mm-211 === Subject: complex problemI have to solve equetions with complex coefficients:T11= - 1.50005 + 5.93649i;G1 = 0.08562 + 0.76492i;T12 = -6.4481 - 2.79474i;G1^2 * S22 + 2*G1*S11 - G1^2 + T11*G1 - T11*S11 = 0andG1*S22 - T12*G1 + S11 + T12*S11 = 0I know the solution wch is:S11 = 0.13272 + 0.80569 iS22 = -0.52264 - 0.13305 iBut from analytical I ïve gotS11 = G1S22 = -1Where I made a mistake? How to get ts solution (S11 = 0.13272 === the computation time of ts Gradient optimization method, O(kN) or O(kN^2)?I am trying to figure out the complexity level of the gradient method for> minimizing an objective function.Assume J is the scalar objection function of N vectors, each vector is of > dimension.> Now we are trying to find a k-dimension vector x such that J is minimized,> e.g. J is the sum of squares of the distances from N points in R^k space to> a line segment,wch> is connected by a known vector a and the unknown point x in R^k space.So given ts definition of J, the way we did is to calculate dJ/dx = 0 and> find the vector x.> However, the expression dJ/dx is not usually in explicit form, i.e, x can> not directly be obtained and> have to be done by some iterative approach.Then for ts kind of optimization problem, how to define the computational> complexity?> It is O(kN) or O(kN^2)?And how to verify ts?It seems to me there are (at least) two questions here.The first is, WhatÍs the operations count for a single gradient descentiteration? Presumably function evaluations and partial derivatives area single operation, else the question is completely fatuous. If youcanÍt figure ts out... IÍll say no more.The second is the far more interesting question, How many iterations(hence, how many function evaluations/§ops) are required to get aniterate witn epsilon of a minimizer? And ts surely has no answer: itÍs unbounded, since the convergence of gradient descent can be asslow as we wish, even for convex functions.A modification of ts: how many iterates does it take to get f(x_k)witn epsilon of the minimum? As I recall, the best possible bound(i.e. in the worst possible case) for convex functions is sometnglike |f(x_k) - f(xmin)| <= c/sqrt(k),so it can take a LONG time. (The constant c involves === no such tng as infinity>> Ah, I remember when I first started programming. All I had was a really>> hot needle and had to write my code directly to the SIMMs. It was a major>> leap forward when I upgraded and could just type copy con file.exe...When I started, SIMMs hadnÍt been invented. We had people who had seen> so many punch cards and papertapes they could read the holes by eye.> I set the bootstrap instructions using an array of toggle switches on> one mainframe we had.Paper cards? Ah, you were lucky! When I first started investigatingprogramming, we had to csel instruction sets into very large rocks andcarry them ourselves to the CPU (Cseling Processors Union) to get thework done. And we were === infinity,,, wibbled:> >> Ah, I remember when I first started programming. All I had was a really>> hot needle and had to write my code directly to the SIMMs. It was a major>> leap forward when I upgraded and could just type copy con file.exe...When I started, SIMMs hadnÍt been invented. We had people who had seen> so many punch cards and papertapes they could read the holes by eye.> I set the bootstrap instructions using an array of toggle switches on> one mainframe we had.Paper cards? Ah, you were lucky! When I first started investigating> programming, we had to csel instruction sets into very large rocks and> carry them ourselves to the CPU (Cseling Processors Union) to get the> work done. And we were thankful...I had ts void, and I thought, let there be light, === is no such tng as infinityStop bragging guys ... please! I feel diminished ... I am only 22! :(Sometimes I get ïOhhh!Í, ïWow!Í, and ïOoooh!Ís for having a 386[Who am I kidding! I *had* it, === no such tng as infinityImam Tashdid ul Alam wibbled:640K ought to be enough for anybody.> -BILL GATES, 1981 [I was born the following year :o]> Well, since I recall a time when Comp Sci students did an entire undergraduate course with 5Mb of diskspace and all their programs had to fit into 120kb, maybe he had a point. Mind you that was on a macne with 60 bit words, and the compilers all used overlays === 4, the associativity of multiplication, is also not redundant. There>exist finite counterexamples called semifields.On the other hand, IÍm suspicious of axiom 5: a * 1 = a. Maybe IÍm missing>sometng obvious, but can there possibly be a finite ring without identity>that has no zero divisors?Z_3 with addition as usual, and ts multiplication table: 0 1 2> --------> 0 |0 0 0 > 1 |0 2 1> 2 |0 1 2fits all the requirementsNo, it doesnÍt; your 2 is the 1 specified in the === doesnÍt; your 2 is the 1 specified in the definition.It looks === field>> No, it doesnÍt; your 2 is the 1 specified in the definition.>It looks like a typo to me.>Well, it wasnÍt. I carefully checked all the conditions, and forgotabout ts === defining a finite field> Z_3 with addition as usual, and ts multiplication table: 0 1 2> --------> 0 |0 0 0 > 1 |0 2 1> 2 |0 1 2fits all the requirementsYour 2 looks very much like an identiy.-- Daniel W. sonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/039 === and others> I am just curious: why are people like you> interested in even acknowleding and s ilk ?> Ts guy has been oopsing on sci.math for ages (at least several> years) I see. Is it not worthwle for everyone to just ignore m at ts point ? > There is real danger of course that someone would take m seriously,> but I feel thats remote.It seems that he has taken to calling universities to complain. > Lets not giving m any more attention. Without attention, he will> shrivel up and disappear.Is there a story or an obvious reason that I am missing ?I made ts same point a wle ago. I tnk you are right. For thosewho feel a need to reply to JSH keep innocent readers from being takenin, I suggest simply replying to everytng he posts with Ts isbullst! If he wants to waste s own time writing long longpostings of incorrect mathematics, professional mathematiciansshouldnÍt have to waste their time replying in detail. On the otherhand, maybe people are compassionate, taking pity on m, and tryingto set m straight in the hopes that someday he might actually dosometng === Ullrich and others Oh yeah, in case youÍre wondering about what Ullrich finds so> offensive that the subject of racial slur came to s mind, I said> that heÍd acted as my lapdog in an instance.> That was YEARS ago.It was three days ago:http://mathquest.com/discuss/sci.math/a/m/580751/580751Try Google withmoron factorization bullstand guess who comes === Google with> moron factorization bullst> and guess who comes R Ramsden (jr@adslate.com)---Eternity is a long time, === to Prof Ullrich and others > I am just curious: why are people like you > interested in even acknowleding and s ilk ? > Ts guy has been oopsing on sci.math for ages (at least several > years) I see. come from the discussions. I have learned quite a bit myself. Alsointeresting results come from it, like the Magidin-McKinnon theorem.One of the tngs I learned was how to deal properly with quadraticfields. > Is there a story or an obvious reason that I am missing ?There is a story indeed. Even for mathematicians it becomesincreasingly difficult to refute JamesÍ arguments because of theway he veils them.-- === interested in even acknowleding and s ilk ?> Ts guy has been oopsing on sci.math for ages (at least several> years) I see. >come from the discussions. I have learned quite a bit myself. Also>interesting results come from it, like the Magidin-McKinnon theorem.>One of the tngs I learned was how to deal properly with quadratic>fields.Hear, hear! As a complete mathematics novice, I have learned enoughmaths from reading detailed and lengthy expositions to JSH that I ameven dabbling with the idea of investigating some maths myself. If theestimable persons the OP addressed had not acknowledged JSH with theirposts, even my (tiny) knowledge would have remained unadded to, and Icould have made a complete and utter pillock of myself quoting m to mymathematical friends (including a teacher/graduate). As it is, I canhave interesting conversations with them and, sticking with sci.mathbecause of JamesÍ lunacy, I can even explain the ïwaiters puzzleÍ topeople (a minuscule victory, but mine own).-- MinSo where === Hammick> 1) int[-inf to inf] cos x^2 dx = sqrt(pi/2)> 2) int[-inf to inf] exp(-x^2) dx = sqrt(pi)> Eqn (2) is easy to show by looking at the square of the left side> (a double integral) and switcng to polar coordinates. I tried the> same approach with (1). ... everyone for the replies. I can use several of those ideas. (IÍmghlights from advanced calculus, === integrals>References regarding FresnelÍs Integralssee Articles 560, 1166, 1169,1172,1323 in the followingnice books : (Library incomplete references are : Cauchy [Comptes Rendus, tom XV, 534,573] Fresnel [Oeuvres, tom I] Gilbert [Memoires couronnes de lÍAcad.de Bruxelles, t.XXXI, 1863] Knockenhauer [,,Die Undulationstheorie des Lichts,...?] Preston[,,Theory of Light , Art.141,...?] Verdet [Oeuvres tom V] = Perhaps help,Alex === mistakes in my corrections. I hate it when I do that.now hereÍs original problems thatÍs been really corrected. Torrey loves Matt on some days, but Torrey hates Matt on other days. Torrey loves Paul on some days, but Torrey hates Paul on other days. Torrey loves Sandy on some days, but Torrey never hates Sandy. Sandy hates Torrey on some days, but Sandy loves Torrey on otherdays. Sandy loves Gary on some days, but Sandy never hates Gary. Matt loves Sandy on some days, but Matt hates Sandy on other days Paul Loves Matt on some days, but Paul hates Matt other days. Paul hates Sandy on some days, but Paul never loves Sandy. Gary hates Sandy on somedays, but Gary never loves Sandy. Gary Loves Torrey on some days, but Gary hates Torrey on other days. I have previously reduced the problem to just hate and loverelationsp. is there anytng wrong with doing that? because IÍmignoring the time/day information. what about the time/love/haterelationsps?is there a way to figure out what pattern they have? with theinformation above, taking only the love and hate relationsps, thereare 32 possibles. what happens if i take into consideration thedays? or is there a way to do that? what other information will oneneed in order to do that? I tnk gary and paul should be nicer to sandy. all in advance. === right. Using rule 9, it is easy to show that everyelement has a multiplicative inverse. Fix a, then a*b must bedifferent for all b (if a*b=a*c, then a*(b-c)=0 violating rule 9 if band c different). Therefore the set {a*b} must be a permutation of{b} (thatÍs where finite comes in), so that there is an element d sothat a*d=1. The same idea works for left inverse Services---------- ** SPEED ** RETENTION ** COMPLETION ** === Big Rip> http://qedcorp.com/destiny/> under constructionDARK MATTERS SURROUND DARK ENERGYwith my comments, additions and physics correctionsTwo big stories from the world of physics may portend the arrival of > new weapons of mass destruction far more powerful and compact than > atomic bombs. In recent years it has been discovered that our universe > is being blown apart by a mysterious anti-gravity effect called dark > energy. Mainstream physicists are scrambling to explain ts mysterious > acceleration in the expansion of the universe. Some physicists even > believe that the expansion will lead to The Big Rip when all of the > matter in the universe is torn asunder - from clusters of galaxies in > appears to be made of two unknowns - roughly 23% is dark matter, an > invisible source of gravity, and roughly 73% is dark energy, an > invisible anti-gravity force. Ordinary matter constitutes perhaps 4 > percent of the universe. Recently the British science news journal New > Scientist revealed that the American military is pursuing new types of > exotic bombs - including a new class of isomeric gamma ray weapons.That was an original idea of mine in 1963 at Cornell and I discussed it > with Hans Bethe. That is one of the reasons Ron Bullough invited me to > Harwell in 1966. No doubt others thought of it but probably later. I > thought of it wle at Tech/Ops in Lexington, Mass working for > Parrant Jr.Unlike conventional atomic and hydrogen bombs, the new weapons would > trigger the release energy by absorbing radiation, and respond by > re-emitting a far more powerful radiation. In ts new category of > gamma-ray weapons, a nuclear isomer absorbs x-rays and re-emits gher > frequency gamma rays. The emitted gamma radiation has been reported to > release 60 times the energy of the x-rays that trigger the effect.Gamma-ray weapons could trigger next arms race 19:00 13 August 03> Exclusive from New Scientist Print Edition. Subscribe and get 4 free > issues.An exotic kind of nuclear explosive being developed by the US Department > of Defense could blur the critical distinction between conventional and > nuclear weapons. The work has also raised fears that weapons based on > ts technology could trigger the next arms race. The explosive works by > stimulating the release of energy from the nuclei of certain elements > but does not involve nuclear fission or fusion. The energy, emitted as > gamma radiation, is thousands of times greater than that from > conventional chemical explosives.> The technology has already been included in the Department of DefenseÍs > Militarily Critical Technologies List, wch says: Such extraordinary > energy density has the potential to revolutionise all aspects of > warfare. Scientists have known for many years that the nuclei of some > elements, such as hafnium, can exist in a gh-energy state, or nuclear > isomer, that slowly decays to a low-energy state by emitting gamma rays. > For example, hafnium-178m2, the excited, isomeric form of hafnium-178, > has a half-life of 31 years. The possibility that ts process could be > explosive was discovered when Carl Collins and colleagues at the > University of Texas at Dallas demonstrated that they could artificially > trigger the decay of the hafnium isomer by bombarding it with low-energy > experiment > released 60 times as much energy as was put in, and in theory a much > greater energy release could be aceved.> http://www.newscientist.com/news/news.jsp?id=ns99994049I was tnking in 1963 of a gamma ray laser pumping a nuclear isomeric > transition. Bethe at the time said it wouldnÍt work and basically > discouraged me working on it.Bekkum continued:In the summer of 2000 I contacted Nick Cook, the former aviation editor > and aerospace consultant to JaneÍs Defence Weekly, the international > military affairs journal. Cook had been investigating black budget > super-secret research into exotic physics for advanced propulsion > technologies.Uh Oh :)I had been monitoring electronic discussions between various American > and Russian scientists theorizing about rectifying the quantum vacuum > for advanced space drive. Several groups of scientists, partitioned > into various research organizations, were exploring what NASA calls > Breakthrough Propulsion Physics - exotic technologies for advanced > space travel to traverse the vast distances between stars. Partly > inspired by the pulp science fiction stories of their youth, and partly > by recent reports of multiple radar tracking tapes of unidentified > objects performing impossible maneuvers in the sky, these scientists > were on a quest to uncover the most likely new physics for star travel. > The NASA program was run by Marc Millis, under the Advanced Space > Transportation Program Office (ASTP). Joe Firmage, a Silicon Valley > entrepreneur, who at age 28 had found risen to CEO of a three billion > dollar internet firm, began to fund research in parallel with NASA. He > red a NASA Ames nano-technology scientist, Creon Levit, to run the > International Space Sciences Organization,Joe did that because I suggested it. I introduced Creon to Joe.Cook was intrigued by the apparent connections between various private > investors, defense contractors, NASA, INSCOM (American military > intelligence), and the CIA. Wle researcng exotic propulsion > technologies Cook had heard rumors of a new kind of weapon, a > sub-quantum atomic bomb, being wspered about in the dark halls of > defense research.I tnk that must have come from me regarding J. P. VigierÍs tight > atomic states with experiments in Beograd by Z. Maric and G. Dragic. > But how did Cook hear about that? We brought Vigier to ISSO in San > Francisco several times along with physicist Gennady Spov from Moscow. > That story with photographs of Vigier and the group is in my > autobiography Destiny Matrix. Dragic A, Maric Z, Vigier JP; Phys. > Lett. A 265 (2000) 163. New quantum mechanical tight bound states and > ïcold fusionÍ. Creon Levit and Vigier met with Maric in Budapest in 2000.Bekkum who is one of my on line students continued:Sub-quantum physics is a controversial re-interpretation of quantum > theory, based on so-called pilot wave theories, where an information > that the predictions of ordinary quantum mechanics could be recast into > a pilot wave information theory. Recently Anthony Valentini of the > Perimeter Institute has suggested that ordinary quantum theory may be a > special case of pilot wave theories, leaving open the possibility of new > and exotic non-quantum technologies. Even thought rumors of a > sub-quantum bomb may be purely fantasy ItÍs not fantasy. It might not work, Maric and Dragic in Beograd, wle > not ostensibly trying to make a weapon by any means, were trying to test > VigierÍs basic theory of the spatially extended electron, wch I tnk > is basically a correct idea and fits my own ideas including why the > electron appears to shrink to less than 10-16 cm under gh resolution > imaging (i.e. scattering) and how the electric charge distribution is > contained by the strongly attractive zero point energy exotic vacuum > dark matter core (Abraham-Becker-Lorentz-Poincare stress problem of > 100 years ago). Ts only works in the Bohm dden or extra variable > interpretation. That is, a classical spatially extended electric charge > distribution is unstable. It explodes under its own self-repulsion. Ts > is why physicists had to postulate a point electron because they did not > understand that the strong gravity attraction of the positive zero point > pressure in a possible state of exotic vacuum would hold the charge > together. As Herbert Frohlich told me at UCSD in La Jolla in 1966 the > basic tng wrong with physics is the idea of the point electron. The > bad idea of the point electron gives the infinite energy in quantum > electrodynamics. Richard Feynman told me in s office at Cal Tech in > 1968 that infinite renormalization is a shell game, and it is a scandal > in physics that no one could do better than what he had done. They did > not know 100 years ago that 1/3 or so of the universe was ts kind of > exotic vacuum. For example, there is a huge sphere of exotic vacuum of > w = -1 positive pressure that holds our galaxy together preventing our > solar system from going off into space on its own. Ts sphere looks > like w = 0 cold dark matter from our vantage point. What works on ts > large scale also works on the small scale of the single electron (and > all the charged lepto-quarks). A neutrino has some mass and is simply a > micro-geon of pure zero point energy with positive pressure. there is no question that physicists seriously contemplate a phase > transition in the quantum vacuum as a real possibility. The quantum > vacuum defies common sense, because empty space in quantum field theory > appear and disappear far too quickly to be detected directly, but their > existence has been confirmed by experiments that demonstrate their > in§uence on ordinary matter.A major component of the physical quantum vacuum consists of virtual > electrons frotng and bubbling at the Fermi surface edge of the Dirac > negative energy sea. Ts is because of the Pauli exclusion principle > that only none or one electron per quantum state. A virtual electron > pops out of the vacuums Fermi surface leaving a hole bend. The hole > is the virtual positron. The result is a virtual electron-positron > pair. However, the virtual electron and the virtual positron attract > because they have opposite charges and they are exchanging virtual > photons. Therefore, some of them form a more stable bound state. An > enormous number of these virtual pairs Bose-Einstein condense into the > same center of mass quantum wave packet forming the Vacuum Coherence > Field (AKA In§ation Field). Ts is a dynamic steady state of > detailed balance in wch there is a continual in§ow and out§ow of > virtual pairs into and out of ts giant quantum or macro-quantum > super§uid. Essentially ts is a vacuum phase transition, similar to > the BCS transition from a normal metal to an electrical superconductor, > from the globally §at micro-quantum electrodynamic vacuum without any > gravity at all to the curved macro-quantum electrodynamic vacuum with > emergent gravity. Einsteins field equation of general relativity can > be derived from the phase wiggles and ripples in the robust stable > macroscopically occupied center of mass quantum wave packet of the bound > state of the virtual electron-positron pair. The exotic vacuum dark > energy and dark matter are simply the amplitude wiggles and ripples of > ts same virtual pair quantum wave packet. The wave packet spreads > over the entire 3D space of the post-in§ationary bubble on wch our > Hubble-horizoned universe is located along with an infinity of parallel > American. If the world hologram idea is correct, take the surface area > of the expanding Hubble sphere that is the causal retarded boundary of > 3D space of our past light cone at Earth and divide it by the quantum of > area. That gives us the number of Bekenstein-Shannon c-bits and > explains the arrow of time (AKA Second Law of Thermodynamics) of > increasing thermodynamic entropy in terms of the dynamical expansion of > the 3D space of the universe. Lenny Susskind calls ts DeSitter Space.Such research should be forbidden!Too late. Pandoras Box is open. Schrodingers Cat has jumped out of it.In the early 1970Ís Soviet physicists were concerned that the vacuum of > our universe was in fact only one possible state of empty space. The > fundamental state of empty space is called the true vacuum. Our > universe was considered to reside in a false vacuum, protected from > the true vacuum by the wall of our world. A change from one vacuum > state to another is known as a phase transition. Ts is analogous to > the transition between frozen and liquid water. Lev Okun, a Russian > physicist and storian recalls Andrei Sakharov, the father of the > Soviet hydrogen bomb, expressing s concern about research into the > phase transitions of the vacuum. If the wall between the vacuum states > was to be breached, calculations showed that an unstoppable expanding > bubble would continue to grow until it destroyed our entire universe! > Sakharov declared Such research should be forbidden! since there was > always the possibility that an experiment might accidentally trigger a > vacuum phase transition.British Astronomer Royal, Sir Rees, Master of Trinity College, > and Director of the Cambridge University Institute of Theoretical > Astronomy on Madingley Road where Stephen Hawking works discusses all > ts in Chapter 9 of s important book Our Final Hour.Could the wall of our universe be breached from witn? The amount of > energy required to punch a hole through the wall appeared to be > enormous, and no known natural physical phenomena, even the most > energetic, had punched through either. A recent report commissioned to > examine potential dangers at the Large Hadron Collider, one of the next > the best of our existing knowledge. Others are not so certain, however. > At least one of the Russian physicists I had corresponded with was said > to have been a former associate of Andrei Sakharov. He strongly nted > at new theories the Russians had developed wch allow for the > manipulation of the fundamental constants of nature, but he never > revealed more than a sketch of s ideas. He claimed that a breakthrough > was witn reach, perhaps witn five years Who was that? Not Ryazanov?Recent theoretical explorations may suggest another approach to the > physics of the vacuum. The invisible gravitating dark matter could be > the other side of the invisible dark energy coin, and that suggests the > possibility of manipulating the vacuum for energy release.Now ts is my original idea that you got from our communications over > the past few years. I am the only physicist in the world today, as far > as I know who has suggested ts and has already published it in my two > books of 2002 so its in the official record at the Library of Congress.If a controllable parameter could be found to mediate the balance > between the invisible dark forces, the result would unleash the vacuum > energy of creation in all of its awful power and majesty. If it were > possible to control the dark sides of the force then spacetime, the > arena where everytng we know takes place, could be bent and twisted > with infinitely greater ease than was ever suspected. Ts would open > PandoraÍs box to everytng from vacuum energy weapons of mass > destruction (capable of destroying the universe!) to spacetime warp > drives and time macnes.Exactly, the above is the thesis of all my books since 2002 at least.A quick survey of the international electronic arcve of physics > papers at www.arXiv.org shows that research into the vacuum of spacetime > for energy production is alive and well.I do not tnk that is true. You need to cite specifics here. There are > lots of §akey new age papers on free energy on the Web written by > people without any real credentials but they are not on www.arXiv.org > wch is not even allowing competent fringe papers in controversial > topics like cold fusion. So what exactly are you tnking of here? > Indeed, Paul Ginsparg, who controls the arcve, does not even allow > Carlos Castro to publish conservative competent papers on Clifford > Algebras wch are not fringe at all!Most authors are independent researchers struggling with limited > funding and resources, yet their theoretical results suggest that > somewhere in Nick CookÍs black world, a major breakthrough has already > taken place. Most likely the United States and Russia are in the lead, > but Cna, France, Ukraine, Iran, India and Saudi Arabia all have > scientists actively pursuing the fundamental physics that determine the > fabric of our reality, and are seeking the theory and the means to > access the enormous energies locked inside of the vacuum since the > creation of the universe. Even if the black budget world has yet to > unleash the enormous potential of vacuum energy, there are signs that > those in power may have begun to take notice. Dr. Harold Puthoff, a > scientist with strong government connections, who has previously worked > on classified projects for the CIA, is a major proponent of vacuum > energy physics. Nick CookÍs book, The Hunt for Zero Point, and s > recent stories on zero point energy in JaneÍs Defence Weekly have also > brought attention to the dangers and military potential of vacuum > research. The American intelligence community financed so-called psycc > spies for over twenty years and through four presidential > administrations. It is ghly unlikely that they would ignore the > potential of the quantum vacuum. Dr. Chapline, of the Lawrence > Livermore National Laboratory, and Dr. Jack Sarfatti in San Francisco, > knew each other in the sixties in La Jolla, have independently proposed > that the quantum vacuum may unstable to the formation of coherent > virtual processes. Sarfatti suggests that gravity is an emergent > property determined by the physics of the vacuum. s idea is to find a > means of directly interacting with the physics of the vacuum that > controls the shape of spacetime. Such a possibility would be consistent > with the reported success of Evgeny> Podkletnov, the Russian scientist who is experimenting with spinning > superconducting disks. PodkletnovÍs most recent papers report the > appearance of a mysterious coherent beam of gravity like radiation > with a measured force of 1000 G. In an interview on BBC radio, Nick > Cook pointed out one immediate application of the Podkletnov beam - the > destruction of missiles and satellites in §ight or in orbit around the > earth. Cook showed the BBC internal documents from Boeing, the American > aerospace contractor, proving their interest in PodkletnovÍs research.Ts beam stuff I am suspicious of. Of course if the experiment is good, > I have to tnk more about it. I am not so sure if Podkletnovs > experiment is any good and has been replicated.The connections between PodkletnovÍs results, and the kind of vacuum > research explored by Sarfatti, beginning in 1999 at the International > Space Sciences Organization are the latest threads in a trail that most > likely originates in cold war disinformation, a game played by East and > West against each other. Glasnost has sfted the balance of > partnersps and the positions of the players, but not the stakes of an > outcome that would leave the world with even more prolific and powerful > weapons of mass destruction.That is true, as shown in Destiny Matrix, however you leave out the > most important evidence -UFOs!whole business are the connections. Although Nick Cook never revealed > the identity of s deep throat contact called Dr. Dan Marckus in > the book The Hunt for Zero Point, there was no question that the > Podkletnov results had played a major part in fitting together the > pieces of the puzzle. The amount of interest was in PodkletnovÍs > reports by NASA, Boeing, and others in the international arena of > aerospace and military research communities was evidence that there was > more here to explore than the latest musings of the intellectual elite. > The truth is that a fundamental theory of gravity at the scales of > subatomic nuclear physics does not exist. The fact is that no one > understands the nature of the gravitational field at very small scales. > In fact gravity has barely been probed much below one millimeter. Every > attempt to unify the physical theories of gravity with the well-known > standard model physics of electromagnetism, and the strong and weak > nuclear forces, has failed. More importantly there has been recent > progress in the exotic areas of mainstream research, such as superstring > theory, wch suggest new kinds of physics, wch might support > explanations for PodkletnovÍs impulse gravity effect. One of the > current fads in theoretical physics involves large extra dimensions of > space that allow a much stronger version of gravity to leak off the > membrane world of our ordinary three dimensions. The large dimensional > picture allows for the well known forces of electromagnetism, and the > strong and weak nuclear forces, to be confined to a three dimensional > brane-world §oating in a gher dimensional space. Gravitons, the > are able to slip off of our brane-world, wch explains why the > gravitational force is so much weaker than the other forces that hold > matter together. Gravitons could be exchanged between our brane-world > and another brane §oating nearby in the same gher dimensions.The Sarfatti picture offers a more direct interaction with the new > physics than the brane world ideas. SarfattiÍs vision is to find a > means of using electromagnetic fields in the Josephson effect to couple > to the virtual electron-positron pair giant coherent condensate > in§ation field inside the vacuum that controls the shape of spacetime > to the real electron pair giant coherent condensate of a control gh > temperature superconductor. UCBs Ray Cao has a similar idea using a > superconductor to transduce electromagnetic far field waves to gravity > waves with gh efficiency conversion. Sarfatti wants to do the same > tng with non-propagating electromagnetic and gravity near fields. One > wonders if the black budget world may have already produced some of the > technology needed to explore and test these new realms.Not a chance. They are clueless about the theory. They are still stuck > in HalÍs PV model and Bernie HaischÍs zero point ideas, wch will > never, in my opinion §y. They are not asking the right questions and do > not have the right idea in their minds. It is my belief, until I see > evidence to the contrary, that I am the only physicist on the planet > today who is doing real theoretical work directly relevant to the > acevement of practical metric engineering anchored in the now observed > reality of dark energy. All my work is public. I would love to be > proved wrong on ts especially by Hal Puthoff, but I am not holding my > breath.;-) Extraordinary claims require extraordinary proof. Everytng > else I have seen is either on the wrong track asking the wrong questions > like the work of Puthoff, Haisch, Ibison & Rueda for example, wch at > least is real physics that has proved itself wrong in IbisonÍs PV > cosmology paper, or else the claims are patently obvious nonsense that > Feynman called Cargo Cult Science. There is also the Russian torsion > work of Akimov and Spov and I am not prepared to make a definitive > statement on that, as the issue is not simple because of several factors > some political. One must be careful there to separate SpovÍs > theoretical work from claims made about practical devices including > weapons applications. I do, however, agree with you that there is a real > issue here as defined in Ch. 9 of ReesÍs Our Final Hour.Hal Puthoff coined the term metric engineering for §ying saucer > technology. Hal has been working on ts problem for many decades and > has held gh USG security clearances and has been privy to reliable > information that the saucers are real and are alien. Otherwise he > would not be working on the problem. However, HalÍs theories, both of > the zero point energy and of the gravity field will not solve the > problem because they are too naively based and do not ask the right > questions. The basic physics required for ts task is way beyond the > depth of Puthoffs self-described engineering approach and can be > found in RovelliÍs new book on quantum gravity.> Hey, I just started reading your stuff, and I find it all very interesting indeed. for posting here.talk about here, from the New York Times, written so the average joe could also enjoy some of these ideas, for anyone who might interested:New === Re: Infinite Galois Groups>> Let K be the field generated by the square roots of the prime numbers.>> What is G(K/Q)? What is G(F/K)?> Ts is the increasing union ofQ < Q(sqrt(2)) < Q(sqrt(2),sqrt(3)) < Q(sqrt(2),sqrt(3),sqrt(5)) < ...In each case, the relative Galois group is Z/2Z, and the Galois group> of the extension Q(sqrt(2),...,sqrt(pn)) over Q> where pn is the n-th prime is equal to (Z/2Z)^n (you would need to> prove ts; see for example Section 6.7 in LangÍs Algebra, 3rd edition).> The maps going down are ... --> Z_2 x Z_2 x Z_2 --> Z_2 x Z_2 --> Z_2 --> 1where each map chops off the last coordinate. The inverse limit is an> infinite product of copies of Z_2.Of course here Galois theory provides the most elegant proofs, butit is worth mentioning that elementary proofs are available, e.g.THEOREM Let Q be a field with 2 != 0 and L = Q(S) be an extension of Qgenerated by n square roots S = {:/a, :/b, ...} of elts a,b,...in Q.If every nonempty subset of S has product not in Q then each successive adjunction Q(:/a), Q(:/a,:/b),...doubles degree, so totally [L:Q] = 2^nHence all of the 2^n such subset products comprise a basis of L over Q.PROOF: by induction on the tower height n = number of root adjunctions.Lemma below => [1,:/a] [1,:/b] = [1,:/a,:/b,:/ab] is a Q-vector spacebasis of Q(:/a,:/b) if (and only if) 1 is the only basis element in Q.Our job is: lift ts to n > 2: [1,:/a] [1,:/b] [1,:/c] ... (2^n elts)n = 1: L = Q(:/a) so [L:Q] = 2 since :/a not in Q by hypothesis.n > 1: L = K(:/a,:/b), K of height n-2. By induction [K:Q] = 2^(n-2)so we need only show [L:K] = 4 since then [L:Q] = [L:K][K:Q] = 4 2^(n-2).The lemma below shows [L:K] = 4 if r = :/a, :/b, :/ab all arenÍt in K,true since induction on K(r) of height n-1 yields [K(r):K] = 2. QEDLEMMA [K(:/a,:/b) : K] = 4 if :/a,:/b,:/ab all arenÍt in K, & 2 != 0.Proof: Let L = K(:/b). [L:K] = 2 via :/b not in K, thus it issufficient to prove [L(:/a):L] = 2. It fails only if :/a in L = K(:/b)and then :/a = r + s :/b for r,s in K; but thatÍs not possible sinceabove^2 -> a = rr + ss b + 2rs :/b contradicts hypotheses as follows:rs != 0 => :/b is in K via solving above for :/b, using 2 != 0 s = 0 => :/a is in K via :/a = r in K r = 0 => :/ab is in K via :/a = s :/b, multiplied by :/b. QEDIn the classical case Q is the field of rationals and the square rootshave radicands being distinct primes. Here it is quite familiar that a product of any nonempty subset of them is irrational since, over a UFD,a product of coprime elements is a square iff each factor is a square(mod units). So the classical case satisfies the theoremÍs hypothesesElementary proofs like that above are often credited to Besicovitch (see below). But I havenÍt seen s paper so I canÍt say for sure whether or not s proof proceeds in a similar way as that above. If a reader has s proof handy, IÍd be very grateful if they couldsummarize it here. Finally, see below for some stronger results.-Bill Dubuque------2,33f 10.0X Besicovitch, A. S. On the linear independence of fractional powers of integers.J. London Math. Soc. 15, (1940). 3--6.------Let a_i = b_i p_i, i=1,...,s , where the p_i are s different primes andthe b_i positive integers not divisible by any of them. The author proves by an inductive argument that, if x_j are positive real roots of x^{n_j} - a_j = 0, j=1,...,s , and P(x_1,...,x_s) is a polynomial withrational coefficients and of degree not greater than n_j - 1 with respect to x_j, then P(x_1,...,x_s) can vanish only if all its coefficients vanish. Reviewed by W. Feller------15,404e 10.0XMordell, L. J.On the linear independence of algebraic numbers.Pacific J. Math. 3, (1953). 625--630.------Let K be an algebraic number field and x_1,...,x_s roots of the equationsx_i^n_i = a_i (i=1,2,...,s) and suppose that (1) K and all x_i are real, or(2) K includes all the n_i th roots of unity, i.e. K(x_i) is a Kummer field.The following theorem is proved. A polynomial P(x_1,...,x_s) with coefficientsin K and of degrees in x_i , less than n_i for i=1,2,...,s , can vanish only ifall its coefficients vanish, provided that the algebraic number field K is suchthat there exists no relation of the form x_1^m_1 x_2^m_2 ... x_s^m_s = a,where a is a number in K unless m_i = 0 mod n_i (i=1,2,...,s) . When K is ofthe second type, the theorem was proved earlier by Hasse [Klassenkorpertheorie,Marburg, 1933, pp. 187--195] by help of Galois groups. When K is of the firsttype and K also the rational number field and the a_i integers, the theorem wasproved by Besicovitch in an elementary way. The author here uses a proofanalogous to that used by Besicovitch [J. London Math. Soc. 15b, 3--6 (1940);these Rev. 2, 33]. Reviewed by H. Bergstrom------46 #1760 12A99Siegel, Carl LudwigAlgebraische Abhaengigkeit von Wurzeln. (German)Acta Arith. 21 (1972), 59--64.------Two nonzero real numbers are said to be equivalent with respect to a realfield R if their ratio belongs to R . Each real number r != 0 determinesa class [r] under ts equivalence relation, and these classes form amultiplicative abelian group G with identity element [1]. If r_1,...,r_hare nonzero real numbers such that r_i^n_i in R for some positive integers n_i(i=1,...,h) , denote by G(r_1,...,r_h) = G_h the subgroup of G generated by[r_1],...,[r_h] and by R(r_1,...,r_h) = R_h the algebraic extension field ofR = R_0 obtained by the adjunction of r_1,...,r_h . The central problemconsidered in ts paper is to determine the degree and find a basis of R_hover R . Special cases of ts problem have been considered earlier by A. S.Besicovitch [J. London Math. Soc. 15 (1940), 3--6; MR 2, 33] and by L. J.Mordell [Pacific J. Math. 3 (1953), 625--630; MR 15, 404]. The principalresult of ts paper is the following theorem: the degree of R_h with respectto R_{h-1} is equal to the index j of G_{h-1} in G_h , and the powers r_i^t(t=0,1,...,j-1) form a basis of R_h over R_{h-1} . Several interestingapplications and examples of ts === Infinite Galois Groups Adjunct Assistant Professor at the University of Montana.>>Let F be the field of lengths constuctible by compass and>>straighthedge. What is G(F/Q)?Well, formally, it is the inverse limit of the Galois Groups of the>> finite Galois subextensions, with the connecting morpsms being the>> corresponding quotient maps.It should be fairly complicated, since it will include the>> subextensions given by the 2^n-th and 3^n-th roots of all integers.>I tnk I am missing sometng here. Why the 3^n-th roots? It is not>possible to construct a segment the length of the cube root of a given>segment with compasses and straightedge.It should be 2^n-th roots; donÍt know where the very selective about what I accept as reality. --- Calvin === Groups>Let F be the field of lengths constuctible by compass and>straighthedge. What is G(F/Q)?Well, formally, it is the inverse limit of the Galois Groups of the> finite Galois subextensions, with the connecting morpsms being the> corresponding quotient maps.It should be fairly complicated, since it will include the> subextensions given by the 2^n-th and 3^n-th roots of all integers.cube root of 3 is === of 3 is constructible?Not with ruler and compass. Any extension field constructed by adjunction of quantities that can be made of ruler and compass either is the same as the previous field or of degree 2 over it. In the end one can only get to fields of degree 2^n over the ground field.That is why it is impossible to trisect the angle with ruler and compass. To do so whould require getting the root of an irreducible cubic === Infinite Galois Groups||>Let F be the field of lengths constuctible by compass and|>straighthedge. What is G(F/Q)?|> |> Well, formally, it is the inverse limit of the Galois Groups of the|> finite Galois subextensions, with the connecting morpsms being the|> corresponding quotient maps.|> |> It should be fairly complicated, since it will include the|> subextensions given by the 2^n-th and 3^n-th roots of all integers.||cube root of 3 is constructible?i guess so. also, itÍs nice to know that you can duplicate the cube(cube root of 2) and trisect a 60-degree angle (9th root of -1).-- === anticlassicalist }{ vi: into the quantum term to exist to mean anytng at all, there must be> sometng that doesnÍt exist.That what doesnÍt exists, exists as a concept.The non-existence does not exist, only if I donÍt know it.But as I know about non-existence, it exists.> It does not matter how the one is distinguished from the other or who> is doing the distinguisng; it does not matter where the line is> drawn between that wch exists and that wch doesnÍt exist, how it> is drawn, who is drawing it, whether it is sharp or fuzzy.> What matters is that for the phrase X exists to mean anytng at> all, there must be some entity Y for wch it is valid to state Y> does not exist.All exists and that what doesnÍt ainÍt.> Otherwise the word exists in the phrase X exists would express> notng whatsoever.Take a deep breath.> Or (in light of the above): ontology is about that wch is> distinguished from that wch does not exist.Ontology imagines reality knowledge.> For a distinction to take place requires that there is more than> possible outcome.Would you explain in view ofmembersp in the universal class?> Q: What is red and invisible?> A: No tomatoes.> Q: But what if theyÍre green and invisible?> A: Then they arenÍt ripe yet.What if IÍm not invited to lunch, do I still enjoythe existence sfted, tossed salad?Riddle of the day: are the unconceived existence === solutionMy differential equations book asks the following:Show that -2x^2y + y^2 = 1 is a solution of the differential equation(in differential form) 2xy dx + (x^2 - y) dy = 0. Ok, thatÍs trivial,one just implicitly differentiates the first equation, etc.They also ask to find at least one explicit solution. Now hereÍs myproblem: What explicit equation satisfies both the equations above?IÍve graphed the implicit solution (first equ above) and get whatlooks like almost y = 2x^2 + 1, and an inverted witch of agnesi with y= -1/((3/2)x^2 + 1). Unfortunately, these are not exact fits.I imagine that there is in fact such an explicit solution to both (onsome interval of definition) and that perhaps there might be a methodto derive it.Any help is greatly appreciated. in === solution> My differential equations book asks the following:Show that -2x^2y + y^2 = 1 is a solution of the differential equation> (in differential form) 2xy dx + (x^2 - y) dy = 0. Ok, thatÍs trivial,> one just implicitly differentiates the first equation, etc.They also ask to find at least one explicit solution. Now hereÍs my> problem: What explicit equation satisfies both the equations above?both? solve the quadratic for y, i get y = x^2 (+/-) sqrt(x^4 + 1).check wch branch is your solution (if any).i have found an even nicer ones -- y = 0, 2x^2! my trick was to notethe right hand side 1 can be replaced by any constant (letÍs see --zero!) and still the implicit differentiation works. of course, it isan explicit solution of the differential equation, not both. i donÍttnk the question wanted both.IÍve graphed the implicit solution (first equ above) and get what> looks like almost y = 2x^2 + 1, and an inverted witch of agnesi with y> = -1/((3/2)x^2 + 1). Unfortunately, these are not exact fits.I imagine that there is in fact such an explicit solution to both (on> some interval of definition) and that perhaps there might be a method> to derive it.too simple a method perhaps.Any help is greatly appreciated. in advance,> === solution>> My differential equations book asks the following:Show that -2x^2y + y^2 = 1 is a solution of the differential equation>> (in differential form) 2xy dx + (x^2 - y) dy = 0. Ok, thatÍs trivial,>> one just implicitly differentiates the first equation, etc.>both? solve the quadratic for y, i get y = x^2 (+/-) sqrt(x^4 + 1).>check wch branch is your solution (if any).>i have found an even nicer ones -- y = 0, 2x^2!Ts is hardly surprising given they are trivial linear combinationsof the two general solutions.-- IÍm not interested in mathematics that might have anytngto do with === Differential equation with implicit solution> My differential equations book asks the following:Show that -2x^2y + y^2 = 1 is a solution of the differential equation> (in differential form) 2xy dx + (x^2 - y) dy = 0. Ok, thatÍs trivial,> one just implicitly differentiates the first equation, etc.They also ask to find at least one explicit solution. Now hereÍs my> problem: What explicit equation satisfies both the equations above?If x = 0 then the DE becomes d/dy(y^2) = 0, wch has a generalsolution y^2 = C. That could be what theyÍre looking for. Butts solution, and the one they gave, are both special casesof the general solution, wch is: y^2 - 2.x^2.y = CIf x != 0, take y = x^2.z in the DE, so dy/dx = x^2.dz/dx + 2xzand the DE becomes, after some manipulation: x.(z - 1)/2.dz/dx + (z - 1)^2 = 1With t = (z - 1)^2 ts becomes: dt/dx + (4/x).t = 4/xMultiplying by x^4 then gives: d/dx(t.x^4) = 4.x^3i.e. t.x^4 = x^4 + Cand the is a long time, especially towards the end. Woody === solution> My differential equations book asks the following:Show that -2x^2y + y^2 = 1 is a solution of the differential equation> (in differential form) 2xy dx + (x^2 - y) dy = 0. Ok, thatÍs trivial,> one just implicitly differentiates the first equation, etc.They also ask to find at least one explicit solution. Now hereÍs my> problem: What explicit equation satisfies both the equations above?IÍve graphed the implicit solution (first equ above) and get what> looks like almost y = 2x^2 + 1, and an inverted witch of agnesi with y> = -1/((3/2)x^2 + 1). Unfortunately, these are not exact fits.I imagine that there is in fact such an explicit solution to both (on> some interval of definition) and that perhaps there might be a method> to derive it.Any help is greatly appreciated. in advance,> RSolving -2x^2y + y^2 = 1 gives two equations for x in terms of y both of wch satisfy -2x^2y + y^2 = 1 and 2xy dx + (x^2 - y) dy = 0.Similarly solving -2x^2y + y^2 = 1 for y in terms of x, gives two === Re: Differential equation with implicit solution> My differential equations book asks the following:Show that -2x^2y + y^2 = 1 is a solution of the differential equation> (in differential form) 2xy dx + (x^2 - y) dy = 0. Ok, thatÍs trivial,> one just implicitly differentiates the first equation, etc.They also ask to find at least one explicit solution. Now hereÍs my> problem: What explicit equation satisfies both the equations above?Maybe that IÍve misunderstood your question, but it seems to me thatall that you are supposed to do is to solve the equation-2x^2*y + y^2 = 1in y. YouÍll get two solutions: y = x^2 + sqrt(x^4 + 1) andy = === trouble. I am working on integrate sqrt ((4x^2) +9)/(x^4) using trigsubstitution.I have: u=2xa=32x/3 = tan thetadx=3/2 sec^2 theta d theta(sqrt (4x^2 + 9))/3S 1/(sqrt (4x^2 +1)) dx= 3/2 S ((sec^2 theta d theta)/sec theta)=3/2 S (sec theta d theta)=3/2 ln abs (sec theta + tan theta) + C=3/2 ln [ abs ((sqrt (4x^2 +9) +2x/3)) +CTngs that are throwing me: 1) the x^4 in the denominator of the problemand 2) the triangle setupI have looked in several books and all over the internet to find a problemlike ts to see if I was on the right track, but none that I found looklike ts. I donÍt know if the answer is right, but it is the best I coulddo with what I know.Any help === ETAsAhQH37BTZh5i2Yb6PwwIFCN28K5rUwIULpT/vcZrvZRuTV+wlIE/ (theta), then express x^4 in terms of thetaalong with the other factors. When you get your trigonometric integrand,it looks complicated; but express it in terms of sines and === integration trouble> . I am working on integrate sqrt ((4x^2) +9)/(x^4) using trig> substitution.I have: u=2x> a=3as you have never used ts last bit (actually, you havenÍt yetintroduced it properly!) I will ignore it.> 2x/3 = tan theta> dx=3/2 sec^2 theta d thetathatÍs right.> (sqrt (4x^2 + 9))/3S 1/(sqrt (4x^2 +1)) dx= 3/2 S ((sec^2 theta d theta)/sec theta)=3/2 S (sec theta d theta)=3/2 ln abs (sec theta + tan theta) + C=3/2 ln [ abs ((sqrt (4x^2 +9) +2x/3)) +CTngs that are throwing me: 1) the x^4 in the denominator of the problem> and 2) the triangle setupwhat did you do here? you cannot just chuck away bits of pieces*under* the integral sign! it is violating the law! remember theintegral rules? i honestly tnk they should add ts one:donÍt do anytng just because you tnk you can do it. name the lawyou are using. if not found, call it quits.there is no rule of throwing away the denominator, by the way.I have looked in several books and all over the internet to find a problem> like ts to see if I was on the right track, but none that I found look> like ts. I donÍt know if the answer is right, but it is the best I could> do with what I know.Any help appreciated! Stacyi donÍt know if anyone is stupid enough to answer your question,because they should not, but i can help you with the triangle.> 2x/3 = tan thetathatÍs right. draw the triangle. the adjacent is 3, opposite is 2x,that makes the hypotenuse sqrt{4x^2 + 9}.[i cannot resist a nt: keep the x^4 bit, convert it, then express === Having integration trouble> . I am working on integrate sqrt ((4x^2) +9)/(x^4) using trig> substitution.> I have:> u=2x> a=3> 2x/3 = tan theta> dx=3/2 sec^2 theta d theta> (sqrt (4x^2 + 9))/3> S 1/(sqrt (4x^2 +1)) dx= 3/2 S ((sec^2 theta d theta)/sec theta)> =3/2 S (sec theta d theta)> =3/2 ln abs (sec theta + tan theta) + C> =3/2 ln [ abs ((sqrt (4x^2 +9) +2x/3)) +C> Tngs that are throwing me: 1) the x^4 in the denominator of the problem> and 2) the triangle setup> I have looked in several books and all over the internet to find a problem> like ts to see if I was on the right track, but none that I found look> like ts. I donÍt know if the answer is right, but it is the best I could> do with what I know.Substitute x = (3/2)*tan(u); dx = (3/2)*sec^2(u) duThen, sqrt(4x^2 + 9)/x^4 = 2*sqrt(x^2 + 9/4) / x^4 = 2*sqrt(9/4(tan^2(u) + 1) / ((81/16)*tan^4(u)) = (3*16/81)*sqrt(sec^2(u)) / tan^4(u) = (3 * 16/81) * sec(u) / tan^4(u)and f(u) du = (48/81)* sec^3(u) / tan^4(u) = cos(u) / sin^4(u) duNow set z = sin(u), du = -dz/cos(u)-- P.A.C. SmithThe vast majority of Iraqis want to live in a peaceful, free world.And we will find these people and we will bring them === integrate [I =] sqrt ((4x^2) + 9)/(x^4) using trig substitution.You can lose all the 4s and 9s by letting y = 2x/3 to give: I/2 = int (sqrt ((y^2 + 1)/y^4)) dyThen take 2y = (z^2 - 1) / z, so that 2.dy/dz = (z^2 + 1) / z^2 and: I/2 = int (1/z + 4z / (z^2 - 1)^2) dz = log(z) + 2 / (z^2 - 1) + C === JSH: Non-uniqueness of factorization <87ad3den4w.fsf@pwumbda.org> <25rZb.64909$KV5.49312@nwrdny01.gnilink.net> <87wu6hd2zr.fsf@pwumbda.org> <5ruZb.63308$IF1.32769@nwrdny03.gnilink.net> Discussion, MS is just erring on the side of caution by deleting a>> string ending with :. One cannot fault them for not knowing who>> JSH is because, after all, heÍs a fan of Java, not C#.>> I guess we disagree on what constitutes caution. Mangling subject>> lines is not cautious. The fact is that, JSH aside, there are plenty>> of opportunities in wch one wants to start a thread with a subject>> of the form, Blah: blah blah blah.>> That the MS coders couldnÍt tnk of any isnÍt surprising. They>> couldnÍt consider that (on rare occasions), lines might>> begin...> Ouch, the mime parsing error is a really horrible tng. I canÍt fathom> that one. They must be working around some horrible kludge that requires> it.I canÍt imagine what kludge would prevent one from reading headers todetermine whether the message is a mime message. I tnk whathappened is ts. Anyone may correct my mistakes, of course.Prior to the widespread use of mime, folks sent messages withuuencoded files in them. The folks on the other end knew to cut theuuenclosure out and decode it to recover the file. When MS designedOE, they remembered ts old practice and realized that their userswould not know what to do if they received a non-mime message with auuenclosure. So, they (stupidly) decided to have their programprotect users by looking for begin and parsing the rest as auuenclosure.Ts is bad, but their implementation is particularly stupid. Why notcheck *first* for a matcng end to the begin before parsing thecontents as an enclosure? Any four-year-old knows thatÍs better.As I said, itÍs not common for the bug to bite unintentionally, but ithappens. I know a guy who uses a software package (an emacs mode, Itnk) that both right and left justifies s email and posts, byadding spacing on each line as needed. Now, we can debate whether heought to use that and whether the result is pleasing, but he uses it.One of s posts is the first (maybe only) unintentional bite of thebegin bug that IÍve seen.> IÍll end ts OT thread by saying that we do not disagree on the meaning of> ïcautionÍ. I was commenting on MS and I *did* say that they ïerredÍ!> FWIW, I tnk user agents should leave the subject field alone.A sensible Re: convention seems perfectly reasonable to me. Too baditÍs beyond the skills found at MS.-- Jesse F. HughesTime and again, story has shown that people who tnk their beliefstrump reality lose, and lose badly. Luckily, I donÍt === JSH: Non-uniqueness of factorization> I canÍt imagine what kludge would prevent one from reading headers to> determine whether the message is a mime message. I tnk what> happened is ts. Anyone may correct my mistakes, of course.> Prior to the widespread use of mime, folks sent messages with> uuencoded files in them.Good call. HereÍs MSÍs page on the === JSH: Non-uniqueness of factorization <87ad3den4w.fsf@pwumbda.org> <25rZb.64909$KV5.49312@nwrdny01.gnilink.net> <87wu6hd2zr.fsf@pwumbda.org> <5ruZb.63308$IF1.32769@nwrdny03.gnilink.net> <87k72gdd8v.fsf@pwumbda.org> Discussion, imagine what kludge would prevent one from reading headers to>> determine whether the message is a mime message. I tnk what>> happened is ts. Anyone may correct my mistakes, of course.>> Prior to the widespread use of mime, folks sent messages with>> uuencoded files in them.> Good call. HereÍs MSÍs page on the error.> http://support.microsoft.com/?kbid=265230ItÍs a beautiful page. HereÍs the workaround for users of OE.To workaround ts problem: * Do not start messages with the word begin followed by two spaces. * Use only one space between the word begin and the following data. * Capitalize the word begin so that it is reads Begin. * Use a different word such as start or commence. Ts is remarkably silly advice. The problem for the OE user who goesto ts page is that heÍs received email (or read a posting) that hecanÍt read. The only workaround that those big brains at MS have form is advice on how to send email/posts that their broken softwarecan read. They give no advice on what to do if you want to readsometng that OE canÍt display.-- Jesse Hughes [I]f gravel cannot make itself into an animal in a year, how could itdo it in a million years? The animal would be dead before it gotalive. --The Creation === factorization Discussion, someone please give me step by step details on how to ignore posts by> > and replies to s posts? I just donÍt have the willpower to> > not read s bullst.> contains jstevh@msn.com in either header or body will be the best.> (Until he changes address of course.)Filtering on the body will require considerably longer times thanfiltering only on the standard headers returned by the news server(including from headers and === everybody,suppose I have a grate, N x M. how many paths are from node (0,0) to >(N-1,M-1) eg. from one corner to the diagonal-opposite one?allowable paths are only with zero or positive change in nodeÍs coordinate.how can I compute ts? I would appreciate any suggestion and help..cheers,>n.You should be able to calculate ts. Take advantage of symmetry.> Start 1 position away from the end and work backwards.Good luckplIf you know basic combinatorics (wch, of course, are proved usinginduction), you can get a much simpler solution than the onesuggested. As a general nt, induction (or itÍs cool teenage brotherthe Zorn Lemma) is a good way to prove sometng, but not a good wayto discover anytng. There are exceptions, of course, but ts isnÍtone of them; recurrence relations here are kinda silly.YouÍre only allowed to move in a positive direction, and only alonggrid lines. So that means you move up n times and right n times, andthe question is how many different orderings of these you can possiblyhave.That is, youÍre going to take exactly 2n steps, half of them north andhalf of them west. So simply pick wch north steps you take. We areselecting n items of 2n, with order not mattering, so the answer is 2nchoose n = === have a grate, N x M. how many paths are from node (0,0) to >(N-1,M-1) eg. from one corner to the diagonal-opposite one?>allowable paths are only with zero or positive change in nodeÍs coordinate.>how can I compute ts? I would appreciate any suggestion and help..You are going to take n steps one way and m steps the other for n+mtotal steps. You just have to count how many ways there are to choosethe n horizontal moves from n+m moves, i.e.: C(n+m, n) = === know the first cohomology group H^1(X,Z) of different setsX. Does any of you know a good website, where I can look these up?Some of the cases are, when X is a point, a circle, a sphere (S^n),the Cantor set and more.I donÍt need to know, how === quasi-unitaryIf x is an element in a C*-algebra, what does it === Few problems about number.Here they are:1. Draw 3 diagrams to show the relationsp among numbers.2. What are the different between number, numeral & digits?3. What are the 3 functions & === are:1. Draw 3 diagrams to show the relationsp among numbers.> 2. What are the different between number, numeral & digits?> 3. What are the 3 functions & numbers?If you tell us what steps youÍve already taken to do your homework, we can give === addition theorem for Weierstrass elliptic function is rathercomplicated, isnÍt it? However after modifying Weierstrass elliptic functionp(x) as p(x+a) where a is constant, the addition theorem is simplified as,p(x+y) = p(x)+p(y)+bp(x)p(y)+c{p(x)p(y)}^2where b and c are constants. Can you believe it? Believe or not, pleasevisit:http://139.134.5.123/tiddler2/cauchy/ cauchyequation.htmthere I explained how to calculate the === my college about Procedures. Please read.Did you take issue with the grading of your homework paper? ItÍsabsurd. If nobody has looked at it, he received a 25% deduction on asingle problem for not writing (g o g) (x) = g(g(x)), (where the ois the composition symbol) and a comment that it was very hard tofollow s work, even though it was quite easy to read. Indeed, itis the TAÍs writing that I find almost impossible to read. At onepoint, several points are taken off for use of notation in the worksection of a problem wch is not a proof (it is a numericalcomputation). In other places points are taken off for no apparentreason. At yet another place he shows that a function is not definedat a certain point and then asserts that it is noncontinuous at thatpoint, at wch the TA takes off points, apparently for not sayingsometng like, Observe that if the function is not defined at tspoint, its limit at ts point cannot equal its (nonexistant) value.As far as I can tell in the cursory glance I gave the test only oneanswer is actually incorrect. You have demonstrated your competence in the subject material ofthe class not only in the practical but in the theoretical. I donÍtreally see what else the teacher could ask for. As for trials and lawyers, IÍm not sure theyÍd do you much good. YouÍd have to dig up a law or sometng on the collegeÍs books againstnot letting you drop a class under certain conditions, and IÍm prettysure they didnÍt write such a tng in. I would argue for more pointswith the teacher, and if that doesnÍt suceed IÍd retake the === about Procedures. Please read.> Did you take issue with the grading of your homework paper? ItÍs> absurd. If nobody has looked at it, he received a 25% deduction on a> single problem for not writing (g o g) (x) = g(g(x)), (where the o> is the composition symbol) and a comment that it was very hard to> follow s work, even though it was quite easy to read. Indeed, it> is the TAÍs writing that I find almost impossible to read. At one> point, several points are taken off for use of notation in the work> section of a problem wch is not a proof (it is a numerical> computation). In other places points are taken off for no apparent> reason. At yet another place he shows that a function is not defined> at a certain point and then asserts that it is noncontinuous at that> point, at wch the TA takes off points, apparently for not saying> sometng like, Observe that if the function is not defined at ts> point, its limit at ts point cannot equal its (nonexistant) value.> As far as I can tell in the cursory glance I gave the test only one> answer is actually incorrect.> You have demonstrated your competence in the subject material of> the class not only in the practical but in the theoretical. I donÍt> really see what else the teacher could ask for.> As for trials and lawyers, IÍm not sure theyÍd do you much good. > YouÍd have to dig up a law or sometng on the collegeÍs books against> not letting you drop a class under certain conditions, and IÍm pretty> sure they didnÍt write such a tng in. I would argue for more points> with the teacher, and if that doesnÍt suceed IÍd retake the class.I looked over the test, and I tnk that the grading was fair. (For example, the 25% was taken off because he didnÍt do part of that question, that is, find the domain.)On the other hand, I do tnk that the test taker does show some competence - he definitely should not drop out of college. He could retake the course - indeed I didnÍt tnk it was bad enough to drop the course, unless he wants a sure A out of the course, wch I donÍt tnk he will get at the end anyway.I donÍt tnk it is worth arguing the W, wch is a fine grade to obtain for a course.Instead of going to lawyers and such like to argue for sometng wch is not worth having, maybe the student should take s test and talk with the teacher about it - not to get the points back, but rather to see how he could do better next time. Wle I tnk that the lost points were in almost all cases fair, they were mostly lost over little tngs. The only question he got completely wrong was 6(b), and I bet most people in the class also got it wrong.I do agree with you that s handwriting was fine. Maybe it was late at night when the grader looked over s work.My advice would have been to continue with the course - do better in later tests, and be happy with === ATTENTION: Open dispute with my college about Procedures. Please read.> Why donÍt you put all ts creative energy into studying?> I thought i mastered the subject. I am tnking of withdrawing from school completely.I am stupid; my teacher has emphasized ts to me. I do not know what to do anymore; everytng is not as great as it used to be. Life is not enjoyable anymore === college about Procedures. Please read.> I am stupidYouÍre not stupid, but your trolling skills need a *lot* of === about Procedures. Please read.> for calling me names. i tnk it === ATTENTION: Open dispute with my college about Procedures. Please read.> for calling me names. i tnk it will help push me off the edgein> the end.And what names, pray tell, did I call === about Procedures. Please read.> i do not respond to many of the posts because my argument is with ts > school and not people of the message board. The ideas that the board gives > search for > statewide policies and procedures regarding academic progressWell, if ts is for real, I really want you to keep us updated on thehorrible lawyers and horrible judges === Re: Math notation question: ï(ï> I just ran across a math question in a review and IÍm totally stumped> by the answer. Then I realized that I may have been misreading the> notation.The statement was:> There exist x, y, and z that are nonzero. Such that 1 ( y>x and xy=z.Does 1 ( y>x mean that both y & x are less than 1? Looking at the> answer that is the only tng that makes sense to me. I just donÍt> understand the notation. for help in advance.Ben..Never seen ts sort of notation in my life. Looks like a typo in thebook or sometng. === question in a review and IÍm totally stumped> by the answer. Then I realized that I may have been misreading the> notation.> The statement was:> There exist x, y, and z that are nonzero. Such that 1 ( y>x and xy=z.The period after nonzero looks wrong. Is it a comma?> Does 1 ( y>x mean that both y & x are less than 1? Looking at the> answer that is the only tng that makes sense to me. I just donÍt> understand the notation.Neither do I. Have you accurately transcribed the === ï(ï> The statement was:> There exist x, y, and z that are nonzero. Such that 1 ( y>x and xy=z. The period after nonzero looks wrong. Is it a comma?Yeah, it shouldnÍt be there. Sorry. See whole whole transcriptionbelow. > Does 1 ( y>x mean that both y & x are less than 1? Looking at the> answer that is the only tng that makes sense to me. I just donÍt> understand the notation. Neither do I. Have you accurately transcribed the _complete_ sentence?If x, y, and z are nonzero numbers such that 1(y>x and xy=z, wch ofthe following CANNOT be true?1) y>z2) y=z3) z=x4) x>z5) z>0I tnk itÍs 4) x>z, but the answer in the review says itÍs 2) y=z.However, when x=1, y=2, and z=2 it meets all of the criteria and y=z,so #2 cannot be the answer. When looking at their explanation for why#2 is the answer they say because all of the other choises CAN betrue #2 must be the correct answer. Wch, to me, is not only a BScop-out, but wrong.Also, in the #4 explanation it says According to the question, Y mustbe greater to or equal to one. Thus *my* question on the notation.Whatever happened to <=? I should also state that nowhere else inthe review does it mention that the ï(ï to <= notation substitutionexists. I tnk === Re: Math notation question: ï(ï The statement was: There exist x, y, and z that are nonzero. Such that 1 ( y>x and xy=z.> The period after nonzero looks wrong. Is it a comma?> Yeah, it shouldnÍt be there. Sorry. See whole whole transcription> below. Does 1 ( y>x mean that both y & x are less than 1? Looking at the answer that is the only tng that makes sense to me. I just donÍt understand the notation.> Neither do I. Have you accurately transcribed the _complete_ sentence?If x, y, and z are nonzero numbers such that 1(y>x and xy=z, wch of> the following CANNOT be true?> 1) y>z> 2) y=z> 3) z=x> 4) x>z> 5) z>0I tnk itÍs 4) x>z, but the answer in the review says itÍs 2) y=z.> However, when x=1, y=2, and z=2 it meets all of the criteria and y=z,> so #2 cannot be the answer. When looking at their explanation for why> #2 is the answer they say because all of the other choises CAN be> true #2 must be the correct answer. Wch, to me, is not only a BS> cop-out, but wrong.Also, in the #4 explanation it says According to the question, Y must> be greater to or equal to one. Thus *my* question on the notation.> Whatever happened to <=? I should also state that nowhere else in> the review does it mention that the ï(ï to <= notation substitution> exists. I tnk itÍs just a shoddily put-together review. again.The question clearly contains a typo. All books have typos, expect themand work around them. I conjecture that the question was meant to be:If x, y, and z are nonzero numbers such that 1>=y>x and xy=z, wch ofthe following CANNOT be true?1) y>z2) y=z3) z=x4) x>z5) z>0The correct answer to ts question is (2). For suppose y=z. Then fromxy=z (and the fact that x,y,z are nonzero), we get x = 1. But thenfrom 1 >= y > x we would get 1 > 1, a contradiction.All the others are possible. For example for (1) let x = -1, y=0.5, andz=-0.5 For (3), let x = 0.5, y=1, and z=0.5. For (4) and (5), letx=1/4, y=1/2, and z=1/8.-Leonard === not 4 ...x=-1, y=-z z < -1then xy = z and x > zIÍm guessing that the type-o was very significant to the problem and IÍd The statement was: There exist x, y, and z that are nonzero. Such that 1 ( y>x and xy=z.> The period after nonzero looks wrong. Is it a comma?> Yeah, it shouldnÍt be there. Sorry. See whole whole transcription> below. Does 1 ( y>x mean that both y & x are less than 1? Looking at the answer that is the only tng that makes sense to me. I just donÍt understand the notation.> Neither do I. Have you accurately transcribed the _complete_ sentence?> If x, y, and z are nonzero numbers such that 1(y>x and xy=z, wch of> the following CANNOT be true?> 1) y>z> 2) y=z> 3) z=x> 4) x>z> 5) z>0> I tnk itÍs 4) x>z, but the answer in the review says itÍs 2) y=z.> However, when x=1, y=2, and z=2 it meets all of the criteria and y=z,> so #2 cannot be the answer. When looking at their explanation for why> #2 is the answer they say because all of the other choises CAN be> true #2 must be the correct answer. Wch, to me, is not only a BS> cop-out, but wrong.> Also, in the #4 explanation it says According to the question, Y must> be greater to or equal to one. Thus *my* question on the notation.> Whatever happened to <=? I should also state that nowhere else in> the review does it mention that the ï(ï to <= notation substitution> exists. I tnk itÍs just a shoddily put-together review.> === algebraGee. for your help. Looks like ts path --- ZornÍs lemma andall that --- is absolutely necessary, I mean I cannot see another way[using first order logic only? or is it still first order?]. But whatconfuses me is: shouldnÍt the same argument apply to the finite basiscase? Why go for the other detailed (Gaussian) technique? We areactually interested in much less information that the algorithm (forelimination) provides, === give me an original (as far as possible)copy of the Shannon/Nyquist Sampling And Reconstruction Theorem.(IÍve obtained ShannonÍs, A Mathematical Theory Of Communicationin the last couple of days, so IÍm hoping that the former will alsobe available, but it hasnÍt appeared to me in any Google === Airy R. Bean was cut from the messthat had once been s life, he managed to utter:>IÍm looking for a URL to give me an original (as far as possible)>copy of the Shannon/Nyquist Sampling And Reconstruction Theorem.>(IÍve obtained ShannonÍs, A Mathematical Theory Of Communication>in the last couple of days, so IÍm hoping that the former will also>be available, but it hasnÍt appeared to me in any Google search)Those intending to respond, should bear in mind that the ïtechnicalÍreading of ts poster, is on a par with s wife ElisabethÍs readingof Womans Weekly (or was it Peoples Friend?).Anytime you are ready for another go at The English Game, old fruit.You be sure and let me know.Old === variableDefinition:A *probability distribution* for a discrete random variable X withvalues{ x_k | k in Z }where Z is the set of integers, is a set of numbers (I tnk they meanreal){ p_k | k in Z }such thatP(X = x_k) = p_k >= 0 andsum_{k = -infty}^infty p_k = 1Definition:If X is a discrete random variable with values { x_k | k in Z }and probability distribution { p_k | l in Z }then the *mean* or *expected value* of X isE(X) = sum_{k = -infty}^infty x_k p_kThen comes the confusing bit:Proposition:Let Y = g(X) be a function on X. Then then *mean* or *expected value*of Y isE(Y) = sum{k = -infty}^infty g(x_k) p_kMy question is, is that a definition? Because, in case of continuousrandom variable, they are clean in saying that it is. However, here,they comment:It is possible to give a general proof of ts proposition, but weshall not do so here.Leaving me in eternal confusion that started the day I was born.When I was born I was so surprized that I couldnÍt talk for a yearand a half--- I honestly donÍt remember who === a random variable> Definition:> A *probability distribution* for a discrete random variable X with> values> { x_k | k in Z }> where Z is the set of integers, is a set of numbers (I tnk they mean> real)> { p_k | k in Z }> such that> P(X = x_k) = p_k >= 0 > and> sum_{k = -infty}^infty p_k = 1Definition:> If X is a discrete random variable with values > { x_k | k in Z }> and probability distribution > { p_k | l in Z }> then the *mean* or *expected value* of X is> E(X) = sum_{k = -infty}^infty x_k p_kThen comes the confusing bit:Proposition:> Let Y = g(X) be a function on X. Then then *mean* or *expected value*> of Y is> E(Y) = sum{k = -infty}^infty g(x_k) p_kMy question is, is that a definition? Because, in case of continuous> random variable, they are clean in saying that it is. However, here,> they comment:It is possible to give a general proof of ts proposition, but we> shall not do so here.It is not a definition. If X is as described, then Y = g(X) is anotherdiscrete random variable, and it has value g(x_k) in the event thatX has value x_k. The reason the proposition is not immediately obviousis that you could have g(x_k)=g(x_j) for different x_k, x_j. Sothe sets {Y = y_k} could be unions of several of the sets {X = x_k},possibly even infinitely many such sets.-- G. A. Edgar === needed from a group theorist by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1LDK3j18310;>>> An exponent E(G) of a group G is the lcm of all the orders o(g),>> g in G.We have that G is cyclic if and only if E(G)=|G|.>> Does anybody have an idea how can we find all n such that>> the exponent of Z_n*,the group of units of Z/Zn is 2?>> E(Z_n*)=2 n=?>Ts just means that x^2=1 for any invertible x in Z_n. Thus n cannot have>any prime factors other than 2 and 3. Write n=(2^s)(3^t). Splitting Z_n as a>product of two rings, the cases are>-- E(Z_(2^s)*)=2 and E(Z_(3^t)*)=1 i.e. s>=1 and t=0 (n=2,4,8,16,32,...)>-- E(Z_(2^s)*)=1 and E(Z_(3^t)*)=2 i.e. s=0 and t=1 (n=3 only)>-- E(Z_(2^s)*)=2 and E(Z_(3^t)*)=2 i.e. s>=1 and t=1 (n=6,12,24,48,...)>I hope I havenÍt blundered somewhere in here :)I still donÍt understand why is n=(2^s)(3^t),but IÍll take that for granted. :)The order of Z_n* is p(n) (Euler function),and since all the orders of the elements are 2, Z_n* is abelian.Can we determine the number of generators of Z_n* ?If there are m generators of Z_n*,then all we have to do is tosolve p(2^s*3^t)=2^m or 2^s*3^(t-1)=2^mTs is all I can do === argument by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1LDK2u18292;> >> My problem is more general. Too many times we get a very complicated and>> fine tuned mechanism that would require a major jump in order to have any>> effectiveness. Fish that use electricity to stun and kill opponents is a>> remarkable one. The amount of electricity generated, and the coordination>> involving it is very big.>The major jump is only in your mind.>> It looks like it would take a big number of mutations to aceve it, with>> any partial mutation or a small effect totally useless. A functional yet>> small weapon of ts kind is even punisng given the amount of energy>> needed for it, and lack of effective results.>Two words: active sensor>http:// www.§mnh.u§.edu/fish/tropical/JSA/gymno.htm< /a>>Next time, do your own research.>Daniel W. son>panoptes@iquest.net>http:// members.iquest.net/~panoptes/B) as Set Derivative by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1LDJwU18174;Define:1) (A-->B) = (ABÍ)Í = AÍ U Band therefore:2) (A-->B)(B-->A) = (AÍ U B)(BÍ U A) = AB U AÍBÍwhere adjacent parentheses refer to set intersection. The symbol for (A-->B)(B-->A) should be familiar to people in Mathematical Logic, or to Logicians if a, b were propositions:3) (a<-->b) = ~(a^~b)^~(b^~a) = (~aVb)(~bVa) = ab V ~a~bwhere ~ (tilde) represents negation (NOT), V disjunction (OR including AND), ^ conjunction (AND). The left-hand-side of (3)is written in Logic a iff b or a if and only if b, and the Set Theory analog is therefore for sets A, B:4) (A<-->B) = (A-->B)(B-->A) = AB U AÍBÍTaking complements of both extremes in (4) yields:5) (A<-->B)Í = AÍB U ABÍwch is the analog for Sets of:6) (fg)Í = fÍg + fgÍfor differentiable functions f, g. Notice that although U in (5) isset Union (corresponds to logical AND/OR), it happens to be adisjoint or mutually exclusive Union since AÍB does not intersectABÍ (it has zero intersection). It is not unusual to write U as+, for example circled +, in such situations with regard to vectorspaces and so on, but I will retain U to be technically correct.Readers can prove as an exercise that if u is a bounded Lebesguemeasure (a Lebesgue measure on bounded sigma algebras, etc.), then:7) u[(A<-->B)Í] = u(AÍB) + u(ABÍ)wch is even more similar to (6), where A, B are sets in the sigma algebra. To assert, then, that (A<-->B) (A if and only ifB or A iff B) is the set analog of fg with regard to complementa-tion and ï as analogous to differentiation, seems quite intuitiveto say the least, although we donÍt assert anytng for gherprimes like A or other contexts at ts point. Obviously, Ais (AÍ)Í = A but f is not f in general unless f is the exponentialfunction exp(x). The exponential function, by the way, is notas trivial as it may seem, and I leave that as an exercise (nt:consider === Non-Derivative Re: (A<-->B) as Set DerivativeDefine:1) (A-->B) = (ABÍ)Í = AÍ U BI presume that ï denotes the complement of a set (in somepresumed universe of discourse X) and AB denotes the intersectionof A and B. Then ts is just X - (A - B) or X(AB)where - or denotes set difference.> and therefore:2) (A-->B)(B-->A) = (AÍ U B)(BÍ U A) = AB U AÍBÍThe complement of the symmetric difference.> where adjacent parentheses refer to set intersection.The symbol for (A-->B)(B-->A) should be familiar to people in> Mathematical Logic, or to Logicians if a, b were propositions:3) (a<-->b) = ~(a^~b)^~(b^~a) = (~aVb)(~bVa) = ab V ~a~bwhere ~ (tilde) represents negation (NOT), V disjunction (OR> including AND), ^ conjunction (AND). The left-hand-side of (3)> is written in Logic a iff b or a if and only if b, and> the Set Theory analog is therefore for sets A, B:4) (A<-->B) = (A-->B)(B-->A) = AB U AÍBÍTaking complements of both extremes in (4) yields:5) (A<-->B)Í = AÍB U ABÍthe symmetric difference of A and B.> wch is the analog for Sets of:6) (fg)Í = fÍg + fgÍNo, it isnÍt. A possible analogue for sets of ts would be(adopting your notation)(A<-->B)Í = (AÍ<-->B) U (A<-->BÍ)or(AB)Í = (AÍB) U (ABÍ),Slight problem .... both of these are false in general.> for differentiable functions f, g. Notice that although U in (5) is> set Union (corresponds to logical AND/OR), it happens to be a> disjoint or mutually exclusive Union since AÍB does not intersect> ABÍ (it has zero intersection). It is not unusual to write U as> +, for example circled +, in such situations with regard to vector> spaces and so on, but I will retain U to be technically correct.> Readers can prove as an exercise that if u is a bounded Lebesgue> measure (a Lebesgue measure on bounded sigma algebras, etc.), then:Seems you donÍt know what Lebesgue measure is(IÍll tell you: itÍs a measure (not bounded) on a specificsigma-algebra on R^n).> 7) u[(A<-->B)Í] = u(AÍB) + u(ABÍ)> wch is even more similar to (6),No, it isnÍt.> where A, B are sets in the> sigma algebra. To assert, then, that (A<-->B) (A if and only if> B or A iff B) is the set analog of fg with regard to complementa-> tion and ï as analogous to differentiation, seems quite intuitiveseems quite stupid, to say the least> to say the least, although we donÍt assert anytng for gher> primes like A or other contexts at ts point.Very sensible.> Obviously, A> is (AÍ)Í = A but f is not f in general unless f is the exponential> function exp(x). or a multiple of the exponential.Of course ts shows the daftness of ts bogus analogy.> The exponential function, by the way, is not> as trivial as it may seem, and I leave that as an exercise (nt:> consider inverse functions).Now what would it mean for the exponential function to beas trivial as it may seem?WhatÍs the point of ts posting anyway, apart from a compulsivedesire to see analogies where none exist?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes s penis for a square root, 88-9Francis Wheen, _How === Non-Derivative Re: (A<-->B) as Set DerivativeDefine:1) (A-->B) = (ABÍ)Í = AÍ U BI presume that ï denotes the complement of a set (in somepresumed universe of discourse X) and AB denotes the intersectionof A and B. Then ts is just X - (A - B) or X(AB)where - or denotes set difference.> and therefore:2) (A-->B)(B-->A) = (AÍ U B)(BÍ U A) = AB U AÍBÍThe complement of the symmetric difference.> where adjacent parentheses refer to set intersection.The symbol for (A-->B)(B-->A) should be familiar to people in> Mathematical Logic, or to Logicians if a, b were propositions:3) (a<-->b) = ~(a^~b)^~(b^~a) = (~aVb)(~bVa) = ab V ~a~bwhere ~ (tilde) represents negation (NOT), V disjunction (OR> including AND), ^ conjunction (AND). The left-hand-side of (3)> is written in Logic a iff b or a if and only if b, and> the Set Theory analog is therefore for sets A, B:4) (A<-->B) = (A-->B)(B-->A) = AB U AÍBÍTaking complements of both extremes in (4) yields:5) (A<-->B)Í = AÍB U ABÍthe symmetric difference of A and B.> wch is the analog for Sets of:6) (fg)Í = fÍg + fgÍNo, it isnÍt. A possible analogue for sets of ts would be(adopting your notation)(A<-->B)Í = (AÍ<-->B) U (A<-->BÍ)or(AB)Í = (AÍB) U (ABÍ),Slight problem .... both of these are false in general.> for differentiable functions f, g. Notice that although U in (5) is> set Union (corresponds to logical AND/OR), it happens to be a> disjoint or mutually exclusive Union since AÍB does not intersect> ABÍ (it has zero intersection). It is not unusual to write U as> +, for example circled +, in such situations with regard to vector> spaces and so on, but I will retain U to be technically correct.> Readers can prove as an exercise that if u is a bounded Lebesgue> measure (a Lebesgue measure on bounded sigma algebras, etc.), then:Seems you donÍt know what Lebesgue measure is(IÍll tell you: itÍs a measure (not bounded) on a specificsigma-algebra on R^n).> 7) u[(A<-->B)Í] = u(AÍB) + u(ABÍ)> wch is even more similar to (6),No, it isnÍt.> where A, B are sets in the> sigma algebra. To assert, then, that (A<-->B) (A if and only if> B or A iff B) is the set analog of fg with regard to complementa-> tion and ï as analogous to differentiation, seems quite intuitiveseems quite stupid, to say the least> to say the least, although we donÍt assert anytng for gher> primes like A or other contexts at ts point.Very sensible.> Obviously, A> is (AÍ)Í = A but f is not f in general unless f is the exponential> function exp(x). or a multiple of the exponential.Of course ts shows the daftness of ts bogus analogy.> The exponential function, by the way, is not> as trivial as it may seem, and I leave that as an exercise (nt:> consider inverse functions).Now what would it mean for the exponential function to beas trivial as it may seem?WhatÍs the point of ts posting anyway, apart from a compulsivedesire to see analogies where none exist?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes s penis for a square root, 88-9Francis Wheen, _How === Set Derivative>Define:>1) (A-->B) = (ABÍ)Í = AÍ U B>and therefore:>2) (A-->B)(B-->A) = (AÍ U B)(BÍ U A) = AB U AÍBÍ[...]>Taking complements of both extremes in (4) yields:>5) (A<-->B)Í = AÍB U ABÍ>wch is the analog for Sets of:>6) (fg)Í = fÍg + fgÍIt also gives other interesting and useful differentiation formulas:(A u B)Í = AÍBÍ = A<-->B AB = A<-->B (AÍ U BÍ)Íwch I suppose tells us that:(f + g)Í = fg - (fÍ + gÍ)Í-- IÍm not interested in mathematics that might have anytngto do with reality. -- Russell Easterly, in === support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1LDJxB18185;>> mathematics is sometng like Religion.>> You can sell your talents if youÍre smart enough to have a bag of tricks>> wch impress employers or use the talent to get a Tece degree. But>the>> spirit of math some transcends ts. (sometng about the Magic feeling>you>> get when you are learning sometng new, or finally see a connection, or>> just being able to design an Excel sheet that will do stuff you would have>> never had the imet ot do otherwise.) I am convinced that there is a great>> number of people , in spirit, who are Mathematicians, but lack background>> and are frustrated or feeL some loss, because of the madness that passes>for>> Education in grades 1-15.> Mathematicians donÍt have to worry about grades 1-15,> since they come out of the countryÍs *education budget*.> Science and Mathematics has been placed in the countryÍs Jesus Freak>Budget.> , Sir: your message that it give me a new way of tnking what is the math. you said the math is like religion. i thought we had different opinion about it. in my opinion I thought math is like religion is the tng that you must stick everyday then you will find sometng special. I thought math is full of magic before but then I found the math is around us. thatÍs one of the important reasons thatÍs why I want to study it by my spare time. it can help you slove the problme , not just some math problme , itÍs a good tnking way that you can get a good decision when you fact to sometng. sir , if you read my message I hope you can give your opinion and told about it details . IÍm interested in your opinion. and also sorry about my bad english. 008(itÍs my name). === support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1LDJxI18192;I am an undergrad who will be graduating ts semester andwill be going to grad school in fall((if they let me))and i have a few questions.I have heard that you should not only tnk about what you likebut also what is still a growing field.I really enjoy analysis.((classical,functional,Fourier..))I really dislike (at least so far) algebra,number theory ,combinatorics.Now my question is : are the above named analysisÍs still a good field to get into or should i look more at say applied analysis.Someone once told me that unless you are a genious that harmonic analysis is not a good field to get into. Is ts because the field is so old and developed? Would the same apply to classical and functional analysis? I have done some PDE theory and enjoyed working in the Sobolev spaces with the functional analysis techniques. If i was to go into PDEÍs would I expect more of the same (wch i enjoyed) or does it turn into a grab bag of techniques ? I have never done any measure theoretic probability but considering my interests would ts be another one to tnk about? Any replies would be === studies> I have heard that you should not only tnk about what you like but also> what is still a growing field. I really enjoy analysis.> ((classical,functional,Fourier..)) That, of course, depends on your intentions... if you want to study mathematics because you love mathematics and yadda yadda, then study what you like most. But if you actually care about, say, landing a career out of your studies, then considering the ïhot topicsÍ would be useful. Some poeple would, of course, say that you should always have a career in mind, but I donÍt tnk a university is supposed to be a job-training institution (but that argument is for a different thread.)> I really dislike (at least so far) algebra,number theory ,combinatorics. Ah, pity... but to each their own.> Now my question is : are the above named analysisÍs still a > good field to get into or should i look more at say applied> analysis.Someone once told me that unless you are a genious> that harmonic analysis is not a good field to get into.> Is ts because the field is so old and developed?> Would the same apply to classical and functional analysis? The pluaral of analysis should be analyses. I would kind of have to agree with what that person told you... standard and classic algebra and analysis are the core fields of math and are populated with some of the best. They are also (I tnk) the oldest fields and most developed, so it takes a lot of (a combination of) work and ingenuity to get somewhere new in those fields. And, of course, since they are so far developed into abstractness, the applications (and hence funding) are harder to come by. Some fields like applied analysis can study problems and write papers with titles that sound applied-like, like physics related topics. The fields you mentioned that you donÍt like (like number theory and combinatorics) have gained a lot of interest from fields like computer science because of the host of algorithmic problems to study in those fields... I was interested in discrete math and optimization, and when I applied for grad school (to math departments) to ended up in a computer science department. I donÍt give a darn about databases, operating systems, grapcs, etc, but I love the math that I can study here and I get a lot more funding here than I any most student I know (and yet I know I am quite inferior to a majority of them.)> I have done some PDE theory and enjoyed working in the> Sobolev spaces with the functional analysis techniques.> If i was to go into PDEÍs would I expect more of the same> (wch i enjoyed) or does it turn into > a grab bag of techniques ? Every field will have its standard techniques for solving problems, but those will only go so far. > I have never done any measure theoretic probability> but considering my interests would ts be another one> to tnk about? Someone else is probably more qualified to reply to ts :P Good luck, and remember that much of the above is just one personÍs === problemWaringÍs problem has to do with the representation of positiveintegers by means of a sum of non-negative k-th powers: the firstquestion arising is wether for a given k there exists a fixed l=l(k)s.t. there exists a solution of x_1^k+...x_l^k=n for all n and itturns out that if all numbers are representable as a sum of k-thpowers, then there is a least l for wch ts is true, usuallydenoted by g(k).In other words, if we let I_l = {n s.t. n=x_1^k+...+x_l^k does *not* have a solution},then g(k) is the least l for wch I_l is the empty set. But again itturns out that a more interesting (and tough!) problem is that offinding the number G(k) given by the least l s.t. I_l is finite.Phrased differently, the first problem is that of finding how manyk-th powers are required for the representation of *all* numbers,whereas the second one is that of finding how many k-th powers arerequired for the representation of *all sufficiently large* numbers.Now I wonder if anytng is known about about the slightly moregeneral problem of finding how many k-th powers are required for therepresentation of *most* numbers, i.e. of finding the least l forwch (1/n)*card I_l intersects {1...n} -> 0as n->infty.Michele-- youÍll see that it shouldnÍt be so. AND, the writting as usuall isfantastic incompetent. To illustrate, i quote:- Xah Lee trolling on clpmisc, perl bug File::Basename and PerlÍs === e^(-(x^2))dx?Let I be the number we want.I^2> = (integral from 0 to infinity e^(-(x^2))dx)> times (integral from 0 to infinity e^(-(y^2))dy)= double integral (0 to infinity in x and y)[e^-(x^2+y^2)]dy*dx(change to polar coordinates)= double integral(0 <= theta <= pi/2)(0 <= r < infinity)of the function: e^(-r^2) times r*dr*d(theta) infinity> = (-1/2)e^(-r^2)| times pi/2> 0> = pi/4.Since I^2 = pi/4,I = (1/2)sqrt(pi).Holy smoke! ThatÍs neat!I have a question ... very silly ... regarding ts business. Howexactly ts substitution tng work?SPECIFICALLY: What are the necessary and sufficient conditions for asuccessful substitution?Is there anytng about multiple valued functions (probably inversefunctions), open or closed domain and ranges, discontinuities, poles,whatever topological ... that I should be aware of wle doing asubstitution? I know for sure whatever can go wrong normally does notgo wrong, or else I would have done very bad in the exams!Take ts one for example ... coordinating (x,y) plane with (r,theta)has sometng special about it that scares me. x runs from -infinityto infinity, so does y, but r goes just half of that range, andtheta is BOUNDED! I know 0 <= theta <= pi sounds fine, and thatomitting the = in <= would do know harm ... et cetera ... but I justwanted a brief review of the whole story.Anyone care to go through the pain? Just to encourage you: rememberhow it felt when it all came together for you the first === infinity e^(-(x^2))dx? Let I be the number we want. I^2> = (integral from 0 to infinity e^(-(x^2))dx)> times (integral from 0 to infinity e^(-(y^2))dy) = double integral (0 to infinity in x and y)[e^-(x^2+y^2)]dy*dx (change to polar coordinates) = double integral(0 <= theta <= pi/2)(0 <= r < infinity) of the function: e^(-r^2) times r*dr*d(theta) infinity> = (-1/2)e^(-r^2)| times pi/2> 0> = pi/4. Since I^2 = pi/4, I = (1/2)sqrt(pi).> Holy smoke! ThatÍs neat!> I have a question ... very silly ... regarding ts business. How> exactly ts substitution tng work?> SPECIFICALLY: What are the necessary and sufficient conditions for a> successful substitution?> Is there anytng about multiple valued functions (probably inverse> functions), open or closed domain and ranges, discontinuities, poles,> whatever topological ... that I should be aware of wle doing a> substitution? I know for sure whatever can go wrong normally does not> go wrong, or else I would have done very bad in the exams!> Take ts one for example ... coordinating (x,y) plane with (r,theta)> has sometng special about it that scares me. x runs from -infinity> to infinity, so does y, but r goes just half of that range, and> theta is BOUNDED! I know 0 <= theta <= pi sounds fine, and that> omitting the = in <= would do know harm ... et cetera ... but I just> wanted a brief review of the whole story.The reason that it goes from 0 to pi2, is if you graph where the functiongoes wle in dxdy you will see that its only in the first quadrant. The ygoes from 0 to positive infinity and the x goes from 0 to positive infinity.The whole first quadrant goes from 0 to pi/2 and thats why the d_theta hasthose limits.> Anyone care to go through the pain? Just to encourage you: remember> how it felt when it all came together for you the first === integrating from 0 to infinity e^(-(x^2))dx?> -Greg R.IÍd suggest trying a power series.How could that possibly === integrating from 0 to infinity e^(-(x^2))dx? -Greg R. IÍd suggest trying a power series.> How could that possibly help?I was trying to say take the power series for e^x, compose it with -x^2 andthen integrate that; since itÍs improper, youÍd probably have to take alimit. IÍve seen sometng like ts done === help me in integrating from 0 to infinity e^(-(x^2))dx?> -Greg R.>> IÍd suggest trying a power series.>> How could that possibly help?>I was trying to say take the power series for e^x, compose it with -x^2 and>then integrate that; since itÍs improper, youÍd probably have to take a>limit. IÍve seen sometng like ts done in my calculus book.As long as you integrate term-by-term from 0 to A and _then_ let A ->infinity ts is legal (from your first post I assumed you meant takethe integral of each term from 0 to infinity - thatÍs going to give nonsense.) But although itÍs legal itÍs not going to help, becausethereÍs no way youÍre going to be able to evaluate the limitof the resulting series.Hmm, I shouldnÍt jump to conclusions. Just because I canÍtdoesnÍt mean you canÍt. WhatÍs the limit as A -> infinity of sum_0^infinity (-1)^n === can I find pics of cavemens language e.g. A pic of a wall inprintTs really isnÍt a math topic, but anyway, try:http://www.photovault.com/Link/Entertainment/Paintings/ Heiroglypcs/EPHVolume01.html-- Wayne Brown (HPCC #1104) | When your tailÍs in a crack, you improvisefwbrown@bellsouth.net | if youÍre good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- Myers Myers, === peak location, peakheight, peak > width, semi-derivative, derivative, integral, semi-integral, convolution, > deconvolution, curve fitting, and separating overlapped === Hausdorff dimension of graph of Weierstrass function.>[...]Indeed, if t is large enough a very crude argument>suffices to show Box = s.Having dusted off a few cobwebs in the lacunary-series> section of my brain, itÍs clear that Box = s for t > 1> and 1 < s < 2.The proof depends on the fact that the supremum of> a lacunary series is larger than a constant times the> sum of the absolute value of the coefficients. Ts> is well-known, for example it can be proved using> Riesz products (see Katznelson). We need a> slightly sopsticated local version, the proof of> wch must be somewhere in the lacunary series> folder in the filing cabinet in my office:Lemma. Suppose L > 1 and a > 0. There exists> C = C(L, a) > 0 such that if f(x) = sum_1^N a_j cos(n_j x),where n_{j+1}/n_j > L for all j, and I is an interval> of length a/n_1, then the sup of |f| on I is larger> than C sum |a_j|.Assuming that, consider a square grid in the plane> with side length h. Take the sum and split into> the sum from 1 to N - 1 plus the tail, where N is> chosen so that the oscillation of the sum from> 1 to N - 1 on any interval of length h is less than> h (use obvious estimates on the derivatives).> If you choose N maximal with ts property> it turns out that h > a/t^N for some constant> a > 0, so the lemma shows that the max of the> tail minus the min of the tail on any interval of> length h is large. Hence the same is true of f> itself, and ts shows that the graph ts a large> number of squares in the grid, forcing Box > s.Ts seems to be similar to FalconerÍs === Weierstrass function.>>[...]>>Indeed, if t is large enough a very crude argument>>suffices to show Box = s.Having dusted off a few cobwebs in the lacunary-series>> section of my brain, itÍs clear that Box = s for t > 1>> and 1 < s < 2.The proof depends on the fact that the supremum of>> a lacunary series is larger than a constant times the>> sum of the absolute value of the coefficients. Ts>> is well-known, for example it can be proved using>> Riesz products (see Katznelson). We need a>> slightly sopsticated local version, the proof of>> wch must be somewhere in the lacunary series>> folder in the filing cabinet in my office:Lemma. Suppose L > 1 and a > 0. There exists>> C = C(L, a) > 0 such that if f(x) = sum_1^N a_j cos(n_j x),where n_{j+1}/n_j > L for all j, and I is an interval>> of length a/n_1, then the sup of |f| on I is larger>> than C sum |a_j|.Assuming that, consider a square grid in the plane>> with side length h. Take the sum and split into>> the sum from 1 to N - 1 plus the tail, where N is>> chosen so that the oscillation of the sum from>> 1 to N - 1 on any interval of length h is less than>> h (use obvious estimates on the derivatives).>> If you choose N maximal with ts property>> it turns out that h > a/t^N for some constant>> a > 0, so the lemma shows that the max of the>> tail minus the min of the tail on any interval of>> length h is large. Hence the same is true of f>> itself, and ts shows that the graph ts a large>> number of squares in the grid, forcing Box > s.>Ts seems to be similar to FalconerÍs proof.IÍm certain it is - I finally got to the library a few daysago, and the proof of the upper bound that Huntgives is _identical_ to the proof I came up withwhen you started ts thread. ThatÍs because theseare the easy bits.I was a little disappointed when I finally looked upHuntÍs paper. Of course itÍs an extremely keen paper,but from the title IÍd got the impression that heÍdfinally figured out the Hausdorff dimension of thesesuckers. IÍve been tnking a little bit about that -of course I havenÍt proved anytng yet, but I alsohavenÍt yet convinced myself that itÍs all that hard...Taking the approach he takes, you just have toshow that the set where f(t) - f(t+h) is small is small.The functionÍs L^2 norm is right, and lacunary seriessort of behave the same === Re: Pi gallons problem - any ideas?hrllo,>> [...]>>V2 = Ah/3 (r^(3/2) - 1) / (r - 1).>>V2 is the volume of that subset of the frustum below a plane tangent to>>the top and bottom circles at opposite points: i.e., the volume of water>>that would remain in a frustum if it were tipped over until the water>>just touched the bottom circle.>>[...] > plippe 92 how to proove the>>V2 = Ah/3 (r^(3/2) - 1) / (r - 1)>>formula ... I didnÍt succeed....>>(V1 is easy, )Jim Ferry replied:[...]An easy way to prove ts formula is to compute the volume of the shaded> region. The shaded region forms a cone (not a right circular cone, but> an oblique, elliptical cone), so its volume is (1/3) base x height.[...] for showing me ts cone that I didnÍt see ...(I should).-- plippe(chepp at free === = (x+8a)(x+b), where ab=1, and considering when (8a + b) isan integer can give you a perspective on what IÍve been saying, Ihope.Basically the mathematical position of the mainstream--why itÍs acore error--is that you can never have a factor like 8 in an arenawhere -1 and 1 are the only integer units because you canÍt in thering of algebraic integers, except for simple radical factors, if youdonÍt have integer ïaÍ and ïbÍ.ThatÍs it. The arguments now should make sense to you as I talkedabout situation where ïaÍ *should* be a factor of 1, as it is infact--in a more inclusive ring.The §aw with the understanding of the ring of algebraic integers is afascinatingly frustrating one to explain!!!Posters found many ways to confuse the issue, and I still find it hardto believe that experienced professors like Arturo Magidin or RickDecker never figured it out.But itÍs up to them to give a full accounting of why theyÍve said whattheyÍve said.IÍm just glad that I finally thought to put out sometng likeP(x) = (x+8a)(x+b), where ab=1, so P(x) = x^2 + (8a + b)x + 8, with an emphasis on when (8a + b) is aninteger.If youÍre STILL confused, I suggest you play with that quadratic for awle. First consider simple cases like I did, like a=b=1, and seewhat happens, and then move on to irrational ïaÍ and ïbÍ, and see whathappens.Oh yeah, I see notng wrong with the label algebraic integers ortheir definition and wouldnÍt want either changed, but what *has* tobe taught along with what has been previously taught are theinteresting features and limitations of the ring.Universities and mathematicians who continue to refuse to tell thetruth, shouldnÍt expect people to be on their side when at ts pointthereÍs just no excuse.If you wish to be a rogue university, or a rogue math professorrefusing to teach mathematics correctly, then I tnk you should bepunished severely, and no one who respects society and progress shouldhave pity on you.It is, after all, your choice.You have === Frustration?> Using P(x) = (x+8a)(x+b), where ab=1, and considering when (8a + b) is> an integer can give you a perspective on what IÍve been saying, I> hope.Yup. It is still wrong. And for much the same reasom, that you, Jimmy, have no clue as to what is really going on in the relevant mathematics. Your abysmal ignorance keeps you from getting anytng more complicated than gh school algebra correct.Your overweening arrogance keeps you trying.The combination leads to your frequent mathematical pratfalls and the resulting temper tantrums.We all get another good laugh.Keep it === P(x) = (x+8a)(x+b), where ab=1, and considering when (8a + b) is>an integer can give you a perspective on what IÍve been saying, I>hope.>Basically the mathematical position of the mainstream--why itÍs a>core error--is that you can never have a factor like 8 in an arena>where -1 and 1 are the only integer units because you canÍt in the>ring of algebraic integers, except for simple radical factors, if you>donÍt have integer ïaÍ and ïbÍ.Exactly what mathematician ever said that? (I mean, what mathematicianever said sometng about a factor like 8, for that matter?)YouÍre projecting your own misconceptions and feeble understandingonto the mathematical mainstream.>ThatÍs it. The arguments now should make sense to you as I talked>about situation where ïaÍ *should* be a factor of 1, as it is in>fact--in a more inclusive ring.>The §aw with the understanding of the ring of algebraic integers is a>fascinatingly frustrating one to explain!!!>Posters found many ways to confuse the issue, and I still find it hard>to believe that experienced professors like Arturo Magidin or Rick>Decker never figured it out.>But itÍs up to them to give a full accounting of why theyÍve said what>theyÍve said.>IÍm just glad that I finally thought to put out sometng like>P(x) = (x+8a)(x+b), where ab=1, >so P(x) = x^2 + (8a + b)x + 8, with an emphasis on when (8a + b) is an>integer.>If youÍre STILL confused, I suggest you play with that quadratic for a>wle. First consider simple cases like I did, like a=b=1, and see>what happens, and then move on to irrational ïaÍ and ïbÍ, and see what>happens.Uh, weÍre _not_ confused. Well, I imagine IÍm not the only one confused about the questionof exactly what your point is here. But weÍre not confused aboutthe math.>Oh yeah, I see notng wrong with the label algebraic integers or>their definition and wouldnÍt want either changed, but what *has* to>be taught along with what has been previously taught are the>interesting features and limitations of the ring.>Universities and mathematicians who continue to refuse to tell the>truth, shouldnÍt expect people to be on their side when at ts point>thereÍs just no excuse.>If you wish to be a rogue university, or a rogue math professor>refusing to teach mathematics correctly, then I tnk you should be>punished severely, and no one who respects society and progress should>have pity on you.>It is, after all, your choice.>You have been repeatedly warned.And youÍre also sounding like a === sha1:1ypuwbB94gHnqTVuUpnp2m7+Fk4=> If you wish to be a rogue university, or a rogue math professor> refusing to teach mathematics correctly, then I tnk you should be> punished severely, and no one who respects society and progress should> have pity on you.What if we suspect that our university is rogue? How can we tell forsure?Do rogue universities §y the Jolly Roger? Do they advertise foremployment opportunities in rogue journals? I know a guy who wishes to be a rogue university. How should wepunish m severely? Who do I call?So many questions... please help...-- IÍve ... contacted [some of the...] ghest I.Q.Ís in the country...IÍve even helped the FBI out a few times... IÍve met at least onegovernor..., a senator... and IÍve had some really good seats atsports games. My experiences are not your experiences. --JSH != === wish to be a rogue university, or a rogue math professor>> refusing to teach mathematics correctly, then I tnk you should be>> punished severely, and no one who respects society and progress should>> have pity on you.>What if we suspect that our university is rogue? How can we tell for>sure? W. Bush has, with the help of JSH compiled a list of RogueUniversities... canÍt tnk of a punch line, well never mind.>Do rogue universities §y the Jolly Roger? Do they advertise for>employment opportunities in rogue journals? >I know a guy who wishes to be a rogue university. How should we>punish m severely? Who do I call?>So many questions... === Using P(x) = (x+8a)(x+b), where ab=1, and considering when (8a + b) is> an integer can give you a perspective on what IÍve been saying, I> hope.> Basically the mathematical position of the mainstream--why itÍs a> core error--is that you can never have a factor like 8 in an arena> where -1 and 1 are the only integer units because you canÍt in the> ring of algebraic integers, except for simple radical factors, if you> donÍt have integer ïaÍ and ïbÍ.> ThatÍs it. The arguments now should make sense to you as I talked> about situation where ïaÍ *should* be a factor of 1, as it is in> fact--in a more inclusive ring.> The §aw with the understanding of the ring of algebraic integers is a> fascinatingly frustrating one to explain!!!> Posters found many ways to confuse the issue, and I still find it hard> to believe that experienced professors like Arturo Magidin or Rick> Decker never figured it out.> But itÍs up to them to give a full accounting of why theyÍve said what> theyÍve said.> IÍm just glad that I finally thought to put out sometng like> P(x) = (x+8a)(x+b), where ab=1,> so P(x) = x^2 + (8a + b)x + 8, with an emphasis on when (8a + b) is an> integer.> If youÍre STILL confused, I suggest you play with that quadratic for a> wle. First consider simple cases like I did, like a=b=1, and see> what happens, and then move on to irrational ïaÍ and ïbÍ, and see what> happens.> Oh yeah, I see notng wrong with the label algebraic integers or> their definition and wouldnÍt want either changed, but what *has* to> be taught along with what has been previously taught are the> interesting features and limitations of the ring.> Universities and mathematicians who continue to refuse to tell the> truth, shouldnÍt expect people to be on their side when at ts point> thereÍs just no excuse.> If you wish to be a rogue university, or a rogue math professor> refusing to teach mathematics correctly, then I tnk you should be> punished severely, and no one who respects society and progress should> have pity on you.> It is, after all, your choice.> You have been repeatedly warned.> Yeah, weÍve been warned by someone who doesnÍt even understand thearguments. Harris, you are a crank, and everyone but you knows it. Plus, howare you going to prove ts math is wrong, say to a judge, if you take === Genetics and Math-AbilityYou avoid the question of why does it get inhereted, indeed why does itexist at all?The evolutionary pressures on our brains were forged in the Savanahas ofAfrica over a million year period. I can understand the evolutionarypressure on being able to count 1,2,many, many many Wildebeests. Butcalculate the surface area integral of the plain on wch they run? You seesome specilaised organ in Nature, and you look for the purpose wch let itevolve. That doesnÍt exist for maths.There is a fasonable theory of intelligence, called the peacocks tailtheory of human intelliegence. The analogy is to the peacockÍs tail, wchhas no functional use, except to attract mates. That alone allows the traitto develop. But do women really want to go to bed with guys who are good atmath? And if that was so, could it have had enough evolutionary pressure tomake sometng that can make http://mathworld.wolfram.com/?.> Just curious:> What is the current scientific opinion regarding how math-ability is> inherited?> Is it, in general, a recessive trait or is it dominate?> (I say recessive.)> I am only half-serious, since math ability is most probably a result> of many traits and experiences and education.> But I was wondering today because, even though there are some in my> family who were math-teachers, almost everyone else had/has little> ability with mathematics...even very little ability as relative to> me...> (ïrelative to meÍ...snicker, snicker...)> ,> === question of why does it get inhereted, indeed why does it> exist at all?The evolutionary pressures on our brains were forged in the Savanahas of> Africa over a million year period. I can understand the evolutionary> pressure on being able to count 1,2,many, many many Wildebeests. But> calculate the surface area integral of the plain on wch they run? You see> some specilaised organ in Nature, and you look for the purpose wch let it> evolve. That doesnÍt exist for maths.The ability to recognise (and perhaps invent) patterns, to observesimilarities, to argue with neighbours are important too;all of these contribute to mathematical ability.Moreover, it is not the inheritance of special abilities, that is reallyneeded; inheritance of a suitable general ability that can be adaptedtowards a special ability (depending on === Genetics and Math-Ability> My wife, her sister and her first cousin are also married to > mathematicians. Obviously thereÍs an inherited tendency to marry > mathematicians.Have you noticed that people whose parentÍs did not have cldren, also === neededIÍm having trouble integrating by parts. For example:I = Integral(x cox(2x) dx)dv = x, v = x^2/2u = cos(2x), du = (1/2)sin(2x)I = ((x^2/2)cos(2x)) - ((1/4)Integral((x^2)sin(2x) dx))Then IÍm not sure what to do because the integral on the right seems just ascomplex as the one I started with...?BTW === ETAtAhRVpGuxBQYt8NzeGNgX1i6q/r8AIVAKDhSeoSmE1/uDnNxUU/Z6pLhHv6 v to sart. If you set u = x and dv= cos 2x dx, tngs should work out nicely.We are motivated to ts substitution, instead of the one you tried, byknowing that dv = x dx will lead to a gher power of x and that doesnot look good for simplifying the === IÍm having trouble integrating by parts. For example:I = Integral(x cox(2x) dx)dv = x, v = x^2/2> u = cos(2x), du = (1/2)sin(2x)I = ((x^2/2)cos(2x)) - ((1/4)Integral((x^2)sin(2x) dx))Then IÍm not sure what to do because the integral on the right seems just as> complex as the one I started with...?> BTW what do we use on here for the integration symbol?Try it the other way round with u = x and dv = cos(2 x)dx.A bit of practice is needed before one gets good at choosing what should be the u and what the dv in applying integration by === case if you swap u and v you should find it easier. (See VirgilÍspost)To avoid getting a more complicated result, select the term in x, (or powersof x), to be u, the one that gets differentiated, and select the trigfunction to be v, wch will be the one integrated. Eventually, (perhaps byrepeated application), the term in powers of x will reduce to 1.[However, your effort is not wasted, because the integral and other termsthat you got by doing it the wrong way, are equal to the results you getwhen doing it as above; thus you get the result for a further new integral.]Ian === by parts. For example:> I = Integral(x cox(2x) dx)> dv = x, v = x^2/2> u = cos(2x), du = (1/2)sin(2x)> I = ((x^2/2)cos(2x)) - ((1/4)Integral((x^2)sin(2x) dx))> Then IÍm not sure what to do because the integral on the right seems> just as complex as the one I started with...?> BTW what do we use on here for the integration symbol?I usr a funcional notation:Int(f(x), x) for indefinite integral and Int(f(x), x, a, b) for definiteintegral between a and b.In order to choose u and dv are two guidelines usually useful:1) You must know how integrate dv (mandatory!)2) du must be simpler that uAnd if you election leads to a more complicate integral, try a reverse one.Sometimes, you can get in the RHS the same integral, multiplied by acoefficient different from 1, and you can move it to LHS.-- Best === example:>I = Integral(x cox(2x) dx)>dv = x, v = x^2/2>u = cos(2x), du = (1/2)sin(2x)>I = ((x^2/2)cos(2x)) - ((1/4)Integral((x^2)sin(2x) dx))>Then IÍm not sure what to do because the integral on the right seems just as>complex as the one I started with...?A good idea when you are stuck with an integration by parts is to lookfor a different choice of u and dv.A better choice is u = x and dv = cos(2x) dx. Then v = sin(2x)/2 anddu = dx. Integration by parts then yields | | x cos(2x) dx | | = x sin(2x)/2 - | sin(2x)/2 dx | = x sin(2x)/2 + cos(2x)/4 + C>BTW what do we use on here for the integration symbol?There is no single way that is used to represent the non-ASCII symbols.As long as others can understand what you meant them to, wchever wayyou choose will probably be fine. If people have a hard time figuringout what you have written, you might want to try sometng else.ASCII art symbols, such as I have used above, are okay, but the readermust be using a monospaced font in order for different rows to line up.I tnk that most people who read scientific newsgroups use such a fontfor just ts reason.Rob son take out the === needed> IÍm having trouble integrating by parts. For example:I = Integral(x cox(2x) dx)nt: INT(u*cosu du) = u*sinu + INT(sinu du) =u*sin u + cosu.Do a differentiation to check and we get u*cosu - sin u - sin u = u*cosuWhen u dv does not work, do v === boundary=----=_NextPart_000_0008_01C3F868.B7918380--------- get a chance to respond sooner to Bob SilvermanÍs note.As Bob points out there was a §aw in my reasoning. I should not have assumed thatthe probability of prime p divides N is independent of the probability that other primesdivide N. I best understand ts by mentally constructing a roulette wheel.The wheel has 6 positions on it numbered 1 to 6. I mark each position that is divisible by 2 in red.I randomly select a number by spining the wheel. The chances of landing on a red position are 1/2.Now, I construct a second wheel identical to the first and mark each position that is divisible by 3 in red.I randomly select a number by spining the wheel. The chances of landing on a red posibition are 2/3.For the two events to be truly independent the wheels would spin independently of each other and I could calculate the joint probability as 1/2 * 2/3. However, the requirement is that the two wheels beconnected somehow so they stop at the same position. As a consequence the events are not independent.I can represent the two wheels as follows:(divisible by 2) bRbRbR(divisible by 3) bbRbbRwhere b is a black position and R is a red position.Notice that the last position for both wheels is R. LetÍs construct the two wheels so they have 12 positions:bRbRbRbRbRbRbbRbbRbbRbbRThe pattern of black and red positions repeats. Whatever we can say about the first set of 6 positions holds for the secondset of 6 positions. If we randomly choose a position (the same position for each wheel because they are connected),what is the probability that both wheels will be black (that is not divisible by 2 and not divisible by 3)?There are only 2 out of 6 ways for both wheels to be black. Interestingly 2/6 equals 1/2 * 2/3. I tnk ts can be generalized ( and I emphaisize *tnk* not *prove* ) that for any number N less than p * q, the probabilityof N not being divisible by both p and q is ((p-1)/p) * ((q-1)/q).Ts appears to continue to hold if we go to 3 (say p, q, and s where N is less than p * q * s). The probability of N not being divisibleby all three appears to equal ((p-1)/p) * ((q-1)/q) * ((s-1/s). Ts is easy to demonstrate if p = 2, q = 3 and s = 5 and p*q*s = 30.* * * * * * * * bRbRbRbRbRbRbRbRbRbRbRbRbRbRbRbbRbbRbbRbbRbbRbbRbbRbbRbbRbbRbbb bRbbbbRbbbbRbbbbRbbbbRbbbbRThe asterisks indicate positions that are all black and there are 8 of them. Doing the above multiplicationgives 8/30.Ts does not make those positions prime. It only means that they are not divisible by 2,3, and 5. Thosepositions could certainly be divisible by numbers greater than 5. The problem is that by consideringnumbers greater than 5 would require a roulette wheel with more positions. In the case of 2,3,5, and 7 thewheel would have to have 210 positions. I could find numbers not divisible by 2,3,5, and 7 using such a wheelbut it would not work for numbers greater than 7. And, therein lies the error in myreasoning. PS as for my statements about having a proof, I apologize. I was got a little too enthusiasitic.Furthermore, your argument is *almost*, but notmine, but part of that seems to have been left out.*Seemingly* the prob that N is prime should beprod(1-1/p) for p < sqrt(N). But there is a subtleproblem that I discussed in the prior post, but wchseems to have ben omitted.Let N be a large but unspecified integer. By thePNT, the probability that it is prime is ~1/log(N).Now consider the following. *IF* (and as we shall see ts is a big if) theprobability that a primep divides N is independent of the probability thatother primes divide N, then indeed the probability that N is prime is product of (1-1/p i) for p i < SQRT(N). However:Consider P = product(p i) for p i < x.As soon as x ~ log(N), the probabilities are no longer independent. Why? Because by the PNTwe havepsi(x) ~ x and thus P > N. i.e. once you get enough primes in theprobability product, then the product of p will exceed N. Now, theprobability of the next prime you add into your product is no longerindependent.What you are attempting is a sieve method proofof the PNT and it canÍt be done. I now have to gettechnical. Sometng known as the fundamentallemma of the sieve gets in the way. You cannot insert SQRT(N) primes into the product, becauseyou lose independence. (also estimation errorstarts to creep in, but I donÍt discuss that here)See Halberstam & Richert Sieve Methods. Be careful; it is a difficult book. I have kicked it across the room more than once in frustration.Also look up MertensÍ theorem in Hardy & Wright.The constant one gets in MertensÍ theorem(e^gamma) differs from the constant in the PNT(wch is 1) precisely because independence islost as soon as one forces the number of primesin the product beyond log(N). Indeed. If one asks, what is the probability thata large integer N is not divisible by any primes less than x where x is SUFFICIENTLY SMALL WITH RESPECT TO N, then the answer is indeedproduct(1-1/p) for p < x. But you canÍt take xbeyond log(N). And to determine the probability that N is prime you need totake x up to sqrt(N).Thus you canÍt really use ts argument to estimatethe probability that N is === Number[Quoting Bob Silverman:]> See Halberstam & Richert Sieve Methods. > Be careful; it is a difficult book. I have kicked it > across the room more than once in frustration.It seems that the world is divided into the haves and the have nots, as ts book seems to be no longer available at all.If anyone has a copy that they no longer care for, IÍm certainly in the market for a copy.Amazon, however, does say:http://www.amazon.com/exec/obidos/tg/detail/-/0123182506/ qid=1077381040/sr=1-4/ref=sr_1_4/104-5847946-5462337?v=glance& s=books<< result short enough for some kind soul to post it, or need I make a trip > to the library? (I have citations.)An elementary proof (no elliptic curves) is based on descent. That> is, assume a solution exists in positive x, y, z, and show that> another one can be found with smaller z. The proof is not> particularly difficult but too long to type out here (one single-space> textbook page). It can be found in many basic number theory books,> e.g. Niven and Zuckerman in the chapter on Diophantine equations, p.> 107 in the 3rd edition.A proof of the Beal Conjecture was posted 2/18/04. Descent is not === Conjecture was posted 2/18/04. Descent is not used.The second of what?Have you won the dosh yet?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes s penis for a square root, 88-9Francis Wheen, _How === ETAsAhRO71Sss8Pu0NTos/ w1OHq82cZgwwIUdz53qtcwP4RJywFO90nMnqpQ1us= have the same parity and thus for someintegers m and n we have:a = m+nb = m-nThen a^2+b^2 = 2*(m^2+n^2) and therefore m^2+n^2 = c^2. The problem ofthree squares in arithmetic progression is thus reduced to the morefamiliar problem of === both x^4 + y^4 = z^2 and> x^4 - y^4 = z^2 can be found in :> http://www.mathpages.com/home/kmath288.htm, IÍll keep the === the anticlassicalist }{ i: linguistic negation)> In message <1039dr5kpjloif1@corp.supernews.com>, galathaea >: >You should have realised that anytng full of -ism and -ist wch is> spewed>: >to sci.lang and sci.logic is not going to contain any physics.>: : (Nor language; I canÍt speak for sci.logic ;-)>:>: IÍm well aware of that. What I was wondering was whether the _OP_>: realises that thereÍs a difference between these isms and physics, and>: what it consists of.Installment vi>>Before you lies the Void...> >details the properties of the closed linear subspace lattice>of a lbert space, detailing the relationsp with observables and>prediction in quantum systems. Ts builds off of the mathematical theoryYou do understand the physical content, do you not?HavenÍt seen any.Dirac compressed quantum mechanics into ~300 pages of terse elegance, > and didnÍt use ontology once.Well then, he was tersely codifying the mathematical structure as agiven, wch is the easier hard direction. Going in the less easyhard direction requires talking about sometng very much likeontology, whether or not we use the word ... like speaking prose, andall that. ItÍs our old school chums deduction and induction.Basically, professor, it goes kind of like ts (music please ...):Models as a Confidence GameIn the beginning we are confronted with a blooming buzzing surface ofsensory confusion. Being cerebral beings, we create a partiallyordered heirarchy of model tngs, purported to be generative of tssurface wch weÍve been presented. The models are essentially,roughly and enoughly, our working ontology.There: IÍve used ontology in a sentence, and it wasnÍt so bad, wasit? ;-)(Richard Herring vigorously spits and loudly calls for a pint).Some of these models; in a model itself received and conceived -- itÍsall heirarccal, you see -- are for us evolutionarily preconceived:in our lizard brains, ancient and tough, a world populated withpersistent three-dimensional objects is enough.Onto these models we multiply and add; mathematical annexes, now morethan a fad, allow us to predict with numeric decision the results ofsome experiments with admirable precision. Ts predictive concisionincreases our confidence that weÍre on to sometnÍ, as we laboriouslypeel back the skins of the onion -- and well it should: a rigorouspredictive structure is quite good.Confidence, confidence ... did I not add in my vision that each modelis invested with Bayesian strength; and though we sometimes forget weshould add at greater length they are all always subject to empricalrevision?Hssst! The doggerels out. But I tnk you may get the idea: In the great chain of model being we may focus on some precisemathematical bits, the triumphs of a natural science known to becorrect in a range of observation with a degree confidence onlypredicated on a few assumptions; like sanity and the persistence oforder. Once given these bits, reasoning from their structure isprimarily downward -- deductive -- wch is sometng wch cansimilarly be carried out with great rigor and confidence. But we maytend to forget that these models were not themselves handed to us on aplatter -- or rather they were, if we adopt the stance that they areknown givens -- but each the product of protracted groping by groupsof the most talented people of their time, there being no mechanisticcrank we can turn to integrate as readily as we differentiate. And ints groping, like it or not, sometng with a strong isometry to theologies will out; a discussion of === Re: Help needed from a group theorist>An exponent E(G) of a group G is the lcm of all the orders o(g),>g in G.We have that G is cyclic if and only if E(G)=|G|. No we donÍt. Try the nonabelian group of order 6.>Does anybody have an idea how can we find all n such that>the exponent of Z_n*,the group of units of Z/Zn is 2? >E(Z_n*)=2 n=?>The solution is n=2,3,4,6,8,12,24 ,but I donÍt have a clue>where did ts come from. Well Z_n* is the direct product of Z_{p^r}* for the prime powerfactors p^r of n, so ts reduces to n = p^r.For that, Z_{p^r} is cyclic for p > 2 or for p^r = 2, 4,whereas Z_{2^r}* = C_2 x C_{2^{r-2}} for r >= 3.So Z_{p^r} has exponent 2 for p^r = 2, 4, 8, or 3, wch meansZ_n has exponent 2 for exactly those values of n that === group theorist> An exponent E(G) of a group G is the lcm of all the orders o(g),> g in G.We have that G is cyclic if and only if E(G)=|G|.> Does anybody have an idea how can we find all n such that> the exponent of Z_n*,the group of units of Z/Zn is 2?> E(Z_n*)=2 n=?Ts just means that x^2=1 for any invertible x in Z_n. Thus n cannot haveany prime factors other than 2 and 3. Write n=(2^s)(3^t). Splitting Z_n as aproduct of two rings, the cases are-- E(Z_(2^s)*)=2 and E(Z_(3^t)*)=1 i.e. s>=1 and t=0 (n=2,4,8,16,32,...)-- E(Z_(2^s)*)=1 and E(Z_(3^t)*)=2 i.e. s=0 and t=1 (n=3 only)-- E(Z_(2^s)*)=2 and E(Z_(3^t)*)=2 i.e. s>=1 and t=1 (n=6,12,24,48,...)I hope I === important is r^2 in statistics?IÍm running a stepwise linear regression on a dependent variable. Mystatistics program created some models wch in wch the independentvariables have t values of more than 3, wch makes them significant on a 1%level. However, the r^2 of those models is never gher than .14. Of howmuch importance is ts r^2?any comments are === exactly is to choose a branch of log z for complex z? IÍd> appreciate if someone help me picture ts. There is an easier way than using a Riemann surface wch is whatthe wolfram reference gives you. See StewartÍs Advanced Calculus === defined as x^p=1The equation x^p=1where solutions w = e^{2pi*k*i/p}are the roots of unity sometimescalled de Moivre numbers.Now I wonder whether there is some math names for M^p=IM and I are nxn matrix, and I is indentity a === logically possible for sometng, a mathematical statement, to be|absolutely undeterminable?Goedel once expressed an interest in developing a concept of absoluteprovability, but so far as I know nobody has done it, so the question issomewhat undefined.I tnk itÍs plausible that certain questions in mathematical logic arein fact undeterminable in a strong sense.Chaitin has defined an uncomputable real number called Omega.Omega is uncompressable in the sense that for each programminglanguage there exists a constant C such that for any N, a programwch correctly computes the first N bits of Omega is at least N-C bitslong itself.One way to construct a program for computing bits of Omegais to have it search for proofs in a given formal system whoseconclusion is that a given bit of Omega is 0 or that it is 1. Theprogram can then output them in order. So any formal systemcapable of proving what the first N bits of Omega are has to beclose to N bits long as well.I tnk ts provides a pretty convincing argument that all but thefirst few bits of Omega are beyond any reasonable determination.There may be longish formal systems that we believe to becorrect as far as their arithmetic consequences go, but presumablybecause of more fundamental grounds. It seems unlikely that wehave a large amount of innate knowledge of mathematics, in thesense of knowing facts of arithmetic that absolutely cannot bejustified except by assuming a lengthy set of axioms. IÍve seenZF condensed to a relatively short (smaller) set of axioms. gheraxioms of infinity donÍt require much more.The theorem mentioned above leaves open the possibility, of course,of an axiom system being short and deciding the n-th bit of Omegafor some large n, without deciding all the previous ones. Omega isdefined as the probability of a Turing macne halting, given acertain way of constructing one at random. It seems implausible tome that a natural axiom system would decide bits very muchbeyond the first one it couldnÍt decide.Ts is just a plausibility argument, of course, and without an actualdefinition of absolute provability thereÍs no === Re: I got low score on math test, please advise me and take a lookassuming that your report about progress feedback is accurate, a> rejection of the request above is a good glimpse into an incompetent> institution. what kind of justification were you offered? some> bureaucratic non-sense?My guess is that justification offered was THE RULE THAT YOU CANÍT DROP A> COURSE AFTER A CERTAIN DATE.in other words, bureaucratic non-sense. donÍt you tnk that such ruleshould have a provision stateing that students need some performancefeedback prior to the drop deadline?i now realise that i mis-read the guyÍs message. i thought he wasdenied to withdraw, when in fact he was denied to withdraw without aW. i take back my comment === test, please advise me and take a look> assuming that your report about progress feedback is accurate, a rejection of the request above is a good glimpse into an incompetent institution. what kind of justification were you offered? some bureaucratic non-sense? My guess is that justification offered was THE RULE THAT YOU CANÍTDROP A> COURSE AFTER A CERTAIN DATE.> in other words, bureaucratic non-sense. donÍt you tnk that such rule> should have a provision stateing that students need some performance> feedback prior to the drop deadline?How do you know that he didnÍt get any performance feedback before thedeadline?He only said that he didnÍt get TS TEST back before the deadline.How do you know if ts is the first test? How do you know if heÍs hadother assignments, or how many?Ts troll has only told us what he wants us to know. And whateverrespect I had for m, I lost when he started sending me === trying to prove the following theorem regarding vector spaces:If u_1, u_2, ..., u_n span the vector space V, then some subset of {u_1,u_2, ..., u_n} is a basis of V.The following is my attempt at a proof and IÍd be very grateful if someonecould verify its correctness. I also use a well-known lemma wch I willrefer to as lemma 1. Ts lemma is:**Lemma 1:Suppose f_1, f_2, ..., f_n span a vector space Q, then one of the followingholds:i). f_1, f_2, ..., f_n is a basis of Q.ii). There exists a j (0 < j <= n), such that removing f_j from f_1, f_2,..., f_n still leaves a spanning set of Q.**I argue by cases. Either V = {0} or V != {0} (where {0} is the setconsisting of the zero-vector, not a scalar 0, and ï!=Í means ïnot equaltoÍ).If V = {0}, then by definition, its basis is the empty set. Clearly theempty set is a subset of {u_1, u_2, ..., u_n}.For the case V != {0}, we aim for a contradiction by assuming that theredoesnÍt exist a subset of {u_1, u_2, ..., u_n} wch is a basis of V.Now since u_1, u_2, ..., u_n span V, one of the following is true by lemma1:i). u_1, u_2, ..., u_n is a basis of V.ii). There exists a j (0 < j <= n), such that removing u_j from u_1, u_2,..., u_n still leaves a spanning set of V.By our assumption, i) is not true, hence ii) must be. We now consider thesubset of {u_1, u_2, ..., u_n}, wch still spans V, formed by removing thesaid element. Therefore ts subset must also satisfy either i) or ii) o§emma 1, and again since it cannot satisfy i), it must satisfy ii). Clearlywe can repeat ts argument until we conclude that the empty set spans V,wch is clearly a contradiction since V != {0}. Hence our originalassumption is === spaces theorem>I am trying to prove the following theorem regarding vector spaces:>If u_1, u_2, ..., u_n span the vector space V, then some subset of {u_1,>u_2, ..., u_n} is a basis of V.> [...]Someone else has verified your proof, but here is a much simpler one: Let A be a maximal linearly independent subset of {u1,..., u_n}. It is farily straightforward to show A spans V.-- Stephen J. Herschkorn === spaces theorem> I am trying to prove the following theorem regarding vector spaces:If u_1, u_2, ..., u_n span the vector space V, then some subset of {u_1,> u_2, ..., u_n} is a basis of V.The following is my attempt at a proof and IÍd be very grateful if someone> could verify its correctness.ItÍs === eigenvalues/vectors by eye-inspection?For simple matrices, such as[6 4; 4 6][a b; b c][a b;-b c][3 2; 2 3]etc...Any === eigenvalues/vectors by eye-inspection?> For simple matrices, such as[...] > [a b;> b c][a b;> -b c][3 2;> 2 3]The eigenvalues are extremely simple:>> A := matrix([[a,b],[b,a]]) +- -+ | a, b | | | | b, a | +- -+>> linalg::charpoly(A,s) 2 2 2 s - (2 a) s + (a - b )>> linalg::eigenvalues(A) {a + b, a - b}>> A := matrix([[a,b],[-b,a]]) +- -+ | a, b | | | | -b, a | +- -+>> linalg::charpoly(A,s) 2 2 2 s - (2 a) s + (a + b )>> === HELL :-) (Was: Re: the anticlassicalist }{ ii: the spectre continues)> [...]> mitch,> I donÍt snip here as a skip. I just wanted to say that I see clearly> you are one who has already learned that the only secret bend any of> ts is the hunt for patterns. Going out and grabbing disparate works to> look for connections, then hunting down connections to those, and building> them into pictures. When the connections become rigorously symbolised, we> have math (or logic, or numerology, or symbology -- whatever you like).> When a model connects ontology to epistemology, a theory is born.Or lack thereof. Paraconsistency does not admit theory.I want to say The grapcal representation init expresses the basis bend a lot of my confusion.There is a tendency for everyone to decompose problems. When they set up afoundation for reasoning, they always seem to do ts in the same way. Well,that is just not a sensible global outlook. The equilibrium state ofcross-purposeful reasoning will end up with a re§ective structure in wch twoalignments will exchange 0Ís and 1Ís across two consistent frameworks.The classical truth table semantics re§ect idempotence in the sense of deliterature, de Morgan idempotence can account for game-theoretic communication(tit-for-tat) whereas Boolean idempotence can merely represent coherent,incomplete, static theories. Just some big mathematics to account for the roleof essence in AristotleÍs delineation between homonymous and synonymous usage.Linguists must have a lot of fun.So, here is another post that is uninteresting or irrelevant or uniformativeor... whatever other criticisms have been thrown my way for lack of a theory. again Galathaea.:-)mitch> Its nice watcng people have fun with their patterns, play with them,> fit the pieces together. Its almost as fun as doing it myself.> I wanted to say ts because I see many on usenet who frown on such> behavior. They seem to look only for some kind of stale acceptance of a> never surprising universe, and look to attack any excitement or creativity> they see.> So I sometimes want to balance the powers, give praise at the> interesting twists and turns I see people playing with. I have a cld like> wonder for cld like wonder.> You even picked up Conway and Sloan... That was the first book I ever> checked out of a university library (I didnÍt understand it then, but I kept> coming back to try again).> --> -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-> === Model theory puzzle> Find a finite system of axioms (feel free to introduce operations,> relations, constants) so that all of its finite models would have a> prime number of elements, and for every prime p thereÍs a model to> that system of size p.Start with the axioms for a field and introduce a trd binary operation ^for exponentiation, satisfying the following axioms:For all y, 0^y=0.For all nonzero x, x^0=1.For all x and for all y not equal to -1, x^(y+1)=x*(x^y)For all nonzero x, we have x^(-1) = 1In the field with p elements, for nonzero x, one defines x^y to be theproduct of z copies of x, where z is the integer between 0 and p-1representing y, with the convention that x^y is 1 for y=0.If a finite field F of characteristic p satisfies these axioms,the last axiom implies that the product of p-1 copies of a nonzero elementx equals 1, wch implies that the field has only p elements.Allan Adlerara@zurich.ai.mit.edu***** ** Disclaimer: I am a guest and *not* a member of the MIT Artificial ** Intelligence Lab. My actions and comments do not re§ect ** in any way on MIT. Moreover, I am nowhere near the Boston ** metropolitan area. === improvement is to note that the even part of R(x),> (R(x)+R(-x))/2, is equal to 1/p(x), thus only the odd part of R(x),> (R(x)-R(-x))/2, needs to be computed:double P(double x)> {long double s,t=0,b=1,pwr=x;> int i;> s=x;> for(i=2;s!=t;i+=2)> { b/=(i+1);> pwr=pwr*x*x;> t=s;> s+=pwr*b;> }> return .5+s*exp(-.5*x*x-.91893853320467274178L);> }Very nice, and a definite improvement*! It speeds up the convergence> considerably wch, quite possibly, accounts for more accurate results.> i.e. w/CVF on Wintel, x p(x)> 0.123 0.5489464510164368> 1.200 0.8849303297782917> 2.400 0.9918024640754040> 6.100 0.9999999994696567> -6.100 0.0000000005303433> -1.100 0.1356660609463827> 7.200 0.9999999999996979* Actually, itÍs an understatement considering the Syzipan effort at> LANL some odd trty years ago. See, netlibfn lib.I translated the C code to Fortran 95, as shown below. With CVF, theresults agree with those of B. Voh. Using the Lahey/Fujitsu Fortran 95compiler with the option -quad, wch extends DOUBLE PRECISION toquadruple precision, the results agree with the true values posted by Marsaglia to at least 18 decimal places, and often the full 20.Translating both the double and long double C variables to DOUBLEPRECISION in Fortran seem wrong to me, although the results are good.What is the proper translation? x p(x) 0.123 0.54894645101643675908 1.200 0.88493032977829173198 2.400 0.99180246407540387056 6.100 0.99999999946965767371 -6.100 0.00000000053034232629 -1.100 0.13566606094638267517 7.200 0.99999999999969893720 elemental function p(x) result(y)implicit nonedouble precision, intent(in) :: xdouble precision :: ydouble precision :: s,t,b,pwrinteger :: it = 0.0d0b = 1.0d0pwr = xi = 2s = xdo if (s == t) exit b = b/(i+1) pwr = pwr*x**2 t = s s = s + pwr*b i = i + 2end doy = 0.5d0 + s*exp(-0.5*x**2 - 0.91893853320467274178d0)end function pprogram xpimplicit noneinteger, parameter :: n = 7double precision, parameter :: x(n) =(/0.123d0,1.200d0,2.400d0,6.100d0, & -6.100d0,-1.100d0,7.200d0/)integer :: idouble precision :: pexternal pdo i=1,n write (*,(f8.3,f24.20)) x(i),p(x(i))end === translated the C code to Fortran 95, as shown below. With CVF, the> results agree with those of B. Voh. Using the Lahey/Fujitsu Fortran 95> compiler with the option -quad, wch extends DOUBLE PRECISION to> quadruple precision, the results agree with the true values posted by> Marsaglia to at least 18 decimal places, and often the full 20.> Translating both the double and long double C variables to DOUBLE> PRECISION in Fortran seem wrong to me, although the results are good.> What is the proper translation?[...]Double in C and Fortran 95 should be the same, ie, 64 bit. LaheyÍs quad is128 bit, and CÍs long double is 80 bit. The only Fortran 95 compilers forWin32 with support for a long double type are the two === Re: ïerfÍ function in C ...> Is there someplace a table available of the true values of the> normal distribution up to 20 places?A 20 decimal place table of normal percentiles was published in JASA Percentiles by S. Wte. It is downloadable === === Re: Great mistakes of the physicistshttp://hamidvansari.topcities.com/ You see yourself ts way,http://www.mazepath.com/uncleal/effete6.jpg The entire remainder of the planet sees you ts way,http://www.mazepath.com/uncleal/effete3.pnghttp://b5. sdvc.uwyo.edu/bab5/snds/argcstpd.wavhttp://w0rli.home.att.net/ youare.swfhttp://www.mazepath.com/uncleal/sunsne.jpg-- Uncle Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for cldren and most mammals)Quis custodiet ipsos custodes? The === (theoretical) CS?> I have been coerced into teacng a Honors course the Fall> (mostly for non-CS freshman/sophomore Honors students). My> idea was to do some Great Ideas/Problems/Puzzles etc. from> computer science -- emphasis on theory/algorithms > and related areas like graph theory/combinatorics.> Of course, the honors college wants a syllabus in one week!> I looked at the book, Great Ideas in CS and though a nice> book, seems a bit light on the theory side ... given that> I want to focus on theory to keep me interested. I saw the > course/web site at CMU Great Ideas in Theoretical Computer Science > and may use that as a guide for some of the course. Examples> of some tngs I might discuss (besides a couple weeks on> basics/definitions/story) include Towers of Hanoi, Byzantine> Generals, voting problems, maybe a gentle discussion> of interactive proofs, prisonerÍs dilemma, game of life, primality> testing, graph coloring ... anytng that can be discussed in a > day or so to folks with no CS background, yet wch has some > theory component to it ... stuff that is surprising or counter-intuitive> is all the better ;)> Anyway, if anyone has any suggestions for material/topics> that I might cover, I would most appreciate it. Any pointers> would be accessible to students would be great (I have a couple)> would be great. in advance. If I get enough interest, I will post a syllabus> (or a link to one) to comp.theory in a week or so.Cp KlostermeyerOne of my favorite topics is Turing macnes and the Halting Problem. It has a very immediate application, accessible to most undergraduatestudents, to wit the difference between syntax and semantics. Compilers, programs themselves, routinely check all programs in alanguage for syntactical correctness. But by the Halting Problemresult, no program can check all programs === ideas in (theoretical) CS?The structure theorem:Any program can be transformed to a program whose structure is formedfrom combining sequences, wle-loops, and an if-then-else structure.The use of invariants, such as loop invariants, in correctness proofs.The structure theorem can be found proven in H. D. Mill et. al.ÍsStructured Programming. Some neat algorithms and ideas along theselines are be in DijkstraÍs A Discipline of Programming. See alsoGriesÍ Science of Programming.By the way, I recently convinced a colleague of mine that ConwayÍsGame of Life was a universal computer. I showed m memory, an ANDgate, and a NOT gate. Maybe somtng like that would be neat aspart of a lecture on the Church-Turing thesis.-- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.htmlr c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but ghly @ r c m unbecoming to reasonable and free men in === course on great ideas in (theoretical) CS?My CS teacher had a very good lecture on The set of all computablefunctions is countable. He told us: Tnk of the content of thememory of a computer, including executable code. It all boils down toa very big integer, doesnÍt it?The Knapsack problem and Travelling Salesman problem, IMHO, are worthyof paying attention. Or are you looking for sometng more dramatic?Of course, one tng you cannot forget to mention is the P=NP === (theoretical) CS?> and may use that as a guide for some of the course. Examples> of some tngs I might discuss (besides a couple weeks on> basics/definitions/story) include Towers of Hanoi, Byzantine> Generals, voting problems, maybe a gentle discussion> of interactive proofs, prisonerÍs dilemma, game of life, primality> testing, graph coloring ... anytng that can be discussed in a > day or so to folks with no CS background, yet wch has some > theory component to it ... stuff that is surprising or counter-intuitive> is all the better ;)> Anyway, if anyone has any suggestions for material/topics> that I might cover, I would most appreciate it. Any pointersin FOCUS about 12 years or so ago that I recall that claimed thefollowing, if I recall correctly: Given any theorem and a proof of that theorem, it is possible to construct a graph with a Hamiltonian circuit, and where given the graph and a Hamiltonian circuit, the proof can be reconstructed, and someone who does not know the proof can verify that the graph has ts property, without gaining any information about the proof itself.(Someone jump in here, please, and correct any mistakes in the above. Idefintely recall the parts below, but IÍm fuzzy on the above. Inparticular, is the graph constructed from the theorem and the proof, or canthe graph be constructed without knowing a proof of the theorem? I.e., forevery theorem, does there exist a graph that contains a Hamiltonian circuitif and only if the theorem is true?)One of the common examples given to show zero-knowledge proofs is theproblem of proving that you know a Hamiltonian circuit on a graph.Combine those, and use a non-interactive proof, and you get the surprisingresult that if you have a theorem and a proof of it, it is possible toprove to others that you indeed have a correct proof, without giving themany information about that proof!Imagine how frustrating that would be if it were actually practical andsomeone did that for an important problem, like one of the Milleniumproblems. (I wonder...would they get the prize?)> would be accessible to students would be great (I have a couple)> would be great.Take a look at The New Turing Omnibus by A. K. Dewdney. The book itselfis probably too lightweight for your needs, but it will give you many topicideas.For sometng different, how about sorting stacks of pancakes? SupposeyouÍve got a stack of N pancakes, each a different size, and youÍd like thestack sorted with bigger ones on the bottom, smaller on top (e.g., like astack from the Towers of Hanoi). The only operation you have is to pick thetopmost k pancakes (1 <= k <= n), and reverse them. So, if your stack is(from top to bottom) 1 5 3 2 4, and you §ipped the top 4, you would have 23 5 1 4. All the usual questions can be asked about ts: average number of§ips required, worst case, etc.Besides being an interesting problem, there is an interesting bit of triviaassociated with it. It is the subject of the one and only scientific paperby Bill Gates: Gates, W. and Papadimitriou, C. Bounds for Sorting by Prefix Reversal. Discrete Math. 27, 47-57, 1979Ts is why Bill Gates has an === course on great ideas in (theoretical) CS?> Zero-knowledge non-interactive proofs could be good.I was surprised more people werenÍt mentioning ts -- interactiveproofs and zero-knowledge proofs are definitely cool.Sometng along (vaguely) similar lines would be probabilisticallycheckable proofs -- there were even gh-level, about the conceptsection and in Discover magazine. Ts was probably in the 1990-1992time range, but I can look up better references if youÍd like.-- === for course on great ideas in (theoretical) CS?Turing macnes (or post production systems or lambda calculus)Church-Turing ThesisThe Universal Turing MacneHalting ProblemAlgorithms and Big-O, P vs. NP conjecture.Information theory, coding theoryPublic-Key EncryptionPseudorandom number generationQuinesetc.For theory applicable to the real world, use Knuth. If youÍre amathematician like me and have your doubts that there is such a tngas a real world, an excellent === for course on great ideas in (theoretical) CS?Distribution: inet> I have been coerced into teacng a Honors course the Fall> (mostly for non-CS freshman/sophomore Honors students). My> idea was to do some Great Ideas/Problems/Puzzles etc. from> computer science -- emphasis on theory/algorithms> and related areas like graph theory/combinatorics. Hey, great! I have one word for you:http://www.discretemath.com/(One of the cool, yet vaguely disturbing, tngs about CMU isthat our CS department gets these really crazy domain names.) ThatÍs the home page for 15-251 Great Theoretical Ideas inComputer Science I, wch is a required course for CS majors atCarnegie Mellon. Now, I donÍt expect that youÍd be able to pulloff such a great job as Steven Rudich does at CMU, but I bet youcan find some neat topics from 15-251Ís syllabus. Computer Science and may use that as a guide for some of the> course. Well, I guess you knew that. Consider it seconded, then. :-D> Examples of some tngs I might discuss (besides a couple weeks on> basics/definitions/story) include Towers of Hanoi, Byzantine> Generals, I hadnÍt heard of Byzantine Generals before. But it made metnk of the Firing Squad, and then the Dining Plosophers: twoother interesting problems.> voting problems, You might even want to tie in the real-life theoretical problemsinvolving voting secrecy and verifiability. What with the U.S.elections coming up, and a lot of states switcng over to buggyvoting software... ;-) But voting algorithms and ArrowÍs Paradox are fun too.> maybe a gentle discussion of interactive proofs, Yes, I seem to recall sometng about zero-knowledge proofs in15-251, like the proof that I know a three-coloring of graph G.If thatÍs what you mean. :)> prisonerÍs dilemma, I recently heard a talk by Christos Papadimitriou in wch hebrought up a prisonerÍs-dilemma style paradox involving routing.Suppose we have a graph, say a network of roads or Internetbandwidth, and we weight the time it takes to traverse an edge bya function of the percentage of traffic thatÍs on it. For instance,a superghway might take constant time to drive along: T(x)=1; wlea back road might get jammed: T(x)=x.Then if we take a bunch of traffic and run it through ts graph: o x /| 1 / | o |0 o | / 1 |/ x oobviously itÍs better for everyone involved if 50% of them go alongthe top and 50% along the bottom -- everyone finishes in 1.5 timeunits. But if we assume that the people are selfish, and will gravitatenaturally toward the fastest route, weÍll find that eventually everyoneis musng along the path x-0-x, and taking 2 time units to finish!So greedy routing ends up hurting, not helping, in ts case. Discuss. ;-)> game of life, primality> testing, graph coloring ... anytng that can be discussed in a> day or so to folks with no CS background, yet wch has some> theory component to it ... stuff that is surprising or counter-intuitive> is all the better ;) Public-key cryptography. Wanna make ïem paranoid? Ken ThompsonÍs On Trusting Trust. Lots of stuff from las Hofstadter might be appropriate. Ditto ditto Ray Smullyan ditto. (or a link to one) to comp.theory in a week or so. === on great ideas in (theoretical) CS?> Anyway, if anyone has any suggestions for material/topics> that I might cover, I would most appreciate it. Any pointers> would be accessible to students would be great (I have a couple)> would be great. Some excellent ideas in computer science are the ideas that lead to algorithms with computationally faster runtimes than an algorithm that a human would take in solving a problem. E.g. If humans are sorting a list of numbers, they usually use a kind of selection or insertion sort. I donÍt see why a human would manually perform quicksort, yet that is one of the more ideal ways of sorting. Another example is matrix multiplication... humans will likely apply the definition of matrix multiplication to find A*B wle sometng simple like StrassenÍs algorithm (wch people would probably not bother doing by hand) leads to a theoretically faster runtime. How about storical advances? Computing macnes before electronics, === in (theoretical) CS?Great ideas, mmmm....?- Reductionism- Existance of Complete problems- Construction with basic elementsMore specifically:- Coding (nearly all) problems using >1 symbols (say 0s and 1s)- Existance of a universal model of computation (Programability; transfering the computational power from structure of the macne to the programs and back)- Rewriting systems; generating a language- P vs. NP (as tractable vs. intractable)- Probabilistic problem solving;- Generating functions (really enjoyable discussion in ConcreteMathematics of Graham, Knuth, Patashnik)- Modeling === Re: Ideas for course on great ideas in (theoretical) CS?> I have been coerced into teacng a Honors course the Fall> (mostly for non-CS freshman/sophomore Honors students). My> idea was to do some Great Ideas/Problems/Puzzles etc. from> computer science -- emphasis on theory/algorithms> and related areas like graph theory/combinatorics.> Of course, the honors college wants a syllabus in one week!> I looked at the book, Great Ideas in CS and though a nice> book, seems a bit light on the theory side ... given that> I want to focus on theory to keep me interested. I saw the> course/web site at CMU Great Ideas in Theoretical Computer Science> and may use that as a guide for some of the course. Examples> of some tngs I might discuss (besides a couple weeks on> basics/definitions/story) include Towers of Hanoi, Byzantine> Generals, voting problems, maybe a gentle discussion> of interactive proofs, prisonerÍs dilemma, game of life, primality> testing, graph coloring ... anytng that can be discussed in a> day or so to folks with no CS background, yet wch has some> theory component to it ... stuff that is surprising or counter-intuitive> is all the better ;)> Anyway, if anyone has any suggestions for material/topics> that I might cover, I would most appreciate it. Any pointers> would be accessible to students would be great (I have a couple)> would be great.> in advance. If I get enough interest, I will post a syllabus> (or a link to one) to comp.theory in a week or so.> Cp KlostermeyerYou might look through the MIT courseware to see if you find anytng ofinterest:http://ocw.mit.edu/OcwWeb/Global/all-courses.htmAls o you seem to be focusing on math. A whole lot of other tngs come tomind. Computer story and pioneers, the basic structure of computers, ghlevel languages and compliers, Turing macnes, the evolution of hardware overtime, the roles of govt., business, and individuals - e.g. ARPA NET, IBM,Apple, === B^2 = C + A^2 - A + B^2 - B A^2 + B^2 = [A+B] + A[A-1] + B[B-1] A^p + B^p = [A+B] + A*[A^(p-1)-1] + B*[B^(p-1)-1] 3^2 + 4^2 = 5^2 3^3 + 4^3 = 5^3 - 34 34 is a Fibonacci number. I wonder if the Fibonacci series can lead to a proof of FermatÍs Last Theorem? 1,2,3,5,8,13,21,34,55,89,144,233,377,610,... 5^2 - 3^2 = 2^4 [2^2 - 1^2]^2 + [2*2*1]^2 = [2^2 + 1^2]^2 5^2 + [2*2*3]^2 = 13^2 Squares of fibonacci numbers give fibonacci numbers or multiples of them: 1^2 + 2^2 = 5 2^2 + 3^2 = 13 3^2 + 5^2 = 34 5^2 + 8^2 = 89 8^2 + 13^2 = 233 But Fibonacci cubes arenÍt so predictable? 1^3 + 2^3 = 9 = 8 + 1 2^3 + 3^3 = 35 = 34 + 1 3^3 + 5^3 = 152 = 144 + 8 5^3 + 8^3 = 637 = 1 + 5 + 21 === Squares of fibonacci numbers give fibonacci numbers or multiples ofthem:Since one is a Fibonacci number, ts isnÍt too === Conjecture/DirichletÍs Geometric TheoremGerry Myerson> The sequence 2^n + 78557 includes no primes.Perhaps you meant 78557*2^n + 1 is always composite?See e.g.http://mathworld.wolfram.com/ === Conjecture/DirichletÍs Geometric Theorem> Gerry Myerson> >>The sequence 2^n + 78557 includes no primes.> Perhaps you meant 78557*2^n + 1 is always composite?Perhaps he meant what he said. 2^n + 78557 is always composite as well, by much the same reasoning used for 78557*2^n + 1, as it is always divisible by at least one of 3,5,7,13,19,37 or 73.I cannot see any reason why 78557 should work in both cases, but it does.Looking at cases like ts leads me to believe that the original conjecture is about as wrong as possible, and almost all pairs of numbers (a,b) have at most a finite number of positive integer n such that a^n+b is prime.On a style sidetrack, I was curious as to what fraction of the pairs of integers (a,b) had a and b relatively prime and a+1 and b-1 relatively prime. If we define ts as the limit of fraction of pairs (a,b) with 0 See e.g.> http://mathworld.wolfram.com/ === Fourth Conjecture/DirichletÍs Geometric Theorem> STOP PRESS!! :-((> IÍve been incredibly stupid!> The generalized form is clearly a load of dingoÍs kidneys: -> take a=5, b=3 for example - a^n+b will _always_ be even!!> DonÍt waste any more time on it...> JThe series a^n+b [hcf(a,b)=1] includes an infinite number of primes [over> all n]In the cold light of day, there _might_ be sometng of interest here- though I donÍt know how much of it has been covered already?If youÍre [still] interested check out:http://bearnol.is-a-geek.com/DGT.txtwhere comments/suggestions/further experimental testing is === asks:>Let F be the field of lengths constuctible by compass and>straighthedge. What is G(F/Q)?Ts is off the top of my head, so I might be wrong about some orall of ts, but here goes.IÍm not sure G(F/Q) has to be normal over Q. If one insists that the lengthshave to be real, it isnÍt (e.g. sqrt(sqrt(2)+sqrt(3)) has sqrt(sqrt(2)-sqrt(3))for a conjugate). That can make tngs a lot more complicated, since onehas to worry about reality at every stage of the construction of a givennumber. For the sake of simplicity, it might be better to first ask for theproperties of the smallest field E containing Q and closed under takingsquare roots.The field E is normal over Q and one can try to look at G(E/Q).Then G(E/Q) is a pro-2-group. Therefore, E is contained in the maximalnormal 2-extension of Q. Conversely, every normal 2-extension of Q isobtained as a tower of quadratic extensions, so it lies in E. Therefore,E is the maximal pro-2-extension of Q.I havenÍt read much about ts, but I tnk that a certain amount isknown about the group G(E/Q). In fact it might be known completely.I tnk Iwasawa and Shafarevich did work on ts and I tnk thereis also sometng by Serre, but I donÍt know the relevant literature.Actually, if b is a complex number constructible by straight edge andcompass and c is its complex conjugate, then b+c and bc are real andconstructible. So, maybe the answer to the original question aboutconstructible lengths is that one gets the maximal real subfieldof E.Ignorantly,Allan Adlerara@zurich.ai.mit.edu***** ** Disclaimer: I am a guest and *not* a member of the MIT Artificial ** Intelligence Lab. My actions and comments do not re§ect ** in any way on MIT. Moreover, I am nowhere near the === Theorems of Calculus,I am a gh school Calculus student and weÍve just covered theFundamental Theorems in class. Wle I understand how to use theequations and donÍt have a problem with doing the assigned problems,IÍm still having trouble logically working out exactly how theseTheorems work. I have a few specific questions, and IÍd appreciate itif someone could tell me where my thought process is wrong or advisehow I might better understand how the Theorems work. In ts post, FT1means The Fundamental Theorem, Part I, and FT2 means TheFundamental Theorem, Part II or The Integral Evalulation Theorem. in advance!1. In FT1, can ïaÍ (the lower limit of integration) be chosen to beany value at all (even if it winds up being greater than or equal tots question is yes, but that seems to cause problems. For example,what if a=4 and we were evaluating the equation when x=4 as well.WouldnÍt the answer then be zero, no matter what f(t) or f(x) is?Consider the same situation again, when we are evaluating when x=4.Then if ïaÍ was chosen to be less than 4, letÍs assume the answer waspositive. Then if ïaÍ was changed to 4, it would be zero and if ïaÍwas changed to greater than four, the answer would be negative. If Icontinue ts train of thought, I realize that if ïaÍ is changed atall the answer would be different, since youÍre measuring a differentarea under the curve. It seems that the f(x) produced by FT1 wouldcompletely depend on that ïaÍ value, and any change in a would resultin a different function f(x). I donÍt understand how ïaÍ can be anyvalue at all and FT1 still yields the same result: f(x).2. When studying FT2, the book defined F(x) as the signed areafunction:F(x) = INT[from a to x](f(t)dt)When evaluating definite integrals, the formula F(b) - F(a) is used,and the subtraction works in a way that it be in the original definition.However, I was wondering if F(x) can be used by itself, for examplewould a question ever give you a function f(x) and ask you to findF(2). Since thereÍs no subtraction, just as in my first question, itseems here that the ïaÍ value would affect the answer.3. IÍve gone through the derivation in my textbook of FT1 and Ibasically understand how they find that theorem, but IÍm still havingtrouble with actually grasping how that equation works. What is wrongwith my thought process in the following: LetÍs say weÍre trying tofind f(2), and we know from FT1 that f(x) = d/dx INT[from a tox](f(t)dt). Therefore, letÍs start by evaluating the integral partwhen x=2. In ts case, the definite integral from a (whatever thatis) to x (wch is 2) is simply a number, since itÍs just giving youthe signed area under the curve between those two points. You thendifferentiate that number (the d/dx part), and, since the derivativeof a constant is zero, the answer for f(2) must be zero. Under tsreasoning, any function comes out to be y=0 at all points. I donÍt seehow you can take a definite integral (wch yields a constant), thendifferentiate and get anytng but zero.However, maybe my reasoning is wrong because weÍre dealing with x sothe answer is really a function, not a constant (but I canÍtunderstand how ts can be the case). Even if we do get a function, itstill doesnÍt make sense to me. First, youÍre integrating f(t)dt, soyou get some function in terms of t. You then differentiate withrespect to x, so youÍd have to use implicit differentiation and youÍdwind up with both a t and an x in the answer. If the result of thatintegral is a function in terms of the variable t, where does the t goin the answer?4. I was trying to visualize what the signed area function F(x)actually looks like, so I considered the two functions 2x and x^2. IfI understand the inverse nature of differentiation and integrationcorrectly, x^2 should be an integral of 2x or the F(x) that weÍredealing with when 2x is our f(x). In ts situation, I see how theright half of the graph makes sense and the quadratic line shows howthe area under the straight line is increasing exponentially, but itdoesnÍt seem to work on the left side of the graph. Here, 2x goesnegative and the x*2 line, it seems, should also be negative sinceweÍre dealing with signed area. The only way I could see ts workingis if the ïaÍ value is zero. Then the area left of the graph would betaken going from negative value finisng with a net positive value.However, ts makes me wonder about that ïaÍ even more. What if a=1?Then the area would be correct for everytng to the right of one butif you tried to find the area between zero and one youÍd be goingbackwards and the answer would be negative even though that portion ofthe graph is above the x-axis. The value chosen for ïaÍ seems to becausing problems in my reasoning no matter how I look at it.I can proceed with my work in the class, but I donÍt like to go aheadon assumptions without really understanding the theorems that IÍmusing, and IÍm pretty confused as to how FT1 and FT2 work with an ïaÍthat can be chosen arbitrarily. IÍd appreciate any input that wouldhelp === Fundamental Theorems of Calculus> ,I am a gh school Calculus student and weÍve just covered the> Fundamental Theorems in class. Wle I understand how to use the> equations and donÍt have a problem with doing the assigned problems,> IÍm still having trouble logically working out exactly how these> Theorems work. I have a few specific questions, and IÍd appreciate it> if someone could tell me where my thought process is wrong or advise> how I might better understand how the Theorems work. In ts post, FT1> means The Fundamental Theorem, Part I, and FT2 means The> Fundamental Theorem, Part II or The Integral Evalulation Theorem.> in advance!1. In FT1, can ïaÍ (the lower limit of integration) be chosen to be> any value at all (even if it winds up being greater than or equal to> ts question is yes, but that seems to cause problems. For example,> what if a=4 and we were evaluating the equation when x=4 as well.> WouldnÍt the answer then be zero, no matter what f(t) or f(x) is?> Consider the same situation again, when we are evaluating when x=4.> Then if ïaÍ was chosen to be less than 4, letÍs assume the answer was> positive. Then if ïaÍ was changed to 4, it would be zero and if ïaÍ> was changed to greater than four, the answer would be negative. If I> continue ts train of thought, I realize that if ïaÍ is changed at> all the answer would be different, since youÍre measuring a different> area under the curve. It seems that the f(x) produced by FT1 would> completely depend on that ïaÍ value, and any change in a would result> in a different function f(x). I donÍt understand how ïaÍ can be any> value at all and FT1 still yields the same result: f(x).2. When studying FT2, the book defined F(x) as the signed area> function:> F(x) = INT[from a to x](f(t)dt)> When evaluating definite integrals, the formula F(b) - F(a) is used,> and the subtraction works in was chosen to be in the original definition.> However, I was wondering if F(x) can be used by itself, for example> would a question ever give you a function f(x) and ask you to find> F(2). Since thereÍs no subtraction, just as in my first question, it> seems here that the ïaÍ value would affect the answer. Yes it does, but some functions are only defined when a is not negative, so itÍs not really a problem. If f(x) = sqrt(x) or log(x), then both a and b have to non-negatve for the problem to even to make sense for an integral of a function over the real numbers.3. IÍve gone through the derivation in my textbook of FT1 and I> basically understand how they find that theorem, but IÍm still having> trouble with actually grasping how that equation works. What is wrong> with my thought process in the following: LetÍs say weÍre trying to> find f(2), and we know from FT1 that f(x) = d/dx INT[from a to> x](f(t)dt). Therefore, letÍs start by evaluating the integral part> when x=2. In ts case, the definite integral from a (whatever that> is) to x (wch is 2) is simply a number, since itÍs just giving you> the signed area under the curve between those two points. You then> differentiate that number (the d/dx part), and, since the derivative> of a constant is zero, the answer for f(2) must be zero. Under ts> reasoning, any function comes out to be y=0 at all points. I donÍt see> how you can take a definite integral (wch yields a constant), then> differentiate and get anytng but zero.> However, maybe my reasoning is wrong because weÍre dealing with x so> the answer is really a function, not a constant (but I canÍt> understand how ts can be the case). Even if we do get a function, it> still doesnÍt make sense to me. First, youÍre integrating f(t)dt, so> you get some function in terms of t. You then differentiate with> respect to x, so youÍd have to use implicit differentiation and youÍd> wind up with both a t and an x in the answer. If the result of that> integral is a function in terms of the variable t, where does the t go> in the answer?4. I was trying to visualize what the signed area function F(x)> actually looks like, so I considered the two functions 2x and x^2. If> I understand the inverse nature of differentiation and integration> correctly, x^2 should be an integral of 2x or the F(x) that weÍre> dealing with when 2x is our f(x). In ts situation, I see how the> right half of the graph makes sense and the quadratic line shows how> the area under the straight line is increasing exponentially, but it> doesnÍt seem to work on the left side of the graph. Here, 2x goes> negative and the x*2 line, it seems, should also be negative since> weÍre dealing with signed area. The only way I could see ts working> is if the ïaÍ value is zero. Then the area left of the graph would be> taken going from right to negative value finisng with a net positive value.> However, ts makes me wonder about that ïaÍ even more. What if a=1?> Then the area would be correct for everytng to the right of one but> if you tried to find the area between zero and one youÍd be going> backwards and the answer would be negative even though that portion of> the graph is above the x-axis. The value chosen for ïaÍ seems to be> causing problems in my reasoning no matter how I look at it.I can proceed with my work in the class, but I donÍt like to go ahead> on assumptions without really understanding the theorems that IÍm> using, and IÍm pretty confused as to how FT1 and FT2 work with an ïaÍ> that can be chosen arbitrarily. IÍd appreciate any input that would> help lead me in the right direction. YouÍre right. the formula doesnÍt work all the time. It only works as the book says it does for absolute functions, where d_n/dx_n[ f(x) ] actually exists for every n = 0,1,2, ... infinity. And === fear, consider tsJames,I finally decided to take a look at sometng that youÍve said, rather thanjust reading the commands from others and assuming that youÍre wrong. Icompletely followed your math up to the b^2-17b+8 = 0, your algebra seemscorrect (assuming that ab = 1) ... then you assume that b is not prime ...you say that fz = b and you substitute fz for b in your equation:f^2z^2 - 17fz + 8 = 0Then you divided by f:fz^2 - 17z + 8/f = 0So far, all is well. But then came ts:> and notice you STILL have that f on the front.what is the ïfrontÍ? do you mean the fact that there is an f that appearson the left hand side of the equation?> I donÍt want to hear that it isnÍt applicable because f isnÍt an> integer, as if you will have to get a polynomial reducible over Q if> you pick the right f, as thatÍs just bogus.What? Why isnÍt f an integer? What are you talking about here? What youdo mean by if you will have to get a polynomial reducible over Q if youpick the right f? What is Q? Why is ts bogus?> The problem is that 17. For it to work, you need to have sometng> even!What is ïitÍ? Why do you need to have sometng even?In short, I donÍt have a clue what ts is all about. I really tried totake you seriously here, but it is just impossible because you are not beingvery precise about the terms that you use. For instance, you say thatsometng is ïbogusÍ ... does that mean that it leads to a contradiction?If so, can you show me the contradiction that it leads to? If I may be sobold as to suggest a style change to you: please try to fight thetemptation to describe tngs in english ... you tend to be imprecise withyour words (i.e. ïbogusÍ, ïthe problem is that 17Í what problem? ïFor it toworkÍ whatÍs ïitÍ?) ... I followed your brush with algebra at the beginning... you could continue to use symbolic arguments through to the end and IÍllbet the average non-mind-reader could follow.> ItÍs so sad that IÍve tried to come up with simple examples for so> long, and posters like Nora Baron, Dik Winter and Arturo Magidin> have *successfully* come back with their own posts and kept winning at> convincing you all that I was wrong!> ItÍs been so frustrating that IÍm terrified that they will just get> away with it again, so hereÍs another angle as I desperately try yet> again to get someone to care about whatÍs mathematically correct> knowing the kind of people who are out there to come right back and> push incorrect math.> Previously I noted that I could use> P(x) = (x+8a)(x+b), and ab = 1, so P(x) = x^2 + (8a + b)x + 8.> And I considered 8a + b = 17, as then you have> 8/b + b = 17, so b^2 - 17b + 8 = 0.> Now imagine *any* non-unit factor f in the ring of algebraic integers,> like 1+i that might be a factor of b, and let b = fz, and substitute> and you get> f^2 z^2 - 17fz + 8 = 0, so> f z^2 - 17z + 8/f = 0> and notice you STILL have that f on the front.> I donÍt want to hear that it isnÍt applicable because f isnÍt an> integer, as if you will have to get a polynomial reducible over Q if> you pick the right f, as thatÍs just bogus.> The problem is that 17. For it to work, you need to have sometng> even!> IÍm so damned tired. I canÍt be sure if anyone will listen to me.> Arturo Magidin or Dik Winter or Nora Baron or Rick Decker will come> back like they have before, now wonÍt they?> TheyÍll come back and post sometng stupid, and wrong, and just plain> evil, and youÍll go along like youÍve done before.> ItÍs so evil, so frustrating. Notng matters in mathematics,> mathematicians donÍt give a damn about the truth.> NOBODY ING CARES ABOUT THETRUTH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!> === decided to take a look at sometng that youÍve said, ratherthan> just reading the commands from others and assuming that youÍre wrong....> So far, all is well. But then came ts:> and notice you STILL have that f on the front.> what is the ïfrontÍ? do you mean the fact that there is an f that appears> on the left hand side of the equation?> I donÍt want to hear that it isnÍt applicable because f isnÍt an> integer, as if you will have to get a polynomial reducible over Q if> you pick the right f, as thatÍs just bogus.> What? Why isnÍt f an integer? What are you talking about here?IÍm not at all sure that Harris mself knows. Unfortunately JSH is quitedishonest, and he uses several M.O.Ís to evade even the clearest refutationsof s claims: turning a constant into a variable, talking about afactorization x=yz without specifying a ring in wch y and z are to beelements, and so on. In the sci.math arcveshttp://mathquest.com/discuss/sci.math/a/you can find === Re: My fear, consider tsJames,One more point (and with all due respect) ... Imagine that I knew the answerto the famous Computer Science problem that asks about the relationspbetween the sets P and NP, imagine, also, that I only know how to speak theword ïblahÍ ... so, in my brain I had ts wonderful answer all worked out,but when I tried to communicate to others all they heard was ïblah blah blahblah ...Í ... do you see that it is not THEIR fault ... nor, really, is itmine, but it would behoove me to learn how to communicate with the otherpeople (as then I could claim the $1M prize that goes along with a solution)... just an analogy that may help. again.> James,> I finally decided to take a look at sometng that youÍve said, ratherthan> just reading the commands from others and assuming that youÍre wrong. I> completely followed your math up to the b^2-17b+8 = 0, your algebra seems> correct (assuming that ab = 1) ... then you assume that b is not prime ...> you say that fz = b and you substitute fz for b in your equation:> f^2z^2 - 17fz + 8 = 0> Then you divided by f:> fz^2 - 17z + 8/f = 0> So far, all is well. But then came ts:> and notice you STILL have that f on the front.> what is the ïfrontÍ? do you mean the fact that there is an f that appears> on the left hand side of the equation?> I donÍt want to hear that it isnÍt applicable because f isnÍt an> integer, as if you will have to get a polynomial reducible over Q if> you pick the right f, as thatÍs just bogus.> What? Why isnÍt f an integer? What are you talking about here? What you> do mean by if you will have to get a polynomial reducible over Q if you> pick the right f? What is Q? Why is ts bogus?> The problem is that 17. For it to work, you need to have sometng> even!> What is ïitÍ? Why do you need to have sometng even?> In short, I donÍt have a clue what ts is all about. I really tried to> take you seriously here, but it is just impossible because you are notbeing> very precise about the terms that you use. For instance, you say that> sometng is ïbogusÍ ... does that mean that it leads to a contradiction?> If so, can you show me the contradiction that it leads to? If I may be so> bold as to suggest a style change to you: please try to fight the> temptation to describe tngs in english ... you tend to be imprecise with> your words (i.e. ïbogusÍ, ïthe problem is that 17Í what problem? ïFor itto> workÍ whatÍs ïitÍ?) ... I followed your brush with algebra at thebeginning> ... you could continue to use symbolic arguments through to the end andIÍll> bet the average non-mind-reader could follow.ItÍs so sad that IÍve tried to come up with simple examples for so> long, and posters like Nora Baron, Dik Winter and Arturo Magidin> have *successfully* come back with their own posts and kept winning at> convincing you all that I was wrong! ItÍs been so frustrating that IÍm terrified that they will just get> away with it again, so hereÍs another angle as I desperately try yet> again to get someone to care about whatÍs mathematically correct> knowing the kind of people who are out there to come right back and> push incorrect math. Previously I noted that I could use P(x) = (x+8a)(x+b), and ab = 1, so P(x) = x^2 + (8a + b)x + 8. And I considered 8a + b = 17, as then you have 8/b + b = 17, so b^2 - 17b + 8 = 0. Now imagine *any* non-unit factor f in the ring of algebraic integers,> like 1+i that might be a factor of b, and let b = fz, and substitute> and you get f^2 z^2 - 17fz + 8 = 0, so f z^2 - 17z + 8/f = 0 and notice you STILL have that f on the front. I donÍt want to hear that it isnÍt applicable because f isnÍt an> integer, as if you will have to get a polynomial reducible over Q if> you pick the right f, as thatÍs just bogus. The problem is that 17. For it to work, you need to have sometng> even! IÍm so damned tired. I canÍt be sure if anyone will listen to me.> Arturo Magidin or Dik Winter or Nora Baron or Rick Decker will come> back like they have before, now wonÍt they? TheyÍll come back and post sometng stupid, and wrong, and just plain> evil, and youÍll go along like youÍve done before. ItÍs so evil, so frustrating. Notng matters in mathematics,> mathematicians donÍt give a damn about the truth. NOBODY ING CARES ABOUT THE>TRUTH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!> === all,given a transfinite sequence {S_a} of reals, for a c implies |S_a + S_b| a .8ecrit dans le message de> all,> given a transfinite sequence {S_a} of reals, for a uncountable ordinal), suppose the sequence is Cauchy (for any epsilon> in R thereÍs countable ordinal c s.t. a,b > c implies |S_a + S_b| <> epsilon), then does {S_a} converge to a real number ?Yes of courseYou should seek info about filters (its a way to better understand limits).If E is complete, every Cauchy filter is convergent; that apply to any typeof convergence you can imagine.You can also seek info about the old fason moore-Smith sequences, a generalisation of sequence you seek, but a little bit obsolete === Re: transfinite sequences> all,given a transfinite sequence {S_a} of reals, for a uncountable ordinal), suppose the sequence is Cauchy (for any epsilon> in R thereÍs countable ordinal c s.t. a,b > c implies |S_a + S_b| <> epsilon), then does {S_a} converge to a real number ?Yes. In fact more is true, the sequence is eventually constant:there is a countable ordinal c such that for all a,b > c,S_a = S_b .> The question is> the generalisation of the usual construction of R from Q : could R be> analogously completed with respect to omega_1 ?No, R is already complete.We could obtain a bounded transfinite sequence for example by> well-ordering (0,1). Ts only provides an existence of such a> sequence : is there a way to specifically construct one - i.e. define> inductively S_a for all non-limit / limit countable ordinals ?Yes, you can construct one, but it will not be Cauchy. for comments,> typek-- G. A. Edgar === Precision Montgomery Multiplication OptimizationCc: Josh Liu precision Montgomery multiplication algorithm can be stated as> follows:$$eqalign {> t &= x y,cr> u &= (t m^prime) bmod b,cr> v &= u m,cr> A &= (t + v / b) bmod m.cr> }$$where $m^prime = - m^{-1} pmod b$ and $b$ is the radix.It would seem that the term$$x y + (x y -m^{-1} bmod b) m$$can be reduced in some way so that only two or even one multiplication is> necessary. Does anyone have any insight on ts. > I did some work in Mathematica and I discovered that$$x y + (x y (-m^{-1} bmod b) m$$Is equivalent to$$- a x y + m lceil (a b + 1) x y / b rceil$$where$$a = (m m^{-1} - 1) / b,$$and $a < m$.I still need some way to perform the operation with one or === sha1:iyg8cEP+inNIAtBjn9tpQfe0cP4=> admits he was wrong] of the 19th with one barely 2 days later with a> revisionist version, in wch you were being stymied by people> winning at convincing sci.math that you were wrong?JSH is part of sci.math. Those bad guys convinced JSH he was wrong.I see no contradictions.-- Jesse HughesBasically there are two angry groups. I am a harsh force ofone. Against me is a society of mathematicians. So far itÍs been adraw. -- JSH gives another display of keen === ING CARES ABOUT THE>> TRUTH!!!!!!!!!!!!!!!!!!!!!!!!![...]Giggle. You donÍt seem to be keeping count, butThere were 239 exclamation marks. === ts>Previously I noted that I could use>P(x) = (x+8a)(x+b), and ab = 1, so P(x) = x^2 + (8a + b)x + 8.>And I considered 8a + b = 17, as then you have>8/b + b = 17, so b^2 - 17b + 8 = 0.assuming and b are reciprocal units in the ring of algebraic integersotherwise you are talking nonsense>Now imagine *any* non-unit factor f in the ring of algebraic integers,>like 1+i that might be a factor of b, and let b = fz, and substitute>and you get>f^2 z^2 - 17fz + 8 = 0, so>f z^2 - 17z + 8/f = 0>and notice you STILL have that f on the front.>I donÍt want to hear that it isnÍt applicable because f isnÍt an>integer, as if you will have to get a polynomial reducible over Q if>you pick the right f, as thatÍs just bogus.>The problem is that 17. For it to work, you need to have sometng>even!The problem is you insistence that 17 can be written as 8a+b where aand b are algebriac integers and a=1/b, all you you do is prove youcannot do tihs. ItÍs a result, not the one you are claiming though.Although you do not explicitly state a and b are algebraic integers,if they arenÍt then ts is vacuous (even more so than before, Imean).>IÍm so damned tired. I canÍt be sure if anyone will listen to me. >Arturo Magidin or Dik Winter or Nora Baron or Rick Decker will come>back like they have before, now wonÍt they?In proving you donÍt know what youÍre talking about? Yes.Once more your manupulations are correct but your conclusions arewrong.>TheyÍll come back and post sometng stupid, and wrong, and just plain>evil, and youÍll go along like youÍve done before.>ItÍs so evil, so frustrating. Notng matters in mathematics,>mathematicians donÍt give a damn about the truth.>NOBODY ING CARES ABOUT THE>TRUTH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! === from the past contestquestions.1. Al and Bob are at opposite ends of a diameter of a silo in theshape of a tall right circular cylinder with radius 150 ft. al is duewest of Bob. Al begins walking along the edge of the silo at 6 ft. persecond at the same moment that Bob begins to walk due east at the samespeed. The value closest to the time in seconds when Al first can seeBob is what? answer: 482. if a, b, c, and d are nonzero numbers such that c and d aresolutions of x^2+ax+b=0 and a and b are solutions of x^2+cx+d, finda+b+c+d. ans: -23. A boat with an ill passenger is 7.5 mi north of a straightcoastline wch runs east and west. A hospital on the coast is 60miles from the point on shore south of the boat. If the boat startstoward shore at 15 mph at the same time an ambulance leaves thehospital at 60 mph and meets the ambulance, what is the total distance(to the nearest 0.5 mile) travelled by the boat the ambulance? ans:62.5*********Could you post each question separately next time . Replying to 3 makes tsa large posting.Let x = 6t feet be the distance travelled by Al (A) and Bob (B) in tseconds. Sketch a tangent to the circle, (representing the cylinder), fromthe centre O, draw in the radius wch is perpendicular to the tangentcontact at C.pi - x/150 is the acute angle BOCThe cosine of ts angle is OC/OB =150/(150+x) = C, sayUse a grapng calculator to graphy = cos (pi - x/150) andy =150/(150+x)With x from - 1 to 300, y=-1.1 to+1.1intersection gives x=28 8.0448 hence t is close to 48[No grapcal calculator allowed? As a very rough method, a careful sketch150 (1 - .342)/.342 =approx 288 ]**************Quadratics can be written x^2 - (sum of roots)+(product of roots) = 0c+d = -acd = ba+b = -cab = db-c-d=-cb=dc=b/d=1a=d/b=1a+b+c+d=-(a+c)=-2****Take meets to mean arrives at the same time, and assume that boatintelligently steered towards optimum angle x west of due south. Sketch tsfor yourself.Let cosine x=c, sine x=s, tangent x=t,Distance travelled by boat = 7.5/c MilesTime taken by boat = 30/c MilesMiles = minutes for ambulance = 60 - 7.5 tEquating x and dividing by 7.54/c = 8 - t4 = 8 c - ss = 4(2c - 1)s^2 = 16(4c^2 - 4c + 1) = 1 - c^265c^2 - 64c + 15 = 0 = (5c - 3)(13c - 5)Shortest time taken by boat corresponds to largest c,so select c = 3/5 wch also gives t = 4/3Distance travelled by boat = (7.5 x 5)/3 = 12.5 milesMinutes taken by boat= 4 x 12.5 miles = 50but ts is same as miles travelled by ambulance, soTotal distance travelled by boat and ambulance = 62.5[ By the way, the other solution to the quadratic also allows boat andambulance to arrive at same time, but c = 5/13 with t = 12/5 lead to a boatdistance of 19.5 miles and an ambulance distance of 78 miles. Ts isinterpreted as the less useful solution where the boat is steered the eastof south === mathematics and common-senseIn sci.logic, <3c65f87.0402200504.46080ddc@ simple mathematical ideas that> lead to a few complexities, but itÍs been fruitful to explain, or try> to explain, as I work to figure out why these simple ideas either> excite derision, anger or confusion, and I tnk I have it figured> out.> IÍm sure many of you are put off by mathematics, so I assure you up> front that what IÍll be talking about will mostly be *very* simple,> and there will only be a few slightly complicated tngs at the end.> First of all, IÍm going to talk about a case where mathematicians gave> up because they couldnÍt see sometng, and assumed that because they> couldnÍt see sometng it didnÍt exist!> You know how with simple quadratics like x^2 + 3x + 2, itÍs easy> enough to see factors of 2 in the roots?> I mean, itÍs just (x^2 + 3x + 2) = (x+2)(x+1), and there they are.> However, if itÍs sometng like x^2 + 7x + 2, you can use the> quadratic formula and get the roots to find> x = (-7 +/- sqrt(41))/2> and who can see factors of 2 in that tng?I tnk part of the problem here -- and IÍll admit to not beingentirely certain where the prime failure occurs -- is that,if one has an integer such as 6, one can uniquely factor it intoprimes:6 = 2 * 3with the note that, since 1 and -1 are units of the integer ring,the factorization is unique up to multiples thereof, so one canalso have6 = (-2) * (-3)Simple as pi(e). Now, if we go into the realm of algebraic integers,though, IÍm not even sure what a prime is therein, although thereare an awful lot of units. In fact, any roots of the equationsx^n + a_{n-1} * x^{n-1} + ... + a_1 * x + 1 = 0x^n + a_{n-1} * x^{n-1} + ... + a_1 * x - 1 = 0where a_i are algebraic integers, will be a unit.(It doesnÍt matter whether the equation is reducible ornot.) I believe Arturo has demonstrated that all unitsmust solve equations of ts form, as well.A pair of such units is 2 + sqrt(3) and 2 - sqrt(3).Multiply them together and one gets 1. The definingequation for ts pair is x^2 - 4x + 1 = 0.Of course not every algebraic number is a unit, fortunately;sqrt(2) in particular is provably not a unit, as its irreducibleequation is x^2 - 2 = 0. But sqrt(2) isnot prime either, as sqrt(2) = 2^(1/4) * 2^(1/4).One might apply a categorization by declaring thatone has algebraic integers of order n, where n isthe degree of the irreducible equation with integercoefficients, of wch the integer is a root. Onethen might be able to prove that, for any algebraicinteger of degree n, there exists a unique (witnunits) factorization into other algebraic integersof degree either < n, or <= n, I donÍt know wch.In such a system, sqrt(2) is an algebraic prime of order 2.Eric W. WeissteinÍs site is of little help here, althoughhe does have an entry:http://mathworld.wolfram.com/ PrimeAlgebraicNumber.htmlbut all it does is define the term; it gives no examples.There is the possibility that all algebraic primes aresolutions of an irreducible equationx^n + a_{n-1} * x^{n-1} + ... + a_1 * x + p = 0x^n + a_{n-1} * x^{n-1} + ... + a_1 * x - p = 0where p is an integer prime. However, such primes aresubject to the order consideration above, since sqrt(2)is such a number.I shall have to ask ts question in a === mathematics and common-sense> Now, if we go into the realm of algebraic integers,> though, IÍm not even sure what a prime is therein, although there> are an awful lot of units.In fact, in the ring of algebraic integers there are no primes, because, for example, for each algebraic integer p, the equation x^2 - p = 0 shows that p factors into +- sqrt(p), wch are also === and common-sense> I know I shouldnÍt try, but the light... itÍs so beautifu...zap!>ThereÍs sometng else important here wch is the ambiguity of the>square root operatorIt is not ambiguous. Sqrt(x) (or x^(1/2) or the notation with the> square root symbol) all define the principal square root (provided x> is a positive real number). So Sqrt(9) = 3. It is not -3 even though> -3 is a square root of 9. Just read about it on the following link (You could also go to the> library and read a decent math-book there):http://mathworld.wolfram.com/SquareRoot.htmlIt doesnÍt matter how much mathematicians try to deny the obvious ifpeople actually *check* what IÍm saying.Like consider P(x) = (x + 8a)(x + b), with ab=1, and 8a + b = 17.If you work that out, then you have a = (17+/-sqrt(257))/16, now thencan you tell me if (17+sqrt(257))/16 or (17-sqrt(257))/16 has 16 as afactor?Mathematicians arguing with me may promptly lash back with, yes, theycan say that it is NOT a factor in the ring of algebraic integers!ThatÍs true, but so what? Now then, I dare any of you to try and seefor yourselves if that 16 should divide back through, so that thenumber is actually more like an integer than a fraction.If that sounds silly consider (1+sqrt(-3))/2 as thatÍs an algebraicinteger and a factor of 1. It turns out that 1+sqrt(-3) has a factorthat is 2.Tnk IÍm crazy here? Well let 2x = 1 + sqrt(-3), then2x - 1 = sqrt(-3), and squaring both sides4x^2 - 4x + 1 = -3, so4x^2 - 4x + 4 = 0, so x^2 - x + 1 = 0,so what I said IS correct. However, notice again that definition ofalgebraic integers as roots of monic polynomials with integercoefficients!If you believe that covers everytng, then youÍre acting on faith,and not logic, as thereÍs just no mathematical reasn for that belief. Any supposed proof depends on circular reasoning.I suggest those of you who donÍt believe me play with the expressionshere.And remember, if mathematicians are teacng wrong mathematics, whatgood does that do anyone? The issue here may seem esoteric to someextent, but ultimately itÍs about the importance of truth in research.Professionals should teach tngs that are === common-sense> Tnk IÍm crazy here? Yes!No doubt about === It doesnÍt matter how much mathematicians try to deny the obvious if> people actually *check* what IÍm saying.People have been checking your claims for years, but finding any of the the rare and tiny nuggets of truth in your claims will === mathematics and common-sense>> I know I shouldnÍt try, but the light... itÍs so beautifu...zap!>>ThereÍs sometng else important here wch is the ambiguity of the>>square root operatorIt is not ambiguous. Sqrt(x) (or x^(1/2) or the notation with the>> square root symbol) all define the principal square root (provided x>> is a positive real number). So Sqrt(9) = 3. It is not -3 even though>> -3 is a square root of 9. Just read about it on the following link (You could also go to the>> library and read a decent math-book there):http://mathworld.wolfram.com/SquareRoot.html>It doesnÍt matter how much mathematicians try to deny the obvious if>people actually *check* what IÍm saying.>Like consider P(x) = (x + 8a)(x + b), with ab=1, and 8a + b = 17.>If you work that out, then you have a = (17+/-sqrt(257))/16, now then>can you tell me if (17+sqrt(257))/16 or (17-sqrt(257))/16 has 16 as a>factor?A factor in what sense?>Mathematicians arguing with me may promptly lash back with, yes, they>can say that it is NOT a factor in the ring of algebraic integers!No, itÍs not a factor, in the algebraic integers. How is that lasngback? Are you claiming that it _is_ a factor in the algebraicintegers?>ThatÍs true, but so what? Now then, I dare any of you to try and see>for yourselves if that 16 should divide back through, so that the>number is actually more like an integer than a fraction.Huh? Exactly what does more like an integer than a fraction mean?>If that sounds silly consider (1+sqrt(-3))/2 as thatÍs an algebraic>integer and a factor of 1. It turns out that 1+sqrt(-3) has a factor>that is 2.>Tnk IÍm crazy here? Could be. You seem to be insisting that youÍre right and everyoneelse is wrong, but youÍre not even giving a _nt_ what it is thatyouÍre right about. Exactly what is it that youÍre asserting, thatyou claim is correct even though the evil establishment denies it?>Well let 2x = 1 + sqrt(-3), then>2x - 1 = sqrt(-3), and squaring both sides>4x^2 - 4x + 1 = -3, so>4x^2 - 4x + 4 = 0, so >x^2 - x + 1 = 0,>so what I said IS correct. _What_ is correct? I canÍt figure out what it is youÍre saying.> However, notice again that definition of>algebraic integers as roots of monic polynomials with integer>coefficients!>If you believe that covers everytng, then youÍre acting on faith,>and not logic, as thereÍs just no mathematical reasn for that belief. >Any supposed proof depends on circular reasoning.>I suggest those of you who donÍt believe me play with the expressions>here.>And remember, if mathematicians are teacng wrong mathematics, what>good does that do anyone? The issue here may seem esoteric to some>extent, but ultimately itÍs about the importance of truth in research.>Professionals should teach tngs that are === common-sense> >>I know I shouldnÍt try, but the light... itÍs so beautifu...zap!>ThereÍs sometng else important here wch is the ambiguity of the>square root operator>>It is not ambiguous. Sqrt(x) (or x^(1/2) or the notation with the>>square root symbol) all define the principal square root (provided x>>is a positive real number). So Sqrt(9) = 3. It is not -3 even though>>-3 is a square root of 9. >>Just read about it on the following link (You could also go to the>>library and read a decent math-book there):>>http://mathworld.wolfram.com/SquareRoot.html> It doesnÍt matter how much mathematicians try to deny the obvious if> people actually *check* what IÍm saying.> Mathematicians try to deny the obvious.And what here is obvious? That it is impossible to define thesquare root as being the value on the principal branch of z^(1/2)?Wow. I guess you really got us there. My calculatorÍs sqrt buttondoesnÍt work, all my software that calculates sqrt will have to bereturned, itÍs such a waste. But, there you go, itÍs so *obvious*that ts is impossible, I suppose you gotta be right, there, ace.> Like consider P(x) = (x + 8a)(x + b), with ab=1, and 8a + b = 17.If you work that out, then you have a = (17+/-sqrt(257))/16, now then> can you tell me if (17+sqrt(257))/16 or (17-sqrt(257))/16 has 16 as a> factor?> So, what does ts P(x) have to do with anytng? YouÍve set up aquadratic equation: 8a + 1/a = 17 8a^2 - 17a + 1 = 0.you thought having an unusual factorization of P(x) = x^2 + 17x + 8would impress everyone. Tnk again.Your question about whether a is divisible by 16 is meaningless unlessyou specify wch ring should contain the result. Surely that ring isnÍtthe ring of algebraic integers.> Mathematicians arguing with me may promptly lash back with, yes, they> can say that it is NOT a factor in the ring of algebraic integers!> I see. Anyone who responds to deny Your Eminence is now lasng back!Is it appropriate to refer to your alleged mathematics as asswipes?You may promptly asswipe back that I am wrong.> ThatÍs true, but so what? Now then, I dare any of you to try and see> for yourselves if that 16 should divide back through, so that the> number is actually more like an integer than a fraction.> True? The number a isnÍt even an algebraic integer. If it were analgebraic integer (meaning, if you were to take another value), thereis an easy test to establish divisibility: find the minimal polynomialof the quotient. There is also a norm function (defined on algebraicnumber fields) that allows one to discriminate in many cases whetheran algebraic integer is or is not divisible by some specific factor.> If that sounds silly consider (1+sqrt(-3))/2 as thatÍs an algebraic> integer and a factor of 1. It turns out that 1+sqrt(-3) has a factor> that is 2.Tnk IÍm crazy here? Well let 2x = 1 + sqrt(-3), then2x - 1 = sqrt(-3), and squaring both sides4x^2 - 4x + 1 = -3, so4x^2 - 4x + 4 = 0, so x^2 - x + 1 = 0,so what I said IS correct. However, notice again that definition of> algebraic integers as roots of monic polynomials with integer> coefficients!Is ts JSH holding s MastersÍ Class on Algebraic Number Theory?Exactly who is the teacher here? Certainly not Mister James Z[1/2] isthe entire set of real numbers Harris?If you believe that covers everytng, then youÍre acting on faith,> and not logic, as thereÍs just no mathematical reasn for that belief. > Any supposed proof depends on circular reasoning.> What do you mean to say here? That itÍs a mistake to tnk that thewhole world of algebraic numbers comes down to whether the number (1 + sqrt(-3))/2is an algebraic integer? Besides yourself, who on earth would tnksuch a tng?Or, are you saying that thereÍs just no mathematical reasn for thebelief that algebraic integers are *defined* by the condition youcited: roots of monic polynomials over Z?Here you are back to your oft-repeated claim of circular reasoning.You seem to tnk that any use of the definition constitutes circularreasoning.A definition, to the extent that it specifies a unique object, issimply a matter of fixing terminology, and its use in an argumentdoesnÍt amount to any sort of §aw. Ts is not a matter of whatyou want to call mathematical reasn or of its absence.that of direct algebraic manipulation. You donÍt seem to recognizethe use of induction, or of any of the standard arguments of algebra.If you care to exbit the argument youÍre referring to, with yourcommentary as to what steps are OK, and what ones constitute circularreasoning, IÍm sure we would all be amused by your §ailing at thatone.> I suggest those of you who donÍt believe me play with the expressions> here.> What expressions? I showed that neither of the roots for a is analgebraic integer. Big deal.As usual, you havenÍt proven a tng.> And remember, if mathematicians are teacng wrong mathematics, what> good does that do anyone? The issue here may seem esoteric to some> extent, but ultimately itÍs about the importance of truth in research.> What about those people who make careers out of false claims about thecorrectness of mathematics? What about people who continually castaspersions on the credibility, competence, and honesty of those whosearguments refute those false claims?> Professionals should teach tngs that are true.> Oh, I get it. ONLY professionals are supposed to tell the truth.Amateurs get a free ride, not only with respect to the level of rigorthat is required, but also with regard to the === Simple idea, mathematics and common-sense sha1:cHVeKlXPA4TotKDPQFZUWYIg9XA=> that of direct algebraic manipulation. You forgot appeal to Dedekind (or Gauss or anyone whose name ismathematically sacred). Ts is such a powerful proof principle thatit doesnÍt even require reading anytng by these authors or anytngabout these authors aside from popular math stories.-- Jesse F. Hughes Radicals are interesting because they were considered ïradicalÍ bymodern === Re: Simple idea, mathematics and common-sense-----BEGIN PGP SIGNED MESSAGE-----Hash: SHA1>It doesnÍt matter how much mathematicians try to deny the obvious if>people actually *check* what IÍm saying.>Like consider P(x) = (x + 8a)(x + b), with ab=1, and 8a + b = 17.>If you work that out, then you have a = (17+/-sqrt(257))/16, now then>can you tell me if (17+sqrt(257))/16 or (17-sqrt(257))/16 has 16 as a>factor?The question seems meaningless, or at least pointless.In order too talk about factor, you need to identify the ring inwch the factorization occurs. You have not done so.If your ring is the ring of reals, then your point is trivial anduninteresting. If your ring is that of the algebraic integers, youwould first need to show that a is an algebraic integer.If your ring is sometng else, you need to say what it is.>Mathematicians arguing with me may promptly lash back with, yes, they>can say that it is NOT a factor in the ring of algebraic integers!>ThatÍs true, but so what? Now then, I dare any of you to try and see>for yourselves if that 16 should divide back through, so that the>number is actually more like an integer than a fraction.What does more like an integer than a fraction mean?Or are you just making tngs up as you go along?-----BEGIN PGP SIGNATURE-----Version: GnuPG v1.2.4 (SunOS)iD8DBQFAN58gvmGe70vHPUMRAq2tAKCiQncwU0RPhaucggEco0T7Frb /iACcC1mu/e5+DUjmmwvqf/Jepyj8RBU==MOSF-----END PGP === common-senseYes, I can talk it all out rigorously and in a heavily mathematical> format,James,I have been following your threads for some time now and I have seen on > more occasions than I can count a request for you to do just ts.Really? > I seem to recall that when you have deigned to acknowledge these requests > it wasnÍt ...[talked] out rigourously and in a heavily mathematical > format it was more of a diatribe against mathematicians and the evil > society that they control.Hey, thereÍs ts silly error, and IÍve explained it, butmathematicians seem either willing to ignore it, or engage in rathervicious personal attacks or arguments wch I see as designed to dethe issue. > I have seen you claim that clearly stated definitions are wrong, that (as > you did in ts post) there is some inherent ambiguity and that > professional mathematicians (as well as the very capable amatuers) are too > stupid to understand but I have *NEVER* seen you post anytng that even I > would consider to be rigourous.Well IÍm tired, so IÍm pusng more on to people like yourself, soconsiderP(x) = (x + 8a)(x + b), where ab=1, and (8a + b) is an odd integer.You should have the expertise to play with that and figure out theerror in core yourself! > Having now said that you can do ts, when can I expect to see it?Play with P(x), and see if youÍre smart enough to see the mistake inreasoning with current teacng about the ring of algebraic integers.> IÍm sure that if I have any difficulty then either you or somebody else > will be able to clarify.Ivan.I certainly hope that a quadratic isnÍt too much for you!If so, then hey, it may bug you, but I may start talking about evilmath society, and people who refuse to acknowledge the truth again!!!ThereÍs no more room for excuses, even for people like you Ivan.Either actually be a mathematician, and acknowledge mathematicaltruth, or just wait for the consequences of being rogue and againstthe best interests of society at large.Yup, if mathematicians refuse to teach correct mathematics, thenthey are fighting against humanity itself.And I tnk, will be treated accordingly, once that becomes commonknowledge.Why fight against humanity, === mathematics and common-sense >> Yes, I can talk it all out rigorously and in a heavily mathematical>> format,James,I have been following your threads for some time now and I have seen>> on more occasions than I can count a request for you to do just ts.Really?Yes, really ... and IÍm not a mathematician so you canÍt even accuse me of lying (since, if I understand your appeals to the gallery correctly, it is only mathematicians that lie and the rest of us donÍt)> >> I seem to recall that when you have deigned to acknowledge these>> requests it wasnÍt ...[talked] out rigourously and in a heavily>> mathematical format it was more of a diatribe against mathematicians>> and the evil society that they control.Hey, thereÍs ts silly error, and IÍve explained it, but> mathematicians seem either willing to ignore it, or engage in rather> vicious personal attacks or arguments wch I see as designed to de> the issue.As another poster has pointed out, you have acknowledged an error, apologised for it and then repeated it!> >> I have seen you claim that clearly stated definitions are wrong, that>> (as you did in ts post) there is some inherent ambiguity and that >> professional mathematicians (as well as the very capable amatuers)>> are too stupid to understand but I have *NEVER* seen you post>> anytng that even I would consider to be rigourous.> Well IÍm tired, so IÍm pusng more on to people like yourself, so> considerNo James, you are *NOT* pusng anytng on to me - I have no desire and I see no need to do your work for you. If you have a problem then *YOU* solve it. That may be asking a bit much since you seem unable to follow and/or acknowledge the repeated assistance that you have been given by others.P(x) = (x + 8a)(x + b), where ab=1, and (8a + b) is an odd integer.You should have the expertise to play with that and figure out the> error in core yourself!Since I donÍt understand what error it is that you claim I canÍt possibly figure it out. *YOU* claim there is an error, therefore *YOU* must prove it. When you have ...[talked it] out rigourously and in a heavily mathematical format IÍm sure your claims will be examined.> >> Having now said that you can do ts, when can I expect to see it?Play with P(x), and see if youÍre smart enough to see the mistake in> reasoning with current teacng about the ring of algebraic integers.Again, ts is *YOUR* problem, not mine.It also isnÍt an answer to the question so I will ask again - when can I expect to see your claims ...[talked] out rigourously and in a heavily mathematical format ?I am content to sit and laugh at you and with everyone else. I especially like it when you use the F word lots and lots of times and even more so when you have yet another go at Professor Ulrich.On the subject of having a go at people, why do you keep referring to Arturo Magidin in your posts as being one of those who oppose you in ts newsgroup? ArenÍt you the one who told, using gutter language, Arturo not to participate any more? And isnÍt it Arturo who honourably kept s word and has since not responded directly to you?Ts is actually a great loss to many of us since Arturo has what I tnk is a gift for explaining tngs in a manner that even I can understand.> >> IÍm sure that if I have any difficulty then either you or somebody>> else will be able to clarify.Ivan.I certainly hope that a quadratic isnÍt too much for you!If so, then hey, it may bug you, but I may start talking about evil> math society, and people who refuse to acknowledge the truth again!!!No, it doesnÍt bug me at all - since I have faced the truth, that I am a JSH addict, I am able to properly enjoy your bizarre rants.ThereÍs no more room for excuses, even for people like you Ivan.People like me? What are people like me like?Either actually be a mathematician, and acknowledge mathematical> truth, or just wait for the consequences of being rogue and against> the best interests of society at large.Ummmmmm James ? ArenÍt *YOU* the one who has repeatedly boasted that you are not a mathematician ?DonÍt you tnk you should really take your own advice and ...be a mathematician, and acknowledge mathematical truth... ?Yup, if mathematicians refuse to teach correct mathematics, then> they are fighting against humanity itself.And I tnk, will be treated accordingly, once that becomes common> knowledge.Why fight against humanity, against progress and truth?Ummmmmm ... I dunno, why ?> === common-sense more occasions than I can count a request for you to do just ts.Really?As you seem to have no connection to what is actually real, rather than what merely strokes your ego, yes, really.> > I seem to recall that when you have deigned to acknowledge these requests > it wasnÍt ...[talked] out rigourously and in a heavily mathematical > format it was more of a diatribe against mathematicians and the evil > society that they control.Hey, thereÍs ts silly error, and IÍve explained it, But you keep persisting in that error even after you have explained it and publicly apologized for it. How is it that you now claim that it was not really an error after:(1) persisting in that error for months despite clear proofs by many posters that it was an error.(2) your own acknowledgement that it was an error.Only by isolating yourself from reality. > mathematicians seem either willing to ignore it, or engage in rather> vicious personal attacks or arguments wch I see as designed to de> the issue.You have a great talent both for seeing what is not there and for not seeing what is there, and there is no health in you.> > I have seen you claim that clearly stated definitions are wrong, that (as > you did in ts post) there is some inherent ambiguity and that > professional mathematicians (as well as the very capable amatuers) are too > stupid to understand but I have *NEVER* seen you post anytng that even I > would consider to be rigourous.> Well IÍm tired, so IÍm pusng more on to people like yourself, so> considerIs ts another of your dream fantasies and non-sequitur mathematics coming up?P(x) = (x + 8a)(x + b), where ab=1, and (8a + b) is an odd integer.You should have the expertise to play with that and figure out the> error in core yourself!The error in the core gift to the world.To the rest of us, is merely a prime example of what not to be.> > Having now said that you can do ts, when can I expect to see it?When your delusion generator gets into gh gear, you are capable of seeing anytng.Play with P(x), and see if youÍre smart enough to see the mistake in> reasoning with current teacng about the ring of algebraic integers.We already see mistakes in reasoning aplenty, but they are all in your reasoning.Why fight against humanity, against progress and truth?We donÍt know why you are compelled to do so. Perhaps it was due to your being badly potty trained, assuming, of course, === idea, mathematics and common-sense>[...]>If so, then hey, it may bug you, but I may start talking about evil>math society, and people who refuse to acknowledge the truth again!!!My god youÍre a ing idiot. It doesnÍt _bug_ people when youtalk about evil math society, it leads to gales of laughter. Youreally havenÍt figured that out yet? Wow.>ThereÍs no more room for excuses, even for people like you Ivan.>Either actually be a mathematician, and acknowledge mathematical>truth, or just wait for the consequences of being rogue and against>the best interests of society at large.>Yup, if mathematicians refuse to teach correct mathematics, then>they are fighting against humanity itself.>And I tnk, will be treated accordingly, once that becomes common>knowledge.>Why fight against humanity, against progress and truth?Yup, every single mathematician on the planet - the people here,the famous ones you harass via email, the professors you visit,the editors of those journals, _every_ _single_ _one_ of themis fighting against humanity, against progress and truth.You really must have no idea how totally wacky that === sha1:n0i5zkVhWVIHUvr8LCUs4iBk9vc= >> Yes, I can talk it all out rigorously and in a heavily mathematical>> format,I have seen you claim that clearly stated definitions are wrong, that (as >> you did in ts post) there is some inherent ambiguity and that >> professional mathematicians (as well as the very capable amatuers) are too >> stupid to understand but I have *NEVER* seen you post anytng that even I >> would consider to be rigourous.> Well IÍm tired, so IÍm pusng more on to people like yourself, so> consider> P(x) = (x + 8a)(x + b), where ab=1, and (8a + b) is an odd integer.> You should have the expertise to play with that and figure out the> error in core yourself!So you *can* talk it all out rigorously, but youÍre too tired.Boy, thatÍs a shame. Maybe a good lie-down, and then you could tackleit tomorrow, right?> I certainly hope that a quadratic isnÍt too much for you!> If so, then hey, it may bug you, but I may start talking about evil> math society, and people who refuse to acknowledge the truth> again!!!Wait, what did the claim that you could add rigor mean? Did it meanthat if posters on sci.math donÍt present your arguments in a rigorousformat, then we get the evil mathematician harangue again? ïCause Ithought it meant that you would present a rigorous argument ifasked[1].Footnotes: [1] Well, I didnÍt really tnk that.-- IÍve been tnking about my problems with getting any kind ofadmission that my math arguments showing the core error in mathematicsare correct, so IÍve gone to marketing books. -- James S. Harris, on when mathematics === common-sense> IÍve had a time explaining some *very* simple mathematical ideas that> lead to a few complexities, but itÍs been fruitful to explain, or try> to explain, as I work to figure out why these simple ideas either> excite derision, anger or confusion, and I tnk I have it figured> out.IÍm sure many of you are put off by mathematics, so I assure you up> front that what IÍll be talking about will mostly be *very* simple,> and there will only be a few slightly complicated tngs at the end.First of all, IÍm going to talk about a case where mathematicians gave> up because they couldnÍt see sometng, and assumed that because they> couldnÍt see sometng it didnÍt exist!> You mean, like when you couldnÍt see DeckerÍs or RamsayÍs factorizations,you assumed they couldnÍt exist?> You know how with simple quadratics like x^2 + 3x + 2, itÍs easy> enough to see factors of 2 in the roots?I mean, itÍs just (x^2 + 3x + 2) = (x+2)(x+1), and there they are. Just like in gh school: Example 1 in the chapter on factoring byinspection ? > However, if itÍs sometng like x^2 + 7x + 2, you can use the> quadratic formula and get the roots to findx = (-7 +/- sqrt(41))/2and who can see factors of 2 in that tng?> Obviously if r1 and r2 are the two roots, r1*r2 = 2, so right there they are. Proving a point: you are really baf§ed by anytng beyondfactoring by inspection. That is all you seem capable of understanding. Even examples like ts one that you createyourself seem beyond your grasp.> ThereÍs sometng else important here wch is the ambiguity of the> square root operator, No, no, no, thatÍs not it at all. The square root operator ISambiguous in a sense, but not for positive numbers. The accepted convention is that if a > 0 , then sqrt(a) is the positive square root. No ambiguity at all. If a < 0, then sqrt(a) = i*sqrt(abs(a)). Here there is ambiguity, because putting i in there begs the question: how do I know whetherit means +sqrt(-1) or -sqrt(-1) ? The answer is, there is no*algebraic* way to tell. There is an implicit agreement that,when two people write i, they both mean the same tng. Thatseems a bit crazy, but oddly it does not lead to problems. Everybody just agrees to it. > wch may sound complicated but itÍs easy to> demonstrate with another root of a quadratic:(1+sqrt(9))/2and you may tnk, silly, why show sqrt(9) when sqrt(9) = 3, but yeah,> thatÍs *one* of its solutions, as sqrt(9) = -3 as well, No - if want -3, you write it as -sqrt(9). ThatÍs the accepted*definition* of the function.> so you have> *two* numbers(1+sqrt(9))/2 = 2 or -1> No, the convention is that sqrt(9) = +3. Yes, 9 has two squareroots. The other one is -sqrt(9) = -3. 9 has two square roots,but the square root *function* is well-defined by the conventiondescribed above. You are confusing the square root *function*with the fact that 9 has two square roots. The latter does notmean that it is impossible to create a well-defined *function*.> as either solution will work. IÍve had people argue with me that by> definition (really by convention) you take the positive root. But> imagine the world of Contrary. On Contrary the mathematicians for> some odd reason *by definition* take the negative! Is mathematics> really changing depending on such decisions.Imagine youÍve forgotten that you can resolve sqrt(9) to 3 or -3, so> you write these numbers like (1+sqrt(9))/2 and (1-sqrt(9))/2 and> mercifully discover that you can get rid of that square root sign by> adding them together, or multiplying them together.Like adding them gives 1, and multiplying them together gives (1 - 9)/4 = -2and your mathematicians scratched their heads and contemplated such> numbers, and decided that there was *no way* to understand factors of> 2 of numbers like(1+sqrt(9))/2 and (1-sqrt(9))/2 except as to consider them to be unique factors of 2, in some kind of> mysterious way.> Ts is pointless blathering.> But wait, thatÍs not a problem here, of course, because you can just> evaluate the square root, but look back now at x^2 + 7x + 2, wherex = (-7 +/- sqrt(41))/2and consider that you *cannot* resolve sqrt(41) in any way that will> help you here with ts question. Why do you need help? Sqrt(41) is perfectly well-defined. Yourfrustration is that when you look at it, you cannot see a 2 in it anywhere. You want to factor by inspection, and here itsimply does not work. Trying to blame that on your imagined ambiguityof the square root function is just irrelevant. > Sure, you can write it out in decimal format with a lot of numbers> after the decimal place. My computer tells me that sqrt(41)> approximately equals 6.4031242374328486864882176746218, but of course,> you can keep going out to infinity trying just to see sqrt(41).If you drop some of those numbers (to make it easier) and move those> decimal places, to get 64031242374, squaring gives> 4099999999957933155876, wch looks VERY CLOSE to 41 * 10^20, and you> can see what our approximations actually are.We approximate irrational numbers like square roots by finding some> REALLY BIG natural number, and moving the decimal place to the left.> Your point being ???> However, you STILL donÍt have a simple idea of where factors of 2 go,> like you had with the easy and you now might tnk comforting example> of(x^2 + 3x + 2) = (x+2)(x+1). Yes! The point being, youÍre still frustrated that youcannot factor by inspection!> Now then numbers like (-7 + sqrt(41))/2 defy our ability to analyze *your* ability to analyze> because we really, really like integers, and you really, really like factoring by inspection,> but to get an integer you> have to use (-7 - sqrt(41))/2, as then you can add them together to> get -7 and multiply them to get 2, but mathematicians could not figure> out a way to handle such numbers in less than pairs! Ridiculous. Irrational square roots were invented to describehypotenuses of triangles. In that case they represented sometngreal and tanle and there was no question of pairs. I believe they wereinvented before negative numbers were invented. > ThatÍs important. Mathematicians could never figure out a way to> handle such numbers in less than pairs.So some of them decided that what they couldnÍt see, wasnÍt> meaningful.> Must you re-write all of story without even bothering tolook it up? When negative numbers were invented, they werevery quickly seen to be ïrealÍ and useful, and MEANINGFUL.The same was true with irrationals, whether sqrt(2), pi, e, orsqrt(41). > ItÍd be kind of like the weird mathematicians, who couldnÍt evaluate> sqrt(9), and were looking at (1+sqrt(9))/2 and (1-sqrt(9))/2 deciding> that where factors of 2 resided was a mystery to them. But we *can*> evaluate sqrt(9), but we *cannot* really evaluate sqrt(41) wch> leaves a mystery with (-7 + sqrt(41))/2 and ((-7 - sqrt(41))/2, and> some mathematicians have decided that thatÍs it.> Of course no one can write out all the digits of sqrt(41). Howeverit is easily seen to be the hypotenuse of a right triangle with legs5 and 4. ThatÍs tanle, real, representable, and useful. The fact that I cannot write out all the digits does not mean that I am in doubtabout whether it is a bit bigger than 6.4 or a bit less that -6.4.I know that it is the former.> For them, thatÍs all you need to know. Here are these numbers where> we canÍt evaluate the square root. Sorry, no go there, they decided,> youÍre stuck, isnÍt it obvious?> Not so. You can evaluate it to your heartÍs content. It hasa real and useful interpretation. If someone offers you sqrt(41)dollars, you expect to get $6.40 or possibly $6.41 if they aregenerous. No mathematician has trouble with it. You, however,clearly do.> You see they decided that the limit on what they could *see* was a> limit on what could mathematically exist for those examples--irational> roots--where they could not see. As far as those mathematicians were> concerned, end of story.> Irrational roots CAN be seen as described above. Theyare real and useful. Far from having trouble with them,mathematicians INVENTED them.> But thatÍs where my story begins.My mathematical research is about how the kind of patterns you see> with integers, like with(x^2 + 3x + 2) = (x+2)(x+1)where one root has all the meaningful factors of 2 *continues* into> realms where we canÍt see it directly because we canÍt get past those> danged irrationals coming in at least pairs!I found an *indirect* way of looking, wch involves putting> expressions like these polynomials IÍve shown here, but more> complicated into a special but VERY simple mathematical tool.For example,a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)can be peered into, by figuring out that its roots can be considered> in the following mathematical structure:(5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)where the aÍs are the roots of that complicated looking cubic that I> started with, and IÍve managed to place them in a structure opposite a> polynomial that has 49 as a factor.In one sense thatÍs easy. You can take just about any expression like> that, focus on some factor of its last term, and build sometng like> what I did. And it being so easy may explain some of the problems IÍm> having with people taking it seriously!What I figured out though is that what mathematicians couldnÍt see> before can now be logically seen, and in fact, the way you divide off> that 49 is to get sometng like(5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 a_3(x) + 7) = 300125 x^3 - 18375 x^2 - 360 x + 22but some vocal mathematicians donÍt like the idea as it contradicts> what theyÍve been taught, and believed because by their tnking and> training, you canÍt get around the barrier.> The opposite view is, you donÍt like it because it violateswhat *youÍve* been taught: factorization by inspection. Mathematicianstake a broader view because they have learned more about factorization. *YouÍre* the one wearing blinders here. You cannotsee around the barrier because you donÍt know enough math.> Notice that I picked two roots arbitrarily and in fact even if you> solve for the roots you canÍt look and tell *wch* roots should be> divided by 7. All you know is that two of them should be, but can> never know *wch* two.> None of them should be. All share factors in common with 7. Nonehas a factor of 7. Just like with r1 = (-7 + sqrt(41))/2 andr2 = (-7 - sqrt(41))/2. Neither r1 nor r2 is divisible by 2. Nor iseither coprime to 2. > Wch makes my job of explaining harder, but itÍs just that ambiguity> tng coming back into place. We like integers. The math doesnÍt> care as it just handles numbers.> ItÍs not at all the supposed ambiguity of square roots or roots in general that is the issue here. The issue is, you cannot acceptthe obvious truth: even when it ts you square in the face.> We canÍt look *directly* at the roots, but we can use logic to> consider them. What we canÍt see directly here *does* exist because> the mathematics says it does.> Ironically, what you say here is correct. Your logic howeveris faulty. You tnk the only possible factors are 7Ís and 1.But just as with r1 = (-7 + sqrt(41))/2 and (-7 - sqrt(41))/2, whereNEITHER ONE is divisible by 2. so with the roots of your cubic:in general, NONE OF THEM IS DIVISIBLE BY 7. ALL of them are divisible by FACTORS of 7.> To some extent, my problems are with people and mathematicians who> trust logic less than what theyÍve been taught and their own personal> sense of what makes since.> You have it so totally backwards. Your problem is that you havenot learned enough math to see that gh-school methods that YOUhave been taught, and your personal sense of what makes since,DO NOT APPLY. You need to learn some gher math to find your wayhere.> In a nutshell thatÍs the basis for the arguments.I say that just because we canÍt see the factors of irrational roots> like we can with rational ones our limitation is not a mathematical> constraint, wle some people, including a lot of very obsessive> posters, refuse to acknowledge the possibility, fighting for their old> view.> Again, you have it utterly backwards. You are fighting for a viewwch might have prevailed centuries before Gauss, Dedekind,and Kummer. You are fighting for good old factorization by inspection, because that is all you were taught to understandin gh school. One would tnk by now, with your knowledge of DeckerÍs example and RamsayÍs example and others, that sometngmight have started to penetrate. Evidently not.> Is it important?Actually, it is important if mathematicians have an error in their> tnking.> THEY DONÍT. YOU DO. HAVENÍT YOU LEARNED *ANYTNG* IN THE LAST WEEK?> It may be the case that tngs thought to have been proven in> mathematics, really havenÍt.ItÍs quite possible that any number of arguments claimed to be proofs> assumed that what they couldnÍt see, could not exist, as ts issue> has been around for over a hundred years.> You are arguing for pre-19th century math, or even Dark Ages math.> Yes, I can talk it all out rigorously and in a heavily mathematical> format, There is not the slightest evidence that you can do that. Allyour attempts at rigorous math have been wretched failures.> but I hope that giving you some idea what all the fighting is> really about, will help in understanding whatÍs going on, as well as> why I donÍt just capitulate.Mathematicians may tnk that seeing is believing, but I tnk in> mathematics, logic is king.> I could not invent a more perfect example of getting tngsbackward. YOU tnk that if you cannot see a factor, it cannotbe there: hence you tnk it MUST be true that 7 factors out oftwo of the linear terms of your cubic. You cannot conceive thatthere might be any other way to do it, a way that you cannot easily see explicitly. But there can. We have told you howrepeatedly. You need some abstract ideas to understand it. ïAbstractÍmeans: sometng BEYOND seeing is believing. You are essentiallystuck with ancient fossil math, like bones in the LaBrea tarpits,obsolete. You yourself provide a perfectly good examply, with the roots of x^2 + 7*x + 2, and THEN YOU MISINTERPRET IT, and go off on ridiculous tangents. You want notng but seeing-is-believingmath, and then you accuse us of your own error! You saw DeckerÍs original example. Evidently you *still* donÍtbelieve it. You saw RamsayÍs example. You know that it showedsometng that was far from obvious. You saw it, but now you seem to be saying you still donÍt believe it. You say logic isking, but when your own logic defies hard evidence, you tnkthe error must be in the evidence! It reminds me of the Rodney King tape. You apparently would haveagreed with the first jury: seeing is not believing! Those policemenwere not beating an unarmed helpless man on the ground with nightsticks! Like them, apparently, you prefer to believe your own preconceptions. HereÍs another example, perhaps more relevant. Consider x^2 - 6*x + 6. There are two roots: r1 = 3 + sqrt(-3) and r2 = 3 - sqrt(-3). Note that r1 * r2 = 6 = 2*3. It is easy to see that r1 has an algebraic integer factorin common with 3. You can see that *BY INSPECTION*, the onlymethod that you actually trust. It is sqrt(3). Right? ItÍs not so easy to see what factor r1 has in common with 2,is it? CanÍt get that one by inspection, right? So if we agreewith your tnking, perhaps we conclude that it must be either2 or 1. Clearly it is not. Nor do we need to invent an object ring,and declare that (3 +sqrt(-3))/2 is an object. 3 + sqrt(-3)DOES have an algebraic integer factor in common with 2. See if you can figure out what it is. Note that real mathematicians do not have trouble with ts. They have ways of finding such tngs that go beyond your own seeing-is-believing method. You need to stretchyour very limited imagination. Learn a little 19th century algebraic number theory and you might be able to see it also. Let us === mathematics and common-sense I had an error in the example at the end - it doesnÍtmake any conceptual difference, but below is the correctedversion, plus another example:> Mathematicians may tnk that seeing is believing, but I tnk in> mathematics, logic is king.> I could not invent a more perfect example of getting tngs> backward. YOU tnk that if you cannot see a factor, it cannot> be there: hence you tnk it MUST be true that 7 factors out of> two of the linear terms of your cubic. You cannot conceive that> there might be any other way to do it, a way that you cannot > easily see explicitly. But there can. We have told you how> repeatedly. You need some abstract ideas to understand it. ïAbstractÍ> means: sometng BEYOND seeing is believing. You are essentially> stuck with ancient fossil math, like bones in the LaBrea tarpits,> obsolete. You yourself provide a perfectly good examply, with > the roots of x^2 + 7*x + 2, and THEN YOU MISINTERPRET IT, and go off > on ridiculous tangents. You want notng but seeing-is-believing> math, and then you accuse us of your own error! You saw DeckerÍs original example. Evidently you *still* donÍt> believe it. You saw RamsayÍs example. You know that it showed> sometng that was far from obvious. You saw it, but now you > seem to be saying you still donÍt believe it. You say logic is> king, but when your own logic defies hard evidence, you tnk> the error must be in the evidence! It reminds me of the Rodney King tape. You apparently would have> agreed with the first jury: seeing is not believing! Those policemen> were not beating an unarmed helpless man on the ground with nightsticks! > Like them, apparently, you prefer to believe your own preconceptions. HereÍs another example, perhaps more relevant. Consider x^2 - 6*x + 6. The corrected version follows: There are two roots: r1 = 3 + sqrt(3) and r2 = 3 - sqrt(3). Note that r1 * r2 = 6 = 2*3. It is easy to see that r1 has an algebraic integer factor> in common with 3. You can see that *BY INSPECTION*, the only> method that you actually trust. It is sqrt(3). Right? ItÍs not so easy to see what factor r1 has in common with 2,> is it? CanÍt get that one by inspection, right? So if we agree> with your tnking, perhaps we conclude that it must be either> 2 or 1. Clearly it is not. Nor do we need to invent an object ring,> and declare that (3 +sqrt(3))/2 is an object. 3 + sqrt(3)> DOES have an algebraic integer factor in common with 2. See if you can figure out what it is. Note that real mathematicians > do not have trouble with ts. They have ways of finding such tngs that > go beyond your own seeing-is-believing method. You need to stretch> your very limited imagination. Learn a little 19th century algebraic > number theory and you might be able to see it also. Let us know if> you canÍt get it.> Here is another example of ts kind: Let Q(a) = a^2 - 4*a + 6.The roots are r1 = 2 + sqrt(-2) and r2 = 2 - sqrt(-2).Again their product is 6 = 2*3. It is clear(by inspection!) that each of them has a factorin common with 2. Not so easily seen are the factors they have incommon with 3. No, Mr. Harris, not 3 itself in eithercase. Galois would not be surprised. Dedekind would notbe surprised. Harris === Re: Simple idea, mathematics and common-sense> IÍve had a time explaining some *very* simple mathematical ideas that> lead to a few complexities, but itÍs been fruitful to explain, or try> to explain, as I work to figure out why these simple ideas either> excite derision, anger or confusion, and I tnk I have it figured> out.> No, actually, you work to excite your own anger and confusion. Plus,ts is all a psychological experiment to you, or maybe itÍs gamestheory, and youÍre gonna sic the generals on us, to kill everyonewho doesnÍt buy your particular crock of sheeeit.> IÍm sure many of you are put off by mathematics, so I assure you up> front that what IÍll be talking about will mostly be *very* simple,> and there will only be a few slightly complicated tngs at the end.> No, actually, itÍs *you* who are put off by mathematics, and who tnksthat mathematicians are welfare leeches.> First of all, IÍm going to talk about a case where mathematicians gave> up because they couldnÍt see sometng, and assumed that because they> couldnÍt see sometng it didnÍt exist!> Ts is even more of a crock than what preceded. Everyone here knowsthat youÍve spent the better part of several years trying to show thatmathematicians couldnÍt perform factorizations of some sorts, andeach and every example you came up with turned out to be incorrect: The standard theory of algebraic integers proved to be able to produce factors that *you* said could not be found.Now, in a stunning display (well, stunning to those who havenÍt seenyour continual use of ts tactic) of revisionism, you have called tsphenomenon (that is, JSH stating sometng could not be done witnstandard mathematics, but could be done by the miracle of Objects,with every case actually being readily handled by standard techniques): a case where mathematicians gave up because they couldnÍt see sometng, and assumed that because they couldnÍt see sometng it didnÍt exist!IÍm wondering how your use of the Big Lie can be justified. Each caseyou said couldnÍt be handled, was in fact handled. Each time, you spentweeks, sometimes months, braying on about how the mathematicians werelying and cheating and engaging in fraud, and how you might just haveto bring suit against one person or another, and how you had calledsome personÍs employer: IÍve deliberately involved an official at s college to take away plausible deniability for s school in a phone call I made months ago.Each time, you were shown to be incorrect, yet you call ts wholeepisode a case where mathematicians gave up because they couldnÍt see sometng, and assumed that because they couldnÍt see sometng it didnÍt exist!Maybe your grasp of the English language isnÍt quite up to thatwe assume for a gh-school graduate. Perhaps none of the aboveseems at all contradictory, that, hey, people who say the truthare really the liars, and frauds, and those who produce theresults are those who ... gave up because they couldnÍt see sometng, and assumed that because they couldnÍt see sometng it didnÍt exist!You see, to my understanding, someone who produces results withstandard techniques *isnÍt* giving up, but itÍs the person whoclaims that the techniques are inadequate who is giving up instead!> You know how with simple quadratics like x^2 + 3x + 2, itÍs easy> enough to see factors of 2 in the roots?I mean, itÍs just (x^2 + 3x + 2) = (x+2)(x+1), and there they are.However, if itÍs sometng like x^2 + 7x + 2, you can use the> quadratic formula and get the roots to findx = (-7 +/- sqrt(41))/2and who can see factors of 2 in that tng?Man, are we back to your old confusion of terminology regarding factors?The values of x you show are divisors of 2, and 2 is a multiple of them.To refer to a factor of 2, one means that the number 2 divides the number under consideration.ThereÍs sometng else important here wch is the ambiguity of the> square root operator, wch may sound complicated but itÍs easy to> demonstrate with another root of a quadratic:(1+sqrt(9))/2and you may tnk, silly, why show sqrt(9) when sqrt(9) = 3, but yeah,> thatÍs *one* of its solutions, as sqrt(9) = -3 as well, so you have> *two* numbers(1+sqrt(9))/2 = 2 or -1as either solution will work. IÍve had people argue with me that by> definition (really by convention) you take the positive root. But> imagine the world of Contrary. On Contrary the mathematicians for> some odd reason *by definition* take the negative! Is mathematics> really changing depending on such decisions.Yes, everyone sees your idiosyncracy for what it is: an inability todeal with definitions. The square root function is no more ambiguousthan the log function. One can well speak of sqrt(4) and -sqrt(4).The fact that you find it incomprehensible only fits in with yourgeneral level of confusion.Imagine youÍve forgotten that you can resolve sqrt(9) to 3 or -3, so> you write these numbers like (1+sqrt(9))/2 and (1-sqrt(9))/2 and> mercifully discover that you can get rid of that square root sign by> adding them together, or multiplying them together.Like adding them gives 1, and multiplying them together gives (1 - 9)/4 = -2and your mathematicians scratched their heads and contemplated such> numbers, and decided that there was *no way* to understand factors of> 2 of numbers like(1+sqrt(9))/2 and (1-sqrt(9))/2 except as to consider them to be unique factors of 2, in some kind of> mysterious way.Yes, letÍs imagine that you have a clue about sometng. Imagination isa wonderful gift, right?But wait, thatÍs not a problem here, of course, because you can just> evaluate the square root, but look back now at x^2 + 7x + 2, wherex = (-7 +/- sqrt(41))/2and consider that you *cannot* resolve sqrt(41) in any way that will> help you here with ts question.> Why should I consider that? sqrt(41) is a fine number, somewhere between6 and 7. On the other hand, I could produce a continued fraction toapproximate the number, if thatÍs what I had in mind. ItÍs theelementary student who finds the square root to be so mysterious, notthe professional.> Sure, you can write it out in decimal format with a lot of numbers> after the decimal place. My computer tells me that sqrt(41)> approximately equals 6.4031242374328486864882176746218, but of course,> you can keep going out to infinity trying just to see sqrt(41).If you drop some of those numbers (to make it easier) and move those> decimal places, to get 64031242374, squaring gives> 4099999999957933155876, wch looks VERY CLOSE to 41 * 10^20, and you> can see what our approximations actually are.We approximate irrational numbers like square roots by finding some> REALLY BIG natural number, and moving the decimal place to the left.> Why not take the continued fraction approximation? It converges fasterthan the decimal expansion.> However, you STILL donÍt have a simple idea of where factors of 2 go,> like you had with the easy and you now might tnk comforting example> of(x^2 + 3x + 2) = (x+2)(x+1).> You just canÍt get past those reducible polynomials, can you?Now then numbers like (-7 + sqrt(41))/2 defy our ability to analyze> because we really, really like integers, but to get an integer you> have to use (-7 - sqrt(41))/2, as then you can add them together to> get -7 and multiply them to get 2, but mathematicians could not figure> out a way to handle such numbers in less than pairs!> I sure donÍt know what you tnk you mean by that. Certainly a personcan handle (-7 + sqrt(41))/2, all by itself. What on earth, aside fromyour own drunken stupor, would suggest otherwise?Why canÍt you multiply the number (-7 + sqrt(41))/2 by 2, add 7, andsquare the result? DoesnÍt that get you an integer? Does it requireyou to pair it with its conjugate? These numbers arenÍt paired forthe reasons you tnk.> ThatÍs important. Mathematicians could never figure out a way to> handle such numbers in less than pairs.> You have no clue about whatÍs important or whatÍs unimportant, letalone whatÍs true or not. After all, you tnk itÍs important to reportus to the FBI. You tnk itÍs important that you got a letter into Timemagazine. You tnk that Z[1/2] is really the full set of real numbers.You tnk that Galois theory is false.> So some of them decided that what they couldnÍt see, wasnÍt> meaningful.> Prove ts remark. Mathematicians donÍt make such evaluations.> ItÍd be kind of like the weird mathematicians, who couldnÍt evaluate> sqrt(9), and were looking at (1+sqrt(9))/2 and (1-sqrt(9))/2 deciding> that where factors of 2 resided was a mystery to them. But we *can*> evaluate sqrt(9), but we *cannot* really evaluate sqrt(41) wch> leaves a mystery with (-7 + sqrt(41))/2 and ((-7 - sqrt(41))/2, and> some mathematicians have decided that thatÍs it.> Oh, back to the Land of Make-Believe, are we?> For them, thatÍs all you need to know. Here are these numbers where> we canÍt evaluate the square root. Sorry, no go there, they decided,> youÍre stuck, isnÍt it obvious?> I really donÍt know what youÍre on about here. I tnk you were stuckwith a four-function calculator, but no toilet paper as a youth, soyou decided to work it out with a calculator when it came to mattersof hygeine. How else to explain the bulk of excrement you smear acrossthe newsgroups?> You see they decided that the limit on what they could *see* was a> limit on what could mathematically exist for those examples--irational> roots--where they could not see. As far as those mathematicians were> concerned, end of story.> You have decided what motivated past generations of mathematicians,without so much as a smattering of knowledge about the field. Inwhat way is that different from the settlers of the US having decidedwhat motivated the Native American? Or, from wtes in the Southern UShaving decided that their slaves really, really preferred slavery?What gives an outsider the right to decide why any group has developedthe way it has?> But thatÍs where my story begins.> Well, hereÍs sometng I agree with: your story begins at the end ofa bogus revision of story, a tale full of trivialization, purposefulmisstatement, and bigotry. That about sets the stage for your greatand glorious research.> My mathematical research is about how the kind of patterns you see> with integers, like with(x^2 + 3x + 2) = (x+2)(x+1)where one root has all the meaningful factors of 2 *continues* into> realms where we canÍt see it directly because we canÍt get past those> danged irrationals coming in at least pairs!> Again with the mis-characterization of algebraic irrationals. There isno truth to the rumor that your understanding of mathematics is deepenough to submerge the average dust mite.> I found an *indirect* way of looking, wch involves putting> expressions like these polynomials IÍve shown here, but more> complicated into a special but VERY simple mathematical tool.For example,a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)can be peered into, by figuring out that its roots can be considered> in the following mathematical structure:(5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)where the aÍs are the roots of that complicated looking cubic that I> started with, and IÍve managed to place them in a structure opposite a> polynomial that has 49 as a factor.In one sense thatÍs easy. You can take just about any expression like> that, focus on some factor of its last term, and build sometng like> what I did. And it being so easy may explain some of the problems IÍm> having with people taking it seriously!What I figured out though is that what mathematicians couldnÍt see> before can now be logically seen, and in fact, the way you divide off> that 49 is to get sometng like(5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 a_3(x) + 7) = 300125 x^3 - 18375 x^2 - 360 x + 22> Has anyone said that 49 *couldnÍt* be divided from the product inthat fason? No. In fact, itÍs *you* who has said that ts isTHE ONLY WAY to form the division. ItÍs *you* who has said thatthere is no way to divide the product (5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7)by 49, and retain algebraic integers as factors.You stated that division as an impossibility, and also stated thatTHE ONLY WAY to do the division is to divide 7 into each of 2 of thefactors.> but some vocal mathematicians donÍt like the idea as it contradicts> what theyÍve been taught, and believed because by their tnking and> training, you canÍt get around the barrier.> What barrier? You havenÍt taken the mathematics, you donÍt know whathas been taught, and yet you continue to pontificate!Say, if I happen to have seen an episode of Sanford and Son, wayback in the 70Ís (or was it the 80Ís?), does that qualify me toexpound on the nature of African American culture? Do I get tosay foÍ szzle and not be laughed out of the ïhood?> Notice that I picked two roots arbitrarily and in fact even if you> solve for the roots you canÍt look and tell *wch* roots should be> divided by 7. All you know is that two of them should be, but can> never know *wch* two.> Such a crock. You can factor 49 arbitrarily in the complex numbers,into 3 factors, and distribute the factors at will. If you care toconstrain the resulting factors, your factorization of 49 willbe constrained, as will the freedom in placing the factors amongthe three parenthetical factors you started with.Where were you, over the past several months, wle that exact factwas explored? Were you too busy working yourself up into a latherover the antics of the palindromic poster? Or, were you getting yourpanties into a pretzel over the algebraic ruminations of the coolMister Winter? If you had followed your own stick to the mathmantra, rather than your 7-year obsession stick it to themathematicians I rather suspect you wouldnÍt have wasted your ownseveral months wning and ranting and threatening and pleading.> Wch makes my job of explaining harder, but itÍs just that ambiguity> tng coming back into place. We like integers. The math doesnÍt> care as it just handles numbers.> What a dope. What a dope. YOU have to rescue mathematics because WElike integers. You fail to recognize that the mathematics that hasbeen develops recognizes the special place occupied by the integers,and has a consistent view across more types of algebraic (and other)structures than you can imagine. You see a polynomial you canÍt seehow to deal out the components of a factorization of 49, and yougo apest about the Evil Empire of Mathematical Cabalists.> We canÍt look *directly* at the roots, but we can use logic to> consider them. What we canÍt see directly here *does* exist because> the mathematics says it does.> Yeah, whatever. Your ability to tell what we can or canÍtdo is not really credible, is it?> To some extent, my problems are with people and mathematicians who> trust logic less than what theyÍve been taught and their own personal> sense of what makes since.> Well, a full consideration of your problems is (as they say) outside thescope of ts work. The people you are having the most difficulty withare quite simply *not* people who regurgitate what theyÍve learned.These are people who are fully competent to develop completely originalmathematics.> In a nutshell thatÍs the basis for the arguments.> Figures youÍd put it into a nutshell, donÍt you tnk?> I say that just because we canÍt see the factors of irrational roots> like we can with rational ones our limitation is not a mathematical> constraint, wle some people, including a lot of very obsessive> posters, refuse to acknowledge the possibility, fighting for their old> view.> Please. You are the one who has claimed sometng to be impossible.Others have shown each and every one of your so-called impossiblefactorizations to be possible witn the ring of algebraic integers.Who is it who is refusing to acknowledge possibility? Who is it whotakes months on end, in the face of direct computations, to cometo the correct conclusion?> Is it important?Actually, it is important if mathematicians have an error in their> tnking.> If? For seven years now, youÍve been taking every seemingdiscrepancy between standard mathematics and your ownmutterings as a sign of the corruption of mathematics. ItdoesnÍt matter that youÍve ALWAYS been wrong, no, becausein your BB-in-a-boxcar Brain, youÍve decided that you maybe right at some point.DidnÍt you write ts little attack: I hereby charge Rick Decker with academic fraud and note that s college is responsible for ts rogue professor. IÍve deliberately involved an official at s college to take away plausible deniability for s school in a phone call I made months ago. s college has been made aware of s behavior. There may also be a civil matter involved at some point in the future.Why not stand up and do sometng, rather than haranguing the worldwith your interminable if theyÍre wrong cant?Show someone wrong (and not yourself), or get off the can.> It may be the case that tngs thought to have been proven in> mathematics, really havenÍt.> Oh, itÍs the world of maybe: It may be the case that monkeys §y out of my ass. You may already be a winner! Aliens may be living among us! Elvis may still be alive!You talk endlessly about evidence (wch, BTW, is 100% against you),concern for the Truth (wch concern of yours has been shown to becynical in every sense but the one wch connects it to Diogenes).> ItÍs quite possible that any number of arguments claimed to be proofs> assumed that what they couldnÍt see, could not exist, as ts issue> has been around for over a hundred years.> You have never read a proof. You have no background to make ts claim,and the evidence is that you will never produce a correct argument.Find the hole. IÍll note that *all* of algebra is under *continual*examination by tens of thousands of students, *each one of whom*would be thrilled beyond comprehension to find a contradiction,ANY proof of a statement and its negation. None has been found,despite ts intense scrutiny by wannabe experts and experts alike.> Yes, I can talk it all out rigorously and in a heavily mathematical> format, but I hope that giving you some idea what all the fighting is> really about, will help in understanding whatÍs going on, as well as> why I donÍt just capitulate.> Bullst. You can make up pseudocrapological jargon, and repeat ituntil the cows come home, but notng you have said is correct. Youcannot fill in the details, because (1) you donÍt have the abilityand (2) you donÍt have the patience.> Mathematicians may tnk that seeing is believing, but I tnk in> mathematics, logic is king.> Crap. What you have seen with your own two eyes contradicts ts: When faced with your claims that certain numbers couldnÍt have common divisors, mathematicians stepped forward to exbit those divisors. You, on the other hand, put those meathooks of yours right over the eyes, and chanted loudly so you wouldnÍt have to face the truth.Why not try to === Algebraic Numbers even exist?Ts is probably a slightly stupid question, but Eric W. WeissteinÍsdefinition of Prime Algebraic Number:http://mathworld.wolfram.com/ PrimeAlgebraicNumber.htmlgives no examples of such a number at all, and thereforeleaves a lot to be desired. I am wondering what such anumber might look like, if one exists at all.An obvious attempt would be sqrt(2), or, for that matter,the nth root of 2. However, if one considers the productsqrt(2) = 2^(1/4) * 2^(1/4)it is clear that sqrt(2) divides neither factor. Therefore,I fail to see how sqrt(2) could be prime.There are other difficulties. The number 2 + sqrt(3) is infact a unit in the algebraic integers. Its reciprocal2 - sqrt(3) is also an algebraic integer. (That these areindeed units is easily proven by multiplying them together;one gets 4 - 3 = 1. It is also clear that both are algebraicintegers.)The only way I personally can resolve ts is to suggest thatprimes have an order designation; ts would mean that theregular ordinary primes are order 1, sqrt(2) is a prime oforder 2, cuberoot 2 is a prime of order 3, etc. Ts basicallymeans that, in any product of two integers of order <= n, if analgebraic prime of order n divides the product, it must divideone of the integers.Or sometng like that. What an ugly mess.Can anyone === error proof, simpler, shorter> It turns out you can prove that thereÍs an error in core with rather> basic math, using a quadratic:LetP(x) = (x+8a)(x+b), and ab = 1, so P(x) = x^2 + (8a + b)x + 8.Thena = 1/b and b = 1/aand to ghlight the simple case let (8a + b) = 9, and substituting> witha = 1/b gives,8(1/b) + b = 9, so 8 + b^2 = 9b, so b^2 - 9b + 8 = (b-8)(b-1) = 0.But why two solutions? Obviously, you can just have a=1/8, and get(x + 8(1/8))(x + 8) = (x + 8)(x + 1)so itÍs trivially easy whatÍs going on with those two solutions.Okay. You have (8a + b) = k, where k is some integer and ab = 1.The solutions to these two equations satisfy 8a^2 - ka + 1 = 0and b^2 - kb + 8 = 0When k = 9, these two equations become (8a - 1)(a - 1) = 0 = (b - 8)(b - 1)so we have two sets of solutions: (a, b) = (1/8, 8) and (1, 1).The other special case is when k = 6, where we have (4a - 1)(2a - 1) = 0 = (b - 4)(b - 2)wch has solutions (a, b) = (1/4, 4) and (1/2, 2)For any other integer value of k we have two irreducible (over Q)polynomials 8a^2 - ka + 1 and b^2 - kb + 8So for these values, the roots b will be algebraic integersand the corresponding roots a wonÍt be. I fail to see anytngremotely surprising about ts. Notice, for later, thatwle a isnÍt an algebraic integer, 8a is, since 8a^2 - ka + 1 = 0so 8(8a^2 - ka + 1) = (8a)^2 - k(8a) + 8 = 0 Solving 8a^2 - 17a + 1 = 0 with the quadratic formula givesa = (17 +/- sqrt(257))/16but itÍs clear that *only* one of those roots can be like 1/8 before,> wle the other is like 8 from before, but *neither* is an algebraic> integer!What does it mean for one of those roots to be like 1/8 or 8?Are you attempting to generalize some property from the special,reducible, case k = 9? > So whatÍs the core error?The assumptions of some mathematicians would mean that P(x) = (x+8a)(x+b),is impossible in an ring where ïaÍ and ïbÍ have properties like> integers, because itÍs impossible in the ring of algebraic integers,> if ab = 1 and (8a + b) is an integer, when the result is a polynomial> irreducible over Q that has ïaÍ as a root!Are you suggesting that under the assumptions1. P(x) = (x + 8a)(x + b)2. 8a + b is a rational integer3. a, b are algebraic integers with ab = 14. P(x) is irreducible over Q (we donÍt actually need ts)That P(x) = 0 does not have both solutions (x = -8a, x = -b)in the algebraic integers? That, of course, is wrongsince weÍve already established that both b and 8a arealgebraic integers (and, in fact, are conjugates whenP is === sha1:ALXHRBwLcG5eHEdQlOcLxUyDEko=> It turns out you can prove that thereÍs an error in core with rather> basic math, using a quadratic:> Let> P(x) = (x+8a)(x+b), and ab = 1, so P(x) = x^2 + (8a + b)x + 8.> Then> a = 1/b and b = 1/aI donÍt know why you write then here, but never mind. LetÍs stickwith ts as given.[...]> That means that ïaÍ *cannot* be an algebraic integer, as algebraic> integers cannot be roots of primitive non-monic polynomials> irreducible over Q!> But notice that ïbÍ IS an algebraic integer as itÍs the root of a> monic polynomial with integer coefficient.Did you just prove the deep and startling fact that sometimes, if b isan algebraic integer, 1/b is *not* an algebraic integer?-- Many argue that its programmers have turned out shoddy programs, but[their] objective is to make profit, not superlative programs perse. By the profit criterion, Microsoft has been one of the greatestcompanies in the story of ts country. -- ADTI defends === problemCould ts not be formulated as a mathematical programmingproblem:Let y(j) in {0,1} mean bus stops at stop jLet x(i,j) in {0, 1} mean person i gets off at stop jLet p(i)=k mean that person i needs to go to stop kThen:no more than z stops: sum(j, y(j)) <= zpassengers only get off when bus stops: x(i,j) <= y(j) for all i,jpassenger has to get off somewhere: sum(j, x(i,j)) = 1objective: min sum(i, [abs(p(i) -sum(j, j*x(i,j))]^3)Ts would be an MINLP. Probably youÍll want to workon the objective (e.g. getting rid of the abs as tsintroduces a kink i.e. discontinuous derivative).---Erwin Kalvelagenerwin@gams.com, http://www.gams.com/~erwin---> Consider a bus route with k stops. When the bus starts its journey, p> persons are on the bus. Person j wants to get off at stop s_j<=k. Also> s_j>=1. However the bus cant stop at more than z < k stops. So a person> who wants to get off at a stop s_j wch is not among the z stops has to get> off at the nearest one and then walk. Let f(x_j) = (x_j)^3 be the distance> person j has to walk to get to s destination, where x is the number of> stops he has to walk to get to s destination. Let T be the sum f(x_1) +> f(x_2) + ... + f(x_p)> The problem is now to construct a method(algorithm) where the bus stops at> stops, so that the T is minimized. Any ideas on === support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1LLaNO23840;I have watched the discussion diverge aymptotically (or is it exponentially?)from my original question : where did the m as a notation for slope (or gradient)come from? What ever happened to my original question? I gather nobody knows the answer. Paul === discussion diverge aymptotically (or is it> exponentially?)from my original question : where did the m as a notation> for slope (or gradient)come from? What ever happened to my original question? > I gather nobody knows the answer. Paul Bruckman> It has been a year or two since ts was asked here.It used to be a regular question.Anyway, to see some info go down to slope at... http://members.aol.com/jeff570/geometry.html-- G. A. Edgar === as Set Derivative by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1LLaN023804;>It also gives other interesting and useful differentiation formulas:>(A u B)Í = AÍBÍ = A<-->B AB = A<-->B (AÍ U BÍ)Í>wch I suppose tells us that:>(f + g)Í = fg - (fÍ + gÍ)Í>-- >IÍm not interested in mathematics that might have anytng>to do with reality. -- Russell Easterly, in sci.mathI was expecting some reply in ts direction, and Toni Lassila fulfils expectations, but notice that I pointed out that A<-->Bis the analog of fg With Respect to ï. In other words, I onlyindicated that (fg)Í goes over to (A<-->B)Í. Before you begindissecting whether fgÍ should go over to sometng by ts (andwe can argue that forever in Heaven, where I expect to be soon atthe age of 65),letÍs take a look at:1) (A + iB)(A + iB)* = (A + iB)(A - iB) = A^2 + B^2where I use the notation A, B in (1) instead of f, g respectively(differentiable functions) to make the point that we can generalizecomplex numbers to sets (had MacLane and LawvereÍs Category Theoryaccepted Generalized Sets instead of Set-trivializing Objects?).Does anybody recognize BornÍs Probability in Mathematical Physicsin (1)? It might be more familiar in:2) ww* = (x + iy)(x + iy)* = (x + iy)(x - iy) = x^2 + y^2where w is the wave function. Of course, we have to define A^2 andA^2 + B^2, and we may end up having to replace + on the right-hand-side of (1) by U, but heck, nobody said it would be easy.Ah yes, what has been accomplished? Well, A + iB is one sided orNon-Symmetric, like A-->B = AÍ U B. (A + iB)* = A - iB is alsoone-sided or Non-Symmetric, but in an opposite direction (forfunctions, f - ig is down versus the up of f + ig, although thedirections only reverse from the tip of f). Notice that A-->B= AÍ U B, wle B-->A = BÍ U A, wch are reverses of a rather cleartype. So we only have an explanation and a relationsp between the QuantumTheory probability in Mathematical Physics and Classical Set andClassical Probability Theory. However, you might try following Max Born or Niels Bohr or Werner Heisenberg, who not only could notexplain their choice of x^2 + y^2 grounded in Classical Set andClassical Probability Theory but stopped looking - perhaps they weretoo old to === soulcan you please help me aceve the answer?Al and Bob are at opposite ends of a diameter of a silo in theshape of a tall right circular cylinder with radius 150 ft. al is duewest of Bob. Al begins walking along the edge of the silo at 6 ft. persecond at the same moment that Bob begins to walk due east at the samespeed. The value closest to the time in seconds === (choogu)>Al and Bob are at opposite ends of a diameter of a silo in the>shape of a tall right circular cylinder with radius 150 ft. al is due>west of Bob. Al begins walking along the edge of the silo at 6 ft. per>second at the same moment that Bob begins to walk due east at the same>speed. The value closest to the time in seconds when Al first can see>Bob is what? answer: 48First, let us compute the distance they travel. When Al has walked ssilo radii around the silo, he has moved s radians as measured from thecenter of the silo. At ts time, Bob is s+1 silo radii from the centerof the silo. When Al can first see Bob, the triangle formed by Bob, Al,and the center of the silo is a right triangle, with Al at the rightangle. Since Al has walked s radians, the angle at the center of thesilo is pi-s radians. Since the side between the center and Al is 1silo radius and the side between the center and Bob is s+1 silo radii,we get the equation 1/(s+1) = cos(pi-s) = -cos(s)Solving ts equation tells us that s is about 1.92029909426 silo radii.Multiply by the silo radius of 150 ft to get 288.044864139 ft and divideby 6 ft/sec to get 48.0074773565 sec.Rob son take out the trash === Al and Bob are at opposite ends of a diameter of a silo in theshape of a tall right circular cylinder with radius 150 ft. al is duewest of Bob. Al begins walking along the edge of the silo at 6 ft. persecond at the same moment that Bob begins to walk due east at the samespeed. The value closest to the time in seconds when Al first can seeBob is what? answer: 482. if a, b, c, and d are nonzero numbers such that c and d aresolutions of x^2+ax+b=0 and a and b are solutions of x^2+cx+d, finda+b+c+d. ans: -23. A boat with an ill passenger is 7.5 mi north of a straightcoastline wch runs east and west. A hospital on the coast is 60miles from the point on shore south of the boat. If the boat startstoward shore at 15 mph at the same time an ambulance leaves thehospital at 60 mph and meets the ambulance, what is the total distance(to the nearest 0.5 mile) traveled by the boatand the ambulance? ans:62.5*********P.S.I posted ts lot last time you askedLet x = 6t feet be the distance travelled by Al (A) and Bob (B) in tseconds. Sketch a tangent to the circle, (representing the cylinder), fromthe centre O, draw in the radius wch is perpendicular to the tangentcontact at C.pi - x/150 is the acute angle BOCThe cosine of ts angle is OC/OB =150/(150+x) = C, sayUse a grapng calculator to graphy = cos (pi - x/150) andy =150/(150+x)With x from - 1 to 300, y=-1.1 to+1.1intersection gives x=28 8.0448 hence t is close to 48[No grapcal calculator allowed? As a very rough method, a careful sketch150 (1 - .342)/.342 =approx 288 ]**************Quadratics can be written x^2 - (sum of roots)+(product of roots) = 0c+d = -acd = ba+b = -cab = db-c-d=-cb=dc=b/d=1a=d/b=1a+b+c+d=-(a+c)=-2****Take meets to mean arrives at the same time, and assume that boatintelligently steered towards optimum angle x west of due south. Sketch tsfor yourself.Let cosine x=c, sine x=s, tangent x=t,Distance travelled by boat = 7.5/c MilesTime taken by boat = 30/c MilesMiles = minutes for ambulance = 60 - 7.5 tEquating x and dividing by 7.54/c = 8 - t4 = 8 c - ss = 4(2c - 1)s^2 = 16(4c^2 - 4c + 1) = 1 - c^265c^2 - 64c + 15 = 0 = (5c - 3)(13c - 5)Shortest time taken by boat corresponds to largest c,so select c = 3/5 wch also gives t = 4/3Distance travelled by boat = (7.5 x 5)/3 = 12.5 milesMinutes taken by boat= 4 x 12.5 miles = 50but ts is same as miles travelled by ambulance, soTotal distance travelled by boat and ambulance = 62.5[ By the way, the other solution to the quadratic also allows boat andambulance to arrive at same time, but c = 5/13 with t = 12/5 lead to a boatdistance of 19.5 miles and an ambulance distance of 78 miles. Ts isinterpreted as the less useful solution where the boat is steered the eastof south === in detail;Check the following process X_t are gales wrt {F_t}1) X_t = B_t + 4t; B_t is Brownian motion 2) X_t = === some kinetics and have come up with tsdifferential eqn from my experiment.dy/dx = 2y-xy^2I canÍt for the life of me see how you can solve it. As far as I seeit using the substitution y=vx doesnÍt help. Can anyone lend me ahand? for your help, it is really really appreciated.Sarah-- submissions: post to k12.ed.math or e-mail to k12math@k12groups.orgprivate e-mail to the k12.ed.math moderator: kem-moderator@k12groups.orgnewsgroup website: http://www.tnkspot.net/k12math/newsgroup charter: === Ellipses & reducing to General Equation thereof -IÍve read about the General Quadratic Equation (Ax^2+By^2+Cx+Dy+E=0) & the equation for an ellipse (x-h)^2+(y-k)^2=0; IÍve seen a number of web pages dealing with the above.But --Say you have an ellipse with center at the origin, foci at (-2,0) and (2,0).Now translate the ellipse 3 units in the positive x direction, and 2 units positve y. Figuring out the quadratic equation is still pretty straightforward, right?Now rotate the ellipse counterclockwise 30 degrees. How do I figure out the === reducing to General Equation thereofCraig Reed> IÍve read about the General Quadratic Equation (Ax^2+By^2+Cx+Dy+E=0) & the> equation for an ellipse (x-h)^2+(y-k)^2=0; IÍve seen a number of web pages> dealing with the above....> Now rotate the ellipse counterclockwise 30 degrees. How do I figure out> the quadratic equation now?ItÍs just a linear change of variables. For a rotation around the origin,replace x and y byx cos G + y sin Gandx sin G - y cos Grespectively, where G === Equation thereof> -IÍve read about the General Quadratic Equation (Ax^2+By^2+Cx+Dy+E=0) & the > equation for an ellipse (x-h)^2+(y-k)^2=0; IÍve seen a number of web pages > dealing with the above.But --Say you have an ellipse with center at the origin, foci at (-2,0) and > (2,0).Now translate the ellipse 3 units in the positive x direction, and 2 units > positve y. Figuring out the quadratic equation is still pretty > straightforward, right?Now rotate the ellipse counterclockwise 30 degrees. How do I figure out > the quadratic equation now?CraigA translation of coordinates (or, equivalently, an opposite translation of the ellipse) can be aceved by replacing x by xÍ+a and y by yÍ+b in the equation of the ellipse.A rotation of coordinates (or, equivalently, an opposite rotation of an ellipse around origin) can be aceved by relacing x by xÍ*a-yÍ*b and replacing y by xÍ*b+yÍ*a in the === study for math?IÍm trying to see if my study habits are part of the reason why I can behot/cold with math. So how exactly do you study for proof-based math classes? Here is what I usually do:Read the chapter laying in bed (hey, I was originally a story major). Sometimes I wonÍt understand a part of the proof of the theorem and IÍll betempted to just skip the theorem entirely. After IÍm done with ts, IÍll goto the homework (ts may not be the same day). If I donÍt remember one of thedefinitions, IÍll have to go back and look at what it says. I also have aproblem with waiting to look at the answer. If I canÍt figure out the problemin about a minute or so, IÍll just look at the back of the book. IÍm not surehow good/bad ts is considered but itÍs a habit IÍve developed nevertheless.Any === math?> definitions, IÍll have to go back and look at what it says. I also have a> problem with waiting to look at the answer. If I canÍt figure out the problem> in about a minute or so, IÍll just look at the back of the book. IÍm not sure> how good/bad ts is considered but itÍs a habit IÍve developed nevertheless.There are problems that you must sweat to solve, and -should- sweat to solve. DonÍt be so quick to look up the answer. If you have trouble with a problem then find a similar problem that you can solve and carry over the skill you gain.Your one minute approach is way to superficial for you to learn anytng..P.S. Do both your reading and your problem solving with pencil and paper at the ready. Ts means you have -sit up- to do the work. It is better posture, easier on your eyes and better for your circulation or breatng. If you have to take a break, do sometng physical like excercise or jog in === Sequences Involving Binary & Prime-Factorization> First, we have the sequence formed by simply lettinga(m) = 1, if the sum of all exponents of the prime-factorization of m> has no carries when summed in in base-2.a(m) = 0, if there are any carries in the summing of the exponents of> the prime-factorization of m.So, for example, a(12) = 1 because 12 = 2^2 *3^1,> and, in base-2, 2 = ï10Í, 1 = ï1Í,> and ï10Í and ï1Í have their ones in different positions.But, > a(24) = 0, because 24 = 2^3 *3^1,> and, in base-2, 3 = ï11Í, 1 = ï1Í,> wch both share a rightmost one.Unless I made an error, ts sequence is not yet in the Encyclopedia> Of Integer Sequences.a(k) -> 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1,...(Defining a(1) = 1 makes b()-recursion below work.)> Now, let b(m) = sum{k=1 to m} a(k).b(m) gives the number of positive integers <= m and with an a() of 1,> obviously.My main math result related to these sequences:(a somewhat interesting method of calculating {b()} using recursion):b(1) = 1;> b(m) => b(§oor(sqrt(m)) + sum{p=primes<=m} b(§oor(sqrt(m/p))),(or, if we choose to define 1 as a prime:b(m) =sum{p=primes<=m} b(§oor(sqrt(m/p))) )> And, unless I made an error, ts sequence is not yet in the> Encyclopedia Of Integer Sequences either.b(k) -> 1, 2, 3, 4, 5, 5, 6, 7, 8, 8, 9, 10, 11, 11, 11, 12, 13,...> I wondered about the asymptotics of {a(k)} and {b(k)}.We can rewrite the b-recursion (wch the original version I give below in copied message), so as to get:b(m) = b(§oor(sqrt(m))) + sum{k=1 to §oor(sqrt(m))} a(k) pi(m/k^2),where pi(x) is the number of primes <= x, for x = positive real.So, very roughly (and shunning rigor),I assume, using ts new recursion, thatb(m) = O(pi(m)) = O(m/ln(m)),using the prime-number theorem (of course). More precisely (and still unrigorously), I get:limit{m -> oo}(b(m) - b(§oor(sqrt(m)))) *ln(m)/(m - sqrt(m))= sum{k=1 to oo} a(k)/k^2 = C,wch of course converges, since the aÍs are each either 0 or 1.So, on average, each a(k) has aC/ln(k) probability of being 1.Am I right in each of my assumptions? And, what is the constant C =sum{k=1 to oo} === Certain Areas: puzzleI must admit that my formula for part-2 is simply a partial sum,although an interesting partial sum, interesting since it is anunexpected alternate representation of the straight-forwardpartial-sum wch gives the area as the sums of squaresÍ areas (takenover the kÍs from part-1).If anyone can find a more simple representation of part-2Ís answer,ts would be interesting.But I will wait a couple days to give answer to ts puzzle.,> You have a FIXED sequence of distinct positive integers, {k}, wch is> to be determined.For all kÍs <= EACH positive integer m, we can take a number n (n <= m) of (geometric) squares,> each of *integer* side-length,where the j_th square is of the largest square> with an integer area that is <= m/(k_j).> But {k} is such that, if we lined the squares up vertically so that> their horizontally-running edges coincide and they do not overlap,> the length from the upper side of the top square to the lower side of> the bottom square will ALWAYS be m.2 questions: 1) Wch integers make up {k_j}?2) Give a formula for the total area of the (n number of) squares.Example:> For m = 4,We have 1, 2, 3 in {k}, but not 4.So, we have the squares, as lined up: ------> ! ! !> ------ k=1, area = 4 <= 4/1> ! ! !> ==----> ! ! k=2, area = 1 <= 4/2> ==> ! ! k=3, area = 1 <= 4/3> === Learning about cryptographyWhat are the basic prerequisites to learning about cryptography and actuallylearning how to do it? About me: No computer science background but a prettygood (abstract) math background. I ask because neither the math nor thecomputer science departments offers a course on cryptography and IÍm not reallysure what I need to learn. Is it sometng you learn in graduate school? Ifso, what do I need to learn? If it is sometng that is strictly computerscience, can you get into a CS grad school with just a math background to learnabout cryptography? I would not mind learning how to program if I need to knowthat. Any book recommendations would be === What are the basic prerequisites to learning about cryptography andactually> learning how to do it? About me: No computer science background but apretty> good (abstract) math background. I ask because neither the math nor the> computer science departments offers a course on cryptography and IÍm notreally> sure what I need to learn. Is it sometng you learn in graduate school?If> so, what do I need to learn? If it is sometng that is strictly computer> science, can you get into a CS grad school with just a math background tolearn> about cryptography? I would not mind learning how to program if I need toknow> that. Any book recommendations would be appreciated.1. See Chapter 2 (IIRC) of the Handbook for Appliced Cryptography (HAC)available online. It is a summary of most of the math that is needed.The math area depends on where you want to be and can include number theory,abstract algebra, probability, coding theory, information theory, ellipticcurves and others.See counterpane.com for a writeup as to why.2. http://www.ciphersbyritter.com/LEARNING.HTM (see writeup plus list ofreferences)3. Some references:[Bressoud89] D. Bressoud. Factorization and primality testing. UndergraduateTexts in Mathematics. Springer-Verlag, 1989.[Cohen93] Henri Cohen. A course in algorithmic algebraic number theory,volume 138 of Graduate Texts in Mathematics. Springer--Verlag, 1993.[Knuth81] D. E. Knuth. The Art of Computer Programming: SeminumericalAlgorithms. Addison-Wesley, 2nd edition, 1981.[Koblitz87b] Neal Koblitz. A course in number theory and cryptography,volume 114 of Graduate Texts in Mathematics. Springer--Verlag, 1987.[Kranakis1986] Evangelos Kranakis. Primality and cryptography. Wiley-TeubnerSeries in Computer Science. Wiley & Sons, 1986.[Riesel85] Hans Riesel. Prime numbers and computer methods forfactorization, volume 57 of Progress in Mathematics. Birkh.8auser, 2ndedition, === responses. IÍm going to try to figure out if itÍs a problemwith how I study or that I am just not quite good enough for more === advancedmath.Subject: Re: e is transcendental (was: classes of TRYING to learn? Or are you >>so closed-minded that you canÍt be bothered?>>So? The real part of exp(i pi) is cos(pi), and its imaginary part>>is sin(pi), so all you are saying is that cos(pi) = -1 and >>sin(pi) = 0, and we were already aware of these facts. There is >>NO reason to conclude that exp(i pi) = 0. > I,have a reason , with my due respect.> Panagiotis Stefanides>>Yes, but you DONÍT tell us what your reason is. You canÍt expect us to >>accept your claims without giving support for those claims. So what >>possible reason could you have for expecting us to agree with your claim >>that exp(i pi) = 0?>The reason is simply that exp(ipi0=-1 should be accompanied>by the statement that ts is the real part solution.>Is it fair?>>No, that is not a fair comment. exp(i pi), the complex number, is>>equal to -1, the complex number. There is no need to appeal to the >>real part. Also, what does exp(i pi) = -1 is the real part solution>>mean? You look like you are using terminology in a manner not >>recognized in mathematics.>> McAnally>>-->e^[i*pi] ,as accepted ,is a phasor. >>In most of the relevant fields of mathematics, e^[i pi] is a complex >>number.>It is only fair to state that its >polar representation is :>e^[i*pi] = MOD 1 , ARG 180 .>>That is Arg 180 degrees, not just Arg 180. And so what? That does >>not lead to your assertion that exp[i pi] = 0, a result for wch >>you have given absolutely no support. Why donÍt you just give up?>> McAnally>> At the moment, they (the Time Lords) are far from being all-powerful.>> ThatÍs why itÍs been left up to me and me and me.>> quote by: Patrick Troughton in The Three Doctors>>------->I, have made myself very explicit.>My original question of the implication>of the imaginary component: e^[ipi]=j*0 (to the related proof)>Complex Notation, chapter 12 ,Electrical Technology 3RD ED.>Edward Hughes Longmans ,page 338:>Stares:> OA*=OB+jOC=OA(COStheta+jSINtheta)>* Symbols representing phasors are printed in bold face italics, wle those representing only magnitudes are printed in ordinary italics,..>Here is very clear the difference between PHASOR and MAGNITUDES>>I know the difference between a complex number and its modulus >>(or magnitude). You donÍt have to explain the difference to>>me.>[wch(MAGNITUDES) I, referred to as REAL PART or IMMAGINAR PART,>>The magnitude of a complex number is generally not equal to either>>its real or imaginary part. How could you claim that it is?>I, gave a reference:>Complex Notation, chapter 12 ,Electrical Technology 3RD ED.>Edward Hughes Longmans ,page 338:>States:> OA*=OB+jOC=OA(COStheta+jSINtheta)>* Symbols representing phasors are printed in bold face italics, wle those representing only magnitudes are printed in ordinary italics,..> What are these magnitudes then ?OB (wch can obviously be either positive, zero or negative) is the real part of the complex number. OC (wch can obviously be either positive, zero or negative) is the imaginary part of the complex number. OA (wch is presumably positive or zero) is the magnitude of the complex number.theta is the argument of the complex number.Note that neither OB nor OC is the magnitude of the complex number.> I should have stated COMPONENTS ].>>Perhaps you mean that (the imaginary part of e^[i pi]) = 0, in wch>>case you are correct, but you have had a lot of problem expressing >>yourself, especially in view of the way that you initially made>>the claim by stating that e^[i pi] = 0, wch you described as the >>imaginary part solution, using a terminology that nobody but you >>knows.> e[i*theta]=COStheta+iSINtheta>>I know that it is true that exp[i theta] = cos(theta)+i sin(theta). >>It follows that exp[i pi] = -1. Incidentally, e[i theta] = i e theta, >>unlike what you have written.>thetas could be given and calculations could be performed>for numerical evaluations.>>And for other results as well.>In books is stated that it is FORMULA>and also terms such as evaluate:(-1+i*sqrt[3])^10>Are these not solutions to problems?>>No. Evaluate (-1+i sqrt(3))^10 is a problem for wch you can>>get a solution using exp(i theta) = cos(theta)+i sin(theta). Ts>>does not mean that exp(i theta) = cos(theta)+i sin(theta) is itself>>a solution. You need a problem before you can describe anytng as>>a solution.>Of course ,ts crops up when theta is substituted by a given angle.Ts has notng to do with your unorthodox usage of the word solution.>>Nobody knows what you mean when you make a statement like exp[i pi] = -1>>is the real part solution of exp[i pi] = -1, or that exp[i pi] = 0>>is the imaginary part solution of exp[i pi] = -1. I asked you to >>explain your terminology but you havenÍt bothered.>I, doubt it but ,still I, exlpained it anyway.>>I did ask you. And you did not explain your bizarre terminology. >.> There>>was no problem, hence there is no solution, whether real part or>>imaginary part.>How else would do You call them,I, said:>I should have stated COMPONENTSI call them the real part of the complex number (or the equation) and the imaginary part of the complex number (or the equation). I do not call them solutions. McAnally At the moment, they (the Time Lords) are far from being all-powerful. ThatÍs why itÍs been left up to me and me and me. quote by: Patrick Troughton in The Three === Natural>> He wonÍt *if* we apply the condition that s input tape has finitely>> many ones, but there is notng about that convention that is forced>> on us. Why not allow that he has a tape with all of N? After all,>> that assumption is not s problem. s problem is only in reasoning>> about whatÍs on the tape after an infinite number of steps.> Yes, of course I agree that a TM can have anytng written on its tape >to start.If thatÍs the case then it would seem to me that it is very easy fora TM to solve the halting problem (although procuring an appropriateinput tape might be a little tricky). What am I === Every Computable Natural <8765e4q2fe.fsf@pwumbda.org> <8JCdncK7SMPAiK7dRVn-vw@comcast.com> <87hdxovh3h.fsf@pwumbda.org> <87vfm4o85i.fsf@pwumbda.org> <87n07fo9qr.fsf@pwumbda.org> <87k72jnxtp.fsf@pwumbda.org> <87ptcau3xj.fsf@pwumbda.org> > Supposedly is the key word here.>> There will be such an N. N could be quite large (it might even equal 3).>> Obviously, the original tape didnÍt contain every natural number.> Yes, it is obvious, and ts is what everyone has been telling you all> along. (In case you forgot, you were the one claiming the existence of> such a tape) How many times do you need to be told that you will not have> a tape with all the natural numbers on it?Who says?Why not?He wonÍt *if* we apply the condition that s input tape has finitelymany ones, but there is notng about that convention that is forcedon us. Why not allow that he has a tape with all of N? After all,that assumption is not s problem. s problem is only in reasoningabout whatÍs on the tape after an infinite number of steps.If you simply say that infinite input tapes donÍt make sense and/orinfinite computations are meaningless, then you avoid the real sourceof RussellÍs error. You rule out s thought experiment by fiat,without asking whether s assumptions are really sensible (they are)and trying to understand s reasoning (wch is, of course,incorrect).-- But he mself was not to blame for s vices. They grew out of a personaldefect in s mother. She did her best in the way of §ogging m wle aninfant... but, poor woman! she had the misfortune to be left-handed, and acld §ogged left-handedly had better be left === Computable Natural <8765e4q2fe.fsf@pwumbda.org> <8JCdncK7SMPAiK7dRVn-vw@comcast.com> <87hdxovh3h.fsf@pwumbda.org> <87vfm4o85i.fsf@pwumbda.org> <87n07fo9qr.fsf@pwumbda.org> <87k72jnxtp.fsf@pwumbda.org> <87ptcau3xj.fsf@pwumbda.org> <87n07cdes0.fsf@pwumbda.org He wonÍt *if* we apply the condition that s input tape has finitely> many ones, but there is notng about that convention that is forced> on us. Why not allow that he has a tape with all of N? After all,> that assumption is not s problem. s problem is only in reasoning> about whatÍs on the tape after an infinite number of steps. Yes, of course I agree that a TM can have anytng written on its tape to start. The problem, of course, is when one talks of an output it implies that a TM has halted (in finite time) and thus only seen a finite amount of that input, so really having an infinite input is pretty useless. And, yes, we can take your earlier suggestions and define a tape cell to be, in a sense, finalised if the TM tape head is guaranteed to never visit that tape cell again... but then that still leads to s example with having none (instead of one) ï0Í symbols left in the end, wch you and I have tried explaining to m. If you have the patience to handle m, IÍll gladly leave ts thread === <8765e4q2fe.fsf@pwumbda.org> <8JCdncK7SMPAiK7dRVn-vw@comcast.com> <87hdxovh3h.fsf@pwumbda.org> <87vfm4o85i.fsf@pwumbda.org> <87n07fo9qr.fsf@pwumbda.org> <87k72jnxtp.fsf@pwumbda.org> <87ptcau3xj.fsf@pwumbda.org> <87n07cdes0.fsf@pwumbda.org> > He wonÍt *if* we apply the condition that s input tape has finitely>> many ones, but there is notng about that convention that is forced>> on us. Why not allow that he has a tape with all of N? After all,>> that assumption is not s problem. s problem is only in reasoning>> about whatÍs on the tape after an infinite number of steps.> Yes, of course I agree that a TM can have anytng written on its tape > to start. The problem, of course, is when one talks of an output it > implies that a TM has halted (in finite time) and thus only seen a finite > amount of that input, so really having an infinite input is pretty > useless.> And, yes, we can take your earlier suggestions and define a tape cell to > be, in a sense, finalised if the TM tape head is guaranteed to never visit > that tape cell again... but then that still leads to s example with > having none (instead of one) ï0Í symbols left in the end, wch you and I > have tried explaining to m.Now, ts is exactly right and the right way to explain it (not thatRussell is ever going to quit commuting quantifiers). I much preferan explanation of why s reasoning is bad than a bald statement thatinfinite input and output is senseless. After all, working theorists*do* sometimes use infinite tapes.> If you have the patience to handle m, IÍll gladly leave ts thread > now.Well, IÍm not taking any responsibility here, but you havenÍt anyeither. On occasion, I decide that responding to Russell is divertingenough that I do so. Then, I get tired of the same old errors in newguises and wander away for a wle. Russell will always be there,making the same mistakes, if I want to return to the same well-treadground. (Geez, putting it ts way, I wonder *why* I ever do return.)Anyway, if you donÍt want to keep going in the thread, donÍt.RussellÍs confusion was here before you (I tnk) and it will be hereafter your gone.-- Yup, as far as IÍm concerned, if you live out your lives smiling theentire time full of pride in your *believed* accomplishments, when younever had any, well thatÍs ok with === Set Contains Every Computable Natural <8765e4q2fe.fsf@pwumbda.org> <8JCdncK7SMPAiK7dRVn-vw@comcast.com> <87hdxovh3h.fsf@pwumbda.org> <87vfm4o85i.fsf@pwumbda.org> <87n07fo9qr.fsf@pwumbda.org> <87k72jnxtp.fsf@pwumbda.org> <87ptcau3xj.fsf@pwumbda.org> > Assumption: There exists a 0 at the end of computation.Of course there is. How can there not be?>> My TM will never overwrite the last 0.> Ts is the source of your error (and just about all your errors.) There> is no last 0 on the *infinite* list, since there will always be a 0> after it.I disagree. It is not the source of s error. It is a consequenceof s error.The source of s error is an obsessive-compulsive disorder wchmanifests itself in commuting quantifiers willy-nilly, especiallychanging (A x)(E y) to (E y)(A x).-- Jesse HughesHow lucky we are to be able to hear how miserable Willie Nelson couldimagine mself to be. -- Ken Tucker === === post>Subject: Re: JSH: Apology to Ramsay, why I post>Message-id: HarrisÍs most >serious error: tnking that because he is Special he can take a short >cut right to the finish line without having to learn or understand much >of anytng. IÍm not saying ts out of jealousy or desire to protect>the Establishment. Just look at s track record. He makes a lifetimeÍs>worth of serious conceptual errors every month or so ... without even>trying. In the end, not one single realized accomplishment, although>I do tnk he has set some kind of gh-water mark for being the most >incredibly successful and annoying jerk on the Internet up to the present >time. The Isaac Newton, the pioneering Daniel Boone of crank/trolls! >Ts alone might eventually get m a spot on Letterman.> Top Ten Reasons Why Should Appear On Letterman10. Mathematicians are evil.9. There is an error in core.8. Mathematicians are assholes.7. No one has ever found a way to count primes using a partial difference equation.6. Mathematicians are lapdogs.5. It turns out that there *has* to be a ring beyond algebraic integers.4. Mathematicians are liars.3. s peers are Gauss and Dedekind.2. Mathematicians are === transcendental> Since e^[iPi]=cosPi+isinPi> or , e^[iPi]=-1+i[0]> then there are two solutions here, to the given equatio:A) e^[ipi]=-1 the real part solution and B) e^[ipi]=i[0] , or e^[ipi]=0 the imaginary part solutio.>>No, the conclusion from e^[iPi]=-1+i[0] is>>Re(e^[iPi]) = Re(-1+i[0]) = -1 AND>>Im(e^[iPi]) = Im(-1+i[0]) = 0>>-- >>Daniel W. son>>panoptes@iquest.net>>http:// members.iquest.net/~panoptes/039 53 36 N / 086 11 55 W>Daniel,>Just saw Your message.>I, .>Ts is it.>Your help is great, and resolves my question completely.I pointed out exactly the same tng, but you never accepeted it when Imade the point. I presume that it means that you wonÍt make any more silly comments along the lines of exp(i pi) = 0. McAnally Despite anytng you may have heard to the contrary, the rain in === Modular Group Maximal?> Is the group SL(n,Z) maximal (as an discrete group) in SL(n,R)? The> answer is yes for n=2 and one would imagine it is also for any n,> however, I donÍt know how to prove it.Density and Maximality of Arithmetic Subgroups, CrelleÍs Journal,1966. Ts has a theorem (7) wch applies to a much more generalsituation (all Chevalley groups in their standard representation and$Z$-points as the arithmetic subgroup), as does PlatonovÍs Survey inRussian Math Surveys from 1982. It would still be interesting to knowif there is an easier proof for SL(n) than the one presented === dwufalzl,I want to find some methods to find the QR factorization of a matrixthat diagonal elements of matrix R are arranged in order of increasingmagnitude. Can anyone show === functionI need to find a superior bound for x, knowing thatG(a,x) / G(a) > 0.001where G(a,x) is the Incomplete Gamma function and G(a) is the usual Gamma functionAny clues? very === simple example, but I donÍt see any reason more > complicated graphs would not have an exponential number of embeddings.Of course I agree that a planar graph may easily have exponentially manyembeddings. Consequently, checking each embedding separately will not makefor a polynomial time isomorpsm algorithm. I was tnking more along thelines of breaking up the graph into 3-connected pieces and dealing witheach of these separately. I have not worked out the details and am notcompletely sure if anyone has. If you are really interested, I can try anddig out some references and/or do some more serious tnking on thesubject.BTW, I would not go as far as proposing a linear time algorithm for generalplanar graphs, but my feeling is that quadratic time should not be === Mathematical Physics (Week 202)Originator: baez@math-cl-n03.math.ucr.edu ( Baez)Also available at http://math.ucr.edu/home/baez/week202.htmlTs WeekÍs Finds in Mathematical Physics - Week 202 Baez Ts week IÍll deviate from my plan of discussing number theory, and instead say a bit about sometng else thatÍs been on my mind lately: structure types. But, youÍll see my fascination with Galois theorylurking beneath the surface.Andre Joyal invented these in 1981 - he called them especes destructure. Basically, a structure type is just any sort of structurewe can put on finite sets: an ordering, a coloring, a partition, orwhatever. In combinatorics people count such structures usinggenerating functions. A generating function is a power series wherethe coefficient of x^n keeps track of how many structures of the givenkind you can put on an n-element set. By playing around with thesefunctions, you can often figure out the coefficients and get explicitformulas - or at least asymptotic formulas - that count the structuresin question.The reason ts works is that operations on generating functions comefrom operations on structure types. For example, in week190, Idescribed how addition, multiplication and composition of generatingfunctions correspond to different ways to get new structure types fromold.JoyalÍs great contribution was to give structure types a rigorousdefinition, and use ts to show that many calculations involvinggenerating functions can be done directly with structure types. Itturns out that just as generating functions form a *set* equipped withvarious operations, structure types form a *category* with a bunch ofcompletely analogous operations. Ts means that instead of merelyproving *equations* between generating functions, we can construct*isomorpsms* between their underlying structure types - wch implysuch equations, but are worth much more. ItÍs like the differencebetween knowing two tngs are equal and knowing a specific reason WHYtheyÍre equal!Of course, ts business of replacing equations by isomorpsms is calledcategorification. In ts lingo, structure types are categorified power series, just as finite sets are categorified natural numbers. A wle back, James Dolan and I noticed that since you can use powerseries to describe states of the quantum harmonic oscillator, you cantnk of structure types as states of a categorified version of tsphysical system! Ts gives new insights into the combinatorialunderpinnings of quantum physics. For example, the discrete spectrum of the harmonic oscillatorHamiltonian can be traced back to the discreteness of finite sets!The commutation relations between annilation and creation operatorsboil down to a very simple fact: thereÍs one more way to put a ball ina box and then take one out, than to take one out and then put one in.Even better, the whole theory of Feynman diagrams gets a simplecombinatorial interpretation. But for ts, one really needs to gobeyond structure types and work with a generalization called stufftypes.IÍve been tnking about ts business for a wle now, so last fall I decided to start giving a year-long course on categorification and quantization. The idea is to explain bunches of quantum theory, quantum field theory and combinatorics all from ts new point of view. ItÍs fun! Derek Wise has been scanning in s notes, and a bunch of people have been putting their homework online. So, you can follow along if you want:1) Baez and Derek Wise, Categorification and Quantization.IÍd like to give you a little taste of ts subject now. But, instead of explaining it in detail, IÍll just give some examples of how structure types yield some far-out generalizations of the concept of cardinality. Ts stuff is a continuation of some themes developed in week144, week185, week190, so IÍll start with a review.Suppose F is a structure type. Let F_n be the *set* of ways we can put ts structure on a n-element set, and let |F_n| be the *number* of ways to do it. In combinatorics, people take all these numbers |F_n| and pack them into a single power series. ItÍs called the generating function of F, and itÍs defined like ts: |F_n| |F|(x) = sum ----- x^n n!It may not converge, so in general itÍs just a formal power series -but for interesting structure types it often converges to an interesting function. WhatÍs good about generating functions is that simple operations on them correspond to simple operations on structure types. We can use ts to count structures on finite sets. Let me remind you how it works for binary trees!ThereÍs a structure type T where a T-structure on a set is a way of making it into the leaves of a binary tree drawn in the plane. For example, hereÍs one T-structure on the set {a,b,c,d}: b d a c / / / / / / / / / / to the choice of different orderings, the number of T-structures on an n-element set is n! times the number of binary trees with nleaves. Annoyingly, the latter number is traditionally called the (n-1)st Catalan number, C_{n-1}. So, we have:|T|(x) = sum C_{n-1} x^nwhere the sum starts with n = 1.ThereÍs a nice recursive definition of T: To put a T-structure on a set, either note that it has one element, in wch case thereÍs just one T-structure on it, or chop it into two subsets and put a T-structure on each one.In other words, any binary tree is either a degenerate tree with just one leaf: Xor a pair of binary trees stuck together at the root: ----- ----- | | | | | T | | T | | | | | ----- ----- / / / /We can write ts symbolically asT = X + T^2HereÍs why: X is a structure type called being the one-element set, + means exclusive or, and squaring a structure type means you chop your set in two parts and put that structure on each part. (I explained these rulesmore carefully in week190.)I should emphasize that the equals sign here is really an *isomorpsm* between structure types - IÍm only using equals because the isomorpsm key on my keyboard is stuck. But if we take the generating function of both sides we get an actual equation, and the notation is set up to make ts really easy:|T| = x + |T|^2In week144 I showed how you can solve ts using the quadratic equation:|T| = (1 - sqrt(1 - 4x))/2. and then do a Taylor expansion to get |T| = x + x^2 + 2x^3 + 5x^4 + 14x^5 + 42x^6 + ... Lo and behold! The coefficient of x^n is the number of binary trees with n leaves! ThereÍs also another approach where we work directly with thestructure types themselves, instead of taking generating functions.Ts is harder because we canÍt subtract structure types, or dividethem by 2, or take square roots of them - at least, not without stretcng the rules of ts game. All we can do is use the isomorpsm T = X + T^2 and the basic rules of category theory. ItÍs not as efficient, but itÍs illuminating. ItÍs also incredibly simple: we just keep sticking in X + T^2 wherever we see T on the right-hand side, over and over again. Like ts:T = X + T^2T = X + (X + T^2)^2T = X + (X + (X + T^2)^2)^2and so on. You might not tnk weÍre getting anywhere, but ifyou stop at the nth stage and expand out what weÍve got, youÍll get the first n terms of the Taylor expansion we had before!At least, you will if you count stages and terms correctly.I wonÍt actually do ts, because itÍs better if you do it yourself. When you do, youÍll see it captures the recursive process of building a binary tree from lots of smaller binary trees. Each time you see a T and replace it with an X + T^2, youÍre really taking a little binary tree: ----- | | | T | | | -----and replacing it with either a degenerate tree with just a single leaf: Xor a pair of binary trees: ----- ----- | | | | | T | | T | | | | | ----- ----- / / / /So, each term in the final result actually corresponds to a specific tree! Ts is a good example of categorification: when we calculate thecoefficient of x^n ts way, weÍre not just getting the *number* of binary planar trees with n leaves - weÍre getting an actual explicit description of the *set* of such trees. Now, what happens if we take the generating function |T|(x) andevaluate it at x = 1? On the one hand, we get a divergent series:|T|(1) = 1 + 1 + 2 + 5 + 14 + 42 + ...Ts is the sum of all Catalan numbers - or in other words, the number of binary planar trees. On the other hand, we can use the formula|T| = (1 - sqrt(1 - 4x))/2to get|T|(1) = (1 - sqrt(-3))/2 It may seem insane to conclude1 + 1 + 2 + 5 + 14 + 42 + ... = (1 - sqrt(-3))/2but Lawvere noticed that thereÍs a kind of strange sense to it.The trick is to work not with generating function |T| but with the structure type T itself. Since |T|(1) is equal to the *number* of planar binary trees, T(1) should be naturally isomorpc to the *set* of planar binary trees. And it is - itÍs obvious, once you tnk about what it really means. The number of binary planar trees is not very interesting, but the set of them is. In particular, if we take the isomorpsmT = X + T^2and set X = 1, we get an isomorpsmT(1) = 1 + T(1)^2wch says a planar binary tree is either the tree with one leaf or a pair of planar binary trees.Starting from ts, we can derive lots of other isomorpsms involving the set T(1), wch turn out to be categorified versions of equations satisfied by the number|T|(1) = (1 - sqrt(-3))/2 For example, ts number is a sixth root of unity. Wle thereÍs no one-to-one correspondence between 6-tuples of trees and the 1 element set, wch would categorify the formula|T|(1)^6 = 1there *is* a very nice one-to-correspondence between 7-tuples of trees and trees, wch categorifies the formula|T|(1)^7 = |T|(1)Of course the set of binary trees is countably infinite, and so is the set of 7-tuples of binary trees, so they can be placed in one-to-one correspondence - but thatÍs boring. When I say very nice, I mean sometng more interesting: starting with the isomorpsmT = x + T^2we get a one-to-one correspondenceT(1) = 1 + T(1)^2wch says that any binary planar tree is either degenerate or a pair of binary planar trees... and using ts we can *construct* a one-to-one correspondenceT(1)^7 = T(1)The construction is remarkably complicated. Even if you do itas efficiently as possible, I tnk it takes 18 steps, like ts:T(1)^7 = T(1)^6 + T(1)^8 = T(1)^5 + T(1)^7 + T(1)^8 . . . = 1 + T(1) + T(1)^2 + T(1)^4 = 1 + T(1) + T(1)^3 = 1 + T(1)^2 = T(1)IÍll let you fill in the missing steps - itÍs actually quite fun if you like puzzles. If you get stuck, you can look up the answer in a couple of different places. Wle Lawvere was the first to figure ts out, the first to write it up was Andreas Blass: 2) Andreas Blass, Seven trees in one, Jour. Pure Appl. Alg. 103 (1995), 1-21. Also available at http://www.math.lsa.umich.edu/~ablass/cat.html ThereÍs also a nice treatment based on more general results here:3) Marcelo Fiore, Isomorpsms of generic recursive polynomialtypes, to appear in 31st Symposium on Principles of Programming Languages (POPL04). Also available at http://www.cl.cam.ac.uk/~mpf23/papers/Types/recisos.ps.gzIn fact, Fiore and Leinster have a nice general theory that explains whythe set T(1) acts so much like a sixth root of unity:4) Marcelo Fiore and Tom Leinster, Objects of categories as complex numbers, available as math.CT/0212377. The idea is that whenever we have an object Z in a distributive category (a category with finite products and coproducts, the former distributing over the latter), and itÍs equipped with an isomorpsmZ = P(Z)where P is a polynomial with natural number coefficients, we can associate to it a cardinality |Z|, namely any complex solution of the equation|Z| = P(|Z|)Wch solution should we use? Well, for simplicity letÍs consider thecase where P has degree at least 2 and the relevant Galois group actstransitively on the roots of ts equation, so all roots are createdequal. Then we can pick *any* solution as the cardinality |Z|. Anypolynomial equation with natural number coefficients satisfied by onesolution will be satisfied by all, so it wonÍt matter wch one wechoose.Now suppose the cardinality |Z| satisfies such an equation:Q(|Z|) = R(|Z|)where neither Q nor R is constant. Then the results of Fiore andLeinster say we can construct an isomorpsmQ(Z) = R(Z)in our distributive category! In other words, a bunch of equations satisfiedby the objectÍs cardinality automatically come from isomorpsms involving the object itself.Ts explains why the set T(1) of binary trees acts like it has cardinality |T|(1) = (1 - sqrt(-3))/2 or equally well,|T|(1) = (1 + sqrt(-3))/2(Since the relevant Galois group interchanges these two numbers, we canuse either one.) More generally, the set T(n) consisting of binary trees with n-colored leaves acts a lot like the number |T|(n). Ts has gotten me interested in trying to find a nice model of what I call the Golden Object: an object G in some distributive categorythatÍs equipped with an isomorpsmG^2 = G + 1The Golden Object doesnÍt fit into Fiore and LeinsterÍs formalism,since ts isomorpsm is not of the form G = P(G) where P has naturalnumber coefficients. But, it still seems that such an object deservesto have a cardinality equal to the golden ratio.James Propp came up with an interesting idea related to the Golden Object:consider what happens when we evaluate the generating function for binary trees at -1. On the one hand we get an alternating sum of Catalan numbers:|T|(-1) = -1 + 1 - 2 + 5 - 14 + 42 + ...On the other hand, we can use the formula|T| = (1 - sqrt(1 - 4x))/2to get|T|(1) = (1 - sqrt(5))/2 wch is -1 divided by the Golden Ratio. Of course, itÍs possible we should use the other sign of the square root, and get|T|(1) = (1 + sqrt(5))/2wch is just the Golden Ratio! Galois theory says these two roots arecreated equal. Either way, we get a bizarre and fascinating formula:- 1 + 1 - 2 + 5 - 14 + 42 + ... = (1 +- sqrt(5))/2 Can we fit ts into some clear and rigorous framework, or is it just nuts?WeÍd like some generalization of cardinality for wch the set of binary trees with -1-colored leaves has cardinality equal to the Golden Ratio. James Propp suggested one avenue. Following SchanuelÍs ideas on Eulercharacteristic as a generalization of cardinality, it makes sense totreat the real line as a space of cardinality -1. Ts will soundcrazy unless you go back and read week147, so please do that!Anyway, using ts idea it seems reasonable to consider the space ofbinary trees with leaves labelled by real numbers as a rigorousversion of the set of binary trees with -1-colored leaves. So, wejust need to figure out what generalization of Euler characteristicgives ts space an Euler characteristic equal to the Golden Ratio.It would be great if we could make ts space into a Golden Object insome distributive category, but that may be asking too much.Whew! ThereÍs obviously a lot of work left to be done here. HereÍssometng easier: a riddle. WhatÍs ts sequence? un, dos, tres, quatre, cinc, seis, set, vuit, nou, deu,...Now IÍd like to mention some important papers on n-categories. Youmay tnk IÍd lost interest in ts topic, because IÍve been talkingabout other tngs. But itÍs not true!Most importantly, Tom Leinster has come out with a big book on n-categories and operads:5) Tom Leinster, gher Operads, gher Categories, Cambridge U. Press, As youÍll note, he managed to talk the press into letting m keep s book freely available online! We should all do ts. Nobody will ever make much cash writing esoteric scientific tomes - it takes so long, you could earn more per hour digging ditches. The only *financial* benefit of writing such a book is that people will read it, tnk youÍre smart, and want to re you, promote you, or invite you to give talks in cool places. So, maximize your chances of having people read your books by keeping them free online! People will still buy the paper version if itÍs any good....And indeed, LeinsterÍs book has many virtues besides being free. Hegracefully leads the reader from the very basics of category theory straight to the current battle front of weak n-categories, emphasizingthroughout how operads automatically take care of the otherwise mind-numbingtcket of coherence laws that inevitably infest the subject. He doesnÍttake well-established notions like monoidal category and bicategoryfor granted - instead, he dives in, takes their definitions apart, and compares alternatives to see what makes these concepts tick. ItÍs ts sort of careful tnking that we desperately need if weÍre ever going to reach thedream of a clear and powerful theory of gher-dimensional algebra. Hedoes a similar careful analysis of operads and multicategories beforepresenting a generalized theory of operads thatÍs powerful enough to supportvarious different approaches to weak n-categories. And then he describesand compares some of these different approaches!In short: if you want to learn more about operads and n-categories, ts is *the* book to read.Leinster doesnÍt say too much about what n-categories are good for, except for a nice clear introduction entitled Motivation for Topologists, where he sketches their relevance to homology theory, homotopy theory, and cobordismtheory. But ts is understandable, since a thorough treatment of theirapplications would vastly expand an already hefty 380-page book, and diffuse its focus. It would also steal sales from *my* forthcoming book on gher-dimensional algebra - wch would be really bad, since I plan to retire on the fortune IÍll make from ts.Secondly, Michael Batanin has worked out a beautiful extension of s ideas on n-categories wch sheds new light on their applications to homotopy theory:6) Michael A. Batanin, The Eckmann-lton argument, gher operads andE_n spaces, available as math.CT/0207281.Michael A. Batanin, the combinatorics of iterated loop spaces, available as math.CT/0301221.Getting a manageable combinatorial understanding of the space of loopsin the spaces of loops in the space of loops... in some space hasalways been part of the dream of gher-dimensional algebra. Thesek-fold loop spaces or have been important in homotopy theory sincethe 1970s - see the end of week199 for a little bit about them.People know that k-fold loop spaces have k different products thatcommute up to homotopy in a certain way that can be summarized bysaying they are algebras of the E_k operad, also called the littlek-cubes operad. However, their wealth of structure is still a bitmind-boggling. James Dolan and I made some conjectures about theirrelation to k-tuply monoidal categories in our paper Categorification (see week121), and now Batanin is making ts more precise using s approach to n-categories - wch is one of the ones described in LeinsterÍs book.ThereÍs also been a lot of work applying gher-dimensional algebra totopological quantum field theory - thatÍs what got me interested in n-categories in the first place, but a lot has happened since then.For a ghly readable introduction to the subject, with tons of greatpictures, try:7) Joacm Kock, Frobenius Algebras and 2D Topological Quantum Field Ts is mainly about 2d TQFTs, where the concept of Frobenius algebrareigns supreme, and everytng is very easy to visualize. When we go up to 3-dimensional spacetime life gets harder, but also more interesting. Ts book isnÍt so easy, but itÍs packed with beautiful math and wonderfully drawn pictures:8) Thomas Kerler and Volodymyr L. Lyubashenko, Non-Semisimple Topological Quantum Field Theories for 3-Manifolds with Corners, Lecture Notes in Mathematics 1765, Springer, Berlin, 2001.The idea is that if we can extend the definition of a quantum fieldtheory to spacetimes that have not just boundaries but *corners*, wecan try to build up the theory for arbitrary spacetimes from itsbehavior on simple building blocks - since itÍs easier to chopmanifolds up into a few basic shapes if we let those shapes havecorners. However, it takes gher-dimensional algebra to describe allthe ways we can stick together manifolds with corners! Here Kerlerand Lyubashenko make 3-dimensional manifolds going between 2-manifoldswith boundary into a double category... and make a bunch of famous3d TQFTs into double functors.Closely related is ts paper by Kerler:9) Thomas Kerler, Towards an algebraic characterization of 3-dimensionalmath.GT/0008204.It relates the category whose objects are 2-manifolds with a circle asboundary, and whose morpsms are 3-manifolds with corners goingbetween these, to a braided monoidal category freely generated by aquasitriangular Hopf algebra object. (IÍm leaving out some fineprint here, but probably putting in more than most people want!) Itcomes close to showing these categories are the same, but suggeststhat theyÍre not quite - so the perfect connection between topologyand gher categories remains elusive in ts important example.Answer to the riddle: these are the Catalan numbers - i.e., the natural numbers as written in Catalan. Ts riddle was taken from the second volume of StanleyÍs book on enumerative combinatorics (see week144).-----------mathematics and physics, as well as some of my research papers, can beobtained athttp://math.ucr.edu/home/baez/For a table of contents of all the issues of Ts WeekÍs Finds, tryhttp://math.ucr.edu/home/baez/twf.htmlA simple jumping-off point to the old issues is available athttp://math.ucr.edu/home/baez/twfshort.htmlIf you just want the latest issue, go === clusteringyour question is not very precise, but ts sounds like a job for a neuralnetwork or self-organizing map. IÍm not have a dataset with large number of features. I would like to reduce> the dimensionality of the dataset first. In order to do that, I am> tnking of clustering the features first, then replace each cluster> with a meta-feature wch best describe the features in that cluster.> The features from the original dataset are not independent to each> other. I would like to have the clusters to be as pure, and as tight> as possible -- All features in a cluster are similar. Not all features> need === Grothendieck pronounced?How is Grothendieck pronounced? for the replies. I have embarrassed myself by accidentallyposting === Grothendieck pronounced? Cox asked> >> How is Grothendieck pronounced?Adding to two excellent explanations of the German pronounciation, oneshould add that, considering that Alexander Grothendieck spent most of slife in France, a frencfied version of the pronounciation should probablyconsidered equally valid. That would apply mostly to the second sillable,wch, I believe, would turn into sometng remotely similar to a nasaltoe. I donÍt speak French, though, so I may be totally and === Grothendieck pronounced?>Hallo,>>How is Grothendieck pronounced?>I tnk grow-ten-deek would have to be an acceptable zeroth-order>approximation. Only the first part, grow, needs some more>adjustments wch I can only try to describe:>- the ïrÍ should not be pronounced as in right; ts sound does not>exist in the German language (at least in the variant that I tend to>speak). What is the rationale for pronouncing === Re: 8th grade> The question of just what geometric facts are commonly, or may in very> good programs be, discussed in elementary and middle schools came up> in a couple of the earlier notes to ts thread. Since IÍm curious> about ts and have no idea about it, I ask for someone to make a list> for us so we can consider it seriously.> > When (in K-12) is the notion of area introduced, and when are area> formulas developed and for what kinds of regions? What early versions or elements of similarity are taught, and at what> grade level? Regardless of whether it is taught specifically, what geometric facts> (invariance of slope of a line, for example) can we pretty much depend> on most kids knowing, and by what age?I have some provisional ideas about the simplest appealing, memorable> and workable forms of some geometric facts wch perhaps we can> discuss later (the road theorem, the sidewalk theorem, and the> boardwalk [or grill] theorem). To thefollowing.Here are the NCTM Principles & Standards for School Mathematics: http://standards.nctm.org/document/index.htmHere are the Mathematics Content Standards for California PublicSchools:http://www.cde.ca.gov/standards/math/In both sets of standards, check out the algebra and geometryexpectations for middle school, and check out what is expected forlate elementary. The NCTM standards have been used as a model for thestate standards of most states outside of CA.I very much would like to see these provisional ideas posted here atk12.ed.math. I tnk that I speak for many when I say that IÍvethoroughly enjoyed the math content that youÍve posted here andelsewhere.Paul-- submissions: post to k12.ed.math or e-mail to k12math@k12groups.orgprivate e-mail to the k12.ed.math moderator: kem-moderator@k12groups.orgnewsgroup website: http://www.tnkspot.net/k12math/newsgroup charter: === the fearsome variable!Learning to use variables was not so bad. Introductory Algebra, naturallyrelying on variables, rectified many basic difficulties about Math andarithmetic. Mathematics: many definitions but less vocabulary;Natural Language: Much more vocabulary, and not too much relies on precisequantities. Many people have difficulty focusing; vocabulary takes care of ts focus. Assoon as more precise clauses and phrases come to issue, many people areuncomfortable because the topic is too technical. In Algebra, one first studies some important number and numer sentenceproperties in a formal and somewhat rigorous way. G C-- submissions: post to k12.ed.math or e-mail to k12math@k12groups.orgprivate e-mail to the k12.ed.math moderator: kem-moderator@k12groups.orgnewsgroup website: http://www.tnkspot.net/k12math/newsgroup charter: === educationI get a certain amount of mileage at the beginning of each yeartelling my classes of eighth grade Algebra students that they are in aclass all about x education. Even the writers of theMcal-Littel text support ts by so frequently using 6x in aproblem. I tnk they followed the Disney animators lead by slippingin in these sorts of in jokes (slow down the Lion King at the pointwhere Simba is so downhearted about not having s Dad around-he §opsto the ground, kicks up some dust wch spells out sex).I tell my young people that I know they tnk about x a lot, but thatother tngs are important too. Nevertheless, x is a part of theirlives and eighth grade is a good time to learn as much as possibleabout it.At least for a couple of weeks, students come to my class curiousabout what IÍm going to say.-- submissions: post to k12.ed.math or e-mail to k12math@k12groups.orgprivate e-mail to the k12.ed.math moderator: kem-moderator@k12groups.orgnewsgroup website: http://www.tnkspot.net/k12math/newsgroup charter: === someone noted, x is a problem right from the first for manystudents.After I teach about the coordinate plane, plotting points, x- and y-value tables (input/output tables, t- tables, whatever), y=mx+b andall that, then in the midst of explaing the slope formula and the riseover the run, then I appeal to the inner superior eighth grader bymentioning that change-in-y over change-in-x can be written delta yover delta x, but only really advanced Algebra 1 students will dothat. Catches about 10% each year.-- submissions: post to k12.ed.math or e-mail to k12math@k12groups.orgprivate e-mail to the k12.ed.math moderator: kem-moderator@k12groups.orgnewsgroup website: http://www.tnkspot.net/k12math/newsgroup charter: === (?) follow-upIn no particular order:In a public intermediate school, even the best teacher will lose from four to eight minutes at the beginning of each period wlestudents take out needed materials, present notes to the teacher(absence slips, grade tracking forms from the counselors/assistantprincipals, etc.), finish (or try to start!) conversations...Thequality of an intermediate school teacher can be judged by how manyand how long students are actively involved in the class. I recall amovement a few years back where teachers were encouraged to teach tothe bell, an idea that swept through a conference for administratorssomewhere. I know many of my colleagues give in for the last tenminutes, particularly after the end of the first semester.I teach as many minutes as I can. It requires a tremendous amount ofenergy. I donÍt know many jobs that require that kind of manicbehavior.Last year I taught two classes of Algebra 1 and four of TransitionalAlgebra. Transitional Algebra is a class for students who did poorly(or at least not well) in 7th grade pre-Algebra. The idea was to giveinstruction in the basics of Algebra wle remediating problems fromseventh grade. Last year there were six Algebra 1 classes on campusand nine of Transitional Algebra.Ts year I teach six classes of Algebra 1. There are ten Algebra 1classes and four Transitional Algebra. The decision was made to givethose kids,many of whom scored in the below basic category on thosetests we all cringe about, a big shove into Algebra 1. ItÍs not clearto me why ts was done, though the post about losing API points makesa scary amount of sense now. Guess who has most of these youngpeople...I do not teach honors students.-- submissions: post to k12.ed.math or e-mail to k12math@k12groups.orgprivate e-mail to the k12.ed.math moderator: kem-moderator@k12groups.orgnewsgroup website: http://www.tnkspot.net/k12math/newsgroup charter: === Journal of Knot Theory and Its Ramifications - Vol 13 No bergv@math.uiuc.edu (Maarten Bergvelt)Journal of Knot Theory and Its RamificationsView table-of-contents and abstracts athttp://www.worldscinet.com/jktr.htmlContents:Finite Type Invariants For Singular Surface Braids Associated WithSimple 1-Handle SurgeriesMasade IwakiriDistances Between The Surgery Slopes Yielding S3 On A Generic LinkE. MayrandMarkov Trace On The YokonumaHecke AlgebraJ. JuyumayaThe LinksGould Invariant Of Closed 3-BraidsAtsus IsiEnumerating The Prime Alternating Knots, Part IStuart Rankin, Ortho Flint And SchermannEnumerating The Prime Alternating Knots, Part IIStuart Rankin, Ortho Flint And SchermannEnumerating The Prime Alternating LinksStuart Rankin And Ortho FlintFor more information, go to === undecided-unixOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)Is there anytng that can be said about the expected revenue of a(first-price) combinatorial auction with randomly formed bids, forinstance, under the following conditions:- n items, m bids- the bundle a bid b is bidding for is randomly generated, say, b is bidding withprobability p independently for every good- the bid amount is a random function, say, for instance, quadratic overthe size of the bundle- winner allocation maximizes sum of bid amounts of accepted bidsAnyone === undecided-unixOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)Is it true that a function in L^2 is always differentiable in the sense ofdistributions? In other words, for any f in L^2, does it exists a functiong in L^1_loc such that: - = for all v in C_0^inftyI could find it in === undecided-unixOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)> Is it true that a function in L^2 is always differentiable in the sense of> distributions? In other words, for any f in L^2, does it exists a function> g in L^1_loc such that: - = for all v in C_0^inftyThese two statements are not equivalent.The answer to the first question is yes (but the derivative is not in L^1_loc),that to the === undecided-unixOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)> Is it true that a function in L^2 is always differentiable in the sense of> distributions? In other words, for any f in L^2, does it exists a function> g in L^1_loc such that: - = for all v in C_0^inftyI could find it in any book. Is there a good reference?You have already posted ts question at the sci.math newsgroup (twice!)and has already === bergv@math.uiuc.edu (Maarten Bergvelt),let G = (V,E) be a graph such that to each vertex and to each edge there is assigned a positive weight. The weight of a clique sums the the weight of all its vertices and edges. Is there some work about the maximum weighted clique problem? IÍm only aware of numerous algorithms dealing with the maximum vertex weighted === bergv@math.uiuc.edu (Maarten Bergvelt)Two independent questions.1) Does there exist an infinite, finitely presented, simple group G,and a two-transitive action of G?2) Does there exist a residually finite, finitely generated group,satisfying a nontrivial relation, but not virtually solvable? in === bergv@math.uiuc.edu (Maarten Bergvelt)>1) Does there exist an infinite, finitely presented, simple group G,...I recall Kenneth S. Brown mentioned to me in the (first half of) theeighties that he discovered such a group (probably published inInventiones)...>...and a two-transitive action of G?...but I do not know about that. Hans Aberg[ModeratorÍs note: maybe the following will help?@incollection {MR94f:20059, AUTHOR = {Brown, Kenneth S.}, TITLE = {The geometry of finitely presented infinite simple groups}, BOOKTITLE = {Algorithms and classification in combinatorial group theory (Berkeley, CA, 1989)}, SERIES = {Math. Sci. Res. Inst. Publ.}, VOLUME = {23}, PAGES = {121--136}, PUBLISHER = {Springer}, ADDRESS = {New York}, YEAR = {1992}, MRCLASS = {20E32 (20F05 57M05)}, === undecided-unixEpigone-thread: strungshunwhongOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)>Of course I agree that a planar graph may easily have>exponentially many embeddings. Consequently, checking each>embedding separately will not make for a polynomial time>isomorpsm algorithm. I was tnking more along the lines of>breaking up the graph into 3-connected pieces and dealing with>each of these separately. I have not worked out the details and>am not completely sure if anyone has....Yes, they have (and they did it just that way). Ts was hotstuff trty years ago. There was even a linear time algorithm: J. E. Hopcroft and J. K. Wong. Linear time algorithm for isomorpsm of planar graphs (preliminary report). In Conference Record of Sixth Annual ACM Symposium on Theory of Computing, pages 172-184, Seattle, Wasngton, 30 April-2 May 1974. http://portal.acm.org/citation.cfm?id=803896&dl=ACM&coll= portalThe authors state that ts is intended as a theoretical result,since the constant factors are worse than you would see in then log n algorithms for reasonable sizes of graphs. Also, tsdoesnÍt seem to provide a canonical form, wch is what you wouldwant for a searchable database of planar graphs. The canonicalform requires sorting, so I used to tnk it couldnÍt possibly bedone in linear time. Now I donÍt know if radix-sort techniquesmight be able to do even that.Dan === Topological Methods in Computer Science, IISecond Announcement: Algebraic Topological Methods in Computer Science, IIDepartment of MathematicsUniversity of Western OntarioLondon, Ontario, CanadaThe main areas to be covered by ts conference include computationalgeometry and topology, networks and concurrency theory. The meetingwill consist of twenty invited lectures, with additional sessions forshorter lectures.The following mathematical scientists have agreed to speak:Saugata Basu (Georgia Tech)Herbert Edelsbrunner (CS, Duke)Robin Forman (Rice)Eric Goubault (Commissariat a lÍEnergieAtomique, France)Maurice Herlihy (CS, Brown)Kathryn Hess (Lausanne)Michael Joswig (Berlin)Robert Kotiuga (Boston Univ.)Dmitry Kozlov (KTH)Reinhard Laubenbacher (Virginia Bioinformatics Institute) Raussen (Aalborg)Vin de Silva (Stanford)Rade Zivaljevic (Belgrade)Afra Zomorodian (Stanford)All main lectures will be held in Room 110 of Middlesex College.There will be a session for contributed talks. Participants who wouldlike to speak in ts session should send a title and abstract fortheir lecture to one of the organizers. The contributed talks will be20-30 minutes in length, as time permits.Middlesex College is the unique building on the campus of theUniversity of Western Ontario having a clock tower, and houses thedepartments of Mathematics, Applied Mathematics and Computer Science.Housing: A block of rooms has been reserved for conferenceparticipants at the Station Park Hotel (242 Pall Mall Street, London,Ontario, Canada). Participants will be making their own reservations,by calling 800-561-4574 or 519-642-4444 or by sending e-mail tohotel@stationparkinn.ca. The rate for participants is $105.00 CDN pernight - and the group booking name is the conference title: AlgebraicTopological Methods in Computer Science II.For alternative housing arrangements, see Dan ChristensenÍs housingweb page http://jdc.math.uwo.ca/hotels.php. Ts page also containslinks to travel directions and restaurant listings.The University of Western Ontario runs a bed and breakfast operationin Essex Hall (a residence on campus) during the summer months. Thecost is between $29.00 and $43.00 CDN per night, and there is ghspeed internet access in the rooms. You can find further informationand make reservations online at http://www.uwo.ca/hfs/cs/.Ts conference has been funded by grants from the National ScienceFoundation and the Fields Institute. Limited financial support may beavailable for conference participants, especially for grad studentsand postdocs.All further conference announcements and information will be availableat the web page http://www.math.uwo.ca/~jardine/at-csII.html.The organizers for ts meeting are:Gunnar Carlsson: gunnar@math.stanford.eduRick Jardine: jardine@uwo.caWe ask that you advise one of us if you intend to come to themeeting. Please also indicate whether or not you require financialsupport === undecided-unixOriginator: bergv@math.uiuc.edu (Maarten Bergvelt), it will take me a wle to absorb your whole Week 202,but can you tell me ts: By structure type, do you meanexactly the same tng as what Joyal and readers of s bookmean by combinatorial species, so that the difference isonly that you prefer a different name, or are you talking === Re: Symbolic Evaluation> couldnÍt find the answer so please bear with me.I want to be able to send an algebraic problem (probably in mathml) to> a web page, have a student simplify that, and then check to see if it> was simplified correctly.Example: 2x + 3y + 6xStudent would then send back an answer, and I (the server) would check> it to see if it were done correctly.I want to generate examples on the §y, so I would have to check it on> the §y. I do not want to hard code problems, or their solutions.So, does anyone know the best way to accomplish ts? in advanceRayHow about WIMS? === Re: rational arithmetic library? <4pkad382oet.fsf@thar.lbl.gov> sha1:rRAA349Fr32Nzkpjte9GaCqpZsU=> taltman> If thereÍs a limit to an implementationÍs ability to store> taltman> rationals (due to the numerator and/or the denominator being> taltman> fixnums), then theyÍre useless for the scientific> taltman> application I had in mind.> ThereÍs always a limit to any implementationÍs ability to > store rationals: resources are not infinite.Oh, IÍm sure. But all I want is sometng where integer division isrepresented as a tuple of bignums in simplified form, and where itpromotes to an integer (fixnum or bignum) when appropriate.Is there a single Scheme that can do ts? I tnk said thatSISC can do ts; am I quoting you === library?>> taltman> If thereÍs a limit to an implementationÍs ability to store>> taltman> rationals (due to the numerator and/or the denominator being>> taltman> fixnums), then theyÍre useless for the scientific>> taltman> application I had in mind.>>ThereÍs always a limit to any implementationÍs ability to >>store rationals: resources are not infinite.> Oh, IÍm sure. But all I want is sometng where integer division is> represented as a tuple of bignums in simplified form, and where it> promotes to an integer (fixnum or bignum) when appropriate.Is there a single Scheme that can do ts? I tnk said that> SISC can do ts; am I quoting you correctly?> Yes, thats exactly what SISC does. The rationals have bignum components, wch are limited to 2^32 bits (i.e. 2^(2^32) is the largest representable number).You should continue searcng though if youÍre doing gh performancecomputation. Wle SISCÍs bignum arithmetic is competitive, the rest of the Scheme system is not compiled, but rather interprets a gh performance intermediate form (micro-expressions). Ts may not matter if === Riccati challenge to MaximaThe Mathematica group automatically censors any post discussing mathsoftware other than Mathematica. The same should be done when someonefrom Wolfram newsgroup:R.Fateman is talking about solved solutions by Maple.He does not say that the solutions have complicated logarithmic> functions over a quartic. After all in the subsequent applications of> methods, the solutions will be over gher order polynomials wch is> even not possible to calculate symbolically.The use of solved solutions by other software by Mr.Fateman is very> deceptive!.Here I have provided solutions of two Riccati differential equations:http://mathforum.org/epigone/sci.math.symbolic/ smaysangdan/xdvrruinodk0@legacyBy the way Mathematica also seems unable to compute my equations.We are in breaking points of new methods of solving in Algebra and> Analysis.Mr. Leo Harten has also ample time to tnk over the difference of> D[x[t],t]+(x[t])^2 == (k*t^2+(6-2*k)*t+k)/(t^2*(-1+t)^2);In[2]:= InputForm[Simplify[x[t] /. DSolve[ode,x[t],t]]]Out[2]//InputForm= > {((3*(1 + Sqrt[1 + 4*k]) + 2*k*(4 + Sqrt[1 + 4*k]))*t^(1/2 + Sqrt[1 + 4*k]) - > 6*(-1 + k)*(2 + Sqrt[1 + 4*k])*t^(3/2 + Sqrt[1 + 4*k]) + > (9 - 9*Sqrt[1 + 4*k] + 6*k*Sqrt[1 + 4*k])*t^(5/2 + Sqrt[1 + 4*k]) - > 2*k*(-2 + Sqrt[1 + 4*k])*t^(7/2 + Sqrt[1 + 4*k]) + > (2*k*(-4 + Sqrt[1 + 4*k]) + 3*(-1 + Sqrt[1 + 4*k]))*Sqrt[t]*C[1] - > 6*(-1 + k)*(-2 + Sqrt[1 + 4*k])*t^(3/2)*C[1] + > 3*(2*k*Sqrt[1 + 4*k] - 3*(1 + Sqrt[1 + 4*k]))*t^(5/2)*C[1] - > 2*k*(2 + Sqrt[1 + 4*k])*t^(7/2)*C[1])/> ((-1 + t)*t*(t^(1/2 + Sqrt[1 + 4*k])*(-4*k*(-1 + t)^2 + > 3*(1 + t)*(-1 - Sqrt[1 + 4*k] + (-1 + Sqrt[1 + 4*k])*t)) + > Sqrt[t]*(4*k*(-1 + t)^2 + 3*(1 + t)*(1 - Sqrt[1 + 4*k] + t + > Sqrt[1 + 4*k]*t))*C[1]))}In[3]:= === Riccati challenge to Maxima> The Mathematica group automatically censors any post discussing math> software other than Mathematica. The same should be done when someone> from Wolfram Research tries to discuss Mathematica here.The Mathematica group mentioned above is undoubtedly comp.soft-sys.math.mathematica.The corresponding Maple group is comp.soft-sys.math.maple. TS group, sci.math.symbolic,is presumably open to discussion of all matters related to === Riccati challenge to Maxima> TS group,> sci.math.symbolic,Wait, you mean ts isnÍt === challenge to MaximaTs newsgroup is unmoderated. The mathematica newsgroupIS moderated, so all submissions need to be approved bySteve Christensen, who as Caesar says, tends to censorany submissions wch compare software.I tnk that DanLÍs submission is about can a CAS solvecertain Riccati equations, and s answer (yes, with Mathematica)is far more informative than many of the other posts!> >>The Mathematica group automatically censors any post discussing math>>software other than Mathematica. The same should be done when someone>>from Wolfram Research tries to discuss Mathematica here.> The Mathematica group mentioned above is undoubtedly comp.soft-sys.math.mathematica.The corresponding Maple group is comp.soft-sys.math.maple. TS group, sci.math.symbolic,is presumably open to discussion of all matters related to symbolic> computation, as the name indicates.Unfortunately it === Riccati challenge to Maxima> The Mathematica group automatically censors any post discussing math> software other than Mathematica. The same should be done when someone> from Wolfram Research tries to discuss Mathematica here.Caesar === to Fermat by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1PLqjv20749; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1PLf1i19602 by legacy.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: recently I have had great success inusing the shareware Fermat (available at http://www.bway.net/~lewis/)in my mathematics research.The short story is that I needed to row-reduce 2000 by 2500 integermatrices where the size of the entries range form 1 to 200 digits. Ofcourse, as soon as I try RowReduce in Mathematica, my computer wouldjust run out of memory. Yes, Fermat was able to do ts rowreduction. Fermat was also great in that it gave progress reports onhow far along the row reduction it was (ts was indispensible so thatI knew when to abort hopelessly long calcultions).So after over a yearÍs worth of work (using Mathematica, Perl, C++,Lisp) to build up ts matrix, Fermat finished the project by rowreducing it, and now we have a new math theorem. (If anyone cares, we now know the dimension of Siegel modular cusp forms in genus 4 === newsgroup:Today I have received the following e-mail from the editor in cef ofthe Central European Journal of mathematics (in Warsaw) about my twopapers.I have written about more clarification, and that I am §exible aboutrewriting the papers.Generally if they have intention of publisng my papers, since themethods are new, I am willing to expand the papers from two topossibly four papers, relative to their detailed instructions.I am also waiting for responds from Indian Dr.Basti,I agree with the opinion to reject your paper due to it present form.Please feel welcomed to resubmit it after === the answer so please bear with me.I want to be able to send an algebraic problem (probably in mathml) to> a web page, have a student simplify that, and then check to see if it> was simplified correctly.Example: 2x + 3y + 6xStudent would then send back an answer, and I (the server) would check> it to see if it were done correctly.I want to generate examples on the §y, so I would have to check it on> the §y. I do not want to hard code problems, or their solutions.So, does anyone know the best way to accomplish ts? in advanceRay;I tnk there is === Re: kudos to Fermat> I just wanted to report that recently I have had great success in> using the shareware Fermat (available at http://www.bway.net/~lewis/)> in my mathematics research.The short story is that I needed to row-reduce 2000 by 2500 integer> matrices where the size of the entries range form 1 to 200 digits. Of> course, as soon as I try RowReduce in Mathematica, my computer would> just run out of memory. Yes, Fermat was able to do ts row> reduction. Fermat was also great in that it gave progress reports on> how far along the row reduction it was (ts was indispensible so that> I knew when to abort hopelessly long calcultions).> Could you give a pointer to the matrix and the time andmemory that Fermat === just wanted to report that recently I have had great success in> using the shareware Fermat (available at http://www.bway.net/~lewis/)> in my mathematics research.The short story is that I needed to row-reduce 2000 by 2500 integer> matrices where the size of the entries range form 1 to 200 digits. Of> course, as soon as I try RowReduce in Mathematica, my computer would> just run out of memory. Yes, Fermat was able to do ts row> reduction. Fermat was also great in that it gave progress reports on> how far along the row reduction it was (ts was indispensible so that> I knew when to abort hopelessly long calcultions).So after over a yearÍs worth of work (using Mathematica, Perl, C++,> Lisp) to build up ts matrix, Fermat finished the project by row> reducing it, and now we have a new math theorem. (If anyone cares, > we now know the dimension of Siegel modular cusp forms in genus 4 and> weight 16.) Yuendo you have a reference one can look at?and why did you not also try matlab and maple? or have you === vs MuPAD by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1RDW6104168; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1RAHai20363 by legacy.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: it may be that MuPAD has done ts right from the start, and the > lack of bugs is a natural consequence of the design. Do you tnk> that is the case, or is it just the same hodge-podge,> but debugged more?I really donÍt know. > Anyone doing serious statistical studies should be> using a different program, I suspect.Yes, but I tnk that a general CAS should cover everyday[university] math.> Left out of the comparison is the speed.I will probably do it in the next version.> I would expect that MuPAD is slower.Yes, I tnk also so.> Also left out is the fact that Maxima is (now) open-sourceGNU Public License mentioned in the introduction covers ts, I tnk.> The quality of the documentation is a serious issue for > many users. Maxima documentation is not so good, but the available> documentation for Macsyma is substantial.I agree, and an extra difficulty comes from the fact that Maxima doesnotinclude all functions and commands of Macsyma.> I suspect that your conclusions would have to be revised if > someone spent a month fixing bugs in MaximaLet us hope that someone does that! === Re: Re: Maxima vs MuPADMacsymaÍs variants do have different strengths/weaknesses. I spent a lot oftime getting the answers to the share library and demo files CORRECTfor later versions of DOE-Macsyma/Paramax.The Symbolics/Macsyma, Inc. Macsyma did get someextensions in the areas where the DOE version was weak. They addedCylindrical Algebraic Decomposition, for example, wch is not present inthe DOE/Paradigmversions. Question 65/66 with tensors are handled in allversions wch have the ctensr/itensr packages. TheDOE/Paradigm versions were stronger in the use of thetranslator/compiler to allow numerical code to run much faster thaninterpreted Macsyma code. Symbolics worked on a NAG link to incorporate fastFortran routines for a variety of tasks, and later added a PDEpackage as well. Schelter added s own features toMaxima for algebra and fast arrays. So I expect thatMaxima can be made better in a few areas quite easily,but adding solutions for inequalities, for example, isanother large project.> for your valuable comments!> it may be that MuPAD has done ts right from the start, and the> lack of bugs is a natural consequence of the design. Do you tnk> that is the case, or is it just the same hodge-podge,> but debugged more?> I really donÍt know.> Anyone doing serious statistical studies should be> using a different program, I suspect.> Yes, but I tnk that a general CAS should cover everyday> [university] math.> Left out of the comparison is the speed.> I will probably do it in the next version.> I would expect that MuPAD is slower.> Yes, I tnk also so.> Also left out is the fact that Maxima is (now) open-source> GNU Public License mentioned in the introduction covers ts, I tnk.> The quality of the documentation is a serious issue for> many users. Maxima documentation is not so good, but the available> documentation for Macsyma is substantial.> I agree, and an extra difficulty comes from the fact that Maxima does> notinclude all functions and commands of Macsyma.> I suspect that your conclusions would have to be revised if> someone spent a month fixing bugs in Maxima> Let us hope that someone does that! I am willing to do the tests> again.> Pekka === am way out of league here, if so, tell me.We developed a library for Windows. That is for use under Visual Basicand C++.andthey are all based on rational number calculation.Functions:http://www.big-numbers.com/forum/ viewtopic.php?t=12Download:http://www.big-numbers.com/forum/ viewtopic.php?t=27&sid=b5588871941e2916514af64e861825fbSite: === newsgroup:I will again try for other Journals (one more time!).It is understandable that the Journal of Algebra from France (the landof Galois) may not publish unusal papers in polynomials.Let them put their records in the story of such === publications. I did post here insci.math.symbolic, you can see no one is taking about new ways ofsolving differential equations and polynomials.These issues are very complex witn academia and I have learned myhard lessons for about two decades.Finally I have to wait and see what it will be the === peer-reviewed publication.But a person like me that tnks that s mathematics remains forcenturies, will not be played around by the current leaders whethertheir tools of value judgments are peer reviewed or not (like it ornot).I will still wait, to === me that tnks that s mathematics remains for> centuries,IÍm pretty sure that arxiv will stay with us for quite a long time. IÍm alsocertain that in the future its successor === it via arxiv.org?I would like to say that, I would possibly write my first lecture noteand publish it myself.The rest of lecture notes follow later.Those are (i.e. arxiv.org) tools of current leaders in value judgmentsand will not be acceptable by === papers> Those are (i.e. arxiv.org) tools of current leaders in value judgments> and will not be acceptable by myself.Maybe IÍm missing sometng, but arxiv.org is not a === e-mail from the Journal ofAlgebra.Certainly they publish papers such that themselves are only leaders!.Looks like eventually I have to write my lecture notes and somehowdisseminate it myself, wch is equally difficult.I will wait until I receive the results of other journals.Looks like they do not want ts kind of mathematics at ts that your manuscripts A class polynomial, Solving polynomials with differential equations and A class polynomial III are not being accepted for publication in the Journal of Algebra.The Journal of Algebra receives many submissions and must select onlythose papers of the most current important === following e-mail from the Journal of> Algebra.> Certainly they publish papers such that themselves are only leaders!.> Looks like eventually I have to write my lecture notes and somehow> disseminate it myself, wch is equally difficult.> I will wait until I receive the results of other journals.> Looks like they do not want ts kind of mathematics at ts time.> that your manuscripts A class polynomial,> Solving polynomials with differential equations and A class> polynomial III are not being accepted for publication in the Journal> of Algebra.> The Journal of Algebra receives many submissions and must select only> those papers of the most current important interest.