs as above Ive decided to simplify> their de?ition slightly by using a slightly different expression:(5(1-sqrt(-27))x/2 + 5c_1(x,y) +7y)(5(1+sqrt(-27))x/2 + 5c_2(x,y) +2y)> > = 7(25x^2 + 30xy + 2y^2) Its fascinating. The factorization looks similar to one with> polynomial factors but there are the terms 5c_1(x,y), and 5c_2(x,y)> where x and y are somehow entangled, but where clearly x and y in the> terms inside of those functions cant have rational or even algebraic> integer degree.--Give the Gift of Trickier Dick Cheeny -- out of of?e, at last!http://www.benfranklinbooks.com/http://www.wlym.com/pages /music.htmlhttp://www.rand.org/publications/randreview/issues/ rr.12.00/http://members.tripod.com/~american_almanachttp:// =>> It looks like there was more to adding in that y as a variable than>> was realized as I saw the usual suspects replying--yet again--trying>> to attack the mathematics, attacking algebra, as if they could still>> convince all of you.>>However, the other shoe fell as I set y=0 giving>>a^2 - xa + 7x^2 = 0,>>so>>a = (1+/-sqrt(-27))x/2,>>and introducing c_1(x,y) and c_2(x,y) I then have>>(5(1-sqrt(-27))x/2 + c_1(x,y) + 7y)(5(1+sqrt(-27))x/2 + c_2(x,y) + 2y)>> = 7(25x^2 + 30xy + 2y^2)>>from my modi?ation of Rick Deckers example.>> Thats an interesting modi?ation, and I address it in my>response to your post 2 hours before.>Youll need to show us what you intend by introducing the cs,>though, before I can give a complete response.While I could have left the cs as above Ive decided to simplify> their de?ition slightly by using a slightly different expression:(5(1-sqrt(-27))x/2 + 5c_1(x,y) +7y)(5(1+sqrt(-27))x/2 + 5c_2(x,y) +2y)> > = 7(25x^2 + 30xy + 2y^2) Its fascinating. The factorization looks similar to one with> polynomial factors but there are the terms 5c_1(x,y), and 5c_2(x,y)> where x and y are somehow entangled, but where clearly x and y in the> terms inside of those functions cant have rational or even algebraic> integer degree.> Oh yeah, and wrong, as I just recently realized. OOPS!James Harris realized. OOPS!THATS the other shoe! (Wait a minute, how many ?other shoes have you dropped today?)James, I repeat my observation that mathematics is simply too dif?ult for you. Please consider taking upan activity better suited to your capabilities. Ive suggested basket-weaving in another post, but perhapsthrift-store mannequin would be an even better ?.--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 =>> >> Its fascinating. The factorization looks similar to one with>> polynomial factors but there are the terms 5c_1(x,y), and 5c_2(x,y)>> where x and y are somehow entangled, but where clearly x and y in the>> terms inside of those functions cant have rational or even algebraic>> integer degree. Oh yeah, and wrong, as I just recently realized. OOPS!Surprise, surprise, surprise. Looks like the math wins, and you lose.AGAIN.-- Wayne Brown (HPCC #1104) | When your tails in a crack, you improvisefwbrown@bellsouth.net | if youre good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock =There are 17 symmetry groups of isometric transformations in the plane (wallpaper groups), ~230 in space (crystallographic space groups), and 4901 in four dimensions. Im looking for an algorithmic means to enumerate these groups in an arbitrary number of dimensions. The references Ive consulted (which are generally focused on crystallography rather than abstract mathematics) all allude to such a method (one calling it the ?method of frobenius), but in practice all of them proceed to derive these groups with the general paradigm of list all you can think of, and then prove that they are distinct. Where can I ?d an algorithmic, or at least deductive, technique?Furthermore, the derivation of the space groups usually starts with a categorization of lattices: triclinic, monoclinic, othorhombic, etc. Is there a general way of constructing this hierarchy of lattices in R^n =>There are 17 symmetry groups of isometric transformations in the plane >(wallpaper groups), ~230 in space (crystallographic space groups), and 4901>in four dimensions. Im looking for an algorithmic means to enumerate these >groups in an arbitrary number of dimensions....>Furthermore, the derivation of the space groups usually starts with a >categorization of lattices: triclinic, monoclinic, othorhombic, etc. Is there >a general way of constructing this hierarchy of lattices in R^n ?I dont know whether this will be what you are looking for.There is a paper by Thurston, Conway and two other authors.It was available as a preprint for a while but I suspect ithas been published by now.It tries to show a new way of enumerating the groups in 3d.There is a comment in the paper that hinted that the authorsperhaps originally thought the method would handle the problemin a very simple and uniform fashion. But by the time they?ished it looks like they ended up having to deal with morespecial cases than they had hoped for.Nothing in this is meant to criticize any of the authorsor their work, they have obviously done far more in a monththan I will ever do in a lifetime. =Carlo Kleber> Working out a solution to a problem I empirically came across with a> numerical series that is given below, for i = 1, 2, 3, ...,(A+B), and> where A is a any given integer, B=1, 2,...,A, and the ratios are all> taken as the ceiling values. if i belongs to the interval [1, A-B] then :> index[i]= ((A-B)/2) + 1, if (A-B) and i are even;> index[i]= ((A-B)/2), otherwise. and if i belongs to the interval [A-B+1, A+B] then :> index[i]= ((i-1)/2), if i belongs to the interval [(A-B+1), A];> index[i]= (2*A-(i-1))/2), otherwise. Example:> Let A = 5 and B = 1, then we have that:> index[1]=2> index[2]=3> index[3]=2> index[4]=3> index[5]=2> index[6]=3 My question is how I can formally prove that this is true for any> integer A under the restrictions I posed above. I have tried many values experimentally, namely A = 1,2,3,...50. They> all worked. But now I need a formal mathematical proof or, at least, a> good idea of how I could claim this result without needing this formal> mathematical proof.> I ?st thought of induction. But the problem is that I have to> consider two independent variables, namely A and B. How can I then> possibly make the inductive assumption that it holds for A and prove> that it holds for A+1, ignoring the value of B? This seems quite> weird.Argue by induction on A+B, maybe. Just as an example, one can prove byinduction that mn=nm for any natural numbers m,n. Writemn = -(m-1)n -m(n-1) +(m-1)(n-1)= -n(m-1) -(n-1)n +(n-1)(m-1) by the inductive hypothesis= nm.But Im not clear on what are the hypotheses vs. what is the conclusion inposting. Are we _de?ing_ index by the ?st two formulas?LH =Larry Hammick> Write> mn = -(m-1)n -m(n-1) +(m-1)(n-1)> = -n(m-1) -(n-1)n +(n-1)(m-1) by the inductive hypothesis> = nm.Oh [muttered oath] that should bemn = -(m-1)n -m(n-1) +(m-1)(n-1) -1= -n(m-1) -(n-1)n +(n-1)(m-1) -1 by the inductive hypothesis= nm.I must learn to be more careful :)LH =I have compiled a list of online literatures, mostly related tocomputing/mathematics/economics stuff, athttp://www.srcf.ucam.org/~dl276/bookmarks.phpComments and additions are welcome!Ry =okay, i am having serious brain freeze right now ...what i am doing wrong below?e^(j*x) = e^(j*(2*pi*x)/(2*pi)) = (e^(2*pi*j))^(x/(2*pi)) = 1^(x/(2*pi)) = 1but e^(j*x) obviously does not equal 1 for all x! => okay, i am having serious brain freeze right now ...what i am doing wrong below?e^(j*x) = e^(j*(2*pi*x)/(2*pi)) = (e^(2*pi*j))^(x/(2*pi)) = 1^(x/(2*pi)) = 1but e^(j*x) obviously does not equal 1 for all x!You used the false identity a^(b*c) = (a^b)^c.Look in your textbook and see if there are CONDITIONS on a,b,c for thisequation.j = (-1)^(1/2) = (-1)^(2*(1/4)) = ((-1)^2)^(1/4) = 1^(1/4) = 1Can you see where I put one over on you? =rizwan> e^(j*x) = e^(j*(2*pi*x)/(2*pi))> = (e^(2*pi*j))^(x/(2*pi))> = 1^(x/(2*pi))I think by j you mean a square root of -1 (electrical engineers notation).But the formulae^(ab) = (e^a)^bis invalid for complex a,b. We can have e^a = e^c without having a=c.LH => okay, i am having serious brain freeze right now ...> what i am doing wrong below? e^(j*x) = e^(j*(2*pi*x)/(2*pi)) = (e^(2*pi*j))^(x/(2*pi)) = 1^(x/(2*pi)) = 1>j = i = sqr -1 ?> but e^(j*x) obviously does not equal 1 for all x!>e^ix isnt injective, e^ix = e^i(x+2pi).Thus the inverse of e^ix, ie e^(i/x) = (e^i)^(1/x) is multi-valued.You chose the wrong value or branch of the inverse.Some of the laws for exponents of reals dont extent to, dont work forcomplex numbers as youve just experienced.Are your brains now thawing out? ;-) =>>[...]>>Well, its possible to show that FOL is compact iff the Heine-Borel> >theorem holds. Heres a sketch of the idea: Let L be an arbitrary> >consistent in?ite subset of (the wffs of) FOL. Let G(L) be the set> >of Goedel numbers of elements of L. Stick a decimal in front of each> >element of G(L). Now you have a set G(L)* contained in [0,1]. G(L)*> >is obviously bounded, and in fact, (modulo some facts about> >deductions) is closed (proving this is tricky). If you can prove this I will be _very_ surprised. For example> it seems incredibly unlikely that no in?ite decimals appear> in the closure.> Well, this objection isnt too hard to get around. Note that G(L) isthe set of Goedel numbers of the wffs in L. Hence, each element inG(L) is a natural number, and thus has ?itely many digits. Thetransformation * (which concatenates each element in its domain to theright of a .) wont change that when producing G(L)*. =>> >> >>[...]>> >> >Well, its possible to show that FOL is compact iff the Heine-Borel>> >theorem holds. Heres a sketch of the idea: Let L be an arbitrary>> >consistent in?ite subset of (the wffs of) FOL. Let G(L) be the set>> >of Goedel numbers of elements of L. Stick a decimal in front of each>> >element of G(L). Now you have a set G(L)* contained in [0,1]. G(L)*>> >is obviously bounded, and in fact, (modulo some facts about>> >deductions) is closed (proving this is tricky). >> >> If you can prove this I will be _very_ surprised. For example>> it seems incredibly unlikely that no in?ite decimals appear>> in the closure. >Well, this objection isnt too hard to get around. Note that G(L) is>the set of Goedel numbers of the wffs in L. Hence, each element in>G(L) is a natural number, and thus has ?itely many digits. The>transformation * (which concatenates each element in its domain to the>right of a .) wont change that when producing G(L)*.Well duh, I understand that every element of G(L)* has only?itely many digits. Thats one reason that it seems veryunlikely to me that G(L)* is compact! (See, if its compactthen its closed. So if the _closure of_ G(L)* containsreals with in?itely many digits then the closure of G(L)*is not G(L)*, so G(L)* is not compact.)>prove. :) You said it was true, but that the proof was tricky. Whatproof did you have in mind?Another thing you should note is that it cannot possiblybe true for _every_ Godel-numbering scheme - if it istrue for some such scheme it depends on the propertiesof that scheme. For example, heres a Godel numberingfor which its false: Enumerate the wffs in alphabeticalorder. Assign 1 to the ?st one, 11 to the second one,111 to the third one, etc. Then G(L)* is the set {.1, .11, .111, ...}which is certainly not compact. (Um: its clearly notcompact _because_ the closure contains an in?itedecimal, namely .111... .)> I know there is some connection, as I have read on the>topic, but I dont remember the exact hypothesis, and hence, I>probably mistated the theorem. In any event, Ill put pencil to paper>and work something out.Probably you should do that before sending the book to a publisher...>Bye. :)>Alex************************David C. Ullrich =bccI need to look at Penroses old paper that started the spin network craze in loop quantum gravity. I actually have it.As I recall the way Penrose did it makes the connection to spintronics qubits obvious hence to quantum computers. This connectionis obscured to my mind at least in the modern developments like with Fotini Markopoulos whose work is interesting,but requires a lot of work to understand.plausible to me since the line and the area are dual in 3D.George Chapline made a bombshell remark about black hole thermodynamics and strings and loops today. I need to understand what he means in more detail. George is a deep thinker and I do not casually dismiss anything he says about physics.Jack,I have previously looked at Baezs Week 62 through Week 65 on the ADELie algebras, and also on John McKays A Rapid Introduction to ADE Theoryathttp://math.ucr.edu/home/baez/ADE.htmlwhat an ALE space is. In Mckays Rapid Introduction he says:There have been many applications of these ideas in various contexts. Letme cite two: Peter Kronheimer has used them in his paper on asymptoticallylocally ?d asymptotically locally euclidean spaces in connection withcosmological geometry.So ALE means asymptotically locally euclidean.McKay doesnt supply a reference to this paper. Actually there are several.I will mention two of them: P.B. Kronheimer, A Torrelli-Type Theorem for Gravitational Instantons,J. Differential Geom. 29, 685-698 (1989) P.B. Kronheimer, Instantons and the Geometry of the Nilpotent Variety,J. Differential Geom. 32, 473-490 (1990) So an ALE space is a type of gravitational instanton -- a vacuumsolution to Einsteins GR ?ld equations, assuming that time t is Wickrotated to it (and in this sense Euclideanized).JS: Ok so the instanton is a solution toRuv = 0with +--- ---> ++++Has anyone studiedRuv - (1/2)Rguv + /guv = 0 ?I suppose, off top of my head, this is what Susskind means when he mentions DeSitter and anti-DeSitter space?That is for / = constant.In my theory / is a variable.SPS: Note: a gravitational instanton is a generalization (by Hawking) of theYang-Mills gauge theory instanton idea. Such an ALE space has an in?ity that looks like SU(2)/Gamma. Thismight relate to the recent claim by Jeff Weeks (et alia) in *Nature* 429,9i.e. it would have the topology of SU(2)/Octahedral Double group. This ODgroup is the McKay group corresponding to the E8 Coxeter graph (or Dynkindiagram). Such a gravitational instanton is very interesting mathematically: amongother things it is a hyper-Kaehler manifold, which means that it has threeindependent complex structures and other powerful properties.subgroup of SU(2)] can act on C^2 = R^4, and generate an ALE spaceC^2/Gamma. Actually, C^2/Gamma is a 4-d variety with a singularity of the typewhich is ADE classi?d by V.I. Arnold (ref: *Singularity Theory* 1981). The ALE space corresponding to each ADE graph is the resolved (ordesingularized) form of C^2/Gamma. This resolution occurs by lifting intothe nilpotent variety which is a substructure of the corresponding ADE-typeLie algebra. There is a close relationship between singularity theory and catastrophetheory, which is described in Arnolds book *Catastrophe Theory* (1986). Infact the proto-ALE space C^2/Gamma is the identity ?er in the catastrophebundle, whose base space is C^r/W, where r is the rank of the ADE-type Liealgebra, and W is the ADE-type re?n group (also called the Coxetergroup or the Weyl group). So here we have a beautiful connection betweenthe McKay group (?ite subgroup of SU(2)) and the re?n group. McKay does not exaggerate when he says: There have been manyapplications of these ideas in various contexts. In fact there are so manyapplications (all of interest to physics) that I have proposed that all ofthese applications be studied together in a program I have been calling ADEXtheory. Here is a partial list of ADE classi?d objects [Some of theseobjects have a somewhat larger classi?tion scheme; however, thesimply-laced Coxeter graphs classi?s the most important sub-class (whichis called simple by V.I. Arnold, who also claims that ADE classi?s allsuch simple objects in mathematics -- an unproven but very suggestiveclaim).]: Lie algebras (& Lie groups) Kac-Moody (in?ite-dim.) algebras Coxeter (re?n) groups (also called Weyl groups) McKay groups (?ite subgroups of SU(2)) Regular polyhedra Hyperspace crystallography Sphere-packing lattices (root lattices) Error-corresting codes Quatizing lattices (weight lattices) Analog-to-digital transforms Conformal ?ld theories (which live on the 2-d string world-sheet) Gravitational instantons (providing a link with Penrose twistors) Singularities of differentible maps Thom-Arnold catastrophes Heisenberg algebras Korweg-deVries hierarchy of non-linear equations Generalized braid groups (related to knots and links) Quivers Caustics Wave frontsAs V.I. Arnold says on p. 92 of *Catastrophe Theory* At ?st glance, functions, quivers, caustics, wave fronts and regularpolyhedra have no connection with each other. But in fact, correspondingobjects bear the same label not just by chance: for example, from theicosahedron one can construct the function x^2 + y^3 + z^5, and from it thediagram E8, and also the caustic and wave front of the same name. To easily checked properties of one of a set of associated objectscorrespond properties of the others which need not be evident at all. Thusthe relations between all the A, D, E-classi?ations can be used for thesimultaneous study of all simple objects, in spite of the fact that theorigin of many of these relations (for example, of the connections betweenfunctions and quivers) remains an unexplained manifestation of themysterious unity of all things.BTW: John McKay in his Rapid Introduction mentions that: Thiscorrespondence has appeared in the book *Roots of Consciousness* as a basisfor an explanation of consciousness! I should mention here that this is a reference to a 40 page paper,Consciousness: a Hyperspace View*, which I published as an appendix to thebook *Roots of Consciousness* by Jeffrey Mishlove (1993). I also published a short summary of this idea Consciousness in *Toward a Science ofConsciousness: the First Tucson Discussions and Debates, ed. by Hameroff,Kaszniak, and Scott, MIT Press (1996).Nuff said!Saul-Paul =Portfolio of PAF as of 8JAN04:BCE 6,500 22.63 $147,095.00 BMY 50 29.38 $1,469.00 MRK 100 48.08 $4,808.00 Q 14,200 4.52 $64,184.00 SBC 12,200 27.59 $336,598.00 realestate land 3APR03 of 3 lots $19,000.science-art of pictures,porcelain etc starting JAN03 for $12,160.realestate land 30JUL03 another lot $11,500.Today I sold some more Qwest for 4.51 per share for a small pro? andsold 2,800shares of SBC at 27.60 for a substantial pro? since SBC has beengoing almost straight up as it hit its ex dividend date on 7 January,but whenever a stock of mine goes up after the ex dividend date I amalways tempted to make a switch out of a sizeable chunk, because thepresumption that it will go down below the price at the ex dividenddate which I remember to be 27. even is a fairly high probability andthat a crossover can be achieved in the short term future and perhapsbuyback more.Stockmarket in that I had stocks not for Crossover technique butrather instead on hopes of a buyought. So I am getting the portfoliointo that of 100% obedience to the Crossover technique. And riddingmyself of con?g and competing strategies.Let the OS do the work.Archimedes Plutoniumwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxies =Friday January type of sequence. It must be good for something. Maybethis group can come up with some ideas.I was labored with written calculations on and off for two months during mySpiritual Trip to India to come up with aworking prototype. I hereby name this type of sequence the EvensonSequence.The sequence is circular; after the end it repeats again from the beginning.The sequence is 96 digits long. The digits of the sequence is made up of 4different entities.For this sequence I use the digits 0,1,2,3. There are 24 of each of the 4digits.Since the 4 digits can be represented by binary maybe there can anencryption use.The Evenson Sequence:0 1 2 3 0 3 1 0 2 3 2 1 0 3 1 32 1 2 0 1 3 1 2 0 2 3 0 1 0 2 03 1 2 3 2 0 1 0 3 2 1 3 1 0 3 02 3 1 3 0 1 2 1 3 2 0 3 0 1 3 23 0 2 1 2 3 1 0 1 2 0 3 2 3 1 21 0 2 1 3 0 3 2 0 2 1 0 1 3 0 2Observations:1) No digit is repeated in the sequence. No group of digits is repeated inthe sequence.2) All possible four-digit combinations (taking rule #1 in account) arerepresented in the forward reading of the sequence.3) All possible four-digit combinations (taking rule #1 in account) arerepresented in the BACKWARD reading of the sequence.Here is a real life example to explain this sequence: Imagine Buddhistmonks chanting a mantra based on the Evenson Sequence.Imagine the monks using four different syllables in the chant. If the monkschant over and over again the sequence using thosesyllables the listener could isolate 96 different four syllable words!Imagine, 96 four-syllable words represented by a 96 single-syllablesequence.I must admit when I experienced the Tibetan monks chanting at the BodhigayaTemple Complex in December of 1999 the ideaintrigued my mathematical mind and I suffered two-months of mind labor?ding the sequence.I hereby introduce my creation to the unwashed masses for critique. Pleasebe gentle. When mentioning my sequence inposts please use my Surname--two months of mental torture is worth at leastthat much. My only hope is Evenson. Evenson.> I invented this new type of sequence. It must be good for something. Maybe> this group can come up with some ideas.> I was labored with written calculations on and off for two months during my> Spiritual Trip to India to come up with a> working prototype. I hereby name this type of sequence the Evenson> Sequence.The sequence is circular; after the end it repeats again from the beginning.> The sequence is 96 digits long. The digits of the sequence is made up of 4> different entities.> For this sequence I use the digits 0,1,2,3. There are 24 of each of the 4> digits.> Since the 4 digits can be represented by binary maybe there can an> encryption use.> The Evenson Sequence:0 1 2 3 0 3 1 0 2 3 2 1 0 3 1 3> 2 1 2 0 1 3 1 2 0 2 3 0 1 0 2 0> 3 1 2 3 2 0 1 0 3 2 1 3 1 0 3 0> 2 3 1 3 0 1 2 1 3 2 0 3 0 1 3 2> 3 0 2 1 2 3 1 0 1 2 0 3 2 3 1 2> 1 0 2 1 3 sequence you have! I put each of the 4-digitnumbers in one column (col A below) and sorted into column B. Thensubtracting the 1st in B from the 2nd, 2nd from 3rd, 3rd from 4th...etc. with the results in column C. An interesting peak occursevery 6th number at about 700. Not sure if this has any connectionwith anything. Good luckBob CarlsonA B C123 121 310 123 22321 132 9313 210 782120 230 201312 310 80230 313 31020 1020 7073123 1021 12010 1030 93213 1203 1731030 1302 992313 1312 10121 2010 6983203 2120 110132 2310 1903021 2312 22310 2313 11203 2321 82312 3021 7001021 3032 113032 3123 91210 3203 801302 3213 10 => The sequence is circular; after the end it repeats again from thebeginning.> The sequence is 96 digits long. The digits of the sequence is made up of4> different entities.> For this sequence I use the digits 0,1,2,3. There are 24 of each of the 4> digits.> Since the 4 digits can be represented by binary maybe there can an> encryption use.> The Evenson Sequence: 0 1 2 3 0 3 1 0 2 3 2 1 0 3 1 3> 2 1 2 0 1 3 1 2 0 2 3 0 1 0 2 0> 3 1 2 3 2 0 1 0 3 2 1 3 1 0 3 0> 2 3 1 3 0 1 2 1 3 2 0 3 0 1 3 2> 3 0 2 1 2 3 1 0 1 2 0 3 2 3 1 2> 1 0 2 1 3 0 3 2 0 2 1 0 1 3 0 20, 1, 2, 3, 0, 3, 1, 0, 2, 3, 2, 1, 0, 3, 1, 3, 2, 1, 2, 0, 1, 3, 1, 2, 0,2, 3, 0, 1, 0, 2, 0, 3, 1, 2, 3, 2, 0, 1, 0, 3, 2, 1, 3, 1, 0, 3, 0, 2, 3,1, 3, 0, 1, 2, 1, 3, 2, 0, 3, 0, 1, 3, 2, 3, 0, 2, 1, 2, 3, 1, 0, 1, 2, 0,3, 2, 3, 1, 2, 1, 0, 2, 1, 3, 0, 3, 2, 0, 2, 1, 0, 1, 3, 0, 2For the skeptical I veri?d his claims about the combinatorics of thissequence. It doesnt appear to be in the OEIS http://www.research.att.com/%7Enjas/sequences/index.htmlbut it could be with some permutation of 0, 1, 2, 3. I tried adding 1 toeach entry but still didnt ?d it.It is reminiscent of de Bruijn sequences, but with restrictions. Maybesomeone studying necklaces has come across such creatures.--Edwin Clark> Observations: 1) No digit is repeated in the sequence. No group of digits is repeatedin> the sequence.> 2) All possible four-digit combinations (taking rule #1 in account) are> represented in the forward reading of the sequence.> 3) All possible four-digit combinations (taking rule #1 in account) are> represented in the BACKWARD reading of the sequence.> Evenson.> I invented this new type of sequence. It must be good for something. Maybe> this group can come up with some ideas.> I was labored with written calculations on and off for two months during my> Spiritual Trip to India to come up with a> working prototype. I hereby name this type of sequence the Evenson> Sequence.The general term for this type of sequence is a ?chain code. Aftermarking the links of a chain with this sequence, by looking at 4successive links you can tell where on the 96-link chain you are. Theyare also used in, e.g. radar because by looking at the signal comingback, you can tell how long it was since you transmitted it.But only when you achieve enlightenment will the chain no longer bindyou.Another term to Google on is Shift register sequence.> My only hope is that I have not recreated the wheel.Is the second man to create the wheel less clever than the ?st?The great wheel turns for us all.-- Palmer>Is the second man to create the wheel less clever than the ?st?No but the guy with the third wheel is unnecessary. => I was labored with written calculations on and off for two months during> my Spiritual Trip to India to come up with a working prototype. I> hereby name this type of sequence the Evenson Sequence.>Have youve not learned a nit from your spiritual trip aboutnon-attachment to the fruits of action?> Here is a real life example to explain this sequence: Imagine Buddhist> monks chanting a mantra based on the Evenson Sequence.> Imagine the monks using four different syllables in the chant.Imagine you chanting your chant or theirs, not for two months, but for twoyears and then speak again unto us unknownings, the prophesies of yourmindfullness.> Imagine, 96 four-syllable words represented by a 96 single-syllable> sequence.>Whats 96? Have you not learned Indian lore about 108 ?> I must admit when I experienced the Tibetan monks chanting at the Bodhigaya> Temple Complex in December of 1999 the idea> intrigued my mathematical mind and I suffered two-months of mind labor> ?ding the sequence.>Ah, they got you to concentrate, at least for two months.Indeed, even looking into the Hare Krishna chant, one sees binary numbers.> I hereby introduce my creation to the unwashed masses for critique.Wash your hands before coming to the table.> Please be gentle. When mentioning my sequence in posts please use my> Surname--two months of mental torture is worth at least that much.Such arrogance merits gentle reprimand.> My only hope is that I have not recreated the wheel.>Squeak, it needs sesame oil.So all crestfallen pride aside, what did you see in their chants? Evenson.> I invented this new type of sequence. It must be good for something.Maybe> this group can come up with some ideas.> I was labored with written calculations on and off for two months duringmy> Spiritual Trip to India to come up with a> working prototype. I hereby name this type of sequence the Evenson> Sequence. The sequence is circular; after the end it repeats again from thebeginning.> The sequence is 96 digits long. The digits of the sequence is made up of4> different entities.> For this sequence I use the digits 0,1,2,3. There are 24 of each of the 4> digits.> Since the 4 digits can be represented by binary maybe there can an> encryption use.> The Evenson Sequence: 0 1 2 3 0 3 1 0 2 3 2 1 0 3 1 3> 2 1 2 0 1 3 1 2 0 2 3 0 1 0 2 0> 3 1 2 3 2 0 1 0 3 2 1 3 1 0 3 0> 2 3 1 3 0 1 2 1 3 2 0 3 0 1 3 2> 3 0 2 1 2 3 1 0 1 2 0 3 2 3 1 2> 1 0 2 1 3 0 3 2 0 2 1 0 1 3 0 2> Observations: 1) No digit is repeated in the sequence. No group of digits is repeatedin> the sequence.> 2) All possible four-digit combinations (taking rule #1 in account) are> represented in the forward reading of the sequence.> 3) All possible four-digit combinations (taking rule #1 in account) are> represented in the BACKWARD reading of the sequence. Here is a real life example to explain this sequence: Imagine Buddhist> monks chanting a mantra based on the Evenson Sequence.> Imagine the monks using four different syllables in the chant. If themonks> chant over and over again the sequence using those> syllables the listener could isolate 96 different four syllable words!> Imagine, 96 four-syllable words represented by a 96 single-syllable> sequence. I must admit when I experienced the Tibetan monks chanting at theBodhigaya> Temple Complex in December of 1999 the idea> intrigued my mathematical mind and I suffered two-months of mind labor> ?ding the sequence. I hereby introduce my creation to the unwashed masses for critique.Please> be gentle. When mentioning my sequence in> posts please use my Surname--two months of mental torture is worth atleast> that much. My only hope is that I> have not recreated the wheel.> in this order? Following the 3observations it can be of any order you want it to be therefore youconsidering to re?e its properties. One possible use perhaps is on statemachines...? => I invented this new type of sequence. It must be good for something.Maybe> The sequence is circular; after the end it repeats again from thebeginning.> The Evenson Sequence:> 0 1 2 3 0 3 1 0 2 3 2 1 0 3 1 3> 2 1 2 0 1 3 1 2 0 2 3 0 1 0 2 0> 3 1 2 3 2 0 1 0 3 2 1 3 1 0 3 0> 2 3 1 3 0 1 2 1 3 2 0 3 0 1 3 2> 3 0 2 1 2 3 1 0 1 2 0 3 2 3 1 2> 1 0 2 1 3 0 3 2 0 2 1 0 1 3 0 2> Observations:> 1) No digit is repeated in the sequence. No group of digits is repeatedin> the sequence.> 2) All possible four-digit combinations (taking rule #1 in account) are> represented in the forward reading of the sequence.> 3) All possible four-digit combinations (taking rule #1 in account) are> represented in the BACKWARD reading of the sequence.> chant over and over again the sequence using those> syllables the listener could isolate 96 different four syllable words!> Imagine, 96 four-syllable words represented by a 96 single-syllable> sequence.>To claim it there must be public publishing and application. For thepublishing put it on a web page. The application is mentioned as phonics butI do not have much view of phonics application.If there is a musical representation then that is not application but musiccopyright.A graphic design representation (like tiling) can be an application... =>The Evenson Sequence:>0 1 2 3 0 3 1 0 2 3 2 1 0 3 1 3>2 1 2 0 1 3 1 2 0 2 3 0 1 0 2 0>3 1 2 3 2 0 1 0 3 2 1 3 1 0 3 0>2 3 1 3 0 1 2 1 3 2 0 3 0 1 3 2>3 0 2 1 2 3 1 0 1 2 0 3 2 3 1 2>1 0 2 1 3 0 3 2 0 2 1 0 1 3 0 2> To claim it there must be public publishing and application. For the> publishing put it on a web page. The application is mentioned as phonicsbut> I do not have much view of phonics application. If there is a musical representation then that is not application butmusic> copyright. A graphic design representation (like tiling) can be an application...>Graphic design can be trademark or copyright. Tiling can be application... =Pardon my ignorance but ... Since the 4 digits can be represented by binary ... ?> Friday January 9, type of sequence. It must be good for something.Maybe> this group can come up with some ideas.> I was labored with written calculations on and off for two months duringmy> Spiritual Trip to India to come up with a> working prototype. I hereby name this type of sequence the Evenson> Sequence. The sequence is circular; after the end it repeats again from thebeginning.> The sequence is 96 digits long. The digits of the sequence is made up of4> different entities.> For this sequence I use the digits 0,1,2,3. There are 24 of each of the 4> digits.> Since the 4 digits can be represented by binary maybe there can an> encryption use.> The Evenson Sequence: 0 1 2 3 0 3 1 0 2 3 2 1 0 3 1 3> 2 1 2 0 1 3 1 2 0 2 3 0 1 0 2 0> 3 1 2 3 2 0 1 0 3 2 1 3 1 0 3 0> 2 3 1 3 0 1 2 1 3 2 0 3 0 1 3 2> 3 0 2 1 2 3 1 0 1 2 0 3 2 3 1 2> 1 0 2 1 3 0 3 2 0 2 1 0 1 3 0 2> Observations: 1) No digit is repeated in the sequence. No group of digits is repeatedin> the sequence.> 2) All possible four-digit combinations (taking rule #1 in account) are> represented in the forward reading of the sequence.> 3) All possible four-digit combinations (taking rule #1 in account) are> represented in the BACKWARD reading of the sequence. Here is a real life example to explain this sequence: Imagine Buddhist> monks chanting a mantra based on the Evenson Sequence.> Imagine the monks using four different syllables in the chant. If themonks> chant over and over again the sequence using those> syllables the listener could isolate 96 different four syllable words!> Imagine, 96 four-syllable words represented by a 96 single-syllable> sequence. I must admit when I experienced the Tibetan monks chanting at theBodhigaya> Temple Complex in December of 1999 the idea> intrigued my mathematical mind and I suffered two-months of mind labor> ?ding the sequence. I hereby introduce my creation to the unwashed masses for critique.Please> be gentle. When mentioning my sequence in> posts please use my Surname--two months of mental torture is worth atleast> that much. My only hope is that I> Evenson.> =Jose Carlos Santos> I would like to know where can I ?d a proof of the following> statement: if M is a real analytic manifold and v is an analytic> vector ?ld on M, then the ?sociated with M is also> analytic.Correcting another slip of mine, the referenceDieudonn.8e, _Treatise on Analysis_, 18.2.12is correct, but thats in volume 4, not volume 2. Nine volumes in all, andnot exactly light reading.LH =>I would like to know where can I ?d a proof of the following>statement: if M is a real analytic manifold and v is an analytic>vector ?ld on M, then the ?sociated with M is also>analytic.> Correcting another slip of mine, the reference> Dieudonn.8e, _Treatise on Analysis_, 18.2.12> is correct, but thats in volume 4, not volume 2. Nine volumes in all, and> not exactly light reading.in Dieudonn.8es treatise, but in the third volume instead of the fourth one.Jose Carlos Santos = I would like to know where can I ?d a proof of the following> statement: if M is a real analytic manifold and v is an analytic> vector ?ld on M, then the ?sociated with M is also> analytic.Dieudonn.8es _Treatise on Analysis_ cost me a bundle, but at times like thisI dont regret it. See Remark 18.2.12 in Volume 2!Dont ask me to explain it. Im not Dieudonn.8e :)LH =>Dieudonn.8es _Treatise on Analysis_ cost me a bundle, Ah, but was it a principal bundle?Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 =Every so often, I think about how much math Ive forgotten (B.S. degree inMathematics from Caltech, followed by 20+ years of work in which I never useany of it, and so most of it is gone), and decide to start over. I pull outmy Herstein, Topics in Algebra, or Apostol, Calculus, open up to page 1,and start from the beginning. Sometimes, for variety, I instead pick up aDover book, such as Clark, Elements of Abstract Algebra. I dilligently gothrough, making sure to do a fair number of the exercises, and usually get afew chapters in before work and other things pull me away long enough that Iforget this stuff again, and eventually the cycle repeats. :-)Ive noticed that in pretty much every math book Ive ever read, even in the early parts of each section, where all there is is easystuff, there are always a couple problems that I just cannot get anywhereon at all.For example, in the Clark book I mention above, the very ?st problem onsubgroups is: Show that a non-empty subset H of a group G is a subgroup of G if and only if a, b member of H implies a*b member of H. (Where b means the inverse of b).Well, I came up completely blank on that one. I dont think Im an idiot.I get my fair share of the problems marked hard in books that rank theirproblems.Is it just me, or does this happen to everyone?-- --Tim Smith =Every so often, I think about how much math Ive forgotten (B.S. degree in>Mathematics from Caltech, followed by 20+ years of work in which I never use>any of it, and so most of it is gone), and decide to start over. I pull out>my Herstein, Topics in Algebra, or Apostol, Calculus, open up to page 1,>and start from the beginning. .snip (a problem from algebra)>Well, I came up completely blank on that one. I dont think Im an idiot.>I get my fair share of the problems marked hard in books that rank their>problems.Is it just me, or does this happen to everyone?>Something similar happens to me all the time. Every so often Ill look back atwas the text. Today, I cant understand a single one of the many (mostlycorrect) proofs I submitted. Herstein? Hopeless. I can, however, recall that the group of quarternionunits provides an answer to some starred problem, but cant for the life of meremember which one. Anyone know the problem Im thinking of?rich =>Every so often, I think about how much math Ive forgotten (B.S. degree in>Mathematics from Caltech, followed by 20+ years of work in which I never use>any of it, and so most of it is gone), and decide to start over. I do the same thing.It happens to me. If you dont use the knowledge often, yourbrain gets ef?ient and deletes the storage. However, relearningis easier the next time around and doesnt take as long./BAH =* Tim Smith> For example, in the Clark book I mention above, the very ?st problem on> subgroups is: Show that a non-empty subset H of a group G is a subgroup of G if> and only if a, b member of H implies a*b member of H. (Where b means> the inverse of b).Actually, this shouldnt be too dif?ult: 1. Use (or prove) the lemma that says: A nonempty subset H of a group G is a subgroup if it is closed on group operation and inverse operation. 2. Use the assumption (a*b in H) to show that the identity element is in H. 3. Show then that H is closed on inverse operation. 4. Finally show that it is closed on group multiplication. 5. QED with lemma from (1).> Is it just me, or does this happen to everyone?Yes, it does, at least to me. When given the solution, I sometimesjust bang my head and sometimes I shrug and think that this was reallya much more dif?ult exercise than it appeared to be.-- Jon Haugsand =On 09 Jan Smith>> For example, in the Clark book I mention above, the very ?st problem on>> subgroups is:>> >> Show that a non-empty subset H of a group G is a subgroup of G if>> and only if a, b member of H implies a*b member of H. (Where b means>> the inverse of b).Actually, this shouldnt be too dif?ult: 1. Use (or prove) the lemma that says: A nonempty subset H of a> group G is a subgroup if it is closed on group operation and inverse> operation. 2. Use the assumption (a*b in H) to show that the identity element> is in H. 3. Show then that H is closed on inverse operation. 4. Finally show that it is closed on group multiplication. 5. QED with lemma from (1).6. Prove the converse by noting that H is a subgroup => b in H for all b in H => a*b in H for all a, b in H.7. See 5. =* Toni Lassila> 6. Prove the converse by noting that H is a subgroup => b in H for> all b in H => a*b in H for all a, b in H.7. See 5.-- Jon Haugsand =I have just attended a course on scheduling for project managers and therethe beta distribution comes up in relation to estimation of task durations.I have looked up in my old probability theory book regarding this and foundthe proof for the fact that the sum of _many_ beta distributions convergesto a gaussian distribution, but neither my beloved book (and some hard coreintegral solvining) or extensive search on the internet has given mewhat I have found (this includes characteristic function of the betadistribution and its density function) I am beginning to believe that thereis no distribution de?ed/named for the sum of two beta distributions, butI might be wrong.Can anyone answer this one?/Torben-- P.S. The views expressed above are my own. =>I have just attended a course on scheduling for project managers and there>the beta distribution comes up in relation to estimation of task durations.>I have looked up in my old probability theory book regarding this and found>the proof for the fact that the sum of _many_ beta distributions converges>to a gaussian distribution,Presumably youre talking about independent random variables with the same parameters for the beta distribution. Then this is just the CentralLimit Theorem - nothing special about beta distribution.> but neither my beloved book (and some hard core>integral solvining) or extensive search on the internet has given me>what I have found (this includes characteristic function of the beta>distribution and its density function) I am beginning to believe that there>is no distribution de?ed/named for the sum of two beta distributions, but>I might be wrong.Im almost certain theres no closed-form general formula for the distribution of the sum (again, assuming independence). In some particular cases (e.g. if the parameters are positive integers) there may be a formula, but its not a standard named distribution.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 =|In is de?itely Quines famous set theory.||Which doesnt say whether it was his ?st.True.New Foundations was introduced by Quine in 1937. The paper was reprintedthe set theory you had in mind was from the 1950s.Some ideas do suffer from being upstaged by more famous ones, though. Asa student, one professor advised me to go back to original papers, sinceoften the paper containing some famous result also contains a number ofother worthwhile ideas that merely never became as famous.|>What youre describing sounds like NFU, New Foundations modi?d to|>permit urelements. The strati?ation is stricter, however; the|>variables in P have to be ranked so that a is a member of b occurs|>only for b of rank exactly one more than the rank of a. ||The one that Im talking about didnt have the exactly one more|restriction.Isnt that too strong? Instead of writing xnotin x, I could write(Ey) y=x and ynotin x, except that strati?ation usually requiresthat a=b be written only for a and b of the same level. But in thepresence of extensionality, y=x is equivalent to (Az) zin x <-> zin y,and (Ey) [(z) zin y<->zin x] & ynotin x satis?s the weaker form ofstrati?ation, doesnt it?|>Also in the presentations Ive seen, the urelements are elements |>which dont have members,||The one that Im talking about satis?d extensionality.So isnt it inconsistent to have both the stronger form of thestrati?d comprehension axiom and extensionality.Id be a little surprised at how little mention this stronger versionof strati?d comprehension gets, if its acceptable.|>I would guess this might make a big difference in NFU, since the |>formula for has no members can be strati?d, and has only |>itself as a member cant.||Well, obviously Quine had reasons for the change. Im wonderingLet us know if you remember or uncover any more details.Keith Ramsay = > in>>In <5uqKb.7862$G1.38799@tor-nn1.netcom.ca>, on said:>> >> >Admitting either the set of all sets or the set of everythingallows>you to select a subset from either that corresponds to the>self-contradictory Russell Set (the set of all sets that are not>elements of themselves).>> >> No. There are set theories in which there is a universal set but>>Russells Paradox does not exist.>> > >Only by doing things like arbitrarily banning self-membership of setsor>> postulating some sort of in?ite hierarchy of sets. To me, it allseems>so>> unnecessary and so inelegant. By simply treating set membership likeany>> other logical predicate of two variables, you can prove that the> so-called>> Russell set cannot exist (my website). The job of any set theory thenis>> make sure that you cannot prove that it does exist. One way -- and I> dont>> know if it is really anything new -- is to have a simpli?d settheory>that>> does not postulate the existence of any particular set or sets. Thisis>the>> approach I have taken in my proof writing program. It seems to workand>> neatly avoid problems like RP that have plagued naive set theory. I> would>be>> very interested to see if you can derive any contradictions from it.> >Dan> > Your system can not have the set of all natural numbers, N. Its true that number theory is not built into my simpli?d set theory,the> way it is in ZF. To develop number theory, I start with a premise> corresponding to Peanos Axioms. (Click on the Numbers menu.)>The Axiom of In?ity is needed to guarantee that N number has a successor? Seeabove.No.I mean the Axiom of In?ity. AoI states there exists a set that is closedunder the successor function. If your set theory doesnt assume theexistence of any set then it doesnt include the AoI.PA does have the AoI.> At any rate, there is no guarantee that anything actually existences inmy> set theory. Nevertheless, I think you should still be able to develop a> large part if not all of mathematical theory using my program.Certainly not all.Your system cant have a set of all natural numbers.Russell- 2 many 2 count = > > >> Shmuel (Seymour J.) 11:05 PM, Dan Christensen said:>> >Admitting either the set of all sets or the set of everything> allows> >you to select a subset from either that corresponds to the> >self-contradictory Russell Set (the set of all sets that are not> >elements of themselves).>> >>No. There are set theories in which there is a universal set but> Russells Paradox does not exist.>> Only by doing things like arbitrarily banning self-membership ofsets> or>>postulating some sort of in?ite hierarchy of sets. To me, it all> seems>> so>>unnecessary and so inelegant. By simply treating set membership like> any>>other logical predicate of two variables, you can prove that the>so-called>>Russell set cannot exist (my website). The job of any set theorythen> is>>make sure that you cannot prove that it does exist. One way -- and I>dont>>know if it is really anything new -- is to have a simpli?d set> theory>> that>>does not postulate the existence of any particular set or sets. This> is>> the>>approach I have taken in my proof writing program. It seems to work> and>>neatly avoid problems like RP that have plagued naive set theory. I>would>> be>>very interested to see if you can derive any contradictions from it.>> >> Dan>> > >Your system can not have the set of all natural numbers, N.> > Its true that number theory is not built into my simpli?d set theory,> the>way it is in ZF. To develop number theory, I start with a premise>corresponding to Peanos Axioms. (Click on the Numbers menu.)> >The Axiom of In?ity is needed to that every natural number has a successor? See> above. No.> I mean the Axiom of In?ity. AoI states there exists a set that is closed> under the successor function. If your set theory doesnt assume the> existence of any set then it doesnt include the AoI.It doesnt.> PA does have the AoI.>That is what I meant to say. You dont need to include the AoI in a theoryof sets. You can introduce it as a premise, along with the the rest of PA.>At any rate, there is no guarantee that anything actually existencesin> my>set theory. Nevertheless, I think you should still be able to develop a>large part if not all of mathematical theory using my program. Certainly not all.> Your system cant have a set of all natural numbers.>Again, to develop number theory in my system, you begin by introducing theequivalent of PA as an initial premise. You dont need to have the naturalnumbers built into your set theory.DanVisit DC Proof Online at http://www.dcproof.com -- FREE download = > > >> >>Shmuel >in>In <5uqKb.7862$G1.38799@tor-nn1.netcom.ca>, on said:> > >Admitting either the set of all sets or the set of everything>allows>>you to select a subset from either that corresponds to the>>self-contradictory Russell Set (the set of all sets that arenot>>elements of themselves).> > No. There are set theories in which there is a universal set but>Russells Paradox does not exist.> >> >>Only by doing things like arbitrarily banning self-membership of> sets>or> postulating some sort of in?ite hierarchy of sets. To me, it all>seems>>so> unnecessary and so inelegant. By simply treating set membershiplike>any> other logical predicate of two variables, you can prove that the>> so-called> Russell set cannot exist (my website). The job of any set theory> then>is> make sure that you cannot prove that it does exist. One way -- andI>> dont> know if it is really anything new -- is to have a simpli?d set>theory>>that> does not postulate the existence of any particular set or sets.This>is>>the> approach I have taken in my proof writing program. It seems towork>and> neatly avoid problems like RP that have plagued naive set theory.I>> would>>be> very interested to see if you can derive any contradictions fromit.>> >>Dan>> Your system can not have the set of all natural numbers, N.> >Its true that number theory is not built into my simpli?d settheory,>the>> way it is in ZF. To develop number theory, I start with a premise>> corresponding to Peanos Axioms. (Click on the Numbers menu.)> >> The Axiom of In?ity is needed to that every natural number has a successor? See>above.> > No.>I mean the Axiom of In?ity. AoI states there exists a set that isclosed>under the successor function. If your set theory doesnt assume the>existence of any set then it doesnt include the AoI. It doesnt. > PA does have the AoI.> > That is what I meant to say. You dont need to include the AoI in a theory> of sets. You can introduce it as a premise, along with the the rest of PA.>> At any rate, there is no guarantee that anything actually existences> in>my>> set theory. Nevertheless, I think you should still be able to developa>> large part if not all of mathematical theory using my program.> > Certainly not all.>Your system cant have a set of all natural numbers.> > Again, to develop number theory in my system, you begin by introducing the> equivalent of PA as an initial premise. You dont need to have the natural> numbers built into your set theory.>I handle number theory in the same way that I imagine you do,say, grouptheory in ZFDan Visit DC Proof Online at http://www.dcproof.com -- FREE download =Can anybody help me with one of the next Integrals:a int(6*x / (x^2 +x+1)^(3/2))andb int( ln(1-x^2) / x^2)andc int(x*ln(x) / (x^2-1)^(1/2))Matlab gives me, except for the last one, nice answers but how do you startsomething like this manually?I cant seem to ?d the right solution either by partitial integration orby replacing parts of the equation (looking in tables).Kimball =Subteam escribi.97 Integrals: a int(6*x / (x^2 +x+1)^(3/2))>I = Int(6*x/(x^2 + x + 1)^(3/2), x)Probably there is something more directly, but lettingx + t = (x^2 + x + 1)^(1/2) > x^2 +2tx + t^2 = x^2 + x + 1 ==>x = (t^2 - 1)/(1 - 2t)dx = -2(t^2 - t + 1)/(1 - 2t)^2you getI = Int(12(t^2 - 1)/(t^2 - t + 1)^2, t)a rational integral than you can manage by standar methods.> b int( ln(1-x^2) / x^2)I = Int(Ln(1 - x^2)/x^2, x)Integrating by parts, withu = Ln(1 - x^2) > du = (-2x/(1 - x^2))dxdv = 1/x^2 > v = -1/xI = -Ln(1 - x^2)/x - 2*Int(1/(1 - x^2), x)and the last integral is easy.> and c int(x*ln(x) / (x^2-1)^(1/2))Againg use integration by parts. Letu = Ln(x) ==> du = (1/x)dxdv = x/(x^2 - 1)^(1/2) ==> v = (x^2 - 1)^(1/2)and let x = cosh(t) in the new integral.-- Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com =I must have been tired ..... I didnt see the diff. by parts.The ?st one lookst ok ..... dont know how you got the idee for using x+ t ..... but checking it in Matlab .... looks ok ... but the answer isterrible to look at ;))Ignacio Larrosa Ca.96estro schreef in> Subteam anybody help me with one of the next Integrals:> > a int(6*x / (x^2 +x+1)^(3/2))> > I = Int(6*x/(x^2 + x + 1)^(3/2), x) Probably there is something more directly, but letting> x + t = (x^2 + x + 1)^(1/2) > x^2 +2tx + t^2 = x^2 + x + 1 ==> x = (t^2 - 1)/(1 - 2t) dx = -2(t^2 - t + 1)/(1 - 2t)^2 you get I = Int(12(t^2 - 1)/(t^2 - t + 1)^2, t) a rational integral than you can manage by standar methods.>b int( ln(1-x^2) / x^2) I = Int(Ln(1 - x^2)/x^2, x) Integrating by parts, with u = Ln(1 - x^2) -Ln(1 - x^2)/x - 2*Int(1/(1 - x^2), x) and the last integral is easy. > and> > c int(x*ln(x) / (x^2-1)^(1/2)) Againg use integration by parts. Let u = Ln(x) ==> du = (1/x)dx dv = x/(x^2 - 1)^(1/2) ==> v = (x^2 - 1)^(1/2) and let x = cosh(t) in the new integral.> -- Ignacio Larrosa Ca.96estro> A Coru.96a (Espa.96a)> ilarrosaQUITARMAYUSCULAS@mundo-r.com =En el escribi.97: I must have been tired ..... I didnt see the diff. by parts. The ?st one lookst ok ..... dont know how you got the idee for> using x + t ..... but checking it in Matlab .... looks ok ... but> the answer is terrible to look at ;))>It is a standard change of variable to solve some irrational integrals.More exactly, if R(x, sqrt(ax^2 + bx + c)) is a rational function of its twoarguments, you can use one or more of this changes in order to rationalizeInt(R(x, sqrt(ax^2 + bx + c)), x):i) If a > 0, sqrt(ax^2 + bx + c) = sqrt(a)x + t (the one I used)ii) If c > 0, sqrt(ax^2 + bx + c) = tx + sqrt(c)iii) If r is a real root of ax^2 + bx + c, sqrt(ax^2 + bx + c) = t(x - r)If more than one of them ara applicable, use the former. The resultantrational integrals may be tedious enough ...Other alternatives are trigonometric or hiperbolic changes.-- Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com our exams a few years ago which simply>>said State and prove Cauchys theorem. Plenty of scope there for>>alternative answers!Take your pick:The intended meaning was none of those!Derek Holt. = >Ive noticed lately that the de?tion of science I picked up from> >Karl Popper (that it is a critical activity including criticism by> >experimentation where possible) must be drastically different than the> >de?tion of science assumed by some of more visible posters here.> >How about a little poll on this.> >In one or two paragraphs in your own words, describe what science> >means to you, and if possible from whom your de?tion is in?d> >by. Nature explained.The process of creating a sparse dictionarywhose nouns (Things) and verbs (Causes and effects)have the highest correlation with observations.The same is also true of philosophy,but science assigns greater value to externalized correlations, whereasphilosophy assigns greater value to internalized correlations.Science = it is reality.Philosophy = I am reality.--Tom Potter http://tompotter.us =Science = it is reality.And objective proposition. One can check.> Philosophy = I am reality.A subjective illusion. Right up there with I am Napoleon. Bishop Berkeley would have approved.Bob Kolker => The GR correction is about 44 microseconds or 13km per day> and would be cumulative. You can hardly call that negligible.>> >> 1. What periodicity of corrections of parameters in GPS?> >The corrections are of the order of nanoseconds per day so>> con?m GR to roughly one part in 10,000 by that simplistic>> comparison. There is much more information available in web>> pages if you want to look for more details.> > This information is incomplete.Yes, thats why I said of the order of.> Whether you can point out the information, which one describes>concrete theoretical model of GPS in detail?I could if I were prepared to dig around on the internet and> look for such a description, but you are quite capable of> doing that for yourself.>> 2. What parameters are adjusted in GPS?>>3. What medial - statistical values have parameters adjusted>> in GPS in each session of corrections?> > I asked about complete set of quantities of adjusted parameters.There are only really two things that can be adjusted, the> location of the craft at any time, given principally by> the ephemeris data, and the time on the on-board clock.> Both are severely constrained because (a) the system must> give accurate range to receivers over the whole surface> of the Earth visible from the craft and (b) many GPS> installations are used at ?ed locations purely as a time> reference. In effect, the same technique that gives accurate> location and time at a receiver using multiple satellites> works in reverse to constrain the parameters on each> satellite to satisfy multiple receivers.> The adjusted parameters are interdependent, therefore theorist can>gain distinguishing sets of quantities of adjusted parameters>depending on arbitrary selected model. The theorist can to play>by quantities of adjusted parameters on the arbitrariness in some>boundaries.There is no room to play at all for the ?theorist since> the equations of GR have no adjustable parameters. All that> can be done in the system is to adjust the ephemeris data> so that it accurately predicts the actual location of the> craft, and then adjust the clocks so they match the required> system time. Those adjustments are many orders smaller than> the tens of microseconds per day difference between orbiting> and ground-based clocks.http://tycho.usno.navy.mil/ptti/ptti2002/paper20.pdf TIMEKEEPING AND TIME DISSEMINATION IN ADISTRIBUTED SPACE-BASED CLOCK ENSEMBLES. Francis Zeta Associates IncorporatedB. Ramsey and S. Stein Timing Solutions CorporationJ. Leitner, M. Moreau, and R. Burns NASA Goddard Space Flight CenterR. A. Nelson Satellite Engineering Research CorporationT. R. Bartholomew Northrop Grumman TASCA. Gifford National Institute of Standards and TechnologyAbstractThis paper examines the timekeeping environments of several orbital and aircraft ?cena rios for application to the next generation architecture of positioning, navigation, timing, and communications. <... snip ...> Real-time hardware performance parameters, environmental factors, orbital perturbations, and the effects of special and general relativity are modeled in this simulator for the testing and evaluation of timekeeping and time dissemination algorithms. 1. INTRODUCTIONTimekeeping and time dissemination among laboratories is routinely achieved at a level of a few nanoseconds at the present time and may be achievable with a precision of a few picoseconds within the next decade. The techniques for the comparison of clocks in laboratory environmentshave been well established. However, the extension of these techniques to mobile platforms and clocks in space will require more complex considerations.A central theme of this paper - and one that we think should receive the attention of the PTTI community - is the fact that normal, everyday platform dynamics can have a measurable effect on clocks that?ride on those platforms. It is generally appreciated that clocks moving on satellites with a high speed and at a high altitude are subject to relativisticeffects.The principal example is the Global 34th Annual Precise Time and Time Interval (PTTI) Meeting Positioning System (GPS), where there is a net secular effect of 38 [Micro]s per day, a periodic effect with an amplitude of up to 46 ns, and a Sagnac effect of typically 100 ns for a stationary receiver on the geoid. 2.3 RELATIVISTIC ALGORITHMS<... snip ...> For measurements with a precision at the nanosecond level, there are three relativistic effects that must be considered [2, 3].First, there is the effect of time dilation. The velocity of a moving clock causes it to appear to run slow relative to a clock on the Earth. GPS satellites revolve around the Earth with an orbital period of 11.967 hours and a velocity of 3.874 km/s. Thus on account of its velocity, a GPS satellite clock appears to run slow by 7 [Micro]s per day.Second, there is the effect of the gravitational redshift, a frequency shift caused by the difference in gravitational potential. (The term redshift is generic regardless of sign, but for a satellite clock the frequency shift is actually a blueshift.) The difference in gravitational potential between the altitude of the orbit and the surface of the Earth causes the satellite clock to appear to run fast.At an altitude of 20,184 km, the clock appears to run fast by 45 [Micro]s per day.The net effect of time dilation and gravitational redshift is that the satellite clock appears to run fast by approximately 38 [Micro]s per day when compared to a similar clock at rest on the geoid, including the effects of the velocity of rotation and the gravitational potential at the Earth’s surface. This is an enormous rate difference for a clock with a precision of a few nanoseconds. To compensate for this large secular effect, the clock is given a fractional rate offset prior to launch of -4.465 [Times]10^-10 from its nominal frequency of exactly 10.23 MHz, so that on average it appears to run at the same rate as a clock on the ground. The actual frequency of the satellite clock prior to launch is thus 10.229 999 995 43 MHz.Although the GPS orbits are nominally circular, there is always some residual eccentricity. The eccentricity causes the orbit to be slightly elliptical. Thus, the velocity and gravitational potential vary slightly over one revolution and, although the principal secular effect is compensated by a rate offset, there remains a small residual variation that is proportional to the eccentricity. For example, with an orbital eccentricity of 0.02, there is a relativistic sinusoidal variation in the apparent clock time having an amplitude of 46 ns at the orbital period. This correction must be calculated and taken into account in theusers receiver.The third relativistic effect is the Sagnac effect. The Sagnac correction in an Earth-Centered Earth-Fixed (ECEF) frame of reference is equivalent to a light-time correction for the receiver motion in an Earth-Centered Inertial (ECI) frame. For a stationary terrestrial receiver on the geoid, the Sagnac correction can be as large as 133 ns (corresponding to a GPS signal propagation time of 86 ms and a velocity of 465 m/s at the equator in the ECI frame). This correction is also applied in the receiver.The analysis of signals used in cross-link ranging, time transfer among satellites and ground stations, and interoperability acrossconstellations involves three steps:(1) a relativistic transformation from the proper time reading of the clock at the transmitter to the coordinate time of transmission in the adopted coordinate system; (2) calculation of the coordinate time of signal propagation, including both relativistic and nonrelativistic effects; (3) a relativistic transformation from the coordinate time of reception to the proper time reading of the clock at the receiver. In addition, all proper times must be corrected for hardware effects, such as signal noise and clock environment.For cross-link ranging, the secular corrections must be modeled, especially for satellite systems that operate at different orbital altitudes. In addition, the relativistic eccentricity correction must be applied by receivers onboard the satellites themselves. Like those applied by ground receivers, these corrections will typically be on the order of tens of nanoseconds. <... snip ...> 4. SATELLITE CLOCK SIMULATIONSTo study the theoretical importance of various hardware-related biases, environmental factors,propagation delays, orbital perturbations, and relativistic effects on satellite clocks, <... snip ...> The key features of the simulation are its inclusion of relativistic effects, hardware noise, and arbitrary platform motion during the exchange. <... snip ...> 7. CONCLUSIONSTo study the theoretical importance of various hardware-related biases, environmental factors,propagation delays, orbital perturbations, and relativistic effects, a mathematical simulator has been formulated to model aircraft ?test and satellite scenarios. The model for timekeeping and time dissemination has been illustrated by clocks onboard the International Space Station, a Molniya satellite, and a geostationary satellite. In addition, a time transfer simulator integrated into the Formation Flying Testbed at the NASA Goddard Space Flight Center is being developed. Real-time hardware performance parameters, environmental factors, orbital perturbations, and the effects of special and general relativityare modeled in this simulator for the testing and evaluation of timekeeping and time dissemination algorithms. Thus we have following ranged system, which one in?s precision of system as a whole:1. various hardware-related biases, 2. environmental factors,3. propagation delays, 4. orbital perturbations, 5. relativistic effects. 0. Whether you can point out the information, which one describes concrete theoretical model of GPS in detail? >;^)1. What periodicity of corrections of parameters in GPS? >;^)2. What complete set of quantities of adjusted parameters are adjusted in GPS?3. What medial - statistical values have parameters adjusted in GPS in =On 9 Jan Timofeev)It is generally appreciated that clocks moving on satellites with >a high speed and at a high altitude are subject to relativistic>effects.The principal example is the Global 34th Annual Precise Time and Time >Interval (PTTI) Meeting Positioning System (GPS), where there is a net secular effect of 38 [Micro]s per day, a periodic effect with an amplitude of up to 46 ns, and a Sagnac effect of typically 100 ns for a stationary receiver >on the geoid. 2.3 RELATIVISTIC ALGORITHMS<... snip ...> >For measurements with a precision at the nanosecond level, there >are three relativistic effects that must be considered [2, 3].>First, >there is the effect of time dilation. The velocity of a moving clock >causes it to appear to run slow relative to a clock on the Earth. >GPS satellites revolve around the Earth with an orbital period of >11.967 hours and a velocity of 3.874 km/s. Thus on account of its >velocity, a GPS satellite clock appears to run slow by 7 [Micro]s per day.>Second, >there is the effect of the gravitational redshift, a frequency shift >caused by the difference in gravitational potential. (The term >redshift is generic regardless of sign, but for a satellite clock >the frequency shift is actually a blueshift.) The difference in >gravitational potential between the altitude of the orbit and the >surface of the Earth causes the satellite clock to appear to run fast.At an altitude of 20,184 km, the clock appears to run fast by 45 [Micro]s >per day.The net effect of time dilation and gravitational redshift is that >the satellite clock appears to run fast by approximately 38 [Micro]s per >day when compared to a similar clock at rest on the geoid, including >the effects of the velocity of rotation and the gravitational >potential at the Earth’s surface. This is an enormous rate >difference for a clock with a precision of a few nanoseconds. To >compensate for this large secular effect, the clock is given a >fractional rate offset prior to launch of -4.465 [Times]10^-10 from its >nominal frequency of exactly 10.23 MHz, so that on average it appears >to run at the same rate as a clock on the ground. The actual >frequency of the satellite clock prior to launch is thus >10.229 999 995 43 MHz.>Load of crap.If the orbiting clocks appear to run slow it is because they DO run slow. Theground observer counts every ?tick emitted per orbit.The GPS clock preset is required simply to compensate for an increase in clockrates due to being in free fall......unless the ?tick-fairies have been active again!Henri Wilson. Any connection between Einsteinian relativity and truth is purely coincidental.www.users.bigpond.com/hewn/index.htm => On 9 (AleksandrTimofeev) >The velocity of a moving clock> >causes it to appear to run slow relative to a clock on the Earth.> >GPS satellites revolve around the Earth with an orbital period of> >11.967 hours and a velocity of 3.874 km/s. Thus on account of its> >velocity, a GPS satellite clock appears to run slow by 7 [Micro]s per day....>The difference in> >gravitational potential between the altitude of the orbit and the> >surface of the Earth causes the satellite clock to appear to run fast.> >At an altitude of 20,184 km, the clock appears to run fast by 45 [Micro]s> >per day.> >The net effect of time dilation and gravitational redshift is that> >the satellite clock appears to run fast by approximately 38 [Micro]s per> >day when compared to a similar clock at rest on the geoid, including> >the effects of the velocity of rotation and the gravitational> >potential at the Earth’s surface. This is an enormous rate> >difference for a clock with a precision of a few nanoseconds. To> >compensate for this large secular effect, the clock is given a> >fractional rate offset prior to launch of -4.465 [Times]10^-10 from its> >nominal frequency of exactly 10.23 MHz, so that on average it appears> >to run at the same rate as a clock on the ground. The actual> >frequency of the satellite clock prior to launch is thus> >10.229 999 995 43 MHz.> Load of crap.The above are merely statements of fact. The clocks are builtto run at 10.22999999543MHz prior to launch so that the signalreceived once they are in orbit is measured as 10.23MHz on theground receivers.> If the orbiting clocks appear to run slow it is because they DO run slow.They run fast, not slow.> The ground observer counts every ?tick emitted per orbit.Well a ground observer can only see any satellite for a fractionof its orbit but if the Earth didnt get in the way, you wouldbe right.> The GPS clock preset is required simply to compensate for an increase inclock> rates due to being in free fall.The change of coordinate rate is correlated to the difference ingravitational potential, not force, but with that clari?ationyou are correct again.> .....unless the ?tick-fairies have been active again!If a satellite signal could be received throughout its orbit theground observer could count every tick emitted, so there are notick fairies.Nothing you have said con?in any way with what was stated,so what were you objecting to?George =Nice page Aleksandr. It makes my point well:> Timekeeping and time dissemination among laboratories is routinely> achieved at a level of a few nanoseconds at the present time> and may be achievable with a precision of a few picoseconds within> the next decade.> [38 [Micro]s per day] .. is an enormous rate> difference for a clock with a precision of a few nanoseconds. To> compensate for this large secular effect, the clock is given a> fractional rate offset prior to launch of -4.465 [Times]10^-10 from its> nominal frequency of exactly 10.23 MHz, so that on average it appears> to run at the same rate as a clock on the ground. The actual> frequency of the satellite clock prior to launch is thus> 10.229 999 995 43 MHz. Although the GPS orbits are nominally circular, there is always> some residual eccentricity. ..> For example, with an orbital eccentricity of 0.02,> there is a relativistic sinusoidal variation in the apparent> clock time having an amplitude of 46 ns at the orbital period.> This correction must be calculated and taken into account in the> users receiver.These are the main effects that are corrected by being builtinto the system. With these in place, the system can run formany days without any ground contact. Regular adjustments aremade but they are at the level of nanoseconds or smaller andare in no way comparable to the primary 38 [Micro]s per day offsetas can be seen from the fact that handling of the 46 ns sinecomponent is built into the receiver.> 0. Whether you can point out the information, which one describes> concrete theoretical model of GPS in detail? >;^)> 1. What periodicity of corrections of parameters in GPS? >;^) 2. What complete set of quantities of adjusted parameters are adjusted> in GPS?> 3. What medial - statistical values have parameters adjusted> in GPS in each session of corrections?Sorry Aleksandr but I have made it clear I do not have theanswers to these questions and I am not going to ?d themfor you. If you are genuinely interested, I am sure you can?d out yourself.It may help to look for pages discussing how long the systemwould maintain a given accuracy in the absence of groundcontact. I know the latest (or maybe next) constellation isspeci?ally intended to extend that period.Happy New YearGeorge =Could somebody point me to a book/books, where I can ?d somethingthat wovens the above topics together ?I am interested of the theoretical foundations of the solvingprocedure of Schr.9adinger equation in condensed matter systems.I am a layman in this ?ld. I am just in it for fun.I am interested in the following questions:1. Why can a differential eqn. like Schr.9adingers be treated as aneigenvalue eqn. ? Can all diff. equations be treated aseigenvalue-eigenvector problems ?2. Why can symmetry be used in the solving procedure ? One see people-expand the the assumed solution in a set of basis functions that havethe same symmetry as the system considered, and solve for thecoef?ients as system of linear equations.(Gauss-elimination) The basis functions of the one-dimesional representaion of the groupthat the system belongs to, are the basis functions that are used. Why? Because the symmtry operators and the hamiltonian operator commute(?????????)Help !3. How does variational theory (and what is it ?) come into thesolving theory of Schr.9adinger eqn. ?4. Perturbation theory in quantum mechanics. Is it one of thesenone-proven things ? Is it used in mechanics ? I never saw it mynewtonian mechanics class.When you suggest books, consider that I have the education of anelectronics engineer. No fancy mathematics courses like functionalanalysis, etc.So maybe you have to suggest a book for the basics also....I dont ask for help ?ding the books because of lazyness. I donthave a decent library near me.Lasse interested in the following questions:1. Why can a differential eqn. like Schr.9adingers be treated as an> eigenvalue eqn. ?An eigenvalue equation is, essentially, an equation of the typelinear operator acting on a vector gives the same vector, multipliedwith a constant. The linear operator in Schroedingers equation is-hbar^2/2m Laplace + V, the vector is the wave function it is acting on.(in coordinate space; if you are familiar with the Dirac notation, itbecomeseven more obvious that Schroedingers (time-independent!) equation is aneigenvalue equation)> Can all diff. equations be treated as> eigenvalue-eigenvector problems ?No, only the ones of the type outlined above (for example, a requirementis that the differential operator has to be linear). > 2. Why can symmetry be used in the solving procedure ?If an equation has a certain symmetry, one can show that the *set* o?s solutions has to have the same symmetry. This in general greatlysimpli?s ?ding the solutions.> One see people> -expand the the assumed solution in a set of basis functions that have> the same symmetry as the system considered,Well, usually the basis functions have *not* the same symmetry as thesystem... (for example, not all wave functions for the H atom arespherically symmetric!)> and solve for the> coef?ients as system of linear equations.(Gauss-elimination)> The basis functions of the one-dimesional representaion of the group> that the system belongs to, are the basis functions that are used. Why?Sorry, Im not very familiar with this...> Because the symmtry operators and the hamiltonian operator commute> (?????????)Help !When one can ?d operators for certain symmetries which commute withthe Hamiltonian, one usually chooses the wave functions to beeigenfunctions to these operators, too. Again, this is done because itusually simpli?s ?ding the solutions (example: for solving the Hatom, it is *very* helpful to try to ?d a wave function which is aneigenstate for angular momentum, too - because then one already knowsthat the wave function has to be a multiple of a spherical harmonic!) > 3. How does variational theory (and what is it ?) come into the> solving theory of Schr.9adinger eqn. ?AFAIK, the Schroedinger equation usually cant be solved by thevariational theory - one can only ?d *approximate* solutions. What itis? Well, the idea (in the case of the Schroedinger equation) is to usesome test functions and calculate the expectation value for the energyfor them. It is easy to prove that these expectation values have to begreater or equal to the ground state energy, hence by looking for thelowest possible energy among all of the test functions, one gets anupper bound for the ground state energy. (this is only the tip of theiceberg, but I think you get the general idea...)> 4. Perturbation theory in quantum mechanics. Is it one of these> none-proven things ?I think a physicist would say that it is proven, but a mathematicianwould disagree. ;-)Short answer: it works, often with astounding accuracy.> Is it used in mechanics?Yes, although in completely different form. Example: for studying thein? of the other planets when trying to calculate the orbit of oneplanet around its sun.> I never saw it my newtonian mechanics class.Well, I think in many mechanics classes, this is considered to be anadvanced topic.> When you suggest books, consider that I have the education of an> electronics engineer. No fancy mathematics courses like functional> analysis, etc.But apparently you have some experience with differential equations andmatrices, dont you? What about the general de?ition of vectorspace? What about the notion of symmetry be used in the solving procedure ? If an equation has a certain symmetry, one can show that the *set* of> its solutions has to have the same symmetry. This in general greatly> simpli?s ?ding the solutions.You might be interested in:Peter Olver, Applications of Lie Groups to Differential Equations,Second Edition, Springer-Verlag, education of an>electronics engineer. No fancy mathematics courses like functional>analysis, etc.O, then Olver probably is not for you (or perhaps it is...).But in case anyone else is interested, i give the reference anyway.Herman Jurjus =>> following sum:>> S(n) = sum {k=1,...,n} n!/(k*(n-k)!)>> Need closed form solution or asymptotic estimations for large n.Mathematica gives a closed form for your sum. Not surprisingly, its in>terms of a hypergeometric function: n HypergeometricPFQ[{1, 1, 1 - n}, {2}, -1]See for more>information, if necessary.Notice that S(n) n --- = > C(n,k) (k-1)! --- k=1 n --- = > (C(n-1,k)+C(n-1,k-1)) (k-1)! --- k=1 n-1 --- = S(n-1) + > C(n-1,k) k! --- k=0 n-1 --- = S(n-1) + (n-1)! > 1/k! --- k=0Therefore, S(n) - S(n-1) 1 1 1 = e (n-1)! - - - ------ - ----------- - ... n n(n+1) n(n+1)(n+2)Thus, if I am not mistaken, asymptotically, 1 1 1 1 1 1 S(n) ~ e (n-1)! ( 1 + --- + --- --- + --- --- --- + ... ) n-1 n-1 n-2 n-1 n-2 n-3 1 2 5 15 52 203 ~ e (n-1)! ( 1 + - + --- + --- + --- + --- + --- + ... ) n n^2 n^3 n^4 n^5 n^6Rob Johnson take out the trash before replying =>) We do the famous triangulation method (Cantor?)>) ) Of course there are a lot of repetitions. We can pass over repreated >) values, or else accept that a rational has several representations. >) However, it means that any positive rational can be represented by an >) unary number.Isnt there also a triangulation method that doesnt have duplicates ?>I seem to recall it, but I dont remember exactly how it went.Its easy to de?e one, but hard to think of an algorithm for going from the unary representation to a rational.It would be relatively easy to determine whether a given representation was the ?st representation in a multi-representation system, though.Gerry Quinn -- http://bindweed.com Screensavers, Games, KaleidoscopesDownload free trial versions => Zero is not needed as a placeholder in the Alternate Number System.> Also in this system in base 1, the digits are ones as required and is> simply a tally system.> In base (b) the digits are 1 to b which is more logical than didgits 0 to> (b-1) in our existing system.> http://my.tbaytel.net/forslund/ans.html>How would the ANS represent fractional numbers in general, and currency inparticular?How would you represent:$0.05$0.10$1.00Pete =you cant defend yourself! that was a good call, continued fractions, sinceany number can be so-represented. ina ny case,it should be obvious that base-1 uses only 0,as a stand-in for the typical counter (pebble);that is, the digit is *only* a place-holder,for the individual things-being-counted. now, if you allow the Liberal Media -- and Howie Troisieme etc.ad vomitorium -- to cover-up for Cheeny,youll probably be counting nothing *but* rocks. > My particular de?ition of the term base-1 is not my own, yet I> cannot recall where else I have seen it used in the sense I use it in> my original post, but I have seen the term used this way in several> different (and reputable) places, I am sure.--Give the Gift of Trickier Dick Cheeny -- out of of?e, at last!http://www.benfranklinbooks.com/http://www.wlym.com/pages /music.htmlhttp://www.rand.org/publications/randreview/issues/ rr.12.00/http://members.tripod.com/~american_almanachttp:// =XML converted to unicode (and then Outlook Express converted it further)AnonMoos schreef in bericht[Thorn][NonBreakingSpace][RegisteredTrademark][ NonBreakingSpace]> View the following string as part of an HTML page in an internationalized> web browser to see this in Hebrew:.be.be[NonBreakingSpace][Eth][CapitalUGrave][ CapitalYAcute][Eth][CapitalYAcute].be[Thorn][NonBreakingSpace] [RegisteredTrademark][NonBreakingSpace] =XML converted to unicode (with unipad)AnonMoos schreef in bericht[ ... ]> View the following string as part of an HTML page in an internationalized> web browser to see this in Hebrew: [Times].85[Times][Times][Thorn]205[Times].85.b9.85[Times][Eth] [Times].85.a6[Times].85[Times].85[Times].85.b9.85[ CapitalOTilde][Times][Eth][Times]>[ ... ] =See my proposed answers below. The comments are all mine, Im not anexpert and these should all be taken with a grain of salt.> In Hebrew, the feminine plural of the foreign loan word, nilpotent, when> written without vowels or other punctation, is writtennylpw_tn_tywtwhere _t is teth. What I would like to know is how it would be written> if one did write the vowels and other punctation. For example:> (0) which of the consonants is dotted?If dotted means having a dagesh, then its just the p. (Comment:in a borrowed word, you would put dots only where they are necessaryfor correct pronunciation, as there are now processes of lostconsonants etc. that may cause dots to appear in a Hebrew word).> (1) is the y following the ?st n retained and the n endowed with a chireq?Yes. (comment: when a word is spelled out completely as you seek, thiswould always be the case as far as I can see).> (2) is there a schwa under the l?Yes. (Comment: Im not sure what else could it be. Theres not vowel,so schwa it is).> (3) is the w following the p replaced by a cholem or is it replaced by> a qamats qatan?Cholem (that it, its retained and gets endowed with a cholem).(Comment: Qamats katan is a rarity, I cant think of any modernborrowed word that would use one).> (4) what is under the ?st teth: a schwa, a segol or a chataf segol?Segol. (comment: theres a vowel, E, right after this, so a schwa ismore or less out of the question. A chataf segol is very, very, veryrarely used on anything that is not Aleph, Heh, Heth or Ayin, and inany event - again - I cant think of a reason to use one in a borrowedword).> (5) is there a schwa under the second n?Yes. (Comment: (2) above).> (6) is the y following the second teth retained and the teth endowed with> a chireq?Yup. (Comment: (1) above).I would also like to know how I might ?ure out examples like this, involving> foreign loan words, on my own, instead of having to ask people in every> instance.Ah, well. Hopefully the comments above provide some guidance. Thereare essentially rules for punctuation (set by Rabbi Eliyahu Bachur, Ibelieve) that indicate whether a short or long (not sure what theproper technical term is in English) punctuation is to be used, whichpretty much settles it in most cases (segol vs. tseire, etc.). Therules are simple and I can repeat them if you like.Im afraid that natives Hebrew speakers pretty much know how topunctuate without having a clearly de?ed rationale. For modern,borrowrd, words, though, this should be easy. Exercise: do pwlynwm,algbra, aydal (the least one is ideal).details - not sure I can deliver, but I can try.Alan =>Never heard this word before. It must be very modern Hebrew or veryAncient>Aramaic. There is the shoresh LVE - to loan, lend, borrow. But that isall I>can help you with.> > YS> > >In Hebrew, the feminine plural of the foreign loan word, nilpotent,when>> written without vowels or other punctation, is written> >nylpw_tn_tywt> >where _t is teth. What I would like to know is how it would be written>> if one did write the vowels and other punctation. For example:>> (0) which of the consonants is dotted?>> (1) is the y following the ?st n retained and the n endowed with a> chireq?>> (2) is there a schwa under the l?>> (3) is the w following the p replaced by a cholem or is it replaced by>> a qamats qatan?>> (4) what is under the ?st teth: a schwa, a segol or a chataf segol?>> (5) is there a schwa under the second n?>> (6) is the y following the second teth retained and the teth endowedwith>> a chireq?> >I would also like to know how I might ?ure out examples like this,> involving>> foreign loan words, on my own, instead of having to ask people inevery>> instance.> >Ignorantly,>> Allan Adler>> ara@zurich.ai.mit.edu> >************************************************************* ***************>> *> *> *>> * Intelligence Lab. My actions and comments do notre?*>> * in any way on MIT. Moreover, I am nowhere near theBoston> *>> * metropolitan area.> *>> *> *> >************************************************************* ***************> > --->Checked by AVG anti-virus system (http://www.grisoft.com). I cannot help with the Hebrew but it may help those outside of> sci.math to know that this is a mathematical term which I do not think> have any use outside of mathematics. It is not in my Concise Oxford> Dictionary and I would be surprised to see it outside of a maths book. So I doubt that it is ancient. More likely is a modern coining. The> spelling above suggests an attempt to mimic the English word. JWikipedia:In mathematics, an element x of a ring R is called nilpotent if there existssome positive integer n such that xn = 0. Examples: This de?ition can beapplied in particular to square matrices. If x is nilpotent, then 1-x is aunit, because xn = 0 entails(1-x) (1 + x + x2 + ... + xn-1) = 1 - xn = 1.More info at:http://en2.wikipedia.org/wiki/NilpotentYS---Checked by AVG anti-virus system (http://www.grisoft.com). = >...... >You are *again* misstating the position. The divisor of >(5 a_1(x,y) + 7y) that is in common with 7 is only a divisor of >the complete expression. The 7y is independent of x. > > Any factor in common with 7 has to show up in the coef?ient of the > term independent of x,Yes, and it does. (Any factor of 7 is also a factor of 7y.) > and it does as 7 is that factor, in an > appropriately inclusive ring, while the ring of algebraic integers by > virtue of its de?ition excludes *any* algebraic integer factor in > common with 7 in this case.Why? You know, this is false, as has been shown already. It is truethat 7 itself is in most cases not a factor, however divisors of 7*can* be factors. > So Im not misstating anything. If as you seem intent on claiming > there is some other factor than 7, dependent on x, or x and y, then it > has to show up as a factor of 7y.Why? (Note however, that whatever that factor is, it indeed *does*show up as a factor of 7y.) >Oh, come on, this is simply too simple... > v1(x, y) = gcd(5 a_1(x, y) + 7y, 7) > v2(x, y) = gcd(5 b_2(x, y) + 2y, 7) > k(x, y) = v1(x, y) * v2(x, y) ; is 7 or a multiple of 7 (%) > g(x, y) = k(x, y) / 7 ; the excess > h1(x, y) = gcd(v1(x, y), g(x, y)) ; split in h1 * h2 where h1 is a > h2(x, y) = g(x, y) / h1(x, y) ; factor of v1 and h2 of v2 > w1(x, y) = v1(x, y) / h1(x, y) > w2(x, y) = v2(x, y) / h2(x, y) >You may verify that for all algebraic integers x and y, the above >functions are algebraic integer functions, that w1(x, y) * w2(x, y) = 7 >and that w1(x, y) is a factor of (5 a_1(x, y) + 7y) and w2(x, y) a factor >of (5 b_2(x, y) + 2y) in the algebraic integers. > >(% Note that for any three algebraic integers p, q and r, gcd(pq, r) >divides gcd(p, r)*gcd(q, r).) > >Now I wonder what you mean with the change of exponent... > > Dividing off 7 from both sides, and check at x=0, as then you should > notice that the terms independent of x are now y, and 2y, so 7 was > divided off from 7y.Pray tell me: *which* of the steps and or de?itions above is wrong?But lets see. I have: (5 a_1(x, y) + 7y)/w1(x, y) and (5 b_2(x, y) + 2y)/w2(x, y)Filling in above, I get (at x = 0): v1(0, y) = 7, v2(0, y) = gcd(y, 7), k(x, y) = 7 * gcd(y, 7), g(0, y) = gcd(y, 7), h1(0, y) = gcd(y, 7), h2(0, y) = 1, w1(0, y) = 7 / gcd(y, 7), w2(0, y) = gcd(y, 7).So in the above expressions, when I set x = 0, I get: (5 a_1(0, y) + 7y)/w1(0, y) = y * gcd(y, 7)and (5 b_2(0, y) + 2y)/w2(0, y) = 2y / gcd(y, 7)So, what is wrong about that? If y is coprime to 7 I get y and 2y.But if y is not coprime to 7 I get something else. I could haverearranged the de?itions such that I would have gotten y and 2yin all cases, but that is not necessary as clearly both terms aboveare algebraic integers for algebraic integer y. > When there were constants you were able to ?e n otion that > somehow, you could divide a constant as a function, but now youve > revealed just how little algebra you really understand by claiming you > can have w1(x,y), w2(x,y), and w3(x,y), as how do you suppose they can > divide 7 from 7y without changing the exponent of y, if they are > functions of x and y?It is *still* not clear what you mean by changing the exponent. > Youre in too deep now to worry about the truth, eh?*Which* of the steps in my de?itions above is wrong? And why?-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ =>...>(5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). >Now for readers wondering about it that expression isnt even>questioned by posters, who are pushing the result that *both*>(5a_1(x,y) + 7y) and (5b_2(x,y) + 2y) share non-unit factors in common>with 7, where those factors vary as a function of x.>However, their position requires that the terms *independent* of x,>which on the left are 7y and 2y, and on the right 7(2y^2) do in fact>have some kind of dependency, to force different factors in common>with 7 as x varies.>You are *again* misstating the position. The divisor of>(5 a_1(x,y) + 7y) that is in common with 7 is only a divisor of>the complete expression. The 7y is independent of x.Any factor in common with 7 has to show up in the coef?ient of the> term independent of x,No. This is only true if the factor divides (5a_1(x,y) + 7y)for all x and y. As the factor varies as a function of x and yit does not have to show up in the coef?ient. > and it does as 7 is that factor, in an> appropriately inclusive ring, while the ring of algebraic integers by> virtue of its de?ition excludes *any* algebraic integer factor in> common with 7 in this case.So Im not misstating anything. If as you seem intent on claiming> there is some other factor than 7, dependent on x, or x and y, then it> has to show up as a factor of 7y.> >Notice that these people dont have the option of now claiming an x>and y dependency as theres no mechanism for w_1(x,y) w_2(x,y) = 7,>where the ws are factors of those terms that vary with x, and y, as>necessarily then the degree of the y exponent in 7y and 2y would>change once you divided by 7, so I think trying to claim a w_1(x,y)>and w_2(x,y) is too wacky for even Dik Winter to try!>Oh, come on, this is simply too simple...> v1(x, y) = gcd(5 a_1(x, y) + 7y, 7)> v2(x, y) = gcd(5 b_2(x, y) + 2y, 7)> k(x, y) = v1(x, y) * v2(x, y) ; is 7 or a multiple of 7 (%)> g(x, y) = k(x, y) / 7 ; the excess> h1(x, y) = gcd(v1(x, y), g(x, y)) ; split in h1 * h2 where h1 is a> h2(x, y) = g(x, y) / h1(x, y) ; factor of v1 and h2 of v2> w1(x, y) = v1(x, y) / h1(x, y)> w2(x, y) = v2(x, y) / h2(x, y)>You may verify that for all algebraic integers x and y, the above>functions are algebraic integer functions, that w1(x, y) * w2(x, y) = 7>and that w1(x, y) is a factor of (5 a_1(x, y) + 7y) and w2(x, y) a factor>of (5 b_2(x, y) + 2y) in the algebraic integers.>(% Note that for any three algebraic integers p, q and r, gcd(pq, r)>divides gcd(p, r)*gcd(q, r).)>Now I wonder what you mean with the change of exponent...Dividing off 7 from both sides, and check at x=0, as then you should> notice that the terms independent of x are now y, and 2y, so 7 was> divided off from 7y.> This is only true if (5a_1(x,y) + 7y) is divided by 7 and (5b_2(x,y) + 2y) is divided by 1. However, this is only oneof an in?ite number of ways to divide the LHS by 7.> When there were constants you were able to ?he notion that> somehow, you could divide a constant as a function, but now youve> revealed just how little algebra you really understand by claiming you> can have w1(x,y), w2(x,y), and w3(x,y), as how do you suppose they can> divide 7 from 7y without changing the exponent of y, if they are> functions of x and y?>You have provided no reason that the exponent of y should not change.True, if you divide the product by the constant 7, then the exponent ofy in the product will not change. But there is no reason that the exponentsof y in individual terms in the factors should not the extent that talking aboutthe exponent of y in non-polynomial functions makes any sense). - William Hughes =>>Id suggest you stop looking at the so-called constant>>terms. Theyre doing you far more harm than good.>Why would you so suggest?>The problem is that you take something like> p P(x) = g1(x)g2(x)>and conclude that you can determine how p divides>into g1 and g2 by looking at the constant terms of>g1 and g2 (which you de?e to be g1(0) and g2(0)).>For polynomial g1 and g2 this works. However, if g1 and g2Well, I found that posters like yourself could get away with> nonsensical statements like that so I added a variable to take away> the phrase constant terms.In this context there is very little difference between aconstant term and a term involving a variable independent of x.Lets add an independent variable y to the above example p P(x,y) = g1(x,y)g2(x,y)Further let g1(x,y) = (c1(x,y) + d1(y)) and g2(x,y) = (c2(x,y) + d2(y)).As y is independent of x, d1(y) does not vary as x varies. Hence we havetwo basic possibilities, (a) there is a (constant) factor of p which divides c1(x,y) for all x and y, and divides d1(y) for all y (b) the common factor of p and g1(x,y) varies with xIf g1 and g2 are polynomials then we know that (a) must be true. We canthen look at the common factors of p and d1 and p and d2 to draw conclusionsabout the possible common factors of p and g1 and p and g2. However,if g1 and g2 are not polynomials, then (b) might be true. We cannotconclude anything about the possible factors of p and g1 and p and g2by looking at the common factors of p and d1 and p and d2. To show that (b) can be true consider the following trivial example.Let g1(x,y) = 1-y for x rational and y for x irrational. Letg2(x,y) = y for x rational and 1-y for x irrational. Let d1(y)=d2(y)=y.Then g1(x,y)g2(x,y) = 2y for all x and y.Thus the product is divisible by 2 for all x and y. This despite the fact thatfor rational x, 2 and g1(x,y) are coprime while for irrational x, 2 and g2(x,y) arecoprime. Furthermore, in general, neither d1(y) nor d2(y) is divisible by 2. I also added that variable to a *quadratic* example given by Rick> Decker, a professor at Hamilton College.Heres the result:(5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) where the as are roots of > > a^2 - (x - y)a + 7(x^2 + xy).Notice that x and y are *independent* variables, as I havent related> them to each other in any way.Now letting x=0, gives a(a + y) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, you havea_2(x,y) = b_2(x,y) - y, so(5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). Now for readers wondering about it that expression isnt even> questioned by posters, who are pushing the result that *both*> (5a_1(x,y) + 7y) and (5b_2(x,y) + 2y) share non-unit factors in common> with 7, where those factors vary as a function of x.You have now added the variable y. So the factors now vary asfunctions of both x and y.> However, their position requires that the terms *independent* of x,> which on the left are 7y and 2y, and on the right 7(2y^2) do in fact> have some kind of dependency, to force different factors in common> with 7 as x varies.Notice that these people dont have the option of now claiming an x> and y dependency as theres no mechanism for w_1(x,y) w_2(x,y) = 7,> where the ws are factors of those terms that vary with x, and y, as> necessarily then the degree of the y exponent in 7y and 2y would> change once you divided by 7This makes no sense. What do you mean by the degree of the y exponent.Why must it change if you divide by 7? Why cant it change? - William Hughes =for the sake of the peanut gallery, as I think thatthis has been done, before,could you take a much simpler example,showing the very same sort of analysis? I dont think that Jimi Harris will ever get it,by trading his tirades on Usenet/googolplex, sincehe wont even study Fermats little theorem (say);I mean, what is the use? yeah, Im glad that you few & proud are holding himto task, and Im not really quali?d ...its just as boring for most of us, as I see, now,as it was for the last nine years of it. its sortof like a soap-opera that has an unlimited, foundation-paid stayon TV; no-one actually has to watch it! Attacking the Algebra that You Should Know and Love! > First, note that 7(25x^2 + 30xy + 2y^2) = 7((x^2 + xy)*5^2 + (x - y)*y*5 + 7*y^2.> Assume a factorization of the form> > (a_1(x)*5 + 7y)*(a_2(x)+ 5 + 7y) = 7((x^2 + xy)*5^2 + (x - y)*y*5 + 7*y^2).> > Thinking of this as a factorization in 5, -7*y/a_1 and -7*y/a_2 > are roots of > > 7*((x^2 + xy)*s^2 + (x - y)y*s + 7*y^2 = 0,> > where s is the variable. This implies that a_1 and a_2 satisfy> > a^2 - (x - y)*a + 7*(x^2 + x*y) = 0.> > This of course means that > > a_1 * a_2 = 7*(x^2 + x*y).> > > Now assume (as is more often the case than not) that x^2 + x*y is> coprime to 7. De?e> > w1 = GCD(a_1, 7) and> > w2 = GCD(a_2, 7).> > This implies that a_1/w1, 7/w1, a_2/s2, and 7/w2 are> all algebraic integers [by the de?ition of GCD in the> ring of algebraic integers]. This in turn implies that > > ((a_1/w1)*5 + (7/w1)*y)*((a_2/w2)*5 + (7/w2)*y) > > is a factorization of> > (x^2 + xy)*5^2 + (x - y)*y*5 + 7*y^2 > > in which all the coef?ients are algebraic integers.> > Now: how might you know that neither w1 nor w2 are > units?> > If w1 or w2 were a unit, then one may assume that the > other one is equal to 7. If w1 = 7, one readily ?ds> that a_1 cannot be an algebraic integer. One therefore> concludes that w1 <> 7 and similarly w2 <> 7. Thus also> neither w1 nor w2 can be units. Note that all of a1, a2, w1, and w2, are functions of > both x and y.> Complicating this thing by adding in the variable y really makes no > difference. The key fact is, YOUR factorization, in which > you choose w1 = 7 and w2 = 1, does not work. [A singular exception occurs> when x = 0.] The de?ition of of w1 and w2 in terms of GCDs as > given above DOES work, whenever x^2 + x*y is coprime to 7, to > give a factorization of your original polynomial in which all the > coef?ients are algebraic integers. This was what you were > aiming for in the ?st place. Contrary to what you have said> repeatedly, it is not impossible. Dik shows that with a little more complication, you can even dispense > with the requirement that x^2 + x*y is coprime to 7. >Theyre attacking the algebra that you should know and love.--Give the Gift of Trickier Dick Cheeny -- out of of?e, at last!http://www.benfranklinbooks.com/http://www.wlym.com/pages /music.htmlhttp://www.rand.org/publications/randreview/issues/ rr.12.00/http://members.tripod.com/~american_almanachttp:// =... > >Therefore, while the sqrt() operator is in place you cannot further > >resolve the expression to see *which* of the solutions is coprime to > >7. > > > Let primes denote complex conjugation and let c denote a root of > a^2 - a + 42 = 0 supposed to be coprime to 7 in the ring A of > algebraic integers. Note that the other root is c. > > If c and 7 are coprime in A then there are elements p and q in A > such that cp + 7q = 1.Be careful here. In James all-inclusive ring of two complex conjugates,one can be a unit while the other is not a unit. A bit strange, and Ido not know whether his ring really does exists, but the prooffails for that. Because if c is a unit, q = 0 and p = inv(c). Butp = inv(c) does not necessarily belong to the ring (conjugation isnot a ring operation). > So 1 = 1 = (cp + 7q) = (cp) + (7q) = cp + 7q = cp + 7q, > and since p and q are in A, c and 7 are coprime in A.Indeed, a valid proof is conjugation is a valid operation in the ring.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ =help -- Jimi Harris is in the Ring of Algebraic Integers, andhe cant get out! > Be careful here. In James all-inclusive ring of two complex conjugates,> one can be a unit while the other is not a unit. A bit strange, and I> do not know whether his ring really does exists, but the proof> fails for that. Because if c is a unit, q = 0 and p = inv(c). But> p = inv(c) does not necessarily belong to the ring (conjugation is> not a ring operation). > So 1 = 1 = (cp + 7q) = (cp) + (7q) = cp + 7q = cp + 7q,> > and since p and q are in A, c and 7 are coprime in A.Indeed, a valid proof is conjugation is a valid operation in the ring.--Give the Gift of Trickier Dick Cheeny -- out of of?e, at last!http://www.benfranklinbooks.com/http://www.wlym.com/pages /music.htmlhttp://www.rand.org/publications/randreview/issues/ rr.12.00/http://members.tripod.com/~american_almanachttp:// =>My teacher assigns me some challenge problems to solve. But it seems>to be too hard for me. Can someone give me some ideas. Thanx !This is one of them :>> 1)The function f(x) has f ?(x) > 0 for all x 0 and the graph is asymptotic to y = ax + b as x. >> Show that f(x) - ax - b has negative derivative for all x > 0,>and that f(x) > ax + b for all x 0.[/quote:b29fe8cde7]Consider the function g(x) = f(x) - ax - b. What is its asymptote?What can you say about g(x)?Rob Johnson take out the trash before replying =You may view my proofs at the Florida State web page:www.math.fsu/Science/Specialized under Goldbach Proof and InEvans. =the URL didnt work. I like it, though,when other people think that Fermat had a Method;he just didnt reveal all of it! perhaps it will ultimately be shown tahtthe last theorem was actually one of his earliest ones;eh?> You may view my proofs at the Florida State web page:> www.math.fsu/Science/Specialized under Goldbach Proof and In> Evans.--Give the Gift of Trickier Dick Cheeny -- out of of?e, at last!http://www.benfranklinbooks.com/http://www.wlym.com/pages /music.htmlhttp://www.rand.org/publications/randreview/issues/ rr.12.00/http://members.tripod.com/~american_almanachttp:// =As a part of the automated decision support system for an empire-based>strategy game, I would like to determine the ?ancial condition of a>player. It should be a simple evaluation, stating how better or worse than>the others the player is (this value is later fuzzi?d).The value should be 0 if the evaluated player has the same amount of money>as each of the other players, non-zero otherwise (negative when hes poorer,>positive when richer). The values should differ for [1, 0, 0, 0] and [5, 4,>4, 4] cases (?st player is being evaluated), i.e. not only a difference,>but also a proportion should count.What equation would you suggest?A simple solution would be to evaluate difference from the average wealth, divided by the average wealth.1, 0, 0, 0Av = 0.25Wealth = 4.0, -1.0, -1.0, -1.05, 4, 4, 4Av = 4.25Wealth = +0.18, -0.06, -0.06, -0.06Gerry Quinn -- http://bindweed.com Screensavers, Games, KaleidoscopesDownload free trial versions =Let me try to pose the problem, I am trying to solve.I have a matrix, A (for simplicity let us assume it is a 2 X 2 matrix), thatis supposed to be symmetric and positive de?ite. There is a differentialequation the numerical solution of which gives me A at a current instant, inthe form Adot = f(x,p)..where x and p are the states of this dynamicalsystem and f is a nonlinear function in x, p. Since A has only three uniqueparameters for this case, I actually obtain, theta_dot = g(x,p), where thetaA(t) thereby enforcing symmetry..... To enforce positive de?iteness, Ineed to do the following,theta(1) > 0, theta(2) > 0 and -sqrt(theta(1)*theta(2)) < theta(3) is supposed to be symmetric and positive de?ite. There is a differential> equation the numerical solution of which gives me A at a current instant, in> the form Adot = f(x,p)..where x and p are the states of this dynamical> system and f is a nonlinear function in x, p. Since A has only three unique> parameters for this case, I actually obtain, theta_dot = g(x,p), where theta> A(t) thereby enforcing symmetry..... To enforce positive de?iteness, I> need to do the following,theta(1) > 0, theta(2) > 0 and -sqrt(theta(1)*theta(2)) < theta(3) <> sqrt(theta(1)*theta(2))...Then by using some sort of a projection algorithm, I could enforce positive> de?iteness in this way, however, this doesnt scale well for higher> dimensioned matrices.I was wondering if people were aware of literature that does something like> this...Any references would be greatly appreciated.L TunesIf everything is real you might want to write A as tilde B B , namely asthe product of a matrix and its transpose. Whatever the resulting equations,A will be guranteed symmetric positive-de?ite. Perhaps you could makeB upper-triangular for simplicity.-- Julian V. NobleProfessor Emeritus of ^^^^^^^^^^^^^^^^^^http://galileo.phys.virginia.edu/~jvn/ God is not willing to do everything and thereby take away our free will and that share of glory that rightfully belongs to us. -- N. Machiavelli, The Prince. =>> >> ... To enforce positive de?iteness, I>> need to do the following,>> >> theta(1) > 0, theta(2) > 0 and -sqrt(theta(1)*theta(2)) < theta(3) <>> sqrt(theta(1)*theta(2))...>> >> Then by using some sort of a projection algorithm, I could enforce positive>> de?iteness in this way, however, this doesnt scale well for higher>> dimensioned matrices.>> >> I was wondering if people were aware of literature that does something like>> this...Any references would be greatly appreciated.If everything is real you might want to write A as tilde B B , namely as>the product of a matrix and its transpose. Whatever the resulting equations,>A will be guranteed symmetric positive-de?ite. Perhaps you could make>B upper-triangular for simplicity.When I looked at this problem, many years ago, I failed to ?d anythingthat wasnt equivalent. Given a symmetric real matrix, the only way tocheck for its positive de?iteness is to do a decomposition. And, asyou point out, if you START with a decomposition, the result is immediate!I was actually interested in positive semi-de?iteness, but mutismutandis.Nick Maclaren. =I have problem that I cant seem to work out. Say there are 136,000cows in my country. On average 10 of them die per year. That makes theodds that one particular cow (lets call her Daisy) will die this yearof 1 in 13,600.What I want to know is what are the odds of Daisy dying tomorrow? Iewhat are the odds of Daisy dying on any particular day throughout theyear?Any help is gratefully received.Joe => I have problem that I cant seem to work out. Say there are 136,000> cows in my country. On average 10 of them die per year. That makes the> odds that one particular cow (lets call her Daisy) will die this year> of 1 in 13,600.What you are citing is the probability.The odds are 13,599 to 1 against.> What I want to know is what are the odds of Daisy dying tomorrow?> Ie what are the odds of Daisy dying on any particular day throughout> the year?We may assume that the average 10 cows would die on different days.(It doesnt make that much difference anyway.)So the probability of a cow dying on a particular day is 10/365.Since the probability of Daisy dying this year is 1/13600,the probability of Daisy dying on a particular day of the year is:(1/13600)*(10/365). => We may assume that the average 10 cows would die on different days.> (It doesnt make that much difference anyway.)> So the probability of a cow dying on a particular day is 10/365.> Since the probability of Daisy dying this year is 1/13600,> the probability of Daisy dying on a particular day of the year is:> (1/13600)*(10/365).That is not the probability for any arbitrary date, since cows (like men)are only permitted to die once. The probability of Daisy dying on day n inthe future should be:(1-p)^n * pwhere p is the probability of death today. We know the probability p thatDaisy dies this year, which of course is related to p by:(1-p)^(366) = (1-p)(1-p) = (1-p)^(1/366)p = 1-(1-p)^(1/366)For small ps it is true that p is _approximately_ p/366, but as p getscloser to 1 this approximation gets worse.(366: Its a leap year!)Of course this all assumes that one cow is as near death as another, no dayis more deadly than another, no cows born, etc.HTH--KCS =Im not sure if anybody would know this information, but here it goes.In the development of the calculus, notation was an important factor. Thetypical notation of d/dt and the integral sign are from Liebntiz, andbecause Newton had such jacked up notation, England fell behind the rest ofcontinental Europe in analysis for over 100 yrs. My question is this: I knowthat Newton used the dot notation for differentiation, but what type ofnotation did he use for the integral/integration? If anybody knows the anseror can point me to a source, I would be most appreciative.TIA,Jack =>Squares of 0.999... tend away from 1There is only one square of 0.999...; perhaps you can tell me why you> used the plural?Square, cube, etc. > But your numerical evidence below supports that the squares of> 0.9, 0.99, 0.999, ... tend TOWARD 1, not AWAY. To see this, all you> need to do is arithmetic. Sequence 0.9^2, 0.9^3, 0.9^4, ... tends toward 0Sequence 0.99^2, 0.99^3, 0.99^4, ... tends toward 0Sequence 0.999^2, 0.999^3, 0.999^4, ... tends toward 0etc., etc., etc.,Sequence 0.999...^2, 0.999...^3, 0.999...^4, ... tends toward 0 Garry Denke, GeologistDenoco Inc. of Texas =>> Squares of 0.999... tend away from 1>There is only one square of 0.999...; perhaps you can tell me why you>used the plural?Square, cube, etc. > >But your numerical evidence below supports that the squares of>0.9, 0.99, 0.999, ... tend TOWARD 1, not AWAY. To see this, all you>need to do is arithmetic. Sequence 0.9^2, 0.9^3, 0.9^4, ... tends toward 0> Sequence 0.99^2, 0.99^3, 0.99^4, ... tends toward 0> Sequence 0.999^2, 0.999^3, 0.999^4, ... tends toward 0> etc., etc., etc.,Sequence 0.999...^2, 0.999...^3, 0.999...^4, ... tends toward 0 > Garry Denke, Geologist> Denoco Inc. of TexasThe sequence 0.999...^2, 0.999...^3, 0.999...^4, ... is term by term identical to the sequence 1,1,1,.... If you are claiming that this sequence tends towrds 0, it can only be because your level of mathematical understanding tends towards 0.If you mean that, for ?ed positive integers m, the sequenceg(n) = (1 - 1/10^m)^n goes towards zero, you have not managed to express that thought clearly enough to be understood.You should try a little less ambiguity if you are to discuss matahematics effectively. =>Sequence 0.999...^2, 0.999...^3, 0.999...^4, ... tends toward 0>0.999... = 1, so your sequence is 1^2, 1^3, 1^4, ...-- Richard-- FreeBSD rules! => Sequence 0.999...^2, 0.999...^3, 0.999...^4, ... tends toward 0> Its hard to guess what proof means in Garryworld. Perhaps the following sequence will help: .9, .99^2, .999^3, .9999^4, ... =*** post for FREE via your newsreader at post.newsfeed.com ***>> Squares of 0.999... tend away from 1> > There is only one square of 0.999...; perhaps you can tell me why you>used the plural? Square, cube, etc. > But your numerical evidence below supports that the squares of>0.9, 0.99, 0.999, ... tend TOWARD 1, not AWAY. To see this, all you>need to do is arithmetic. Sequence 0.9^2, 0.9^3, 0.9^4, ... tends toward 0> Sequence 0.99^2, 0.99^3, 0.99^4, ... tends toward 0> Sequence 0.999^2, 0.999^3, 0.999^4, ... tends toward 0> etc., etc., etc., Sequence 0.999...^2, 0.999...^3, 0.999...^4, ... tends toward 0Herc already showed you how 0.999... squared was equal to one. Yoursequence tends to 1 because each element in the series is not differentfrom 1. karl mhttp://www.newsfeed.com - The #1 Newsgroup Service in the World!-----== 100,000 Groups! - 19 Servers! - Unlimited Download! =----- =En el Denke escribi.97:> Squares of 0.999... tend away from 1>> There is only one square of 0.999...; perhaps you can tell me why you>> used the plural? Square, cube, etc.pows or powers, then. Im not native english speaking ...>> But your numerical evidence below supports that the squares of>> 0.9, 0.99, 0.999, ... tend TOWARD 1, not AWAY. To see this, all you>> need to do is arithmetic. Sequence 0.9^2, 0.9^3, 0.9^4, ... tends toward 0> Sequence 0.99^2, 0.99^3, 0.99^4, ... tends toward 0> Sequence 0.999^2, 0.999^3, 0.999^4, ... tends toward 0> etc., etc., etc., Sequence 0.999...^2, 0.999...^3, 0.999...^4, ... tends toward 0Yes, Lim((1 - 10^k)^n, n, inf) = 0. And what?Otherwise, Lim((1 - 10^k)^n, k, inf) = 1And if both, k and n, go to in?ity simultanously? It depend on how go toin?ity k and n ... In a simlar way that happen with k/n.-- Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com =Assuming your post is not a troll,In order to obtain a contradiction?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) => Squares of 0.999... tend away from 1.99999.... is a single number, so there is only one square of it. Talking > about Squares of 0.999... is a bit of a non-starter.You seem to be interested in the sequence .9^2, .99^2, .999^2, ... Clearly > these numbers increase, so how can they tend away from 1? Your own > computations show they get closer to 1.Assuming your post is not a troll, start with the fact that the sequence > .9, .99, .999, ... tends to 1. Next observe that if 0 < x < 1, then 1 - x^2 > = (1+x)(1-x) < 2(1-x). Therefore the numbers 1-.9^2, 1-.99^2, 1-.999^2, > ... are, respectively, less than 2(1-.9), 2(1-.99), 2(1-.999), ... and the > latter sequence tends to 0. Therefore the former sequence tends to 0, which > is the same as saying .9^2, .99^2, .999^2, ... tends to 1.Sequence .9^2, .9^3, .9^4, ...Sequence .99^2, .99^3, .99^4, ...Sequence .999^2, .999^3, .999^4, ...Sequence .9999^2, .9999^3, .9999^4, ...Sequence .99999^2, .99999^3, .99999^4, ...Sequence .999999^2, .999999^3, .999999^4, ...Sequence .9999999^2, .9999999^3, .9999999^4, ...Sequence .99999999^2, .99999999^3, .99999999^4, ...Sequence .999999999^2, .999999999^3, .999999999^4, ...Sequence .9999999999^2, .9999999999^3, .9999999999^4, ...Sequence .99999999999^2, .99999999999^3, .99999999999^4, ...Sequence .999999999999^2, .999999999999^3, .999999999999^4, ...Sequence .9999999999999^2, .9999999999999^3, .9999999999999^4, ...etc., etc., etc.,tend to 0.Does sequence .999...^2, .999...^3, .999...^4, ... tend to 0 or to 1?Garry Denke, GeologistDenoco Inc. of Texas =Does sequence .999...^2, .999...^3, .999...^4, ... tend to 0 or to 1?Since .999... = 1 in your question, your sequence is 1,1,1,..., and doesnt tend anywhere, it stays put at 1. =>> Squares of 0.999... tend away from 1>.99999.... is a single number, so there is only one square of it. Talking >about Squares of 0.999... is a bit of a non-starter.>You seem to be interested in the sequence .9^2, .99^2, .999^2, ... Clearly >these numbers increase, so how can they tend away from 1? Your own >computations show they get closer to 1.>Assuming your post is not a troll, start with the fact that the sequence >.9, .99, .999, ... tends to 1. Next observe that if 0 < x < 1, then 1 - x^2 >= (1+x)(1-x) < 2(1-x). Therefore the numbers 1-.9^2, 1-.99^2, 1-.999^2, >... are, respectively, less than 2(1-.9), 2(1-.99), 2(1-.999), ... and the >latter sequence tends to 0. Therefore the former sequence tends to 0, which >is the same as saying .9^2, .99^2, .999^2, ... tends to 1.Sequence .9^2, .9^3, .9^4, ...> Sequence .99^2, .99^3, .99^4, ...> Sequence .999^2, .999^3, .999^4, ...> Sequence .9999^2, .9999^3, .9999^4, ...> Sequence .99999^2, .99999^3, .99999^4, ...> Sequence .999999^2, .999999^3, .999999^4, ...> Sequence .9999999^2, .9999999^3, .9999999^4, ...> Sequence .99999999^2, .99999999^3, .99999999^4, ...> Sequence .999999999^2, .999999999^3, .999999999^4, ...> Sequence .9999999999^2, .9999999999^3, .9999999999^4, ...> Sequence .99999999999^2, .99999999999^3, .99999999999^4, ...> Sequence .999999999999^2, .999999999999^3, .999999999999^4, ...> Sequence .9999999999999^2, .9999999999999^3, .9999999999999^4, ...> etc., etc., etc.,tend to 0.Does sequence .999...^2, .999...^3, .999...^4, ... tend to 0 or to 1?> Sequence 1, 1, 1, ... tends to 1.See interchange of limits in a real analysis text to see whylimit 0 cannot be deduced from the previous lines. => Squares of 0.999... tend away from 1.99999.... is a single number, so there is only one square of it. Talking > about Squares of 0.999... is a bit of a non-starter.What would be a more accurate term for a starter?You seem to be interested in the sequence .9^2, .99^2, .999^2, ... Clearly > these numbers increase, so how can they tend away from 1? Your own > computations show they get closer to 1.I am interested in the sequence .9^2, .9^3, .9^4, ...and the sequence .99^2, .99^3, .99^4 ... and the sequence .999^2, .999^3, .999^4 ...etc., etc., etc. > Assuming your post is not a troll, start with the fact that the sequence > .9, .99, .999, ... tends to 1. Next observe that if 0 < x < 1, then 1 - x^2 > = (1+x)(1-x) < 2(1-x). Therefore the numbers 1-.9^2, 1-.99^2, 1-.999^2, > ... are, respectively, less than 2(1-.9), 2(1-.99), 2(1-.999), ... and the > latter sequence tends to 0. Therefore the former sequence tends to 0, which > is the same as saying .9^2, .99^2, .999^2, ... tends to 1.Do you agree the sequence .9^2, .9^3, .9^4, ... tends to 0?Do you agree the sequence .99^2, .99^3, .99^4 ... tends to 0?Do you agree the sequence .999^2, .999^3, .999^4 ... tends to 0?etc., etc., etc.If so, do you agree that 0.999... = 0.999... and is not equal to 1? Garry Denke, GeologistDenoco Inc. of Texas =>> Squares of 0.999... tend away from 1>.99999.... is a single number, so there is only one square of it. Talking >about Squares of 0.999... is a bit of a non-starter.What would be a more accurate term for a starter?Why dont you de?e what you mean by 0.999... for a starter. Iveasked you several times, but all you provide is a truncated list drawnfrom the BEGINNING of 0.999... each time. sha1:4qzrHmXcYqozsdFZ9AAixT/23Hk == Do you agree the sequence .9^2, .9^3, .9^4, ... tends to 0?> Do you agree the sequence .99^2, .99^3, .99^4 ... tends to 0?> Do you agree the sequence .999^2, .999^3, .999^4 ... tends to 0?> etc., etc., etc.This is a basic theorem taught to Calc II students.> If so, do you agree that 0.999... = 0.999... and is not equal to 1?Of course not. The sum from 0 through in?ity of a^n is given by1/(1-a), where 0> >>Squares of 0.999... tend away from 1>> >> .99999.... is a single number, so there is only one square of it. Talking >> about Squares of 0.999... is a bit of a non-starter.>What would be a more accurate term for a starter?>> >> You seem to be interested in the sequence .9^2, .99^2, .999^2, ... Clearly >> these numbers increase, so how can they tend away from 1? Your own >> computations show they get closer to 1.>I am interested in the sequence .9^2, .9^3, .9^4, ...>and the sequence .99^2, .99^3, .99^4 ... >and the sequence .999^2, .999^3, .999^4 ...>etc., etc., etc.This would only be a problem if limits commute. They need not commute.There is no reason to expect that lim_{x -> a} lim_{y -> b} f(x,y) = lim_{y -> b} lim_{x -> a} f(x,y).You seem to be demanding that limits commute in spite of the fact that such an expectation is wrong.In other words, your objections are based on a false premise.David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills. =>> Squares of 0.999... tend away from 1>.99999.... is a single number, so there is only one square of it. Talking >about Squares of 0.999... is a bit of a non-starter.What would be a more accurate term for a starter?Saying what you mean would help.Do you mean that the sequence .9^2, .99^2, .999^2, ..., does not converge to 1? If you do, you are wrong, as any fool (except you) can plainly see.>You seem to be interested in the sequence .9^2, .99^2, .999^2, ... Clearly >these numbers increase, so how can they tend away from 1? Your own >computations show they get closer to 1.I am interested in the sequence .9^2, .9^3, .9^4, ...> and the sequence .99^2, .99^3, .99^4 ... > and the sequence .999^2, .999^3, .999^4 ...> etc., etc., etc.OK, you have, essentially, f(m,n) = (1 - 1/10^m)^n, where m is the number of nines in the decimal number, and n is the power it is being raised to.Now for any ?ed positive integer, n, lim_{m -> oo} f(m,n) = 1and for any ?ed positive integer, m, lim_{n -> oo} f(m,n) = 0.You seem to think that the second of these limits disproves the ?st, but you are wrong. Again.>Assuming your post is not a troll, start with the fact that the sequence >.9, .99, .999, ... tends to 1. Next observe that if 0 < x < 1, then 1 - x^2 >= (1+x)(1-x) < 2(1-x). Therefore the numbers 1-.9^2, 1-.99^2, 1-.999^2, >... are, respectively, less than 2(1-.9), 2(1-.99), 2(1-.999), ... and the >latter sequence tends to 0. Therefore the former sequence tends to 0, which >is the same as saying .9^2, .99^2, .999^2, ... tends to 1.Do you agree the sequence .9^2, .9^3, .9^4, ... tends to 0?> Do you agree the sequence .99^2, .99^3, .99^4 ... tends to 0?> Do you agree the sequence .999^2, .999^3, .999^4 ... tends to 0?> etc., etc., etc.See above. That lim_{n -> oo} (1 - 1/10^m)^n = 0 has no effect on the fact than lim_{m -> oo} (1 - 1/10^m)^n = 1.You are con? the effect of changing one variable with the effect of changing the other. It dont work that way.The quality of your mathematics leads me to ask, hesitantly, what sort of dowsing rods do you use to ?d your oilIf so, do you agree that 0.999... = 0.999... and is not equal to 1?I never agree to nonsense, except possibly on April ?st. Garry Denke, GeologistAnd non-mathematician> Denoco Inc. of Texas => I think you mean that each Square from your list is adding digits away from> its trailing one, which is illustrated by your example.Wrong. I dont. The squares of 0.999... tend toward 0, not toward 1 .9^2 = .81 .9^3 = .729 .9^4 = .6561 .99^2 = .9801 .99^3 = .970299 .99^4 = .96059601 .999^2 = .998001 .999^3 = .997002999 .999^4 = .996005996001 .9999^2 = .99980001 .9999^3 = .999700029999 .9999^4 = .9996000599960001 .99999^2 = .9999800001 .99999^3 = .999970000299999 .99999^4 = .99996000059999600001 .999999^2 = .999998000001 .999999^3 = .999997000002999999 .999999^4 = .999996000005999996000001 .9999999^2 = .99999980000001 .9999999^3 = .999999700000029999999 .9999999^4 = .9999996000000599999960000001 .99999999^2 = .9999999800000001 .99999999^3 = .999999970000000299999999 .99999999^4 = .99999998000000059999999600000001 .999999999^2 = .999999998000000001 .999999999^3 = .999999997000000002999999999 .999999999^4 = .999999996000000005999999996000000001 .9999999999^2 = .99999999980000000001 .9999999999^3 = .999999999700000000029999999999 .9999999999^4 = .9999999996000000000599999999960000000001 .99999999999^2 = .9999999999800000000001 .99999999999^3 = .999999999970000000000299999999999 .99999999999^4 = .99999999996000000000059999999999600000000001 .999999999999^2 = .999999999998000000000001 .999999999999^3 = .999999999997000000000002999999999999.999999999999^4 = .999999999996000000000005999999999996000000000001[snip rest for brevity]Prove the squares of 0.999... tend toward 1or please stop claiming that 0.999... = 1 Garry Denke, GeologistDenoco Inc. of Texas =>I think you mean that each Square from your list is adding digits away from>its trailing one, which is illustrated by your example.Wrong. I dont. The squares of 0.999... tend toward 0, not toward 1(...snip a list that does NOT represent, and can NEVER represent, thetotality of 0.999... ) > Prove the squares of 0.999... tend toward 1> or please stop claiming that 0.999... = 1 The squares of 0.999... are ALL EQUAL TO ONE. Herc showed you thisalready!karl m toward 1> or please stop claiming that 0.999... = 1 > Garry Denke, Geologist> Denoco Inc. of TexasLet a_n = 1 - 10^(-n). Then a_0 = 0, a_1 = .9, a_2 = .99, etc. If wetake the sequence {(a_0)^2, (a_1)^2, (a_2)^2, (a_3)^2...} we canexamine what happens to the squares of .9999...999 as n grows toin?ity.We have that lim (n -> in?ity) (a_n)^2 = lim (n -> in?ity) 1 - 2 *10^(-n) + 10^(-2n) = 1 - lim [2 * 10^(-n) + 10^(-2n)] = 1 - 0 = 1.Notably, you dont seem to be interested in only the squares of thenumbers, since you include also the fourth power. Maybe you shouldstate more clearly precisely what you sha1:Vwih4Znk8NG9QAWsOJ8jmnem4MA == Wrong. I dont. The squares of 0.999... tend toward 0, not toward 1Sigh. Powers of any number strictly between zero and one converge tozero. This fact is well known. You have hardly proven that the *sum*of these terms converges to zero. Worse, the sum of powers is notequal to the power of the sum, so your computation is also irrelevant.In particular, any series whose sum is 1, and whose terms arepositive, will exhibit the property that powers of the terms convergeto zero. The powers of the sum, however, are all uniformly 1.--Len =>> I think you mean that each Square from your list is adding digits away from>> its trailing one, which is illustrated by your example.>Wrong. I dont. The squares of 0.999... tend toward 0, not toward 1By your own calculations, .9^2, .99^2, .999^2, is an *increasing* sequence, getting closer to 1. Simple observation shows that,> .9^2 = .81> .99^2 = .9801> .999^2 = .998001> .9999^2 = .99980001> .99999^2 = .9999800001> .999999^2 = .999998000001> .9999999^2 = .99999980000001> .99999999^2 = .9999999800000001> .999999999^2 = .999999998000000001> .9999999999^2 = .99999999980000000001> .99999999999^2 = .9999999999800000000001> .999999999999^2 = .999999999998000000000001David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills. =>I think you mean that each Square from your list is adding digits away from>its trailing one, which is illustrated by your example.Wrong. I dont. The squares of 0.999... tend toward 0, not toward 1Since 0.999... is just one number, it only has one square, which does not tend to be any different than its one value.What you might possibly mean is that the sequence0.9^2, 0.99^2, 0.999^2, ..., tends towards zero, not one. But you would then be wrong.If what you might mean above were true, then f(x) = x^2 would have to be discontinuous at x = 1, which means that polynomials of degree greater than one need not be continuous or differentiable or integrable, and all of calculus would collapse. The mathematical and engineering worlds would then burn you in ef?y. Or perhaps not in ef?y. => >I am soliciting some advice and opinions about how I should interact with my> >granddaughter.> > >First some history. I am a 72 year old Mathematician. I got my degree in> >Recursive Function theory in 1955. I was a bit of a prodigy in that by the time> >I graduated high school I know nearly all the Mathematics I would need for a> >Ph.D (I had done little research; I did not yet have my thesis).> > >My daughter is now nearly 40 years old. When she was growing up, I taught her> >elementary Mathematics, in part, as a form a high quality father-daughter> >interaction. The effect of this was that when she was in the 9th grade she took> >AP Calculus and aced the test.> >I have often wondered whether it would have made much of a difference in her> >life if she learned Calculus, three years later, the usual time for bright kids.> >She never learned much more Mathematics and eventually went to Caltech, where> >she got a degree in Molecular Biology.> > >Her daughter is now seven years old. She is also intellectually precocious, I> >am wondering how I should interact with my granddaughter .> > >Her arithmetic skills are now good enough that I could begin to teach her> >beginning high school Algebra, but I dont know to what end. No matter what I> >do or dont do, she will eventually know high school Algebra.> > >Perhaps the best legacy I could leave her, in the few years I may have left, is> >an an understanding and love for the beauty of Mathematics. But I dont know> >how to impart that.> > >Three other things Ive thought of teaching her are:> > Beginning Number Theory (divisibility, primes, modulo arithmetic ...)> > Combinatorics (how many subsets does a ?ite set have, how many> permutations, ...).> > Very elementary group theory, with an eye toward being able to produce a> rigorous proof.> > >I would very much appreciate some opinions. Also any suggestions about text> >books, outlines of material, etc. would be most welcome.I would suggest teaching really basic mathematics. This> does not include such things as calculus. Also, it would> be better that she learn things well, NOT as the schools> teach them, in classes which are far too slow for anyone> with ability. Is there ANYTHING which the present schools> teach even reasonably well?Concentrate on concepts, not manipulation. It looks like> most of what you have listed is manipulation; this is> easily learned after the concepts, but the other way is> not; it is HARD. Learning computational calculus does> not help with the concepts, nor does learning arithmetic,> nor algebra as it is usually taught, and the good Euclid> course almost does not exist.I would suggest you start with logic, including the> restricted predicate calculus. This is what my son did> at ages 5-6, using the two books of Suppes, while also> studying algebra. However, I believe that you will ?d> my late wifes logic book easier, although many of the> applications may not be appropriate. Frankly, I believe> that this book, somewhat rewritten, should be a part of the> regular elementary school curriculum for all those with> conceptual ability. However, start with the linguistic> notion of variables, which is more general. If you have> dif?ulty in getting a copy (it is out of print), let> me know; I am in the process of ?ding out how to make> it generally available. My daughter did it later, and> by not using mnemonics, she never had any problem with> using any notation for variables in any ?ld.> This is a question, not a suggestion. Have you heard of a logic gamecalled Wff ?n Proof? A game could perhaps be more instructive thaninstruction.David Ames> I would also suggest at least the ?st part of Landaus> _Foundations of Analysis_, which consists of deriving the> properties of the positive integers and positive rationals> from the Peano Postulates. I believe that this, rewritten,> should be no later than the primary grades. But anyone who> does not understand induction does not understand integers,> and this includes most college mathematics majors, and > while Landau may be short on verbiage, the approach is > very detailed, especially for the positive integers. There> is a logical lack, which you might catch, but if you assume> that addition and multiplication exist, it goes away.In both of these, if she has dif?ulty with producing > proofs, do not make too big an issue early. Learning more> will make proofs easier. Foundations belong in kindergarten, not in graduate school. =>And a related question: If I wanted to understand Baroque>music, I should go to school and study astrophysics, right?Rather, thats what you should study if you want to understand the Music> of the Spheres.Just started a blog on this:Music of the Spheres weblog: http://esolutionswork.com/Designer/ =Aristotle will not help you, here, Im afraid;the father of science, my butt! I want to thank your *current* PM (alias,in the Harry Potter PS books, as the Muggles Liaisonto the Minister of Misinformation [of the real goment])for his policy that echoes Dame Maggies:Hey, George; lets you and Saddam ?ht! >Rather, thats what you should study if you want to understand the Music>of the Spheres.Just started a blog on this:> Music of the Spheres weblog: http://esolutionswork.com/Designer/--Give the Gift of Trickier Dick Cheeny -- out of of?e, at last!http://www.benfranklinbooks.com/http://www.wlym.com/pages /music.htmlhttp://www.rand.org/publications/randreview/issues/ rr.12.00/http://members.tripod.com/~american_almanachttp:// =So what youre saying is the sequence .9^2, .9^3, .9^4 ....9^2 = .81.9^3 = .729.9^4 = .6561etc the sequence .99^2, .99^3, .99^4 ....99^2 = .9801.99^3 = .970299.99^4 = .96059601etcthe sequence .999^2, .999^3, .999^4 ... .999^2 = .998001.999^3 = .997002999.999^4 = .996005996001etc tend to 1 making the sequence .999...^2, .999...^3, .999...^4 tend to 1.Is that what you are saying?Garry Denke, GeologistDenoco Inc. of Texas =>>Sorry, I didnt get your point. Do you mean you cant add a 9 at the end >>of 0.99... because then you get 1?>>You cant add a 9 at the end of 0.99.. because it has no end.>But you did add a 5 at the end of 0.99.. Whys that allowed? There aint >no difference.If you read carefully, he added a ?5 at the end of an arbitrary, but> FINITE, sequence of ?9. The sequence 0.999... has no end. karl mIMHO, I think Jon was a bit tongue in cheek in his post.Rickwork@ostrander.de =* Tim Wouters>> Sorry, I didnt get your point. Do you mean you cant add a 9 at the end >> of 0.99... because then you get 1?>You cant add a 9 at the end of 0.99.. because it has no end.But you did add a 5 at the end of 0.99.. Whys that allowed? There aint > no difference.No, I cant. But I was just fooling around, and I thought I made itclear with the following:> (To be sure: :-), but I remember seeing this argument before.)Sorry to confuse you.-- Jon Haugsand = .9^2 = .81 .99^2 = .9801 .999^2 = .998001 .9999^2 = .99980001 .99999^2 = .9999800001 .999999^2 = .999998000001 .9999999^2 = .99999980000001 .99999999^2 = .9999999800000001 .999999999^2 = .999999998000000001 .9999999999^2 = .99999999980000000001 .99999999999^2 = .9999999999800000000001 .999999999999^2 = .999999999998000000000001 .9999999999999^2 = .99999999999980000000000001 .99999999999999^2 = .9999999999999800000000000001 .999999999999999^2 = .999999999999998000000000000001 .9999999999999999^2 = .99999999999999980000000000000001 .99999999999999999^2 = .9999999999999999800000000000000001 .999999999999999999^2 = .999999999999999998000000000000000001 .9999999999999999999^2 = .99999999999999999980000000000000000001 .99999999999999999999^2 = .9999999999999999999800000000000000000001Each position after the decimal point eventually becomes a 9 and staysthat way.-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W => No, .999... is not equal to 1, therefore any operation> involving .999... may not substitute 1 therefor. At least, one can fashion a semi-consistent math system> therewith, although it would take a lot of work and probably> would not be worth it.(For starters, if .999... != 1, what is .999... - (.999... * 10 - 9)> equal to? Answer: 0.000...9. This sort of thing will get rather> weird very quickly.)Of course, if one cubes .999... one can do something like the following:.9 * .9 * .9 = .729.99 * .99 *.99 = .729 + 3*.0729 + 3*.00729 + .000729> = .729 + .2187 + 0.02187 + .000729> = .970299Note the 7 in the 10^-2 spot..999 * .999 * .999 = .970299 + 3*.0088209 + 3*.00008019 + .000000729> = .970299 + .0264627 + .00024057 + .000000729> = .997002999and yet another 7, this time in the 10^-3 spot..9999 * .9999 * .9999 = .997002999 + 3*.0008982009 + 3*.0000008091 +> .000000000729> = .997002999 + .0026946027 + 3*.0000024273 +> .000000000729> = .999700029999and theres that 7 again. Is that second term always going to be> .000...0002something ?Of course its probably simpler to abstract the problem and> equate .999...9 = 1 - 10^(-n), and then(1 - 10^(-n))^3 = 1 - 3*10^(-n) + 3*10^(-2*n) - 10^(-3*n)which proves that (.999...9)^3 = .999...99700...0299...9, in a> simpler fashion than your equations below, or for that matter> my algebra.So far, no problem here. But one cannot conclude that(.999...)^3 = .999...99700...0299...unless one ascribes a de?itive meaning to digit expansions> with in?ite ellipses and de?es their proper addition,> subtraction, etc. It is naive to conclude that the> arithmetic operation10 * (.999...)^3 - 9 - (.999...)^3 = 9.999...97000...2999...990 - 9 - .999...99700...0299...999> = -.000...02699...7300...009.makes any sense at all.I have no idea how to write it in a way that anyone here would accept,but perhaps there is a way to write the squares of 0.999... tendingtoward 0, not toward 1, settling the matter. Simple arithmetic says... 0.9^2 = .81 0.9^3 = .729 0.9^4 = .6561 0.99^2 = .9801 .099^3 = .970299 0.99^4 = .96059601 0.999^2 = .998001 0.999^3 = .997002999 0.999^4 = .996005996001 0.9999^2 = .99980001 0.9999^3 = .999700029999 0.9999^4 = .9996000599960001 0.99999^2 = .9999800001 0.99999^3 = .999970000299999 0.99999^4 = .99996000059999600001 0.999999^2 = .999998000001 0.999999^3 = .999997000002999999 0.999999^4 = .999996000005999996000001 0.9999999^2 = .99999980000001 0.9999999^3 = .999999700000029999999 0.9999999^4 = .9999996000000599999960000001 0.99999999^2 = .9999999800000001 0.99999999^3 = .999999970000000299999999 0.99999999^4 = .99999998000000059999999600000001 0.999999999^2 = .999999998000000001 0.999999999^3 = .999999997000000002999999999 0.999999999^4 = .999999996000000005999999996000000001 0.9999999999^2 = .99999999980000000001 0.9999999999^3 = .999999999700000000029999999999 0.9999999999^4 = .9999999996000000000599999999960000000001 0.99999999999^2 = .9999999999800000000001 00.99999999999^3 = .999999999970000000000299999999999 0.99999999999^4 = .99999999996000000000059999999999600000000001 0.999999999999^2 = .999999999998000000000001 0.999999999999^3 = .9999999999970000000000029999999999990.999999999999^4 = .999999999996000000000005999999999996000000000001etc., etc., etc....the squares are heading toward 0. Garry Denke, GeologistDenoco Inc. of Texas => ...the squares are heading toward 0. Garry Denke, Geologist> Denoco Inc. of TexasThe rockheads posting here seem to be headed for an IQ of 0.You are ?dling with f(m,n) = (1-1/10^m)^n.Varying the value of n is irrelevant in considering how the function behaves as one varies m.For any ?ed n, the limit is 1 as m increases without bound.Changing n is irrelevant.Since g_n(x) = x^n is continuous at all positive real x and for all positive integers n, thenLim_{m -> oo} g(1-1/10^m) = g( Lim_{m -> oo} 1-1/10^m)Thus Lim_{m -> oo} (1-1/10^m)^n =( Lim_{m -> oo} 1-1/10^m )^nand your raising to powers is of no consequence. => In this case there _is_ a _completely_ _standard_ de?ition of what> these symbols mean already. An author is not required to restate> every standard de?ition he uses. But the de?ition of what these> symbols mean _has_ been given in this thread, several times by> several people.A number is not a character string - this is not quibbling, this is> the whole point. One more time: .999... is not a character string,> it is a number. It is _true_ that a real number can have more than> one representation as a character string. So what? Were not talking> about what the notation _should_ mean, were talking about what it _does_> mean. .999... is not a symbol at all, it is a number. .999... is a symbol> for this number. It _is_ a symbol for the _sum_ of a certain > in?ite series. That means that it _is_ a symbol for number.With all due respect, you completely miss my point. I dont know howplainly I can state it. The problem or question isnt what thestandard meaning of .999 or .999 is or should be. The problemis that in the proof given that .999=1, the .999 is used in twodifferent ways: as a number and as an in?ite series with a limit of1. Those are the two cases that I listed. I am not asking which iscorrect or standard. I am simply saying that the proof to which Iresponded is inconsistent as to which it is, with adverseconsequences.If .999 and 1.000 are numbers in the proof, then two differentnumbers are equal and there is no longer a unique representation for anumber. (And I said that Im talking about the literalrepresentation, not expressions in general. Im not denying that1+1=2. My gawd.)If .999 and 1.000 are in?ite series in the proof, then thereferences to subtracting one from the other, and looking for a numberinbetween them, dont make any sense. Those are clearly references tothe 9s as being digits of a real number.Enough time has been spent on this high school math exercise. GoogleGroups has much more interesting topics than this, which is where Iwill now turn my attention (if anyone is looking for me.)Charlie Volkstorf> ************************David C. Ullrich => If .999... and 1.000... are numbers in the proof, then two different> numbers are equal and there is no longer a unique representation for a> number.No, they are not two different numbers; they are two different namesfor the same number.And yes, you are quite right that there is not a unique representationfor a number.Thomas => With all due respect, you completely miss my point. I dont know how> plainly I can state it. The problem or question isnt what the> standard meaning of .999? or .999? is or should be. The problem> is that in the proof given that .999?=1, the .999? is used in two> different ways: as a number and as an in?ite series with a limit of> 1. No, it is used in only one way: as a number, namely, whatever number aparticular in?ite series converges to.It *always* refers to the number, and not the particular series.Thomas => In this case there _is_ a _completely_ _standard_ de?ition of what>> these symbols mean already. An author is not required to restate>> every standard de?ition he uses. But the de?ition of what these>> symbols mean _has_ been given in this thread, several times by>> several people.>> >> A number is not a character string - this is not quibbling, this is>> the whole point. One more time: .999... is not a character string,>> it is a number. It is _true_ that a real number can have more than>> one representation as a character string. So what? Were not talking>> about what the notation _should_ mean, were talking about what it _does_>> mean. .999... is not a symbol at all, it is a number. .999... is a symbol>> for this number. It _is_ a symbol for the _sum_ of a certain >> in?ite series. That means that it _is_ a symbol for number.With all due respect, you completely miss my point. I dont know how>plainly I can state it. The problem or question isnt what the>standard meaning of .999 or .999 is or should be. The problem>is that in the proof given that .999=1, the .999 is used in two>different ways: as a number and as an in?ite series with a limit of>1. Thats _not_ two different ways. The sum of an in?ite series_is_ a number.>Those are the two cases that I listed. I am not asking which is>correct or standard. I am simply saying that the proof to which I>responded is inconsistent as to which it is, with adverse>consequences.If .999 and 1.000 are numbers in the proof, then two different>numbers are equal and there is no longer a unique representation for a>number. As I said yesterday: Thats correct, the decimal representation of areal number is _not_ unique. This is a well-known fact. A questionI asked yesterday which you declined to answer: So what?> (And I said that Im talking about the literal>representation, not expressions in general. Im not denying that>1+1=2. My gawd.)If .999 and 1.000 are in?ite series in the proof, then the>references to subtracting one from the other, and looking for a number>inbetween them, dont make any sense. You dont seem to have read my post past the ?st spot where youall the proofs that have been posted here are correct - many ofthem are nonsense. Do you have any complaint with the detailedproof that 0.999... = 1 that I gave in the post youre replying to?>Those are clearly references to>the 9s as being digits of a real number.Enough time has been spent on this high school math exercise. Giggle. Yes, its just high-school math. Weve been trying tohelp you get your high-school math straight.>Google>Groups has much more interesting topics than this, which is where I>will now turn my attention (if anyone is looking for me.)Charlie Volkstorf> ************************>> >> David C. Ullrich************************David C. Ullrich <732gvv0t3iolt0299fc94rr8ucf4lrml53@4ax.com> <38iivvk4pk1fdf4bk8f3skg5p42h5740tb@4ax.com> <4pgqvvgsepbgnrrg5dnr5e6k6jkduivklr@4ax.com> => With all due respect, you completely miss my point. I dont know how> plainly I can state it. The problem or question isnt what the> standard meaning of .999... or .999... is or should be. The problem> is that in the proof given that .999...=1, the .999... is used in two> different ways: as a number and as an in?ite series with a limit of> 1.Then your point is wrong, wrong, wrong..999.... is a number. Namely, it is the sum of the series which hasbeen repeatedly mentioned here. It is *not* the series itself, butthe sum of the series. That sum is a real number, so there is nocontradiction at all in claiming(1) .999... is a real number;(2) .999... is the sum of the in?ite series;(3) .999... is not an in?ite series.-- Yup, as far as Im concerned, if you live out your lives smiling theentire time full of pride in your *believed* accomplishments, when younever had any, well thats ok with me. --James Harris, a man of remarkable accomplishments. => In this case there _is_ a _completely_ _standard_ de?ition of what>> these symbols mean already. An author is not required to restate>> every standard de?ition he uses. But the de?ition of what these>> symbols mean _has_ been given in this thread, several times by>> several people.>> >> A number is not a character string - this is not quibbling, this is>> the whole point. One more time: .999... is not a character string,>> it is a number. It is _true_ that a real number can have more than>> one representation as a character string. So what? Were not talking>> about what the notation _should_ mean, were talking about what it _does_>> mean. .999... is not a symbol at all, it is a number. .999... is a symbol>> for this number. It _is_ a symbol for the _sum_ of a certain >> in?ite series. That means that it _is_ a symbol for number.With all due respect, you completely miss my point. I dont know how>plainly I can state it. The problem or question isnt what the>standard meaning of .999 or .999 is or should be. The problem>is that in the proof given that .999=1, the .999 is used in two>different ways: as a number and as an in?ite series with a limit of>1. Those are the two cases that I listed. I am not asking which is>correct or standard. I am simply saying that the proof to which I>responded is inconsistent as to which it is, with adverse>consequences.If .999... and 1.000... are numbers in the proof, then two different>numbers are equal and there is no longer a unique representation for a>number. (And I said that Im talking about the literal>representation, not expressions in general. Im not denying that>1+1=2. My gawd.)Well then, how do handle the the fact that all of 1.0, 1e0, and 1.00 are differently-spelled literals (in the computer-language sense) that denote the same number? If you admit that thats OK then on what basis do you exclude .99... from the set of allowable literals that denote it too?BTW, Davids comment about expressions is exactly the point, despite your my gawd dismissal. In maths, unlike computer programming, there is little intrinsic difference between a literal and any other sort of expression -- they both denote (somewhat like evaluate you allow for such things in your ontology).The key difference is between *representation* (a concrete string of characters or bits) and *denotation* (a purely abstract Platonic value).>If .999 and 1.000 are in?ite series in the proof, then the>references to subtracting one from the other, and looking for a number>inbetween them, dont make any sense. Those are clearly references to>the 9s as being digits of a real number.The .999... is a literal which is the syntactic abbreviation for an in?ite series, which is one of many series, it is meaninful to talk both about the individual terms (concrete objects in the representation) and the abstract value of the whole series (or of the individual terms). You might ?d it helpful to think of it as a kind of type conversion in programming terms.>Enough time has been spent on this high school math exercise.I do recall that in school we coverred numbers in different bases, which provides a whole additional set of different concrete numerals for the same abstract number: 10, A_16, 1010_2, 31_3, etc. etc.> Google>Groups has much more interesting topics than this, which is where I>will now turn my attention (if anyone is looking for me.)Dont let the door hit you on the way out...>Charlie Volkstorf>> >> David C. Ullrich-- ---------------------------| B B aa rrr b || BBB a a r bbb | | B B a a r b b | | BBB aa a r bbb | ----------------------------- =Okay I have never been very good with math and hopfully someone outthere can help. Here is the situation:I Have a total of ?e different items: I am shipping out bowes with acombination of these different items and need to prepare all thedifferent combinations before hand. So lets name the items:1. tshirt2. glove3. hat4. ring5. necklacenow a package can have either all ?e items or a choice of lets saynumber 1,2,4 or 1,2 or 1,4,5 or just 2 and so on... I think you problyget the jist of what im asking.What are the total number of combinations for the items obviouslywithout just rearanging thier order. and what is the formula to do so?Happy New Year =>Okay I have never been very good with math and hopfully someone out>there can help. Here is the situation:I Have a total of ?e different items: I am shipping out bowes with a>combination of these different items and need to prepare all the>different combinations before hand. So lets name the items:>1. tshirt>2. glove>3. hat>4. ring>5. necklacenow a package can have either all ?e items or a choice of lets say>number 1,2,4 or 1,2 or 1,4,5 or just 2 and so on... I think you probly>get the jist of what im asking.What are the total number of combinations for the items obviously>without just rearanging thier order. and what is the formula to do so?You are asking how many different packages there are. Each packagedetermines a subset of {1,2,3,4,5} (given by which items are in thebox), and you are allowing any subset except for the emptysubset. There are 2^5 = 32 possible subsets, one of which is empty. Sothere are 31 different possible combinations/boxes.-- what I accept as reality. --- Calvin (Calvin and Hobbes) hope I have chosen the correct newsgroup for this posting. If not pleaselet me know.I am trying to ?d the inde?ite integral ofSin(x)/xI have been unable to ?d it in any tables. My best guess wouldbe to divide it up into1/x * Sin(x)and integrate it by parts, however this doesnt always producea simpler expression.Any suggestions, or answers ?Thomas Christensen =I hope I have chosen the correct newsgroup for this posting. If not please>let me know.I am trying to ?d the inde?ite integral of>Sin(x)/xI have been unable to ?d it in any tables. My best guess would>be to divide it up into>1/x * Sin(x)and integrate it by parts, however this doesnt always produce>a simpler expression.Any suggestions, or answers ?Thomas Christensen>Its called the sine integral and there is no primitive in terms ofusual functions see http://mathworld.wolfram.com/SineIntegral.html = >I hope I have chosen the correct newsgroup for this posting. If notplease> >let me know.> >I am trying to ?d the inde?ite integral of> >Sin(x)/x> >I have been unable to ?d it in any tables. My best guess would> >be to divide it up into> >1/x * Sin(x)> >and integrate it by parts, however this doesnt always produce> >a simpler expression.> >Any suggestions, or answers ?> >Thomas Christensen> Its called the sine integral and there is no primitive in terms of> usual functions see http://mathworld.wolfram.com/SineIntegral.htmlI would also look at the MacLaurin Series.David Moran => This may sound simple, but its not my ?ld of expertise. I wonder if> anyone can help me create a unique number from two others.Lets say I have two columns of numbers:col_A col_B> 1 4> 2 2> 3 3> 4 1... but I know I wont get true duplicates such as:> col_A col_B> 1 4> 1 4How can I create a unique number from the two that will not be> produced by another pair?The best Ive come up with so far is adding a prime to the second> column, but Im not sure if this produces unique results every time:a * (b+3)Am I just working to get to the point where its highly unlikely Ill> suggestionsPossibly I do not fully understand the problem, but whats wrong with N = 10(ABS(A)) + ABS(B) =>>How can I create a unique number from the two that will not be>>produced by another pair?If you know that both numbers are positive integers, then> the following formula will work.f(A,B) = (A+B-1)*(A+B-2)/2 + BIn fact, this is a bijection between ordered pairs of positive> integers and positive integers.heres another way f(A,B) = 2^A (2B + 1)(for A and B >= 0). every number has a max power of 2 possibly 0, dividethat out and you get some odd number. This is not as nice a bijection, in that it bounces all over the place, but it is easier to see that it must be a bijection.Any others?Mitch Harris => index.htmlWatch for a superior proof from Harris any day...Mark Folsom =|>I got confused by the laymans explanation on the claymath.org link (the|>claymath.org after I posted. They talk about setting up an imaginary|>rubber band on the surface and then asking if it can be deformed|>smoothly onto a point. They did give me the impression this was a|>question on a 3-sphere. So maybe they should re-write their|>introduction to make that clearer. |> |> Milnor), and it is a little sloppy:[...]|> So to make things clearer, I would replace essentially characterized|> by essentially determined and actually *state* what the|> corresponding question is. Something like ...asked the analogous|> question for the three dimensional sphere: is the three dimensional|> sphere determined by the property of simple connectivity?That would have only cost them one sentence...|> of sending in corrections and suggestions. I guess I could post on the|> Clay Math forum.Did I understand correctly, _they_ are the ones running the prize?|>The PDF did make it absolutely|>clear: is every simply connected 3-domain homologous to a 3-sphere.|> |> Instead of homologous you want homeomorphic. There are in fact|> many homology 3-spheres, 3-manifolds that have the same homology as a|> 3-sphere. One usually doesnt say homologous even in this context|> however. One usually says have the same homology or something to|> that effect.Yes, should have realised that... I do know what a homologouscollapse means (from astrophysics).-- cu,Brucedrift wave turbulence: http://www.rzg.mpg.de/~bds/ =In message <3ffd8c6d@news3.accesscomm.ca>, John Sefton >>machine,>> but he has not answered that.>> I conclude that my conjecture was right.>> Franz>I dont see you advancing any ideas, Sir.>Perhaps some conjectures on Physics from your>quarter rather than dissing those who>continue to try to ?ure things out?>I remember reading one interesting idea>from you several (8?) years ago but it>was miniscule in scope. Come on! You still>got it in you to wonder and theres more>data all the time. The more we know the>less we know we know. Step up to the>plate!! Is anything jelling in that worthy>cranium or is it all ossifying? :-)>John>The real problem is that you and your dis-functional friends need to post this crap to alt.physics.egomaniacs.mentalmeandering. None of your years of usenet babble has produced anything of worth.For Christs (or Einsteins) sake, get out and do something constructive, even if its only to learn not to cross-post inappropriately.Note; This is a rant. I do not expect considered responses, nor will antithesis of hills =If in a Erathostenes Sieve of 210 columns we eliminate the columnsmultiples of 2,3,5 and 7, it remains 48 columns that can containprimes.QUESTION. - If the primes are replaced by points, what are thepatterns that are prohibited, that is : impossible to be produced?(The matrix have 48 columns but can be prolonged to any quantity ofrows)L.Rodriguez =To preface, I am aware that halting predicate of machine z on any> input is undecidable, and also halting of any machine on given input> x.However, I want to write an algorithm that will tell me for _given_> machine numbered m, and _given_ input k, whether H(m,k) is true.So I am confused, what is the answer to the yes/no question: is> H(17,19) true ?> Alex.The question as to whether a given program (including its input) willhalt is a very interesting one. We are ?ing (making constant) boththe program and its input. Thus, whether it halts or not is aconstant: yes or no. Consider the 4 possibilities:1. It does halt and we can prove it.2. It does halt and we cant prove it.3. It doesnt halt and we can prove that it doesnt.4. It doesnt halt and we cant prove that it doesnt.By prove it, I mean using a sound system of deduction: if we provesomething then it must be true. We can only prove true statements(but cant necessarily prove all true statements.)These are the only possibilities, prima facie. However, can # 2actually occur? No! If it does halt, then we can easily prove itsimply by simulating its execution. So # 2 is out.How about # 4? Well, if # 4 never occurs, then we are left with only# 1 and # 3. But if only # 1 and # 3 can occur, then we can alwaysprove whether it halts or not, which violates Turings 1937 theoremthat the Halting Problem is unsolvable. Thus # 4 must occur.It is easy to come up with instances where # 1 and # 3 occur. Thuseither # 1, # 3 or # 4 may actually occur.Then the answer to your question is: It might halt but sometimes, whenit doesnt halt, we cant tell.As far as the fact that there is an algorithm to output yes andthere is an algorithm to output no goes: This certainly doesntrefute Turings 1937 result. I would say that there are at least 3ways to look at this: (1) Its not really a solution to the HaltingProblem because it has no input. (2) Its not a solution becausethere are two answers, and we must have a single answer that works inall cases. (3) This is the most interesting to me: We can outputyes and we can output no, but we cant output no with a proofthat the answer is no. Why? Because of the proof above that wecant always prove that it doesnt halt when in fact it doesnt. Thuswe reject an output of no that is not preceded by a proof that theanswer is in fact no.Other interesting questions: What can we say about case # 4? Whatnecessary or suf?ient conditions can we give? Can we ever provethat a particular H(x,y) is case # 4? When can we prove that itisnt? Generalize these questions by considering each subset of 1-4instead of just # 4. Any takers?Charlie VolkstorfCambridge, MAPS For a completely formal proof of the unsolvability of the HaltingProblem, and other theorems from the Theory of Computation, seehttp://www.arxiv.org/html/cs.lo/0003071 => 1. It does halt and we can prove it.> 2. It does halt and we cant prove it.> 3. It doesnt halt and we can prove that it doesnt.> 4. It doesnt halt and we cant prove that it doesnt. Other interesting questions: What can we say about case # 4? What> necessary or suf?ient conditions can we give? Can we ever prove> that a particular H(x,y) is case # 4? When can we prove that it> isnt? Generalize these questions by considering each subset of 1-4> instead of just # 4. Any takers?>I can expound on condition #3.A TM will not halt if it has no halt states.A halt state is a state that has no instruction for some input.This can be determined by examining the state transitiontable which is ?ite.A TM will halt if all of the halt states are unreachable.A state is unreachable if it is not the initial state andno other state changes to the unreachable state.Again, this can be determined from the state table.Russell- 2 many 2 count => 1. It does halt and we can prove it.>2. It does halt and we cant prove it.>3. It doesnt halt and we can prove that it doesnt.>4. It doesnt halt and we cant prove that it doesnt.> > Other interesting questions: What can we say about case # 4? What>necessary or suf?ient conditions can we give? Can we ever prove>that a particular H(x,y) is case # 4? When can we prove that it>isnt? Generalize these questions by considering each subset of 1-4>instead of just # 4. Any takers?> >I can expound on condition #3.> A TM will not halt if it has no halt states.> A halt state is a state that has no instruction for some input.> This can be determined by examining the state transition> table which is ?ite.A TM will halt if all of the halt states are unreachable.> A state is unreachable if it is not the initial state and> no other state changes to the unreachable state.> Again, this can be determined from the state table.There are lots of other ways in which a halt state is unreachable.And there are TM which will halt for some initial tapes but not others.The halting problem is not so trivial as you would have it.> Russell> - 2 many 2 count> > =I came across 2 questions I had that maybe someone can answer :1) If A is an n x n matrix, and ||A|| is de?ed as the smallest number csuch that ||Ax|| <= c||x|| for all x in C^n, the n-fold complex numbers,then why is it true that a sequence of matrices A_m converges to a matrix Aif and only if ||A_m - A|| --> 0?2) Consider an in?ite series whose terms are matrices : A_0 + A_1 + A_ 2+ ... If the sum (from m = 0 to in?ity) of ||A_m|| < in?ity, thenthe in?ite series of matrices is said to converge absolutely. Why is ittrue that if a series converges absolutely, then the partial sums of theseries form a cauchy sequence?Moshe =I came across 2 questions I had that maybe someone can answer :1) If A is an n x n matrix, and ||A|| is de?ed as the smallest number c> such that ||Ax|| <= c||x|| for all x in C^n, the n-fold complex numbers,> then why is it true that a sequence of matrices A_m converges to a matrix A> if and only if ||A_m - A|| --> 0?2) Consider an in?ite series whose terms are matrices : A_0 + A_1 + A_ 2> + ... If the sum (from m = 0 to in?ity) of ||A_m|| < in?ity, then> the in?ite series of matrices is said to converge absolutely. Why is it> true that if a series converges absolutely, then the partial sums of the> series form a cauchy sequence?> What have you done so far? In both cases, they are fairly simple ifyou know the proper background. => 1) If A is an n x n matrix, and ||A|| is de?ed as the smallest number c> such that ||Ax|| <= c||x|| for all x in C^n, the n-fold complex numbers,> then why is it true that a sequence of matrices A_m converges to a matrix A> if and only if ||A_m - A|| --> 0?Well, ||A_m - A|| -> 0 is often the de?ition of the convergence of a sequence of matrices. Whats your de?ition?> 2) Consider an in?ite series whose terms are matrices : A_0 + A_1 + A_ 2> + ... If the sum (from m = 0 to in?ity) of ||A_m|| < in?ity, then> the in?ite series of matrices is said to converge absolutely. Why is it> true that if a series converges absolutely, then the partial sums of the> series form a cauchy sequence?Start with ||A + B|| <= ||A|| + ||B||. Then go back to calculus and review the proof that absolute convergence implies convergence. =I have to build a context free grammar that generates the following language : { a^i b^j c^k | i,j,k >= 0 and k <= i+j }I thought to split the problem in 6 possible subproblems (S1, ... , S6).S1 generates only a^iS2 generates only b^jS3 generates only a^i b^jS4 generates only a^i c^kS5 generates only b^j c^kS6 generates only a^i b^j c^kS -> S1 | S2 | S3 | S4 | S5 | S6 | epsilonS1 -> a S1 | epsilonS2 -> b S2 | epsilonS3 -> S1 S2S4 -> A a B cS5 -> D b E cA -> A a | epsilonB -> a B c | epsilonD -> D b | epsilon E -> b E c | epsilonNow the problem is that I dont know how to write S6. DGA =Recall that a set A of nonnegative integers is said to be an additivebasis of order h if every nonnegative integer is the sum of h (notnecessarily distinct) elements of A. A is said to be a basis if itis a basis of order h for some ?ite h.For example, the set of all nonnegative integers is a basis of order 1and the set of nonnegative squares is a basis of order 4.More generally, suppose A is a basis of ?ite order *and* that A haspositive lower density (i.e., lim inf A(n)/n > 0, where A(n)= |Aintersect {1, .., n}|). Then it can be shown that for each k >= 1, theset of kth powers of elements of A is also a basis of ?ite order.(See, e.g. [Theorem 11.14 (p. 373), 1].)Now suppose merely that A is an additive basis. Are there any weakersuf?ient conditions in the literature for, say, A^2:={a^2: a in A},to be a basis of ?ite order? It does not seem easy (at least to me)to even construct an example of a basis of order 2 for which A^2 isnot a basis of ?ite order. Can anyone point me to papersconstructing such an example? Is it true that for every pair of h, k >1, there is an asymptotic basis of order h whose kth powers are not abasis?- Paul[1] M. B. Nathanson, Elementary methods in number theory, Springer,New York, 2000; MR 2001j:11001 =Whats fascinating is that you can also consider what happens when y=0> to separate out further, so letting y=0, I havea^2 - xa + 7x^2 = 0,soa = (1+/-sqrt(-27))x/2,and introducing c_1(x,y) and c_2(x,y) I then have(5(1-sqrt(-27))x/2 + 5c_1(x,y) + 7y)(5(1+sqrt(-27))x/2 + 5c_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). >Do you still want a factorization in algebraic integers?Do you still want the ?st factor to be divisible by 7?Do you still want the second factor to be coprime to 7?Do you want this factorization to hold for any rational integers x, y?Do you want the cs to be algebraic integers for any rational integersx and y?How do you know that the cs even exist so that the factorizationmeets your answers to the above questions?> Whats fascinating here is that completely broken is *any* idea that> theres a variable factor in common with 7 that divides differently> dependent on the value of x, No, that idea is just as true as it was before.Rick =I read in The Book of Prime Number Records (1rt ed.) that Kn.9adelconsidered (in 1953) the following set: C_k = {n composite > k | if 1= 1 ?ed. (C_1 is Charmichael numbers)He proved that C_k is in?ite.I have several questions about this:1) Why its necessary that n > k? If we de?e C_k = {n composite | if1=1 we can de?e K_{b,c} = {n composite | if 1I am familiar with Euler theorems and the Jacobi/Legendre symbols. >However this question still eludes me.Prove for any _positive_ integer m;x^2 =/= -m mod 4m-1First, m is a square modulo 4m-1: verify that (2m-1)^2 = m (mod 4m-1).Since gcd(m,4m-1) = 1, if -m were also a square, then -1 would be asquare.But 4m-1 is congruent to 3 modulo 4; hence it is divisible by at leastone prime p which is congruent to 3 modulo 4. Anything which is asquare modulo 4m-1 must also be a square modulo p. Is -1 a squaremodulo p?-- what I accept as reality. --- Calvin (Calvin and Hobbes) Im not sure what a should be...In mechanics of materials, there are two approaches, Thin shell andThick shell for membranes loaded by internal pressure P. In theformer,(r_1-r_2) is set approximately to thickness t. The percentageelongation (strain)of radius = delta R /R (same applies forcircumference) = P R (1-mu)/(2 t E) where E and mu are materialproperties called Youngs modulus and Poissons ratio respy., and R->(r_1+r_2)/2 -> even r_1 or r_2; changes in thickness of shell aresmall and are usually neglected from an engineering standpoint.The latter approach is governed by Lames equations, pays moreattention to how strain varies through the thickness of shell, notjust what happens at inner radius r_1.Total volume of the shell shrinks, unless mu is 0.5 (happens forrubber). => > >> > > >> Well what the did [you] may well be right on this one... mean,> >> >> if it didnt mean that your comments increased the likelihood that the> >> >> original [im]poster was right?> >> >>if i had written> >> >>you could be correct...> >> > I see no practical difference.> > in that case, i have a question for you - how often do you get>involved in pointless discussions?At least as often as I talk to you. and what do you gain from it? besides looking like a fool?> >> >> >>geez, i would hate to assume that you are having problems with> >> >> >>existential quanti?rs. is that it?> >> >> > >> >> Well, I think I know a thing or two about logic, but please enlighten> >> >> >> me about what error I made regarding existentials.> >> >> >> >>well, ?st tell me how much logic you know. can you handle venn> >> >>diagrammes?> >> > >> I have more than suf?ient background in logic, I am sure. Honest.> >> >>evidence?> > I have a piece of paper saying that I have a PhD in Logic,> >> Computability and Methodology.> > well, well. i guess you might have well proven that it is possible to>have a PhD in logic while at the same time being very careless when>examining a colloquial statement.> > how good are you in your ?ld?>would i be able to ?d publications under your name?Who cares?apparently, you do.> The question was whether I would understand your tutelage> on the existential quanti?r. Now that I hope I am up to your high> standards, please enlighten me.you have already been. you made redundant assumptions and i dispelled them.> >> >> >> Did you mean only this?> >> >> >> >>what else could i have meant?> >> >> >> >>do you have further objections?> >> >> >> >> >> (1) Some adjuncts are incompetent.> >> >> >> (2) Arturo is an adjunct.> >> >> >> -------> >> >> >> Therefore, it is possible that Arturo is incompetent.> >> > >> I just want to be clear: The argument above is the argument you know> >> >> claim to have advocated?> >> >>what do you think i said?> >> >>no baseless assumptions.> > If I was certain, I wouldnt have asked.> > judgeing from your initial reaction, seems that you jumped the gun a>bit too soon.Maybe. Why dont you explain what you meant then, jackass.i have already explained. tell me what still confuses you?> All these lame attempts at sophistry and Socratic/Freudian techniques> (What do you think I meant? How would that make you feel?) are> painful and dull.but they shoulnt. painful and dull is your apparent stubbornness.> >> Why wont you answer the question?> > i answered your question long ago. go up the thread.No, you didnt. You have never stated what your comments were> supposed to mean. i think i did. i recall you refusing to consider them.> >> nor advocated the above argument.> > here is what i meant:> > arturo magidin is ADJUNCT assistant professor. ADJUNCTS are the>pariahs>in the academic caste system. so whomever jstevh@yahoo.com is, it may>well be right on this one...> > let me know what you think i said. OK?That doesnt tell one a goddamn thing what you meant, since it gives> no further clue of the relevance between Arturos position and the> comment that the poster is likely right.bingo! it took you a bit long, but congrats.> Youre a complete moron. If you want to insult Arturo, then have the> balls to admit it.i do not want to insult arturo, in fact i never did.> I will drop out of this conversation.thats a problem entirely between you and your psychiatrist. =PaulI just read Ohanian & Ruf?is 1.9 and it completely supports my position and not yours IMHO.My position is:1. Einsteins original equivalence principle ~ 1915 is ?e and there is no internal contradiction in the concepts of General Relativity.Locally the gravity g-force is indistinguishable from an inertial force that in Newtons mechanics would be an artifact of a non-inertial accelerating frame like Coriolis and centrifugal forces for observers stuck on surface of rotating Earth for the former or in one of those Coney Island rides in the latter case.Whether or not the local curvature tensor vanishes is irrelevant to the above.2. Yes, you can use an idea droplet with zero surface tension to locally measure components of the 4th rank curvature tensor. Those dropets are in free fall mind you. Logically they are equivalent to g-force on either! Thats exactly why the shape of droplet deforms!I see no internal consistency at all and I see no need to writeConnection ?ld = Real Gravity Tensor Part + Fake Gravity Non-Tensor Part.You CAN do that, but the distinction has no utility at the next curvature tensor level where the two pieces above mix together!3. You need to invoke curved space-time globally in order to preserve the consistency of 1 as is shown clearly in Hawkings example inThe Universe in a Nutshell with two back-to-back accelerating observers on opposite sides of the Earth maintaining a constant spacelike separation.General Relativity is battle tested in its classical domain and needs no improvements there.Yilmazs model and Puthoffs PV model both fall under the Less with more category IMHO.If it aint broke, dont ? it.The foundational issues are:4. Quantum gravity5. Emergence of classical curved space-time from some kind of substratum. Different views of the substratum are:I. spin-networks (quantum computers) -> spin foams in loop quantum gravityII. vibrations of strings of pure energy Brian Greens Elegantly Tailored Fashion Show on NOVA ;-) See Vatican scene in Felinis ROMA. :-)III. Macro-quantum vacuum coherence (ODLRO) General Relativity and Dark Energy/Matter emergent More is different (PW Anderson)as in Andrei Sakharov, S. Adler, G. Chapline, G.E. Volovik and my own simpler dynamical instability QED model for in?. => What I like about adding to Rick Deckers example is that it makes it> easy.In case you missed it, Rick Decker is a professor at Hamilton College> (do a web search to ?d out about his school) who posted a> *quadratic* in an attempt to refute some of my conclusions.Well, I found his example grabbed me, for various reasons, and ?ally> found that I could modify it slightly to suit my purposes, and now> con?ent in your understanding--even as undergrads--of independence> between variables, I feel I can ?ally show you how I have been> defending mathematics against people working to undermine it!Heres the quadratic that Decker gave:(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30xy + 2y^2) where the as are roots of > > a^2 - (x - 1)a + 7(x^2 + x).And my modi?ation was just to add in the variable y, to get(5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) where the as are roots of > > a^2 - (x - y)a + 7(x^2 + xy).Notice that x and y are *independent* variables, as I havent related> them to each other in any way.That second polynomiala^2 - (x - y)a + 7(x^2 + xy)is the polynomial being analyzed by considering(5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2).You ?d it from7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7y(x - y)(5) + 7^2 y^2where whats on the right side is just a regrouping of terms from> whats on the left, as you can ?d out by simplifying the right side.Notice that with(5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) it *looks* like you have 7y and 7y as terms independent of x on the> left side, but that contradicts with 7(2y^2) on the right!So what gives?Well the problem is that x is in the way, and a simple technique in> analysis is to clear out one variable by setting it to 0.Now letting x=0, gives a(a + y) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, you havea_2(x,y) = b_2(x,y) - y, so(5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). And now everything matches up correctly.Obviously, now the coef?ients of the terms that have y but dont> have x as a factor are revealed, and not surprisingly, 7y(2y) matches> up with what you see on the right side with 7(2y^2).So if x does not equal y then it follows that(5a_1(x,y)/7 + y)(5b_2(x,y) + 2y) = 25x^2 + 30xy + 2y^2but if x-y=0, then you havea^2 + 7(x^2 + xy), de?ing the as, and you can directly calculate that a = +/- sqrt(-14)y, and picking> a_1(y,y) = sqrt(-14)y, you have(5 sqrt(-14)y + 7y)(5 sqrt(-14)y + 7y) = 7(25y^2 + 30 y^2 + 2y^2).Now considering when x does not equal y its easy enough to see that> regardless of anyones feelings on the matter, its not possible for> the terms *independent* of x, to actually have a dependency on x.So, with> >{***} (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2)if (5b_2(x,y) + 2y) had any factors in common with 7, then those> factors would have to be a factor of 2y, as that term is both visible> and independent of x.> No - that doesnt follow. Consider, for example, 5(3x + 11y) + 2y The ?st term, 5(3x + 11y) in general is coprime to 33. The second term, 2y, also in general is also coprime to 33.But the whole thing can be re-written as 15x + 57y = 3(5x + 19y)which has a factor (3) in common with 33, regardless of the values of x and y. This shows the error in your logic here quite clearly, I think.> So you can see my conclusion about what happens when you divide both> sides by 7, when x does not equal y, follows from some basic algebra.Whats fascinating is that you can also consider what happens when y=0> to separate out further, so letting y=0, I havea^2 - xa + 7x^2 = 0,soa = (1+/-sqrt(-27))x/2,and introducing c_1(x,y) and c_2(x,y) I then have(5(1-sqrt(-27))x/2 + c_1(x,y) + 7y)(5(1+sqrt(-27))x/2 + c_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). Whats fascinating here is that completely broken is *any* idea that> theres a variable factor in common with 7 that divides differently> dependent on the value of x, as you have the lead coef?ients1-sqrt(-27))/2 and 1+sqrt(-27))/2 which are, of course, constant.> Clearly c_1(x,y) must equal 5*[(x - y - sqrt((x - y)^2 - 28*(x^2 + xy))/2] - 5*(1 - sqrt(-27))x/2 This however is easily seen to be false. Therefore your claims that the ?st factor of {***}must be divisible by 7 are false. Nora B.> Therefore > What I like here is that quadratics are simpler than cubics to show> the result, the example I modi?d was presented ?st by Rick Decker,> a professor at Hamilton College, and my conclusion follows from> independence between independent variables.That is to doubt me here you need to doubt that concept in mathematics> itself--the concept of independence between variables.Of course, the problem for the old view on algebraic integers, is that> if you supposed that you were in the ring of algebraic integers> (notice I made no mention of a ring before now) then you run into a> problem speci? to that ring, where dividing both sides by 7 pushes> you out of the ring of algebraic integers.It might seem esoteric and distant as a problem, but my discovery of> that problem is actually a fascinating story, and now you can see how> mathematicians *worldwide* answer the question:What do you do when your paradigm is forced to shift by a discoverer> who thought outside of the box?Want more?Then read my other threads on sci.math or check out my blog archives:James Harris =Of course, the problem for the old view on algebraic integers, is that> if you supposed that you were in the ring of algebraic integers> (notice I made no mention of a ring before now) then you run into a> problem speci? to that ring, where dividing both sides by 7 pushes> you out of the ring of algebraic integers.Not for the examples weve been looking at. Stripped to the basics,we have a homogeneous quadratic polynomial P(x, y) with rational integercoef?ients, with each coef?ient divisible by some integer m.[In our example, P(x, y) = 7(25x^2 + 30xy + 2y^2), and m is 7.][P must have a speci? form, but well skip that for now.]Then we ?d two functions a_1(x, y) and a_2(x, y) which havealgebraic integer values for every rational integer pair (x, y) so thatfor each rational integer pair (x, y), P(x, y) factors over thealgebraic integers as P(x, y) = (n*a_1 + m*y)(n*a_2 + m*y)[In our example n = 5 and, as mentioned above, m = 7.][The a(x, y) functions also have a speci? form, but thatsnot important for this summary.]All of this is beyond argument. You then claim that for thisfactorization, one of the terms must be divisible by thecontent, m.[In our example, you claim 5a_1 + 7y must be divisible by 7.]This is simply not true in general. However, for all but afew values of x and y, what is true is that in the algebraicintegers1. NEITHER of n*a_1 + m*y, n*a_2 + m*y are divisible by m, and2. BOTH factors have non-unit divisors which also divide m.Heres the moral of this story:--------------------------------------------------------- The only way we are pushed out of the ring of algebraicintegers is when you insist that, in our example,one of 5a_1 + 7y or 5a_2 + 7y is divisible by 7. If youdrop that insistence, then you can ?d a validfactorization in the algebraic integers.----------------------------------------------------- ----While I have the ?some things are worth noting:1. While there is nothing inherently wrong with introducing the extra variable y, it does nothing to the exposition except make it less clear. Any problems we had without the ys we will still have with y = 1.2. Any variable transformations like b_2 = a_2 + y will in general change the look of the factorization, but will have no effect on the values of the factorization. For example, using the transformation above will change 5a_2 + 7y to 5b_2 + 2y, but the values of the two expressions will be the same. Therefore, it makes no sense to deduce any properties from the constant terms in this case.Rick =Does anyone know the de?ition of the Spectralradius of a realoperator, or even better: Knows where I can download some litteratureabout spectral-theory for Real Banach Algebras, if there is such athing. (There is lots of stuff about complex banach algebras, but Iam not interested in that!)If you want me to be more precise, here is the problem:Let K be a compact subset of R^k. Let T:C(K) -> C(K) be an operator,where C(K) is the set of continuous functions from K to R. In theof T, without de?ing it.Mogens =I am working on the program where is necessary to calculate hugeamount of numbers in always the same sequence, before the programactually opens.In other words, these number have to be calculated and saved intomemory before the program starts. I was trying to use number pi, but to calculate 100 million decimalplaces of number pi would take forever and no one would want to waitthat long. On the other hand I cant afford to include the whole 100million dec. places along with the software, that would take too muchof the space and too much time to download. Is there any other number like PI, which I could calculate faster?Serge-------------------------------------------------- -------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **---------------------------------------------------------- http://www.usenet.com =: Its the same machine.: Assume I have a TM that always ?ds a natural number larger than: any natural number on any given ?ite tape.: This same TM will try to ?d the largest natural on an in?ite tape.: The TM has no way of knowing the tape is in?itely long.: This TM will always say there is a natural number larger than: any natural it has found on the tape.What is the output of the following program if it never reachesthe end of input? while (!end_of_input) { input x; if (x>max) max=x; } print max;What do you think the output of this program would be? while (true) { n=n+1; } print done;Stephen <87smiudtrj.fsf@phiwumbda.org> <3ffadc1f$19$fuzhry+tra$mr2ice@news.patriot.net> =In jesse@phiwumbda.org (Jesse F. Hughes) said:>I dont see that an inconsistency in ZFC would necessarily yield an> >inconsistency in, say, Peano arithmetic.Because, having modelled ZFC in PA, an inconsistency in ZFC can be> translated back into an inconsistency in PA.> ZF with the axiom of in?ity replaced by its negation is equivalent toPA.-- G.C. =>>Also, thank you to R3769, Michel Hack and Jesse F. Hughes for correcting>>my sloppy thinking concerning the notion of inconsistency with regard to>>mathematics. I now understand that an inconsistency in ZFC does not>>mean an inconsistency in mathematics. > Well, now we must be a bit careful. Shmuel Metz informs me that> Goedel proved the relative consistency of ZFC to PA --- so that *if*> PA is consistent, *then* ZFC is consistent. Hence, an inconsistency> in ZFC would in fact yield an inconsistency in PA --- not due to the> simple fact that PA has a model in ZFC, but to the more surprising> (reported) fact that ZFC has a model in PA.Umm. Are you sure youre remembering that correctly? ZFC without the Axiom of In?ity has a model in PA, but I dont think ZFC itself does. I could of course be wrong (a far from uncommon occurence), but in this particular case I would be surprised were that so. Peano arithmetic doesnt seem strong enough to have a model of <87smiudtrj.fsf@phiwumbda.org> <3FF9DD0F.CC42A336@mdli.com> <87isjmyzmw.fsf@phiwumbda.org> =>> >> >Also, thank you to R3769, Michel Hack and Jesse F. Hughes for correcting>my sloppy thinking concerning the notion of inconsistency with regard to>mathematics. I now understand that an inconsistency in ZFC does not>mean an inconsistency in mathematics. >> >> >> Well, now we must be a bit careful. Shmuel Metz informs me that>> Goedel proved the relative consistency of ZFC to PA --- so that *if*>> PA is consistent, *then* ZFC is consistent. Hence, an inconsistency>> in ZFC would in fact yield an inconsistency in PA --- not due to the>> simple fact that PA has a model in ZFC, but to the more surprising>> (reported) fact that ZFC has a model in PA. Umm. Are you sure youre remembering that correctly? Im fairly sure that Im remember what Shmuel said and am stating itcorrectly. I dont know anything about the claim than Ive saidabove.My apologies to Shmuel if Ive misrepresented things.-- Jesse HughesSo far as this negative attitude toward life is concerned, Buddhismis merely Taoism a little touched in its wits. -- Lin Yutang, /My Country and My People/ => at 07:55 PM, r3769@aol.com (R3769) said:>Pardon my ignorance, but what does modelled ZFC in PA mean?>>You can encode statement about ZFC as statements about naturals, in>>such a way that a proof in ZFC translates to a proof in PA. As a>>result, if there is an inconsistency in ZFC then there is also an>>inconsistency in PA. Google for relative consistency or for G.9adel.> Ok. Section 3.6 of Endertons A Mathematical Introduction to> Logic describes this process (near as I can ?ure). However, the> *relative* inconsistency of PA isnt really what we are concerned> about. For mathematics to be doomed, you still need to show the> inconsistency of ALL other set theories, ST, where ST can be> modelled in PA. Because (we assume) ZFC is inconsistent, there is> NO possibility of ?ding some other consistent ST, such that ST> can be modelled in PA?It wouldnt matter if you could model a consistent theory in PA. presumably in a natural way, so that any proof of P & ~P in ZFC wouldyield a proof of Q & ~Q for some formula Q of PA.-- Many argue that its programmers have turned out shoddy programs, but[their] objective is to make pro?, not superlative programs perse. By the pro? criterion, Microsoft has been one of the greatestcompanies in the history of this country. -- ADTI defends Microsoft => at 07:55 PM, r3769@aol.com (R3769) said:>Pardon my ignorance, but what does modelled ZFC in PA mean?>>You can encode statement about ZFC as statements about naturals, in>such a way that a proof in ZFC translates to a proof in PA. As a>result, if there is an inconsistency in ZFC then there is also an>inconsistency in PA. Google for relative consistency or for G=F6del.> Ok. Section 3.6 of Endertons A Mathematical Introduction to>> Logic describes this process (near as I can ?ure). However, the>> *relative* inconsistency of PA isnt really what we are concerned>> about. For mathematics to be doomed, you still need to show the>> inconsistency of ALL other set theories, ST, where ST can be>> modelled in PA. Because (we assume) ZFC is inconsistent, there is>> NO possibility of ?ding some other consistent ST, such that ST>> can be modelled in PA?It wouldnt matter if you could model a consistent theory in PA.=20=20>No they dont. Ill try to convince you of this two ways. First note that ifevery proof in ZFC translates to a proof in PA then every proof in ZF can alsobe so translated. By any reasonable de?ition of translate, we can alsotranslate from PA back to ZFC, ZF, and PA. So every theorem of ZFC is also atheorem of ZF. At this point the only real problem is that we all know the Cin ZFC stands for Choice, and not, say, Cohen. So it is at least *plausible*that not every proof in ZFC translates into a proof in PA, right?At this point I should probably stop writing, because I dont really understandthe subject well enough to explain it clearly and thus convince you that infact not all proofs in ZFC translate to proofs in PA. However, as near as I can ?ure, yes it is true that you can encode statementsabout ZFC as statements about the naturals but NOT in the way Shmuel Metzsuggests. That there is an interpretation, f, from the theory cn(PA) to thetheory ZFC is a fact. By cn(PA) I mean the consequences of PA, i.e. thosesentences in PA that are true in every model of PA. So to encode a theorem inZFC, say s, as a sentence in PA is easy: let t=f^(-1)(s). Why does this work? Because interpretations are, by de?ition, 1-1. Shmuel Metz asserts that t isa theorem of PA, hence t is in cn(PA), hence f is onto ZFC, hence f is abijection. My (very fuzzy, at this point) understanding is that Godel(?)proved something akin to the opposite: there is NO such *bijective* f, unlessZFC is inconsistent. Perhaps my understanding is incorrect.In any event, Ill have to take a closer look at this over the next few years,and try to develop a deeper understanding of this very interesting subject.rich => presumably in a natural way, so that any proof of P & ~P in ZFC would> yield a proof of Q & ~Q for some formula Q of PA.This line I do not understand. Gentzen proved consistency of PA.But he did it in ZF... using induction up to epsilon_0.So how is the above reduction possible? <87u135nv7i.fsf@phiwumbda.org> =>> presumably in a natural way, so that any proof of P & ~P in ZFC would>> yield a proof of Q & ~Q for some formula Q of PA. This line I do not understand. Gentzen proved consistency of PA.> But he did it in ZF... using induction up to epsilon_0.> So how is the above reduction possible?This isnt a result with which I am really familiar, so I cant helpyou. Shmuels response to one of my own postings is about all I knowabout it.I havent a clue about how Goedels proof proceeds. Sorry. I wouldappreciate a gentle outline of how the translation goes myself(gentler than go to the library and see the original sources).-- All intelligent men are cowards. The Chinese are the worlds worst?hters because they are an intelligent race[...] An average Chinesechild knows what the European gray-haired statesmen do not know, thatby ?hting one gets killed or maimed. -- Lin Yutang =>I havent a clue about how Goedels proof proceeds. Sorry. I would>appreciate a gentle outline of how the translation goes myself>(gentler than go to the library and see the original sources).>Whats so wrong with For the ?st time in the human history, learning is *easy*-- just aclick (or two) away! I want to know how to bottle feed my three baby goats with two hands while mywifes always hungry dog watches, drooling. Goedel *is* easy by comparison. ;-)rich => >Any reasonable de?ition of computable that I know of will>> allow me to ?d the smallest member of a ?ite set.>> Why would a computer assume that an algorithm that>> works on any ?ite set will not work on some in?ite sets?> > This requires that the computer be able to compare>sizes of members of the set, which is not necessarily the case.OK> One can describe sets of numbers in a way that does not allow of>comparisons of those numbers for size.Can you describe such sets to a Turing Machine?I have noticed that many of the refutations other people have given> rely on induction. I am assuming that a TM can perform an in?ite> number of operations in ?ite time. Lets call this in?ite computation.It appears that induction and in?ite computation come to different> conclusions on some problems. Is there any way to formalize the> differences?>When you can produce an exemplar of such a machine, perhaps we will begin to take its possibility more seriously.There are a lot of conjectures whose truth or falsehood are not now known. There are some of them whose truth may never be known. Indeed, there may even be some of them whose truth is in some serious sense unknowable. Let a numbers binary expansion have digit values based on the truth or falsehood of a sequence of such unresolved conjectures and put your TM to work on ?ding its position on the number line. Russell> - 2 many 2 count> > => >Yes, I am assuming that the computer can perform an in?ite>> number of operations in ?ite time.> > So where can we get one?There is no theoretical limit on how fast something can be computed.> (At least there isnt a limit I have seen.)IBM is saying they have already exceeded Moores law.> Russell> - 2 many 2 countIn order for a TM, or any computer, to complete in?itely many operations in ?ite time, it must be able to complete all but ?itely many operations in an arbitrrily short time. IBM isnt even in the running for that kind of speed.All computers built so far, or even projected by serious designers, have some positive minimum time, however small, for their fastest operation. As long as that situation exists, your TM is hopeless.In?itely many repetitions of any ?ed positive quantity, however small, add up to in?ity. = > >> Yes, I am assuming that the computer can perform an in?ite>>number of operations in ?ite time.> >So where can we get one?> > There is no theoretical limit on how fast something can be computed.>(At least there isnt a limit I have seen.)> > IBM is saying they have already exceeded Moores law.> > Russell>- 2 many 2 count> > In order for a TM, or any computer, to complete in?itely many> operations in ?ite time, it must be able to complete all but ?itely> many operations in an arbitrrily short time. IBM isnt even in the> running for that kind of speed. All computers built so far, or even projected by serious designers, have> some positive minimum time, however small, for their fastest operation.> As long as that situation exists, your TM is hopeless.>Not so.There are already examples of computers that are in?itely fast.For example, there are analog computers that will computethe instantaneous derivative of an input signal. The speed of lightis the only real limitation on the speed of these machines.(The signal has to go through the machine.)Russell- 2 many 2 count => > >>Yes, I am assuming that the computer can perform an in?ite> number of operations in ?ite time.>> >> So where can we get one?> >There is no theoretical limit on how fast something can be computed.>> (At least there isnt a limit I have seen.)> >IBM is saying they have already exceeded Moores law.> >Russell>> - 2 many 2 count> > In order for a TM, or any computer, to complete in?itely many>operations in ?ite time, it must be able to complete all but ?itely>many operations in an arbitrrily short time. IBM isnt even in the>running for that kind of speed.> > All computers built so far, or even projected by serious designers, have>some positive minimum time, however small, for their fastest operation.>As long as that situation exists, your TM is hopeless.> >Not so.> There are already examples of computers that are in?itely fast.> For example, there are analog computers that will compute> the instantaneous derivative of an input signal. The speed of light> is the only real limitation on the speed of these machines.> (The signal has to go through the machine.)TMs are distinctly NOT analogue, and even analogue computers must output only ?itely many items in ?ite time.You are stuck with an unfeasible assumption.> Russell> - 2 many 2 count> > =>There are already examples of computers that are in?itely fast.>For example, there are analog computers that will compute>the instantaneous derivative of an input signal. The speed of light>is the only real limitation on the speed of these machines.>(The signal has to go through the machine.) TMs are distinctly NOT analogue, and even analogue computers must output> only ?itely many items in ?ite time.The distinction between analog and digital is not as sharp as one mightthink.Almost anything that can be computed by a sequential digital computercan also be computed by a Boolean circuit.> You are stuck with an unfeasible assumption.Do mathematicians care if a theory is physically realizable?My idea of in?ite computation is no less feasible thanusing induction over all of the natural numbers.Russell- 2 many 2 count =>There are already examples of computers that are in?itely fast.>> For example, there are analog computers that will compute>> the instantaneous derivative of an input signal. The speed of light>> is the only real limitation on the speed of these machines.>> (The signal has to go through the machine.)> > TMs are distinctly NOT analogue, and even analogue computers must output>only ?itely many items in ?ite time.The distinction between analog and digital is not as sharp as one might> think.As far as thinking goes, speak for yourself only.> Almost anything that can be computed by a sequential digital computer> can also be computed by a Boolean circuit.How are Boolean circuits analog?> You are stuck with an unfeasible assumption.Do mathematicians care if a theory is physically realizable?> My idea of in?ite computation is no less feasible than> using induction over all of the natural numbers.Do you assert that Turing would accept TMs such as you propose?> Russell> - 2 many 2 count> =>> at 07:55 PM, r3769@aol.com (R3769) said:>Pardon my ignorance, but what does modelled ZFC in PA mean?>>You can encode statement about ZFC as statements about naturals, in>>such a way that a proof in ZFC translates to a proof in PA. As a>>result, if there is an inconsistency in ZFC then there is also an>>inconsistency in PA. Google for relative consistency or for G.9adel.>Ok. Section 3.6 of Endertons A Mathematical Introduction to Logic>describes>this process (near as I can ?ure). However, the *relative* inconsistency>of>PA isnt really what we are concerned about. For mathematics to be doomed,>you>still need to show the inconsistency of ALL other set theories, ST, where ST>can be modelled in PA. >Because (we assume) ZFC is inconsistent, there is NO possibility of ?ding>some other consistent ST, such that ST can be modelled in PA? >Another question: Do you mean to say that ZFC is inconsistent implies ZF isinconsistent or that ZF cant be modelled in PA?And now that I am really confused, I think Ill go feed my goats.rich = >>Many people would argue that every real number is computable.> Many...> > Turing believed that numbers like PI and e were computable. Probably, he believed this because numbers like PI and e are> computable.Maybe.> Now, how does that support the claim, Many people would argue that> every real number is computable?I should have said many people would argue that every de?able realnumber is computable. There could be an in?ite number of unde?ablereals.Russell- 2 many 2 count <3ffadc52$20$fuzhry+tra$mr2ice@news.patriot.net> <1g75muz.1d66n4i14gjvdlN%panoptes@iquest.net> <1g75yr4.1si6z861lg1554N%panoptes@iquest.net> <87brpe1wtx.fsf@phiwumbda.org> => I should have said many people would argue that every de?able real> number is computable. I suppose that many people would argue simply false things. ChaitinsOmega is an interesting de?able number which is provably notcomputable. See .> There could be an in?ite number of unde?able reals.You think? Yeah, I guess there *could* be.(Just to make myself perfectly clear: There is no question to whetheror not there is an in?ite number of unde?able reals[1]. Thereare.Footnotes: [1] Well, one question, perhaps. One might need to be careful tospecify what one means when he calls a real de?able.-- Im talking about mathematics--hard, brutal, extreme ... pushing yourmind beyond the limits to understand what no one else can becausetheyre afraid to risk it all, to lose their freaking worthless mindsin the push to know. --James Harris, for the Nike Derivator = > I should have said many people would argue that every de?able real>number is computable. I suppose that many people would argue simply false things. Chaitins> Omega is an interesting de?able number which is provably not> computable. See .This page mentions hypercomputers, which are similar to my ideaof in?te computation.http://en2.wikipedia.org/wiki/Hypercomputer>There could be an in?ite number of unde?able reals. You think? Yeah, I guess there *could* be. (Just to make myself perfectly clear: There is no question to whether> or not there is an in?ite number of unde?able reals[1]. There> are.> Footnotes:> [1] Well, one question, perhaps. One might need to be careful to> specify what one means when he calls a real de?able.>I know I am just digging myself deeper, but I like to argue.I know the standard proof.One shows the de?able numbers are countable, the real numbersare uncountable and concludes there are an in?ite number ofunde?ed reals. (Its horrible that I can recite these proofs, now.)The obvious ?th is argument is that every real number hasa de?ition. A real number is de?ed as the limit of a Cauchysequence of rationals (or a Dedekind cut). Either there are anuncountable number of such de?itions or the real numbers are countable.The computable numbers are supposed to be countable.Even so, the computable numbers include all of rationalsand all of the de?able irrationals.It is hard to imagine that the limit of a Cauchy sequencewould be uncomputable. Doesnt this mean thatevery real number is computable?Russell- 2 many 2 count => The computable numbers are supposed to be countable.> Even so, the computable numbers include all of rationals> and all of the de?able irrationals.What is your de?ition of de?able in this context?> It is hard to imagine that the limit of a Cauchy sequence> would be uncomputable. Doesnt this mean that> every real number is computable?You have a most ?e imagination, it easily swallows camels but then strains at gnats. => > >>Many people would argue that every real number is computable.>> Many...> >Turing believed that numbers like PI and e were computable.> > Probably, he believed this because numbers like PI and e are>computable.Maybe.> Now, how does that support the claim, Many people would argue that>every real number is computable?I should have said many people would argue that every de?able real> number is computable. There could be an in?ite number of unde?able> reals.> Russell> - 2 many 2 countHow does unde?able differ from uncomputable? =>I should have said many people would argue that every de?able real>number is computable. There could be an in?ite number of unde?able>reals.> How does unde?able differ from uncomputable?I am not sure it does.How would one prove that an unde?able real number isactually a real number?Russell- 2 mnay 2 count =>I should have said many people would argue that every de?able real>> number is computable. There could be an in?ite number of unde?able>> reals.> How does unde?able differ from uncomputable?I am not sure it does.> How would one prove that an unde?able real number is> actually a real number?>Unde?able is your word, not mine, so explaining it is your problem, not mine. Russell> - 2 mnay 2 count> > => How would one prove that an unde?able real number is> actually a real number?By noting that the de?ition of real number does not require anythingthat corresponds to de?able.-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W Im an informative broadcaster.Thats what they *all say., you beat me to it.> By the way, get outta alt.atheism. Nobody here believes *anybody is> psychic so youre wasting bandwidth.Nobody buys Hercs claims on sci.skeptic, either, but you know how crossposting loons are.-Chris Krolczyk Helping or Hurting? = I have been reading some of you works on the mind Dr. Sarfatti andin one particular paper on your QED website about the Mind-Matterinterface via the Bohmian Q.P. shows some terms/concepts that are notexactly described and de?ed. It was an interesting read though Istill do not quite fully understand what you are getting at and ithas not been my ?st time that I have read this paper.JS: You do not have enough background.See ?st if you can understand that book, which is pre-requisite.You must crawl before you can dance. ZD: Speci?ally what is the back action that you keep referring too?JS: Read p. 30 and 14.6 of the above. Note their statement (I paraphrase)Unlike other ?lds the quantum information ?ld in orthodox quantum theoryhas no sources. The quantum potential is fragile hence signal locality.Google Antony Valentini and read his papers. Beyond quantum theory iswhen the quantum information ?ld has sources. Also the giant macro-quantuminformation ?lds are local and very much like classical force ?lds with someimportant differences.the matter/mass/?ld ect. back on the mind.The giant DeBroglie MACRO-QUANTUM q-bit ground state condensate in theliving brain IS the MIND!ZD: It is this superposingof this wave and intent retarded waves from the mind that createconsciousness. Could you de?e this back action in terms ofphysical principles?JS: Read p. 30 and 14.6 of the above.Yes, no ACTION without REACTION.See also Shelley Goldsteins recent paper on how presponse advanced effect comes in.An advanced from-the-future micro-causal arrow and a retarded from the past MACRO-arrow of time.This idea needs further development, i.e. putting Wheeler-Feynman together with Bohms ontology.See Dick Biermans experiments on presponse.If I am correct, the presponse curve of the subject should weaken as the subject gets increasing sedation withanaesthetics.ZD: You have the equation: Phi-> <-X - (Phi->X)=CJS: Phi = BIT (pilot wave)X = IT (extra variable)-> = ACTION<- = REACTION ZD: What does this mean? Is Phi-><-X the conscious intent acting onmatter and back and the last term is the unconscious intent?JS: No.BIT <- ITis matter-geometry generating inner consciousness in the BIT mind-?ld.BIT -> IT is how the mind-?ld MOVES matter in space.The two act together in a self-organizing loop of conscious intent.ZD: Also what is the difference between the active q-bit waves and theinactive q-bit waves?JS: You do not have enough background.See ?st if you can understand that book, which is pre-requisite.You must crawl before you can dance.ZD: And ?ally is Phipps version of special relativity the same as theLorentzian interperatation?JS: No. Phipps is plain wrong period. He is a nice man and a good writer, but he is wrong.Phipps does not believe length contraction although he accepts time dilation.I think he gave up on his idea. It never got anywhere really except for a non-replicated empirical claim about lack of Thomas precession as I recall. Einstein wins!ZD: i.e.- An absolute frame of reference wherethings DO happen simultaneously.JS: General relativity allows that in special situations of high symmetry like the FRW cosmological metric of standard model.The laws of GR are locally frame independent, but the solutions are not. In the FRW metric de?e a GLOBAL cosmic time with global simultaneity, i.e. a preferred 3D space-like foliation of 4D space-time in which all points on the preferred space-like hypersurface see absolute temperature. This is a practical way to navigate when you use star gate time travel tunnels and weightless warp drive by metrically engineering the zero point quantum pressures of the physical vacuum via a kind of Josephson effect IMHO, i.e. interfering the virtual vacuum condensate with a real ground state condensate using variations on the Bohm-Aharonov dynamical phase and Berry topological phase effects. probability problem as part of ajob-interview pre-screen. After giving a totally wrong answer (?causeI wasnt thinking), I gave the following answer. However, theinterviewer said that my answer was still incorrect.Fair enough; I ?d probability unintuitive. But now my curiosity ispeaked: What is the right answer? Question: You have a jar with 1000 coins in it, 999 are normal, butone is two-headed (i.e both sides are heads). You choose a coin atrandom and toss it 10 times. With each toss the coin comes up heads,so you get 10 consecutive heads. What is probability that you havechoosen the two-headed coin? My answer:P(two headed) = 1 - P(normal coin *and* 10 heads in a row) = 1 - .999 * (1/2)^10 = 1 - .999 * (1/1024) = 1 - .000976 = 0.999024 = 99.9024%If this is wrong, what is the right answer? How do you calculate it?Stuart =Consider P an n-set. U and V are the power set of P, that is,the set of all subsets of P with 2^n elements.Lets consider a simple membership graph with P, U and V as verticeswhich has the following property: - there are only edges pairwise between P and V (lets call this subgraphL) and U and V (lets call this subgraph R) but not between P and U. - there is an edge from p (in P) to v (in V) iff p is in v. - there is an edge from v (in V) to u (in U) iff v is a subset of u.Consider the following 3edge layout problems:1. Suppose we impose the constraint that the in-degree of any vertex in U is less than m, how to layout edges such that: for each u in U, each p in u can ?d a path to u via some v in V Whats the maximum out degree of any vertex in P?2. Suppose now we consider the opposite, lets keep the max. out-degree of any vertex in P be k, whats the max. in-degree of vertices in U?3. Suppose now there is no constraint, whats the minimal max. degree ofvertices in P union U?Does anyone see something similar before? Please give me the reference.Any suggestion of the approach I can take would be very welcome?Aldar =has anyone read this book and would care to comment?Is it useful and/or a good read?http://www.cwru.edu/artsci/math/wells/pub/abouthbk.html== ==Let: G(k) =sum{1<=j<=k, GCD(k,j)=1} 1/j.(G(k) = the sum of the reciprocals of the positive integers which arecoprime with k and are <= k.)And let H(k) = sum{j=1 to k} 1/j,the k-th harmonic number (the sum of the reciprocals of all thepositive integers <= k).2 misc sum identities:sum{k=1 to n} G(k) =sum{m=1 to n} (([n/m]+1)H([n/m]) -[n/m]) mu(m) /m,where [n/m] = ? /m),and mu() is the Mobius (Moebius) function.And,sum{k|n} G(k) =sum{k|n} phi(k) H(n/k) /k =(1/n) sum{k=1 to n} GCD(k,n) H(GCD(k,n)).(phi(k) is the number of positive integers <= k and coprime with k, asalways.)I may have made a mistake, and the above results are meaningless, butthey should be posted anyway. ==>Harris, you really are an insufferable fool. Almost all your>statements about mathematical usage are obviously wrong, even to a>casual onlooker such as myself. Polynomials are not in rings, they>are _over_ rings; this is not just a quibble over prepositions, the>point is that a polynomial over the integers has coef?ients which>are integers, but is not itself an integer.> > Well I think the telling point for readers is you began with an> insult.> > That you have a weak case is obvious, so you apparently wished to try> and hurt my feelings with an opening insult, as you cant win on the> logic.[...]I think your detractors are winning on the evidence, for exampleall those factored polynomials in agreement with the M and M theorem.David Bernier =>[megasnip]>> Which makes no sense at all. Youre saying that you cant comprehend>> how, say, given P(x) in the ring of integers, I can talk of P(x) being>> an integer, and this problem for you prevents you from advancing>> through the entire paper??!!!>>Harris, you really are an insufferable fool. Almost all your>statements about mathematical usage are obviously wrong, even to a>casual onlooker such as myself. Polynomials are not in rings, they>are _over_ rings; this is not just a quibble over prepositions, the>point is that a polynomial over the integers has coef?ients which>are integers, but is not itself an integer.> > Well I think the telling point for readers is you began with an> insult.But ended with mathematics.> > That you have a weak case is obvious,Only to you.> so you apparently wished to try> and hurt my feelings with an opening insult, as you cant win on the> logic.The logic is two sentences later. Didnt you get that far?> Given P(x) = x+1--and I notice you deleted out that example which I> used repeatedly--in the ring of integers, your argument would mean> that you cant in any way say that P(x) is an integer.If P(x) is a polynomial, you cant. This is what Dot meanswhen she says you arent using polynomial in the standardsense. If P(x) is a polynomial, x is a placeholder, a freevariable. Depending on x, the value of P(x) could be any real number, or indeed any complex number.And you havent pinned it down by saying in the ring of integers,because there are at least two very different interpretationsof that statement.1. x is con?ed to the integers. In that case it istrue that (x+1) is an integer. But for most such expressions,it doesnt make sense to talk about roots or algebraicintegers. For instance if P(x) = x^2-2, x in Z, then theresno such thing as a root of P(x).2. P(x) is in the ring of polynomials with integer coef?ients.This is what people have been guessing you meant. In fact,sometimes you DO seem to mean that, just not in this post.A polynomial with integer coef?ients cant be said tobe an integer, as it is a mapping from complex to complex.Neither the input nor the output is necessarily aninteger. And in most cases, the roots certainly are not integers,nor are they even necessarily real.>An old, but wonderfully simple, book on this subject is A Concrete>Approach to>Abstract Algebra by W. W. Sawyer (>40yrs ago). Looking for his>de?ition of a polynomial I chanced on an interesting discussion on>the distinction you are unable to grasp between a polynomial as a sort>of generalized arithmetic, and a polynomial as a formal object.>Curiously enough, he describes these as being the viewpoints of>bright pupils and dull pupils -- and would you believe[!!] -- your>view is that of the bright pupils.> > Logical fallacy is appeal to authority.Going to a standard reference for de?ition of terms isntappeal to authority. - Randy = [.newsgroups trimmed.]>2. P(x) is in the ring of polynomials with integer coef?ients.>This is what people have been guessing you meant. In fact,>sometimes you DO seem to mean that, just not in this post.>A polynomial with integer coef?ients cant be said to>be an integer, as it is a mapping from complex to complex.No, no, no, no, no. A polynomial is NOT the same as a polynomialfunction. There is a canonical map from the ring of polynomials withcoef?ients in R (R a ring), R[x], to the ring of functions from R toR; but this map is not always surjective, and it is not alwaysinjective. The ring R[x] may be de?ed in any number of ways; oneis to de?e it as the collection of all almost-null sequenceswith values in R with pointwise addition, and de?e themultiplication as we know. The other is to de?e it as the set of allformal sums a_0 + ... + a_n*x^n where x is a formal symbol and n is anatural number, a_i in R for each i, and de?e the addition andmultiplication as we know, etc.Given an element of R[x], a_0+...+a_n*x^n, we get a function from R->Rby the rule r|-> a_0 + a_1*r + ... + a_n*r^n. This gives a map fromR[x] to the set of functions R->R.The map is not always injective: the polynomial x^2+x and the zeropolynomial give the same function over F_2. The map is not alwayssurjective: any function Q->Q which is zero at an in?ite number ofrationals and nonzero at an in?ite number of rationals is not apolynomial function.And even when the map is injective, it is important to keep thedistinction clear. Particularly in the context where James is tryingto go: for example, consider the polynomials 2 and x^2+x in Z[x]. Ifwe consider them as functions, then note that the value of thepolynomial function 2 always divides the corresponding value of thepolynomial function x^2+x in the integers. However, the polynomial 2does NOT divide the polynomial x^2+x in Z[x]. = reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A mans 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 ?ures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the inde?ite 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 [.newsgroups trimmed.]> > > >2. P(x) is in the ring of polynomials with integer coef?ients.> >This is what people have been guessing you meant. In fact,> >sometimes you DO seem to mean that, just not in this post.> >A polynomial with integer coef?ients cant be said to> >be an integer, as it is a mapping from complex to complex.> > No, no, no, no, no. A polynomial is NOT the same as a polynomial> function. There is a canonical map from the ring of polynomials with> coef?ients in R (R a ring), R[x], to the ring of functions from R to> R; but this map is not always surjective, and it is not always> injective. [Snip informative tutorial]this stuff enough times and, unlike James, it may sink in.Also unlike James, I read what you write.I dont know whether it is deliberate or not, but Ithink that almost everything James proves eventuallyrelies on an ambiguity, on a phrase which can be interpretedone way early in the proof, but which needs to be interpretedanother way in order to draw the conclusion. The shiftingmeanings of factor are a prime (sorry) example. Thusit would be shooting himself in the foot were he toactually specify the de?ition of anything. And Isuspect he knows that.In this thread, by polynomial in the ring of integershe appears to be saying polynomial function evaluated withinteger arguments, not polynomial with integer coef?ients.I dont think he has understood a single comment aboutrings and factorizations, and I think he just keeps repeatingthe ring is algebraic integers whenever anybody asks himany question about rings, hoping that they will ?d theanswer they seek in some interpretation of that statement,though he has no idea himself what he means with that phrase. - Randy =>> [.newsgroups trimmed.]>> >> >> >2. P(x) is in the ring of polynomials with integer coef?ients.>> >This is what people have been guessing you meant. In fact,>> >sometimes you DO seem to mean that, just not in this post.>> >A polynomial with integer coef?ients cant be said to>> >be an integer, as it is a mapping from complex to complex.>> >> No, no, no, no, no. A polynomial is NOT the same as a polynomial>> function. There is a canonical map from the ring of polynomials with>> coef?ients in R (R a ring), R[x], to the ring of functions from R to>> R; but this map is not always surjective, and it is not always>> injective. >>[Snip informative tutorial]>>this stuff enough times and, unlike James, it may sink in.>Also unlike James, I read what you write.>>I dont know whether it is deliberate or not, but I>think that almost everything James proves eventually>relies on an ambiguity, on a phrase which can be interpreted>one way early in the proof, but which needs to be interpreted>another way in order to draw the conclusion. The shifting>meanings of factor are a prime (sorry) example. Thus>it would be shooting himself in the foot were he to>actually specify the de?ition of anything. And I>suspect he knows that.Undoubtedly. I have long been of the opinion that if he ever getsaround to actually being careful with his terms and determiningexactly what each thing is supposed to be in his polynomialfactorization arguments, one of two things will occur: either theresult will be trivially correct, but it will not apply in thesituation he is trying to apply it; or else it will be triviallyfalse, because the things he claims exist do not have the propertieshe thinks they have.My guess is that he thinks about the theorem with examples that makesit trivally true and inapplicable, and then tries to apply it bythinking of it in the false setting. 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 mans 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 ?ures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the inde?ite 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 => > Your belief here is irrelevant as a proof begins with a truth then> proceeds by logical steps to a conclusion which then must be true. > Challenging a paper requires that you ?d that it did not begin with> a truth, or that you ?d a break in the logical chain.> Actually, almost the opposite is true. You can challenge the *proof* of a fact by ?ding a false premise or logical ?ut the result may be true independent of these errors, if one can ?d another proof.To challenge a statement, you need to show that it is false. To do this, you need not _ever_ see the proof. If you can draw false conclusions from a result, it is false. The proof is absolutely irrelevant if it is incorrect.If I claim that the product of two algebraic integers is an algebraic integer, and give a proof (A. Magidin and I, if not others, have done so on sci.math recently, proofs can be found in, say Dummit & Footes _Abstract Algebra_ or Atiyah and Macdonalds _Commutative Algebra_), if someone ?ds a ? the proof, this does _not_ invalidate the result. The result is invalidated when someone produces two explicit algebraic an airtight argument for the existence of two such algebraic integers, without explicitly writing them down.That said, a _courteous_ thing to do when one has found an erroneous statement is to help the author of that statement ?d the ? his premeses or logic... Michael A. Van OpstallPadelford C-113opstall@math.washington.eduhttp://www.math.washington.edu /~opstall/> NIST is trying to do is only a small portion of the