mm-42 === That might be good enough. Of course that's probably related> to the impossibility of reliably defining things such as> .999... - .999...99800...001 = .000...?!?!?!...> I wouldn't go so far as to say it's impossible. After all, nonstandard> analysis makes sense of infinitely large integers. But then, the> hyperintegers of NSA are not the same as the transfinite ordinals, and I> am not aware that anyone has tried to use digit strings indexed by the> hyperintegers to define values in the hyperreals.> NSA presumably routinely works with very large (1,024-bit)> integers, but they *are* integers; the challenge is mostly> in defining operations therewith in terms of arrays> of smaller integers, sort of like defining multidigit> arithmetic in terms of the times table, except that> instead of a 10 [Times] 10 matrix one can memorize by rote,> one has a 65536 [Times] 65536 affair that is implemented by the> microprocessor's multiply instruction.1024 bits is not enough to hold a hyperinteger.My NSA is nonstandard analysis. Is yours perhaps the National SecurityAgency?-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === In sci.logic, Dave Seamanon Mon, 12 Jan 2004 00:36:30 +0000 (UTC):> In sci.logic, Dave Seaman> That might be good enough. :-) Of course that's probably related> to the impossibility of reliably defining things such as> .999... - .999...99800...001 = .000...?!?!?!...>:-)> I wouldn't go so far as to say it's impossible. After all, nonstandard> analysis makes sense of infinitely large integers. But then, the> hyperintegers of NSA are not the same as the transfinite ordinals, and I> am not aware that anyone has tried to use digit strings indexed by the> hyperintegers to define values in the hyperreals.> NSA presumably routinely works with very large (1,024-bit)> integers, but they *are* integers; the challenge is mostly> in defining operations therewith in terms of arrays> of smaller integers, sort of like defining multidigit> arithmetic in terms of the times table, except that> instead of a 10 [Times] 10 matrix one can memorize by rote,> one has a 65536 [Times] 65536 affair that is implemented by the> microprocessor's multiply instruction.> 1024 bits is not enough to hold a hyperinteger.> My NSA is nonstandard analysis. Is yours perhaps the National Security> Agency?> Yeppers; apology for the miscommunication. :-) Obviouslythe governmental NSA is more interested in encryption thaninfinities. :-)I'm not very familiar with nonstandard analysis, althoughit's obvious Garry Denke has a very nonstandard methodof analyzing infinite decimals... ;-)-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === >So we have,>y1 = a1 + a2 + ....... + aN>y2 = 2.a1 + 2^2.a2 + ...... + 2^N.aN> ......................>yN = N.a1 + N^2.a2 + .......+ N^N.aN>Now if we take the mean square expectation on this equation we get>1.(sig)^2 = E[a1^2] + E[a2^2] + ..... +E[aN^2]>2.(sig)^2 = 2^2.E[a1^2] + 2^4.E[a2^2] + ..... +2^2N.E[aN^2]>.........................>So we get a set of N linear equations. We wish to solve for E[a1^2], >E[a2^2],....,E[aN^2]. Basically, we wish to obtain a closed form>expression for E[aj^2] which boils down to finding the inverse of the>matrix:> -- -->| 1 1 . . . 1 |>| 2^2 2^4 . . . 2^2n |>| . . . . . . |>| n^2 n^4 . . . n^2n | > -- --You might look at the preprint of Bender, Brody and Meister, Inverse of a Vandermonde Matrix, at Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === > The pdf of the sum of 2 Beta rv's (the question posed above) will > generally not be close to Normal. One can derive the characteristic > function of the sum, and then invert it numerically to derive the pdf,> for given parameter values. > For example, here is a quick 2-line derivation and pdf plot using> mathStatica: [...]Well, if numerical approximations are allowed, here's a short construction in Octave which amounts to the same thing (since theconvolution is implemented via the fast Fourier transform). x = 0:0.01:1; f = 1/beta(a, b) * x.^(a-1) .* (1-x).^(b-1); f2 = conv(f, f); f2 = f2 / sum(f2*0.01); plot([0:0.01:2], f2);It's more than 2 lines, but the upside is that you don't have tobuy somebody's book and/or program to do it.Happy new year,Robert Dodier--If I have not seen as far as others, it is because giants were standing on my shoulders. -- Hal Abelson === >What is the current status of chaos theory? A decade ago it was supposed >to be The Next Big Thing.Yeah, well, that kind of thing is notoriously hard to predict, longterm.Lee Rudolph === :>What is the current status of chaos theory? A decade ago it was supposed :>to be The Next Big Thing.: Yeah, well, that kind of thing is notoriously hard to predict, longterm.ROTFLMFAO.=)Justin === >Sorry, I meant to ask: what day of the year is the longest?>This question is usually given a wrong or incomplete answer.> All the days are 24 hours except when a leap second is added.> If you mean to ask which day has the longest daylight at a given> location, that would be the day of the summer solstice. >That is true if you define the day to be from sunrise to sunset.>But if you define the day to be from sunrise to sunrise, then>the day is longest when the Earth is closest to the Sun, which>happens around December 23 (a little after winter solstice).>I define the day to be from midnight to midnight. Perhaps you meant to>ask about the solar day, which is the time between transits of the>sun. However, your answer is incorrect. Perihelion occurs in early>January (around January 8, IIRC), and therefore the answer is *not* the>perihelion date. In general, the solar day is longer than average near>the solstices (both of them) and shorter than average near the>equinoxes.>The length of the solar day is determined by the sun's daily movement>in right ascension, which determines the difference between the>sidereal day and the solar day. What's confusing you is that the sun>moves fastest *in longitude* at perihelion, but fastest movement in>longitude does not equal fastest movement in right ascension. The>obliquity of the ecliptic causes the fastest movement in right>ascension to occur near the two solstices, despite the fact that one of>them happens to be near aphelion.> Why is Dec. 19 the longest solar day instead of perihelion Jan. 4th?It is neither. For the years in which I have computed it, the longestsolar day has been either Dec. 22 or Dec. 23. That is, it seems to fallwithin a day or so of the December solstice.The reason is that the length of the solar day is affected more by thesolstice than it is by perihelion, as I explained.>The longest solar day is presently achieved near the December solstice,>because that one is closer to perihelion than the June solstice is.>Because of precession of the equinoxes, there will come a time when>perihelion is not particularly close to either solstice, and therefore>nowhere near the longest solar day.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === >Why, that's absurd.> Whatever you are drinking, it is doing funny things to your synapses.I'm glad you find it entertaining.Virgil, my impression of your attitude is that your opinion is amethod of determining comparisons of infinite set sizes besides usingcardinality would be hard to do correctly and easy to err in using.This is from my having read your many statements saying thatconstructions of comparative set sizings for infinite sets would berife with complications.We've had some discussion of uniform probability distributions overinfinite sets. Whether or not you agree, as has been statedelsewhere, that they are easy to construct or assume, we can makeobservations about what properties they would have. For example,given a uniform probability distribution over the integers, theprobability of a sample being an even integer could only beinfinitesimally different than one half.In comparing the integers to the rationals, where the integers areuniformly dense in the rationals, in the sense that any fixed lengthinterval over the rationals contains the same, plus/minus one, amountof integers as another fixed length interval over the rationals, it'sfair to say that the probability of a random rational being an integeris near zero, yet at the same time it is not zero. Also it is notless than every positive rational number. Given the unit interval ofthe rationals, [0,1), the probability of a value selected at randomfrom it being an integer is the same as what would be the value of theleast rational, in terms of dividing the interval into all thepossible neighborhoods of each rational in that interval.That would seem to be invariant with respect to the density of therationals. In a way it is a circular definition, the least rational,or via extension the least real, or least real components of ahypercomplex number.Comparing infinite sets may well be more complicated than bijectioncan be shown with naturals or not, that's because it can tell us manymore things about the sets in questions than countable oruncountable. For example in terms of infinitesimal analysis, thatbeing very much equivalent to the integral calculus, the area of a 1x1square is 1 square unit.Argue that there are more rationals on [0,2] than [0,1]. There areabout twice as many, no more than infinitesimally different thantwice, as the rationals are uniformly dense among the reals, and twois twice one.About the axiom-free logic, basically it says no axioms, yet equalityholds only for equal things and not not p does not necessarily equalp. I assert that no axioms are required, and that of all thepossible set constructions from the empty set that there areinfinitely many sets of statements free of contradiction among whichany is adequate to state and prove any logical statement that agreeswith all other accepted, non mutually contradictory, logicalstatements, for example, 2+2=4.Ceci n'est pas un pipe. - MagritteRoss === > Imagine yourself in space and look at this third rock from Sun, Earth.> Known most intelligent living being named human being rules this> Planet, 6 billion in number.[snip]Did somebody replace your wog brain with a smear of loose dog? > Where things have gone wrong?Ecce homo, Friedrich Nietzsche. It's a book - a devastatingcritique of modernity. Yeah well... Nietsche did not have computers,flush toilets, deodorant, or alt.binary.pictures.erotica.indian-asian Go spend your time more productively. > rather than trying to understand our Nature and God.[snip]foreskin bit and all religion is summarized. God is brain puke - aState lottery for the devout, a voluntary tax on the stupid.The simple measure of a country is how many immigrants want in... andhow many citizens want out. How many American immigrants aredemanding entry into India?--Uncle Alhttp://www.mazepath.com/uncleal/qz.pdfhttp:// www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) === On Sun, 11 Jan 2004 17:34:55 -0800, Uncle Al Imagine yourself in space and look at this third rock from Sun, Earth.> Known most intelligent living being named human being rules this> Planet, 6 billion in number.>[snip]>Did somebody replace your wog brain with a smear of loose dog?> Where things have gone wrong?>Ecce homo, Friedrich Nietzsche. It's a book - a devastating>critique of modernity. Yeah well... Nietsche did not have computers,>flush toilets, deodorant, or alt.binary.pictures.erotica.indian-asian >Go spend your time more productively.I'm nearly certain both Nietsches had flush toilets, at least in theirlater years (as in after Elizabeth came back to Germany to look afterFriedrich after he went mad).> rather than trying to understand our Nature and God.>[snip]>foreskin bit and all religion is summarized. God is brain puke - a>State lottery for the devout, a voluntary tax on the stupid.>The simple measure of a country is how many immigrants want in... and>how many citizens want out. How many American immigrants are>demanding entry into India?>-->Uncle Al>http://www.mazepath.com/uncleal/qz.pdf>http:// www.mazepath.com/uncleal/eotvos.htm> (Do something naughty to physics) === > On Sun, 11 Jan 2004 17:34:55 -0800, Uncle Al Imagine yourself in space and look at this third rock from Sun, Earth.>Known most intelligent living being named human being rules this>Planet, 6 billion in number.>[snip]>Did somebody replace your wog brain with a smear of loose dog?>Where things have gone wrong?>Ecce homo, Friedrich Nietzsche. It's a book - a devastating>critique of modernity. Yeah well... Nietsche did not have computers,>flush toilets, deodorant, or alt.binary.pictures.erotica.indian-asian>Go spend your time more productively.I'm nearly certain both Nietsches had flush toilets, at least in their> later years (as in after Elizabeth came back to Germany to look after> Friedrich after he went mad).German flush toilet technology's never been something to brag about, unlessyou're a smuggler.Jim === >On Sun, 11 Jan 2004 17:34:55 -0800, Uncle Al Imagine yourself in space and look at this third rock from Sun, Earth.> Known most intelligent living being named human being rules this> Planet, 6 billion in number.>[snip]>Did somebody replace your wog brain with a smear of loose dog?>Where things have gone wrong?>Ecce homo, Friedrich Nietzsche. It's a book - a devastating>critique of modernity. Yeah well... Nietsche did not have computers,>flush toilets, deodorant, or alt.binary.pictures.erotica.indian-asian>Go spend your time more productively.>I'm nearly certain both Nietsches had flush toilets, at least in their>later years (as in after Elizabeth came back to Germany to look after>Friedrich after he went mad).> German flush toilet technology's never been something to brag about, unless> you're a smuggler.Yeah... That central pedestal to display your droppings high and dryprior to a dismal flush. I anticipated somebody unfamiliar with areal world German facility to protest error. A lifetime of Germanwater closets would drive anybody with a sense of personal hygiene todeclare civilization was doomed. God has been remarkably silent aboutclever ways to deal with raw sewage. Who flushes in heaven?Note that the US Enviro-whiner 3-gallon flush (and smaller) is ahorror. We had a plumber replace the flush mechnism on one of ourAmerican Standard 5-gallon jobs. He offered us $300 cash for ourpudgy porcelain goddess plus a brand new eco-john free, with freeinstallation. No way! 5-gallon jobs are being smuggled in fromCanada as you read this. They go for $1000+ each.There's no way a manly US turd will fit through a Euro-trash tinkler. Might as well dig a slit latrine in the backyard and have done withit.-- Uncle Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals)Quis custodiet ipsos custodes? The Net! === > On Sun, 11 Jan 2004 17:34:55 -0800, Uncle Al Imagine yourself in space and look at this third rock from Sun, Earth.>Known most intelligent living being named human being rules this>Planet, 6 billion in number.>[snip]>Did somebody replace your wog brain with a smear of loose dog?>Where things have gone wrong?>Ecce homo, Friedrich Nietzsche. It's a book - a devastating>critique of modernity. Yeah well... Nietsche did not have computers,>flush toilets, deodorant, or alt.binary.pictures.erotica.indian-asian>Go spend your time more productively.>I'm nearly certain both Nietsches had flush toilets, at least in their> later years (as in after Elizabeth came back to Germany to look after> Friedrich after he went mad).>German flush toilet technology's never been something to brag about, unless>you're a smuggler.> Yeah... That central pedestal to display your droppings high and dry> prior to a dismal flush.Those basins die out. I seem to remember that they were introduced becausesome bigwig (king or emperor) was a bit obsessed with checking the quality ofhis sausage and since no one else knew about water toilets they just saw thiskind of basin as normal due to lack of water toilet experience.Lots of Greetings!Volker === On Sun, 11 Jan 2004 23:00:31 -0800, Clave> On Sun, 11 Jan 2004 17:34:55 -0800, Uncle Al Imagine yourself in space and look at this third rock from Sun, Earth.>Known most intelligent living being named human being rules this>Planet, 6 billion in number.>[snip]>Did somebody replace your wog brain with a smear of loose dog?>Where things have gone wrong?>Ecce homo, Friedrich Nietzsche. It's a book - a devastating>critique of modernity. Yeah well... Nietsche did not have computers,>flush toilets, deodorant, or alt.binary.pictures.erotica.indian-asian>Go spend your time more productively.I'm nearly certain both Nietsches had flush toilets, at least in their> later years (as in after Elizabeth came back to Germany to look after> Friedrich after he went mad).>German flush toilet technology's never been something to brag about, unless>you're a smuggler.Austria:Flush toilets appeared in our homes only after WW2.At least on the countryside, where I grew up.They interviewed me, because they could not understandwhy there always remained some water in the bowl,they wanted to make sure and asked me how toilets are doing in the new public school buildingwhere I was due to learn the alphabet.w. === > Let me try to pose the problem, I am trying to solve.> I have a matrix, A (for simplicity let us assume it is a 2 X 2 matrix), that> is supposed to be symmetric and positive definite. There is a differential> equation the numerical solution of which gives me A at a current instant, in> the form Adot = f(x,p)..where x and p are the states of this dynamical> system and f is a nonlinear function in x, p. Since A has only three unique> parameters for this case, I actually obtain, theta_dot = g(x,p), where theta> A(t) thereby enforcing symmetry..... To enforce positive definiteness, I> need to do the following,> theta(1) > 0, theta(2) > 0 and -sqrt(theta(1)*theta(2)) < theta(3) <> sqrt(theta(1)*theta(2))...> Then by using some sort of a projection algorithm, I could enforce positive> definiteness in this way, however, this doesnt scale well for higher> dimensioned matrices.> I was wondering if people were aware of literature that does something like> this...Any references would be greatly appreciated.> L TunesIs matrix A obtained by solving numerically a Riccati equation? === Is matrix A obtained by solving numerically a Riccati equation?Firstly thanks for all the responses.This sort of a problem normally arises in control of mechanical systems,wherein the equations of motion of the system is written in the second orderform as, M xddot + C xdot + K x = F.To command a specific response of the system, i.e. x(t), one needs togenerate a control (Force/Torque) input that achieves this. It can be shownthat the Control input is a function of the Mass term (M). However, in mostpractical situations, it is common to assume that the true value of M is notknown so the control algorithm uses an estimate of M, which would be asolution to a nonlinear differential equation. Since Mass is a symmetricpositive definite quantity, it then becomes important to ensure that thisproperty is preserved in the solution (estimate).I think the factorization of M in terms of a triangular matrix and it'stranspose seems like the first approach I would like to work with.L Tunes === > Is there an efficient way to solve the following problem:> Given a,b,eps>0 find all positive neps> of being integer.> For example, let a=sqrt(2), b=sqrt(3) and eps=10^-k. I've compiled a table>of> n that work for various small values of k. Is it complete?> [The numbers are shortened: 297..072(26) is the 26 digit number starting>with> 296 and ending with 072]> k=13> 297..072(26), 617..409(26), 915..158(26)> k=14> 394..552(28), 483..617(28), 685..472(28), 774..537(28)> k=15> 135..884(30)> k=16> 218..750(32)> k=17> 173..971(34)> k=18> 506..352(36), 546..517(36), 972..568(36)> k=19> 131..132(38)> k=20> 239..576(40), 296..519(40), 904..493(40), 961..436(40)>Just an idea: Use continued fractions, see>Best Rational Approximations to Real Numbers by R. Knott>http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/ cfINTRO.html#convergents>find the simultaneous approximations I listed. What I am trying to figure outis how good the algorithm is. My guess is I have missed a few n. I know lattice reduction methods can usually find one solution but can't figureout how to tweak them to find multiple solutions. Surely there must be away...rich <3ffc9a68$7$fuzhry+tra$mr2ice@news.patriot.net> <4000a20b$18$fuzhry+tra$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === In , on 01/10/2004 at 11:56 PM, Dan Christensen said:>Who said anything about doing without infinite sets? If you want to>develop number theory in my system, for example, you simply start>with a premise corresponding to Peano's Axioms which includes an>axiom of infinity.Do that and it's a different system.>What do you do in ZF if you want to develop group theory?I write out a definition and start proving theorems.>Wouldn't you simply start by assuming the group axioms?No. I don't need to if I already have ZF.>Indeed. The only advantage my system offers is its simplicity.Its simplicity is an illusion if you need to complexify it to actuallyapply it.>While it may suit your purposes, for them, ZF is needlessly>complicated and overly abstract.Is it any more complex than yours, once you're done adding axioms?-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === > In , on 01/10/2004> at 11:56 PM, Dan Christensen said:>Who said anything about doing without infinite sets? If you want to>develop number theory in my system, for example, you simply start>with a premise corresponding to Peano's Axioms which includes an>axiom of infinity.Do that and it's a different system.>Yes, it is different.>What do you do in ZF if you want to develop group theory?I write out a definition and start proving theorems.>In much the same way, you can develop number theory in my system. The groupaxioms define one system. Peano's axioms define another. There somesimilarities. Both postulate the existence of a set with one or morefunctions defined on it. Why should they necessarily be treated differentlyby set theory?>Wouldn't you simply start by assuming the group axioms?No. I don't need to if I already have ZF.>Call them definitions if you like.>Indeed. The only advantage my system offers is its simplicity.Its simplicity is an illusion if you need to complexify it to actually> apply it.>Have you tried to use my system? You click on Peano's Axioms (on the Numbersmenu) to generate the required premise (or definition if you like). Thisdefines n, 1 and a successor function on n. Then you start developing numbertheory.>While it may suit your purposes, for them, ZF is needlessly>complicated and overly abstract.Is it any more complex than yours, once you're done adding axioms?>What sense, for example, are most people to make of the set{e, {e}} where e is the empty set? Who would ever guess that it is thenumber 2?My system, on the other hand, makes use of everyday concepts like firstand next (as defined by PA). In my opinion, it is much more intutitive, ifsomewhat less austere than ZF. Do have a look at it.DanVisit DC Proof Online at http://www.dcproof.com -- FREE download <40009798$9$fuzhry+tra$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === In , on 01/11/2004 at 04:14 PM, Joona I Palaste said:>I can't see how using the notation I defined above, fU = {f(x) | x in>U}, fU and f(U) could be identical by definition.Because in conventional notation we write f(U) for {f(x)|x in U}.And, yes, that is an abuse of notation, but it is nearly universal.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === Shmuel (Seymour J.) Metz scribbled the following:> In , on 01/11/2004> at 04:14 PM, Joona I Palaste said:>I can't see how using the notation I defined above, fU = {f(x) | x in>U}, fU and f(U) could be identical by definition.> Because in conventional notation we write f(U) for {f(x)|x in U}.> And, yes, that is an abuse of notation, but it is nearly universal.But in the question I was asking, I was not using that conventionalnotation. Sheesh, is David Ullrich the only one to understand me?-- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland ---------- http://www.helsinki.fi/~palaste --------------------- rules! --------/It was, er, quite bookish. - Horace Boothroyd === >notation. Sheesh, is David Ullrich the only one to understand me?Dave got it right. The standard notation in logic is to definef[U]={f(x):xin U}. In situations where U may be both a subsetof X and an element of X, it is quite possible for f(U), the imageof the element, and f[U], the collection of images of elements of U,to be different. In fact, it is likely that f(U) is not a subset of X,while f[U} will always be.On the other hand, this possibility does not arise very often inmath outside of logic. Hence, for most people, f(U) is used for f[U].This is standard notation outside of logic.--Dan GrubbX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === In , on 01/11/2004 at 05:57 PM, gds@best.cut.here.com said:>Considering that mathematics was (and is) used to gain insight into>practical problems (e.g. engineering), I don't see why it is wrong to>teach applications. I didn't sy that it was wrong, just that it didn't do what he wasasking for. But paintings *ARE* used to cover cracks: would you saythat it would be right to use that as the basis for teaching ArtAppreciation?>IMHO, applications reinforce the theory.A view not shared by, e.g., G. H. Hardy.>Why not useful, fascinating, and fun?Nothing, as long as the first is kept firmly subordinate to the otherwhether she can apply something a decade from now. She *is* likely tobe interested in whether it is fun. -- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === >I didn't sy that it was wrong, just that it didn't do what he was>asking for. But paintings *ARE* used to cover cracks: would you say>that it would be right to use that as the basis for teaching Art>Appreciation?Of course not. Your reply to Diego implied that (engineering)math applications are in some way analogous to covering cracks withartwork. That is a very limiting assessment of the value ofapplications, imho.Here's another example. If someone wants to impart a love andunderstanding of music to someone else, they might:* have them learn an instrument and how to read music* have them learn the history of music* take them to concerts* listen to recordings with themLikewise, for mathematics, there are various approaches: one need notstrictly teach theory. If she really wants to do theory, it willbecome apparent.>Nothing, as long as the first is kept firmly subordinate to the other>whether she can apply something a decade from now. She *is* likely to>be interested in whether it is fun.What basis do you have of determining that she wouldn't findapplications fun right now?--gregbogds at best dot com X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === In , on 01/11/2004>I had thought that words transliterated into Hebrew strictly used>tet.How is nylpw_tn_tywt a word transliterated into Hebrew?nylpw_tn_tywt would be a tranliteration, but what you gave wasderived from that, and such a derivation has to adhere to the rules ofDiqduq.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === [snip]> I had thought that words transliterated into Hebrew strictly used tet.> Certainly not. For words of English origin, th becomes Tav and t>becomes Teth. Case in point: Mathematics -> mathematika,>mem-tav-mem-teth-yod-qoph-heh.>Alan> Are there any other examples of this occurance?> YSYou mean, examples of t->Teth and th->Tav? Of course, plenty.pathetic, aesthetic, telepathy, metallic, meta-(whatever),symmetry, protein, intelligence, theory, theoretic, myth,Thales, Plato, Aristotle, technical are all words (or names)that have been carried into Hebrew, and in all of those cases the ruleis observed.This is just off the top of my head, I can continue on and on, but theimportant point is that there's no single case I'm aware of where thisrule is broken.Alan === > [snip]>I had thought that words transliterated into Hebrew strictly usedtet.>Certainly not. For words of English origin, th becomes Tav and t> becomes Teth. Case in point: Mathematics -> mathematika,> mem-tav-mem-teth-yod-qoph-heh.>Alan>Are there any other examples of this occurance?>YSYou mean, examples of t->Teth and th->Tav? Of course, plenty.> pathetic, aesthetic, telepathy, metallic, meta-(whatever),> symmetry, protein, intelligence, theory, theoretic, myth,> Thales, Plato, Aristotle, technical are all words (or names)> that have been carried into Hebrew, and in all of those cases the rule> is observed.This is just off the top of my head, I can continue on and on, but the> important point is that there's no single case I'm aware of where this> rule is broken.>Very interesting indeed. I am looking at the commonalities of these words,most seem derived from the Greek. I am wondering if this is a key factor ofthis rule.Any clues appreciated.YS---Checked by AVG anti-virus system (http://www.grisoft.com). <_3oMb.4416$i4.1565@newsread1.news.atl.earthlink.net> === [th-> taf, t->tet]Very interesting indeed. I am looking at the commonalities of these> words, most seem derived from the Greek. I am wondering if this is a> key factor of this rule. Any clues appreciated.The undageshed taf was pronounced th in ancient Hebrew (just as theundageshed dalet was pronounced dh, the voiced version of th foundin either). Samaritans still retain these pronunciations.Len. === >[th->taf, t->tet]>Very interesting indeed. I am looking at the commonalities of these>words, most seem derived from the Greek. I am wondering if this is a>key factor of this rule. Any clues appreciated.The undageshed taf was pronounced th in ancient Hebrew (just as the> undageshed dalet was pronounced dh, the voiced version of th found> in either). Samaritans still retain these pronunciations.Len.>Yemenites as well.YS---Checked by AVG anti-virus system (http://www.grisoft.com). === Slightly OT,but does anyone publish maths in Hebrew nowadays?Are school maths textbooks normally in Hebrew?-- Timothy Murphy tel: +353-86-2336090, +353-1-2842366 === On Mon, 12 Jan 2004 18:32:57 +0000, Timothy Murphy>Slightly OT,>but does anyone publish maths in Hebrew nowadays?Nope. Occasionally in yiddish.>Are school maths textbooks normally in Hebrew?Again, yiddish.>-- >Timothy Murphy >tel: +353-86-2336090, +353-1-2842366 === >Slightly OT,>but does anyone publish maths in Hebrew nowadays?> Nope. Occasionally in yiddish.Is that true?I thought Yiddish was deprecated, at least in Israel's early years. >Are school maths textbooks normally in Hebrew?> Again, yiddish.I should of course have said maths textbooks in Israelbut I assume that was understood.-- Timothy Murphy tel: +353-86-2336090, +353-1-2842366 === On Mon, 12 Jan 2004 19:48:52 +0000, Timothy Murphy>Slightly OT,>but does anyone publish maths in Hebrew nowadays?> Nope. Occasionally in yiddish.>Is that true?>I thought Yiddish was deprecated, at least in Israel's early years.Not at all. It is the language of choice in 'Israel'. >Are school maths textbooks normally in Hebrew?> Again, yiddish.>I should of course have said maths textbooks in Israel>but I assume that was understood.Indeed. >-- >Timothy Murphy >tel: +353-86-2336090, +353-1-2842366 <3ffa32dd$8$fuzhry+tra$mr2ice@news.patriot.net> <8765fp8z1v.fsf@becket.becket.net> <3ffc9d5e$9$fuzhry+tra$mr2ice@news.patriot.net> <87lloj9qj9.fsf@becket.becket.net> <87hdz79ppk.fsf@becket.becket.net> <4000aafc$21$fuzhry+tra$mr2ice@news.patriot.net> <87llofyvmu.fsf@becket.becket.net>X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === In <87llofyvmu.fsf@becket.becket.net>, on 01/10/2004 at 08:08 PM, tb+usenet@becket.net (Thomas Bushnell, BSG) said:>Um, no. The Greeks didn't have anything like decimal expansions.is irrational so that a modern reader would understand it. I couldhave written 2^(1/2) is irrational and it would have been just asalien to the ancient Greek perspective for other reasons.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === > In <87llofyvmu.fsf@becket.becket.net>, on 01/10/2004> at 08:08 PM, tb+usenet@becket.net (Thomas Bushnell, BSG) said:>Um, no. The Greeks didn't have anything like decimal expansions. is irrational so that a modern reader would understand it. I could> have written 2^(1/2) is irrational and it would have been just as> alien to the ancient Greek perspective for other reasons.Yes, the point isn't how you write one-half. What does infinityhave to do with irrationality? There are only three connections I canthink of:* the rigorous sense in which irrational numbers have infinite decimal expansions. I thought this was what you meant, which was why I said that, ignorant of decimal expansions, the Greeks wouldn't have thought this was any kind of connection; or* the vague sense in which an irrational ratio is measured by a zero-length commensurate unit repeated infinitely many times.* the rigorous analogue of the preceding where we speak of the limit of an approximate ratio as the commensurate unit approaches zero (and the number of units approaches infinity). This of course was also foreign to ancient mathematicians.Perhaps I have missed one; I was trying to understand what you meantfrom a very brief statement. What connection between irrationalnumbers and infinity do you think the Greeks might have noticed?Thomas <3ffa32dd$8$fuzhry+tra$mr2ice@news.patriot.net> <8765fp8z1v.fsf@becket.becket.net> <3ffc9d5e$9$fuzhry+tra$mr2ice@news.patriot.net> <87lloj9qj9.fsf@becket.becket.net> <4000a9c3$20$fuzhry+tra$mr2ice@news.patriot.net> <87hdz3yvik.fsf@becket.becket.net>X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === In <87hdz3yvik.fsf@becket.becket.net>, on 01/10/2004 at 08:11 PM, tb+usenet@becket.net (Thomas Bushnell, BSG) said:>I'm not sure how you find it to be vague to say without limits at>the same time as you assert that anything can be limited. Then read it again.>So, Aristotle said something incorrect in one area, he must be crazy>about math? So you're still cheating on your income tax?I gave an example. write what you really meant.>I gave several examples:No: you slung words around in a fashion that had no semantic content.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === > write what you really meant.So the concept of infinite physical extension has no meaning? I seephysicists use the the concept all the time; you think they aretalking nonsense? Or you didn't bother to read that part of my post?It's very easy to say what you write has no meaning, especially ifyou simply skip huge chunks of text and offer no actual criticism.I fear you have the disease that everything outside your own area ofexpertise is regarded as either trivial or nonsense.(Your insistence on the details of Hebrew spelling, on another thread,makes your casually incorrect Latin [and apparent misunderstanding ofwhat de gustibus non disputandum est really means] all the morelaughable.)For extra credit, you can say what QED stands for, and means.Thomas === > this function f brings sex in the city.> How can I prove now |p(z)|<=M|z|^n?> Could you explain little more? Do you know the maximum modulus theorem? === > Why do you always take out the things out of context? As G.9adel and > Cohen established, there are undecidable statements. Concluding > whether the reals are countable or not it probably is one of them.> By the way, someone asked what the heck G.9adel has to do with this. > Here you can find the answer. I hope you don't have to regret it> It might be undecidable to you, but it is decidable, and has been decided, by anyone who really knows what is going on. Cantor created a definition of countability of a set, and proved, despite your inability deal with his proofs, that the set of reals does not satisfy that definition. Therefore, the set of reals is not a countable set. Your inability to grasp this must be attributed to either ignorance or stupidity or some combination of the two. === >On Wed, 7 Jan 2004 17:50:17 +0000 (UTC), nico80@jazzfree.com (Niclas>I think we have to center the tread, because at this moment it looks >like the chats in the auditorium before the concert.>POINTS>Point 1:>The surjection f: N -> R is not possible because the definition of >naturals and reals do not allow it.>This needs proof. Although it can easily be proved, so in some>sense it's true. In exactly the same way it is true to say that>sqrt(2) is irrational because the defintion of the rationals>do not allow it to be rational. So if Point 1 makes the standard>proof of the uncountability of R invalid it also makes the standard>proof of the irrationality of sqrt(2) invalid. (And it makes _any_>proof by contradiction invalid.)> If sqrt(2) cannot be rational by definition, what else it can be? > Irrational. Good. But how do you determine whether sqrt(2), or any other real, is rational or not? By declaration, as you seem to want, or by testing whether it satisfies the definition?Mathematicians universally vote for testing whether it satisfies the definition. You apparently think you have a more reliable pipeline to the truth that avoids proofs. > And now> If N <-> R cannot be by definition, what else it can be? NOTHING. Go > with an optionless premise to a proof by contradiction and youll > get NOTHING (or rubbish).If I have a new set S, the only way to test it for countablility is to test whether it satisfies the definition.If you think you can do better, tell me whether the set S I am thinking of is countable or not, then I will tell you what set I am thinking of. === >I think we have to center the tread, because at this moment it looks >like the chats in the auditorium before the concert.Well, you postings are certainly out of tune, but no one else's seem to be.>POINTS>Point 1:>The surjection f: N -> R is not possible because the definition of >naturals and reals do not allow it.>So what is it about the definitions of the naturals and the reals that>allows you to confidently state that such a surjection is not possible?> This question has been answered in my latest version.It has not been answered to the satisfaction of anyone but the Foz.>Obviously we cannot use the definitions of the naturals and the rationals >to prove that there is no surjection f : N -> Q, since we know that there>is such a surjection. This means that you cannot use the argument thatr>N is a proper subset of R (N is a proper subset of Q and there exists a >surjection f : N -> Q).> There is a thread dedicated to this stuff and, as you can see, it is > not so clear that Q is countable. I also have some arguments against > this proof, but if you don't mind I will leave untreated the theme at > moment.It is clear that there are injections from N to Q, f(n) = n/1 is one such, so that |N| <= |Q|.It is clear that there are injections from Q to N, for example,g(p/q) = 2^|p|*3^q, for p < 0, or g(p/q) = 5^p*7^q, for p >= 0,where p/q is the unique standard form (lowest terms) for each rational,so that |Q| <= |N|.But this clearly proves |N| = |Q|, contrary to your assertion that the matter is not clear. >Point 2:>Cantor clearly stated that we can say that an infinite set K is >countable if and only if it is possible to establish a one-to-one >correspondence between K and N, i.e. f: N <-> K.>Point 3:>As the one-to-one correspondence N <-> R is not possible (Point 1), >Your claim is to the effect that it is trivial to demonstrate that>such a one-to-one correspondence does not exist. Let's see your>trivial proof.> As probably you have already seen, it is a really trivial proof. > You can find a less annoying version of the whole proof in the thread > *A mathematical proof of the Cantors goof?* There you dont need to > play the role of Mr. Na.95ve if you want to refute it.Your proofs are is trivial that they are vaporware. You have not shown nor referenced a valid proof of your naive claims, but merely reiterated them without proof or valid support.>the criterion settled by Cantor (Point 2) is not valid to resolve >whether R is countable or not.Which one of his several proofs is invalid? All you have extablished so far is that you don't like Cantor, not that there is any logical fault in his work.>Point 1 means that R is not countable, so it does provide an answer,>except that you have not deemed it necessary to provide the trivial>proof that makes the non-existence of a surjection trivially obvious >to you, but not to others. Point 2 is valid. It is the DEFINITION>of countability of a set. If you can show that R does not satisfy >Cantor's criterion for countability as espoused in Point 2, then R>is not countable. What is so difficult about this point?> Perhaps the analogy in the latest version it will help you to grasp > the idea I want to transmit.Is there an idea in all those words? If so I missed it along with everyone else.In other words, we cannot use this >criterion with R. (See note below)>Point 1 implies that R is uncountable. What more do you want?> Once again, the analogy may help you to see the difference.Analogy is not proof. Give us proof or get the to a nunnery.>Point 4: >If a proof (any) comes to the conclusion that R is not countable >because it is not possible to accomplish the criterion established in >Point 2,>In other words, we conclude that R is uncountable because it fails >the criterion for countability.> YESExactly how does it fail to do so? Vague claims without details are not mathematically acceptable. The Cantor definition of countable for Q wouold require the existence of a surjectiive mapping from N to Q. Thus the uncountability must require proof that no such injective map exists.But you refuse to allow that existence/non-existence to be put to the test. Why?>then the proof is useless because that criterion is not >valid for that purpose (Point 3).>Yes, it is. It is the DEFINITION of countability of a set. It lays>down the one and only criterion for countability. R fails that >criterion, so R does not satisfy the definition for countability,>so R is not countable.> NO. If for you *the money only can be counted if and only is it is > counted by hand* is a definition, for me is a CRITERION that > obviously fails in some cases.There is a difference. The only way of _counting_ the umber of members in a set is to create injective and/or surjective mappings with other sets of know counts. The set, N, of naturals and certain of its finite subsets are our standard sets, at least up to and including N itself.If you e forbid us from considering mappings between a given set and N or any of N's subsets, how do you propose that we count anything?>Note that the proof could be >correct and the conclusion too.>DEFINITION>Useless proof: A correct proof that proves nothing.>You have not demonstrated why we should accept that there is a >trivial proof of the nonexistence of a surjection f : N -> R.> Now I didNot so. Word count high, content zero.>QUESTION>Does somebody know any other criteria (different from the one >established in Point 2) to resolve whether R is countable or not?>It is Cantor's definition for countability. This means that any>infinite set which satisfies the criterion is countable, and any>infinite set which does not satisfy the criterion is uncountable.> NO. It is Cantors criterion. If there are not other criteria, then > this is another problem.That may be a problem for you, But not for us, but then you here show that you have a talent for making problems where none need exist. If >you know one of them, I would like to know it please (but first be >sure that it doesn't involve implicitly the criterion of Point 2).>NOTE: >Someone could interpret that if N <-> R is not possible, then R is >uncountable. This conclusion is true, but in this case, it will imply >that R is uncountable because we haven't got a suitable tool to count >its elements, since the properties of N and R are incompatible. It >will never imply that R has more elements than N.Cantor denotes the cardinality of set S to be a |S|, and defines an order on cardinals by |A| <= |B| if and only if there is an injective mapping f:A -> B. He then goes on to say that |A| = |B| if both |A| <= |B| and |B| <= |A|, and that |A| < |B| if both |A| <= |B and not |A| = |B|. Finally, he establishes that this is a total order relation on sets, at least in what we might nowadays call the category of set and functions.>We know that R has more elements than N since there exists an injection>f : N -> R, but there is no surjection f : N -> R. That is ALL that>you need.> Probably you are already thinking that this not as obvious as it > looks like.What may not be obvious to you, need not be obscure to anyone else.> Nicolas de la Foz>David McAnally> Despite anything you may have heard to the contrary,> the rain in Spain stays almost invariably in the hills.> === > Let p and q be distinct primes. Suppose that H is a proper subset of the> integers and H is a group under addition that contains exactly 3 elements of> the set {p, p+q, pq, p^q, q^p}. Determine which fothe following are the 3> elements in H.> a) pq, p^q, q^p> b) p+q, pq, p^q> c) p, p+q, pq> d) p, p^q, q^p> e)p, pq, p^q> The answer turns out to be e) but it took me forever to figure this out. > Since> it was on a GRE math practice exam, it must be fair game for a regular exam. > My question is this: Is there any fast way to know that the answer is e) as> opposed to the others? I would have had to skip the question on a real exam> because it took me far too long.Every additive subgroup of the integers is a principal ideal, consisting of all integer multiples of a single fixed value. If that value is p, then p*q abd p^q are such multiples of p, but q^p and p + q are not. === >Let H(n) = sum{k=1 to n} 1/k, >the n_th harmonic number.>Prove, for each m = positive integer,>sum{k=1 to m-1} H(k) k! (m-k)!>is congruent to>(1/2) m! H(floor(m/2))) (-1)^m >(mod (m+1)) .>(I do not know if the result is trivial {or am certain it is>correct}.)> We can rewrite the above so it becomes, perhaps to some reading this,> more simply stated and more natural.> (Take sum from 1 to m, instead of from 1 to m-1)> sum{k=1 to m} H(k) k! (m-k)! is congruent to> m! *h(m) (mod(m+1)),> where h(m) = sum{k=0 to floor((m-1)/2)} 1/(m -2k) .> Now, h(m) => sum{k=0 to floor((m-1)/2)} 1/(m -2k)> is interesting in itself, for it is (analogously to> double-factorials) the sum of the reciprocal of EVERY-OTHER positive> integer <=m> (even or odd, depending on m).> Neither sequence:> m!*h(m) > nor> m!!*h(m)> nor> sum{k=1 to m} H(k) k! (m-k)!> seems to be in the EIS either, if I calculated correctly by hand.> (EIS: http://www.research.att.com/~njas/sequences/index.html#L)By the way, the sequence of {m! h(m) (mod {m+1})} -> 1, 1 ,0, 3, 4, 2,.. (?)does not seem to be in the EIS either.Is there a closed form for this sequence? ...(...which also equals{sum{k=1 to m} H(k) k! (m-k)! (mod {m+1})})> Leroy> Quet>By the way, for those interested in harmonic-number math puzzles>(limits, this time, instead of congruences), here is a link to a>recent sci.math math-puzzle no one has yet answered.>(I am including the link because I am cross-posting this to>rec.puzzles, as well as to sci.math.)>http://groups.google.com/groups?dq=&start=200&hl=en &lr=&ie=UTF-8&group=sci.math&safe=off&selm=b4be2fdf .0401041840.3c577d6c%40posting.google.com>thanks,>Leroy Quet === Being tired of lurking, on Sun, 11 Jan 2004 07:11:23 -0600, StarshineMoonbeam posted:> Buchanan(?????????@?????????.???) dropped a +5 bundle of words...> Dr. Flonkenstein, let's talk. I've been concerned about you ever since> you flung yourself headfirst into alt.usenet.kooks sometime around Sun,>Being tired of lurking, on Sun, 11 Jan 2004 07:04:01 +0000, Robert>Buchanan posted:>Dr. Flonkenstein, let's talk. I've been concerned about you ever>since you flung yourself headfirst into alt.usenet.kooks sometime>Being tired of lurking, on Sat, 10 Jan 2004 18:32:36 +0000, Robert>Buchanan posted:>, let's talk. I've been concerned about you ever since you flung>yourself headfirst into alt.usenet.kooks sometime around Sat, 10>in alt.usenet.kooks:> in alt.usenet.kooks:> Now that you've followed Flonky and me to AUK, have you met>Twonky yet? You both have a somewhat>frothy-yet-incomprehensible style, y'know.>i don't follow nobody. this is rec.arts.poems.>Where was rec.arts.poems in your crosspost between AUK,> alt.flame and AAVFFF a couple of days ago, ying-a-ling?>i don't follow nobody. this is internet. when i see fcukt posts>i comment.>I see. The fact that your first post outside the poetry groups>was a reply to me is a *coincidence*. That's *very* believable.> there is no coincidence. the reason coins are coined because the> queens are vain. for the same townken when i see your name, my> brain registered a possibility of vanity. thus i came i post and> yabounce.40 po stin> g.google.com>you people keep pour in because of the traffic. even the>aliens can see the bits going back and forth in copper wires.>i try to give you different names at least in the same post,>peter jolly sandwich.>Why are you so reluctant to talk about the aliens? oh i am not. i always focus on my point. oh well, let's talk about> the aliens.> the aliens are mathmatical probabilities. people say, if we are> this much smarter than other creatures, there must be creatures> much smarter than us. so some of those belief had 'transcended' to> worldly forms, such as kings, queens, vassals, dukes and peons who> believed them. of course, with help of weapons and crook warlords,> who are later renamed 'joint chiefs of staff'. delutional idealogies wouldn't had survived rational exercises of> history and time, if not for the re-enforcement of this and that> emergency over-management. in the end, the myth won, truth fail> because voters all want to be dukes. only they don't understand> that was only a sink.> just like you can't have more> doctors+nurses+adjusters+legislator>total patients, too many> managers of truth and reconciliation's will branch off minor truth> that eventually combined adjust the trunk truth. in doubt of> democracy, these alien wannabes figured if we create something> cool to think about, more people will come to party with us, thus> we could charge cover, or sell chapbooks.> life is so boring. people want to be excited by truth or fictions> alike, whatever turns them on. so the aliens came. brought some> spices from out-mongolia. got humans hooked then slink the price> tag, which treatens the greenspan economy. how would he image the> inflation by design that brings total amount of global account to> four folds in every thirty years will be offset by some alien> afghanistan chronical or comlumbia silver mountains in the> imagination.> i am sorry, yarn, after that, life gets even more boring.> day-in-day-out you spice the bland orbitals with AI, or narcotic> cosmology. so finally people learned if aliens were going to tell> us something, they wouldn't channel through the democrazy.> personally i think teh really smart aliens are way too far out> there near where happiness is at. and we had mostly integrated> dumb aliens that were near by, thus there is this mathmatical> improbability zone of aliens visiting us for the sake of proving> to us of their existance. what do you drink?>WOW. He's got my vote.>Didn't I told you he was a kook?>Sure, but I didn't expect this Wollmann-style screed. What a nutbar.>Honestly, it's the first time I encounter this kind of Eddie dribble>in his postings. Normally he just makes a fool out of himself with>typical childish oneliners.> I want more of it!>BTW: I liked delutional idealogies most of all!> My favorite part was the mathmatical probability aliens.> What were the odds that they existed?1 out of the number of years the Goldberg conjecture will be proved.-- mhm 27x12smeeter #28Usenet Valhalla Circle #19 & #21CEO Alcatroll Labs Inc. === >I have had many students over the years who have misplaced the >instruction books for their calculators. A relatively recent solution >to their dilemma is to visit the website of the calculator manufacturer. > On many such sitets, it is possible to download an instruction book, >probably as a pdf file, for any - or at least most - models of calculator.>I am not familiar with Casio, so I can't give exact details on it.>Bob Lindsayhttp://world.casio.com/edu/download/en/manual/ JimX-Received: (from approve@localhost) === Going from a solid to a liquid is 'melting'; from a liquid to a solid is 'freezing'; from a liquid to a gas is 'evaporating'; from a gas to a liquid is 'condensing'; from a solid to a gas is 'subliming'. What does one call the transition from a gas to a solid? === |Going from a solid to a liquid is 'melting'; from a liquid to a solid is|'freezing'; from a liquid to a gas is 'evaporating'; from a gas to a liquid|is 'condensing'; from a solid to a gas is 'subliming'.| What does one call the transition from a gas to a solid? It's also called subliming.By the way, this is a somewhat peculiar question to post tosci.math-- why not sci.physics or sci.chem?Keith Ramsay === > Going from a solid to a liquid is 'melting'; from a liquid to a solid is > 'freezing'; from a liquid to a gas is 'evaporating'; from a gas to a liquid > is 'condensing'; from a solid to a gas is 'subliming'.> What does one call the transition from a gas to a solid? > Frost?X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0C1kI105105; === >Don't you mean (2^k)^k = 2^2k, assuming that ^ means exponentation?(2^k)^k = 2^(k*k) = 2^k for k infinite cardinalX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0C1kIq05101; === Let x,y be elements of a group G. Prove that if xy=yx then rank(xy)=LCM(rank(x),rank(y)).I know how to do it when LCD(rank(x),rank(y))=1, but have no idea how to expand the proof to be good for the more general case. === > Let x,y be elements of a group G. Prove that if xy=yx then rank(xy)=LCM(rank(x),rank(y)).> I know how to do it when LCD(rank(x),rank(y))=1, but have no idea how to expand the proof to be good for the more general case.On the assumption that rank means order (the order of an element gis represented as |g|) then the proof is as follows:Let n be any positive number, then, since xy=yx, (xy)^n = x^n y^n Now, the order of a group element g is the smallest integer m suchthat g^m = e, where e is the identity of the group. But for any n x^n y^n = eonly if n divides both |x| and |y|. Thus, the |xy| is the smallestnumber that both |x| and |y| divides, which is the LCM(|x|,|y|).So,|xy| = LCM(|x|,|y|). The rest is obvious if you can see that, lettingL = LCM(|x|,|y|), x^L = y^L = e.Patrick === Should we throw this in the FAQ? Seems to come up a lot...> Let x,y be elements of a group G. Prove that if xy=yx then rank(xy)=LCM(rank(x),rank(y)).This is false.> I know how to do it when LCD(rank(x),rank(y))=1, but have no idea how to expand the proof to be good for the more general case.>On the assumption that rank means order (the order of an element g>is represented as |g|) then the proof is as follows:>Let n be any positive number, then, since xy=yx, (xy)^n = x^n y^n >Now, the order of a group element g is the smallest integer m such>that g^m = e, where e is the identity of the group. But for any n x^n y^n = e>only if n divides both |x| and |y|. Thus, the |xy| is the smallest>number that both |x| and |y| divides, which is the LCM(|x|,|y|).>So,>|xy| = LCM(|x|,|y|). The rest is obvious if you can see that, letting>L = LCM(|x|,|y|), x^L = y^L = e.Alas, no. You have shown that (xy)^L = e, but you have not shown thatL is the smallest integer that will work. In fact, that statement isfalse in general: if, say, x=y^{-1}, then xy= e, so |xy| = 1, butL=|x|=|y|. It is true that if xy=yx, and both x and y have finite order, then Gmust contain an element of order LCM(|x|,|y|), and this element is theproduct of a power of x by a power of y, but it need not be equal toxy. Seehttp://groups.google.com/groups?selm=bstfds%242e63%241% 40agate.berkeley.edufor one way to answer that question.-- === ============================================ === ===It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) === ======================================= === =========Arturo Magidinmagidin@math.berkeley.edu===>Let x,y be elements of a group G. Prove that if xy=yx then rank(xy)=LCM(rank(x),rank(y)).>I know how to do it when LCD(rank(x),rank(y))=1, but have no idea how to expand the proof to be good for the more general case.Could you please define rank(x).Derek Holt. === >Let x,y be elements of a group G. Prove that if xy=yx then>rank(xy)=LCM(rank(x),rank(y)).>I know how to do it when LCD(rank(x),rank(y))=1, but have no idea how to>expand the proof to be good for the more general case.What happens if xy=yx=e, i.e. if y = x^{-1}?David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills. === > 1> 1 1 1> 1 2 3 2 1> 1 3 6 7 6 3 1> , where each element of row j is the sum of the three elements above it in> row j-1 (null entries = 0).> I want to show that the sum of each row is 3 ^ (i - 1), where i = 1 for> the top row in the picture.Consider a generic row i: A B C D ...The next row, i+1 will be: A A+B A+B+C B+C+D C+D+E ...Note that each element of row i contribues to 3 elements of i+1. So, whenyou add up the elements of row i+1, you will have 3 A's, 3 B's, and so on,so the sum is: 3A+3B+3C+...which is just 3 times the sum of row i. Since the first row sums to 3^0,the successive row sums are 3^1, 3^2, etc.-- --Tim Smith === > I'm having some problems trying to come up with an induction proof for the>following triangle (not the Pascal triangle):> 1> 1 1 1> 1 2 3 2 1>1 3 6 7 6 3 1>, where each element of row j is the sum of the three elements above it in>row j-1 (null entries = 0).> I want to show that the sum of each row is 3 ^ (i - 1), where i = 1 for the>top row in the picture.> I'm fairly new with induction and quite clumsy. It seems to me that I>would want to show that each each element is the sum of three elements in>the preceding row first.??? But that's how you _defined_ your triangle. There's nothing to prove.> But, my attempts get sloppy quickly when trying to>explain what a null entry is in this 'picture of numbers'.So put in the zeros explicitly.... 0 0 0 1 0 0 0 ...... 0 0 1 1 1 0 0 ...... 0 1 2 3 2 1 0 ...etcSince each entry in row i contributes to three entries in row i+1,the sum of the entries in row i+1 (which is finite since there areonly finitely many nonzero entries) is 3 times the sum of the entriesin row i.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Where can I find a readable account (not some differential form formulationon complex varieties full of theorems without use) of how to tackle 2Dvector calculus, especially so as to integrate in the complex plane, namely,integrate on the plane (not on a contour as one usually find it done when itcomes to integration in C) for a function f(z,z*) stuff of the like(@/@z + @/@z*)f(z,z*)dz dz*@^2/@z@z* f(z,z*) dz dz*(2D, or complex, divergence and laplacian)I am turning everything to real variables,@/@z -> (1/2)(@/@x -i @/@y), dz dz* -> 2dx dybut I feel it lacks the power of complex tools.--(with @ partial derivative, z* complex conjugate of z, and so on) === [deleted]>Could you send me that program ASAP?What's wrong with this picture? I see that today the math forum has dumped another batch of queued,outdated posts to sci.math, including some _very_ outdated posts. === > I am working on the program where is necessary to calculate huge> amount of numbers in always the same sequence, before the program> actually opens.> In other words, these number have to be calculated and saved into> memory before the program starts. I was trying to use number pi, but to calculate 100 million decimal> places of number pi would take forever and no one would want to wait> that long. On the other hand I can't afford to include the whole 100> million dec. places along with the software, that would take too much> of the space and too much time to download. > Is there any other number like PI, which I could calculate faster?> Serge> Since you are looking for a quickly calculated sequence ofpsuedo-random integers, perhaps you could use something which, oneach iteration, simply applies an appropriate transform to all of theearlier-calculated terms so as to get the new terms, doubling thenumber of terms on each iteration.(But each iteration does not take a constant amount of time tocalculate, of course.) For example, the m_th string A(m) isA(m-1), A(m-1)*(a(m)+(-1)^a(m)),where a(m) is the m_th term of the sequence (of A(m-1)).I have no idea if this example even looks semi-random to human eyes,and it certainly is not unpredictable under most circumstances, butmaybe something along this line would still be of use to you.thanks,Leroy Quet === Let a(m,n) be a doubly-indexed sequence of reals in the interval (0,1) with the following properties:- For each fixed m, the sequence a(m,n) is strictly decreasing in n and approaches 0 as n approaches infinity; and- for each fixed n, the sequence a(m,n) is strictly increasing in m and approaches 1 as m approaches infinty.Examples: b^(n/m), (1-b^m)^n for fixed b in (0,1). This question is inspired by the thread on powers of 0.999...Given x in (0,1), do there necessarily exist increasing sequences b and c of naturals such that a(b_k, c_k) approaches x as k approaches infinity?Some sufficient conditions for this to be true:- For all x in (0,1) and positive integers j and k, there exist m and n such that min(m,n) >= j and |a(m,n) - x | < 1/k.- lim _m max _n a(m,n) - a(m, n+1) = 0- lim_n max_m a(m+1, n) - a(m,n) = 0.Thus, the conclusion is true for the two given examples.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === |Let a(m,n) be a doubly-indexed sequence of reals in the interval |(0,1) with the following properties:|- For each fixed m, the sequence a(m,n) is strictly decreasing in |n and approaches 0 as n approaches infinity; and|- for each fixed n, the sequence a(m,n) is strictly increasing in |m and approaches 1 as m approaches infinty.[...]|Given x in (0,1), do there necessarily exist increasing sequences b |and c of naturals such that a(b_k, c_k) approaches x as k |approaches infinity?No. Define f(t) to be t/2 for 0 Let a(m,n) be a doubly-indexed sequence of reals in the interval > (0,1) with the following properties:> - For each fixed m, the sequence a(m,n) is strictly decreasing in > n and approaches 0 as n approaches infinity; and> - for each fixed n, the sequence a(m,n) is strictly increasing in > m and approaches 1 as m approaches infinty.> Examples: b^(n/m), (1-b^m)^n for fixed b in (0,1). This question is > inspired by the thread on powers of 0.999...> Given x in (0,1), do there necessarily exist increasing sequences b > and c of naturals such that a(b_k, c_k) approaches x as k > approaches infinity?Yes. (1) While term <= x, add 1 to the increasing index. (2) While term >= x, add 1 to the decreasing index.(3) Goto (1).This will converge to any x such that 0 < x < 1.This is similar to showing that a sequence that is conditionally , but not absolutely, convergent can be rearranged to converge to any real value whatsoever. === Let a(m,n) be a doubly-indexed sequence of reals in the interval > (0,1) with the following properties:> - For each fixed m, the sequence a(m,n) is strictly decreasing in > n and approaches 0 as n approaches infinity; and> - for each fixed n, the sequence a(m,n) is strictly increasing in > m and approaches 1 as m approaches infinty.> Examples: b^(n/m), (1-b^m)^n for fixed b in (0,1). This question is > inspired by the thread on powers of 0.999...> Given x in (0,1), do there necessarily exist increasing sequences b > and c of naturals such that a(b_k, c_k) approaches x as k > approaches infinity?>Yes.No.>(1) While term <= x, add 1 to the increasing index. >(2) While term >= x, add 1 to the decreasing index.>(3) Goto (1).>This will converge to any x such that 0 < x < 1.Nope.>This is similar to showing that a sequence that is conditionally , but >not absolutely, convergent can be rearranged to converge to any real >value whatsoever.************************David C. Ullrich === > Let a(m,n) be a doubly-indexed sequence of reals in the interval > (0,1) with the following properties:> - For each fixed m, the sequence a(m,n) is strictly decreasing in > n and approaches 0 as n approaches infinity; and> - for each fixed n, the sequence a(m,n) is strictly increasing in > m and approaches 1 as m approaches infinty.> Examples: b^(n/m), (1-b^m)^n for fixed b in (0,1). This question is > inspired by the thread on powers of 0.999...> Given x in (0,1), do there necessarily exist increasing sequences b > and c of naturals such that a(b_k, c_k) approaches x as k > approaches infinity?No, because you can define an appropriate sequence of functions f_n, each with a fixed gap in its range. For example, let g(x) = x/[3(x+1)] for x in [0,oo). For n = 1, 2, ... define f_n(x) = x/3n, 0 <= x <= n, f_n(x) = 2/3 + g(x-n), x > n. Then each f_n is strictly increasing to 1 at infinity, and the f_n's are pointwise strictly decreasing to 0. The range of each f_n omits (1/3,2/3). So a(m,n) = f_n(m) is an example. === >Let a(m,n) be a doubly-indexed sequence of reals in the interval >(0,1) with the following properties:>- For each fixed m, the sequence a(m,n) is strictly decreasing in >n and approaches 0 as n approaches infinity; and>- for each fixed n, the sequence a(m,n) is strictly increasing in >m and approaches 1 as m approaches infinty.>Examples: b^(n/m), (1-b^m)^n for fixed b in (0,1). This question is >inspired by the thread on powers of 0.999...>Given x in (0,1), do there necessarily exist increasing sequences b >and c of naturals such that a(b_k, c_k) approaches x as k >approaches infinity?> No, because you can define an appropriate sequence of functions f_n, each > with a fixed gap in its range. For example, let g(x) = x/[3(x+1)] for x in > [0,oo). For n = 1, 2, ... define f_n(x) = x/3n, 0 <= x <= n, f_n(x) = 2/3 > + g(x-n), x > n. Then each f_n is strictly increasing to 1 at infinity, and > the f_n's are pointwise strictly decreasing to 0. The range of each f_n > omits (1/3,2/3). So a(m,n) = f_n(m) is an example.Easier example: Let f be any strictly increasing function on the integers with f(n) -> 0 as n -> -oo and f(n) -> 1 as n -> +oo. Set a(m,n) = f(m-n). Then a(m,n) satisfies the decreasing/increasing criteria. Because the range of f is a discrete subset D of (0,1), the only x's in (0,1) that are limits in the sense specified lie in D. (Furthermore, any discrete D in (0,1) that accumulates at both 0 and 1 will be the range of some f as above.) === On Mon, 12 Jan 2004 00:20:05 -0800, The World Wide Wade> Let a(m,n) be a doubly-indexed sequence of reals in the interval >(0,1) with the following properties:>- For each fixed m, the sequence a(m,n) is strictly decreasing in >n and approaches 0 as n approaches infinity; and>- for each fixed n, the sequence a(m,n) is strictly increasing in >m and approaches 1 as m approaches infinty.>Examples: b^(n/m), (1-b^m)^n for fixed b in (0,1). This question is >inspired by the thread on powers of 0.999...>Given x in (0,1), do there necessarily exist increasing sequences b >and c of naturals such that a(b_k, c_k) approaches x as k >approaches infinity?> No, because you can define an appropriate sequence of functions f_n, each > with a fixed gap in its range. For example, let g(x) = x/[3(x+1)] for x in > [0,oo). For n = 1, 2, ... define f_n(x) = x/3n, 0 <= x <= n, f_n(x) = 2/3 > + g(x-n), x > n. Then each f_n is strictly increasing to 1 at infinity, and > the f_n's are pointwise strictly decreasing to 0. The range of each f_n > omits (1/3,2/3). So a(m,n) = f_n(m) is an example.>Easier example: Let f be any strictly increasing function on the integers >with f(n) -> 0 as n -> -oo and f(n) -> 1 as n -> +oo. Set a(m,n) = f(m-n). >Then a(m,n) satisfies the decreasing/increasing criteria. Because the range >of f is a discrete subset D of (0,1), the only x's in (0,1) that are limits >in the sense specified lie in D. (Furthermore, any discrete D in (0,1) that >accumulates at both 0 and 1 will be the range of some f as above.)Even easier example: Let a(m,n) = 0 if n > m, 1 otherwise.************************David C. Ullrich === > Even easier example: Let a(m,n) = 0 if n > m, 1 otherwise.Not so fast. Those numbers are not in (0,1) nor are they strictly decreasing/increasing as the OP intended. === On Mon, 12 Jan 2004 09:53:32 -0800, The World Wide WadeEven easier example: Let a(m,n) = 0 if n > m, 1 otherwise.>Not so fast. Those numbers are not in (0,1) nor are they strictly >decreasing/increasing as the OP intended.Um...************************David C. Ullrich === I would be very much obliged if someone could point me to a proof of thefollowing:If S=f(x,y) is a surface, C is a level curve (horizontal cross section) of S,and P is a point on C, then the directional derivative of f at P is zero if thedirection in question is tangent to C.Muchas gracias.Peace === > If S=f(x,y) is a surface, C is a level curve (horizontal cross section) of S,> and P is a point on C, then the directional derivative of f at P is zero if > the> direction in question is tangent to C.Suppose C is parameterized as C(t). Then f(C(t)) is constant, so df(C(t))/dt = 0. But by the chain rule, df(C(t))/dt = . Therefore the directional derivative of f in the direction of C'(t) is 0, and C'(t) is tangent to C at C(t). === > I would be very much obliged if someone could point me to a proof of the> following:> If S=f(x,y) is a surface, C is a level curve (horizontal cross section) of S,> and P is a point on C, then the directional derivative of f at P is zero if > the> direction in question is tangent to C.> Muchas gracias.> Peace> A casual geometric argument:The directional derivative measures how fast that direction will take one up or down hill on the surface, but moving along a level curve doesn't move either up or down. === > Two days ago I had an interesting discussion with my coworker at the Math> dept of the U of Crete about continuity.The discussion was interesting, because he is a computer scientist, so our> views on continuity had some substancial differences. The main argument is> presented below:1) Mathematician: The brain can perceivecontinuity (perhaps intuitively,> but it can).> 2) Computer Scientist: The brain cannot perceive continuity, cause all> actions in the universe are quantized.As another poster suggested, you should read the latest issue of SciAm.Some quantum loop theories quantize time and space.Plank's time is about 10^(-44) second.This issue also comes up in the geometry newsgroups.If a point on a line has zero extent how can even aninfinite number of them have a non-zero extent?I don't find continuity to be very intuitive.What bothers me the most is that it shouldreally be called discontinuity.Between any two non-equal real numbers there arean infinite number of real numbers.This means no real number has a nearest neighboror even a set of nearest neighbors.There is no way to connect any real numberto any other.For example, let's say I want to represent themotion of the center of gravity of a rabbit.I want to put together a set of all the pointsthe rabbit passes through and I want toorder this set in the same order the rabbitpassed through the points.Standard set theory says this is impossible.There is no way I can order the set to representthe order the rabbit went through each point.I could assume the rabbit moves along somemapping of the rational numbers. That wouldbe weird. The rabbit could be half way throughthe course and then go back almost to thebeginning.I guess this is no stranger than the quantummechanic analysis of the probability thatthe rabbit is at some point on the course.(Of course, there is a non-zero probabilitythe rabbit isn't on the course.)Russell- Zeno was right. Motion is impossible. === > I don't find continuity to be very intuitive.> What bothers me the most is that it should> really be called discontinuity.> Between any two non-equal real numbers there are> an infinite number of real numbers.> This means no real number has a nearest neighbor> or even a set of nearest neighbors.> There is no way to connect any real number> to any other.In terms of the Dedekind construction, every non-empty set of reals that has an upper bound has a least upper bound, but this is not true for the Integers or rationals, or algebraics. If a set of reals has an upper bound then the greatest lower bound of its upper bounds is its own least upper bound, so that there are no holes or discontinuities in the set of reals. === Russell Easterly [snip]As another poster suggested, you should read the latest issue of SciAm.> Some quantum loop theories quantize time and space.> Plank's time is about 10^(-44) second.> Russell> - Zeno was right. Motion is impossible.--Ioannis Galidakishttp://users.forthnet.gr/ath/jgal/------------------- -----------------------Eventually, _everything_ is understandable === I'm trying to make a counter that works in native base three mode./* 000,001,002,010,011,020,021,022,100,101,102,110,111,112,120,121 ,122,200,201,202,210,211,212,220,221,222,2000................. .............*/ And so on without any base conversion, just a plain recursive loopdealing with the integers within the base [0,1,2] and their multiples[10,100,1000,.....]Is such a counter possible to make? Or is it impossible to build arecursive base three counter?JT === In sci.math, JTon 11 Jan 2004 21:33:33 -0800<2ecfd00b.0401112133.21ec1229@posting.google.com>:> I'm trying to make a counter that works in native base three mode.> /* 000,001,002,010,011,020,021,022,100,101,102,110,111,112,120,121 ,122,200,201,202,210,211,212,220,221,222,2000................. .............*/> And so on without any base conversion, just a plain recursive loop> dealing with the integers within the base [0,1,2] and their multiples> [10,100,1000,.....]> Is such a counter possible to make? Or is it impossible to build a> recursive base three counter?> JTConceptually, one could do things like this.digit_array number = 0;for(;;){ int digit_index = 0; while(number[digit_index] == '2') { number[digit_index] = '0'; digit_index ++; } if(number[digit_index] == 'blank') number[digit_index] = '1'; else number[digit_index] ++; println(number);}with the caveat that number[digit_index] refers to thedigit_index'th power of 3 -- in other words, the digitsare stored kinda backwards -- and println() is appropriatelydefined.Obviously this can be extended to any base by replacing the '2'with an appropriate value.This is very close to a C++ implementation, in fact;one need merely typedef digit_array to std::vectorand make some tiny modifications to the 'blank' test andthe initial assignment. Of course the above is not recursive,but one could do something like this:procedure loop_forever(){ for(;;) { countup(number, 0); println(number); }}procedure countup(digit_array number, int index){ if(number[index] == '2') { number[index] = '0'; countup(number, index+1); } else if (number[index] == 'blank') number[index] = '1'; else number[index] ++;}which *is* recursive and appears to satisfy your problem nicely.-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === | I'm trying to make a counter that works in native base three mode.|/* 000,001,002,010,011,020,021,022,100,101,102,110,111,112,120,121 ,122,200,201,202,210,211,212,220,221,222,2000................. .............*/|| And so on without any base conversion, just a plain recursive loop| dealing with the integers within the base [0,1,2] and their multiples| [10,100,1000,.....]|| Is such a counter possible to make? Or is it impossible to build a| recursive base three counter?Here is a very simple one written in REXX:_________________________________________________________ ____________/**/ arg n .; if n=='' then n=1000 do j=0 to n if verify(3456789,j,'match')==0 then say j end___________________________________________________________ __________You could write one in C just as easily (or PL/1), or Fortran. TheVERIFY function looks for a 3,4,5,6,7,8, or 9 digit in J and if thereisn't one, the SAY does a write to the screen. _________________Gerard S. <40014E86.4060308@comcast.net> <20040111063024.S12460@agora.rdrop.com> <4001A856.6040809@comcast.net> === >explain this. I looked at another text and this interesting topic>wasn't even discussed. Maybe I need a better book.>You're welcome. Don't despair quite yet, did he leave such proofs as>exercises for the student in the problem section?Don't see it in the exercises for this and the next section. He may get> to it later. I'll be OK with the generous help I get from this group.> Did the author explain why the d-open balls for metric or pseudo-metric is>a topology?Yup, very interesting, too. > The hard part is to show the intersection of two balls is>open.He does that indirectly by showing every set is T-open iff it is an> element of the topology T. So d-open balls are T-open, and are members> of the topology T; and the intersection of two of them is still a member> of the topology T and therefore open.>Oh sure, the topology generated by the open balls. The real problem is toshow that the topology has a base of open balls. For that you need showthe intersection of two balls is a union of balls. Thus every open set isa union of open balls. === I would like to implement itself since i am going to makemodifications to the counter for working with primes in base three.I've started coding thus coming so far as below but the incrementalcounter won't work. I would very much appreciate if any math inclinedperson could run the code in browser and apply a quick fix. === Suppose all Turing Machines with n+1 states (1 halt state) ,{0,1} astape symbols, and 1^n (11...111) as input string for all of them.f(n):N -> N is the maximum number of 1's on these TMs after halt. Isf(n) Computable? Is it Decidable?Thanx in Advance,Behrang. === > Suppose all Turing Machines with n+1 states (1 halt state) ,{0,1} as> tape symbols, and 1^n (11...111) as input string for all of them.> f(n):N -> N is the maximum number of 1's on these TMs after halt. Is> f(n) Computable? Is it Decidable?It is not computable, because it implies that all Turing machines with n+1states and alphabet {0,1} will halt when given 1^n as input. If for any n,TM_n does not halt, f(n) is undefined. Determining whether an arbitrary TMhalts on a given input is undecidable. Note: even if you take the maximumnumber of 1s outputted by just the halting TMs, it is also undecidable,since you may never know whether a TM will output a number of ones largerthan all other TMs with n+1 states unless it halts. But you can't determineif it is just taking a long time to halt, or if it is looping.If you restrict your set of TMs to deciders, then it is computable. Sincethere are a finite number of n+1 state TMs for every n, you could simply runthem all on 1^n and take the maximum.> Thanx in Advance,> Behrang.l8r, Mike N. Christoff === > I met the following problem i one book,which can be solved either by>measure theory or by Lebesque Dominated Convergence theorem,but the>author says it is too difficult to handle without these means .I think>I need help on it.>Let fn:[0,1]--->[0,1] are continuous functions and fn--->0 for each x>in [0,1].Then Int(fn(x)dx,0,1)--->0!>Define f_n for n >= 2 by 2> f (x) = n x for x in [0,1/n]> n> = n(2 - nx) for x in [1/n,2/n] = 0 for x in [2/n,1]>He did say fn:[0,1]--->[0,1].Oops... my mistake. In that case, the function f(x) = 1 which is mostdefinitely in L^1[0,1] dominates each f_n. Sorry for the interruption.Rob Johnson take out the trash before replying === > I met the following problem i one book,which can be solved either by>measure theory or by Lebesque Dominated Convergence theorem,but the>author says it is too difficult to handle without these means .I think>I need help on it.>Let fn:[0,1]--->[0,1] are continuous functions and fn--->0 for each x>in [0,1].Then Int(fn(x)dx,0,1)--->0!> It is false as stated. See if you can find a counterexample.Perhaps you didn't see the -->[0,1]? This makes the f_n bounded.similar result (f_n continuous, {f_n(x)}_n decreases to 0 for each x==> int_0^1 f_n(x) dx --> 0) using purely elementary arguments. Thatis, non-measure-theoretic arguments.--Ron Bruck === >I met the following problem i one book,which can be solved either by>measure theory or by Lebesque Dominated Convergence theorem,but the>author says it is too difficult to handle without these means .I think>I need help on it.>Let fn:[0,1]--->[0,1] are continuous functions and fn--->0 for each x>in [0,1].Then Int(fn(x)dx,0,1)--->0!>It is false as stated. See if you can find a counterexample.> Perhaps you didn't see the -->[0,1]? This makes the f_n bounded.> similar result (f_n continuous, {f_n(x)}_n decreases to 0 for each x> ==> int_0^1 f_n(x) dx --> 0) using purely elementary arguments. That> is, non-measure-theoretic arguments.> --Ron BruckThis one isn't in fact very hard. It's a fairly easy proof that if f_n are continuous functions which decrease pointwise to zero then they tend uniformly to zero. Therefore the integral tends to zero.Here's a quick proof:Let e > 0. Consider X_n = f_n^{-1}( [e, infinity) ). These are closed subsets of [0, 1] with X_n containing X_m for m > n (as the functions decrease). Suppose they were all non-empty. Then their intersection would be non-empty by compactness. Let x lie in the intersection. For all n f_n(x) >= e, so f_n(x) does not tend to zero. This is a contradiction.Hence there is some n such that X_n is empty. i.e. for all x in [0, 1] |f_n(x)| < e. Because the f_n are decreasing this means for all m >= n we have for all x |f_n(x)| < e. Hence f_n -> 0 uniformly.David === > I met the following problem i one book,which can be solved either by> measure theory or by Lebesque Dominated Convergence theorem,but the> author says it is too difficult to handle without these means .I think> I need help on it.> Let fn:[0,1]--->[0,1] are continuous functions and fn--->0 for each x> in [0,1].Then Int(fn(x)dx,0,1)--->0!>It is false as stated. See if you can find a counterexample.> Perhaps you didn't see the -->[0,1]? This makes the f_n bounded.> similar result (f_n continuous, {f_n(x)}_n decreases to 0 for each x> ==> int_0^1 f_n(x) dx --> 0) using purely elementary arguments. That> is, non-measure-theoretic arguments.> --Ron Bruck-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === > I met the following problem i one book,which can be solved either by> measure theory or by Lebesque Dominated Convergence theorem,but the> author says it is too difficult to handle without these means .I think> I need help on it.> Let fn:[0,1]--->[0,1] are continuous functions and fn--->0 for each x> in [0,1].Then Int(fn(x)dx,0,1)--->0!> MladenThe uniform convergence of (f_n) is the only interesting conclusion with such hypothesis. The result Int(fn(x)dx,0,1)--->0 is obvious if you know that.GS === > I met the following problem i one book,which can be solved either by> measure theory or by Lebesque Dominated Convergence theorem,but the> author says it is too difficult to handle without these means .I think> I need help on it. > Let fn:[0,1]--->[0,1] are continuous functions and fn--->0 for each x> in [0,1].Then Int(fn(x)dx,0,1)--->0! >The uniform convergence of (f_n) is the only interesting conclusion with > such hypothesis. The result Int(fn(x)dx,0,1)--->0 is obvious if you >know that.No, you don't know that: the sequence (f_n) does not necessarily convergeuniformly. Try e.g. f_n(x) = exp(-n^3 (x-1/n)^2).Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === > I met the following problem i one book,which can be solved either by>measure theory or by Lebesque Dominated Convergence theorem,but the>author says it is too difficult to handle without these means .I think>I need help on it.>Let fn:[0,1]--->[0,1] are continuous functions and fn--->0 for each x>in [0,1].Then Int(fn(x)dx,0,1)--->0!>The uniform convergence of (f_n) is the only interesting conclusion with > such hypothesis. The result Int(fn(x)dx,0,1)--->0 is obvious if you >know that.> No, you don't know that: the sequence (f_n) does not necessarily converge> uniformly. Try e.g. f_n(x) = exp(-n^3 (x-1/n)^2).> Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2> OMG of course I made a mistake. It's really obvious it can't work (with your counterexample or any other counterexamples i can construct in my mind). The usual arguments are equicontinuity or one of the two Dini's theroem to imply the uniform convergence. === Clevers Kristof schrieb:> But I was wondering if we know by proof that f.i. the 5X+1 ( or 7X+1, ...)> analogue problem has divergent sequences.> Look at mensanators place for instance of make use of google. He posted(think he will come up soon in this thread...:-) )Gottfried === Hey,thx for your response.What o you mean with mensanators? I can't find any info or google hits onthat subjectthx in advGottfried Helms schreef in bericht> Clevers Kristof schrieb:> But I was wondering if we know by proof that f.i. the 5X+1 ( or 7X+1,...)>analogue problem has divergent sequences.> Look at mensanators place for instance of make use of google. He posted> (think he will come up soon in this thread...:-) )>Gottfried> === > Hey,> thx for your response.> What o you mean with mensanators? I can't find any info or google hits on> that subjectQ. Why is mensanator like the Queen?A. Neither is a subject.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === > The page with links to mathematical departments in the world> http://www.math.hr/~duje/mathdept.html> now contains links to 1003 departments from 93 countries.> Any comments and suggestions are very welcomed.> You seem to have missed Germanys University of Ulm.http://www.mathematik.uni-ulm.deA serious omission, I feel for some reason.HTH,Michael.-- &&&&&&&&&&&&&&&&#@#&&&&&&&&&&&&&&&&Dr. Michael UlmFB Mathematik, Universitaet Rostockmichael.ulm@mathematik.uni-rostock.de === > The page with links to mathematical departments in the world> http://www.math.hr/~duje/mathdept.html> now contains links to 1003 departments from 93 countries.> Any comments and suggestions are very welcomed.> Andrej DujellaI haven't looked elsewhere, but the entries for Israel require somerevision. The entry for Ben Gurion refers to the CS, rather than themath, department; Math is at http://www.math.bgu.ac.il/. Also missingNice job!Alan === > Herr Weyl you must explain me one thing-what is> What about Chapman inequality?Is it named after Robin Chapman?Mr Matey, please explain one thing -- what is theChapman inequality ?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === > Just out of curiosity, with all these unproven conjectures about...> what happens when the conjecture is proven to be correct? Does the> newly declared theorem retain the name of the person originating the> conjecture, or will it get named after the person(s) who proves it?> I suppose some conjectures are so famous one would think the theorem> would retain the name of the person originating the conjecture. Say> maybe one day we will come to know of the Poincare Theorem.> It is interesting to note, that historically there wasn't much> consistency in the naming rights of various theorems etc. Fermats> famous Last Theorem, was hardly a theorem at all, but only a> conjecture (until it was proven). The fact it was called a theorem, it> just one of those things that stuck. Now that Wiles had given a proof> of the theorem (conjecture) now surely it is officially a theorem.> This may not be a good example, but it strikes me that the> Fermat-Wiles Theorem would be a more appropriate name. Maybe it is> already called that, that I am not sure of.> In general is there an agreed concensus as to naming rights in> mathematics, and/or is there a body that governs naming rights?Regarding PROVED former hypotheses' names:http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe= off&frame=right&th=c30d7b5734b0ffab&seekm=b4be2fdf .0112081349.42c845f9%40posting.google.com#link1(I suggested replacing hypothesis in a hypothesis' name withHYPERthesis when it is proved true {or when proved false, perhaps aswell}...{suggested this kind of as a joke})Regarding (generally) naming theorems, etc, after mathematicians, Ifind the fact humorous that there are SO MANY mathematical items namedafter Euler.One particular/peculiar example is Euler numbers, which are not tobe confused with Eulerian numbers...As for my own results...I have so many myself, I would not know whichwould best be named Quet's theorem...;)(I guess a good start for determining which theorem of mine should be*the* Quet's theorem would be to name such a theorem of mine which hasa decent amount of originality relatively. Yet I myself am not the oneto determine this...)thanks,Leroy Quet === > Regarding (generally) naming theorems, etc, after mathematicians, I> find the fact humorous that there are SO MANY mathematical items named> after Euler.The most amusing thing is that if it has anything to do withmathematics, Gauss probably invented it, or found proofs for it, onlyhe seldom tells anyone until after someone else makes a similarannouncement. === > What (if at all) the existence of higher ordinal infinities (such as strongly inaccessible cardinals and higher) imply in terms of CH or GCH ?As far as we know: nothing.-- Aatu Koskensilta (aatu.koskensilta@xortec.fi)Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Let (X, d) be Haussdorf compact metric space, T a transformation on Xsuch that family {T, T^2, T^3, ...} is equicontinuous. Suppose T issurjective. Does T need to be homeomorphism? If yes, is it true thatfamily {..., T^-2, T^-1, Id, T, T^2, ...} is equicontinuous?I guess both answers are yes. Please excuse bothering you, couldn'tfind any book on this problem.Mateusz Kwasnicki === I have been very curious about the existence of a way to divide a realnumber set into two sets as follows..Let R be a real number set [0,1].Let A and B be two real number sets which don't have common points with eachother satisifying the followingconditions:1)AUB = R2)When an interval I [p,q] within R where p,q which saitisfy 0<=p I have been very curious about the existence of a way to divide a real> number set into two sets as follows..> Let R be a real number set [0,1].> Let A and B be two real number sets which don't have common points with each> other satisifying the following> conditions:> 1)AUB = R> 2)When an interval I [p,q] within R where p,q which saitisfy 0<=p |q - p| < d, are selected arbitrarily, make it possible so that> we can always find both points with real number concentration , in other> words aleph 1, belonging to A and points belonging to B for any d(delta).> In other words, when a E A and a E I , and b E B and b E I, make it> possible so that there are many a,b with aleph1 concentration within I at> the same time for any d.> a E A means A contains a as an element.> The problem is :> Does such a way to divide R into two sets, A and B exist?> If we can do this, we can make R divide densely into two sets.> For reference,when A is a rational number set and B is a irrational number> set, the condition 2) cannot be satisfied - not with aleph1 concentration> for this case - for rational number set is not with aleph1 concentration..So> the way to divide by rational-irrational cannot be the way to satisfy the> above conditions.> G.C> That such a partition is potentially possible can be shown (see below) although I do not know of any specific construction.The reals, R, form a vector space over the rationals, Q, of uncountable dimension. Let B be a basis of that vector space, i.e., a minimal set of reals such that every real is a finite linear combination of elements of B elements with rational coefficients. B is clearly uncountable. Partition B into C and B-C, with both uncountable, then I belive that, where the span is the set of oall finite linear combinations with rational coefficients, G = span(C) and H = R-span(C) partition R as you desired. === > I have been very curious about the existence of a way to divide a real> number set into two sets as follows..> Let R be a real number set [0,1].> Let A and B be two real number sets which don't have common points with each> other satisifying the following> conditions:> 1)AUB = R> 2)When an interval I [p,q] within R where p,q which saitisfy 0<=p|q - p| < d, are selected arbitrarily, make it possible so that> we can always find both points with real number concentration , in other> words aleph 1, belonging to A and points belonging to B for any d(delta).> In other words, when a E A and a E I , and b E B and b E I, make it> possible so that there are many a,b with aleph1 concentration within I at> the same time for any d.> a E A means A contains a as an element.> The problem is :> Does such a way to divide R into two sets, A and B exist?The cardinality of the real numbers is c = 2^aleph_0, which is known asthe cardinality of the continuum. It is unknown whether c = aleph_1,where aleph_1 is defined as the cardinality of the first uncountableordinal. In fact, this proposition is known as the Continuum Hypothesis,and is known to be independent of the axioms of set theory. If we assumethe axiom of choice, then c >= aleph_1.> If we can do this, we can make R divide densely into two sets.> For reference,when A is a rational number set and B is a irrational number> set, the condition 2) cannot be satisfied - not with aleph1 concentration> for this case - for rational number set is not with aleph1 concentration..So> the way to divide by rational-irrational cannot be the way to satisfy the> above conditions.> G.CI will define A and B in a manner that is slightly reminiscent of theCantor set. Let A consist of all those reals having a base-3representation that uses only finitely many 1's. Let B be the complementof A. Then both A and B are dense in the line, and the respectiveintersections with any nondegenerate interval have cardinality c.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === On Sat, 10 Jan 2004 04:52:00 -0800, zeynep kabuloglu>I'm interested in doing some animations in maya for which time will have>slowed significantly and speeds will be fast enough that I was intrigued at>the possibility of doing modeling over Minkowskian space-times. I have readI only gave a quick glance to the rest of your post, due to timeconstraints, but I remember having seen a POVray patch some time agojust to trace scenes in a (flat) Minkowskian space-time. This commentmay not be relevant, but in case it is you should be able to retrievethe proper link with google and, say, relativistic raytracingpovray.HTH,Michele-- > Comments should say _why_ something is being done.Oh? My comments always say what _really_ should have happened. :)- Tore Aursand on comp.lang.perl.misc === : I only gave a quick glance to the rest of your post, due to time: constraints, but I remember having seen a POVray patch some time ago: just to trace scenes in a (flat) Minkowskian space-time. This comment: may not be relevant, but in case it is you should be able to retrieve: the proper link with google and, say, relativistic raytracing: povray.different renderers to check out, including the one with POVray. All of theeffects I was curious about in the flat case seem to be available, includingdoppler, aberation, Terrel rotation, etc. That certainly will save me somework and I thank you immensely!-- -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-galathaea: prankster, fablist, magician, liar === >The above are merely statements of fact. The clocks are built>to run at 10.22999999543MHz prior to launch so that the signal>received once they are in orbit is measured as 10.23MHz on the>ground receivers.so that the signal received once they are in orbit is measured> as 10.23MHz on the ground receivers.Really??? >;^)The satellite is in a state of continuous motion always.> Why do you have forgotten about Doppler effect?It doesn't work quite like that.The P-code (military) pseudo random pattern is transmittedwith 10.23Mbps, but this bit rate isn't normally measuredfrom the ground.The important point is that the 10.23MHz signal isthe frequency standard of the satellite clocks.That is, the satellite clock counts 10 230 000 cycles foreach second it advances. The clock time is transmitted.(Coded as a bit pattern modulated on two carriers.)We know the frequency must be right since the satelliteclocks stay in synch with the GPS time, not becausethe frequency is measured from the ground.Paul === On Mon, 12 Jan 2004 11:55:20 +0100, Paul B. Andersen The above are merely statements of fact. The clocks are built>to run at 10.22999999543MHz prior to launch so that the signal>received once they are in orbit is measured as 10.23MHz on the>ground receivers.so that the signal received once they are in orbit is measured> as 10.23MHz on the ground receivers.Really??? >;^)The satellite is in a state of continuous motion always.> Why do you have forgotten about Doppler effect?>It doesn't work quite like that.>The P-code (military) pseudo random pattern is transmitted>with 10.23Mbps, but this bit rate isn't normally measured>from the ground.>The important point is that the 10.23MHz signal is>the frequency standard of the satellite clocks.>That is, the satellite clock counts 10 230 000 cycles for>each second it advances. The clock time is transmitted.>(Coded as a bit pattern modulated on two carriers.)>We know the frequency must be right since the satellite>clocks stay in synch with the GPS time, not because>the frequency is measured from the ground.>Paul>It is all explained by the fact that lioght speed is source dependent.Henri Wilson. Einsteinian relativity is the arse end of physics.www.users.bigpond.com/hewn/index.htm === It is all explained by the fact that lioght speed is source dependent.Not according to the Sagnac experiment.George === On Fri, 9 Jan 2004 21:05:16 -0000, George Dishman On 9 Jan 2004 04:31:26 -0800, a_n_timofeev@my-deja.com (Aleksandr>Timofeev)>The velocity of a moving clock>causes it to appear to run slow relative to a clock on the Earth.>GPS satellites revolve around the Earth with an orbital period of>11.967 hours and a velocity of 3.874 km/s. Thus on account of its>velocity, a GPS satellite clock appears to run slow by 7 [Micro]s per day.>...>The difference in>gravitational potential between the altitude of the orbit and the>surface of the Earth causes the satellite clock to appear to run fast.>At an altitude of 20,184 km, the clock appears to run fast by 45 [Micro]s>per day.>The net effect of time dilation and gravitational redshift is that>the satellite clock appears to run fast by approximately 38 [Micro]s per>day when compared to a similar clock at rest on the geoid, including>the effects of the velocity of rotation and the gravitational>potential at the Earth’s surface. This is an enormous rate>difference for a clock with a precision of a few nanoseconds. To>compensate for this large secular effect, the clock is given a>fractional rate offset prior to launch of -4.465 [Times]10^-10 from its>nominal frequency of exactly 10.23 MHz, so that on average it appears>to run at the same rate as a clock on the ground. The actual>frequency of the satellite clock prior to launch is thus>10.229 999 995 43 MHz.> Load of crap.>The above are merely statements of fact. The clocks are built>to run at 10.22999999543MHz prior to launch so that the signal>received once they are in orbit is measured as 10.23MHz on the>ground receivers.If the orbiting clocks appear to run slow it is because they DO run slow.>They run fast, not slow.Yes I know.I said if. It dosn't matter for the purpose of the argument, anyway.The ground observer counts every 'tick' emitted per orbit.>Well a ground observer can only see any satellite for a fraction>of its orbit but if the Earth didn't get in the way, you would>be right.The GPS clock preset is required simply to compensate for an increase in>clock> rates due to being in free fall.>The change of coordinate rate is correlated to the difference in>gravitational potential, not force, but with that clarification>you are correct again.Is it just a matter of definition then? No I don't think so. The clocks arerelieved of their own mechanical strain due to gravitational force.They run slightly faster when in free fall. The fact that the change is of thesame order as the so called GR correction is purely coincidental......unless the 'tick-fairies' have been active again!>If a satellite signal could be received throughout its orbit the>ground observer could count every tick emitted, so there are no>tick fairies.It doesn't have to be seen over the whole orbit. The orbiting clock (OC) knowshow many ticks it emits per orbit. The GO also knows, simply by reading theOC's time. The orbit itself provides a common clock.>Nothing you have said conflicts in any way with what was stated,>so what were you objecting to?The claim that relativity is incorporated into the GPS system. It isn't.>George>Henri Wilson. Any connection between Einsteinian relativity and truth is purely coincidental.www.users.bigpond.com/hewn/index.htm === > On Fri, 9 Jan 2004 21:05:16 -0000, George Dishman On 9 Jan 2004 04:31:26 -0800, a_n_timofeev@my-deja.com (Aleksandr>Timofeev)>The velocity of a moving clock>causes it to appear to run slow relative to a clock on the Earth.>GPS satellites revolve around the Earth with an orbital period of>11.967 hours and a velocity of 3.874 km/s. Thus on account of its>velocity, a GPS satellite clock appears to run slow by 7 [Micro]s per day.>...> The difference in>gravitational potential between the altitude of the orbit and the>surface of the Earth causes the satellite clock to appear to run fast.>At an altitude of 20,184 km, the clock appears to run fast by 45 [Micro]s>per day.>The net effect of time dilation and gravitational redshift is that>the satellite clock appears to run fast by approximately 38 [Micro]s per>day when compared to a similar clock at rest on the geoid, including>the effects of the velocity of rotation and the gravitational>potential at the Earth’s surface. This is an enormous rate>difference for a clock with a precision of a few nanoseconds. To>compensate for this large secular effect, the clock is given a>fractional rate offset prior to launch of -4.465 [Times]10^-10 from its>nominal frequency of exactly 10.23 MHz, so that on average it appears>to run at the same rate as a clock on the ground. The actual>frequency of the satellite clock prior to launch is thus>10.229 999 995 43 MHz.> Load of crap.>The above are merely statements of fact. The clocks are built>to run at 10.22999999543MHz prior to launch so that the signal>received once they are in orbit is measured as 10.23MHz on the>ground receivers.>If the orbiting clocks appear to run slow it is because they DO runslow.>They run fast, not slow.Yes I know.I said if. It dosn't matter for the purpose of the argument, anyway.I assumed it was a typo but sometimes what looks like a typocan be deliberate so I try to just note them in passing justin case.>The ground observer counts every 'tick' emitted per orbit.>Well a ground observer can only see any satellite for a fraction>of its orbit but if the Earth didn't get in the way, you would>be right.>The GPS clock preset is required simply to compensate for an increasein>clock>rates due to being in free fall.>The change of coordinate rate is correlated to the difference in>gravitational potential, not force, but with that clarification>you are correct again.Is it just a matter of definition then? No I don't think so. The clocksare> relieved of their own mechanical strain due to gravitational force.> They run slightly faster when in free fall.I thought this looked more deliberate but it could also havebeen just a slip. Let me put it in Newtonian terms to keep itsimple, gravitational force varies as r^-2 while potentialvaries as r^-1. The change of rate varies as r^-1, not r^-2.> The fact that the change is of the> same order as the so called GR correction is purely coincidental.> .....unless the 'tick-fairies' have been active again!>If a satellite signal could be received throughout its orbit the>ground observer could count every tick emitted, so there are no>tick fairies.It doesn't have to be seen over the whole orbit. The orbiting clock (OC)knows> how many ticks it emits per orbit. The GO also knows, simply by readingthe> OC's time.Sure, but I was agreeing with you anyway.> The orbit itself provides a common clock.Sorry, it doesn't, the duration of the orbit is not the samemeasured on the ground as in orbit.>Nothing you have said conflicts in any way with what was stated,>so what were you objecting to?The claim that relativity is incorporated into the GPS system. It isn't.I assume that was a typo and you meant The claim that acorrection for relativistic effects ..., you don'tincorporate relativity. That is untrue as you can see fromthe quote above:To compensate for this large secular effect, the clock is givena fractional rate offset prior to launch of -4.465 [Times]10^-10 fromits nominal frequency of exactly 10.23 MHz, so that on averageit appears to run at the same rate as a clock on the ground.Regardless of your views, the clocks are built with a correctionterm applied that is not included in ground-based clocks.George === HenriWilson skrev i melding> The fact that the change is of the> same order as the so called GR correction is purely coincidental.Of course a coincidence to a precision of 10^-12 can easily happen.Paul === > 2. Why can symmetry be used in the solving procedure ?>If an equation has a certain symmetry, one can show that the *set* of>its solutions has to have the same symmetry....> Do you know the name of the person who proved that ?No, sorry. That type of proof can be found in many standard book ofQuantum Mechanics, but AFAIK, there is no particular name associatedwith it.I'll give you a simple example here. Look at the one-dimensional,time-independent Schroedinger equation -hbar^2/2m psi''(x) + V(x) psi(x) = E psi(x)and imagine that you've got a potential which is an even function of x:V(-x) = V(x).Now simply change x to -x everywhere in the equation: -hbar^2/2m psi''(x) + V(-x) psi(-x) = E psi(-x)Using the fact that the potential is even, this becomes -hbar^2/2m psi''(x) + V(x) psi(-x) = E psi(-x)Hence obviously psi(-x) is also a solution to this equation. This leavesyou with two possibilities:1) There is no degenaracy, psi(x) is the only solution - then,obviously, psi(-x) has to be identical to psi(x), or, in other words,the only solution function has the same symmetry as the differentialequation itself.2) There is a degeneracy, psi(-x) is different from psi(x). But then atleast the *set* of all solutions has the same symmetry as thedifferential equation itself: choosing an arbitrary solution and turningx into -x gives you another solution.[snip]> Because the symmtry operators and the hamiltonian operator commute> (?????????)Help !>When one can find operators for certain symmetries which commute with>the Hamiltonian, one usually chooses the wave functions to be>eigenfunctions to these operators, too. Again, this is done because it>usually simplifies finding the solutions (example: for solving the H>atom, it is *very* helpful to try to find a wave function which is an>eigenstate for angular momentum, too - because then one already knows>that the wave function has to be a multiple of a spherical harmonic!)> Any references about the theory of the commutation ?I don't think there is an entire theory about commutation; but lots ofinformation about commutation can be found in most books on QuantumMechanics or Linear Algebra.If you have got any specific questions, feel free to ask!> 3. How does variational theory (and what is it ?) come into the> solving theory of Schr.9adinger eqn. ?>AFAIK,> AFAIK ?? You've lost me.Sorry. As Far As I Know. (a standard abbreviation used on usenet -well, at least AFAIK ;-) )>the Schroedinger equation usually can't be solved by the>variational theory - one can only find *approximate* solutions. What it>is? Well, the idea (in the case of the Schroedinger equation) is to use>some test functions and calculate the expectation value for the energy>for them. It is easy to prove that these expectation values have to be>greater or equal to the ground state energy, hence by looking for the>lowest possible energy among all of the test functions, one gets an>upper bound for the ground state energy. (this is only the tip of the>iceberg, but I think you get the general idea...)> Easy introductory text?Not in English, sorry. I've studied physics in Germany and have usedmostly German books...The only English book on QM I can think of in the moment are by Sakuraiand by Dirac - but both are not by any means introductory texts... :-(If you want to see the above mentioned proof that the expectation valuefor the energy is always at least as great as the ground state energy, Ican give it to you - it's very short, requires only a few lines.However, it requires some familarity with the concept of completesystems of functions. But judging from what you say below, you appearto know this concept...[snip]> When you suggest books, consider that I have the education of an> electronics engineer. No fancy mathematics courses like functional> analysis, etc.>But apparently you have some experience with differential equations and>matrices, don't you? What about the general definition of vector>space? What about the notion of differential operators?> Sure Bjoern, I've read some intriductory quantum mechanics books, and> they often have a section of functions whicj are orthogonal to eachOh, nice! So you have some of the basic knowledge for Quantum Mechanicsavailable.> A differntial operator must be that I can write> a diff. eqn. like (d^2/dt^2)y + y=0, like (D^2 +1)*y=0.Right. (although I wouldn't write * in the last equation)> I've calculated these type of equations in the calculus courses. But I> have no clue of the behindlying theory. It is the theory I want to dig> into....So you need information about mathematics, not about physics?Have you seen the post by Herman Jurjus, where he recommended a book?Bye,Bjoern === [snip]>When one can find operators for certain symmetries which commute with>the Hamiltonian, one usually chooses the wave functions to be>eigenfunctions to these operators, too. Again, this is done because it>usually simplifies finding the solutions (example: for solving the H>atom, it is *very* helpful to try to find a wave function which is an>eigenstate for angular momentum, too - because then one already knows>that the wave function has to be a multiple of a spherical harmonic!)Any references about the theory of the commutation ?Not sure if this is what you want to know, here, butperhaps you're interested in the 'Poisson bracket' (see Google).[snip]> I've calculated these type of equations in the calculus courses. But I> have no clue of the behindlying theory. It is the theory I want to dig> into....Then i'd say: Lie groups, Lie algebras and the Peter Olver book.But don't expect easy goings, here. Most of the texts are eitherfocussed on theory, or on applications, without making theconnection between the two explicit.BTW, if you want to dig into theory, functional analysis isquite worthwhile (and doesn't need to be *that* hard).Herman Jurjus === A mathematically-inclined glove-fetishist has a number of pairs of gloves ofidentical design, but of several (at least three) different colors. She hasat least three pairs of each color. In the dark she can distinguish thehandedness of a glove, but not its color. Unfortunately, she keeps thegloves jumbled up in a drawer in an unlit cellar.She knows that if she takes out 21 gloves, in the dark, she can be sure ofgetting at least one pair.What is the maximum number of pairs of gloves that she might have?-- Clive Toothhttp://www.clivetooth.dk === > A mathematically-inclined glove-fetishist has a number of pairs of gloves of> identical design, but of several (at least three) different colors. She has> at least three pairs of each color. In the dark she can distinguish the> handedness of a glove, but not its color. Unfortunately, she keeps the> gloves jumbled up in a drawer in an unlit cellar.> She knows that if she takes out 21 gloves, in the dark, she can be sure of> getting at least one pair.> What is the maximum number of pairs of gloves that she might have?If she has exactly 3 pairs in each of exactly 3 colors, where did she find those last 3 gloves in the dark? === > A mathematically-inclined glove-fetishistWhat a wonderful introduction to a puzzle!-- Richard Heathfield : binary@eton.powernet.co.ukUsenet is a strange place. - Dennis M Ritchie, 29 July 1999.C FAQ: http://www.eskimo.com/~scs/C-faq/top.htmlK&R answers, C books, etc: http://users.powernet.co.uk/eton === > A mathematically-inclined glove-fetishist has a number of pairs of gloves of> identical design, but of several (at least three) different colors. She has> at least three pairs of each color. In the dark she can distinguish the> handedness of a glove, but not its color. Unfortunately, she keeps the> gloves jumbled up in a drawer in an unlit cellar.> She knows that if she takes out 21 gloves, in the dark, she can be sure of> getting at least one pair.> What is the maximum number of pairs of gloves that she might have?(Rot-13):Gjragl-svir.Fur unf fvk cnvef va rnpu bs guerr pbybhef, naq frira cnvef va n sbhegu pbybhe.Ol gnxvat gjragl yrsg-unaq tybirf fur pna or fher bs univat ng yrnfg bar bs rnpu pbybhe. Gur gjragl-svefg vf n evtug-unaq tybir, juvpu vf obhaq gb zngpu.-- Alec McKenziemckenzie@despammed.com === > A mathematically-inclined glove-fetishist has a number of pairs of gloves of>identical design, but of several (at least three) different colors. She has>at least three pairs of each color. In the dark she can distinguish the>handedness of a glove, but not its color. Unfortunately, she keeps the>gloves jumbled up in a drawer in an unlit cellar.>She knows that if she takes out 21 gloves, in the dark, she can be sure of>getting at least one pair.>What is the maximum number of pairs of gloves that she might have?(Rot-13):Gjragl-svir.> Fur unf fvk cnvef va rnpu bs guerr pbybhef, naq frira cnvef va n sbhegu> pbybhe.> Ol gnxvat gjragl yrsg-unaq tybirf fur pna or fher bs univat ng yrnfg bar> bs rnpu pbybhe. Gur gjragl-svefg vf n evtug-unaq tybir, juvpu vf obhaq> gb zngpu.>You can do three better if you use one fewer color and the same strategy.KevObgeneralization: What the maximum number of pairs if it takes 22 glovesto assure getting a pair. === Is there any list of known undecidable problems, like those for> NP-Complete problems (Garey-Johnson, etc)?Also, unlike NP-complete problems, for which Garey & Johnson give a >diverse catalogue, a very large class of undecidable problems is covered >by Rice's Undecidability Theorem; ...> Rice's Theorem covers only a narrow (though infinite!) class of> 10th problem have nothing to do with Rice's Theorem.Well, I wouldn't say -nothing-. Just that it is not at all obvious that they are connected.> What I need is a comprehensive, categorized list.The only places I know of with lists are:1) For formal languages/automata, there's a table in Hopcroft & Ullman.2) For logic, there's Boerger, Graedel, Gurevich, The Classical DecisionProblem. They give the minimal logics, with restrictions on the syntax ofquantifiers, that are undecidable.As to the rest of CS/Logic, there is quite a bit (the classic Church's Thesis canonical problems: Post Correspondence Problem, Rewrite systems (string (semi-Thue/type-0) and term), lambda calculus, recursive functions)As to the rest of math, there's quite a bit more: algebra (e.g. word problems), analysis (solving PDE's, equality involving elementary functions), topology (er..all I can think of are things that reduce to word problems).> I think, it is important to have such a list. If I cannot find> anything, I'll probably try to compile one and put it on the web, and> ask people to add their undecidable problems.Yes, that would be a Good Thing. In a related matter, since most of these problems reduce to the halting problem for TMs, I wonder if in those other problem areas, there are corresponding Rice's theorems (e.g. a nontrivial property of a given algebra (with given word presentation) is undecidable)Mitch Harris === > As to the rest of math, there's quite a bit more: algebra (e.g. word> problems), analysis (solving PDE's, equality involving elementary > functions), topology (er..all I can think of are things that reduce to > word problems).Topology? Interesting! Could you name an undecidable problem intopology?> In a related matter, since most of these problems reduce to the halting > problem for TMs, I wonder if in those other problem areas, there are > corresponding Rice's theorems (e.g. a nontrivial property of a given > algebra (with given word presentation) is undecidable)As far as I've seen, any reduction from halting problem consists of akind of simulation of Turing machines (or equivalents) in theenvironment of that problem. Once we get such a simulation, we canapply Rice's theorem to the environment. But it is not clear to mewhether the nontrivial properties of the languages make any usefulsense in such an environment.Let's see:- For PDEs, The nontrivial properties of the languages, are probablytranslated to nontrivial properties of the trajectories, but do notinvolve all of them.- For diophantine equations, it is translated to the nontrivialproperties of the solution space.and so.Siamak === >As to the rest of math, there's quite a bit more: algebra (e.g. word>problems), analysis (solving PDE's, equality involving elementary >functions), topology (er..all I can think of are things that reduce to >word problems).> Topology? Interesting! Could you name an undecidable problem in> topology?> A quick check with google: google four manifold undecidableCheck out these links:http://graphics.stanford.edu/~afra/goodies/markov.pdfIn this paper, Markov reduces the problem of determining whether two4-manifolds are homeomorphic to the word problem for finitely-presentedgroups. If I recall correctly, he had already shown this problem to beundecidable.http://at.yorku.ca/cgi-bin/amca/cajy-25In this paper (abstract only at this site), Chernavsky addressesa collection of (seemingly) related decidability issues. ... the rest deleted ...> SiamakDale === > Is there an easy way to trisect a line segment, i.e. split it into> three equal portions, with just straightedge and compass?> If not an easy way, is there any way at all?> Jonathan ChristensenGiven the line C to be cut, attach a line segment Aso that CA is an angle.Reproduce A n-times, marking off the end points of each segment.At the end of the repeated A's, connect with the remainingend of C, so you now have a triangle. Call this line T.Now from each end-segment of the repeated A's, draw a lineparallel to T, and crossing C.C will now be n-sected.You can prove that you have done so by taking advantageof similar triangles (which was used in the constructionof the n-section.) === > Evil is the pursuit of ignorance, challenging new truths to hold on to> an old, and comfortable point of view. It's also stupid as change is> inevitable, and the energy of the evil person is wasted.> However, in spite of its stupidity evil poses challenges, which the> discoverer is ever tasked with fighting through, including handling> those who fight for evil in their attempts to maintain their own> comfort against knowledge.> That fight is one of the continuing burdens of the Universe's first,> greatest, and last fighting force.It's SuperCrank! Faster than a speeding slide rule! Cue cheesy music. === |Do you make use of some computing package for this? I was going to let people guess which one it was, butI guess that's a little silly. I was consideringimplementing the algorithm from Henri Cohen's book oncomputational algebraic number theory, but then itoccurred to me that it'd be a lot easier to use thepackage a group of people including him developed:Pari-GP. My Christmas present to myself. :-) It'savailable for free. Someone else guessed Magma, andalthough I haven't used it I believe it would alsodo the job fine.I would challenge James Harris to figure out how touse the thing. The stuff I've been doing really onlyrequires a few of the functions, primarily functionsfor working in number fields. The neat thing is thateven though Pari-GP is doing some fancy tricks togenerate the answers for me, I can show that theanswers are correct using much more elementarycalculations, the kind that one could mainly do withjust a calculator.Keith Ramsay === Just for your enjoyment, if you haven't actually seen thefollowing problems, I'd thought I'd share them with you.1. Given two non-intersecting circles of different radii, Construct all the lines tangent to both circles using straight edge and compass.2. Two pairs of lengths a,b and y,z are the measurements of a trapezoid's sides. Each pair describes opposite sides. Reconstruct the original trapezoid using just straight edge and compass. Infinite operations are not allowed. === > Just for your enjoyment, if you haven't actually seen the> following problems, I'd thought I'd share them with you.> 1. Given two non-intersecting circles of different radii,> Construct all the lines tangent to both circles using> straight edge and compass.> 2. Two pairs of lengths a,b and y,z are the measurements of> a trapezoid's sides. Each pair describes opposite sides.> Reconstruct the original trapezoid using just straight> edge and compass. Infinite operations are not allowed.Oh yeah, in order to simplify 2, assume a & b are the measuresof the parallel sides. === Could anyone give a (Hamel) basis of the reals over the rationals?How can these be classified?Rolf Bardeli === >Could anyone give a (Hamel) basis of the reals over the rationals?No. If the Axiom of Determinacy is consistent, and Ibelieve even otherwise, there are models where thereare no Hamel bases for the reals over the rationals.If the reals can be well-ordered, a Hamel basis canbe the set of all reals which are not finite linearcombinations of previous reals (in the well-ordering). >How can these be classified?This question is unclear. -- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University === > Could anyone give a (Hamel) basis of the reals over the rationals?Such a basis is necessarily uncountable. Since the set if real numbersthat can be defined by finite symbol strings in any given language isnecessarily countable, any Hamel basis must contain numbers that arealmost all indescribable in any given language.The usual way to demonstrate that a Hamel basis exists depends on theaxiom of choice.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === On Mon, 12 Jan 2004 13:34:40 +0100, Rolf Bardeli>Could anyone give a (Hamel) basis of the reals over the rationals?Give in the sense of say exactly what it is? No, you need theAxiom of Choice to show there is such a thing, and it doesn'tgive the basis explicitly.>How can these be classified?>Rolf Bardeli************************David C. Ullrich === matt grime I like larger and smaller, you may use finer or coarser >The larger or smaller semantics are dodgy cos, intuitively, a large >number of sets in the topology's base, means 'the open sets can be >smaller'. Example, discrete is the finest which is your largest. yet >the smallest set, the single point, is open.Others use stronger and weaker. >Anyway, the point I tried to get across, not very well, is that you >need to say more than just there is a larger topology, as this is >trivially true, but that there is some kind of 'smallest larger' >topology.I said largest that assured the designated limits of a bunch of sequences. >Anyway, where are you going with this?I'm looking for couterexample or proof to a proposal madeby Stephen J. Herschkorn in thread 'powerset topology'.Another claims above a certain cardinal less than Aleph_omega0 theproposal doesn't hold. I've yet to understand how this was proven.---- === > matt grime I like larger and smaller, you may use finer or coarser>The larger or smaller semantics are dodgy cos, intuitively, a large>number of sets in the topology's base, means 'the open sets can be>smaller'. Example, discrete is the finest which is your largest. yetthe smallest set, the single point, is open.> Others use stronger and weaker.>Anyway, the point I tried to get across, not very well, is that you>need to say more than just there is a larger topology, as this is>trivially true, but that there is some kind of 'smallest larger'>topology.> I said largest that assured the designated limits of a bunch of sequences.But surely any larger topology will also assure it? === >I said largest that assured the designated limits of a bunch of sequences.> But surely any larger topology will also assure it?>Let the designated limits for a,a,a,... be b, the space S = {a,b}. Noother limits are disignated. Then the indiscrete space and the strongerSierpinski topology with base sets {a} and {a,b} assures a,a,a,... withlimit b. However the yet larger discrete topology will not assurea,a,a,... with limit b. Nor does the larger than indiscrete topology withbase sets {b}, {a,b} assure a,a,a,... with limit b. So the largesttopology for {a,b} assuring a,a,a,... with limit b is the first Sierpinskitopology as mentioned above. === >I said largest that assured the designated limits of a bunch of sequences.> But surely any larger topology will also assure it?Let the designated limits for a,a,a,... be b, the space S = {a,b}. No> other limits are disignated. Then the indiscrete space and the stronger> Sierpinski topology with base sets {a} and {a,b} assures a,a,a,... with> limit b. However the yet larger discrete topology will not assure> a,a,a,... with limit b. Nor does the larger than indiscrete topology with> base sets {b}, {a,b} assure a,a,a,... with limit b. So the largest> topology for {a,b} assuring a,a,a,... with limit b is the first Sierpinski> topology as mentioned above.Looks like I was misunderstanding what you meant by 'assure' then === > The double limit, lim_{(m,n)->(+oo,+oo)} (1 - 1/10^m)^n, does not exist. > For it to exist it would be necessary at least for both the single > limits above to have the same value, which they do not.To fit into your scheme of things, yes.> The limit as m goes to +oo is relevant the meaning of 0.999..., whereas > the limit as n goes to +oo is not.Thus it is back to inventing a meaning for 0.999... which isconsistent and commutable. An 0.999... definition causing nocontradictions.Garry Denke, GeologistDenoco Inc. of Texas === >The double limit, lim_{(m,n)->(+oo,+oo)} (1 - 1/10^m)^n, does not exist.> And if I write unicorn does that mean that a unicorn must exist?>For it to exist it would be necessary at least for both the single >limits above to have the same value, which they do not.> To fit into your scheme of things, yes.To fit into any reasonable scheme of things, yes. But your schemes defy reason.>The limit as m goes to +oo is relevant the meaning of 0.999..., whereas >the limit as n goes to +oo is not.> Thus it is back to inventing a meaning for 0.999... which is> consistent and commutable. An 0.999... definition causing no> contradictions.The general definition causes no problems except in what passes for your mind, it is no problem for anyone else.I begin to think your problems more suseptible to medication than reason.> Garry Denke, Geologist> Denoco Inc. of Texas === > The double limit, lim_{(m,n)->(+oo,+oo)} (1 - 1/10^m)^n, does not exist.> For it to exist it would be necessary at least for both the single > limits above to have the same value, which they do not.> To fit into your scheme of things, yes.That's funny. In another thread you expressed the opinion that theiterated limits should commute. Others have pointed out that they donot, in fact, commute. Sounds like you are the one who is attempting tobend the facts to fit your scheme of things.> The limit as m goes to +oo is relevant the meaning of 0.999..., whereas > the limit as n goes to +oo is not.> Thus it is back to inventing a meaning for 0.999... which is> consistent and commutable. An 0.999... definition causing no> contradictions.A contradiction is a statement of the form P and not P. The statementab does not equal ba is not a contradiction. Anyway, it's not thedefinition of 0.999... that is causing the limits not to commute; it'sthe definition of the limit concept itself. And no, that does not meanlimits are contradictory, any more than matrices or quaternions arecontradictory.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === > The double limit, lim_{(m,n)->(+oo,+oo)} (1 - 1/10^m)^n, does not exist.> it does not exist in the sense that it is not independent of the choicesmade: you must specify the way in which you let n and m go to infinity.Different choices lead to different answers.> For it to exist it would be necessary at least for both the single > limits above to have the same value, which they do not.> To fit into your scheme of things, yes.and what's your scheme of things? > The limit as m goes to +oo is relevant the meaning of 0.999..., whereas > the limit as n goes to +oo is not.> Thus it is back to inventing a meaning for 0.999... which is> consistent and commutable. An 0.999... definition causing no> contradictions.There is no contradiction. As it has just been explained to you what'sgoing on, the only place left for a contradiction is in yourunderstanding, not in mathematics. Why must the definitions you seek becommutable? Assuming we decipher commutable correctly.> Garry Denke, Geologist> Denoco Inc. of Texas === > lim(n->+oo) lim(m->+oo) (1-1/10^m)^n = 1 (1)>Why do you prefer the above as opposed to the below? > lim(n->+oo) (1-1/10^m)^n = 0> This holds for each m > 0, and therefore> lim(m->+oo) lim(n->+oo) (1-1/10^m)^n = 0 (2) Comparing (1) and (2), we see that the limits do not commute.They should commute. Why the contradiction? > It's not a matter of prefering one limit to the other. It's simply a> matter of understanding what the notation 0.999... means. Since> 0.999..., by definition, means lim (m->+oo) (1-1/10^m), it follows that> lim (n->+oo) 0.999...^n = lim (n->+oo) lim (m->+oo) (1-1/10^m)^n, which> is the limit that appears in (1).So it's the definition causing the contradiction.Garry Denke, GeologistDenoco Inc. of Texas === > lim(n->+oo) lim(m->+oo) (1-1/10^m)^n = 1 (1)> Why do you prefer the above as opposed to the below? > lim(n->+oo) (1-1/10^m)^n = 0>This holds for each m > 0, and therefore> lim(m->+oo) lim(n->+oo) (1-1/10^m)^n = 0 (2)>Comparing (1) and (2), we see that the limits do not commute.> They should commute. Why should they commute?> Why the contradiction?What contradiction? The way it is is the way it is! That the world does not wag in the way you want is your problem, not ours. Only a madman expects the world to change for his benefit. It's not a matter of prefering one limit to the other. It's simply a>matter of understanding what the notation 0.999... means. Since>0.999..., by definition, means lim (m->+oo) (1-1/10^m), it follows that>lim (n->+oo) 0.999...^n = lim (n->+oo) lim (m->+oo) (1-1/10^m)^n, which>is the limit that appears in (1).> So it's the definition causing the contradiction.No! It is some sort of short circuit in your head that causes you to imagine contradictions where there are none.If you stand on a hillside, do you expect the same rate of incline in all directions? As a supposed geologist, you should know better. The incline depends on the direction you take.Then if you have a function of two variables, such as f(x,y) = (1 - 1/10^x)^y, why do you expect the same behaviour from one variable as from the other? The behavior depends on which variable you vary. Otherwise, why bother with two variables? === > They should commute. Why the contradiction? Would you also expect the limits to commute in the following case?lim(x->0) lim(y->0) arctan((x-y)/(x+y))lim(y->0) lim(x->0) arctan((x-y)/(x+y))-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === > lim(n->+oo) lim(m->+oo) (1-1/10^m)^n = 1 (1)>Why do you prefer the above as opposed to the below? > lim(n->+oo) (1-1/10^m)^n = 0> This holds for each m > 0, and therefore> lim(m->+oo) lim(n->+oo) (1-1/10^m)^n = 0 (2)> Comparing (1) and (2), we see that the limits do not commute.> They should commute. Why the contradiction? What contradiction? There are all kinds of things in mathematics that donot commute. Multiplication of matrices or quaternions, to name twoexamples. Are matrices and quaternions contradictory?Learning which things commute and which things don't is a part ofmathematics.> It's not a matter of prefering one limit to the other. It's simply a> matter of understanding what the notation 0.999... means. Since> 0.999..., by definition, means lim (m->+oo) (1-1/10^m), it follows that> lim (n->+oo) 0.999...^n = lim (n->+oo) lim (m->+oo) (1-1/10^m)^n, which> is the limit that appears in (1).> So it's the definition causing the contradiction.What contradiction? Which definition? (Not the definition of 0.999...,in case you are wondering.)-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === > lim(n->+oo) lim(m->+oo) (1-1/10^m)^n = 1 (1)>Why do you prefer the above as opposed to the below? > lim(n->+oo) (1-1/10^m)^n = 0> This holds for each m > 0, and therefore> lim(m->+oo) lim(n->+oo) (1-1/10^m)^n = 0 (2)> Comparing (1) and (2), we see that the limits do not commute. They should commute. Why the contradiction? > No, limits should not in general commute. By considering a matrix withinfinitely many rows and columns you should be able to construct doublyindexed sequences with lim_m (lim_n a(n,m)) being vastly different fromlim_n (lim_m a(n,m)) at will.> It's not a matter of prefering one limit to the other. It's simply a> matter of understanding what the notation 0.999... means. Since> 0.999..., by definition, means lim (m->+oo) (1-1/10^m), it follows that> lim (n->+oo) 0.999...^n = lim (n->+oo) lim (m->+oo) (1-1/10^m)^n, which> is the limit that appears in (1).> So it's the definition causing the contradiction.> Garry Denke, Geologist> Denoco Inc. of Texas === In sci.math, Garry Denkeon 11 Jan 2004 09:13:34 -0800<54056cf9.0401110913.29f0219d@posting.google.com>:> What is a double limit?> lim(m,n->+oo) (1 - 1/10^m)^n> Does the above have an assigned value?Unfortunately, no.> lim(m->+oo) (1-1/10^m)^n = 1> Why do you prefer the above as opposed to the below? > lim(n->+oo) (1-1/10^m)^n = 0Depends on which direction you want to go. For most,.999... = 1. However, as n -> +oo, (.999...9)^n -> 0for any finite length of digit string, but 1^n -> 1and (1.000...01)^n -> +oo. One can also do thingslike lim (n->+oo) (1 + 1/n)^n = eor lim(n->+oo) (1 - 1/n)^n = 1/e.If there's a resolution to this dilemma, I've missed it.But that by itself does not prove that the Cauchy sequenceimplicitly inherent in .999... (or for that matter1.000...1) does not have the limit 1.> Garry Denke, Geologist> Denoco Inc. of Texas-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === >Yeah, is that really true? Can anybody suggest any books, or (if its>not too much to ask) an outline of a proof? That's really cool. >Can't think of a counterexample for class one either...This is what puzzled me. I can never remember what thoseBaire classes are. But it's certainly true that if f is Lebesguemeasureable then there is a sequence of continuous functionsconverging to f ae...Ah, just realized that it does not _follow_ from that that iff is Lebesgue measurable then f = g ae, where g = lim g_n_everywhere_, g_n continuous. Duh, glad you mentionedyou don't see a counterexample for class one - not thatI have a proof or counterexample, but I was puzzled whythe positive result for class one was not obvious fromwell-known facts, and you're saying this jogged meinto seeing why it's not.A start of a proof for class two: The characteristic functionof a closed set is the pointwise limit of a sequence ofcontinuous functions (class one?) hence the characteristicfunction of an F-sigma is class two(?). That gives theresult for the characteristic function of a measurableset - I bet you can prove it for general measurablefunctions by thinking about pointwise limits of(lower?) semicontinuous functions.>I just read a section in a real analysis book that states that up to a set of>measure zero any L.measurable function on the line is equal to a function of>Baire class two.************************David C. Ullrich === On Sun, 11 Jan 2004 13:44:18 +0000 (UTC), nico80@jazzfree.com (Nicolas>I would like to know if this is a mathematical proof of the Cantors >goof, or conversely it is just another mathematical goof about the >Cantors proof.That's a lie. It's clear from your behavior in other threads thatyou have no interest in having your mathematical errors corrected.[...]************************David C. Ullrich === > The goofs are all those of the people who assert that Cantor goofed.Not goofs. They are not goofs at all, but fools, and fools with agreat need to be the Next Great Wonder in mathematics, by taking theeasy road.As I noted previously, the crank psychology doesn't allow forprogressive extension of knowledge, preferring to be the agent ofTotal Revolution. Like most Total Revolutionaries, the proponentshave only a vague idea of what they will replace the status quo by.By contrast, real scholarship (even that which gets the labelrevolutionary) is nearly always progressive, not overturning thepast but extending it, changing how it is viewed, but nearly alwaysleaving its correctness more or less intact.Thomas === On Sun, 11 Jan 2004 16:09:42 -0500, walala Dear all,>Is there any function that relates log(x+y) with log(x) and log(y)...Yes, if f(s,t) = log(exp(s) + exp(t)) then log(x+y) = f(log(x), log(y)).>In an optimization problem, I basically need to seek the minization of the>function of the form log(x+y)... but in order for further mathematical>manipulation, I need to break log(x+y) up into log(x) and log(y),>right?>-Walala>************************David C. Ullrich === > Dear all,> Is there any function that relates log(x+y) with log(x) and log(y)...The function may need to be continuously defined in +Inf/-Inf rangeof argument. If we have to express log (4 -2)in terms of log (4) andlog (-2), real log is not defined for negative arguments. Can 1/(x+y)be defined in terms of 1/x and 1/y ? For Sin (or any trig function),Exp etc.,there is no such problem. === >Can 1/(x+y) be defined in terms of 1/x and 1/y ? Of course. Call these a,b, and c respectively. Then a = 1/(1/b+1/c)=b/(b+c) .Naturally, some division is required, e.g. 1/5 is not in the ringZ[1/2,1/3].Note to Original Poster: in the same way, you can compute the function fof the subject line: f(b,c)=log(exp(b)+exp(c)). There is no other solution(what did you expect?) although I supposed you could write this functionf in different ways, e.g f(b,c)= b + log(1 + exp(c-b)) .>For Sin (or any trig function),Exp etc.,there is no such problem.Really? If cos(x)=0 and cos(y)=0, what is cos(x+y)?dave === > Dear all,> Is there any function that relates log(x+y) with log(x) and log(y)...> In an optimization problem, I basically need to seek the minization of the> function of the form log(x+y)... but in order for further mathematical> manipulation, I need to break log(x+y) up into log(x) and log(y),> right?> -Walala> Choose the 'f' of the title line to bef(a,b) = log(exp(a)+exp(b))-- Use of tools distinguishes Man from Beast. And UNIX users from WINDOZE lusers. === > You might consider that:d(log(x+y))/dx = d(log(x+y))/dy = 1/(x+y)It's unclear what kind of optimization problem you want to solve.> Without any more information than that, log(x+y) clearly doesn't> have any local minimum.Dear Guillaume,The reason I need to break log(X+Y) up is because the problem is tominimize (X+Y-C)^2where C is known. So it is something like quadratic optimization, evenworse, X is defined to be a^x, Y is defined to be a^y, C is defined to bea^c, where c is known. Hence the true variables are x, y.The problem becomesminimize (a^x+a^y-a^c)^2where a is the common base for all these terms...If I take log a, it will make things a lot easier, that's why I want tobreaklog_a_(a^x+a^y-a^c)up...Can you give me some more ideas?-Walala === >Dear all,>Is there any function that relates log(x+y) with log(x) and log(y)...>In an optimization problem, I basically need to seek the minization ofthe>function of the form log(x+y)... but in order for further mathematical>manipulation, I need to break log(x+y) up into log(x) and log(y),>right?>-Walala> A simple-minded idea: minimizing x+y will minimize log(x+y).> - Ken>Dear Ken,The reason I need to break log(X+Y) up is because the problem is tominimize (X+Y-C)^2where C is known. So it is something like quadratic optimization, evenworse, X is defined to be a^x, Y is defined to be a^y, C is defined to bea^c, where c is known. Hence the true variables are x, y.The problem becomesminimize (a^x+a^y-a^c)^2where a is the common base for all these terms...If I take log a, it will make things a lot easier, that's why I want tobreaklog_a_(a^x+a^y-a^c)up...Can you give me some more ideas?-Walala === >Dear all,>Is there any function that relates log(x+y) with log(x) and log(y)...>In an optimization problem, I basically need to seek the minization ofthe>function of the form log(x+y)... but in order for further mathematical>manipulation, I need to break log(x+y) up into log(x) and log(y),>log(x+y)=log(x)+log(1+y/x) might help.But for minimization it is better to set the derivatives to zero> rather than do such transformations...Arnold Neumaier>Dear Arnold,I want to discuss further about this with you:The reason I need to break log(X+Y) up is because the problem is tominimize (X+Y-C)^2where C is known. So it is something like quadratic optimization, evenworse, X is defined to be a^x, Y is defined to be a^y, C is defined to bea^c, where c is known. Hence the true variables are x, y.The problem becomesminimize (a^x+a^y-a^c)^2where a is the common base for all these terms...If I take log a, it will make things a lot easier, that's why I want tobreaklog_a_(a^x+a^y-a^c)up...Can you give me some more ideas?-Walala === >The reason I need to break log(X+Y) up is because the problem is to>minimize (X+Y-C)^2>where C is known. So it is something like quadratic optimization, even>worse, X is defined to be a^x, Y is defined to be a^y, C is defined to be>a^c, where c is known. Hence the true variables are x, y.>The problem becomes>minimize (a^x+a^y-a^c)^2>where a is the common base for all these terms...>If I take log a, it will make things a lot easier, that's why I want to>break>log_a_(a^x+a^y-a^c)>up...>Can you give me some more ideas?>Huh? Clearly, the function is minimized for any X and Y such that X + Y = C. Are x and y constrained?-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === >The reason I need to break log(X+Y) up is because the problem is to>minimize (X+Y-C)^2>where C is known. So it is something like quadratic optimization, even>worse, X is defined to be a^x, Y is defined to be a^y, C is defined to be>a^c, where c is known. Hence the true variables are x, y.>The problem becomes>minimize (a^x+a^y-a^c)^2>where a is the common base for all these terms...>If I take log a, it will make things a lot easier, that's why I want to>break>log_a_(a^x+a^y-a^c)>up...>Can you give me some more ideas?> Huh? Clearly, the function is minimized for any X and Y such> that X + Y = C. Are x and y constrained?-- > Stephen J. Herschkorn herschko@rutcor.rutgers.edu>Dear Stephen,As you guessed, x and y are constrained...Sorry I did not make it clear... x and y are integer and are overdertermined by more equations than unknown variables... hence there is noexact solution...and there are a bunch of such variables X and Y's... hencea bunch of x and y's ... finally what I want is to minimize the MSE for abunch of such X and Y'si.e.minimize sum{ (a^x_i+a^y_i-a^c_i)^2 , i from 0 to N} ...-walala === >The reason I need to break log(X+Y) up is because the problem is to>minimize (X+Y-C)^2>where C is known. So it is something like quadratic optimization, even>worse, X is defined to be a^x, Y is defined to be a^y, C is defined to be>a^c, where c is known. Hence the true variables are x, y.>The problem becomes>minimize (a^x+a^y-a^c)^2>where a is the common base for all these terms...>If I take log a, it will make things a lot easier, that's why I want to>break>log_a_(a^x+a^y-a^c)>up...>Can you give me some more ideas?>Huh? Clearly, the function is minimized for any X and Y such>that X + Y = C. Are x and y constrained?>-- >Stephen J. Herschkorn herschko@rutcor.rutgers.edu> Dear Stephen,> As you guessed, x and y are constrained...> Sorry I did not make it clear... x and y are integer and are over> dertermined by more equations than unknown variables... hence there is no> exact solution...and there are a bunch of such variables X and Y's... hence> a bunch of x and y's ... finally what I want is to minimize the MSE for a> bunch of such X and Y's> i.e.> minimize sum{ (a^x_i+a^y_i-a^c_i)^2 , i from 0 to N} ...What is it that you really want to solve(objective, constraints, ranges of variables)?Revealing only small bits at a time will not giveyou a satisfying answer.You are well advised to learn AMPL or GAMS to create anintelligible problem description.Arnold Neumaier === What is this theorem called?If f: R->R is a continuous function, and f(a)<0 and f(b)>0, then thereexists a c between a and b so that f(c)=0.-- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland ---------- http://www.helsinki.fi/~palaste --------------------- rules! --------/We sorcerers don't like to eat our words, so to say. - Sparrowhawk === > What is this theorem called? If f: R->R is a continuous function, and> f(a)<0 and f(b)>0, then there exists a c between a and b so that f(c)=0.The people who told you Bolzano's Theorem are right.The people who told you Intermediate Value Theorem are unconsciouslygeneralizing, and then telling you the name of the generalized theorem.Bolzano's is that if a continuous function changes sign on an interval, thanit must be zero somewhere on the interval.IVT is that if a continous function is f(a) at a, f(b) at b, then on [a,b]it takes on every value in [f(a), f(b)]. Bolzano's Theorem is a trivialconsequence of the IVT.-- --Tim Smith === > What is this theorem called? If f: R->R is a continuous function, and> f(a)<0 and f(b)>0, then there exists a c between a and b so that f(c)=0.> The people who told you Bolzano's Theorem are right.> The people who told you Intermediate Value Theorem are unconsciously> generalizing, and then telling you the name of the generalized theorem.> Bolzano's is that if a continuous function changes sign on an interval, than> it must be zero somewhere on the interval.> IVT is that if a continous function is f(a) at a, f(b) at b, then on [a,b]> it takes on every value in [f(a), f(b)]. Bolzano's Theorem is a trivial> consequence of the IVT.I would have stated that the other way around: the IVT is a trivialconsequence of Bolzano's theorem, in somewhat the same way that the MVTis a trivial consequence of Rolle's theorem.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === >What is this theorem called?>If f: R->R is a continuous function, and f(a)<0 and f(b)>0, then there>exists a c between a and b so that f(c)=0.Usually that theorem (but also sometimes a trivially moregeneral theorem) is called the Intermediate Value Theoremin (American) English-language textbooks.Lee Rudolph === Intermediate Value Theorem> What is this theorem called?> If f: R->R is a continuous function, and f(a)<0 and f(b)>0, then there> exists a c between a and b so that f(c)=0.> === > What is this theorem called?> If f: R->R is a continuous function, and f(a)<0 and f(b)>0, then there> exists a c between a and b so that f(c)=0.I've been told this is called Darboux property of continuous functions.Mateusz === > What is this theorem called?> If f: R->R is a continuous function, and f(a)<0 and f(b)>0, then there> exists a c between a and b so that f(c)=0.Intermediate Value Theorem-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Robin Chapman scribbled the following:> What is this theorem called?> If f: R->R is a continuous function, and f(a)<0 and f(b)>0, then there> exists a c between a and b so that f(c)=0.> Intermediate Value TheoremI have been told that the Intermediate Value Theorem is:If f: R->R is a derivable function, and aRobin Chapman scribbled the following:> What is this theorem called?> If f: R->R is a continuous function, and f(a)<0 and f(b)>0, then there> exists a c between a and b so that f(c)=0.Intermediate Value Theorem>I have been told that the Intermediate Value Theorem is:>If f: R->R is a derivable function, and abetween a and b so that f(b)-f(a) = f'(ksi)*(b-a).Really? The absolutely standard name for _that_ theorem isMean Value Theorem.(there's a result which people sometimes might call the intermediatevalue theorem for derivatives: If f is differentiable, f'(a) < 0 andf'(b) > 0 then there exists x between a and b such that f'(x) = 0.That is, a derivative satisfies the conclusion of the IntermediateValue Theorem (the result you asked about) even though itneed not be continuous.)>Are there more specific names for these theorems?************************David C. Ullrich === >Robin Chapman scribbled the following:>What is this theorem called?>If f: R->R is a continuous function, and f(a)<0 and f(b)>0, then there>exists a c between a and b so that f(c)=0.>Intermediate Value Theorem>I have been told that the Intermediate Value Theorem is:>If f: R->R is a derivable function, and abetween a and b so that f(b)-f(a) = f'(ksi)*(b-a).> Really? The absolutely standard name for _that_ theorem is> Mean Value Theorem.I'm not sure about that - here folks usually call it Lagrange's theorem.Felix. === On 12 Jan 2004 10:17:51 -0800, felixg@tx.technion.ac.il (FelixRobin Chapman scribbled the following:>What is this theorem called?>If f: R->R is a continuous function, and f(a)<0 and f(b)>0, then there>exists a c between a and b so that f(c)=0.Intermediate Value Theorem>I have been told that the Intermediate Value Theorem is:>If f: R->R is a derivable function, and abetween a and b so that f(b)-f(a) = f'(ksi)*(b-a).> Really? The absolutely standard name for _that_ theorem is> Mean Value Theorem.>I'm not sure about that - here folks usually call it Lagrange's theorem.Really? Where is that? I've never heard it called Lagrange's Theorem,not even once. Well, until just now of course...>Felix.************************David C. Ullrich === >Robin Chapman scribbled thefollowing:> What is this theorem called?> If f: R->R is a continuous function, and f(a)<0 and f(b)>0, thenthere> exists a c between a and b so that f(c)=0.> Intermediate Value Theorem>I have been told that the Intermediate Value Theorem is:>If f: R->R is a derivable function, and abetween a and b so that f(b)-f(a) = f'(ksi)*(b-a).>Really? The absolutely standard name for _that_ theorem is>Mean Value Theorem.I'm not sure about that - here folks usually call it Lagrange's theorem.Maybe there, but here we try to avoid that because there are too manytheorems it could apply to. Too bad we don't do the same with all thoseEuler's theorems.I suspect that Joona Palaste's confusion of Mean with Intermediate is atranslation error and not an understanding error. Intermediate meansin-between and Mean means average, which of course will be between thetwo extremes, but will be something more as well (depending on which meanyou mean).Jon Miller === Jonathan Miller scribbled the following:>Robin Chapman scribbled the> following:> What is this theorem called?> If f: R->R is a continuous function, and f(a)<0 and f(b)>0, then> there> exists a c between a and b so that f(c)=0.> Intermediate Value Theorem>I have been told that the Intermediate Value Theorem is:>If f: R->R is a derivable function, and abetween a and b so that f(b)-f(a) = f'(ksi)*(b-a).>Really? The absolutely standard name for _that_ theorem is>Mean Value Theorem.> I suspect that Joona Palaste's confusion of Mean with Intermediate is a> translation error and not an understanding error. Intermediate means> in-between and Mean means average, which of course will be between the> two extremes, but will be something more as well (depending on which mean> you mean).You are correct that it's a translation error. In Finnish the theoremabove about derivable functions and f'(ksi) is v.8aliarvolause,translating directly to intermediate value theorem. I didn't knowmean and intermediate meant so different things in English.That's where the confusion came from.-- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland ---------- http://www.helsinki.fi/~palaste --------------------- rules! --------/That's no raisin - it's an ALIEN! - Tourist in MTV's Oddities === > Robin Chapman scribbled the following:> What is this theorem called?> If f: R->R is a continuous function, and f(a)<0 and f(b)>0, then there> exists a c between a and b so that f(c)=0.> Intermediate Value Theorem> I have been told that the Intermediate Value Theorem is:> If f: R->R is a derivable function, and a between a and b so that f(b)-f(a) = f'(ksi)*(b-a).> Are there more specific names for these theorems?Check out intermediate value, mean value and Rolle theorems. === >Robin Chapman scribbled the following:> What is this theorem called?> If f: R->R is a continuous function, and f(a)<0 and f(b)>0, then there> exists a c between a and b so that f(c)=0.Intermediate Value Theorem>I have been told that the Intermediate Value Theorem is:>If f: R->R is a derivable function, and abetween a and b so that f(b)-f(a) = f'(ksi)*(b-a).>Are there more specific names for these theorems?That would be the Mean Value Theorem. Or one of them at least. Don'teven get started on the multiple Fundamental Theorems of Calculus. === >What is this theorem called?>If f: R->R is a continuous function, and f(a)<0 and f(b)>0, then there>exists a c between a and b so that f(c)=0.Bolzano's Theorem. === I am learning combinotics now. One of exercise is to create 6*6 magiccub, but I don't know where to related algorithm in the internet?Could you help me? === >Last night, at about 00.55 AM, I went up on terrace and with full moon>right over my head, I tossed one rupee coin.>It was Head.>Today morning, I tossed again one rupee coin in blue sky.>Again, it was Head.>So God is telling me to file one more patent application. I have>already filed 7 patent applications.> No, God is telling you that Rosencrantz and Guildenstern are dead.>----------------------------->In the end, R&G resign themselves to their fate, although Guildenstern>says, There must have been a moment, at the beginning, when we could>have said -- no. But somehow we missed it. Perhaps.>-----------------------------They are still very dead.--Dr.Postman USPS, MBMC, BsD; Disgruntled, But UnarmedMember,Board of Directors of afa-b, SKEP-TI-CULT member #15-51506-253.Shake it like a polaroid picture. - Andre 3000 of Outkast === >Last night, at about 00.55 AM, I went up on terrace and with full moon>right over my head, I tossed one rupee coin.>It was Head.>Today morning, I tossed again one rupee coin in blue sky.>Again, it was Head.>So God is telling me to file one more patent application. I have>already filed 7 patent applications.>No, God is telling you that Rosencrantz and Guildenstern are dead.>----------------------------->In the end, R&G resign themselves to their fate, although Guildenstern>says, There must have been a moment, at the beginning, when we could>have said -- no. But somehow we missed it. Perhaps.>-----------------------------> They are still very dead. -----------------------------------------R&G inhabit a world completely beyond their comprehension. Unsure ofwhere they are going (and even of who they are and where they comefrom), they depend upon others to give their lives meaning. Whileawaiting instructions, they fall back upon games -- word play andsimple wagers -- that rarely achieve their intended goals.Instructed by the King and Queen to glean what afflicts poor Hamlet,the boys attempt to cross-examine the prince but end up only moreconfused. Neither do they have the wit to see their own deathsforetold when the Player and his Tragedians rehearse the melodramatictragedy, The Murder of Gonzago , which includes the execution of twosmiling accomplices -- friends -- courtiers -- two spies whoaccompany a prince to England, only to be betrayed by a purloinedletter.-----------------------------------------Oh, yes. I got my instructions, script to enter in this new characteron universal stage. I have done lot of rehearsal in real life and Iam aware of my melodramatic tragedy. That is why I am shouting that,those rehearsals were setup, a plot. I was controlled to do thoserehearsals in real life.In those rehearsals, I never tried to commit suicide. This is why Ican't die. So I will have to play this character.Whatever I do, it is already scripted anyway.-Abhi. === A recent thread on sci.math reminded me of something I'vewondered about for some time. Consider the following twofacts:(i) Every once in a while on usenet I hear someone say[someone] proved ZF|-Con(PA), using induction up toepsilon_0.(ii) The fact that ZF|-Con(PA) is obvious: it's clear that thefinite ordinals form a model of PA.While (i) and (ii) are not mathematically inconsistent,they're real-world inconsistent - if ZF|-Con(PA) isreally as simple as suggested in (ii) then nobody wouldtalk about who proved it (also I don't see what the reasoningin (ii) has to do with induction up to epsilon_0.)What gives? Is (i) a misquotation, or are things not assimple as it seems in (ii) or what?************************David C. Ullrich === David C. Ullrich says...>A recent thread on sci.math reminded me of something I've>wondered about for some time. Consider the following two>facts:>(i) Every once in a while on usenet I hear someone say>[someone] proved ZF|-Con(PA), using induction up to>epsilon_0.>(ii) The fact that ZF|-Con(PA) is obvious: it's clear that the>finite ordinals form a model of PA.>While (i) and (ii) are not mathematically inconsistent,>they're real-world inconsistent - if ZF|-Con(PA) is>really as simple as suggested in (ii) then nobody would>talk about who proved it (also I don't see what the reasoning>in (ii) has to do with induction up to epsilon_0.)The interesting thing about (ii) is that it doesn't really usemuch ZFC at all. You can prove Con(PA) in *PA*, if you add theprinciple of epsilon_0 induction. That is, you come up with acoding #alpha of the ordinals less than epsilon_0 as naturals(not hard) and define a recursive binary relation lt(x,y) whichholds if and only if for some ordinals alpha and beta,x = #alpha and y = #beta and alpha < beta. That isn't hard, either.Then you add the following induction schema to PA: all y (all x ( lt(x,y) -> Phi(x) ) -> Phi(y)) -> all y Phi(y)Then this new theory, PA + epsilon_0 induction, is able to prove con(PA).The proof is *proof-theoretic* rather than *model-theoretic*. Basically,you assign an ordinal less than epsilon_0 to each proof in PA. Then youprove (by induction) that there is no proof of a contradiction.I think this was due to Gentzen.--Daryl McCulloughIthaca, NY === > Then this new theory, PA + epsilon_0 induction, is able to prove con(PA). Actually it is an esential feature of Gentzen's proof that it is formalizablein PRA+epsilon_0 induction. That is, only quantifier-free induction is used.So the proof is perfectly finitary except for the use of epsilon_0 induction. === On 12 Jan 2004 09:35:26 -0800, torkel@sm.luth.se (Torkel Franzen)> Then this new theory, PA + epsilon_0 induction, is able to prove con(PA).Actually it is an esential feature of Gentzen's proof that it is formalizable>in PRA+epsilon_0 induction. That is, only quantifier-free induction is used.I can't figure out what you mean by quantifier-free induction. (No doubt it should be clear and I'm just being slow...)>So the proof is perfectly finitary except for the use of epsilon_0 induction.************************David C. Ullrich === David C. Ullrich says...>On 12 Jan 2004 09:35:26 -0800, torkel@sm.luth.se (Torkel Franzen)Then this new theory, PA + epsilon_0 induction, is able to prove con(PA).Actually it is an esential feature of Gentzen's proof that it is formalizable>in PRA+epsilon_0 induction. That is, only quantifier-free induction is used.>I can't figure out what you mean by quantifier-free induction. >(No doubt it should be clear and I'm just being slow...)Induction proves a formula of the form forall x, Phi(x). I believethat Torkel is talking about whether the formula Phi(x) containsquantifiers or not. If Phi(x) is a primitive-recursive predicate, thenit can be written with no quantifiers (or at least, no unboundedquantifiers). Of course, the terminology is a little strange, becausethe theorem being proved, forall x, Phi(x) certainly has quantifiers.--Daryl McCulloughIthaca, NY === On 12 Jan 2004 06:58:12 -0800, daryl@atc-nycorp.com (Daryl McCullough)>David C. Ullrich says...>A recent thread on sci.math reminded me of something I've>wondered about for some time. Consider the following two>facts:>(i) Every once in a while on usenet I hear someone say>[someone] proved ZF|-Con(PA), using induction up to>epsilon_0.>(ii) The fact that ZF|-Con(PA) is obvious: it's clear that the>finite ordinals form a model of PA.>While (i) and (ii) are not mathematically inconsistent,>they're real-world inconsistent - if ZF|-Con(PA) is>really as simple as suggested in (ii) then nobody would>talk about who proved it (also I don't see what the reasoning>in (ii) has to do with induction up to epsilon_0.)>The interesting thing about (ii) is that it doesn't really use>much ZFC at all. You can prove Con(PA) in *PA*, if you add the>principle of epsilon_0 induction. That is, you come up with a>coding #alpha of the ordinals less than epsilon_0 as naturals>(not hard) and define a recursive binary relation lt(x,y) which>holds if and only if for some ordinals alpha and beta,>x = #alpha and y = #beta and alpha < beta. That isn't hard, either.>Then you add the following induction schema to PA: all y (all x ( lt(x,y) -> Phi(x) ) -> Phi(y)) > -> all y Phi(y)>Then this new theory, PA + epsilon_0 induction, is able to prove con(PA).>The proof is *proof-theoretic* rather than *model-theoretic*. Basically,>you assign an ordinal less than epsilon_0 to each proof in PA. Then you>prove (by induction) that there is no proof of a contradiction.>I think this was due to Gentzen.about (i) is that it doesn't really use much ZFC at all then thismakes sense to me. ???************************David C. Ullrich === David C. Ullrich says...>(i) Every once in a while on usenet I hear someone say>[someone] proved ZF|-Con(PA), using induction up to>epsilon_0.>(ii) The fact that ZF|-Con(PA) is obvious: it's clear that the>finite ordinals form a model of PA.>...The interesting thing about (ii) is that it doesn't really use>much ZFC at all.>about (i) is that it doesn't really use much ZFC at all then this>makes sense to me. ???Yes, I meant (i) instead of (ii). I don't think that (ii) uses inductionin any meaningful way.--Daryl McCulloughIthaca, NY === > A recent thread on sci.math reminded me of something I've> wondered about for some time. Consider the following two> facts:> (i) Every once in a while on usenet I hear someone say> [someone] proved ZF|-Con(PA), using induction up to> epsilon_0.This is Gentzen's theorem> (ii) The fact that ZF|-Con(PA) is obvious: it's clear that the> finite ordinals form a model of PA.Yes, obvious.> While (i) and (ii) are not mathematically inconsistent,> they're real-world inconsistent - if ZF|-Con(PA) is> really as simple as suggested in (ii) then nobody would> talk about who proved it (also I don't see what the reasoning> in (ii) has to do with induction up to epsilon_0.)> What gives? Is (i) a misquotation, or are things not as> simple as it seems in (ii) or what?Gentzen's theorem doesn't use all of ZF. Essentially (and I don'tunderstand this at all well) all it uses is the fact that epsilon_0is a well-ordered set.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === On Mon, 12 Jan 2004 14:51:57 +0000, Robin ChapmanA recent thread on sci.math reminded me of something I've> wondered about for some time. Consider the following two> facts:> (i) Every once in a while on usenet I hear someone say> [someone] proved ZF|-Con(PA), using induction up to> epsilon_0.>This is Gentzen's theorem(ii) The fact that ZF|-Con(PA) is obvious: it's clear that the> finite ordinals form a model of PA.>Yes, obvious.While (i) and (ii) are not mathematically inconsistent,> they're real-world inconsistent - if ZF|-Con(PA) is> really as simple as suggested in (ii) then nobody would> talk about who proved it (also I don't see what the reasoning> in (ii) has to do with induction up to epsilon_0.)> What gives? Is (i) a misquotation, or are things not as> simple as it seems in (ii) or what?>Gentzen's theorem doesn't use all of ZF. Essentially (and I don't>understand this at all well) all it uses is the fact that epsilon_0>is a well-ordered set.So my assumption that I must be confused was wrong, fine.(Some people can't get _anything_ right...)************************David C. Ullrich === > Could a proof of the Goldbach Conjecture be worth of a Fields prize? > Don't you know the regulations for your own prize?Fields Medals are restricted to those under age 40. === |hundreds of spams every day.I'm not. I get less than ten a day, and they are almost alleasily recognizable as spam. Most of them now end ingarbage intended to circumvent spam filters. Most of therest give rather implausible subject lines.|There is nothing wrong with protecting I have no problem with that. I mainly want to say it's notalways as bad as it sounds.Keith Ramsay === > |hundreds of spams every day. I'm not. I get less than ten a day, and they are almost all> easily recognizable as spam. Most of them now end in> garbage intended to circumvent spam filters. Most of the> rest give rather implausible subject lines.> |There is nothing wrong with protecting > I have no problem with that. I mainly want to say it's not> always as bad as it sounds.> Keith RamsayThat worked for a while with very few errors (i.e., very few until spammers find a way to defeat it. Which they probably will. Managing spam is becoming an academic discipline of its own. and Anti-Spam' to be held in Mountain View, CA, July 30-Aug 1. Nora B. === |I've been waiting for someone who really knows this stuff to speak|up, but nobody has (or if anyone has, he's also included disclaimers|about how he could be wrong...)You know, it would have been a little more likely that I would've beenreading this if somebody could have renamed the thread something a littlemore descriptive than rationals are uncountable. I mean, just becausesomeone decides to name a thread something like Yer Bleedin' Doggiesor Help or Proof of 2+2=5 doesn't mean it has to stay that way....One can prove Con(PA) in ZFC. Someone mentioned Gentzen's proof (whichformalizes easily enough in ZFC). That's fine. The gist of it goeslike this. Gentzen reformulated PA in an equivalent form in sequentcalculus. A sequent is written as A1,...,An |- B1,...,Bm where the A_iand the B_i are formulas (they can have free variables) and it standsfor the statement (A1&...&An)->(B1 or B2 or ... or Bm) (which in turnmeans that this holds for all possible assignments to the free variables).He made the axiom of induction into a deductive rule. The cute thing aboutthe sequent calculus is that almost all of the rules satisfy the so-calledsubformula property: every subformula of sequents to which the rule isapplied is also a subformula of the consequent. So everything in the proofis a subformula of the conclusion (possibly with variables renamed!),unless there is an application of the cut rule: A1,...,An |- B1,...,Bm,B C1,...,Cr,B |- D1,...,Ds ---------------------------------------------------- A1,...,An,C1,...,Cr |- B1,...,Bm,D1,...,Dsin which the cut formula B disappears. The cut rule is very powerfulin a sense; the maximum number of steps required to write an n-step proofwithout using cuts is some fast-growing function of n... which I forget.On the other hand, cuts can be eliminated in principle. Much of the coreof Gentzen's proof is showing how to eliminate cuts from proofs in hisformulation of PA.If cuts can be eliminated, then PA is consistent. A proof of acontradiction amounts to a proof of |- in the system (with nothingon either side of the turnstile). (As I recall, a proof of |- 1=0 can bemade a proof of |- in this system.) But there is of course no cut-freeproof of |-, since there are no subformulas of to work with.People like to say Gentzen proved induction on epsilon_0 implies theconsistency of PA. That's because he used induction on epsilon_0 to provethat his cut-elimination procedure terminates. Each proof-tree can beassigned a different ordinal number a2>...>a_k. You can compare them by using the fact that the term withthe highest power of w dominates. So ordinals under epsilon_0 can bewritten as things like w^w^w^w^3 + w^w^(w^5+4*w^3+126) + 12*w^3 + 2.Gentzen's proof is sort of overkill for this context, however. It'svaluable mainly for giving more information than just proving that theconsistency of PA follows from the axioms of ZFC. If you want an easierproof in ZFC that PA is consistent, just use the fact that ZFC can provethat PA has a model. You can define the set of true statements about thenatural numbers in ZFC, and prove that they include the axioms of PA, areclosed under deduction, and don't include 1=0. That may seem a littlesuspicious, but it's a perfectly valid proof of the consistency of PA(with the details filled in). Truth of statements in the language of PAcan be defined all right-- in ZFC.Since Con(PA) can be proven in ZFC, if it were also possible to proveCon(PA)->Con(ZFC) in ZFC, then it would be possible to prove Con(ZFC)in ZFC. But if it were, then there would be a contradiction in ZFC. That'sbecause by Goedel's second incompleteness theorem, any system that isstrong enough in a specific sense (satisfied by PA and ZFC) and provesits own consistency is inconsistent. (Someone already suggested thisargument in this thread. Perhaps they just didn't state it confidentlyenough? Well, it's correct.)The same holds for Con(PA)->Con(ZF), since it's possible to prove in PAor even less that Con(ZF)<->Con(ZFC). Con(ZFC)->Con(ZF) is obvious. Goedelproved that Con(ZF)->Con(ZFC), and his proof requires only elementaryassumptions.Now we have time for some questions.Shmuel (Seymour J.) Metz asked:|We need to distinguish three separate issues.|| 1. Is it possible to formulate all statements of ZF in PA, such that | proofs in ZF translate into proofs in PA?Yes, you could translate X as (an arithmetic encoding, like Goedel's,of) X is a theorem of ZF. That's sort of uninteresting, though.By (the proof of) Goedel's completeness theorem, to each recursivelyenumerable set of axioms, we can associate a computable binary tree,such that if the tree is finite the axioms are inconsistent, and ifthe tree is infinite, each infinite branch corresponds to a model. Thishas been used to show that there's always a model which can be definedin arithmetic. For instance, the leftmost infinite branch defines a model.If we apply this to a system like ZF (assuming it's consistent), the modelwe get is highly nonstandard; the statements of arithmetic true in themodel are definable by an arithmetic formula, so they are generallyincorrect. (The set of statements of arithmetic that are true cannot bedefined by a formula of arithmetic.) But it's a model of ZF at least. Wecould use this to translate the statements of ZF into statements of PA.Each theorem of ZF would translate into something whose truth follows inPA *from the assumption that ZF is consistent*.| 2. Using such a formulation, is it possible to prove that|con(PA)=>con(ZF)?Not in ZFC, unless ZFC is inconsistent. In a stronger system than ZFC,typically Con(ZF) is a theorem already, which makes it somewhat besidethe point to prove Con(PA)->Con(ZF).| 3. Did Kurt G.9adel actually publish such a proof and, if so, | in which paper.Naturally not.Goedel proved Con(ZF)->Con(ZFC) by defining a class called L, known asGoedel's class of constructible sets. The sets in L form a hierarchyindexed by ordinals. Once we have the sets L_b for b|I've been waiting for someone who really knows this stuff to speak>|up, but nobody has (or if anyone has, he's also included disclaimers>|about how he could be wrong...)>You know, it would have been a little more likely that I would've been>reading this if somebody could have renamed the thread something a little>more descriptive than rationals are uncountable. I mean, just because>someone decides to name a thread something like Yer Bleedin' Doggies>or Help or Proof of 2+2=5 doesn't mean it has to stay that way....Right. Otoh I just started a new thread with a question - I gave thethread an appropriate title. And you've answered the question Iasked in that thread here in _this_ thread! Just goes to show ya...>One can prove Con(PA) in ZFC. Someone mentioned Gentzen's proof (which>formalizes easily enough in ZFC). That's fine. The gist of it goes>like this. Gentzen reformulated PA in an equivalent form in sequent>calculus. A sequent is written as A1,...,An |- B1,...,Bm where the A_i>and the B_i are formulas (they can have free variables) and it stands>for the statement (A1&...&An)->(B1 or B2 or ... or Bm) (which in turn>means that this holds for all possible assignments to the free variables).>He made the axiom of induction into a deductive rule. The cute thing about>the sequent calculus is that almost all of the rules satisfy the so-called>subformula property: every subformula of sequents to which the rule is>applied is also a subformula of the consequent. So everything in the proof>is a subformula of the conclusion (possibly with variables renamed!),>unless there is an application of the cut rule: A1,...,An |- B1,...,Bm,B C1,...,Cr,B |- D1,...,Ds> ----------------------------------------------------> A1,...,An,C1,...,Cr |- B1,...,Bm,D1,...,Ds>in which the cut formula B disappears. The cut rule is very powerful>in a sense; the maximum number of steps required to write an n-step proof>without using cuts is some fast-growing function of n... which I forget.>On the other hand, cuts can be eliminated in principle. Much of the core>of Gentzen's proof is showing how to eliminate cuts from proofs in his>formulation of PA.>If cuts can be eliminated, then PA is consistent. A proof of a>contradiction amounts to a proof of |- in the system (with nothing>on either side of the turnstile). (As I recall, a proof of |- 1=0 can be>made a proof of |- in this system.) But there is of course no cut-free>proof of |-, since there are no subformulas of to work with.>People like to say Gentzen proved induction on epsilon_0 implies the>consistency of PA. That's because he used induction on epsilon_0 to prove>that his cut-elimination procedure terminates. Each proof-tree can be>assigned a different ordinal number step replaces a proof with a proof corresponding to a smaller ordinal.>It's possible to define epsilon_0 inside PA, but it's not possible to>prove that it's an ordinal. Everything in the proof is elementary enough>aside from the induction to epsilon_0.>What's epsilon_0? It's the limit of w, w^w, w^w^w, w^w^w^w, ... where w>(an ASCII approximation to lowercase greek omega) is the first infinite>ordinal. It's the smallest ordinal satisfying epsilon_0 = w^epsilon_0. So>what's w^a mean, for an ordinal a? The ordinals less than w^a can be>written in Cantor normal form, i.e. n1*w^a1+n2*w^a2+...+n_k*w^a_k where>a1>a2>...>a_k. You can compare them by using the fact that the term with>the highest power of w dominates. So ordinals under epsilon_0 can be>written as things like w^w^w^w^3 + w^w^(w^5+4*w^3+126) + 12*w^3 + 2.>Gentzen's proof is sort of overkill for this context, however. It's>valuable mainly for giving more information than just proving that the>consistency of PA follows from the axioms of ZFC. If you want an easier>proof in ZFC that PA is consistent, just use the fact that ZFC can prove>that PA has a model. You can define the set of true statements about the>natural numbers in ZFC, and prove that they include the axioms of PA, are>closed under deduction, and don't include 1=0. That may seem a little>suspicious, but it's a perfectly valid proof of the consistency of PA>(with the details filled in). Truth of statements in the language of PA>can be defined all right-- in ZFC.Fabulous. This seemed clear to me - I assumed I must beoverlooking something because if it were that simple thenwhy would people talk about what Gentzen did?Question: Why the C? Does either Gentzen's proof or theobvious one depend on AC?>Since Con(PA) can be proven in ZFC, if it were also possible to prove>Con(PA)->Con(ZFC) in ZFC, then it would be possible to prove Con(ZFC)>in ZFC. But if it were, then there would be a contradiction in ZFC. That's>because by Goedel's second incompleteness theorem, any system that is>strong enough in a specific sense (satisfied by PA and ZFC) and proves>its own consistency is inconsistent. (Someone already suggested this>argument in this thread. Perhaps they just didn't state it confidently>enough? Well, it's correct.)I stated it - the reason for the lack of confidence was I assumed Imust be confused about ZF |- Con(PA), for reasons mentionedabove. But you say I wasn't confused, I was confused about myconfusion. Fine.>The same holds for Con(PA)->Con(ZF), since it's possible to prove in PA>or even less that Con(ZF)<->Con(ZFC). Con(ZFC)->Con(ZF) is obvious. Goedel>proved that Con(ZF)->Con(ZFC), and his proof requires only elementary>assumptions.>Now we have time for some questions.>[...]>I think the main source of uncertainty that would follow from the>discovery of an inconsistency in conventional mathematics would come>simply from a sense that the mathematical community was not as reliable>as had been thought. There are some people here who don't seem to think themathematical community is all that reliable as it is...Overall I just get the sense that the damage would>be generally reparable, however, by some sort of reformulation.>Keith Ramsay************************David C. Ullrich === On Sun, 11 Jan 2004 13:42:49 -0500, Shmuel (Seymour J.) Metz>In <61706c80.0401100837.7618ab15@posting.google.com>, on 01/10/2004> at 08:37 AM, hack@watson.ibm.com (Michel Hack) said:>I enjoy Shmuel's postings -- usually he sets people straight. In>this case however I think he misremembers what Goedel did.>G.9adel did several things. One of them was to translate staements about>apparently stronger systems into statements about naturals. That lead,>among other things, to his incompleteness result, but I'm pretty sure>that he also proved con(PA) => con(ZF).Huh. That's a _far_ cry from what you've been saying, which is thathe _did_ prove this...You should really give a specific reference instead of just advisingpeople to study Godel. Or you could note the following (all ofwhich I'm pretty sure is correct): Con(PA) is a theorem of ZF,so if Godel proved that Con(PA) implies Con(ZF) then Con(ZF)is also a theorem of ZF. But Godel _did_ prove that if Con(ZF)is a theorem of ZF then ZF is inconsistent...************************David C. Ullrich <3ffd3edf$21$fuzhry+tra$mr2ice@news.patriot.net> <61706c80.0401100837.7618ab15@posting.google.com> <40019929$48$fuzhry+tra$mr2ice@news.patriot.net> === > You should really give a specific reference instead of just advising> people to study Godel.But, but, but... he *did* give a specific reference!,----[ <3ffd41a6$22$fuzhry+tra$mr2ice@news.patriot.net> ]| Well, it takes more machinery. Read a Scientific American papaer| called Goedel's Proof for background, then read On the Consistency| of the Axiom of Choice and the Generalized Continuum Hypotheses.`------ Jesse F. Hughes[Mathematical] society has evolved far enough away from mainstreamsociety that it has become rogue, and now is willing to push its needsagainst that of the majority. -- James S. Harris X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === In <20040111005239.29555.00002102@mb-m02.aol.com>, on 01/11/2004 at 05:52 AM, r3769@aol.com (R3769) said:>Yes, I agree the argument is flawed. What I can't figure out is how>theorems in ZFC translate into theorems of ZF, yet axioms can't be so>translated.Axioms can be translated. The translations may look highly artificial.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org <87smiudtrj.fsf@phiwumbda.org> <3ffadc1f$19$fuzhry+tra$mr2ice@news.patriot.net> <87smisrg0b.fsf@phiwumbda.org> <3ffd41a6$22$fuzhry+tra$mr2ice@news.patriot.net> <871xqa1n9s.fsf@phiwumbda.org> <40009ae6$11$fuzhry+tra$mr2ice@news.patriot.net> <87smim7rqf.fsf@phiwumbda.org> X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oftX-Terminate: SPA(GIS) === In , on 01/11/2004 at 06:40 AM, David C. Ullrich said:>I've been waiting for someone who really knows this stuff to speak>up, but nobody has (or if anyone has, he's also included disclaimers>about how he could be wrong...) So, although I'm not an expert, I'll>say what I think happened here.We need to distinguish three separate issues. 1. Is it possible to formulate all statements of ZF in PA, such that proofs in ZF translate into proofs in PA? 2. Using such a formulation, is it possible to prove thatcon(PA)=>con(ZF)? 3. Did Kurt G.9adel actually publish such a proof and, if so, in which paper.Most of the discussion has centered on question 1. And, indeed,discussion of question 2 would seem to be meaningless without firstdispensing with question 1. -- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === On Sun, 11 Jan 2004 18:59:41 -0500, Shmuel (Seymour J.) Metz>In , on 01/11/2004> at 06:40 AM, David C. Ullrich said:>I've been waiting for someone who really knows this stuff to speak>up, but nobody has (or if anyone has, he's also included disclaimers>about how he could be wrong...) So, although I'm not an expert, I'll>say what I think happened here.>We need to distinguish three separate issues.1. Is it possible to formulate all statements of ZF in PA, such that > proofs in ZF translate into proofs in PA?Of course it is. Jesse has given a trivial way to do this, whichdoesn't do what you actually meant. It's also well-knownand very easy to show (if one leaves out the details) thatwhat you meant by (1) is possible.> 2. Using such a formulation, is it possible to prove that>con(PA)=>con(ZF)?3. Did Kurt G.9adel actually publish such a proof and, if so, > in which paper.>Most of the discussion has centered on question 1. And, indeed,>discussion of question 2 would seem to be meaningless without first>dispensing with question 1. Question 1 has been dispensed with. Nobody has any doubts aboutquestion 1. But you're the only person who has expressed anyconfidence that the answer to question 2 is yes - nobody elsehas any idea what 1 has to do with 2. (You could try readingdon't see what 1 has to do with 2).I haven't read the entire thread carefully, but I don't recallseeing any response from you to the following. I thinkthat it's all true - which statement below seems falseto you:(i) Con(PA) is a theorem of ZF.(ii) So if Godel proved what you say then Con(ZF) is a theorem of ZF.(iii) Godel proved that if Con(ZF) is a theorem of ZF then ZF isinconsistent.************************David C. Ullrich === On Mon, 12 Jan 2004 06:51:16 -0600, David C. Ullrich>[...]>I haven't read the entire thread carefully, but I don't recall>seeing any response from you to the following. I think>that it's all true - which statement below seems false>to you:>(i) Con(PA) is a theorem of ZF.>(ii) So if Godel proved what you say then Con(ZF) is a theorem of ZF.>(iii) Godel proved that if Con(ZF) is a theorem of ZF then ZF is>inconsistent.Ah. Btw, Google for Godel, or rather Google for ZF Con(PA) doessuffice to verify the one bit above I was not absolutely certain of,at least to the extent that finding a statement in someone's onlinelecture notes proves it's correct. That would be (i). If youhttp://www.google.com/search?q=ZF+Con(PA)&hl=en&lr=&ie=UTF- 8&start=10&sa=N(Agent truncates the url at the '(' - put the whole thing into a webbrowser)one of the hits iswww.calvin.edu/~rpruim/courses/m381/F98/notes.ps, which says quite explicitly We shall show that ZF |- Con(PA).(ii) should be clear, and Google for Godel really _should_suffice for (iii).>David C. Ullrich************************David C. Ullrich <87smiudtrj.fsf@phiwumbda.org> <3ffadc1f$19$fuzhry+tra$mr2ice@news.patriot.net> <87smisrg0b.fsf@phiwumbda.org> <3ffd41a6$22$fuzhry+tra$mr2ice@news.patriot.net> <871xqa1n9s.fsf@phiwumbda.org> <40009ae6$11$fuzhry+tra$mr2ice@news.patriot.net> <87smim7rqf.fsf@phiwumbda.org> <4001e36d$17$fuzhry+tra$mr2ice@news.patriot.net> === > In , on 01/11/2004> at 06:40 AM, David C. Ullrich said:>I've been waiting for someone who really knows this stuff to speak>up, but nobody has (or if anyone has, he's also included disclaimers>about how he could be wrong...) So, although I'm not an expert, I'll>say what I think happened here.We need to distinguish three separate issues. 1. Is it possible to formulate all statements of ZF in PA, such that > proofs in ZF translate into proofs in PA? 2. Using such a formulation, is it possible to prove that> con(PA)=>con(ZF)? 3. Did Kurt G.9adel actually publish such a proof and, if so, > in which paper.Most of the discussion has centered on question 1. And, indeed,> discussion of question 2 would seem to be meaningless without first> dispensing with question 1. (1) is trivial as stated. Translate every statement of ZF as 0=0.But the fact that you're now doubting (1) suggests that David is beingtoo charitable in his guess where things went wrong.-- Jesse F. Hughes I'm better than you, and you know it. -- James Harris <87smiudtrj.fsf@phiwumbda.org> <3ffadc1f$19$fuzhry+tra$mr2ice@news.patriot.net> <87smisrg0b.fsf@phiwumbda.org> <3ffd41a6$22$fuzhry+tra$mr2ice@news.patriot.net> <871xqa1n9s.fsf@phiwumbda.org> <40009ae6$11$fuzhry+tra$mr2ice@news.patriot.net> <87vfni7sqx.fsf@phiwumbda.org>X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === In <87vfni7sqx.fsf@phiwumbda.org>, on 01/11/2004 at 10:13 AM, jesse@phiwumbda.org (Jesse F. Hughes) said:>Son, you shouldn't be insulting. -- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org <87smiudtrj.fsf@phiwumbda.org> <3ffadc1f$19$fuzhry+tra$mr2ice@news.patriot.net> <87smisrg0b.fsf@phiwumbda.org> <3ffd41a6$22$fuzhry+tra$mr2ice@news.patriot.net> <871xqa1n9s.fsf@phiwumbda.org> <40009ae6$11$fuzhry+tra$mr2ice@news.patriot.net> <87vfni7sqx.fsf@phiwumbda.org> <4001e487$18$fuzhry+tra$mr2ice@news.patriot.net> === > In <87vfni7sqx.fsf@phiwumbda.org>, on 01/11/2004> at 10:13 AM, jesse@phiwumbda.org (Jesse F. Hughes) said:>Son, you shouldn't be insulting. >Was my tone insulting? Maybe, but not intentionally.In any case, the big and important difference is that I never toldanybody that what I was saying is such a well-known fact that he canI assume you've since googled? Any luck?right? Had you actually *read* it in the past, or had you decidedthat was just a good guess for where this result might be?-- ...you are around so that I have something else to do when I'm notfiguring something important out. I was especially intrigued on thisiteration by cursing, which I think I'll continue at some later dateas it's so amusing. --- James S. Harris <87smiudtrj.fsf@phiwumbda.org> <3ffadc1f$19$fuzhry+tra$mr2ice@news.patriot.net> <87smisrg0b.fsf@phiwumbda.org> <3ffd41a6$22$fuzhry+tra$mr2ice@news.patriot.net> <871xqa1n9s.fsf@phiwumbda.org> <40009ae6$11$fuzhry+tra$mr2ice@news.patriot.net> <87smim7rqf.fsf@phiwumbda.org>X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === In <87smim7rqf.fsf@phiwumbda.org>, on 01/11/2004 at 10:35 AM, jesse@phiwumbda.org (Jesse F. Hughes) said:>Maybe you should be less condescending PKB. What does the consistency of the axiom of choice have to do withthis? was a condescending question.>How is it that, of all the participants in this thread, you are the>only one who claims Goedel proved Con(PA) -> Con(ZFC)? This is an>honest question. Either I am mistaken or I am not. Neither is grounds for>This result, if true, is surely *at least* as>important and interesting as the consistency of the axiom of choice.Perhaps; I regard the AOC as an importatnt working tool, and it'sconsistency as being rather important.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === On Sun, 11 Jan 2004 19:08:50 -0500, Shmuel (Seymour J.) Metz>In <87smim7rqf.fsf@phiwumbda.org>, on 01/11/2004> at 10:35 AM, jesse@phiwumbda.org (Jesse F. Hughes) said:>[...]>How is it that, of all the participants in this thread, you are the>only one who claims Goedel proved Con(PA) -> Con(ZFC)? This is an>honest question. >Either I am mistaken or I am not. Neither is grounds forHow is this a misrepresentation? You've been claiming thatGodel proved Con(PA) -> Con(ZF). It's well known that he_did_ prove Con(ZF) -> Con(ZFC).>This result, if true, is surely *at least* as>important and interesting as the consistency of the axiom of choice.>Perhaps; I regard the AOC as an importatnt working tool, and it's>consistency as being rather important.************************David C. Ullrich <87smiudtrj.fsf@phiwumbda.org> <3ffadc1f$19$fuzhry+tra$mr2ice@news.patriot.net> <87smisrg0b.fsf@phiwumbda.org> <3ffd41a6$22$fuzhry+tra$mr2ice@news.patriot.net> <871xqa1n9s.fsf@phiwumbda.org> <40009ae6$11$fuzhry+tra$mr2ice@news.patriot.net> <87smim7rqf.fsf@phiwumbda.org> <4001e592$19$fuzhry+tra$mr2ice@news.patriot.net> === > In <87smim7rqf.fsf@phiwumbda.org>, on 01/11/2004> at 10:35 AM, jesse@phiwumbda.org (Jesse F. Hughes) said:>Maybe you should be less condescending PKB. What does the consistency of the axiom of choice have to do with> this? was a condescending question.It was *not* intended as a condescending question. You werethe consistency of the axiom of choice. It is a natural and honestquestion: what *does* the consistency of the axiom of choice have todo with Con(PA) -> Con(ZFC)?>How is it that, of all the participants in this thread, you are the>only one who claims Goedel proved Con(PA) -> Con(ZFC)? This is an>honest question. Either I am mistaken or I am not. Neither is grounds forapologize.>This result, if true, is surely *at least* as>important and interesting as the consistency of the axiom of choice.Perhaps; I regard the AOC as an importatnt working tool, and it's> consistency as being rather important.No doubt it's important. But, if Con(PA) -> Con(ZFC), that would berather important, too, perhaps even more important, because it wouldtell us that our set theory (including, say, the axiom of infinity) isjust as reliable as our theory of natural numbers.-- Jesse HughesYes, I'm one of those arrogant people who tries to be quotable.There is actually at least one person who quotes me often. -- James Harris === >Rationals are Uncountable>Then how do you explain the numerous _injective_ mappings one may>construct from Q to N?I can't.Maybe set theory is inconsistent.>For example:>An arbitary rational may be represented uniquely as p/q, where, WLOG, we>may require that:> p is an integer, and> q is a positive integer, and> p and q have no common factors except 1 and> if p = 0 then q = 1>Define f : Q -> N by> f(p,q) = 2^p*5^q if p is positive or zero, and> f(p/q) = 3^(-p)*5^q if p is negative.>It is trivial to show that f is injective, thus the cardinality of the>rationals cannot be larger than that of the naturals.My proof is easily modified to show that the naturals are uncountable.Showing that the natural numbers and the rationals have the samecardinality doesn't help.>And, of course, the trivial injective mapping g:N -> Q : n -> n/1>completes the issue.Not really.> We already know the rationals are countable. The idea is to find the> defect(s) in his proof.Turing gives a proof that the diagonal of all computable numbers can notbe computable. He states:8. Application of the diagonal process.It may be thought that arguments which prove that the real numbers are notenumerable[5] would also prove that the computable numbers and sequencescannot be enumerable . It might, for instance, be thought that the limit ofa sequence of computable numbers must be computable. This is clearly onlytrue if the sequence of computable numbers is defined by some rule.Or we might apply the diagonal process. If the computable sequences areenumerable, let In be the n-th computable sequence, and let Yn(m) be them-th figure in In. Let J be the sequence with 1 - Yn(n) as its n-th figure.Since J is computable, there exists a number K such that 1 - Yn(n) = YK(n)all n. Putting n = K, we have 1 = 2YK(K), i.e. 1 is even. This isimpossible. The computable sequences are therefore not enumerable.The fallacy in this argument lies in the assumption that J is computable. Itwould be true if we could enumerate the computable sequences by finitemeans, but the problem of enumerating computable sequences is equivalent tothe problem of finding out whether a given number is the D.N of acircle-free machine, and we have no general process for doing this in afinite number of steps. In fact, by applying the diagonal process argumentcorrectly, we can show that there cannot be any such general process.I find it interesting that Turing shows set theory is inconsistent (1 iseven) if itis possible to compute a diagonal like number for the set of allcomputable numbers.I have shown that even finding a diagonal like number for the set of allnatural numbers is enough for set theory to be inconsistent.Russell- 2 many 2 count === >Rationals are Uncountable> Then how do you explain the numerous _injective_ mappings one may> construct from Q to N?> I can't.> Maybe set theory is inconsistent.There is a good deal better evidence that you are.> For example:>An arbitary rational may be represented uniquely as p/q, where, WLOG, we> may require that:> p is an integer, and> q is a positive integer, and> p and q have no common factors except 1 and> if p = 0 then q = 1>Define f : Q -> N by> f(p,q) = 2^p*5^q if p is positive or zero, and> f(p/q) = 3^(-p)*5^q if p is negative.>It is trivial to show that f is injective, thus the cardinality of the> rationals cannot be larger than that of the naturals.> My proof is easily modified to show that the naturals are uncountable.Since the definition of countability requires that the naturals be countable, how are you going to do that?> Showing that the natural numbers and the rationals have the same> cardinality doesn't help.It certainly doesn't help your thesis, but it proves the set of rationals satisfies the definition of countability.> And, of course, the trivial injective mapping g:N -> Q : n -> n/1> completes the issue. Not really.Yes really. At least to all those who have a clue.>We already know the rationals are countable. The idea is to find the>defect(s) in his proof.> Turing gives a proof that the diagonal of all computable numbers can not> be computable. [snipped as irrelevant to the countablility of the rationals]> === > In <97adneZXNdeRbWCi4p2dnA@comcast.com>, on 01/08/2004> at 04:04 PM, Russell Easterly said:>There is no theoretical limit on how fast something can be computed.There is, however, a theoretical limit one whether something that> doesn't satisfy the definition of a TM is a TM.I can program a Turing machine to perform the operations I describe.In <462dnZeiCeE1bGCi4p2dnA@comcast.com>, on 01/08/2004> at 04:11 PM, Russell Easterly said:>I am assuming that a TM can perform an infinite>number of operations in finite time.Such a device might have a role in the theory of computation, but it> would not be a turing machine and theorems about Turing Machines would> not apply to it.I only make this assumption to satisfy those that claim the TM willnever produce an output. I think this argument is a cop out.If we assume a TM can only perform a finite number ofoperations then my claims are trivially true.The rationals must be uncountable for a system that can onlyperform a finite number of computations.If a TM can only perform a finite number of operations thenno TM can compute PI or any other irrational.Even some base 2 rational numbers would be uncomputable.A finite TM could never compute the set of all natural numbers.>It appears that induction and infinite computation come to different>conclusions on some problems.No. The different conclusions come from applying things to contexts> where they are inapplicable.Finding the largest member of a set seems applicable to what isessentiially Cantor's diagonal proof applied to the natural numbers.Russell- 2 many 2 count === >I am assuming that a TM can perform an infinite>number of operations in finite time.>Such a device might have a role in the theory of computation, but it>would not be a turing machine and theorems about Turing Machines would>not apply to it.> I only make this assumption to satisfy those that claim the TM will> never produce an output. I think this argument is a cop out.> If we assume a TM can only perform a finite number of> operations then my claims are trivially true.> The rationals must be uncountable for a system that can only> perform a finite number of computations.But countability is not defined in terms of TMs, so how are TMs relevant to determining countability?> If a TM can only perform a finite number of operations then> no TM can compute PI or any other irrational.> Even some base 2 rational numbers would be uncomputable.> A finite TM could never compute the set of all natural numbers.Exactly!>It appears that induction and infinite computation come to different>conclusions on some problems.>No. The different conclusions come from applying things to contexts>where they are inapplicable.> Finding the largest member of a set seems applicable to what is> essentiially Cantor's diagonal proof applied to the natural numbers.Only in the sense that they are both like looking in a dark room for a black cat that isn't there. It is the failure to find what isn't there that is significant in all three cases. === > Any non-negative integer can be written as a finite sum > a1*1! + a2*2! + a3*3! + ... + an*n! > where each ak is an integer such that 0 <= ak <= k. If trailing zeroes > are removed from the sum, this representation is unique. (Note that the > case of 0 ends up with zero terms.)Right. > Any rational in [0,1) can be written as a finite sum > b1/2! + b2/3! + b3/4! + ... + bn/(n+1)! > where each bk is an integer such that 0 <= bk <= k. If trailing zeroes > are removed from the sum, this representation is unique. (Note that the > case of 0 ends up with zero terms.) > This proof is trickier. It would probably be necessary to use induction > on n to show that all multiples of 1/(n+1)! in [0,1) can be represented > uniquely (up to trailing zeroes) with at most n terms, and note that any > rational in [0,1) has some factorial that can be used as a denominator. > (Uniqueness would require either induction on the trailing terms or a > combinatorial argument.)I think it can be done easier, because we do not actually need uniqueness.Start with finding the smallest integer n such that for the rational r,r * (n + 1)! is integral. Write r as cn/(n + 1)!. As we can write cn ascn/(n+1) + cn%(n+1) (/ is integer division, % is remaindering), we cannow define c_n-1 = cn/(n+1) and bn = cn%(n+1), and we haver = c_n-1/n! + bn/(n+1)!. bn satisfies the requirement and c_n-1 < n!.Iterate through, we end with some c1 < 2!. This algorithm gives foreach rational in the range a unique representation of the desired form,and that is all we need (that there are possibly other representationswith the same value is irrelevant). > And then the bijection consists of transferring the coefficients from > one sum to the other.A bijection is not needed. For a well-ordering we only need an injectioninto a well-ordered set. Let's have a set S and a well-ordered set W.Let us also have an injection f: S -> W. We say s1 < s2 when f(s1) < f(s2).This gives a well-ordering on S. Because given any subset T of S, we canlook at f(T). That one has a first element, say w. So the first elementof T is the pre-image of w (which exists and is in T). And as far as Ican see, well-ordering was all that Russell claimed.That the representation is actually unique is indeed a bit trickier.The simplest way to do that is by showing that the difference of twonumbers in the representation can only be 0 when the difference ofall coefficients is 0.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === > A bijection is not needed. For a well-ordering we only need an injection> into a well-ordered set.Oh, right, because any subset of a well-ordered set is well-ordered withthe same ordering, and we can use the range of the injection to get abijection.> And as far as I> can see, well-ordering was all that Russell claimed.Well, he started with a list of the rationals in [0,1), i.e., a mappingfrom the naturals to those rationals. By the time you prove that youcan reverse that to get an injective mapping, you have proven abijection. > That the representation is actually unique is indeed a bit trickier.> The simplest way to do that is by showing that the difference of two> numbers in the representation can only be 0 when the difference of> all coefficients is 0.Yes, there are a couple variations on this, one of which is based oninduction on the number of potentially different trailing terms.-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Base ! has two representations for any rational just like fixed radix bases.0.123456... base ! = 1.0... base !Only one of these representations has a finite number of non-zerocoefficients.Russell- 2 many 2 count === > Base ! has two representations for any rational just like fixed radix bases. > 0.123456... base ! = 1.0... base ! > Only one of these representations has a finite number of non-zero > coefficients.Eh? My posting was about numbers in the interval [0,1). So I have *no*idea how your post here relates to that. Pray find the two representationsof 1/2 in base !.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === > Eh? My posting was about numbers in the interval [0,1). So I have *no*> idea how your post here relates to that. Pray find the two representations> of 1/2 in base !.That would be 0.1 and 0.023456...This is comparable to the 0.50000... and 0.49999... issue with base 10.You can reduce the last digit by 1 (borrowing as needed), use themaximum possible value for each of the infinite quantity of digits afterthat, and you have a Cauchy sequence. (The convergence is based on 1/n!rather than on 0.1^n .)This is why I made a point of specifying that each sequence ofcoefficients was finite. The main point is that you don't have to worryabout two infinite representations being equal if you can exclude allinfinite representations a priori. -- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === > Base ! has two representations for any rational just like fixed radixbases. 0.123456... base ! = 1.0... base !> Only one of these representations has a finite number of non-zero> coefficients.Eh? My posting was about numbers in the interval [0,1). So I have *no*> idea how your post here relates to that. Pray find the tworepresentations> of 1/2 in base !.Sorry.My comment was about base !, not about the injection.1/2 =0.1000... base !0.0234567... base !This is equivalent to saying:0.5000... base 10 = 0.4999... base 10Russell- 2 many 2 count === > My latest brainchild, as minor as most of the others, is this> mini-theorem. Let s_0 and s_1 be nonnegative integers, not both zero.> Define inductively s_{n+2} = | s_{n+1} - s_n |> for n >= 0. Show that the sequence (s_n) eventually looks like> 0, x, x, 0, x, x, 0, x, x, ...> and x = gcd(s_0,s_1).> For example, if s_0 = 69 and s_1 = 39, the sequence runs> 69, 39, 30, 9, 21, 12, 9, 3, 6, 3, 3, 0, 3, 3, 0, ...I think this is fairly easy to verify,as the sequence is bounded and so must become periodic.In fact s_n >= 0 for all n, ands_n <= max(s_{n-1},s_{n-2}).It follows thatmax(s_{n+1},s_n} <= max(s_{n-1},s_{n-2})with strict inequality unless some s_n = 0.Hence some s_n = 0, and your conclusion follows.-- Timothy Murphy tel: +353-86-2336090, +353-1-2842366 === Timothy Murphy> My latest brainchild, as minor as most of the others, is this>mini-theorem. Let s_0 and s_1 be nonnegative integers, not both zero.>Define inductively s_{n+2} = | s_{n+1} - s_n |>for n >= 0. Show that the sequence (s_n) eventually looks like>0, x, x, 0, x, x, 0, x, x, ...>and x = gcd(s_0,s_1).I think this is fairly easy to verify,> as the sequence is bounded and so must become periodic.> In fact s_n >= 0 for all n, ands_n <= max(s_{n-1},s_{n-2}).It follows thatmax(s_{n+1},s_n} <= max(s_{n-1},s_{n-2})with strict inequality unless some s_n = 0.> Hence some s_n = 0, and your conclusion follows.Yes, it's not tough. I started with the observation that gcd(s_n,s_{n+1}) isconstant, rather than the fact that (s_n) is bounded. Then it goes much thesame way.LH === after many hours in the library without big successes, I hope you can help me.I have two problems:1.) Given a Galton-Watson-tree it's a well known theorem that the population will die out, if the expected numer of descendants is less than one. But are there any theorems about the distribution of the distinction in time? I mean, can we say anything about the probability that a population dies out after n generations? (Of course, the distribution of the number of descendants is given.)2.) The Propp-Wilson-algorithm (coupling from the past - CFTP) samples exactly on a finite space. But what about infinite (discrete) spaces? I know from the construction of the algorithm that it doesn't work without any modifications, but I was quite suprised I haven't even found CFTP the infinite space was the next logical step for me.I'm looking forward to your ideas and advices.Greetz, Alex === >It took about 22 seconds to print out 1,000,000 primes. How much faster is>it able to be?> also the program produces 10,000,000 primes in 22.27 seconds> The program stores all the primes in a bit array and can be accessed to see if it is prime or not.Rodriguez answer:My idea is to print in the screen the primes as points or astherics inthe 48 columns that can contain primes from p = 11 to 211 includingthe colums under 121,143,169,187 and 209.Then to frame hypothesis about the impossible configurations. Forexample:The maximum consecutive chain in columns is 10 because that chain isthe maximum arithmetical progression with difference = 210.Now. What is the longest row with consecutive points?If someone wants the program presented as a game, please write me andI send my game STARRY.EXE. In order to avoid any problem I'll send thefile STARRY wi1thout .EXEX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0C2eos09616; === >It took about 22 seconds to print out 1,000,000 primes. How much faster is>it able to be?> also the program produces 10,000,000 primes in 22.27 seconds>I think you will find that print out is what takes the time. The program stores all the primes in a bit array and can be accessed to see > if it is prime or not.>I guess the question is then: Print to where? At that time is must be the screen, however, anoher question then comes up, print them to the screen for what reason? === >I am trying to write a piece of software quite like Granular Synthesis>Demo 1 in http://www.dcs.gla.ac.uk/~jhw/audioclouds/ . > Hmm, that looks like a very interesting project! It looks difficult> but well worth solving -- that's what I call a good problem.>I was wondering if anyone could explain the difference between a>probabilty distribution and a probabilty density?> Well, a probability distribution is essentially a function that > tells you how much stuff is within a given region. It turns out> that a very convenient way to construct such functions is to work> instead with the probability density -- that's a function that says> here the stuff is piled up so deep, and there it's so deep, and> stuff in a region -- voila, the probability distribution.>And if you have a probabilty distribution can a density be obtained>from it and if so how?> Yes, you can get the density by differentiating the distribution.> Whether it's easier to go from density to distribution or vv depends> on the details of the problem. >Also the main bit i am stuck on is how a density can be conditioned>(in the example given it is conditioned by the x and y positions of a>mouse) how would you go about doing such a thing.> By conditional on the mouse position they just mean that the > density is a function of the mouse position -- conceptually there's> a different density for every mouse position. Of course, for > position that are close together, the densities will be very similar;> the trick is to figure out just how the density should vary. so-called Gaussian mixtures to model the probability density of > different sounds. You'll find a lot of stuff on the web if you > search for Gaussian mixture.> I disagree with the other posters who say you need to study up on> probability theory before getting started. I think you should spend> some time working up a preliminary solution, then go study, then > come back and throw away the preliminary solution and start over:> you won't recognize the answer if you don't know what is the question.> For what it's worth,> Robert Dodiermixtureshow to you obtain a gaussian mixture if you already have a number ofgaussian probabilty distributions?the stuff on the web that i have found doesnt explain it very well. Isthere any books or sites that you could recommend? === > mixtures> how to you obtain a gaussian mixture if you already have a number of> gaussian probabilty distributions?> the stuff on the web that i have found doesnt explain it very well. Is> there any books or sites that you could recommend?A good book is Finite Mixture Models by Geoffrey McLachlan and David PeelWiley (2000)A site with links to mixture software ishttp://www.maths.uq.edu.au/~gjm/other.html . === >I am interested in the sequence a^2, a^3, a^4, ...>for a = .9, a = .99, a = .999, through a = .999... You'll have to specify what you mean by a=.9, a=.99, a=.999, througha=.999.... If you'd said a=.9, a=.99, a=.999, through a=.99999999I would have understood you, but the sequence .9, .99, .999, ... doesn'tinclude .999..., so it's like saying a=1, a=2, a=3, through a=-7.For each of a=.9, a=.99, a=.999, ... the sequence a^2, a^3, a^4, ...tends to zero, but for a=.999... it is a the constant sequence 1, 1, ...-- Richard-- FreeBSD rules! === Pardon my simplistic view but there truly seems to be a connectionbetween everything that exists in nature, and to a good extent,everything created by man that is perceived to be pleasant to the eyeor ear. Everything in nature appears to have some connection to Piand/or phi (Fibonacci sequence - Golden Ratio). There also seems tobe a link between double toroids, fractals, and images created basedon Phi. Two toroids linked like chain links form the image ofsplitting cells, two hydrogen atoms and, I'll bet my last buck, thestructure of the universe.Well with the big mystery solved what do we do next? === What the connection is and why it exists? ;)There's always work to be done. === > so, when he says that they are the same (dasselbe), he is wrong?I have seen that passage translated as the same as and as just. Godel did have a tendency of being a bit informal in the introductoryparts of his paper. Maybe someone who speaks German can help? All Ican say is that it is a number vs. a sequence consisting of only thatnumber.Charlie Volkstorf === I'm writting a program that will try to solve function using Newtons method.Can You please look at my program algorith draft and check if everyting is correct / corract mistakes if any?For 1-dimensional, we got F(x)=... we got x[k] (k-th step of iteration)and we want to calculate x[k+1].If x[0] is chosen correctly then for big k, F(x[k]) will be close to 0.0 lim k->oo F(x[k]) =~ 0The basic formula is: x[k+1] = x[k] + F(x)/F'(x)F'(x) is gradient / derrivent (is it good English name?) or F in point x. I use discrete forward derrivent, F'(x) =(F(x)+F(x-h))/h where h is very smallSo finaly formula for 1-D function would be x[k+1] = x[k] + F(x) / ( (F(x)+F(x+h))/h )Btw, what is formula for inverting 2-d wectorw = | a b | | c d |w^-1 = ... ?TIA-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~l-.~~~~~~~~~~~~~~~~~~~GG- 1175498 ____| ]____, Rafal 'Raf256' Maj X-( * )Rafal(at)Raf256(dot)com ,---------- === > I'm writting a program that will try to solve function using Newtons> method.> Can You please look at my program algorith draft and check if everyting is> correct / corract mistakes if any?For 1-dimensional, we got F(x)=... we got x[k] (k-th step of iteration)> and we want to calculate x[k+1].If x[0] is chosen correctly then for big k, F(x[k]) will be close to 0.0> lim k->oo F(x[k]) =~ 0The basic formula is:> x[k+1] = x[k] + F(x)/F'(x)No, x[k+1] = x[k] - F(x)/F'(x)Please don't post in HTML.-- Clive Toothhttp://www.clivetooth.dk === TheLastDanishPastry@yahoo.com > x[k+1] = x[k] + F(x)/F'(x)> No, x[k+1] = x[k] - F(x)/F'(x)also for> 1 | d -b |> w^(-1) = ------- | |> ad - bc | -c a |to Toni Lassila. Btw, what does it mean that matrix is singular - it conatins zero, or all zeros? How can we calculate inverse of singular matrix (or is it an invalid operation)?> Please don't post in HTML.I posted in plain-text,-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~l-.~~~~~~~~~~~~~~~~~~~GG- 1175498 ____| ]____, Rafal 'Raf256' Maj X-( * )Rafal(at)Raf256(dot)com ,---------- === > TheLastDanishPastry@yahoo.com> x[k+1] = x[k] + F(x)/F'(x)>No, x[k+1] = x[k] - F(x)/F'(x)>also for> 1 | d -b |> w^(-1) = ------- | |> ad - bc | -c a |to Toni Lassila. Btw, what does it mean that matrix is singular - it> conatins zero, or all zeros? How can we calculate inverse of singular> matrix (or is it an invalid operation)?The short answer is that a 2x2 matrix is singular iff ad-bc = 0.Singular means that there is no inverse (although severalgeneralized inverses exist). === >Btw, what is formula for inverting 2-d wector>w = | a b |> | c d |>w^-1 = ... ?If w is a non-singular 2x2 matrix then: 1 | d -b |w^(-1) = ------- | | ad - bc | -c a | === I'm trying to generalize Newtons method for n-dimensional functions.The formula is thenw[k+1] = w[k] + F(w[k]) / F'(w[k])where F' is an gradient (or derrivential?) of F.w[k+1] = w[k] + F(w[k]) * (F'(w[k])) ^ -1w[k+1] = w[k] + F(w[k]) * A ^ -1now, how do I calculate A - the vector of gradient? Is it a vector, scalar or matrix?AFAIR it was something like, assuming w = [x,y,z]A = F(x+h,y ,z ) - F(x,y,z) / h, F(x ,y+h,z ) - F(x,y,z) / h F(x ,y ,z+h) - F(x,y,z) / hYes?W is a vector, F(w[k]) is a... scalar? and A is... matrix?-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~l-.~~~~~~~~~~~~~~~~~~~GG- 1175498 ____| ]____, Rafal 'Raf256' Maj X-( * )Rafal(at)Raf256(dot)com ,---------- === > Greetings math lovers,> I was given the following probability problem as part of a> job-interview pre-screen. After giving a totally wrong answer ('cause> I wasn't thinking), I gave the following answer. However, the> interviewer said that my answer was still incorrect.> Fair enough; I find probability unintuitive. But now my curiosity is> peaked: What is the right answer? > Question: You have a jar with 1000 coins in it, 999 are normal, but> one is two-headed (i.e both sides are heads). You choose a coin at> random and toss it 10 times. With each toss the coin comes up heads,> so you get 10 consecutive heads. What is probability that you have> choosen the two-headed coin? > My answer:> P(two headed) = 1 - P(normal coin *and* 10 heads in a row)> = 1 - .999 * (1/2)^10> = 1 - .999 * (1/1024)> = 1 - .000976> = 0.999024> = 99.9024%> If this is wrong, what is the right answer? How do you calculate it?> StuartHere is yet another solution method, which does not (explicitly) use Bayes' Formula.Let:P1 = prob of choosing two-headed coin and getting 10 heads = 1/1000P2 = prob of choosing fair coin and getting 10 heads = (999/1000)*(1/1024)PA = conditional prob of two-headed coin given 10 heads (unknown)PB = conditional prob of fair coin given 10 heads (unknown)Then solve the following system of equations for PA and PB:PA + PB = 1PA/PB = P1/P2This gives PA = 0.5062, as advertised.Tom Larson === >Greetings math lovers,>I was given the following probability problem as part of a>job-interview pre-screen. After giving a totally wrong answer ('cause>I wasn't thinking), I gave the following answer. However, the>interviewer said that my answer was still incorrect.>Fair enough; I find probability unintuitive. But now my curiosity is>peaked: What is the right answer? >Question: You have a jar with 1000 coins in it, 999 are normal, but>one is two-headed (i.e both sides are heads). You choose a coin at>random and toss it 10 times. With each toss the coin comes up heads,>so you get 10 consecutive heads. What is probability that you have>choosen the two-headed coin? >My answer:>P(two headed) = 1 - P(normal coin *and* 10 heads in a row)> = 1 - .999 * (1/2)^10> = 1 - .999 * (1/1024)> = 1 - .000976> = 0.999024> = 99.9024%>If this is wrong, what is the right answer? How do you calculate it?>Stuart>Nice math, Stuart, but wrong.>Instead of saying there is one two-headed coin,>say there is one red coin. >If you pick one coin at random out of 1000 (without>looking), the probability that you picked the red>coin is 1/1000.>You can toss the coin all day long and this doesn't>change.>Get it?True Stuart's math is wrong, but your example is not an accuraterepresentation of the problem. The question is not pick a coin atrandom; what is the chance that it is the two-headed coin? In thatquestion, we are not given the same information as pick a coin atrandom; flip it ten times; it comes up heads each time; what is thechance that it is the two-headed coin? Information changes estimatesof probability.It is true that, once picked, flipping the coin does not change itstwo-headedness, but noting the outcome of those flips does change ourestimates of which coin we picked. Consider the problem pick a coinat random; flip it once; it comes up tails; what is the chance that itis the two-headed coin? Certainly flipping the coin does not changeits identity, but noting the outcome of that one flip definitely saysthat the probability that we have chosen the two-headed coin is 0 andnot 1/1000.There are two ways that we can see the behaviour seen in the originalproblem:1. We pick the two-headed coin and flip it 10 times getting all heads.2. We pick a normal coin and flip it 10 times getting all heads.Out of 1024000 trials, we would expect to see 1024 instances of thebehaviour seen in 1 since the probability of choosing the two-headedcoin is 1/1000 and the probability of getting 10 heads in a row withthat coin is 1.Out of those same 1024000 trials, we would expect to see 999 instancesof the behavior seen in 2 since the probability of choosing a normalcoin is 999/1000 and the probability of getting 10 heads in a row withthat coin is 1/1024.Thus, out of those 1024000 trials, we expect 1024 of them to have beenthe result of having picked the two-headed coin and 999 of them to havebeen the result of having picked a normal coin. Therefore, given thatwe are in one of those two cases, the probability that we have pickedthe two-headed coin is 1024/(1024+999) = 1024/2023.Rob Johnson take out the trash before replying === [.snip.]>I agree that such an argument might not go through in the Harris>ring, but might it not fail for a more basic reason? It is not>obvious that his ring would be a B'ezout domain.>That is something I am also wondering about. If it is a ring between the algebraic integers and the algebraicnumbers then I think it will be. But whether the ring satisfies thatcondition is by no means clear.Let D be a Bezout domain (an integral domain in which any finitelygenerated ideal is principal), let K be its field of fractions, andlet R be a ring contained between D and K. Let x and y be elements ofR. We want to show that the ideal R(x,y) is principal.Since x,y in K, there exist a,b,c,d in D such that x = a/b andy=c/d. Moreover, since D is Bezout, we may assume thatD(a,b)=D(c,d)=(1). Let D(ad,cb) = D(g). The claim is that R(x,y) = R(g/bd).First, we prove that g/bd lies in R(x,y). Note thatx = a/b = (ad)/(bd)y = c/d = (cb)/(bd)There exist r, s, in D such that r(ad) + s(cb) = g. Therefore,rx + sy = r(ad)/bd + s(cb)/bd = g/(bd); therefore, g/bd lies inR(x,y).Conversely, to prove that R(x,y) is contained in R(g/bd), it is enoughto show that both x and y are R-multiples of g/bd. And indeed: sincead,cb lie in D(g), there exist u,v in D such that ad=gu, cb=gv, sou*(g/bd) = (ug)/(bd) = (ad)/(bd) = a/b = xv*(g/bd) = (vg)/(bd) = (cb)/(bd) = c/d = y.Therefore, R(x,y) is principal. Therefore, R is a Bezout domain. So any ring between the ring of algebraic integers and the ring/fieldof algebraic numbers will be a Bezout domain.-- === ============================================ === ===It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) === ======================================= === =========Arturo Magidinmagidin@math.berkeley.edu===> I don't know if you noticed or not, but I've been in an argument that> has gone on for quite some time about a problem with the ring of> algebraic integers. Luckily for me, I've found the way to end the> arguing, by relying on your understanding of *independent* variables. > Note that you have x and y below where x is INDEPENDENT of y, and vice> versa. My hope is that such a simple concept that should be familiar> to all, will end the arguing.> I've been using a modification of an example put forward by Rick> Decker, a professor at Hamilton College.> Consider,> 7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7(xy - y^2)(5) + 7^2 y^2> so> (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) > where the a's are roots of > a^2 - (x - y)a + 7(x^2 + xy).> I have a challenge for you. Factor the quadratic539*x^2+462*xy+198*y^2 using your method. Your method IS applicableto this polynomial. === >I don't know if you noticed or not, but I've been in an argument that>has gone on for quite some time about a problem with the ring of>algebraic integers. Luckily for me, I've found the way to end the>arguing, by relying on your understanding of *independent* variables. >Note that you have x and y below where x is INDEPENDENT of y, and vice>versa. My hope is that such a simple concept that should be familiar>to all, will end the arguing.>I've been using a modification of an example put forward by Rick>Decker, a professor at Hamilton College.>Consider,>7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7(xy - y^2)(5) + 7^2 y^2>so>(5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) >where the a's are roots of a^2 - (x - y)a + 7(x^2 + xy).> I have a challenge for you. Factor the quadratic> 539*x^2+462*xy+198*y^2 using your method. Your method IS applicable> to this polynomial.>James is AFK temporarily. However, I don't understand the purposeof your question. Just replace my 7 and 5 by 11 and 7, respectively.Everything then goes through as before, save that one + becomes a -.Rick === On Sun, 11 Jan 2004 18:47:23 +0000, John Roberts-Jones>...>If c and 7 are coprime in A then there are elements p and q in A>such that cp + 7q = 1.>Be careful here. In James' all-inclusive ring of two complex conjugates,>one can be a unit while the other is not a unit. A bit strange, and I>do not know whether his ring really does exists, but the proof>fails for that. Because if c is a unit, q = 0 and p = inv(c). But>p' = inv(c)' does not necessarily belong to the ring (conjugation is>not a ring operation).> So 1 = 1' = (cp + 7q)' = (cp)' + (7q)' = c'p' + 7'q' = c'p' + 7q',>and since p' and q' are in A, c' and 7 are coprime in A.>Indeed, a valid proof is conjugation is a valid operation in the ring.>Actually, I thought I was careful to restrict my argument to the>algebraic integers, the ring in which James alleged coprimality.>I do not know whether James alleges coprimality in that ring. He is>pretty unclear about that.>For once, he was reasonably explicit, in the thread title and in>the antepenultimate paragraph of the post starting this thread:>Of course the problem with the ring of algebraic integers is the>implication from previous interpretations (here I rely on what I've>heard primarily from sci.math posters like Arturo Magidin, Nora>Baron, and Dik winter) is that *both* roots have non-unit factors in>common with 7 in the ring of algebraic integers.> I agree that such an argument might not go through in the Harris>ring, but might it not fail for a more basic reason? It is not>obvious that his ring would be a B'ezout domain.>That is something I am also wondering about. But I think James' ring>does not even need to be really be Bezout, but what he utters about>his ring is not even sufficient to clear this. If I remember right,>it has been shown that you can get rings with only one of two complex>conjugates adjoined to the integers.> James has on several occasions lowered his shield of vagueness>sufficiently to allow identification of a specific element of his>ring that is not an algebraic integer.>I have done a few times too, and each time I could not get through>the conjugation. In James' case you need many complex conjugates>of which only one is included in the ring. Does that lead to a>contradiction? I have no idea. It may be that it is possible to>make a choice in each case such that you get a complete ring. On>the other hand, that might be impossible.> Each time, people have>jumped up and down pointing out the unintended consequences if that>number's (algebraic) conjugates were also included.>Yup.> The usually>implicit appeal to the symmetry rendering conjugates algebraically>indistinguishable probably went way over James' head. Certainly,>he has always run away from the questions raised, but his silence>might be explained as withdrawal of his examples after realising>their inapplicability, perhaps due to some ineffable subtlety>visible only to himself.>I think this is *not* the case. Note his current insistance where>(I presume in his ring) one of two complex conjugate numbers is>divisible by 7 and the other is coprime to 7, where you can not tell>which is which due to the duality of the sqrt operator.>I think I see what you mean. Perhaps James accepts that neither of>the zeros, a and a', of x^2 - x + 42 is coprime to 7 in the ring A>of algebraic integers, but is asserting that in his ring, H, a is>divisible by 7 and a' is coprime to 7.>In A, choose v = gcd {6, a} and w = gcd {7, a} so that a = vw.>Then a' = v'w' and ww'=7u for some unit u in A. >If A is a subring of H then a' = v'w' in H, and if a' is coprime >to 7 in H then w' must be a unit in H and w/7 = u/w' is in H. >It is then clear that 7 is not a unit in H, since >1 = a + a' = vw + v'w' = 7v(u/w') + v'w' >and v'w' is coprime to 7 in H.>At first sight, James would have no problems in working in one of>the rings formed by adjoining a or a', but not both, to A.Should of course be adjoining a/7 or a'/7 ....>However, if he also adjoins quotients arising from equations other>than the only one he is really interested in, all he achieves is to>increase the potential for undesirable interactions. >But it is typical of James that, instead of adjoining just the>elements he needs, he throws in everything but the kitchen sink and>no one can say what are the properties of the ensuing monstrosity.> I am therefore reluctant to deduce any>properties of his ring from those examples.>The problem with James' proposition is that he needs a ring where>only two of three factors are divisible by 7 and the third is>co-prime to it. It is my opinion that *if* such a ring does>exist (i.e. where complex conjugates are not necessarily both>element of the ring), that might be a proof of FLT for n=3 (I should>read through James' proof to see whether that is true, but I am>reluctant to do so until the existance of such a ring is established).>The sad thing is that the inverse is not true (I think). So proving>the existence of such a ring is harder than proving FLT for n = 3.>I have not had time to look at any cubic examples.>John Roberts-JonesJohn Roberts-Jones === > ... >I do not know whether James alleges coprimality in that ring. He is >pretty unclear about that. > For once, he was reasonably explicit, in the thread title and in > the antepenultimate paragraph of the post starting this thread: >Of course the problem with the ring of algebraic integers is the >implication from previous interpretations (here I rely on what I've >heard primarily from sci.math posters like Arturo Magidin, Nora >Baron, and Dik winter) is that *both* roots have non-unit factors in >common with 7 in the ring of algebraic integers.Yes, indeed. He never acknowledges when non-unit factors are actuallyshown. But his idea about coprimeness is not exactly the same as themathematical idea about coprimeness. >I think this is *not* the case. Note his current insistance where >(I presume in his ring) one of two complex conjugate numbers is >divisible by 7 and the other is coprime to 7, where you can not tell >which is which due to the duality of the sqrt operator. > I think I see what you mean. Perhaps James accepts that neither of > the zeros, a and a', of x^2 - x + 42 is coprime to 7 in the ring A > of algebraic integers, but is asserting that in his ring, H, a is > divisible by 7 and a' is coprime to 7.Yup, something like that. > At first sight, James would have no problems in working in one of > the rings formed by adjoining a or a', but not both, to A. > However, if he also adjoins quotients arising from equations other > than the only one he is really interested in, all he achieves is to > increase the potential for undesirable interactions. It does not appear he is willing to. The number of equations he isinterested in is already so large that the potential for undesirableinteractions are already too large to handle. > But it is typical of James that, instead of adjoining just the > elements he needs, he throws in everything but the kitchen sink and > no one can say what are the properties of the ensuing monstrosity.But that is just James.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Yes definite March 14 - 22.re: London pre- Vigier V sort of.BTW Second week March 14 to March 22 is looking better for me than the first week March 8 - 13.JS:5. Since Einstein's GR comes from locally gauging ONLY the 4-parametertranslation subgroup of the Poincare group, of course Einstein's earlyformulation of the equivalence principle was only an approximaterotational degrees of freedom.PZ: Exactly.JS: Again hardly news. The Ruffini book looks interesting BTW and I will read it cover to cover.Einstein's guv field of curvedspace-time with the symmetric connection force field is simply thecompensating gauge force field needed to restore the now local gaugesymmetry which is equivalent to local general coordinatetransformations.PZ: Whether this gauge symmetry has a physical basis is a *physical* question, and not justa formal mathematical one.JS: It is physical. It is clearly physical in my vacuum coherence model that builds upon Hagen Kleinert's metric elasticity model.guv(Einstein Locally Curved Space-Time) = Special Relativity Metric (Globally Flat Space-Time) + Elasticity Strain TensorWhat is strained? It is the World Crystal Lattice. When it gets over-strained it fractures into fault lines of curvature disclination defects and also maybe fault lines of torsion dislocation defects on scales large compared to the unit cell.The physical distortion field of the stressed World Crystal is my Bohm Pilot Wave Constraint:Distortion Field = (Loop Gravity Quantum of Area)(Partial Derivative of Goldstone Phase of Vacuum Coherence)This equation does for the elastic-plastic World Crystal Lattice what Bohm'sv(IT) = (h/m)Grad(Phase of Pilot BIT Wave)Note thatWitten's alpha' = hc/(String Tension) = Loop Gravity Quantum of Area from Spin Networks of combinatorial pre-geometry (Penrose).Quantum Gravity Area = (Loop Quantum of Area)(QED electron-photon dimensionless coupling)What stresses the the World Crystal causing it to distort, warp and finally buckle and then fracture?Exotic vacuum zero point pressure causes it!All ordinary matter are simply micro-geons of STRONGLY ATTRACTIVE positive zero point pressure string vortex cores with quantized trapped gauge force fluxes where/zpf = (Loop Gravity Quantum of Area)^-1[(Loop Gravity Quantum of Area)^3/2|Vacuum Coherence|^2 - 1Note that the Vacuum Coherence drops to zero inside the stringy vortex core whose diameter is the coherence length. The gauge force fluxes vary on the scale of the Meissner effect penetration depth like in a Type II superconductor.* This essentially qualitatively conceptually solves the 100 + year old Abraham-Becker-Lorentz extended electron problem that Johnny Glogower and I struggled with without success in Walter Breen's group in 1954. It took me 50 years to solve this problem. It was the first serious physics problem I worked on at age 14. Johnny was 13, but he had already mastered Synge & Schild's Tensor Calculus a book with a rough orange hard cover.PZ: If the gravitational and inertial fields are not even locallycompletely physically equivalent as Einstein supposed, then the gauge symmetry yourefer to may not have any deep physical meaning.JS: You are very wrong there IMHO. Way off the mark. Ruffini does not say that? Does he? I would be surprised. The local gauge principle is perhaps the most useful principle in modern theoretical physics.PZ: The point is Einstein was convinced that it did have a deep meaning; he believed thathis strict gravitational-inertial equivalence principle *explained* the Eotvos proportionalityof the gravitational and inertial masses. It is very important to understand this aspect ofEinstein's reasoning IMO.Otherwise, you cannot possibly understand his theory of general relativity. This isexactly what Einstein said to von Laue when responding to probing criticism of histheory in the early 1920s:... what characterizes the existence of a gravitational field from the empiricalstandpoint is the non-vanishing of the [components of the affine connection],not the non-vanishing of the [components of the Riemann tensor].JS: There you have it! That's a point I was making to you. I was not aware of Einstein's remark you cite.PZ: If one doesnot think in such intuitive [anschaulich] ways, one cannot grasp why somethinglike curvature should have anything at all to do with gravitation. In any case, norational person would have hit upon anything otherwise.The key to the understanding of the equality of gravitational mass and inertial masswould have been missing.-- A. Einstein, quoted in How Einstein Discovered General Relativity: A With Some Contemporary Morals, J.J. Stachel (1986)).JS: Ruffini calls Einstein a Sleep Walker in Arthur Koestler's sense. BTW I met Koestler at the Uri Geller Bohm Birkbeck tests. Went to his home on Montpelier Square for drinks ... See Martin Gardner's Magic and Paraphysics.PZ: If you want to say that this very strong Einsteinian hypothesis of equivalence can nowonly be regarded as a heuristic tool of limited depth and applicability, then fine, I agreewith that. Eddington, for one, took this position in 1921, and in fact this is what I havebeen saying all along.JS: So that's what you have been saying? Why didn't you say so? That's basically what I have been saying!Who's on first?http://www.baseball-almanac.com/humor4.shtmlI'm Costello BTW ;-)If, in addition, you locally gauge the 6-parameterLorentz subgroup, then you get an additional torsion force field, i.e.an anti-symmetric piece of the connection field. This will obviouslymodify the predictions of GR for the tidal torques on extended testPZ: Yes, this is a good point. Now, are such effects *necessarily* scale-sensitive?JS: Define scale-sensitive.6. Hagen Kleinet shows that tidal curvature, the basis for gravitywaves, comes from stringy (vortex core if one puts in a vacuumcoherence O(1) ODLRO parameter) disclination defects in a Plancklattice in 4D.PZ: He shows that you can *get* tidal curvature from disclination defects in his 4DPlanck lattice model. Of course that doesn't in itself prove that this is the truephysical origin of tidal curvature, or even that such a lattice actually exists.JS: If it walks like a Duck, and talks like a Duck ...There is no torsion field in 1915 GR, but if there wasone in Nature, as Akimov & Shipov claim in Moscow, then it wouldcorrespond to dislocation lattice defects. The different 4D discreteworld crystal symmetry groups of the tiled unit cell are differentphysical vacuum structures quite obviously. This is 4D (and maybe 11D)crystallography with additional supersymmetry matrix dimensions).PZ: Right. Physics has moved forward since 1915-16. Einstein's equivalence principlenow looks rather naive as a serious physical model. However, it was undeniablyfruitful from a purely heuristic standpoint.JS: OK7. There are still two more subgroups left in the 15 parameterConformal Group. You will get new compensating gauge force fieldphysics if you locally gauge the 1-parameter dilation subgroup and alsolocally gauge the 4-parameter special conformal transformation ofconstant acceleration boosts for hyperbolic motion in SpecialRelativity e.g. MTW has a chapter on this.PZ: This kind of formalistic groupology needs to be well-anchored in actual physicsand actual physical reasoning IMHO.But still, this looks interesting.JS: Read any book on QCD's SU(3) and the electro-weak U(1)xSU(2) with elegant fashion show of weaves of the fabric of reality from spinnetworks to pre-geometric spin foams to quantized Area operators andworld holograms et-al I find Penrose saying:The algebra I have used to treat linear displacements and rotationstogether, or linear and angular momentum together, I call the algebraof twistors. I have used the term 'twistor' to denote a 'spinor' forthe six-dimensional (++----) pseudo-orthogonal group O(2,4).9. It is obvious to any School Boy that O(2,4) is simply Lorentz SRO(1,3) with string world sheet O(1,1) fiber. Therefore, string theoryis inherent here.PZ: Wouldn't it be more correct to say that certain aspects of the *mathematicalstructure* of string theory is inherent?JS: OK. I am still seeing this Grand Synthesis Through The Glass Darkly on a Foggy Day in London Town March 2004.BTW Second week March 14 to March 22 is looking better for me than the first week March 8 - 13.Penrose continues:The twistor group is the (++--) pseudo-unitary group SU(2,2) which islocally isomorphic with O(2,4). In turn O(2,4) is locally isomorphicwith the fifteen-parameter (local) conformal group of space-time. Undera conformal transformation of space-time the twistors will transformlinearly according to a representation of the group SU(2,2). p. 175Combinatorial Space-Time Quantum Theory and Beyond (Cambridge,1971).Suddenly the connections among1. Quantum loop gravity of spin networks --> foams with quantized areaetc. operators2. Local gauge invariance3. String theory4. Twistors & Conformal Groupare becoming clearer.PZ: If so, then at least something is becoming clearer. :-)JS: http://www.classicalmidiconnection.com/cgibin/x.cgi?midi/n3/ zsunrise.midPenrose's twistor is a SU(2,2) object with 4 complex components with the appropriate Hermitian invariant of signature ++-- , which is obviously not positive definite. Penrose also gives a detailed geometric picture of twistors in globally flat space-time in other papers. The rest mass. Finite rest mass from composite systems of zero rest mass.Dilation invariance is not quite satisfied. p. 176Two null twistors intersect when their Hermitian product vanishes (orthogonal).Some null twistors describe infinity rather than actual null lines.Non-null twistors have parity helicity RH or LH as their Hermitian norm is positive or negative respectively.A non-null twistor is a twisted system of null geodesics in globally flat Minkowski space-time.The Hermitian norm is twice the spin. Therefore, a null twistor is a zero rest mass object of zero spin. The norm must be quantized.Non-null twistors of non-zero spin are not localized to a single null geodesic.One can define 4-momenta Pu and angular momenta Muv to the twistor Z.The twistor components define a 8D phase space in which globally flat Minkowski space is the configuration space.Z and Z* obey a boson commutation relation like creation and destruction operators.Also their anti-commutator is like a quantum harmonic oscillator, hence quantized Hermitian norms.The 15 generators of the Conformal Group's Lie Algebra are written in terms of the twistor components. Let a & b denote twistor components, a,b, = 0,1,2,3The Conformal Group's infinitesimal generators of the Lie Algebra areZaZ*b - (1/4)Kb^aZ^cZ*cK^ba is Kronecker delta* is complex conjugateRemember basically previously Penrose used 2-component spinors in a kind of quantum computer combinatorial pre-geometry qubit spin-network to get a relational space-like 3D hypersurface to emerge in the limit of several chunks of macroscopically large qubit networks.Penrose wants to use twistors to get time emergent. There is no gravity curvature as yet. Penrose is simply here in 1971 trying to get globally flat Minkowski space-time to emerge. So I use Penrose's model here as a plug-in to my model just as I used Hagen Kleinert's.My starting point is emergent globally flat space-time that I then show is unstable to the formation of curved space-time from vacuum coherence of the zero point vacuum fluctuations of ALL quantum fields. Residual zero point fluctuations at different scales form the exotic vacua of dark energy AND dark matter. If my theory is correct no dark matter detector will ever click with The Right Stuff only with false positives that will not explain Omega(Dark Matter) ~ 0.3 at the FRW large scale of cosmology.Null twistors of zero spin (scalar fields) carry over directly to curved space-time but not non-null twistors which appear to be intrinsically quantum gravitational in Penrose's view 33 years ago when I attended his twistor seminar at Birkbeck.Now just use null twistors.Orthogonality as intersection of null twistors does NOT carry over into curved space-time as gravity.A symplectic (phase space) structure of the exterior AREA two-form dZ^a/dZ*a survives passage from globally flat to locally curved (space-time version of local gauging of internal phase symmetries from extra space dimensions of hyperspace where the compensating gauge force fields ((connection fields for parallel transport in the extra dimensions)) restore the initial global symmetries).Null twistors form a 7D real dimensional manifold. The symplectic manifold is 8D real dimensional so the non-null twistors creep in through the back door.Penrose defines a Poisson Bracket {PSI , CHI} for sets of complex holomorphic functions PSI(Z,Z*), CHI(Z,Z*) ... of the twistors that he says supports the symplectic phase space structure for curved spacetime. This is where I thought one would need adaptive wavelet transform generalizations of rigid Fourier transforms.{PSI , CHI} = i[PSI,Z^a CHIZ*a - CHI,Z^a PSIZ*a]Space-time curvature as a canonical transformation in phase space mixing twistor position with twistor momentum Penrose speculated in 1971. === Consider P an n-set. U and V are the power set of P, that is,the set of all subsets of P with 2^n elements.Let's consider a simple membership graph with P, U and V as verticeswhich has the following property: - there are only edges pairwise between P and V (let's call this subgraphL) and U and V (let's call this subgraph R) but not between P and U. - there is an edge from p (in P) to v (in V) iff p is in v. - there is an edge from v (in V) to u (in U) iff v is a subset of u.Consider the following 3edge layout problems:1. Suppose we impose the constraint that the in-degree of any vertex in U is less than m, how to layout edges such that: for each u in U, each p in u can find a path to u via some v in V What's the maximum out degree of any vertex in P?2. Suppose now we consider the opposite, let's keep the max. out-degree of any vertex in P be k, what's the max. in-degree of vertices in U?3. Suppose now there is no constraint, what's the minimal max. degree ofvertices in P union U?Does anyone see something similar before? Please give me the reference.Any suggestion of the approach I can take would be very welcome?Aldar === The value should be 0 if the evaluated player has the same amount ofmoney as each of the other players, non-zero otherwise (negative whenhe's poorer, positive when richer). The values should differ for [1,0, 0, 0] and [5, 4, 4, 4] cases (first player is being evaluated),i.e. not only a difference, but also a proportion should count.What equation would you suggest?You might create an index of each player versus the group average. This gives us (using the mean):Situation A:Wealth: [1, 0, 0, 0] mean= 0.25Status: [4, 0, 0, 0]Situation B:Wealth: [ 5, 4, 4, 4] mean= 4.25Status: [1.2, 0.9, 0.9, 0.9]-Will Dwinnellhttp://will.dwinnell.com === > We may assume that the average 10 cows would die on different days.>(It doesn't make that much difference anyway.)>So the probability of a cow dying on a particular day is 10/365.>Since the probability of Daisy dying this year is 1/13600,>the probability of Daisy dying on a particular day of the year is:>(1/13600)*(10/365).That is not the probability for any arbitrary date, since cows (like men)> are only permitted to die once. The probability of Daisy dying on day nin> the future should be:(1-p')^n * p'Even that is an assumption about the distribution of deaths throughout theyear. In practice, the distribution of deaths assumed is chosen forcomputational convenience. Probably not a good assumption for canine oreven bovine deaths, but works adequately for human deaths. And, sincethere's a fair amount of money riding on payments as a result of humandeaths, you can bet that if it made a significant difference (in terms ofReturn on Equity), insurance companies would take it into account whenpricing.Jon Miller === > JS: The following statements are true:> 1. To a good approximation the non-tensor connection field g-force on a> where> m(passive) = m(active)> The approximation is two-fold> A. Not near a space-time singularity, i.e. not falling behind the event> horizon of a black hole.[snip]I enjoy your posts.But is it true that you never reply to a response?Double-A === $V06.2333@newssvr27.news.prodigy.com:> On Sunday, January 11, 2004, at 01:57 PM, Paul Zielinski > JS: The following statements are true:> 1. To a good approximation the non-tensor connection field g-> force on a ....Where are you going to go on vacation this year Jack? I was thinking about going to either the Caribbean or to the French Polynesian Islands. I would really like to see Bora-Bora as many have said it is the beautiful spot on the planet.r-- Nothing beats the bandwidth of a station wagon filled with DLT tapes. === Also available at http://math.ucr.edu/home/baez/week201.htmlJanuary 10, 2004This Week's Finds in Mathematical Physics - Week 201John Baez Lately James Dolan and I have been studying number theory. I used to *hate* this subject: it seemed like a massive waste of time. Newspapers, magazines and even lots of math books seem to celebrate the idea of people slaving away for centuries on puzzles whose only virtue is that they're easy to state but hard to solve. For example: are any odd numbers the sum of all their divisors? Are there infinitely many pairs of primes that differ by 2? Is every even number bigger than 2 a sum of two primes? Are there any positive integer solutions to x^n + y^n = z^nfor n > 2? My response to all these was: WHO CARES?! Sure, it's noble to seek knowledge for its own sake. But working on a math problem just because it's *hard* is like trying to drill a hole in a concrete wall with your nose, just to prove you can! If you succeed,I'll be impressed - but I'll still wonder why you didn't put all thatenergy into something more interesting. Now my attitude has changed, because I'm beginning to see that behind these silly hard problems there lurks an actual *theory*, full of deep ideas and interesting links to other branches of mathematics, including mathematical physics. It just so happens that now and then this theoryhappens to crack another hard nut. I'd known for a while that something like this must be true: after all, when Andrew Wiles proved Fermat's Last Theorem, even the newspapers admitted this was just a spinoff of something more important, namely a special case of the Taniyama-Shimura Conjecture. They said this had something to do with elliptic curves and modular forms, which are very nice geometrical things that show up all over in complex analysis and string theory. Unfortunately, the actual statement of this conjecture seemed impenetrable - it didn't resonate with things I understood. In fact, the Taniyama-Shimura Conjecture is part of a big *network* of problems that are more interesting but harder to explain than the flashy ones I listed above: problems like the Extended Riemann Hypothesis, the Weil Conjecture (now solved), the Birch-Swinnerton-Dyer Conjecture,and bigger projects like the Langlands Program and developing the theory of motives. And these problems rest on top of a solid foundation of beautiful stuff that's already known, like Galois theory and class field theory, and stuff about modular forms and L-functions.As I'm gradually beginning to understand little bits of these things, I'm getting really excited about number theory... so I'm dying to *explain* some of it! But where to start? I have to start with something basic thatunderlies all the fancy stuff. Hmm, I think I'll start with Galois theory.As you may have heard, Galois invented group theory in the process ofshowing you can't solve the quintic equationax^5 + bx^4 + cx^3 + dx^2 + ex + f = 0by radicals. In other words, he showed you can't solve this equation by means of some souped-up version of the quadratic formula that just involves taking the coefficients a,b,c,d,e,f and adding, subtracting, multiplying, dividing and taking nth roots. The basic idea is something like this. In general, a quintic equation has 5 solutions - and there's no best one, so your formula has got to be a formula for all five. And there's a puzzle: how do you give one formula for five things? Well, think about the quadratic formula! It has that plus or minus in it, which comes from taking a square root. So, it's really a formula for *both* solutions of the quadratic equation. If there were a formula for the quintic that worked like this, we'd have to get all 5 solutions from different choices of nth roots in this formula.Galois showed this can't happen. And the way he did it used *symmetry*!Roughly speaking, he showed that the general quintic equation is completely symmetrical under permuting all 5 solutions, and that this symmetry group - the group of permutations of 5 things - can't be built up from the symmetrygroups that arise when you take nth roots. The moral is this: you can't solve a problem if the answer has some symmetry, and your method of solution doesn't let you write down an answer that has this symmetry! An old example of this principle is the medieval puzzle called Buridan'sAss. Placed equidistant between two equally good piles of hay, thisdonkey starves to death because it can't make up its mind which alternativeis best. The problem has a symmetry, but the donkey's method of solution doesn't, so it's stuck.Buridan's ass would also get stuck if you asked it for *the* solutionto the quadratic equation. Galois proof of the unsolvability of the quintic by radicals is just a more sophisticated variation on this theme. (Of course, you *can* solve the quintic if you strengthen your methods.)A closely related idea is Curie's principle, named after Marie's husband Pierre. This says that if your problem has a symmetry and it is a unique solution, the solution must be symmetrical. For example, if some physical system has rotation symmetry and it has aunique equilibrium state, this state must be rotationally invariant.Now, in the case of a ferromagnet below its Curie temperature, the equilibrium state is *not* rotationally invariant: the little magnetizedelectrons line up in some specific direction! But this doesn't contradict Curie's principle, since there's not a unique equilibrium state - there are lots, since the electrons can line up in any direction. Physicists use the term spontaneous symmetry breaking when any *one* solution of a symmetric problem is not symmetrical, but the whole set of them is. This is precisely what happens with the quintic, or even the quadratic equation.While these general ideas about symmetry apply to problems of all sorts,their application to number theory kicks in when we apply them to *fields*.A field is a gadget where you can add, subtract, multiply and divideby anything nonzero, and a bunch of familiar laws of arithmetic hold, which I won't bore you with here. The three most famous fields are the rational numbers Q, the real numbers R, and the complex numbers C. However, there are lots of other interesting fields. Number theorists are especially fond of algebraic number fields. An algebraic number is a solution to a polynomial equation whose coefficients are rational numbers. You get an algebraic number field by taking the field of rational numbers, throwing in finitely many algebraic numbers, and then adding, subtracting, multiplying and dividing them to get more numbers until you've got a field. For example, we could take the rationals, throw in the square root of 2, and get a field consisting of all numbers of the forma + b sqrt(2)where a and b are rational. Notice: if we add, multiply, subtract ordivide two numbers like this, we get another number of this form.So this is really a field - and it's called Q(sqrt(2)), since we use round parentheses to denote the result of taking a field and extending it by throwing in some extra numbers.More generally, we could throw in the square root of any integer n, and get an algebraic number field called Q(sqrt(n)), consisting of allnumbersa + b sqrt(n)where a and b are rational. If sqrt(n) is rational then this field isjust Q, which is boring. Otherwise, we call it a quadratic number field.Even more generally, we could take the rationals and throw in asolution of any quadratic equation with rational coefficients. But it's easy to see that this doesn't give anything beyond fields likeQ(sqrt(n)). And that's the real reason we call these the quadratic number fields.There are also cubic number fields, and quartic number fields,and quintic number fields, and so on. And others, too, where wethrow in solutions to a whole bunch of polynomial equations!Now, it turns out you can answer lots of classic but rather goofy-sounding number theory puzzles like which integers are a sum of two squares? by converting them into questions about algebraic number fields. And the good part is, the resulting questions are connected to all sorts of other topics in math - they're not just glorified mental gymnastics! So, from a modern viewpoint, a bunch of classic number theory puzzles are secretly just tricks to get certain kinds of people interested in algebraic number fields. But right now I *don't* want to explain how we can use algebraic number fields to solve classic but goofy-sounding number theory puzzles.In fact, I want to downplay the whole puzzle aspect of number theory.Instead, I hope you're reeling with horror at thought of this vast complicated wilderness of fields containing Q but contained in C.First there's a huge infinite thicket of algebraic number fields...and then, there's an ever scarier jungle of fields that contain transcendental numbers like pi and e! I won't even talk about *that* jungle, it's so dark and scary. Physicists usually zip straight past this whole wilderness and work with C. But in fact, if you stop and carefully examine all the algebraic number fields and how they sit inside each other, you'll find some incredibly beautiful patterns. And these patterns are turning out to be related to Feynman diagrams, topological quantum field theory, and so on...However, before we can talk about all that, we need to understand thebasic tool for analyzing how one field fits inside another: Galois theory!A function from a field to itself that preserves addition, subtraction,multiplication and division is called an automorphism. It's justa *symmetry* of the field. But now, suppose we have a field K which contains some smaller field k. Then we define the Galois group of K over k to be the group of all automorphisms of K that act as the identity on k. We call this groupGal(K/k)for short.The classic example, familiar to all physicists, is the Galois group of the complex numbers, C, over the real numbers, R. This group has two elements: the identity transformation, which leaves everything alone, and complex conjugation, which switches i and -i. Since the only group with 2elements is Z/2, we haveGal(C/R) = Z/2Where does complex conjugation come from? It comes from the fact thatC from R by throwing in a solution of the quadratic equation x^2 = -1.We say C is a quadratic extension of R. But as soon as we throw in one solution of this equation, we inevitably throw in another, namely its negative - and there's no way to tell which is which. And complex conjugation is the symmetry that switches them! Note: we know that i and -i are different, but we can't tell which is which! This sounds a bit odd at first. It's a bit hard to explainprecisely in ordinary language, which is part of why Galois had to invent group theory. But it's fun to try to explain it in plain English... so let me try. The complex numbers have two solutions tox^2 = -1.By convention, one of them is called i, and the other is called-i. Having made this convention, there's never any problem tellingthem apart. But we could reverse our convention and nothing wouldgo wrong. For example, if the ghost of Galois wafted into your officewherever there had been i, everything in these books would still be true!Here's another way to think about it. Suppose we meet some extraterrestrials and find that they too have developed the complex numbers by taking the real numbers and adjoining a square root of -1, only they call it @. Then there would be no way for us to tell if their @ was our i or our -i. All we can do is choose an arbitraryconvention as to which is which.Of course, if they put their @ in the lower halfplane when drawing the complex plane, we might feel like calling it -i... but here we are secretly making use of a convention for matching their complex plane with ours, and the *other* convention would work equally well! If they drew their real line *vertically* in the complex plane, it would be more obvious that we need a convention to match their complex plane with ours, and that there are two conventions for doing this, both perfectly self-consistent. If you've studied enough physics, this extraterrestrial scenario should remind you of those thought experiments where you're trying toexplain to some alien civilization the difference between left andright... by means of radio, say, where you're *not* allowed to referto specific objects you both know - so it's cheating to say imagine you're on Earth looking at the Big Dipper and the handle is pointing down; then Arcturus is to the right. If the laws of physics didn't distinguish between left and right, you couldn't explain the difference between left and right without cheating like this, so the laws of physics would have a symmetry group with two elements: the identity and the transformation that switches left and right. As it turns out, the laws of physics *do*distinguish between left and right - see week73 for more on that. But that's another story. My point here is that the Galois group of C over R is a similar sort of thing, but built into the very fabric of mathematics! And that's why complex conjugation is so important. I could tell you a nice long story about how complex conjugation isrelated to charge conjugation (switching matter and antimatter) and also time reversal (switching past and future). But I won't!Here's another example of a Galois group that physicists should like. Let C(z) be the field of rational functions in one complex variable z -in other words, functions likef(z) = P(z)/Q(z)where P and Q are polynomials in z with complex coefficients. You can add, subtract, multiply and divide rational functions and get other rational functions, so they form a field. And they contain C as a subfield, because we can think of any complex number as a *constant* function. So, we can ask about the Galois group of C(z) over C.What's it like?It's the Lorentz group!To see this, it's best to think of rational functions as functions noton the complex plane but on the Riemann sphere - the complex plane together with one extra point, the point at infinity. The only conformal transformations of the Riemann sphere are fractional lineartransformations: az + bT(z) = ------ cz + dSo, the only symmetries of the field of rational functions that act as the identity on constant functions are those coming from fractional transformations, like this:f |-> fT where fT(z) = f(T(z)) If you don't follow my reasoning here, don't worry - the details aren't hard to fill in, but they'd be distracting here. The last step is to check that the group of fractional linear transformations is the same as the Lorentz group. You can do this algebraically, but you can also do it geometrically by thinking of the Riemann sphere as the heavenly sphere: that imaginary sphere the stars look like they're sitting on. The key step is to check this remarkable fact: if you shoot past the earth near the speed of light, the constellations willlook distorted by a Lorentz transformation - but if you draw lines connectingthe stars, all the *angles* between these lines will remain the same; only their *lengths* will get messed up! Moreover, it's obvious that if you rotate your head, both angles and lengthson the heavenly spehere are preserved. So, any rotation or Lorentz boost gives an angle-preserving transformation of the heavenly sphere - that is, a conformal transformation! And this must be a fractional linear transformation.Summarizing, the Galois group of C(z) over C is the Lorentz group, ormore precisely, its connected component, SO_0(3,1):Gal(C(z)/C) = SO_0(3,1).We've talked about the Galois group of C(z) over C and the Galois groupof C over R. What about the Galois group of C(z) over R? Unsurprisingly,this is the group of transformations of the Riemann sphere generated by fractional linear transformations *and* complex conjugation. And physically,this corresponds to taking the connected component of the Lorentz groupand throwing in *time reversal*! So you see, complex conjugation is relatedto time reversal. But I promised not to go into that....I've been talking about Galois groups that physicists should like, butyou're probably wondering where the number theory went! Well, it's all part of the same big story. In number theory we're especiallyinterested in Galois groups likeGal(K/k)where K is some algebraic number field and k is some subfield of K.For starters, consider this example:Gal(Q(sqrt(n))/Q)where sqrt(n) is irrational. I've already hinted at what this group is! Q(sqrt(n)) has sqrt(n) in it, so it also has -sqrt(n) in it, and there's an automorphism that switches these two while leaving all the rational numbers alone, namelya + b sqrt(n) |-> a - b sqrt(n) (a,b in Q)So, we have:Gal(Q(sqrt(n)))/Q) = Z/2just like the Galois group of C over R. To get some bigger Galois groups, let's take Q and throw in a primitive nth root of unity. Hmm, I may need to explain what that means. There are n different nth roots of 1 - but unlike the two square roots of -1, these are not all created equal! Only some are primitive.For example, among the 4th roots of unity we have 1 and -1, which are actually square roots of unity, and i and -i, which aren't. A primitive nth root of unity is an nth root of 1 that's not an kth root for anyk < n. If you take all the powers of any primitive nth root of unity,you get *all* the nth roots of unity. So, if we take some primitive nth root of unity, call it1^{1/n}for lack of a better name, and extend the rationals by this number, we get a fieldQ(1^{1/n})which contains all the nth roots of unity. Since the nth roots of unityare evenly distributed around the unit circle, this sort of field is calleda cyclotomic field, for the Greek word for circle cutting. In fact,one can apply Galois theory to this field to figure out which regularn-gons one can construct with a ruler and compass!But what's the Galois groupGal(Q(1^{1/n})/Q)like? Any symmetry in this group must map 1^{1/n} to some root of unity, say 1^{m/n} - and once you know which one, you completely know the symmetry. But actually, this symmetry must map 1^{1/n} to some *primitive*root of unity, so m has to be relatively prime to n. Apart from that, though, anything goes - so the size of Gal(Q(1^{1/n})/Q)is just the number of m less than n that are relatively prime to n. And if you think about it, these numbers relatively prime to n are just the same as elements of Z/n that have multiplicative inverses! So if you thinksome more, you'll see thatGal(Q(1^{1/n})/Q) = (Z/n)*where (Z/n)* is the multiplicative group of Z/n - that is, the elements of Z/n that have multiplicative inverses, made into a group via multiplication! This group can be big, but it's still abelian. Can we get some nonabelianGalois groups from algebraic number fields?Sure! Let's say you take some polynomial equation with rational coefficients, take *all* its solutions, throw them into the rationals - and keep adding, subtracting, multiplying and dividing until you get some field K. This K is called the splitting field of your polynomial. But here's the interesting thing: if you pick your polynomial equation at random, the chances are really good that it has n different solutions if the polynomial is of degree n, and that *any* permutation of these solutions comes from a unique symmetry of the field K. In other words:barring some coincidence, all roots are created equal! So in general we haveGal(K/Q) = S_nwhere S_n is the group of all permutations of n things.Sometimes of course the Galois group will be smaller, since our polynomialcould have repeated roots or, more subtly, algebraic relations betweenroots - as in the cyclotomic case we just looked at. But, we can already start to see how to prove the unsolvability of the general quintic! Pick some random 5th-degree polynomial, let K be itssplitting field, and noteGal(K/Q) = S_5Then, show that if we build up an algebraic number field by startingwith Q and repeatedly throwing in nth roots of numbers we've already got, we just can't get S_5 as its Galois group over the rationals! We've already seen this in the case where we throw in a square root of n, or an nth root of 1. The general case is a bit more work. But instead of giving the details, I'll just mention a good textbook on Galois theory for beginners:1) Ian Stewart, Galois Theory, 3rd edition, Chapman and Hall, New York, 2004.Ian Stewart is famous as a popularizer of mathematics, and it showshere - he has nice discussions of the history of the famous problemssolved by Galois theory, and a nice demystification of the Galois'famous duel. But, this is a real math textbook - so you can really learn Galois theory from it! Make sure to get the 3rd edition, sinceit has more examples than the earlier ones.Having given Ian Stewart the dirty work of explaining Galois theory inthe usual way, let me say some things that few people admit in a firstcourse on the subject. So far, we've looked at examples of a field k contained in some biggerfield K, and worked out the group Gal(K/k) consisting of all automorphisms of K that fix everything in k. But here's the big secret: this has NOTHING TO DO WITH FIELDS! It worksfor ANY sort of mathematical gadget! If you've got a little gadget k sitting in a big gadget K, you get a Galois group Gal(K/k) consisting of symmetries of the big gadget that fix everything in the little one. But now here's the cool part, which is also very general. Any subgroup of Gal(K/k) gives a gadget containing k and contained in K: namely, the gadget consisting of all the elements of K that are fixed by everything in this subgroup.And conversely, any gadget containing k and contained in K gives a subgroup of Gal(K/k): namely, the group consisting of all the symmetries of K that fix every element of this gadget. This was Galois' biggest idea: we call this a GALOIS CORRESPONDENCE.It lets us use *group theory* to classify gadgets contained in one and containing another. He applied it to fields, but it turns out to be useful much more generally. Now, it would be great if the Galois corresondence were always a perfect 1-1 correspondence between subgroups of Gal(K/k) and gadgets containingk and contained in K. But, it ain't true. It ain't even true when we're talking about fields! However, that needn't stop us. For example, we can restrict ourselves to cases when it *is* true. And this is where the Fundamental Theorem of Galois Theory comes in! It's easiest to state this theorem when k and K are algebraic number fields, so that's what I'll do. In this case, there'sa 1-1 correspondence between subgroups of Gal(K/k) and extensions of k contained in K if:i) K is a finite extension of k. In other words, K is a finite-dimensionalvector space over k.ii) K is a normal extension of k. In other words, if a polynomial with coefficients in k has no roots in k, but one root in K, then all its roots are in K.For general fields we also need another condition, namely that K be a separable extension of k. But this is automatic for algebraic numberfields, so let's not worry about it. At this point, if we had time, we could work out a bunch of Galois groupsand see a bunch of patterns. Using these, we could see why you can'tsolve the general quintic using radicals, why you can't trisect the angle or double the cube using ruler-and-compass constructions, and why you candraw a regular pentagon using ruler and compass, but not a regular heptagon.Basically, to prove something is impossible, you just show that some numbercan't possibly lie in some particular algebraic number field, because it's the root of a polynomial whose splitting field has a Galois group that's fancier than the Galois group of that algebraic number field. For example, ruler-and-compass constructions produce distances that lie in iterated quadratic extensions of the rationals - meaning that you just keep throwing in square roots of stuff you've got. Doubling the cube requires getting your hands on the cube root of 2. But the Galois group of the splitting field of x^3 = 2has size divisible by 3, while an iterated quadratic extension has a Galoisgroup whose size is a power of 2. Using the Galois correspondence, we seethere's no way to stuff the former field into the latter.But you can read about this in any good book on Galois theory, so I'd ratherdive right into that thicket I was hinting at earlier: the field of ALL algebraic numbers! The roots of any polynomial with coefficients in this field again lie in this field, so we say this field is algebraically closed. And since it's the smallest algebraically closed field containing Q, it's called the algebraic closure of Q, or Qbar for short - that is, Q with a bar over it. This field Qbar is huge. In particular, it's an infinite-dimensional vector space over Q. So, condition i) in the Fundamental Theorem of Galois Theory doesn't hold. But that's no disaster: when this happens, we just need to put a topology on the group Gal(K/k) and set up the Galois correspondence using*closed* subgroups of Gal(K/k). Using this trick, every algebraic number fieldcorresponds to some closed subgroup of Gal(Qbar/Q).So, for people studying algebraic number fields, Gal(Qbar/Q)is like the holy grail. It's the symmetry group of the algebraic numbers, and the key to how all algebraic number fields sit inside each other!But alas, this group is devilishly complicated. In fact, it has literally driven men mad. One of my grad students knows someone who had a breakdown and went to the mental hospital while trying to understand this group!(There may have been other reasons for his breakdown, too, but as readersof E. T. Bell's book Men in Mathematics know, the facts should never getin the way of a good anecdote.) If Gal(Qbar/Q) were just an infinitely tangled thicket, it wouldn't be so tantalizing. But there are things we can understand about it! To describethese, I'll have to turn up the math level a notch...First of all, an extension K of a field k is called abelian if Gal(K/k) is an abelian group. Abelian extensions of algebraic number fields can beunderstood using something called class field theory. In particular, the Kronecker-Weber theorem says that every finite abelian extension of Q is contained in a cyclotomic field. So, they all sit inside a field calledQcyc, which is gotten by taking the rationals and throwing in *all* nth roots of unity for *all* n. SinceGal(Q(1^{1/n})/Q) = (Z/n)*we know from Galois theory that Gal(Qcyc/Q) must be a big group containing all the groups (Z/n)* as closed subgroups. It's easy to see that (Z/n)* isa quotient group of (Z/m)* if m is divisible by n; this lets us take the inverse limit of all the groups (Z/m)* - and that's Gal(Qcyc/Q). This inverse limit is also the multiplicative group of the ring Z^, the inverse limit of all the rings Z/n. Z^ is also called the profinite completion of the integers, and I urge you to play around with it if you never have! It's a cute gadget.In short:Gal(Qcyc/Q) = Z^*and if we stay inside Qcyc, we're in a zone where the pattern of algebraicnumber fields can be understood. This stuff was worked out by people likereciprocity theorem, laid in place by Emil Artin in 1927. In a certain sense Qcyc is to Qbar as homology theory is to homotopy theory: it's all about *abelian* Galois groups, so it's manageable. People now use Qcyc as a kind of base camp for further expeditions intothe depths of Qbar. In particular, since Q is contained in Qcyc and Qcyc is contained in Qbarwe get an exact sequence of Galois groups:1 -> Gal(Qbar/Qcyc) -> Gal(Qbar/Q) -> Gal(Qcyc/Q) -> 1So, to understand Gal(Qbar/Q) we need to understand Gal(Qcyc/Q),Gal(Qbar/Qcyc) and how they fit together! The last two steps are not so easy. Shafarevich has conjectured that Gal(Qbar/Qcyc) is the profinite completion of a free group, say F^. This would give1 -> F^ -> Gal(Qbar/Q) -> Z^* -> 1but I have no idea how much evidence there is for Shafarevich's conjecture,or how much people know or guess about this exact sequence.More recently, Deligne has turned attention to a certain motivic versionof Gal(Qbar/Q), which is a proalgebraic group scheme. This sort of grouphas a *Lie algebra*, which makes it more tractable. And there are a bunchof fascinating conjectures about this Lie algebra is related to the Riemannzeta function at odd numbers, Connes and Kreimer's work on Feynman diagrams,Drinfeld's work on the Grothendieck-Teichmueller group, and more! I really want to understand this stuff better - right now, it's a completemuddle in my mind. When I do, I will report back to you. For now, though,let me give you some references. For two very nice but very different introductions to algebraic number fields, try these:2) H. P. F. Swinnerton-Dyer, A Brief Guide to Algebraic Number Theory,Cambridge U. Press, Cambridge 2001.3) Juergen Neukirch, Algebraic Number Theory, trans. Norbert Schappacher,Springer, Berlin, 1986. Both assume you know some Galois theory or at least can fake it.Neukirch's book is good for the all-important analogy between Galois groups and fundamental groups, which I haven't even touched upon here!Swinnerton-Dyer's book has the virtue of brevity, so you can see theforest for the trees. Both have a friendly, slightly chatty style thatI like. For Shafarevich's conjecture, try this:4) K. Iwasawa, On solvable extensions of algebraic number fields,Ann. Math. 58 (1953) 548-572.For Deligne's motivic analogue, try this:5) Pierre Deligne, Le groupe fondamental de la droite projectivemoins trois points, in Galois Groups over Q, MSRI Publications 16 (1989),79-313.This stuff has a lot of relationships to 3d topological quantum fieldtheory, braided monoidal categories, and the like... and it all goesback to the Grothendieck-Teichmueller group. To learn about this group 6) Leila Schneps, The Grothendieck-Teichmueller group: a survey,in The Grothendieck Theory of Dessins D'Enfants, London Math. SocietyNotes 200, Cambridge U. Press, Cambridge 1994, pp. 183-204.To hear and watch some online lectures on this material, try:7) Leila Schneps, The Grothendieck-Teichmuller group and fundamental groups of moduli spaces, MSRI lecture available athttp://www.msri.org/publications/ln/msri/1999/vonneumann/ schneps/1/Grothendieck-Teichmueller group and Hopf algebras,MSRI lecture available at http://www.msri.org/publications/ln/msri/1999/vonneumann/ schneps/2/For a quick romp through many mindblowing ideas which touches on thismaterial near the end:8) Pierre Cartier, A mad day's work: from Grothendieck to Connes and Kontsevich - the evolution of concepts of space and symmetry, Bulletin of the AMS, 38 (2001), 389 - 408. Also available at http://www.ams.org/joursearch/index.htmlFor even more mindblowing ideas along these lines:9) Jack Morava, The motivic Thom isomorphism, talk at the Newton Institute,December 2002, also available at math.AT/0306151Quote of the week:Paris, 1 June - A deplorable duel yesterday has deprived the exact sciences of a young man who gave the highest expectations, but whose celebrated precosity was lately overshadowed by his political activities. The young Evariste Galois... was fighting with one of his old friends,a young man like himself, like himself a member of the Society ofFriends of the People, and who was known to have figured equally ina political trial. It is said that love was the cause of the combat.The pistol was the chosen weapon of the adversaries, but because oftheir old friendship they could not bear to look at one another and left their decision to blind fate. - Le Precursor, June 4 1832---------------------------------------------------------- -------------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 This Week's 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 tohttp://math.ucr.edu/home/baez/this.week.html