An efficient way to compute the variance uses the raw-score formula of statistics. If you take the defn. of variance formula and expand it out and replace mu with its estimate (sample mean) you get the raw-score formula.In terms of computation, you will only require the sum of the sample and the sum of the squares of the sample along with the number of samples. Clay > Dear all, In one of my program, I need to compute the variance of a 8x8 data matrix as > fast as possible... Any fast algorithm with lowest complexity? By the way, since what I actually need is an indication of signal local > activity, is there any other measure which can also be a indicator of local > activity and is less complex than computing variance? > -Walala ==== > In one of my program, I need to compute > the variance of a 8x8 data matrix as > fast as possible... What does it means ? You just have 16 samples of a signal and you want the variance ? For me it is not clear why deviation from the mean is a good indicator of activity. Obviously, there is no fast algorithm for computing a sum of 16 terms ! BUT if you compute the variance on a moving windows, there is a very fast algorithm, that you can figure out pretty easily. Your just have to perform : 1. compute mean m and replace your image I by M=(I-m)^2 2. compute moving average along X direction. This can be done by a recursive algorithm, because if S(k) is the moving average and F(k) is the initial function (e.g. a row of M), you have S(k) = S(k-1)+F(k)-F(k-9) (here I assume window of size 8) 3. compute moving average along Y direction. All this stuff needs about O(n) operations where n is the number of pixels in the original image and it's independant of the size of the window. Gabriel ==== > You just have 16 samples of a signal and you want the variance ? Sorry, I meant 64 samples. > Obviously, there is no fast algorithm for computing a sum > of 16 terms ! Sorrrry once again 8x8=64 not 16 ... G. ==== myfile.> DON02. Calc.Factors.FermatMers.Mersenne.CARMP163.SPMP163 subject:Mersenne numbers, Mp163 > D.Calc.Factors.FermatMers.Mersenne.CARMP163.SPMP163 We look at the usual long multiplication of, for example, 123*48. 123 x 48 ------- 984 +492 ------- 5904. However, I wish to concentrate just on the tens and units figures of the resulting product. This is called 'multiplying modulo 100.' That is, any whole number of 100s can be thrown away in the process. Then, (100+23)*48 == 48*100+23*48 (is congruent to) == (20+3)*(40+8) == 2*4*100+3*40+8*20+3*8 == 120+160+24 == 104 == 04 modulo 100. In calculating the remainder of product (integers w*z) after dividing by positive integer, t, it can often be made easier by taking away convenient multiples of the modulus, t, from w or z respectively before executing the multiplication. Let w = at+b, say, and z = ct+d, say. Then, wz = (at+b)(ct+d) = at(ct+d) +b(ct+d) = t(act +ad +bc) +bd == bd modulo t. Bd should usually be less than w*z, but it may require to be reduced further. I photographed car number plate 'MP163' at Hanson St, Newtown, Wellington on today, 26-5-2002. This number plate could be shorthand for Mersenne numbers, 2^prime-1. (Figure 2 raised to a prime index, subtract 1.) The world's largest known prime number in 2001 often is 2^13.4million-1. (Greater than 13 million 2s (Dominion Post, NZ Herald, GIMPS great internet mersenne prime search.) By the way, the 40th known Mersenne prime exponent is a twin prime, 10x(23x63)^2 +/-1. And a previous record holder was virtually a palindrome, Table Mountain, increasing decreasing. Exponent (1*2*34*44432+1)= 3021377. [sic.] _____ / A theorem of mathematics number theory says if (a is a positive integer ).. a^prime-1 has factor/s they must be of the form 2*k*p+1. However, the Penguin Dictionary of Curious and Interesting Numbers (1997), entry 28, table of perfect numbers, does not give M_163 as a Mersenne prime. Therefore, I have written myprogram 'powabc1' in BBC interpreted Basic64, which tests for just such factors. (On Acorn A5000 computer, UK 1990, RISC OS-reduced instruction set computer, 4 Megabytes RAM randomaccess memory.) In this case, several programs found 150287 and 704161, etc. divide M_163. Or 2^163 == 1 modulo 150287. Have I found a factor of M_163? The following, I believe, shows that I have. Check. 2^163= 2^3*(2^20)^8 (is congruent to) == 8*1048576^8 2^20==1048576-6*150287 == 146854 mod 150287 == -3433. Squaring 2^40= 11785489 == 63103 Squaring 2^80==3,981,988,609 ==134544== -15743 Squaring 2^160==247,842,049 ==18786 2^163==8*18786 ==150288==1 mod 150287. Q.e.d. By the way, my methods found (the least prime) factor 1580,187,223 of (10^9)^(10^9)+3. (sci.math sensation 2001.) This is, digit 1 followed by 9 billion zeroes. Surpassed only by Graham's number. I suspect gigaplex = 10^billion. (There is a typo error in Penguin Dictionary of Curious and Interesting Numbers, 1997.) yours sincerely, / Donald S. McDonald (Wellington, New Zealand) ==== There was a time when people spoke their minds, and were not afraid to offend - and that since then, too many truths have been buried. Mark Kurlansky, 1968: The Year That Rocked The World (Ballantine, 2004) thanks to Frank Lauria In 1968 I was in La Jolla CA at UCSD during time Greg Benford describes in Timescape living on Bonair Street Wind an Sea Beach near Unicorn Theater with Ken Kesey's Merry Prankster Bus parked nearby frequently. I was also teaching at San Diego State with Fred Alan Wolf. ... PZ: I can't imagine anything more wrong-headed. And you say Rovelli is a big shot? That is why I say Rovelli's position is incoherent. JS: Is coherence in the mind of the beholder? PZ: If you want to treat g_uv as a physical field, based on its dynamical character (i.e. matter-dependence), then the natural thing to do is separate the background generalized Minkowski kinematical metric (which is NOT matter dependent) as chronogeometric, and treat the gravitational g_uv alone as physical -- by Rovelli's own argument. JS: That will not work. It will not tip the light cones. It will not give the correct bending of light. Also you are too vague on what you mean by matter on the RHS of Guv(Einstein) = -(alpha)(alpha')Tuv(Matter) alpha = e^2/hc ~ 1/137 alpha' = 8pi(Witten's reciprocal string tension) alpha' ~ (10^-32 cm)^2 in a common convention. I also include on RHS tuv(Vacuum) ~ [(alpha)(alpha')]^-1/zpfguv /zpf = Lp^-1[Lp^3|Vacuum Coherence|^2 - 1] Repulsive dark energy is /zpf > 0 Attractive dark matter is /zpf < 0 FRW Omega(Dark Energy + Dark Matter) ~ 0.96 With Omega(Total) = 1 i.e. FLAT SPACE Preferred foliation is where CMB is maximally isotropic to ~ 10^-5. This allows accurate navigation with weightless warp drive and traversable wormholes both supported by configurations of dark energy and dark matter exotic w = -1 vacua. PZ: Yet he doesn't even mention this. JS: There was a time when people spoke their minds, and were not afraid to offend - and that since then, too many truths have been buried. Mark Kurlansky, 1968: The Year That Rocked The World ... JS: Also at quantum level Rovelli mention's Dirac's insight that the Heisenberg Picture is better than Schrodinger Picture and that time as in the flow of time's arrow is not dynamical time but is a statistical thermodynamic construct. PZ: Sure, if you want to quantize everything and abandon time. Unruh's for example, and some of the others. Unruh has some important ideas however. PZ: But then I fail to see any material distinction between relabeling the bare unindividuated spacetime points (passive diffeomorphism), on the one hand, and shifting all physical fields (including the GR metric field) with respect to such a raw manifold (active diffeomorphism), on the other. Kretschman's point appears to be fully valid in both cases: how can physics in general -- any sensible physics -- possibly depend on a mere re-labeling of raw unindividuated spacetime points; or, for that matter, on a common shift of all physical systems, including the *physical* metric field g_uv, with respect to such a set of unindividuated points? JS: Admittedly a sticky wicket that I also need to understand more deeply. PZ: It's an artificial model that has nothing to do with physical relativity IMO. JS: I found this remark by Goldstein helpful: In the ADM formulation 4-diffeomorphism invariance amounts to the requirement that one ends up with the same space-time, up to coordinate transformations, regardless of which path in multi-fingered space-time is followed, i.e. which lapse function N is uses. p.278 OK so this idea goes back to the archetypal notions of classical thermodynamics with a state function, to holonomic integrability of equality of mixed partial derivatives with not multiply-connected manifolds, to a closed Cartan exterior differential form on a cycle (no boundary), no topological defects and all that. The passive coordinate transformations are like EM gauge transformation (e/c)Au -> (e/c)Au + hChi,u in a fixed gauge constraint. PZ: Rotating a system in isotropic space is connected with a true physical symmetry of the system including the vacuum in which it is embedded. Yes, the symmetry of the system Hamiltonian under such a transformation is an active transformational symmetry that is characteristic of the particular system -- unlike the invariance of the proper formal expression of any physical law under a *mere* coordinate transformation. JS: Note also 12.2.2 p.279 alluding to your digging up Kretchmann from Dr. Frankenstein's favorite graveyard. ;-) The fundamental symmetry at the heart of general relativity is its invariance under general coordinate transformations of spacetime. It is important to stress that almost any theory can be formulated in such a 4-diffeomorphism invariant manner by adding further structure to the theory (e.g. a preferred foliation of spacetime as a dynamical object). General relativity has what is sometimes called serious diffeomorphism invariance, meaning that it involves no spacetime structure beyond the 4-metric and, in particular, singles out no special foliation of spacetime. Goldstein and Teufel then knock standard QGrav including even Ashtekar -> Loop Spin Foams that are perhaps near to being falsified by NASA's EINSTEIN in Jack better check this story out: Gary S. Bekkum ... How to get localization in space and the flow of time as we experience it in our immediate inner consciousness has nothing to do with the particular local coordinate representation like r and t in, for example, K = e^2GM/c^2r dr/dt = c/K for null geodesic and in his Tables generally in the context of potentially practical metric engineering of the guv field using the EM Au field in spite of the enormous gravity string tension ~ c^4/G ~ 10^19 Gev per 10^-33 cm. PZ: You are blocking his actual definition of r. Classic operationalism does not apply to a theory of this type. That is a critical point. Miss that and of course nothing makes sense. JS: I am not ready to renounce PW Bridgman's Operationalism. Indeed, nothing Hal Puthoff says about the foundations of his PV makes any sense to my mind. If it ain't broke, don't fix it. Of course I am not a doctrinaire positivist like Stephen Hawking proudly proclaims he is. ... JS: Correct, with the proviso that matter includes both real, i.e. on mass shell, sources as well as virtual, i.e. off mass shell sources. The virtual sources divide into two classes: I. Non-exotic near EM field Fuv giant coherent quantum states of virtual photons that contribute to Omega(Matter) of the FRW metric and to Tuv in Einstein's local field equation. II. Exotic vacuum w = -1 zero point stress-energy density local tensor ~ (String Tension)/zpfguv for both repulsive dark energy /zpf > 0 of negative pressure and attractive dark matter /zpf < 0 of positive pressure. These exotic vacuum virtual sources contribute Omega(Exotic Vacua) ~ 0.96 to Omega(Total) = 1 in our large-scale spatially flat post-inflationary local Level I Hubble sphere brane world as in Lenny Susskind's megalopolis Landscape subject to the natural selection of the Weak Anthropic Principle (WAP) OK, I think I made an error above including brane worlds in sense of parallel worlds? Here is why I think I made an error (If I did so did Hawking and Scientific American in their popular science reports): D-Branes are extended surfaces without edges. In order that the black hole be a localized object, it is assumed that our ordinary four dimensions (three space and one time) are all orthogonal to these D-Brane surfaces ... Thus to us these D-Branes would look as though they were located at a point (or at least a very small region) of our observable three dimensions of space. W. G. Unruh p. 168 Black holes, dumb holes, and entropy, i.e. the D-Branes are in the compactified Calabi-Yau space. I have to look again at Hawking's The Universe in a Nutshell that seems to give the wrong idea here? Perhaps I mis-remember? Note also Ed Witten's formula generalizing Heisenberg's quantum uncertainty principle, i.e. eq. (5.9) p. 136 Delta X > h/DeltaP + alpha'(DeltaP)/h The second gravity-string source of uncertainty should give the irreversible statistical arrow of time not found when alpha' = 0 i.e. infinite string tension, or infinite space-time stiffness of action without reaction as is also found in the signal locality of orthodox quantum theory in sense of Antony Valentini's papers. Mass without mass, but with a strong micro-gravity G* ~ 10^40G on 1 fermi scale. The wormhole has an attractive dark matter exotic vacuum core where Vacuum Coherence --> 0 just like inside a quantized vortex of circulation in superfluid HeII. Therefore, in the core /zpf ~ - (1 fermi)^-2 w = -1 Therefore the quantum pressure is positive and the exotic vacuum core gravitates as Grad^2V(Exotic Vacuum) ~ -c^2(1 fermi)^-2 this prevents the spread out electric charge from exploding and also compensates any quantized rotation centrifugal forces. For now keep charge and rotation zero for simplicity. That is I here only model a spin 0 neutral micro-geon. That means I would need a high-power graviton laser as the Heisenberg uncertainty scattering probe microscope. That's OK since this is only a gedankenexperiment. The SSS metric then has the factor 1 - 2GM/c2r Where G(mass density) is replaced by c^2/zpf GM is replaced by ~ c^2/zpfR^3 for a micro-geon of size R r is DeltaX as in Witten's formula above with DeltaP as the scattering momentum transfer between gravitons and the micro-geon. Therefore, the Schwarzschild factor is (neglecting factors of 2, pi etc all lumped into dimensionless parameter b 1 - bc^2/zpfR^3/DeltaX The critical value of Delta X is then the event horizon where the Schwarzschild factor vanishes - can it be reached? This micro-geon is a solution of the exotic vacuum local field equation Guv(Einstein) + /zpfguv = 0 At critical Delta X, the geon looks like a POINT PARTICLE from the huge space-warp induced by the probe's momentum transfer Delta P. C is the circumference dC/dR = 2pi (1 - bc^2/zpfR^3/DeltaX)^1/2 Where one considers the radial size fixed at R the scale of the throat of the wormhole. So when can we have DeltaX = bc^2/zpfR^3 ? Note that alpha' may be large of order (10^-11 cm)^2 not (10^-32 cm)^2 as in Witten's idea. ==== Interesting! Jack, Superstring theory does not view the Planck scale of spacetime as a quantum foam but rather as the Planck scale is approached the dimensionality of spacetime goes to 10-d. In string theory the spacetime does fluctuate but this fluctuation is harmonic rather than chaotic. In fact the assumption that the harmonics of the strings absorbs all the quantum fluctuation can be used to derive the dimensionality of spacetime. Also Lorentz invariance is assumed in this calculation. This was first done by L. Brink and H.B. Nielsen in 1973 (A Simple Physical Interpretation of the Critical Dimension of Space-time in Dual Models, *Physics Letters* 45B:4 (1973) 332-336. This paper is also included in the anthology edited by John Schwartz, *Superstrings: the First 15 Years of Superstring Theory, Vol. 1* (World Scientific, 1985). [Note that Dual Models is the old name for string theory, when it was still evolving away from the terminology of s-matrix theory. However, the Mandelstam labels of S, T, and U duality have resurfaced in membrane theory!] I mentioned the Brink-Nielsen view of string theory in my paper Notes on pdf.] Also on page 6 of F.W. Stecker's pdf paper [referred to in the NASA report], he says We note that there are variants of quantum gravity and large extra dimension models which do not violate Lorentz invariance and for which the constraints considered here do not apply. BTW: In the Acknowledgments on page 8, Serge Rudaz is one of four people thanked for helpful discussions. I remember meeting Serge Rudaz in seminars. He was then a young physics student (at Cornell?) who was an old acquaintance of yours. He told us about instantons -- which was a new idea then. I have not heard of him since 1976 -- until seeing this acknowledgement! Nuff said! Saul-Paul JS: Yes, Serge was with me at UCSC in Summer 1973 when I went to see Jean Cocteau's Orphee on campus with Helen Quinn (who was close to having he baby at that time) and I think Serge and a few others. This was days before I went to SRI to meet with Puthoff and Targ on the tape you have. The story is in the book Destiny Matrix. ---------- Therefore quantum foam is suspect. Also Bohm's quantum potential view of vacuum fluctuations is relevant in context of the recent paper from Teheran. I need to follow the experiment below more carefully. Jack better check this story out: Gary S. Bekkum ==== i have realised a few mathematical functions?, properties?, lately and was wondering about them, such as whether they are already known, or are even important. if anyone is interested, hit me up,AIM:heartxenocide ==== I really hope no one takes this as spam, I'm a student member of the MAA, and I've posted here a few, brief times, however since I'm still a lower division undergraduate most of the subject matter is far above my head, so I'm not a regular (yet!) With that said, I have created some math t-shirt designs that just make me smile and I'd like to share them with the math community, with the hopes that they delight someone else as much as they delight me. the url is http://www.cafeshops.com/subjective and the designs are under Math and Science (click on the Einstein picture) All comments, critiques, commendations, cat-calls and creative input ==== lurking travis > I really hope no one takes this as spam, I'm a student member of the > MAA, and I've posted here a few, brief times, however since I'm still > a lower division undergraduate most of the subject matter is far above > my head, so I'm not a regular (yet!) ... > the url is http://www.cafeshops.com/subjective and the designs are > under Math and Science (click on the Einstein picture) Baseball is ninety percent mental. The other half is physical. -- Yogi Berra LH ==== I am stuck in high school maths mode, and can't seem to get into university level maths. This might be because I am entirely self taught, but I don't know. Does anyone else have this problem? I am at a level where I understand most high school maths, and I've studied Calculus Made Easy. It would be helpful to have some kind of way to check my knowledge. Does anyone know of any good textbooks that cover high school maths with worked exercises, and any texts that help the transition from high school maths to the more exciting stuff at university level? A book that takes time to explain things, point out applications and say why rather than just how. I am trying to understand maths, not just learn some techniques or shortcuts. By the way, I would prefer internet resources and small paperback books. There's no way I can afford college textbooks unfortunately. Johnathan ==== > I am stuck in high school maths mode, and can't seem to get into > university level maths. This might be because I am entirely self > taught, but I don't know. Does anyone else have this problem? Most people who went to high school have this problem. You have my sympathies. > I am at a level where I understand most high school maths, and I've > studied Calculus Made Easy. It would be helpful to have some kind of > way to check my knowledge. Does anyone know of any good textbooks that cover high school maths > with worked exercises, There are lots of books like that. Several of the Schaum's outlines books cover what is (really) high school math with lots and lots of worked examples. The explanations in most of the Schaum's books are not half bad as well. > and any texts that help the transition from > high school maths to the more exciting stuff at university level? A > book that takes time to explain things, point out applications and say > why rather than just how. The best book I have seen of this genre you desire is Basic Mathematics by Serge Lang (Springer-Verlag), which teaches all of secondary mathematics from a university mathematics viewpoint. It has answers to selected exercises, some worked examples, and an excellent overall perspective on how concepts fit together. I am enjoying using Lang's book with my son as he studies math at home. Hope this helps! Good luck. ==== I am stuck in high school maths mode, and can't seem to get into > university level maths. This might be because I am entirely self > taught, but I don't know. Does anyone else have this problem? I am at a level where I understand most high school maths, and I've > studied Calculus Made Easy. It would be helpful to have some kind of > way to check my knowledge. Does anyone know of any good textbooks that cover high school maths > with worked exercises, and any texts that help the transition from > high school maths to the more exciting stuff at university level? A > book that takes time to explain things, point out applications and say > why rather than just how. I am trying to understand maths, not just > learn some techniques or shortcuts. By the way, I would prefer internet resources and small paperback > books. There's no way I can afford college textbooks unfortunately. Johnathan You could check out: http://store.doverpublications.com/by-subject-science-and-mathematics-mathem atics-calculus.html they are cheap and useful. You could also try Schaum's outlines; they are also very good. For web content, try: http://www.math.temple.edu/~cow/ If you run a google search, you will find a plethora of math related websites. Many, many, many more than I care to list. Make google your best friend. You can find anything you could possibly want to know using google in about 30 seconds. Lurch ==== > > I am stuck in high school maths mode, and can't seem to get into > university level maths. This might be because I am entirely self > taught, but I don't know. Does anyone else have this problem? > > I am at a level where I understand most high school maths, and I've > studied Calculus Made Easy. It would be helpful to have some kind of > way to check my knowledge. > > Does anyone know of any good textbooks that cover high school maths > with worked exercises, and any texts that help the transition from > high school maths to the more exciting stuff at university level? A > book that takes time to explain things, point out applications and say > why rather than just how. I am trying to understand maths, not just > learn some techniques or shortcuts. > > By the way, I would prefer internet resources and small paperback > books. There's no way I can afford college textbooks unfortunately. > > Johnathan > > You could check out: > http://store.doverpublications.com/by-subject-science-and-mathematics-mathema tics-calculus.html > > they are cheap and useful. You could also try Schaum's outlines; they are > also very good. > > For web content, try: > > http://www.math.temple.edu/~cow/ > > If you run a google search, you will find a plethora of math related > websites. Many, many, many more than I care to list. Make google your best > friend. You can find anything you could possibly want to know using google > in about 30 seconds. > > Lurch In particular, include ext:pdf site:.edu in your searches. You'll just get postscript notes, usually written up by professors. 'cid ==== In particular, include ext:pdf site:.edu in your searches. You'll > just get postscript notes, usually written up by professors. 'cid That will actually give you adobe acrobat files. If you want postscript, try ext:ps site:.edu. Or don't, unless you know a plug-in for IE on the PC which allows it to read postscript files (I don't). ==== > > In particular, include ext:pdf site:.edu in your searches. You'll > just get postscript notes, usually written up by professors. > > 'cid That will actually give you adobe acrobat files. If you want postscript, try > ext:ps site:.edu. Or don't, unless you know a plug-in for IE on the PC > which allows it to read postscript files (I don't). > Try Ghostscript (Post-script viewer) for IE and windows. ==== There was a time when people spoke their minds, and were not afraid to offend - and that since then, too many truths have been buried. Mark Kurlansky, 1968: The Year That Rocked The World (Ballantine, 2004) thanks to Frank Lauria In 1968 was in La Jolla CA at UCSD during time Greg Benford describes in Timescape living on Bonair Street Wind an Sea Beach near Unicorn Theater with Ken Kesey's Merry Prankster Bus parked nearby frequently. I was also teaching at San Diego State with Fred Alan Wolf. OK, now I have looked at the rest of Rovelli's argument, I'll work through this one more time. PZ: Rovelli's position is very odd. First he says: ...Of course, nothing [in GR] prevents us... from singling out the gravitational field as 'the more equal among equals', and declaring that location is absolute in GR, because it can be defined with respect to it. (p 108) JS: But he rejects that Paul. He says doing that misses the great Einsteinian insight. The great Einsteinian insight being the conditional and physical nature of the gravitational metric field -- which conflicts with the great Einsteinian insight strict equivalence. The opposite of a profound truth is another profound truth. -- Niels Bohr What Rovelli doesn't seem to understand is that this all makes perfect sense once you give up strict equivalence and distinguish the background and physical metrics. JS: I do not understand this distinction. Please give more details what you mean. Have you read pp. 112 - 114 that completely demolishes Hal Puthoff' s use of dr/dt = c' = c/K radial null geodesic in his Tables. PZ: It does no such thing. I would not even characterize pp 112-114 as an argument. It is simply a sketch of a model in which *everything* is quantized except the raw manifold. JS: It shows no intrinsic meaning to Puthoff's r and t as he means it in his Tables. PZ: He wants to throw away time in order to keep a unified g_uv. Why do you think this is an argument against c' = c/K? As far as I can see it is simply a different theory. Rovelli briefly considers an alternative approach in which we retain the Minkowski background of standard QFT. But there he has g_uv = n_uv + fluctuations which makes no sense to me. He does not seem to understand the distinction between kinematical g_uv and dynamic gravitational g_uv. He cannot get his head over the unified metric. What does he mean by fluctuations? JS: What do you mean by kinematical g_uv and dynamic gravitational g_uv apart from Ruvwl = 0 in the former and not in the latter. JZ: Sounds like sheer nonsense to me. Nonsense because it is divorced from the physical fundamentals. JS: No argument from me on that one. PZ: This of course is exactly what the classic Einstein chronogeometric model does, going all the way back to special relativity. JS: I think you are misreading Rovelli. PZ: The Einsteinian model is a chronogeometric model, in which the metric g_uv reflects the fundamental structure of spacetime. JS: If you mean, for example, dT = goo^1/2dt dR = grr^1/2dr Then I agree that dT and dR are physical and dt and dr are not. However, to get a gravity shift of light frequency do dT'/goo'^1/2 = dT/goo^1'2 treating dt as a kind of nonlocal invariant. If you mean more than this, then explain with detailed examples. PZ: That is the great Einsteinian insight -- which is. unfortunately, based on strict Einstein equivalence, which is fictitious. JS: Again I really do not understand what you mean by this sentence. PZ: Now Rovelli wants to pretend that the great Einsteinian insight is something else entirely -- that the gravitational field is a physical field. Loony tunes. He is tying himself up in knots. JS: Yes, in global special relativity, NO in local general relativity. PZ: In Einstein general relativity, the unified metric g_uv represents the fundamental structure of spacetime, and in inseparable from it. That's Einstein. It doesn't even make sense to me to say that this chronogeometric model holds only locally. PZ: That is precisely what distinguishes the Einsteinian from the Lorentzian model. The transformational and metric structure is not, in the classic Einsteinian model, separable from spacetime itself. But then he says: There is no absolute referent of motion in GR; the dynamical fields move with respect to each other. (p 108) JS: Again you are misreading. There is no contradiction here in Rovelli's argument. PZ: This is like arguing that everything is relative because everything is relative to the absolute, including the absolute. Sounds like Lewis Carroll. JS: ;-) PZ: None of this has anything to do with Einstein general relativity, which treats the inertial field as real. The unified metric derives its physical meaning from this equivalence. Rovelli ignores all this and simply takes the unified Einstein g_uv as given. PZ: This second assertion seems to approach the metric field of GR as just another field, which is very close to and even indistinguishable from the physical rubber rod and clock model of PV and Yilmaz -- just another physical field which happens to have metric properties. And of course such a field is fully relational with respect to unindividuated spacetime points on a raw manifold stripped of all coordinate systems, transformation properties, and metrics. JS: Yes on just another field. But NO that it's like PV and Yilmaz. Not true at all because, at least in PV, Hal uses an absolute non-dynamical background global Minkowski space PZ: That is what Rovelli *should* be doing, but he doesn't even consider this possibility. He seems to think you can treat unified g_uv as a physical field. JS: Why do you think you cannot? PZ: I can't imagine anything more wrong-headed. And you say Rovelli is a big shot? That is why I say Rovelli's position is incoherent. JS: Is coherence in the mind of the beholder? PZ: If you want to treat g_uv as a physical field, based on its dynamical character (i.e. matter-dependence), then the natural thing to do is separate the background generalized Minkowski kinematical metric (which is NOT matter dependent) as chronogeometric, and treat the gravitational g_uv alone as physical -- by Rovelli's own argument. JS: That will not work. It will not tip the light cones. It will not give the correct bending of light. Also you are too vague on what you mean by matter on the RHS of Guv(Einstein) = -(alpha)(alpha')Tuv(Matter) alpha = e^2/hc ~ 1/137 alpha' = 8piWitten's reciprocal string tension) alpha' ~ (10^-32 cm)^2 in a common convention. I also include on RHS tuv(Vacuum) ~ [(alpha)(alpha')]^-1/zpfguv /zpf = Lp^-1[Lp^3|Vacuum Coherence|^2 - 1] Repulsive dark energy is /zpf > 0 Attractive dark matter is /zpf < 0 FRW Omega(Dark Energy + Dark Matter) ~ 0.96 With Omega(Total) = 1 i.e. FLAT SPACE Preferred foliation is where CMB is maximally isotropic to ~ 10^-5. This allows accurate navigation with weightless warp drive and traversable wormholes both supported by configurations of dark energy and dark matter exotic w = -1 vacua. PZ: Yet he doesn't even mention this. JS: There was a time when people spoke their minds, and were not afraid to offend - and that since then, too many truths have been buried. Mark Kurlansky, 1968: The Year That Rocked The World PZ: So he has not made any argument at all against a bimetric approach. He has simply ignored it, even though his own argument points to it. JS: with a literal meaning for coordinates r & t, otherwise his formula dr/dt = c/K is physically meaningless, which it is in GR as explained by Rovelli on pp. 112 - 114. PZ: Only if you insist exclusively on operational meaning -- which would also knock out F = ma, which by a similar argument is empirically meaningless. JS: Yes, until you also add things like F = - GMmr/r^3 F = eE + e(v/c)xB etc. That is, Hal's use of engineering is precisely missing the third step on Rovelli's p. 108. PZ: Rovelli's third step is merely a sketch of a proposal to quantize everything, while retaining a unified g_uv treated in its entirety as a physical field. This is not an argument. Rovelli's argument for his proposal, such as it is, is to be found on p 109: In my *opinion*, this is the right way to go. Well, in my *opinion*, it's fundamentally wrong-headed. In addition, Rovelli has grossly mischaracterized the Feynman-type alternative -- which shows that he has not understood it; although he is right that under this alternative, the GR conceptual revolution is indeed dis-valued, but to a much greater extent than he seems to realize. JS: Also p. 112 Recall that Einstein described his great intellectual struggle to find GR as 'understanding the meaning of coordinates'. This is my key objection to Puthoff's and Ibison's use of r ant t in their PV modeling. They do not IMHO understand the meaning of coordinates: in GR. Neither does Yilmaz apparently. PZ: I have no idea of what you are getting at here. Einstein's point about coordinates depends entirely on his strict equivalence thesis, which is untenable. JS: I guess I have never understood your idea here. Can you try to explain it again with complete clarity? PZ: Once you abandon strict equivalence, the unification of gravitational and inertial g_uv is formal and arbitrary. It has no deep physical meaning. Then we recover chronogeometric background coordinates together with a physical rubber-rod- and-clock non-inertial g_uv. Rovelli doesn't seem to understand that. JS: Neither do I. pp 112 - 114 make this clear IMHO. PZ: This is not an argument -- it's merely a proposal: In my opinion.... JS: Also at quantum level Rovelli mention's Dirac's insight that the Heisenberg Picture is better than Schrodinger Picture and that time as in the flow of time's arrow is not dynamical time but is a statistical thermodynamic construct. PZ: Sure, if you want to quantize everything and abandon time. Unruh's for example, and some of the others. Unruh has some important ideas however. PZ: But then I fail to see any material distinction between relabeling the bare unindividuated spacetime points (passive diffeomorphism), on the one hand, and shifting all physical fields (including the GR metric field) with respect to such a raw manifold (active diffeomorphism), on the other. Kretschman's point appears to be fully valid in both cases: how can physics in general -- any sensible physics -- possibly depend on a mere re-labeling of raw unindividuated spacetime points; or, for that matter, on a common shift of all physical systems, including the *physical* metric field g_uv, with respect to such a set of unindividuated points? JS: Admittedly a sticky wicket that I also need to understand more deeply. PZ: It's an artificial model that has nothing to do with physical relativity IMO. JS: I found this remark by Goldstein helpful: In the ADM formulation 4-diffeomorphism invariance amounts to the requirement that one ends up with the same space-time, up to coordinate transformations, regardless of which path in multi-fingered space-time is followed, i.e. which lapse function N is uses. p.278 OK so this idea goes back to the archetypal notions of classical thermodynamics with a state function, to holonomic integrability of equality of mixed partial derivatives with not multiply-connected manifolds, to a closed Cartan exterior differential form on a cycle (no boundary), no topological defects and all that. The passive coordinate transformations are like EM gauge transformation (e/c)Au -> (e/c)Au + hChi,u in a fixed gauge constraint. PZ: Rotating a system in isotropic space is connected with a true physical symmetry of the system including the vacuum in which it is embedded. Yes, the symmetry of the system Hamiltonian under such a transformation is an active transformational symmetry that is characteristic of the particular system -- unlike the invariance of the proper formal expression of any physical law under a *mere* coordinate transformation. JS: Note also 12.2.2 p.279 alluding to your digging up Kretchmann from Dr. Frankenstein's favorite graveyard. ;-) The fundamental symmetry at the heart of general relativity is its invariance under general coordinate transformations of spacetime. It is important to stress that almost any theory can be formulated in such a 4-diffeomorphism invariant manner by adding further structure to the theory (e.g. a preferred foliation of spacetime as a dynamical object). General relativity has what is sometimes called serious diffeomorphism invariance, meaning that it involves no spacetime structure beyond the 4-metric and, in particular, singles out no special foliation of spacetime. Goldstein and Teufel then knock standard QGrav including even Ashtekar -> Loop Spin Foams that are perhaps near to being falsified by NASA's EINSTEIN in Jack better check this story out: Gary S. Bekkum PZ: However, moving *everything physical* -- including the unified g_uv -- along the raw spacetime manifold is quite another kettle of fish. It is devoid of physical content IMHO. But again, a space-time frame of reference is not *merely* a coordinate transformation, since it represents the *motion* of a possible observer. Thus there is no *a priori* reason why physical laws should take the same form in different frames of reference, contrary to Einsteinian doctrine. Of course they *may* as a matter of fact, but that is very different from insisting a priori that they always *should*. That's yet another Einsteinian red herring. PZ: What is physically significant here is not this abstract and artificial definition of diffeomorphism with respect to a raw manifold conceived as an unindividuated point set, but a shift of physical fields with respect to the metric-transformational structure of spacetime, which of course in GR depends on the distribution of matter. JS: Precisely, yes that is the idea I think. It's a return to I think Leibniz's relationism? PZ: Descartes. He is almost literally putting Descartes before the horse of physical relativity. JS: ;-) How to get localization in space and the flow of time as we experience it in our immediate inner consciousness has nothing to do with the particular local coordinate representation like r and t in, for example, K = e^2GM/c^2r dr/dt = c/K for null geodesic and in his Tables generally in the context of potentially practical metric engineering of the guv field using the EM Au field in spite of the enormous gravity string tension ~ c^4/G ~ 10^19 Gev per 10^-33 cm. PZ: You are blocking his actual definition of r. Classic operationalism does not apply to a theory of this type. That is a critical point. Miss that and of course nothing makes sense. JS: I am not ready to renounce PW Bridgman's Operationalism. Indeed, nothing Hal Puthoff says about the foundations of his PV makes any sense to my mind. If it ain't broke, don't fix it. Of course I am not a doctrinaire positivist like Stephen Hawking proudly proclaims he is. PZ: So it appears that Rovelli ends up in a position where he is essentially arguing that the matter- dependence of the metric field implies that the GR metric is really a physical metric, and the GR metric field is to be regarded as a physical field like any other physical field. That is, the *unified* metric field. That is Rovelli's tacit, yet arbitrary, assumption IMO. Then, the physics depends only on the relative disposition of the various physical fields with respect to each other. JS: Correct, with the proviso that matter includes both real, i.e. on mass shell, sources as well as virtual, i.e. off mass shell sources. The virtual sources divide into two classes: I. Non-exotic near EM field Fuv giant coherent quantum states of virtual photons that contribute to Omega(Matter) of the FRW metric and to Tuv in Einstein's local field equation. II. Exotic vacuum w = -1 zero point stress-energy density local tensor ~ (String Tension)/zpfguv for both repulsive dark energy /zpf > 0 of negative pressure and attractive dark matter /zpf < 0 of positive pressure. These exotic vacuum virtual sources contribute Omega(Exotic Vacua) ~ 0.96 to Omega(Total) = 1 in our large-scale spatially flat post-inflationary local Level I Hubble sphere brane world as in Lenny Susskind's megalopolis Landscape subject to the natural selection of the Weak Anthropic Principle (WAP) OK, I think I made an error above including brane worlds in sense of parallel worlds? Here is why I think I made an error (If I did so did Hawking and Scientific American in their popular science reports): D-Branes are extended surfaces without edges. In order that the black hole be a localized object, it is assumed that our ordinary four dimensions (three space and one time) are all orthogonal to these D-Brane surfaces ... Thus to us these D-Branes would look as though they were located at a point (or at least a very small region) of our observable three dimensions of space. W. G. Unruh p. 168 Black holes, dumb holes, and entropy, i.e. the D-Branes are in the compactified Calabi-Yau space. I have to look again at Hawking's The Universe in a Nutshell that seems to give the wrong idea here? Perhaps I mis-remember? Note also Ed Witten's formula generalizing Heisenberg's quantum uncertainty principle, i.e. eq. (5.9) p. 136 Delta X > h/DeltaP + alpha'(DeltaP)/h The second gravity-string source of uncertainty should give the irreversible statistical arrow of time not found when alpha' = 0 i.e. infinite string tension, or infinite space-time stiffness of action without reaction as is also found in the signal locality of orthodox quantum theory in sense of Antony Valentini's papers. PZ: But while this may allow or even imply a relational view of the raw spacetime manifold, in the Cartesian sense, it is clearly not physical general relativity, in the classic Einsteinian sense, which requires fundamental identification of the inertial and gravitational metrics. JS: I do not understand what you just said. EEP in every significant sense still holds IMHO. PZ: You are in a loop. JS: Ground Hog Day. Help let me out of my bottle Oh Thief of Baghdad. The Magic Flying Carpet is obviously the weightless Alcubierre timelike geodesic faster than light warp drive powered by dark energy and dark matter exotic vacua configurations. See my animated picture of this in http://qedcorp.com/APS/StarGate1.mov PZ: EEP is NOT Einstein equivalence. EEP is merely a correspondence principle. Einstein strict equivalence is NOT valid. And even EEP is not strictly valid as advertised (see e.g. Ohanian and Ruffini Ch. 1). given guv field still works for example. PZ: This simply reflects weak equivalence -- equality of gravitational and inertial mass. That does not imply Einstein equivalence. It works whether we use a unified g_uv or go to a bimetric formalism, and regardless of whether we consider inertial forces as real or fictitious. JS: The so-called fictitious or inertial forces, e.g. G-Force when a jet takes off (0 to 120 mph in 2 sec) from an aircraft carrier are physically real. I can tell you that from first-hand experience. The use of fictitious like the use of hidden variable are unfortunate choices of words. Extra variable is better for the latter. PZ: So this is not a valid argument IMO. It is an example of the classical logical fallacy affirming the antecedent. JS: The tidal stretch-squeeze works etc. PZ: Of course. So? The point is that you can always tell the difference between a real gravity field and an inertial field. They are NOT completely physically equivalent. JS: If you mean Ruvwl =/= 0 locally for a real gravity field, and is zero for an inertial field, I agree. However, I do not see how you can write guv = guv(inertial) + guv(real) That is, apart from guv = Globally Flat Metric + du,v + dv,u I do not claim we can physically distinguish the two terms on RHS nor is the second term small compared to first. The Bohm pilot constraint for the extra variable, i.e. actual distortion of Hagen Kleinert's elastic-plastic 4D world crystal lattice i.e. density of vortex line topological defects of curvature disclination and possibly torsion dislocations is du = Lp^2(Goldstone Phase of Vacuum Coherence),u ,u is ordinary partial derivative. Compare my equation here to Goldstein's eq. 12.4 on p. 282. PZ: Consequently, unified g_uv has no deep physical meaning or necessity -- as Feynman argued. PZ: Which all seems to contradict his statement, ...Of course, nothing [in GR] to prevents us... from singling out the gravitational field as 'the more equal among equals', and declaring that location is absolute in GR, because it can be defined with respect to it. JS: No, you have simply misread the context of Rovelli's remark. Read it again more carefully. This is not a valid objection at all to Rovelli's argument IMHO. The way I read him, his argument is splendidly globally self-consistent. PZ: The way I read him, he doesn't actually *have* an argument. He simply *prefers* to quantize space and time intervals rather than unpack Einstein's unified g_uv. PZ: So IMO Rovelli's position is incoherent. If anything, he is arguing for an *alternative* non-Einsteinian interpretation of GR in which matter-dependent g_uv is to be treated as a *physical field*, which means it can in principle be distinguished from the kinematical inertial field with its trivially valid transformable metric tensor. JS: We are back to Square One. The Rock has rolled back down on Sisyphus. I cannot pinpoint the precise misconception that is driving your position here. PZ: It is no wonder you are having this difficulty, since it is *Rovelli's* misconception that is causing the problem. Rovelli doesn't seem to understand that recognition of the dynamical character of non-inertial g_uv fundamentally distinguishes it from the kinematical g_uv, which latter doesn't depend on the distribution of matter. So the natural thing to do is to split unified g_uv and thus clearly separate the dynamical and kinematical metric fields. JS: Show me an algorithm to do that in any problem. PZ: Yet he doesn't even consider this. In other words, Rovelli is really a *Yilmazian*. JS: Not in my book. PZ: Actually, you are right -- because he does not follow the natural development of his own argument to its logical conclusions. If he did, his first alternative (p 109) would fall squarely in the Feynman-Yilmaz category, with rubber rods and clocks and a flat background inertial metric. JS: It seems to me the objective is to eliminate any flat background non-dynamical metric. Nondynamical means ACTION WITHOUT REACTION. Note, I have it formally in my equations but it is not physically measurable separately. Indeed, the pre-inflationary vacuum is the perfectly flat world crystal with no vortex string topological defects of effective multiple-connectivity in the space of the vacuum coherence order parameter which is zero everywhere prior to the inflationary vacuum phase transition like a normal metal going super-conducting. PZ: Obviously it is a field that depends on the distribution of matter, if that's what he means. But it is the only field that is defined in terms of a spacetime metric, so it is obviously unique in that sense. JS: Read Rovelli's paper in the book. Then tell me what you think. PZ: I read it. Several times. I've been scratching my head for several days now. See above. ==== > I can't find a proof of the (multivariable) chain rule that makes sense. Assume g is differentiable at x and f is differentiable at g(x), where differentiable means in the total derivative sense. Let dg(x) and df(g(x)) denote the associated linear differential approximants. Then f(g(x+h)) - f(g(x)) = df(g(x))[g(x+h)) - g(x)] + o(g(x+h)) - g(x)) = df(g(x))[dg(x)(h) + o(h)] + o(g(x+h)) - g(x)) = df(g(x))[dg(x)(h)] + df(g(x))[o(h)] + o(g(x+h)) - g(x)). The first term in the last sum is what we want to see, namely the composition of df(g(x)) and dg(x) evaluated at h. So we're done if we can show both df(g(x))[o(h)] and o(g(x+h)) - g(x)) are o(h), which is not too hard. ==== > I can't find a proof of the (multivariable) chain rule that makes sense. Assume g is differentiable at x and f is differentiable at g(x), where >differentiable means in the total derivative sense. Let dg(x) and >df(g(x)) denote the associated linear differential approximants. Then f(g(x+h)) - f(g(x)) = df(g(x))[g(x+h)) - g(x)] + o(g(x+h)) - g(x)) = df(g(x))[dg(x)(h) + o(h)] + o(g(x+h)) - g(x)) = df(g(x))[dg(x)(h)] + df(g(x))[o(h)] + o(g(x+h)) - g(x)). The first term in the last sum is what we want to see, namely the >composition of df(g(x)) and dg(x) evaluated at h. So we're done if we can >show both df(g(x))[o(h)] and o(g(x+h)) - g(x)) are o(h), which is not too >hard. Fine, but she said I want to see a real proof that uses definitions and little greek letters... I suppose we can leave it to the reader to change a few letters to greek to make things easier to follow. But if she's never seen this proof then possibly she's never seen the definition: (Def: o(h) means some function e(h) with the property that e(h)/||h|| -> 0 as h -> 0.) Def: f: R^n -> R^m is _differentiable_ at x if there exists a linear map T: R^n -> R^m such that f(x + h) = f(x) + Th + o(h); if so then T = df(x). (Then we should also point out how the notation she may be looking at follows from this: If df(x) exists then the matrix corresponding to that linear map has the partial derivatives of the components of f for its entries, and the formula you see for the chain rule in some calculus books is just giving the product of two matrices.) ************************ David C. Ullrich ==== > > I can't find a proof of the (multivariable) chain rule that makes sense. >>Assume g is differentiable at x and f is differentiable at g(x), where >>differentiable means in the total derivative sense. Let dg(x) and >>df(g(x)) denote the associated linear differential approximants. Then >> f(g(x+h)) - f(g(x)) = df(g(x))[g(x+h)) - g(x)] + o(g(x+h)) - g(x)) >> = df(g(x))[dg(x)(h) + o(h)] + o(g(x+h)) - g(x)) >> = df(g(x))[dg(x)(h)] + df(g(x))[o(h)] >> + o(g(x+h)) - g(x)). >>The first term in the last sum is what we want to see, namely the >>composition of df(g(x)) and dg(x) evaluated at h. So we're done if we can >>show both df(g(x))[o(h)] and o(g(x+h)) - g(x)) are o(h), which is not too >>hard. > > Fine, but she said I want to see a real proof that uses > definitions and little greek letters... > How many students have you found who've said that? Not many? It reads to me as though this person is taking a second year course at an American University. The text books taught from can be pretty dire when it comes to formal proof. Indeed one teaches three 'different' (at least) chain rules for special cases, rather than just _the_ chain rule. I'm sure that the course instructor would be happy to provide some way for the OP to satisfy her curiosity, even if it's pointing her towards the library and a relevant book. But just in case, as you're the Analyst round here, which ones would be useful in demonstrating that actually you don't want too many greek letters floating around? Is Rudin any good? Lot's of grad programs seem to prefer it. > I suppose we can leave it to the reader to change a few letters > to greek to make things easier to follow. But if she's never seen > this proof then possibly she's never seen the definition: > > (Def: o(h) means some function e(h) with the property > that e(h)/||h|| -> 0 as h -> 0.) > > Def: f: R^n -> R^m is _differentiable_ at x if there exists a > linear map T: R^n -> R^m such that > > f(x + h) = f(x) + Th + o(h); > > if so then T = df(x). > > (Then we should also point out how the notation she may > be looking at follows from this: If df(x) exists then the matrix > corresponding to that linear map has the partial derivatives > of the components of f for its entries, and the formula you > see for the chain rule in some calculus books is just > giving the product of two matrices.) > > > ************************ > > David C. Ullrich ==== > > I can't find a proof of the (multivariable) chain rule that makes sense. >>Assume g is differentiable at x and f is differentiable at g(x), where >differentiable means in the total derivative sense. Let dg(x) and >df(g(x)) denote the associated linear differential approximants. Then >> f(g(x+h)) - f(g(x)) = df(g(x))[g(x+h)) - g(x)] + o(g(x+h)) - g(x)) >> = df(g(x))[dg(x)(h) + o(h)] + o(g(x+h)) - g(x)) >> = df(g(x))[dg(x)(h)] + df(g(x))[o(h)] >> + o(g(x+h)) - g(x)). >>The first term in the last sum is what we want to see, namely the >composition of df(g(x)) and dg(x) evaluated at h. So we're done if we can >show both df(g(x))[o(h)] and o(g(x+h)) - g(x)) are o(h), which is not too >hard. >> >> Fine, but she said I want to see a real proof that uses >> definitions and little greek letters... >> >How many students have you found who've said that? Not many? It reads to me as though this person is taking a second year course at an >American University. The text books taught from can be pretty dire when it >comes to formal proof. Indeed one teaches three 'different' (at least) chain >rules for special cases, rather than just _the_ chain rule. I'm sure that the course instructor would be happy >to provide some way for the OP to satisfy her curiosity, even if it's >pointing her towards the library and a relevant book. But just in case, as >you're the Analyst round here, which ones would be useful in demonstrating >that actually you don't want too many greek letters floating around? Um, all the comments about greek letters have been tongue in cheek. I don't know what text would be useful for demonstrating that you don't want too many greek letters floating around, because the idea that you don't want too many greek letters floating around is news to me. I suspect that you were being facetious, actually asking about demonstrating something else. But I can't figure out what the something else might be. >Is >Rudin any good? Lot's of grad programs seem to prefer it. >> I suppose we can leave it to the reader to change a few letters >> to greek to make things easier to follow. But if she's never seen >> this proof then possibly she's never seen the definition: >> >> (Def: o(h) means some function e(h) with the property >> that e(h)/||h|| -> 0 as h -> 0.) >> >> Def: f: R^n -> R^m is _differentiable_ at x if there exists a >> linear map T: R^n -> R^m such that >> >> f(x + h) = f(x) + Th + o(h); >> >> if so then T = df(x). >> >> (Then we should also point out how the notation she may >> be looking at follows from this: If df(x) exists then the matrix >> corresponding to that linear map has the partial derivatives >> of the components of f for its entries, and the formula you >> see for the chain rule in some calculus books is just >> giving the product of two matrices.) >> >> >> ************************ >> >> David C. Ullrich ************************ David C. Ullrich ==== Need help on math, visit www.helptosolve.com Just try. Want to help others with math, visit www.helptosolve.com and join the team. ==== > >>Dean said that Northeasterners don't talk about religion. >>That's not true, because I know of Northeasterners who spend almost >>all day talking about The Virgin Mary, Hearing Confession, who's Rabbi >>is where, what Temple this person goes to, I have another Bar Mitzvah, >>Episcopal Parishoners this, American Baptists convention that. >>What Northeast is Dean from ?? > > > What planet are you from? We have people from every planet on Earth in California. --- Former governor Gray Davis -- http://hertzlinger.blogspot.com ==== > Many of the problems with Bush's agenda is the dishonest > reinterpretation of reality that is done so often by the > Democrats. Citations, please. ==== > Dean said that Northeasterners don't talk about religion. > > That's not true, because I know of Northeasterners who spend almost > all day talking about The Virgin Mary, Hearing Confession, who's Rabbi > is where, what Temple this person goes to, I have another Bar Mitzvah, > Episcopal Parishoners this, American Baptists convention that. > > What Northeast is Dean from ?? What does this have to do with mathematics? -- http://hertzlinger.blogspot.com X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft ==== >Dean said that Northeasterners don't talk about religion. K3wl? How is that on topic for sci.math, sci.skeptic, alt.atheism or alt.christnet? Of the 5 groups that you posted to, only alt.politics.democrats is relevant; clearly your intent was to troll for a crss-posted flame war. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== > > Dean said that Northeasterners don't talk about religion. > > Maybe you should furnish an actual quote before you build a castle in the > air upon what you think he might have said. > > ''My father used to tell us how much strength he got from religion, but > we didn't have Bible readings. There are traditions where people do > that. We didn't,'' he said. ''People in the Northeast don't talk about > their religion. It's a very personal private matter, and that's the > tradition I was brought up in.'' Howard Dean, to Boston Globe last > week. Quoted in countless other papers. You gotta interpret that in context. It means northeastern politicians don't talk to the papers about their religion (they leave that to preachers). It was definitely more polite than what I would have said, which would be, I didn't come here today to talk about religion, so what's your next question? ==== > >> Dean said that Northeasterners don't talk about religion. >> >> Maybe you should furnish an actual quote before you build a castle in the >> air upon what you think he might have said. >> >> ''My father used to tell us how much strength he got from religion, but >> we didn't have Bible readings. There are traditions where people do >> that. We didn't,'' he said. ''People in the Northeast don't talk about >> their religion. It's a very personal private matter, and that's the >> tradition I was brought up in.'' Howard Dean, to Boston Globe last >> week. Quoted in countless other papers. You gotta interpret that in context. It means northeastern politicians >don't talk to the papers about their religion (they leave that to >preachers). It was definitely more polite than what I would have said, >which would be, I didn't come here today to talk about religion, so >what's your next question? By the same logic the previous poster used, Bush has also told lies. President Bush's compassionate agenda resonates with the people of both New York and California, -Tracey Schmitt,spokeswoman for the Bush-Cheney '04 campaign That's not true, because I know of Californians who are not impressed with Bush's compassionate agenda. ----- Yang a.a. #28 a.a. pastor #-273.15, the most frigid church of Celcius nee Kelvin EAC Econometric Forecast and Socerey Division Proudly plonked by Lani Girl and Crazyalec The Bush 'balanced' budget: -525 billion and worsening The Bush 'economic' policy: -3 million jobs and counting The Bush Iraq lie: -472 GIs, one friend's co-worker's son and mounting Having Bush fuck up my country: Worthless ==== > Dean said that Northeasterners don't talk about religion. We don't, certainly not when compared to other parts of the country. Most people I know consider it an impolite subject, at least in public & amongst strangers. ==== > Dean said that Northeasterners don't talk about religion. > > Maybe you should furnish an actual quote before you build a castle in the > air upon what you think he might have said. > ''My father used to tell us how much strength he got from religion, but > we didn't have Bible readings. There are traditions where people do > that. We didn't,'' he said. ''People in the Northeast don't talk about > their religion. It's a very personal private matter, and that's the > tradition I was brought up in.'' Howard Dean, to Boston Globe last > week. Quoted in countless other papers. I heard my father once say that people in the South are very conservative religiously. Well I live in Atlanta now and am friends with some folk at the Unitarian Church. My father. What a dirty, rotten liar. ==== |Did you see my reply to the OP? Yes, although the contents were not still clear in mind when I last Amanda's acquaintance objected to her proof based on its being a proof by contradiction. I guess I think the distinction between a proof by proving the contrapositive and a proof by contradiction is fuzzy enough not to make this strictly incorrect. For the question of whether the proof is valid, I think you'd agree it's also not a distinction that could be relevant. People consider the proof okay not because it's crossed this fuzzy line; people consider the proof okay because we're considering a context where proof by contradiction and the whole complex of related forms of reasoning are all accepted. You had two main reasons why proof by contradiction was considered worse: |(i) it's easy to give a wrong proof by contradiction, where |the contradiction arises from some error | |(ii) if you give a direct proof that A implies B, by assuming |A and then deducing B, the steps in the proof can give |some insight into what A really entails - you show A implies |B by showing A implies C and C implies B, and along the |way you've shown two facts that might be interesting and |useful elsewhere, that A implies C and that C implies B. |You don't get this sort of bonus from a proof by contradiction, |since in the course of the proof you're assuming things |which it turns out never actually hold. I think the practice of constructive proof may shed some light on this. The conclusion of a constructive proof tells you more about the nature of the proof than the conclusion of a classical proof, generally. In particular, when the conclusion is negative, the proof was a proof by contradiction (essentially) and when the conclusion is positive it wasn't. I can't tell whether the circumstances reducing the risk of (i) and enhancing the prospects of (ii) are different from the circumstances where the theorem (as it would be stated in constructive terms) is more solidly stated. In constructive mathematics one has the general advice of Bishop to cut down on negations. If we seek to tighten up our theorems by getting rid of flabby negations, then we also as a side-effect refrain from proofs by contradiction. Lots of negations and implications can be upgraded by supplying them with more solid content. Bishop has a paper which uses Goedel's Dialectica paper for this. Goedel has a method of rewriting statements which puts them into a form (there exists w) (for all t) P(w,t) where w ranges over a set of possible witnesses to the truth of the statement, and t ranges over a set of possible cases. If this statement is false, one would like to do more than just negate it; one would like to say something about the function f:w->t for which P(w, f(w)) fails for all w. A proof of the negative statement implicitly provides such an f, as witness. Even when a statement is negative, then, there can be implicit solid content in the construction demonstrating the fact, in terms of information on and around this f. I get the impression that in cases where a theorem is negative, but has relatively tangible constructive content, one is not so especially prone to err in the way described in (i), because one is regarding the construction in the proof (of f, say), which actually *does* exist, and trying to see whether *it* is as advertised. In the proof that there are infinitely many primes, to pick a simple example, the engine is the method for getting a new prime out of a finite set of old primes. An argument which focusses on this kind of concrete construction doesn't seem to suffer from the kind of problem that some attempts at proof by contradiction do. I'm going on impressions, though, and it seems like it would take somewhat tricky casework to get a more solid sense for what kinds of proof do and do not tend to suffer from these problems. Keith Ramsay ==== |Did you see my reply to the OP? Yes, although the contents were not still clear in mind when I last Amanda's acquaintance objected to her proof based on its being a >proof by contradiction. Although I don't think that even he was saying the proof was therefore _wrong_, just that it was not the best possible proof. (Cf for example the title of the thread; it's about finding a beautiful proof, not a _valid_ proof...) >I guess I think the distinction between a >proof by proving the contrapositive and a proof by contradiction is >fuzzy enough not to make this strictly incorrect. I'm honestly not sure what you mean by that sentence, but regarding the minor premise, I don't see what's fuzzy about the distinction. To prove A implies B: (a) Assume ~B. Prove ~A or (b) Assume A and ~B. Prove P and ~P for some P. Some proofs of A implies B have the form (a); some have the form (b). What's fuzzy here? Oh. You're claiming that the distinction is fuzzy enough that those of us who are saying he was wrong in stating the proof was a proof by contradiction are not on solid ground? I disagree. Um, lemme put that a little more strongly: that's not so. The proof Amanda gave is of the form (a), not of the form (b). I mean that's just a mathematical _fact_. You can say you don't see why it matters, fine. But you say above you think the distinction between (a) and (b) is fuzzy - I don't see any fuzziness, and it's simply a fact that (a) is an outline of the proof she gave, for appropriate A and B, while (b) simply isn't. >For the question >of whether the proof is valid, I think you'd agree it's also not a >distinction that could be relevant. People consider the proof okay >not because it's crossed this fuzzy line; people consider the proof >okay because we're considering a context where proof by contradiction >and the whole complex of related forms of reasoning are all accepted. Yes of _course_. Of the people in this thread who do see the difference between (a) and (b) above, some of whom have been saying that (a) is preferable, _none_ of them have stated or even hinted that (b) is invalid. >You had two main reasons why proof by contradiction was considered >worse: |(i) it's easy to give a wrong proof by contradiction, where >|the contradiction arises from some error >| >|(ii) if you give a direct proof that A implies B, by assuming >|A and then deducing B, the steps in the proof can give >|some insight into what A really entails - you show A implies >|B by showing A implies C and C implies B, and along the >|way you've shown two facts that might be interesting and >|useful elsewhere, that A implies C and that C implies B. >|You don't get this sort of bonus from a proof by contradiction, >|since in the course of the proof you're assuming things >|which it turns out never actually hold. I think the practice of constructive proof may shed some light on >this. The conclusion of a constructive proof tells you more about >the nature of the proof than the conclusion of a classical proof, >generally. In particular, when the conclusion is negative, the proof >was a proof by contradiction (essentially) and when the conclusion >is positive it wasn't. I can't tell whether the circumstances reducing the risk of (i) and >enhancing the prospects of (ii) are different from the circumstances >where the theorem (as it would be stated in constructive terms) is >more solidly stated. In constructive mathematics one has the general advice of Bishop >to cut down on negations. If we seek to tighten up our theorems by >getting rid of flabby negations, then we also as a side-effect >refrain from proofs by contradiction. Lots of negations and implications can be upgraded by supplying them >with more solid content. Bishop has a paper which uses Goedel's >Dialectica paper for this. Goedel has a method of rewriting >statements which puts them into a form (there exists w) (for all t) P(w,t) where w ranges over a set of possible witnesses to the truth of >the statement, and t ranges over a set of possible cases. If this >statement is false, one would like to do more than just negate it; >one would like to say something about the function f:w->t for which >P(w, f(w)) fails for all w. A proof of the negative statement >implicitly provides such an f, as witness. Even when a statement >is negative, then, there can be implicit solid content in the >construction demonstrating the fact, in terms of information on >and around this f. I get the impression that in cases where a theorem is negative, >but has relatively tangible constructive content, one is not so >especially prone to err in the way described in (i), because one >is regarding the construction in the proof (of f, say), which >actually *does* exist, and trying to see whether *it* is as >advertised. In the proof that there are infinitely many primes, >to pick a simple example, the engine is the method for getting >a new prime out of a finite set of old primes. An argument which >focusses on this kind of concrete construction doesn't seem to >suffer from the kind of problem that some attempts at proof by >contradiction do. I'm going on impressions, though, and it seems like it would >take somewhat tricky casework to get a more solid sense for what >kinds of proof do and do not tend to suffer from these problems. The context here, or at least the context I had in mind, was proofs by students in classes where half the point to the class is learning to read and write proofs (beginning algebra, analysis, topology classes are often in this category). In that context I can tell you from experience that (i) is a real danger. Of course it's much less of a problem when we're talking about proofs in the real world written by grownups - one has an idea what _sort_ of contradiction to expect, so when one gets a totally irrelevant contradiction one looks for the error instead of saying qed. >Keith Ramsay ************************ David C. Ullrich ==== > > For the greater cogency and obviousness in your paper THERE > SHOULD BE a TABLE with componentries of system, which one are > sorted out on their influence to precision of system GPS as a whole. > > Such a table was posted by Sam Wormley on the 22nd, see > > Then any layman can see that we can neglect neglible small > relativistic corrections as contrasted to by other factors > defining and restricting limiting accuracies GPS as a whole. > > > VI. Summary > > Excluding the deliberate degradation of SA, the dominant error source > for satellite ranging with single frequency receivers is usually the > ionosphere. It is on the order of four meters, depending on the > quality of the single-frequency model. For dual-frequency (P/Y-code) > receivers (which eliminate SA) the Standard Error Model of Table I > has one principal change (in addition to the elimination of the SA > error). The ionospheric error is reduced from four meters to about > one meter. > > > The GR correction is about 44 microseconds or 13km per day > and would be cumulative. You can hardly call that negligible. 1. What periodicity of corrections of parameters in GPS? 2. What parameters are adjusted in GPS? 3. What medial - statistical values have parameters adjusted in GPS in each session of corrections? -- Aleksandr ==== > > For the greater cogency and obviousness in your paper THERE > SHOULD BE a TABLE with componentries of system, which one are > sorted out on their influence to precision of system GPS as a whole. > > Such a table was posted by Sam Wormley on the 22nd, see > > Then any layman can see that we can neglect neglible small > relativistic corrections as contrasted to by other factors > defining and restricting limiting accuracies GPS as a whole. > > VI. Summary > > Excluding the deliberate degradation of SA, the dominant error source > for satellite ranging with single frequency receivers is usually the > ionosphere. It is on the order of four meters, depending on the > quality of the single-frequency model. For dual-frequency (P/Y-code) > receivers (which eliminate SA) the Standard Error Model of Table I > has one principal change (in addition to the elimination of the SA > error). The ionospheric error is reduced from four meters to about > one meter. > > The GR correction is about 44 microseconds or 13km per day > and would be cumulative. You can hardly call that negligible. 1. What periodicity of corrections of parameters in GPS? The corrections are of the order of nanoseconds per day so confirm GR to roughly one part in 10,000 by that simplistic comparison. There is much more information available in web pages if you want to look for more details. > 2. What parameters are adjusted in GPS? > 3. What medial - statistical values have parameters adjusted > in GPS in each session of corrections? Happy New Year George ==== To base 10: STEP 1: For the characteristic: carry out repeated division of the number by 10 until the result falls below 10 and count the number of times. STEP 2: For the mantissa: raise the result to the power of 10 using multiplication only, then divide by 10 again as in Step 1. Continue to Step 2 for further digits. This process generates the logarithm digit by digit, to the same accuracy as the other arithmetical functions. Also works efficiently in binary. This has almost certainly been tried before. I would be interested in any ==== To base 10: >STEP 1: For the characteristic: carry out repeated division of the number by >10 until the result falls below 10 and count the number of times. >STEP 2: For the mantissa: raise the result to the power of 10 using >multiplication only, then divide by 10 again as in Step 1. Continue to Step >2 for further digits. >This process generates the logarithm digit by digit, to the same accuracy as >the other arithmetical functions. Also works efficiently in binary. >This has almost certainly been tried before. I would be interested in any > See Textbook of Algebra v1 by Chrystal pgs 513 to 519 where some of the more elementary, although at the same time more laborious, approximative methods that might be used to compute logarithms are described. Yours seems to be one of them. rich ==== equation. The equation is Ax=0 A is a very large sparse matrix which have 1473-by-1473. Sum(x_i) is 1.0 All x are not zero. Please, would you like to let me know how to solve the simultaneous equation by fortran. Additionally, My major is not mathmatics. Thus, it is difficult to solve the equation in a method of pseudoinverse, SVD, etc.. Please, let me know that, easily and in detail. ==== > equation. The equation is Ax=0 A is a very large sparse matrix which have 1473-by-1473. > Sum(x_i) is 1.0 > All x are not zero. Please, would you like to let me know how to solve the simultaneous > equation by fortran. Additionally, My major is not mathmatics. Thus, it is difficult to > solve the equation in a method of pseudoinverse, SVD, etc.. Please, > let me know that, easily and in detail. > Well, I haven't dealt with matrices that large, but maybe you could try Cramer's rule. As for the code, you are on your own, I don't know Fortran. You could try: http://www.library.cornell.edu/nr/bookfpdf.html Lurch ==== > The equation is >> Ax=0 >> A is a very large sparse matrix which have 1473-by-1473. >Well, I haven't dealt with matrices that large, but maybe you could try >Cramer's rule. You know, I think all of us have a tendency to think we can answer more questions well than we really can. I don't know a whole heckuva lot about numerical analysis, but this strikes me as a really, really bad answer. For one thing, Cramer's rule gives an explicit solution to Ax=b when A is invertible; the OP's matrix A is surely not invertible, lest the only solution be x = 0. For another thing, Cramer's rule expresses the answer in terms of (here) 1474 determinants of 1473x1473 matrices. How do you propose that the determinants be evaluated? A naive algorithm sums 1473! products and introduces the possibility of tremendous quantities of round-off error and massive cancellations. Even using row operations, while much faster, is still time consuming and prone to error if the entries are not, say, small integers. As useful as Cramer's rule can be for theoretical work, I imagine it's essentially _never_ used for computing numerical solutions to linear equations with more than a handful of variables. So what response might be better for the OP? It's probably true that any good matrix package can handle your matrix (a couple thousand rows is not considered huge these days) but since you say your matrix is sparse, you would probably benefit from using routines specifically written for this common special case. Here is a link which was posted the other day in sci.math.num-analysis (which is a much more appropriate group for technical questions of this type): http://vlsicad.cs.ucla.edu/sparse.html Of course the equation Ax=0 has no solution with sum(x_i)=1 if A is nonsingular, so it would be good to think about how you know that your problem has a solution. More generally, it is likely to help those who are helping you if you can indicate where your matrix comes from, since that may signal the use of special routines e.g. for solving differential equations numerically. dave ==== >> A is a very large sparse matrix which have 1473-by-1473. That's actually pretty small. >Well, I haven't dealt with matrices that large, but maybe you could try >Cramer's rule. *splorf* That's the punchline to some jokes I know. Really bad idea. It has exponential complexity, while most numerical algorithms are n^3 or thereabouts. > Of course the equation Ax=0 has no solution with sum(x_i)=1 if A is > nonsingular, So I guess the matrix is singular, but you can add that sum condition as a last row. Introduce a dummy extra variable, and you have a square system again. (Or am I overlooking something?) > More generally, it is likely to help > those who are helping you if you can indicate where your matrix comes > from, since that may signal the use of special routines e.g. for > solving differential equations numerically. Yep. And with that information post on sci.math.num-analysis where I'm sure someone will have seen your problem before. V. -- homepage: cs utk edu tilde lastname ==== > >> The equation is >> Ax=0 >> A is a very large sparse matrix which have 1473-by-1473. >Well, I haven't dealt with matrices that large, but maybe you could try >Cramer's rule. You know, I think all of us have a tendency to think we can answer > more questions well than we really can. I don't know a whole heckuva lot > about numerical analysis, but this strikes me as a really, really bad > answer. For one thing, Cramer's rule gives an explicit solution to > Ax=b when A is invertible; the OP's matrix A is surely not > invertible, lest the only solution be x = 0. For another thing, > Cramer's rule expresses the answer in terms of (here) 1474 determinants > of 1473x1473 matrices. How do you propose that the determinants be > evaluated? A naive algorithm sums 1473! products and introduces the > possibility of tremendous quantities of round-off error and massive > cancellations. Even using row operations, while much faster, is still > time consuming and prone to error if the entries are not, say, > small integers. As useful as Cramer's rule can be for theoretical work, > I imagine it's essentially _never_ used for computing numerical solutions > to linear equations with more than a handful of variables. So what response might be better for the OP? It's probably true that any good matrix package can handle your > matrix (a couple thousand rows is not considered huge these days) > but since you say your matrix is sparse, you would probably benefit > from using routines specifically written for this common special case. > Here is a link which was posted the other day in sci.math.num-analysis > (which is a much more appropriate group for technical questions > of this type): > http://vlsicad.cs.ucla.edu/sparse.html Of course the equation Ax=0 has no solution with sum(x_i)=1 if A is > nonsingular, so it would be good to think about how you know that > your problem has a solution. More generally, it is likely to help > those who are helping you if you can indicate where your matrix comes > from, since that may signal the use of special routines e.g. for > solving differential equations numerically. dave Hey, it's an open forum! I consider this NG like an online math club. I offer suggestions, and they aren't always useful, or correct. None of this stuff is being published, so relax. He isn't paying me for my help; so, he is free to take my advice, or not. It is up to him. I also provided a link to an online book of Numerical analysis in Fortran, which should answer his question. Lurch ==== The equation is >> Ax=0 >> A is a very large sparse matrix which have 1473-by-1473. >Well, I haven't dealt with matrices that large, but maybe you could try >>Cramer's rule. You know, I think all of us have a tendency to think we can answer >more questions well than we really can. [...] I have an explanation for that... hmm, no I don't. ************************ David C. Ullrich ==== I am looking for a formula for the calculation of the midpoint for the smallest, an area A circumscribing circle. Is there like for triangles a connection between the center of gravity and the center of that circle? Are any formulas for such a circle known? Carolin Hau§ner ==== Caro > I am looking for a formula for the calculation of the midpoint for the > smallest, an area A circumscribing circle. Is there like for triangles > a connection between the center of gravity and the center of that > circle? Are any formulas for such a circle known? The circle which meets the three corners of a triangle is called the circumcircle of the triangle, and the center of that circle is the circumcenter of the trinangle. http://mathworld.wolfram.com/Circumcenter.html The circumcenter may be outside the triangle. But see also http://mathworld.wolfram.com/Incenter.html To get formulas for the cartesian coordinates of the circumcenter, write down two linear equations, one for each of two perpendicular bisectors. The incenter lies on both of those lines. LH ==== hi wikipedia needs our help to carry on: http://www.wikimedia.org/letter.html thanks to all bye max ==== What's happened to you ? I get up early every morning to read your diatribes and I haven't seen anything for a few days now. Does this mean I have to learn maths if I want to be a part of this newsgroup ? Does this mean I have to rely on comics for my cackles each morning ? I hope the FBI didn't turn on you and that the Evil Mathematical Establishment (tm) haven't imprisioned you in an infinite series ! Please hurry back, my humour is suffering and my education lacking - at least I was being *forced* to learn maths while you posted for no other reason than trying to understand the people who replied to you! Ivan. So this poster who's opinion, by his own admissions in terms of mathematics is worthless, still feels that his opinion is of value, clearly because he's at least learned the true nature of the math community. JSH writing about me in the Focus on point of dispute, more math thread ... I've been villified by James, does this make me famous? ==== the application of nonlinear systems (particularly catastrophe theory) to human behavior. Through the years I have struggled with balancing the criticisms of catastrophe methodology and the heuristic merits of the theory. I've read most of the catastrophe literature and theory and singularities. Although the proponents of catastrophe theory are also somewhat convincing, I question its applicability and implementation in psychological research. I'm at a point where I need to make a decision: Do I drop this line of research or do I continue working toward the development of catastrophe applications (e.g., write grants to develop software for analyzing catastrophes/bifurcations in psychological data)? Kate ==== > > the application of nonlinear systems (particularly catastrophe > theory) to human behavior. Through the years I have struggled with > balancing the criticisms of catastrophe methodology and the heuristic > merits of the theory. I've read most of the catastrophe literature and > theory and singularities. > > Although the proponents of catastrophe theory are also somewhat > convincing, I question its applicability and implementation in > psychological research. I'm at a point where I need to make a > decision: > > Do I drop this line of research or do I continue working toward the > development of catastrophe applications (e.g., write grants to develop > software for analyzing catastrophes/bifurcations in psychological > data)? > > Kate It would seem that this is more a question for your advisor/committee/ faculty to help you answer. IMO, as a grad student it is generally advisable to do something that gives you a high probability of producing a respectable dissertation ASAP. A speculative fishing expedition (i.e. one where you're not sure there are any fish in the lake as opposed to the standard risk of just not catching any of fish there are) is not what you should undertake early in your career. However, if you still want to pursue it, ask around about funding prospects, that is is anyone supporing this type of work? Maybe there is all kinds of homeland securtiy money being thrown at this area in an attempt to figure out terrorists or something (don't laugh, they're spending billions on a missle defense system that hasn't worked yet). This is where faculty should be able, through their connections, to help you with information on research and funding trends. But unless your advisor knows about this area and is supportive of the idea, you're asking for problems, IMO. Good luck, Russell ==== > > the application of nonlinear systems (particularly catastrophe > theory) to human behavior. Through the years I have struggled with > balancing the criticisms of catastrophe methodology and the heuristic > merits of the theory. I've read most of the catastrophe literature and > theory and singularities. > > Although the proponents of catastrophe theory are also somewhat > convincing, I question its applicability and implementation in > psychological research. I'm at a point where I need to make a > decision: > > Do I drop this line of research or do I continue working toward the > development of catastrophe applications (e.g., write grants to develop > software for analyzing catastrophes/bifurcations in psychological > data)? > > Kate It would seem that this is more a question for your advisor/committee/ > faculty to help you answer. IMO, as a grad student it is generally > advisable to do something that gives you a high probability of > producing a respectable dissertation ASAP. A speculative fishing > expedition (i.e. one where you're not sure there are any fish in the > lake as opposed to the standard risk of just not catching any of fish > there are) is not what you should undertake early in your career. > However, if you still want to pursue it, ask around about funding > prospects, that is is anyone supporing this type of work? Maybe > there is all kinds of homeland securtiy money being thrown at > this area in an attempt to figure out terrorists or something (don't > laugh, they're spending billions on a missle defense system that > hasn't worked yet). This is where faculty should be able, through > their connections, to help you with information on research and > funding trends. But unless your advisor knows about this area and > is supportive of the idea, you're asking for problems, IMO. OTOH, if you're really interested in finding the truth, regardless of funding, then do what you know is right. Bob Muncaster (http://www.math.uiuc.edu/~muncast/) has in the past studied some applications of math to political science and sociology, and I assume psychology is not too different. I know he's into dynamical systems, so he should be able to answer your questions. Jon Miller ==== > the application of nonlinear systems (particularly catastrophe > theory) to human behavior. Through the years I have struggled with > balancing the criticisms of catastrophe methodology and the heuristic > merits of the theory. I've read most of the catastrophe literature and > theory and singularities. Although the proponents of catastrophe theory are also somewhat > convincing, I question its applicability and implementation in > psychological research. I'm at a point where I need to make a > decision: Do I drop this line of research or do I continue working toward the > development of catastrophe applications (e.g., write grants to develop > software for analyzing catastrophes/bifurcations in psychological > data)? Kate Why study pseudo-science anyway? ==== } Why study pseudo-science anyway? All the better to refute asinine one-liners, my dear. ==== > } Why study pseudo-science anyway? All the better to refute asinine one-liners, my dear. > Are you claiming it isn't a pseudo-science? Lurch ==== > > the application of nonlinear systems (particularly catastrophe > theory) to human behavior. Through the years I have struggled with > balancing the criticisms of catastrophe methodology and the heuristic > merits of the theory. I've read most of the catastrophe literature and > theory and singularities. > > Although the proponents of catastrophe theory are also somewhat > convincing, I question its applicability and implementation in > psychological research. I'm at a point where I need to make a > decision: > > Do I drop this line of research or do I continue working toward the > development of catastrophe applications (e.g., write grants to develop > software for analyzing catastrophes/bifurcations in psychological > data)? > > Kate > > Why study pseudo-science anyway? Which one? ;-) Russell ==== > Does the natural density of all positive integers that are the sum of two odd > primes = 1/2? How about the sum of two squares? The sum of two squarefree > integers? ************************************************** If the Goldbach conjecture is true, then the answer to your first question is yes. If a number is the sum of two squares then it can't have a prime factor that is 3 modulo 4 and whose highest power is odd. Since the series 1 + 1/3 + 1/7 + 1/11 + 1/19 + . . . diverges, this would imply that the natural density of the numbers that are the sum of two squares is 0. ( [ (1+1/3)(1+1/7)(1+1/11) . . . ] ^ (-1) = 0.) _________________________________________________________ Eric J. Wingler (wingler@math.ysu.edu) Dept. of Mathematics and Statistics Youngstown State University One University Plaza Youngstown, OH 44555-0001 330-941-1817 ==== >> Does the natural density of all positive integers that are >the sum of two odd >> primes = 1/2? How about the sum of two squares? The sum of >two squarefree >> integers? ************************************************** >If the Goldbach conjecture is true, then the answer to your >first question is yes. If a number is the sum of two squares then it can't have a >prime factor that is 3 modulo 4 and whose highest power is >odd. Since the series 1 + 1/3 + 1/7 + 1/11 + 1/19 + . . . >diverges, this would imply that the natural density of the >numbers that are the sum of two squares is 0. > ( [ (1+1/3)(1+1/7)(1+1/11) . . . ] ^ (-1) = 0.) I give up. Supposing that everything you say is true, which I imagine it is, how does it follow that the sums of two squares have density 0? (My guess is that you deduce somehow that the sum of the reciprocals of the sums of two squares is finite. If so that's interesting because the sum of 1/(j^2 + k^2) is infinite...) _________________________________________________________ Eric J. Wingler (wingler@math.ysu.edu) >Dept. of Mathematics and Statistics >Youngstown State University >One University Plaza >Youngstown, OH 44555-0001 >330-941-1817 > ************************ David C. Ullrich ==== Let T be the set of integers that can be written as a sum of two squares. I don't think that the sum of 1/n for n in T converges. What I get from a rough L-series calculation (which could probably be made rigorous if it were earlier in the evening) is that the Dirichlet series D(T,s) = sum over n in T of 1/n^s diverges as s --> 1+, but that it blows up more-or-less like 1/sqrt(s-1). This implies that the Dirichlet density of T is DirichDen(T) := (lim as s-->1+ of D(T,s)*(s-1)) = 0. Presumably the natural density will also be zero, though since D(T,s) won't have an analytic continuation in a neighborhood of s=1, one can't simply apply a standard Tauberian theorem to get the result. Joe Silverman > >> Does the natural density of all positive integers that are > the sum of two odd >> primes = 1/2? How about the sum of two squares? The sum of > two squarefree >> integers? > >************************************************** >If the Goldbach conjecture is true, then the answer to your >first question is yes. > >If a number is the sum of two squares then it can't have a >prime factor that is 3 modulo 4 and whose highest power is >odd. Since the series 1 + 1/3 + 1/7 + 1/11 + 1/19 + . . . >diverges, this would imply that the natural density of the >numbers that are the sum of two squares is 0. > ( [ (1+1/3)(1+1/7)(1+1/11) . . . ] ^ (-1) = 0.) > > I give up. Supposing that everything you say is true, which > I imagine it is, how does it follow that the sums of two squares > have density 0? > > (My guess is that you deduce somehow that the sum of the > reciprocals of the sums of two squares is finite. If so that's > interesting because the sum of 1/(j^2 + k^2) is infinite...) > > _________________________________________________________ > >Eric J. Wingler (wingler@math.ysu.edu) >Dept. of Mathematics and Statistics >Youngstown State University >One University Plaza >Youngstown, OH 44555-0001 >330-941-1817 > > > ************************ > > David C. Ullrich ==== >>If a number is the sum of two squares then it can't have a >>prime factor that is 3 modulo 4 and whose highest power is >>odd. Since the series 1 + 1/3 + 1/7 + 1/11 + 1/19 + . . . >>diverges, this would imply that the natural density of the >>numbers that are the sum of two squares is 0. >> ( [ (1+1/3)(1+1/7)(1+1/11) . . . ] ^ (-1) = 0.) >I give up. Supposing that everything you say is true, which >I imagine it is, how does it follow that the sums of two squares >have density 0? Let S(p) be the set of positive integers that are either not divisible by p or are divisible by p^2. The natural density of S is 1 - 1/p + 1/p^2. Any positive integer that is the sum of two squares must be in S(p) for all primes p = 3 mod 4. If these primes are p_1, p_2, p_3, ..., the natural density of intersection_{j=1}^n S(p_j) is product_{j=1}^n (1 - 1/p_j + 1/p_j^2) = product_{j=1}^n exp(-1/p_j) (1+O(1/p_j^2)) <= C exp(-sum_{j=1}^n 1/p_j)) which goes to 0 as n -> infinity because the series sum_{j=1}^infty 1/p_j diverges. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== >>If a number is the sum of two squares then it can't have a >prime factor that is 3 modulo 4 and whose highest power is >odd. Since the series 1 + 1/3 + 1/7 + 1/11 + 1/19 + . . . >diverges, this would imply that the natural density of the >numbers that are the sum of two squares is 0. > ( [ (1+1/3)(1+1/7)(1+1/11) . . . ] ^ (-1) = 0.) >I give up. Supposing that everything you say is true, which >>I imagine it is, how does it follow that the sums of two squares >>have density 0? Let S(p) be the set of positive integers that are either not divisible by p >or are divisible by p^2. The natural density of S is 1 - 1/p + 1/p^2. Ah. That's the bit that hadn't clicked - couldn't see where the minus signs in the expansion of 1/(1+1/p) were going to come in, but there it is. Duh. >Any positive integer that is the sum of two squares must be in S(p) for >all primes p = 3 mod 4. If these primes are p_1, p_2, p_3, ..., the >natural density of intersection_{j=1}^n S(p_j) is >product_{j=1}^n (1 - 1/p_j + 1/p_j^2) > = product_{j=1}^n exp(-1/p_j) (1+O(1/p_j^2)) > <= C exp(-sum_{j=1}^n 1/p_j)) >which goes to 0 as n -> infinity because the series sum_{j=1}^infty 1/p_j >diverges. Robert Israel israel@math.ubc.ca >Department of Mathematics http://www.math.ubc.ca/~israel >University of British Columbia >Vancouver, BC, Canada V6T 1Z2 ************************ David C. Ullrich ==== THe cost of a long-distance telphone call is determined by aflat fee for teh first 5 minutes and a fixed amount for each additional minute. If a 15- minute telephone call costs $3.25 and a 23-minute call costs $5.17, find the cost of a 30-minute call. I was thinking of setting it up as a system of equations but im not really sure, can someone explain how i would set this problem up, cause alwyas get stumped on these type of questions. TIA ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- ==== > THe cost of a long-distance telphone call is determined by aflat fee for > teh first 5 minutes and a fixed amount for each additional minute. If a 15- > minute telephone call costs $3.25 and a 23-minute call costs $5.17, find > the cost of a 30-minute call. > I was thinking of setting it up as a system of equations but im not > really sure, can someone explain how i would set this problem up, cause > alwyas get stumped on these type of questions. TIA News==---- > http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 > ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- Let x = the flat fee and y = the fixed amount after the first five ,then 5x + 18y = 5.17 5x + 10y = 3.25 You get x = .17 and y = .24. Lurch ==== > THe cost of a long-distance telphone call is determined by aflat fee for >> teh first 5 minutes and a fixed amount for each additional minute. If a >15- >> minute telephone call costs $3.25 and a 23-minute call costs $5.17, find >> the cost of a 30-minute call. Let x = the flat fee >and y = the fixed amount after the first five >,then 5x + 18y = 5.17 >5x + 10y = 3.25 You get x = .17 and y = .24. Lurch > No, a flat fee is fixed and pays for anything up to including 5 minutes, not a per minute fee. You should have x in the equations instead of 5x. The first five minutes cost .85 (the flat fee) and the additional minutes are .24 each. A 2 minute call would also cost .85 because you don't apply the flat fee on a per minute basis. --Lynn --Lynn ==== >> THe cost of a long-distance telphone call is determined by aflat fee for >> teh first 5 minutes and a fixed amount for each additional minute. If a >15- >> minute telephone call costs $3.25 and a 23-minute call costs $5.17, find >> the cost of a 30-minute call. >Let x = the flat fee >and y = the fixed amount after the first five >,then > >5x + 18y = 5.17 >5x + 10y = 3.25 > >You get x = .17 and y = .24. > >Lurch > > No, a flat fee is fixed and pays for anything up to including 5 > minutes, not a per minute fee. You should have x in the equations > instead of 5x. The first five minutes cost .85 (the flat fee) and the > additional minutes are .24 each. A 2 minute call would also cost .85 > because you don't apply the flat fee on a per minute basis. --Lynn --Lynn > Let c=cost m=cost per minute over 5 and x=call length c=.85+m(x-5) c=.85+m(x-5) c=.85+mx-5m, x>5 As Lynn said, anything below 5 remains .85 A 15 minute call costs $3.25 so x=15 and c=3.25 3.25=.85+15m-5m 2.4=10m m=$.24 So Now we know m and our formula becomes c=.24x-.35 So when x=30, c=6.85. Therefore a 30 minute call costs $6.85 David Moran ==== Can the equation for c be transformed into an elliptic curve? > The Barcelona conjecture: > > Let c=(x+y+z)^p/(pxyz2^p) > > for integer c,x,y,z and p prime greater than or equal to 5, the > Barcelona conjecture is that no solutions exist with gcd(c,xyz)=1 (no > c exist that shares no factor with x or y or z). > I haven't seen this conjecture before, but compare the Beal conjecture: > www.math.unt.edu/~mauldin/beal.html > > Yes, I was aware of the Beal conjecture and briefly attempted to prove > it also :) > > The reason you hadn't seen the Barcelona conjecture is that it is > virtually unknown outside of this newsgroup since I've only posted it > here and in the research math group. I came up with it while > attempting to prove FLT using elementary techniques ( ok, once we are > done laughing the question remains - who hasn't? I mean even JSH keeps > on trying.) > > In some respects it should be easier to prove FLT using the Barcelona > conjecture as the latter places less restrictions on the value of c - > maybe even Fermat was working on this approach as it only requires > elementary methods, though I really doubt it as it isn't documented in > Ribenboim's book on FLT. > > If anyone can lend me a hand I'm trying to get a grip on how FLT was > tied to elliptic curves, my goal being to see how feasible it is to > apply the same methods to the Barcelona conjecture - for the moment it > is way too difficult for me. > ==== BTW; did anyone here bother to check out David Sereda, and of his http://www.ufonasa.com/ EVIDENCE the case for NASA UFOs In some recent additions to my MAZDA like Internal Rocket Rotary Combustion Engine (IRRCE sfc = 15+KW/kg), there seems we also have ourselves a wee bit of lunar He3 to burn off. http://guthvenus.tripod.com/gv-h2o2-irrce.htm The taking of lunar He3 is first come first served, that could be the likes of China or Russia because, we're obviously not smart enough. Venus still offers life; via moon He3 could help turn the trick I've got a few more words of wisdom to offer on behalf of the ARTEMIS PROJECT (lunar He3) http://guthvenus.tripod.com/gv-lse-he3.htm The rest of this report isn't entirely related to energy so much as it relates to truth or consequences. Such as for this next topic of there being other life NOT as we know it on Venus, that's obviously opposing the sorts of anti-humanity folks that couldn't care less if our entire world was destroyed by their resident warlord. Perhaps these folks can get their next level of future funding from the same source as Bush, Salem Laden. As for making policy look like happenstance, and/or vice versa, is key to snookering folks. http://guthvenus.tripod.com/moon-04.htm Though as for we humans need not, and perhaps we should not venture ourselves much beyond Venus L2 (VL2). Wouldn't want to contaminate a perfectly good planet with our inferior DNA nor lack of morals, especially of this group that's bashing honest research just out of spite. Besides, their stealth donkey-carts could be far more lethal than what our WMD donkey-carts can manage. As far as our human physiology being adaptable to pressure. Under such pressure things are not nearly as hot as we've been told, and you wont need but a fraction of a percent of O2. Of course, that degree of adaptation might have to be accommodated at a modus rate of a few bars per day. http://guthvenus.tripod.com/venus-air.htm I have a few other recent/ongoing comments on H2O2/C12H26 and of He3: http://guthvenus.tripod.com/gv-irrce.htm http://guthvenus.tripod.com/gv-hybrid-irc.htm http://guthvenus.tripod.com/gv-cm-ccm-01.htm http://guthvenus.tripod.com/gv-lm-1.htm http://guthvenus.tripod.com/radio-maybe.htm ==== Portfolio of PAF as of 28DEC03 BCE 400 21.90 $8,760.00 BLS 50 27.92 $1,396.00 BMY 100 27.80 $2,780.00 Q 50,000 3.92 $196,000.00 SBC 11,600 25.68 $297,888.00 realestate land 3APR03 of 3 lots $19,000 science-art of pictures,porcelain etc starting JAN03 for $12,160. realestate land 30JUL03 another lot $11,500. and Wyeth at a small profit and with the proceeds bought 200 more shares of BCE. Sold DT because it is hard to play DT as a Crossover switching campaign. Sold Verizon because it is overpriced compared to SBC. And sold Wyeth because I was never really aware of the history of phen fen (excuse the spelling) until I saw a PBS program on that drug and the history of the company of Wyeth in relation to that drug. After seeing the history of phen fen, I could not help but think that Wyeth is a company that fosters a culture of money grubbing ever more than it fosters a culture that they would seek health medicines to make the world better. For a company to know that phenfen was dangerous in Europe and yet bring that drug over into the USA to sell as a diet drug, even knowing that it had a past history of dangerousness, tells me that Wyeth entire corporate upper management needs to be fired. And I do not want to own a drug company of that sort of history of placing money-grubbing over that of good science. Wyeth should emulate Johnson & Johnson or Merck as companies that get the science before they ever think about money. I am wanting and trying to get the PAF portfolio positioned for the year 2004 such that it follows the VonNeumann Game Theory of the Optimal Strategy for Playing the StockMarket. The key and central theme of the OS of Stockmarket is switching campaigns of the Crossover. Through these many years of playing the stockmarket since 1978, I beleive I have found the OS of StockMarket and am trying to arrange the portfolio so that I can faithfully abide by this OS for 2004 and many years beyond. Archimedes Plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies ==== > Portfolio of PAF as of 28DEC03 > BCE 400 21.90 $8,760.00 > BLS 50 27.92 $1,396.00 > BMY 100 27.80 $2,780.00 > Q 50,000 3.92 $196,000.00 > SBC 11,600 25.68 $297,888.00 > realestate land 3APR03 of 3 lots $19,000 > science-art of pictures,porcelain etc starting JAN03 for $12,160. > realestate land 30JUL03 another lot $11,500. > In my typing above I made an error in the date for the above took place today which is 29DEC03 and not 28. Since I archive nearly every one of my posts and knowing the error I changed it in my original post to read like this: Portfolio of PAF as of 29DEC03 [sic] I believe this is the standard means of correcting a writing when the author or editor knows of a mistake in the original and just goes ahead and changes it to the true facts for today is not 28DEC03 when the above transactions took place and so I will use in future this method of [sic] And I must write a little more because a post is wasted if it is just a few lines. Tonight on TV was the Antiques Roadshow which I often catch while eating supper. And I seem to like the British Antique Roadshow more than the USA because they seem to garner more art pictures such as that nurse in WW1. Pictures seem to make the best singular sights on TV. And pictures are the artform that catches the eyes the most. However, there is 3-dimensional art which allows also a sight but also the sense of touch and feel such as porcelain and porcelain is not as sightful as art on 2-D TV screen. Question: in all the time I have seen Antiques Roadshow, I have never seen any Royal Copenhagen porcelain. Is there is bias against RC blue flute??? Question: I collect some RC and have collected some German porcelain that mimics or imitates RC blue. But instead of the blue flower of RC it is almost perfect circles of the chrysanthenum flower. It is made in Germany and one dish is signed PM, which I think is Paul Muller. I have a small dish and 2 eggcups and 2 salt shakers. It has Germany on bottom. Is anyone familar with German porcelain that mimiced or imitated Royal Copenhagen blue pattern? I searched on the Internet and have never found a solid reference to German that imitates RC blue pattern. Annoyance of Antiques Roadshow: I believe that great furniture should look bright and not dark. I buy white oak that is close to being white. I never buy any furniture that is dark, or stay away from it. So I think that on Antiques Roadshow whenever they show old furniture, it is invariably vary dark brown to black and I think those are ugly, and depressing to live with. So, I am in favor of recasting the entire antique business when it comes to furniture. I do not care how old or rare a piece of furniture is, because if it is dark in color it is depressing and it should be low in monetary value. So I am in favor of taking any old furniture and taking away the surface coating and or painting the surface with a white like color. Antiques dealers are falsely perpetuating a myth that original paint and original surface is more money. But who wants to live with ugly dark depressing furniture. So all great furniture is not how old it is nor how skillful building or the artwork, if it is dark brown to black, it is depressing and ugly and should be recast into a glimming white such as a white birch finish. Whenever I see an interior of a home that has wood panelling and has dark furniture, my first impulse is to get out a nice bucket of semigloss white paint and have at it. I have never seen any piece of furniture on the Antiques Roadshow that I would make room for in my home simply because they are depressingly too dark in color. Perhaps in the Arizona desert or in Death VAlley desert would a home be okay for this dark furniture. Because most of the furniture of the past was dark in color that most of these pieces should be of a very low price, and much lower than new furniture that is whitish colored surface. Archimedes Plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies ==== Here I thought in this GOOGLE site, of whatever God was restricted and/or moderated to NASA/NSA/DoD, now DHS. Seems there's some folks that certainly talk and act the part like God, as they certainly are those willing and able to terminate whatever life as we know it, especially mine. Though I guess I'm just happy that I'm not one of those nice Cathars having to dodge another round of exterminations by the Pope, though our resident warlord has certainly been doing his fair share of mastering carnage based upon lies similar to but not even nearly as good as what the Catholic church used to justify their actions. Although, we've got the so what's the difference as our ultimate qualifier that trumps anything Pope. BTW; did anyone here bother to check out David Sereda, and of his http://www.ufonasa.com/ EVIDENCE the case for NASA UFOs In some of my recent additions to the MAZDA like Internal Rocket Rotary Combustion Engine (IRRCE sfc = 15+KW/kg), there seems we also have ourselves a wee bit of lunar He3 to burn off. http://guthvenus.tripod.com/gv-h2o2-irrce.htm The taking of lunar He3 is first come first served, that could be the likes of China or Russia because, we're obviously not smart enough. Venus, not Mars, still offers life; via moon He3 could help turn the trick I've got a few more words of wisdom to offer on behalf of the ARTEMIS PROJECT (lunar He3) http://guthvenus.tripod.com/gv-lse-he3.htm The rest of this report isn't entirely related to energy so much as it relates to truth or consequences. Such as for this next topic of there being other life NOT as we know it on Venus, that's obviously opposing the sorts of anti-humanity folks that couldn't care less if our entire world was destroyed by their resident warlord. Perhaps these folks can get their next level of future funding from the same source as Bush, Salem Laden. As for making policy look like happenstance, and/or vice versa, is key to snookering folks. http://guthvenus.tripod.com/moon-04.htm Though as for we humans need not, and perhaps we should not venture ourselves much beyond Venus L2 (VL2). Wouldn't want to contaminate a perfectly good planet with our inferior DNA nor lack of morals, especially of this group that's bashing honest research just out of spite. Besides, their stealth donkey-carts could be far more lethal than what our WMD donkey-carts can manage. As far as our human physiology being adaptable to pressure. Under such pressure things are not nearly as hot as we've been told, and you wont need but a fraction of a percent of O2. Of course, that degree of adaptation might have to be accommodated at a modus rate of a few bars per day. http://guthvenus.tripod.com/venus-air.htm I have a few other recent/ongoing comments on H2O2/C12H26 and of He3: http://guthvenus.tripod.com/gv-irrce.htm http://guthvenus.tripod.com/gv-hybrid-irc.htm http://guthvenus.tripod.com/gv-cm-ccm-01.htm http://guthvenus.tripod.com/gv-lm-1.htm http://guthvenus.tripod.com/radio-maybe.htm ==== Dear all, I want to ask what is the inverse operation of Kroneck product? More specifically, we know that matrix operation A*X*B=kron(A, B')*vec(X) where kron is the Kronecker product of matrices as defined in matlab; vec(X) is the stacked vector version of matrix X. All matrices are square... Now I want to reverse the operation, suppose I have a big matrix C, how to find A and B to get C=kron(A, B')? Under what condition these A and B cannot be found? Then how to find them approximately, i.e., optimal in the mean-square sense or under other criteria? That's to say, find A and B, such that kron(A, B')=C1 where C1 is a reasonably good approximation to C? -Walala ==== walala: Maybe if you gave the homework questions verbatim from the Prof, we could be of more help. It seems your interpretation of your tasks is confusing some people. Then again, if you really knew the questions, you wouldn't be posting for answers. Jim > Dear all, I want to ask what is the inverse operation of Kroneck product? More specifically, we know that matrix operation A*X*B=kron(A, B')*vec(X) where kron is the Kronecker product of matrices as defined in matlab; vec(X) > is the stacked vector version of matrix X. All matrices are square... Now I want to reverse the operation, suppose I have a big matrix C, how to find A and B to get C=kron(A, B')? Under what condition these A and B cannot be found? Then how to find them > approximately, i.e., optimal in the mean-square sense or under other > criteria? That's to say, find A and B, such that kron(A, B')=C1 where C1 is > a reasonably good approximation to C? > -Walala ==== > walala: Maybe if you gave the homework questions verbatim from the Prof, we could be > of more help. It seems your interpretation of your tasks is confusing some > people. Then again, if you really knew the questions, you wouldn't be > posting for answers. Jim > Dear all, > > I want to ask what is the inverse operation of Kroneck product? > > More specifically, we know that matrix operation > > A*X*B=kron(A, B')*vec(X) > > where kron is the Kronecker product of matrices as defined in matlab; > vec(X) > is the stacked vector version of matrix X. All matrices are square... > > Now I want to reverse the operation, suppose I have a big matrix C, > > how to find A and B to get C=kron(A, B')? > > Under what condition these A and B cannot be found? Then how to find them > approximately, i.e., optimal in the mean-square sense or under other > criteria? That's to say, find A and B, such that kron(A, B')=C1 where C1 > is > a reasonably good approximation to C? > > -Walala > Dear Jim, This really isn't my professor's homework problem... our school is now in break so there are no damn homeworks right now... This is a problem I am currently interested in... please tell me which part of my description confuses you? Then I can make myself clearer... Rgs, -Walala ==== How would I compute the probability of a draw in tac tac toe (noughts and crosses) assuming each player was playing randomly (ie no thought involved). What about the probability that player 1 wins? -- Interbang change nospam to optimusprime to reply ==== >How would I compute the probability of a draw in tac tac toe (noughts >and crosses) assuming each player was playing randomly (ie no thought >involved). What about the probability that player 1 wins? Assuming X goes first and that the players alternate until the grid is full, the number of final complete grids can be found by computing 9C4 - this is the number of ways of choosing the 4 cells that get Os Of course the 9C4 includes many games that are merely rotations or reflections of each other. It also ignores the fact that many of the games would really be over before the grid is full. But you may find that ignoring such issues provides the most straightforward approach. In particular, all the positions among the 9C4 can be regarded as equally likely, which will not be the case if you consider positions at the real end of the game. What you have to do next is find the number of drawn positions, i.e. the number of filled-in grids which do not contain a winning line either for X or for O. You could approach this by enumerating the places in the grid where a winning line can arise and then considering the possible arrangements of the other symbols. You will need to be very careful to avoid counting twice the positions with more than one winning line. Hope this helps ==== > >How would I compute the probability of a draw in tac tac toe (noughts >and crosses) assuming each player was playing randomly (ie no thought >involved). What about the probability that player 1 wins? > > Assuming X goes first and that the players alternate until the grid is > full, the number of final complete grids can be found by computing 9C4 > - this is the number of ways of choosing the 4 cells that get Os > > Of course the 9C4 includes many games that are merely rotations or > reflections of each other. It also ignores the fact that many of the > games would really be over before the grid is full. But you can not ignore that. Many of the 9C4 include patterns where both are winners (actually more than you would think). A simple program gave me the following: played weighed perc final perc games 255168 362880 15120 1st 131184 212256 58.5% 7440 49.2% draw 46080 46080 12.7% 1920 12.7% 2nd 77904 104544 28.8% 1440 9.5% both - - - 4320 28.6% The first three columns are when the games are actually played, the last two are when you look only at the final (completely filled) grid. Off-hand I have no idea how to do this theoretically. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== > >> >> > > >> So you know that tan' pi/4 = whatever. >> Now what does that *mean*? >> (Can you recall the *definition* of derivative?) > > It means that you can apply l'Hopital's rule without knowing in > advance what lim_{x -> pi/4} (tan(x) - 1)/(x - pi/4) is. > > Try again! That's no definition of derivative. > > How do you use L'H's rule to calculate this limit without > already knowing what it is? >> >> The definition of derivative of tan(x) may be deduced from the >> derivatives of sin(x) and cos(x) and the quotient rule for >> derivatives, the derivative of the numerator and denominator of >> (tan(x) - 1)/(x - pi/4) can be found without being aware that >> it is a difference quotient. >> >> If f(x) = tan(x) - pi/4 = sin(x)/cos(x), then >> f'(x) = (cos(x)*cos(x) - sin(x)(-sin(x))/cos(x)^2 + 0 >> = 1/cos(x)^2 >> >> Excellent: you can compute derivatives. Alas you seem >> to have forgotten what they are. :-( >> >> You have (d/dx)(tan x - pi/4) = 1/cos^2 x [dunno what the pi/4 is doing, >> but it's irrelevant anyway]. Now what does that mean? >> >> (Heavy hint: apply the definition of derivative.) >> >> and if g(x) = x -pi/4 then g'(x) = 1 >> >> Any point to this? >> >> Since f(pi/4) = g(pi/4) = 0 but g'(pi/4) and f'(pi/4) are >> continuous and nonzero at x = pi/4, L'Hopital applies. >> >> But why waste time using L'H? > > It is not a matter of whether it is optimal to do it the way I > showed, but merely a question of whether it is possible, and I have > shown it to be possible. No you didn't --- you used the limit itself in the process of applying L'H :-( -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== > [snip] >>In his Commutative Algebra, D. Eisenbud has another definition: >>Definition 2. Let R be a commutative ring. An element p of R is prime if >> p is not a unit and the following is true: If a and b are elements of >>R such that p divides ab, then p divides a or b. >>Is Definition 2 common in the commutative algebra community? >> It is common in algebraic number theory as well. > > Then I must wonder what the common definition of divides is. According > to the definition which I typically use, 0 divides 0. And then according > to Definition 2, we would have 0 being a prime, which surely we don't > want. Not necessarily --- the zero ideal isn't prime in every ring. Def 2 just says that p is prime iff p generates a prime ideal. Of course, in algebra, it's prime ideals, not prime elements, that are of major interest. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== bcc Dear Gary, Jack and Tony : worked for some time on the notion of Self Referential Noise as models of reality. Wheelers's spacetime FOAM is just another interpretation of Noise. You could take that Noise as your absolute frame of reference. JS: The new EINSTEIN results seem to argue against this model? I am not sure of that of course. Sirag says the foam is not really randomly chaotic but coherently harmonic and that sets the number of extra dimensions in Calabi-Yau space? On the other hand Christian Beck seems to agree with you and says he can determine the 25 epicycles of the standard model (some say only 17) from chaotic strings. This would seem choose a definite location on Susskind's Landscape in opposition to the WAP ideas of chaotic papers and books by Lee Smolin. CC: One day you may want to look at : Carlos Castro `` The String Uncertainty Relations Follow from the New Relativity Principle . Foundations of Physics. {bf 8} ( 2000 ) page 1301. for a way to derive the stringy uncertainty relations from first principles. JS: Is the claim being made that the new term beyond Heisenberg uncertainty is the source of irreversibility as in the arrow of time? Remember Hawking talks about a new source of uncertainty although Susskind seems to think that is wrong? Note also Ed Witten's formula generalizing Heisenberg's quantum uncertainty principle, i.e. eq. (5.9) p. 136 Delta X > h/DeltaP + alpha'(DeltaP)/h The second gravity-string source of uncertainty should give the irreversible statistical arrow of time not found when alpha' = 0, i.e. infinite string tension, or infinite space-time stiffness of action without reaction as is also found in the signal locality of orthodox quantum theory in sense of Antony Valentini's papers. Note the conformal look w = 1/z + alpha'z ==== > proofs > > > why do authors usually consider the proofs so important in mathematical > texts ? Because that is what mathematics is without. > Isn't it usually better when only a few people read and check the proofs, No. If you refuse study proofs then you are cravenly deferring to authority. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== > proofs >> >> >> why do authors usually consider the proofs so important in mathematical >> texts ? Because that is what mathematics is without. Is this sentence a word without or something? >> Isn't it usually better when only a few people read and check the proofs, No. If you refuse study proofs then you are cravenly deferring to authority. ************************ David C. Ullrich ==== > > >> proofs >> >> >> why do authors usually consider the proofs so important in mathematical >> texts ? > >Because that is what mathematics is without. > > Is this sentence a word without or something? It seems that we apply a read What I mean (RWIM) module to what we see. For me, it changed without to about when I read Robin Chapman's post, until David Ullrich's post called attention to the problem. I bet it workd for Chapman too, when he was trying to proofread. > >> Isn't it usually better when only a few people read and check the proofs, > >No. If you refuse study proofs then you are cravenly deferring to authority. My sentiment has long been that if you don't work through the proof, you haven't really grasped the meaning of the theorem. I am not as rigid aout this as I once would have liked to be (life is finite!) but I do try to understand something of how the prof works. Also see my .sig quote. -- Chris Henrich Those readers who are unacquainted with the mathematical technicalities will find that they can manage quite well by ignoring them. -- John Nash ==== : > ----------------------------- <^> <(åáåÀåá)> : <^> ----------------------------- : > : > : > Any number having this form can be produced by an RM. : > Some examples of computable numbers: : > : > (0) : > (1) : > 11(0) : > 01(110) : > : > The length of the initial string plus the length of the repeating string : > must be less than or equal to the number of states of the RM that : > produced the number. : > : > : > I don't see this. Here is a simplified conceptual diagram of a Russell : Machine. : > : > s1 s2 s6 : > 0 s5 : > --------------------- : > 1 s3 : > s4 s7 : > s8 : > : > There is one piece of information missing, which state s8 goes to. : > Add to this s8 -> s7 : > : > This RM will output 0 0 1 1 0 0 1 1 1 1 1 1 1 1 : > notation is 0 0 1 1 0 0 (1 1) : > : > If it can't halt then the states get reused. : : I assume that s1->s2->s3->s4->s5->s6->s7->s8->s8. : The ouput is 001100(1). Then it DOES halt, DUMBASS. s8 *IS* a halt state. Are you now going to retract your claim that RMs don't halt because they don't have halt states? ==== : > What kinds of numbers are we talking about here? : > Rational? Real? Complex? Supernatural? Infinitesimal? : > : I have made the proof a little more rigorous. But you haven't answered the question. WHAT IS a number??? : First, I define how to compute a computable number. That is ENTIRELY premature. FIRST, you have to define NUMBER! : I do this with a very non-standard type of Turing machine. : Let's call these Russell machines (RM). : A number is computable if it can be represented by : the infinite binary string produced by a RM. This is just stupid. A TM can do anything you need to do. : A RM has a single, infinitely long tape. : Every position of the tape is initially set to 0. : A RM starts at the leftmost end of the tape. : RMs can perform two operations, labeled 0 and 1. : : 0 = move write head one position to the right : 1 = write a 1 to current location and move one position right : : RMs never read from the tape because they would always read 0. : RMs never halt because they have no halt states. The non-halting is stupid. It basically makes it impossible to have a well-defined result. If you ever DO have a well-defined result, observers can always back-allege that the machine actually DID halt, as soon as the result BECAME well-defined. You can allege that it is still running but WE can allege that what it is now doing no longer deserves to be CALLED running. : The state of an n-state RM can be specified with : (1 + Ceil( log2( n ) ) ) bits where the first bit represents the : operation to perform and the next Ceil(log(2(n)) bits : represent which state to switch to. This is ungrammatical. If the bits represent which state to switch to then they DO NOT represent the state you are CURRENTLY in! But you SAID the state of an n-state RM can be specified with.... I guess specified is just ambiguous. : An n-state RM can be fully specified by listing all states : which requires n * (1 + Ceil(log2(n))) bits. OK. : Some examples of RMs and their output: : : States RM Output : 1 0 (0)... : 1 1 (1)... : 2 01 11 01(1)... : 2 11 00 (10).... : 3 001 010 100 (001)... : : The output of a RM has the following form: : There is a finite, possibly empty initial string that is written once, : and a finite, never empty repeating string that is written over and over. : A computable number can be written as the initial string followed by : the repeating string in parenthesis. In other words, this thing only produces rational numbers. That is a stupid limitation. : Any number having this form can be produced by an RM. : Some examples of computable numbers: : : (0) : (1) : 11(0) : 01(110) : : The length of the initial string plus the length of the repeating string : must be less than or equal to the number of states of the RM that : produced the number. : : Now, prove that no set can contain every RM computable number. But it can; this is just the rational numbers. They are a set. : The standard diagonal argument has a problem - it is usually : impossible to determine if the diagonal number produced is : computable. No, it isn't; that is NOT the problem. I just threw the rest of it away. Bothering to create an alternative paradigm IS STUPID. You should have STUCK to TMs. You relieve yourself of the burden of explanation to everybody (of what non-standard thing you mean) -- everybody ALREADY knows what a TM is. And whenEVER you are asked a direct question, JUST ANSWER IT, if you are going to respond. WHAT KIND OF *NUMBERS* are you talking about (for judging computability)? Natural? Real? Complex? Rational? Supernatural? Infinitesimal? *ANSWER* THE FUCKING *QUESTION*, DUMBASS! ==== > : Let S be the set of all computable numbers > : and assume S is countable. > > I have made the proof a little more rigorous. > > First, I define how to compute a computable number. > I do this with a very non-standard type of Turing machine. > Let's call these Russell machines (RM). > A number is computable if it can be represented by > the infinite binary string produced by a RM. [...] Let C be the set of all RM-computable binary strings, according to your definition. To show that C is countable, it suffices to demonstrate a one-to-one function f : C -> N, where N is the set of natural numbers { 0, 1, 2, ... }. This is not difficult. First, let's agree that any RM M with n states can be uniquely encoded as a natural number. How we choose to do so isn't important, as long as we agree upon the rule -- so let's say for argument's sake that we use your idea as a basis. To encode M, write 1, followed by n zeroes, followed by another 1, and then a sequence of n strings with ceil(lg(n)) + 1 bits each, encoding the transition rules in ascending numerical order as you described. Treat the resulting string as a natural number written in binary, and let that be the encoding for M. Whatever this number is, I'll denote it E(M). It should be obvious that if M1 and M2 are distinct RM's, then E(M1) <> E(M2). Now for any RM-computable string c., let RM(c) be the set of all RM's that compute c. There must be AT LEAST one member of this set, since c is RM-computable*. Let M(c) be the unique element of RM(c) that fulfils these two properties: 1. M(c) in RM(c) 2. For all M in RM(c), E(M(c)) <= E(M). In other words, M(c) is the RM for c that maps to the smallest natural number among all the RM's that compute c, under the RM-encoding rule. It's not important that we pick this one in particular, but doing so eliminates all ambiguity. So: We can now define a one-to-one function f : C -> N as follows: f(c) := E(M(c)) Quickly, let's argue that this is one-to-one: Let c1 and c2 be RM-computable strings with c1 <> c2. Then M(c1) <> M(c2)**, and by the definition of E, E(M(c1)) <> E(M(c2)), so we're all set. This suffices to show that the false diagonalization proof you sketched out is inapplicable here. -M * And, actually, there are infinitely many RM's for any RM-computable string. The proof is left as an exercise for the reader. ;) ** I hold this truth to be self-evident. But if you disagree, it's not so terrible to prove it from your RM definitions. P.S.- I mean for <> to denote the does not equal relation. P.P.S.- This level of detail is overkill for just getting the intuition, but imprecision can be even MORE confusing, I find. -- http://www.dartmouth.edu/~sting/ | Dartmouth College, Hanover, NH, USA ==== in message > : Let S be the set of all computable numbers > : and assume S is countable. > > I have made the proof a little more rigorous. > > First, I define how to compute a computable number. > I do this with a very non-standard type of Turing machine. > Let's call these Russell machines (RM). > A number is computable if it can be represented by > the infinite binary string produced by a RM. [...] Let C be the set of all RM-computable binary strings, according to your > definition. To show that C is countable, it suffices to demonstrate a > one-to-one function f : C -> N, where N is the set of natural numbers { > 0, 1, 2, ... }. This is not difficult. I didn't think so until a few days ago. It's harder than it looks. > So: We can now define a one-to-one function f : C -> N as follows: f(c) := E(M(c)) Quickly, let's argue that this is one-to-one: Let c1 and c2 be > RM-computable strings with c1 <> c2. Then M(c1) <> M(c2)**, and by the > definition of E, E(M(c1)) <> E(M(c2)), so we're all set. Let x be the string produced by the comparator. By construction, x was not produced by f(). E() depends on f(). E(M(x)) is undefined. x represents a RM computable natural number. Russell - 2 many 2 count ==== > Let C be the set of all RM-computable binary strings, according to your > definition. To show that C is countable, it suffices to demonstrate a > one-to-one function f : C -> N, where N is the set of natural numbers { > 0, 1, 2, ... }. This is not difficult. > > I didn't think so until a few days ago. > It's harder than it looks. > > So: We can now define a one-to-one function f : C -> N as follows: > > f(c) := E(M(c)) > > Quickly, let's argue that this is one-to-one: Let c1 and c2 be > RM-computable strings with c1 <> c2. Then M(c1) <> M(c2)**, and by the > definition of E, E(M(c1)) <> E(M(c2)), so we're all set. > > Let x be the string produced by the comparator. I'm sorry, perhaps I missed something. What comparator? > By construction, x was not produced by f(). > E() depends on f(). E(M(x)) is undefined. > x represents a RM computable natural number. Setting aside for the moment that I do not know what string you mean by x, I can safely assert that E does not depend upon f. I chose E to be a simple encoding rule, so that we could be easily convinced that it is a one-to-one mapping from RM's to natural numbers. Its definition does not depend upon f's definition. What do you mean by an RM-computable natural number? Your definition of RM-computability talks about binary strings of infinite length. Due ot their regular structure, you can ENCODE such a string uniquely as a natural number. That's essentially what I proved above. But you did not show a one-to-one mapping from RM-computable strings to natural numbers in your original definition! The obvious mapping -- just dropping the parentheses and gluing the bits together) doesn't quite work, because (for example) all these distinct RM-computable strings map to the same number: 010(10), 0010(10), 00010(10), ... If you don't like referring to a specific RM, an alternate proof formulation would be to first prove that RM-computable strings have the structure you described: wx+ where w in (0+1)* and x in (0+1)(0+1)*. Once you prove that, you can encode any RM-computable string in binary by mapping 0 => 0, 1 => 11, and . to 10 and writing .w.x (the dots guarantee a unique natural number interpretation even when w begins with one or more zeroes). Again, the choice of code is unimportant except insofar as we have to agree what it is. -M -- http://www.dartmouth.edu/~sting/ | Dartmouth College, Hanover, NH, USA ==== Let x initally be (0). > If s has the form 111...111(0) and the length of the initial > segment of s is longer or equal to the length of the initial > segment of x then take the initial segment of s, append a 1, and > make this new string the initial segment of x. Examples: s x > (0) 1(0) > 1(0) 11(0) > 11(0) 111(0) Prove that x differs from every member of S. x differs from every s that does not end with (0). > x differs from every s with an inital segment not of the form > 111.111. x differs from every s with form 111...111(0). > Therefore, x differs from every member of S. > Well ... let S be the set of al RM computable numbers of the form >> 1*(0) (any number of sequential ones followed by a repeating of >> zeros). In this way, you will not be able to compute a x which >> differs from every possible member of S (at least not using your >> technique). > Why not? > x will differ from every member of S and x will be a RM computable > number. > Another way to look at his : >> if all the numbers in a set S or of the form 111...11(0), you can define a >> set S' (containing natural numbers) such that an element of S which has a >> sequence of m 1's corresponds to an element m of S'. Your x' would then be >> m+1. Now if you consider the set S I used in my previous post, it would >> correspond tot the entire set of natural numbers. As for each m, contained >> in this set, m+1 is also contained, there cannot exist an x' which does not >> belong to this set. Earlier in this post, someone said the natural numbers are computable. Let's use your set as an example. We can interpret the strings right to left. Now, each binary string represents a natural number in base 1. Let's call the set S. S(0) = (0) S(1) = (0)1 S(2) = (0)11 ... I give a constructive method of creating a RM computable number, x, that is not in S. How can my method fail to produce x? You assume that every string that corresponds to a finite number is in S. This is not true if my proof is correct. I can only think of two reasons why I would not be able to compute x. 1) S contains a member with infinitely many 1's 2) S contains a member with so many 1's that adding one more 1 results in an infinitely long string of 1's. If you can think of another reason why x can not be computed, please post it. Can you show that x is impossible to compute? Russell - 2 many 2 count ==== > >> Let x initally be (0). >> If s has the form 111...111(0) and the length of the initial >> segment of s is longer or equal to the length of the initial >> segment of x then take the initial segment of s, append a 1, and >> make this new string the initial segment of x. >> Examples: >> s x >> (0) 1(0) >> 1(0) 11(0) >> 11(0) 111(0) >> Prove that x differs from every member of S. >> x differs from every s that does not end with (0). >> x differs from every s with an inital segment not of the form >> 111.111. x differs from every s with form 111...111(0). >> Therefore, x differs from every member of S. >> Well ... let S be the set of al RM computable numbers of the form > 1*(0) (any number of sequential ones followed by a repeating of > zeros). In this way, you will not be able to compute a x which > differs from every possible member of S (at least not using your > technique). > Why not? >> x will differ from every member of S and x will be a RM computable >> number. > Another way to look at his : > if all the numbers in a set S or of the form 111...11(0), you can > define > a > set S' (containing natural numbers) such that an element of S which > has a sequence of m 1's corresponds to an element m of S'. Your x' > would then > be > m+1. Now if you consider the set S I used in my previous post, it > would correspond tot the entire set of natural numbers. As for each > m, > contained > in this set, m+1 is also contained, there cannot exist an x' which > does > not > belong to this set. > > Earlier in this post, someone said the natural numbers are computable. > Let's use your set as an example. We can interpret the strings right > to left. > Now, each binary string represents a natural number in base 1. > Let's call the set S. > > S(0) = (0) > S(1) = (0)1 > S(2) = (0)11 > ... > > I give a constructive method of creating a RM computable number, x, > that is not in S. How can my method fail to produce x? Because every x would be of the form (0)11..1, and because of the definition of S, any such RM number is contained in S. > You assume that every string that corresponds to a finite number is in > S. This is not true if my proof is correct. That was how S was defined : it contains all the strings of the form (0)11..1 (this can be considered as all unary representations of natural numbers). > I can only think of two reasons why I would not be able to compute x. > > 1) S contains a member with infinitely many 1's > 2) S contains a member with so many 1's that adding > one more 1 results in an infinitely long string of 1's. > > If you can think of another reason why x can not be computed, > please post it. 3) for any s in S, s-with-a-1-added-to-the-back is also contained in S > > Can you show that x is impossible to compute? As there can not exist an element which is both member of a set and not contained in the set, yes. > > > > Russell > - 2 many 2 count > > > -- Pento De wereld was soep, en het denken meestal een vork, tot smakelijk eten leidde dat zelden. - H. Mulisch ==== > > I give a constructive method of creating a RM computable number, x, > that is not in S. How can my method fail to produce x? Because every x would be of the form (0)11..1, and because of the > definition of S, any such RM number is contained in S. The definition of S leads to contradiction. S can not contain every natural number. > You assume that every string that corresponds to a finite number is in > S. This is not true if my proof is correct. That was how S was defined : it contains all the strings of the form > (0)11..1 (this can be considered as all unary representations of natural > numbers). I show that there exists an RM computable number of the form (0)11...1 that is not in set S. I can even show how this number can be constructed by a Turing machine. RM's are a subset of TMs. Any RM can be emulated by a universal Turing machine (UTM). We are only concerned with the subset of RM's that output an initial, finite and contiguous string of 1's followed by an infinite string of 0's. The UTM can generate the output tape for each of these RM's. These tapes are then read by a second TM I will call a Comparator (CTM). The CTM compares each input tape with the CTM's output tape. If the input tape has a longer initial string of 1's, the CTM rewinds and 1. After all of the RM tapes have beed read by the CTM we examine the output of the CTM. This tape must contain the representation of a natural number. Every tape read by the CTM represents a natural number. The CTM output tape contains the representation of the successor of some member of S. The successor of a natural number is a natural number. S can not contain a representation of every natural number. > I can only think of two reasons why I would not be able to compute x. > > 1) S contains a member with infinitely many 1's > 2) S contains a member with so many 1's that adding > one more 1 results in an infinitely long string of 1's. > > If you can think of another reason why x can not be computed, > please post it. 3) for any s in S, s-with-a-1-added-to-the-back is also contained in S Not true. I show how to construct such a number that is not in S. > Can you show that x is impossible to compute? As there can not exist an element which is both member of a set and not > contained in the set, yes. The definition of S leads to contradiction. S can not exist. Russell - 2 many 2 count ==== > > > I give a constructive method of creating a RM computable number, x, > that is not in S. How can my method fail to produce x? You have yet to say What KIND of numbers RM numbers CAN OR CAN'T be. Are they natural? Rational? Real? Complex? Supernatural? Infinitesimal? IT MATTERS. ANSWER the question! > > Because every x would be of the form (0)11..1, and because of the > definition of S, any such RM number is contained in S. > > The definition of S leads to contradiction. No, it doesn't. > S can not contain every natural number. It can and it does. YOU defined it. Do you have any idea how STUPID this makes you look, defining a set and then saying that it can't contain things it obviously contains? > > You assume that every string that corresponds to a finite number is in > S. No, we don't assume it, we just NOTICE it. You defined S as the class of all outputs of RMs that printed some natural number of 1s and then stopped. > This is not true if my proof is correct. Ergo, your proof is not one. > That was how S was defined : it contains all the strings of the form > (0)11..1 (this can be considered as all unary representations of natural > numbers). YOU defined S that way. So S contains all natural numbers. > I show that there exists an RM computable number of the form (0)11...1 > that is not in set S. No, you don't. > I can even show how this number can be constructed > by a Turing machine. No, you can't. > RM's are a subset of TMs. No, they're not. For one thing, you alleged that they never halted. Are you going to retract that? > Any RM can be emulated by a universal Turing machine (UTM). > We are only concerned with the subset of RM's that output an initial, > finite and contiguous string of 1's followed by an infinite string of 0's. Since RM output tapes were supposed to START OUT as all 0's, can't the the RM just HALT after it gets through printing all its 1's? > The UTM can generate the output tape for each of these RM's. And put it where, exactly? TM's, by default, only have ONE output tape. > These tapes are then read by a second TM I will call a Comparator (CTM). No, they aren't. There are an infinite number of these tapes and the CTM never finishes reading them all. > The CTM compares each input tape with the CTM's output tape. The CTM does NOT HAVE an output tape, other than its input tape. That's just how TMs ARE DEFINED. ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- > RM's are a subset of TMs. > Any RM can be emulated by a universal Turing machine (UTM). > We are only concerned with the subset of RM's that output an initial, > finite and contiguous string of 1's followed by an infinite string of 0's. > So you're not proving there is no complete RM list, 1st you are proving there is no complete restricted RM list? Is this based on modelling the diagonal number from your previous post? and then prove that it is not in my list. > > 0.000... > 0.1000... > 0.11000... > 0.111000... > 0.1111000... > 0.11111000... > 0.111111000... > ... > > Good luck trying to prove the diagonal number is > not in the list using a countable number of operations. Seems too easy, surely? If I understand you correctly the diagonal > number is: 0.11111.... Every number in your list consists of a finite string of 1s followed > by zeros (000...). The diagonal number doesn't. Ergo it isn't in the > list. I can find an entry in my list that matches the diagonal number to any finite number, n, of positions. This is true for all n. My list is infinite. There is at least one entry in my list that has a 1 in every finite position. How can you prove the diagonal number is not in my list? Russell convenience, I've changed the layout very slightly... > 0.111000... > 0.1111000... > 0.11111000... > 0.111111000... ^-------- Herc ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- > RM's are a subset of TMs. > Any RM can be emulated by a universal Turing machine (UTM). > We are only concerned with the subset of RM's that output an initial, > finite and contiguous string of 1's followed by an infinite string of 0's. > > So you're not proving there is no complete RM list, 1st you are > proving there is no complete restricted RM list? This is the only subset of RM's anyone has concerns about. I think everyone agrees my method works for all the other RMs. > Is this based on modelling the diagonal number from your previous post? Partly. > and then prove that it is not in my list. > > 0.000... > 0.1000... > 0.11000... > 0.111000... > 0.1111000... > 0.11111000... > 0.111111000... > ... > > Good luck trying to prove the diagonal number is > not in the list using a countable number of operations. > > Seems too easy, surely? If I understand you correctly the diagonal > number is: > > 0.11111.... The diagonal method can be converted into the computable number proof. Of course, when I responded to your thread I was arguing against Cantor's diagonal proof. I am arguing for it in this thread. Forcing the diagonal number to be computable shows that invoking infinity is like killing a fly with a machine gun. All we really need to prove it that the diagonal has more 1's than any of the real numbers in the list above. The diagonal can have a finite number of 1's and still not be in the list. I suspect that the diagonal proof can be modified to show that the rational numbers are uncountable. Let R be the set of all rational numbers, r, such that 0 <= r < 1. If R can be ordered such that the diagonal is a rational number, the diagonal proof shows the rationals to be uncountable. Most people will argue that the diagonal of this set must be an irrational number. I have never seen a proof of this. It is easy to come up with a list of rationals that have a rational diagonal. The set I give above is one such set. Russell - 2 many 2 count ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- > Is this based on modelling the diagonal number from your previous post? Partly. > and then prove that it is not in my list. > > 0.000... > 0.1000... > 0.11000... > 0.111000... > 0.1111000... > 0.11111000... > 0.111111000... > > 0.11111.... The diagonal method can be converted into the computable number proof. So basically you're trying to compute 0.1.. zero point one recurring. To show the diagonal (not on the list) computable. So, I will define another method for finding a computable number that is not in a particular set of numbers. Let S be a set and every member of S be a RM computable number. Let s be a member of S. Let x be a RM computable number not in S. Let x initally be (0). If s has the form 111...111(0) and the length of the initial segment of s is longer or equal to the length of the initial segment of x then take the initial segment of s, append a 1, and make this new string the initial segment of x. Examples: s x (0) 1(0) 1(0) 11(0) 11(0) 111(0) Prove that x differs from every member of S. 1 x differs from every s that does not end with (0). 2 x differs from every s with an inital segment not of the form 111.111. 3 x differs from every s with form 111...111(0). 4 Therefore, x differs from every member of S. Prove that x is a RM computable number. OK 1-3 are the properties of x. but 3 seems blatently wrong. Since everything is computable, lets see the actual number not on the list. Lets compute S for real. The point of RMs seems to be to remove hypothetical numbers in the proof. So do all the actual calculations. What is the number not on the list? Every contender for x is in the list. You've proven that x is different to ONE member of the list. Herc ==== > 1 x differs from every s that does not end with (0). > 2 x differs from every s with an inital segment not of the form 111.111. > 3 x differs from every s with form 111...111(0). > 4 Therefore, x differs from every member of S. Prove that x is a RM computable number. > OK 1-3 are the properties of x. but 3 seems blatently wrong. Line 3 is correct. x must differ from every string with form 111...111(0). Specificaly, x has more initial 1's than any string like this in S. And x has a finite number of 1's. Saying x has an infinite number of 1's is overkill. I only need to show that x is longer than any string in S. I don't have to assume x has infinite length. x is exactly one 1 longer than some finite string of 1's in S. x must have a finite string of 1's. > Since everything is computable, lets see the actual number not on the list. > Lets compute S for real. The point of RMs seems to be to remove hypothetical > numbers in the proof. So do all the actual calculations. What is the number not on the list? Every contender > for x is in the list. x represents a relatively small natural number. For example, x=RM(i) and i >> |x|. Are you sure you want me to post x? > You've proven that x is different to ONE member of the list. x differs from every member of S. See above. The proof shows that it is impossible for a TM to count to infinity. Imagine a TM that counts how many tapes it has read. Even if this TM reads an infinite number of tapes, the TM will tell us it has read a finite number. Similarly, a TM will say that the length of any tape is finite. Russell - 2 many 2 count ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- Saying x has an infinite number of 1's is overkill. > I only need to show that x is longer than any string in S. ok by 'any' here you mean all, not exists one string. but this is trivial, just order S and add one more element. its a strange way to define a new element. N <-bijection-> S <-bijection-> X <-element of X not in S > I don't have to assume x has infinite length. > x is exactly one 1 longer than some finite string of 1's in S. > x must have a finite string of 1's. > it has to be longer than the *largest* string in S S is almost identical to X x represents a relatively small natural number. > For example, x=RM(i) and i >> |x|. > Are you sure you want me to post x? > Noone can follow the proof without. Haven't you just proven all finite lists are incomplete? Herc ==== > Let R be the set of all rational numbers, r, such that 0 <= r < 1. > If R can be ordered such that the diagonal is a rational number, > the diagonal proof shows the rationals to be uncountable. And if my grandmother had wheels, she'd be a wagon. There are several ways to put that subset of the rational numbers into one-to-one correspondence with the positive integers. > Most people will argue that the diagonal of this set must be > an irrational number. I have never seen a proof of this. > It is easy to come up with a list of rationals that have a rational > diagonal. The set I give above is one such set. Of course, that set was not all of the rationals in that range. Here is an example of the beginning of a complete list: 0 1/2 1/3 2/3 1/4 3/4 1/5 2/5 3/5 4/5 1/6 5/6 1/7 ... Identifying a rational given its position (or vice-versa) is probably going to require generating the list up to that position, but this list is obviously complete. Therefore, the set of rational numbers in [0,1) is countable. You want a proof that any such complete list has an irrational diagonal? Okay: Assume that a complete list with a rational diagonal can be found. The diagonalization process would generated a rational number not in the list. This would contradict the assumption. Therefore, either the list is incomplete or the diagonal is not rational. (My list is complete, so the diagonal must not be rational. Your list had a rational diagonal, but was obviously incomplete.) When applied to an allegedly complete list of the reals in [0,1), the diagonally-generated number does not have the option of being other than real, so the only possible conclusion is that the list was incomplete. -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W ==== > Let R be the set of all rational numbers, r, such that 0 <= r < 1. > If R can be ordered such that the diagonal is a rational number, > the diagonal proof shows the rationals to be uncountable. And if my grandmother had wheels, she'd be a wagon. There are several ways to put that subset of the rational numbers into > one-to-one correspondence with the positive integers. > Most people will argue that the diagonal of this set must be > an irrational number. I have never seen a proof of this. > It is easy to come up with a list of rationals that have a rational > diagonal. The set I give above is one such set. Of course, that set was not all of the rationals in that range. Of course. > Here is an example of the beginning of a complete list: 0 > 1/2 > 1/3 > 2/3 > 1/4 > 3/4 > 1/5 > 2/5 > 3/5 > 4/5 > 1/6 > 5/6 > 1/7 > ... Identifying a rational given its position (or vice-versa) is probably > going to require generating the list up to that position, but this list > is obviously complete. Therefore, the set of rational numbers in [0,1) > is countable. Maybe. That is the point of the proof. > You want a proof that any such complete list has an irrational diagonal? > Okay: Assume that a complete list with a rational diagonal can be found. The > diagonalization process would generated a rational number not in the > list. This would contradict the assumption. Therefore, either the list > is incomplete or the diagonal is not rational. (My list is complete, so > the diagonal must not be rational. Your list had a rational diagonal, > but was obviously incomplete.) How are you proving your list is complete? To prove that the set contains all rational numbers you will have to show there is no way to order the set such that the diagonal is rational. There are a lot of ways to order a set of rationals. Russell - 2 many 2 count ==== > How are you proving your list is complete? Any rational in [0,1) can be uniquely written as a ratio between two coprime nonnegative integers m/n. That is an entry among the first n(n-1)/2 + 1 entries on the list, although specifying the exact location involves a summation on the Euler phi-function. Anyway, that entry on the list corresponds to no other rational. (This last proviso is moderately important, because some people like to present complete lists of reals in which a given list entry can have more than one real number associated with it.) > To prove that the set contains all rational numbers > you will have to show there is no way to order > the set such that the diagonal is rational. I just proved that the set contains all rational numbers in [0,1). To prove that a given integer is even, is it necessary to show both that its units digit in base two is 0 and that its units digit in base ten is in {0,2,4,6,8}? > There are a lot of ways to order a set of rationals. If you are talking about the complete set of rationals, the number of ways is the same as the number of real numbers. If you think a relevant ordering exists, either point out a flaw in my proof or describe the ordering. Otherwise, you might as well quibble with a proof that the sum of any finite set if even numbers is not odd by pointing out that there are a lot of ways to choose a set of even numbers. -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W <-sOdnVAq4PkNeXGiRVn-sA@comcast.com> <8aGdnb0k8ZEW73CiRVn-jA@comcast.com> <1g6p2lp.1nq0zj71e3df92N%panoptes@iquest.net> <9tCdnYk507_tD3KiRVn-hQ@comcast.com> ==== > Let R be the set of all rational numbers, r, such that 0 <= r < 1. > If R can be ordered such that the diagonal is a rational number, > the diagonal proof shows the rationals to be uncountable. > > And if my grandmother had wheels, she'd be a wagon. > > There are several ways to put that subset of the rational numbers into > one-to-one correspondence with the positive integers. > Here is an example of the beginning of a complete list: > > 0 > 1/2 > 1/3 > 2/3 > 1/4 > 3/4 > 1/5 > ... > > Identifying a rational given its position (or vice-versa) is probably > going to require generating the list up to that position, but this list > is obviously complete. Therefore, the set of rational numbers in [0,1) > is countable. Maybe. That is the point of the proof. et cetera, et cetera. It's equivalent to the abbreviation Q.E.D. There's no reason to say Maybe. [I repeat my opinion that you should search Google before continuing to post in this thread. Pedagogy is fun, and I'll be the first to admit I have benefitted greatly from it, but it tends to clutter up the groups if it's let run amok, IMHO.] > How are you proving your list is complete? Actually, he didn't -- he just said, It's obvious, and moved on. A rigorous proof would use the traditional diagonal method -- NOT Cantor's diagonalization, but a different kind of diagonal analogy I think attributed to Godel. It involves laying out the plane of (positive) rational numbers as follows: 1/1--1/2 1/3--1/4 ... .' .' .' 2/1 2/2 2/3 2/4 ... | .' .' 3/1 3/2 3/3 3/4 ... .' 4/1 4/2 4/3 4/4 ... | ... ... You'll need a fixed-width font to see the ASCII-art line that I've drawn zigzagging across the diagonals of the grid. This line obviously passes through every number in the grid. And every positive rational number is *somewhere* on that grid, as you can see from the way it's laid out. So the line passes through every positive rational number, in sequence. Straighten out the line, and remove duplicates, and add zero and the negative rationals, and you're done -- you have a complete list of all the rationals. It begins 0, 1, -1, 1/2, -1/2, 2, -2, 3, -3, 1/3, -1/3, 1/4, -1/4, 2/3, ... and continues to infinity. Now match up the elements of that list, one-to-one, with the positive integers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, ... Ta-da! [The next bit has been re-ordered for clarity.] > To prove that the set contains all rational numbers > you will have to show there is no way to order > the set such that the diagonal is rational. > There are a lot of ways to order a set of rationals. > You want a proof that any such complete list has an irrational diagonal? > Okay: > > Assume that a complete list with a rational diagonal can be found. The > diagonalization process would generated a rational number not in the > list. This would contradict the assumption. Therefore, either the list > is incomplete or the diagonal is not rational. [Q.E.D.] -Arthur ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- > RM's are a subset of TMs. > Any RM can be emulated by a universal Turing machine (UTM). > We are only concerned with the subset of RM's that output an initial, > finite and contiguous string of 1's followed by an infinite string of > 0's. > > So you're not proving there is no complete RM list, 1st you are > proving there is no complete restricted RM list? This is the only subset of RM's anyone has concerns about. > I think everyone agrees my method works for all the other RMs. > Is this based on modelling the diagonal number from your previous post? Partly. > and then prove that it is not in my list. > > 0.000... > 0.1000... > 0.11000... > 0.111000... > 0.1111000... > 0.11111000... > 0.111111000... > ... > > Good luck trying to prove the diagonal number is > not in the list using a countable number of operations. > > Seems too easy, surely? If I understand you correctly the diagonal > number is: > > 0.11111.... The diagonal method can be converted into the computable number proof. Of course, when I responded to your thread I was arguing against Cantor's > diagonal proof. I am arguing for it in this thread. Forcing the diagonal number to be computable shows that invoking > infinity is like killing a fly with a machine gun. All we really need to prove it that the diagonal has more 1's than > any of the real numbers in the list above. The diagonal can have > a finite number of 1's and still not be in the list. I suspect that the diagonal proof can be modified to show that > the rational numbers are uncountable. Let R be the set of all rational numbers, r, such that 0 <= r < 1. > If R can be ordered such that the diagonal is a rational number, > the diagonal proof shows the rationals to be uncountable. Most people will argue that the diagonal of this set must be > an irrational number. I have never seen a proof of this. This is my argument against the diagonal. I can find a rational on the list of computables that equals any specific representation (finite) of the 'irrational' diagonal construct number. If the rational number equals the irrational number then its the same number. I'm pretty sure Cantor falls on its head, I can give an algorithm for all numbers UTM(Z), and increasing portions of the list it makes. That is all anyone can ask, and from that noone can make an 'infinite' diagonal construct at all in practicality. Every number is on the list because the most they can differ by is (0..9)/oo. Herc > It is easy to come up with a list of rationals that have a rational > diagonal. The set I give above is one such set. > Russell > - 2 many 2 count > ==== > >> I give a constructive method of creating a RM computable number, x, >> that is not in S. How can my method fail to produce x? >> Because every x would be of the form (0)11..1, and because of the >> definition of S, any such RM number is contained in S. > > The definition of S leads to contradiction. > S can not contain every natural number. > >> You assume that every string that corresponds to a finite number is >> in S. This is not true if my proof is correct. >> That was how S was defined : it contains all the strings of the form >> (0)11..1 (this can be considered as all unary representations of >> natural numbers). > > I show that there exists an RM computable number of the form (0)11...1 > that is not in set S. I can even show how this number can be > constructed by a Turing machine. > > RM's are a subset of TMs. > Any RM can be emulated by a universal Turing machine (UTM). > We are only concerned with the subset of RM's that output an initial, > finite and contiguous string of 1's followed by an infinite string of > 0's. > > The UTM can generate the output tape for each of these RM's. > These tapes are then read by a second TM I will call a Comparator > (CTM). > > The CTM compares each input tape with the CTM's output tape. > If the input tape has a longer initial string of 1's, the CTM rewinds > additional 1. > > After all of the RM tapes have beed read by the CTM we examine the > output of the CTM. This tape must contain the representation of a > natural number. > > Every tape read by the CTM represents a natural number. > The CTM output tape contains the representation of the successor > of some member of S. The successor of a natural number is a > natural number. > > S can not contain a representation of every natural number. > >> I can only think of two reasons why I would not be able to compute >> x. >> 1) S contains a member with infinitely many 1's >> 2) S contains a member with so many 1's that adding >> one more 1 results in an infinitely long string of 1's. >> If you can think of another reason why x can not be computed, >> please post it. >> 3) for any s in S, s-with-a-1-added-to-the-back is also contained in >> S > > Not true. I show how to construct such a number that is not in S. > >> Can you show that x is impossible to compute? >> As there can not exist an element which is both member of a set and >> not contained in the set, yes. > > The definition of S leads to contradiction. > S can not exist. OK, consider the set of natural numbers, N. Also consider the operation + 1. Is it not true that for each n in N, n + 1 is also in N ? Now, consider the function f : N -> RM-numbers : f(0) = (0) f(1) = (0)1 f(2) = (0)11 ... and so on, with f(n) corresponding to its own RM-number for each natural number n. Clearly, not all the different RM-numbers are reached by this function f. Consider the set F = f(N) of numbers which are. For each element s of F, we can find a unique natural number n such that s = f(n). This means we can give the inverse function of f, call this g. Now, for each s in F, there can not be an x which is s followed by an extra 1, which is not already in F : this would mean that g(x)=g(s)+1, with g(s) a natural number, would not be a natural number. If you still consider this line of thinking to be wrong, please give me the exact x for which value it goes wrong. -- Pento De wereld was soep, en het denken meestal een vork, tot smakelijk eten leidde dat zelden. - H. Mulisch ==== OK, consider the set of natural numbers, N. > Also consider the operation + 1. Is it not true that for each n in N, n + 1 is also in N ? Maybe. I might not be the person you want to ask this question to. My proof shows that no set can contain every computable natural number. If N does not contain every computable natural number, how can we say N contains all natural numbers? > Now, consider the function f : N -> RM-numbers : f(0) = (0) > f(1) = (0)1 > f(2) = (0)11 > ... > and so on, with f(n) corresponding to its own RM-number for each natural > number n. Clearly, not all the different RM-numbers are reached by this function f. > Consider the set F = f(N) of numbers which are. For each element s of F, we > can find a unique natural number n such that s = f(n). This means we can > give the inverse function of f, call this g. > Now, for each s in F, there can not be an x which is s followed by an extra > 1, which is not already in F : this would mean that g(x)=g(s)+1, with g(s) > a natural number, would not be a natural number. Since g() is the inverse of f(), and x was not generated by f(), g(x) may not be defined. x is RM computable so there exists a RM that will output it. This means f() can not emulate every possible RM. Russell - 2 many 2 count ==== > OK, consider the set of natural numbers, N. >> Also consider the operation + 1. >> Is it not true that for each n in N, n + 1 is also in N ? Maybe. >I might not be the person you want to ask this question to. >My proof shows that no set can contain every computable >natural number. If N does not contain every computable >natural number, how can we say N contains all natural numbers? (I haven't been following this thread, so please make allowances if what I say here is either redundant or irrelevant.) The standard definitions of computable are equivalent to computable by some Turing machine. Now clearly, each particular natural number K can be generated by some TM; indeed, it's even computable by a finite-state machine (with K states). Furthermore, the infinite sequence of all natural numbers together (e.g., in base 1, 010110111011110111110...) can be generated by a (non-halting) TM -- the algorithm repeatedly copies the last string of 1's (which can be done finitely by temporarily replacing the most recent 1 copied with some other symbol, to keep track of how much has already So in what sense isn't every natural number computable? [snip] -- --------------------------- | B B aa rrr b | | BBB a a r bbb | | B B a a r b b | | BBB aa a r bbb | ----------------------------- ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- > >> OK, consider the set of natural numbers, N. >> Also consider the operation + 1. >> Is it not true that for each n in N, n + 1 is also in N ? > >Maybe. >I might not be the person you want to ask this question to. >My proof shows that no set can contain every computable >natural number. If N does not contain every computable >natural number, how can we say N contains all natural numbers? (I haven't been following this thread, so please make allowances if what > I say here is either redundant or irrelevant.) The standard definitions of computable are equivalent to computable > by some Turing machine. Now clearly, each particular natural number K > can be generated by some TM; indeed, it's even computable by a > finite-state machine (with K states). Furthermore, the infinite sequence of all natural numbers together > (e.g., in base 1, 010110111011110111110...) can be generated by a > (non-halting) TM -- the algorithm repeatedly copies the last string of > 1's (which can be done finitely by temporarily replacing the most recent > 1 copied with some other symbol, to keep track of how much has already So in what sense isn't every natural number computable? > Check out this 3 state TM that counts from 0 upwards in binary. Every second position on the tape represents the number. Herc knock knock knock on the door, feet still bleeding ==== > message > >> OK, consider the set of natural numbers, N. >> Also consider the operation + 1. >> Is it not true that for each n in N, n + 1 is also in N ? > >Maybe. >I might not be the person you want to ask this question to. >My proof shows that no set can contain every computable >natural number. If N does not contain every computable >natural number, how can we say N contains all natural numbers? (I haven't been following this thread, so please make allowances if what > I say here is either redundant or irrelevant.) The standard definitions of computable are equivalent to computable > by some Turing machine. Now clearly, each particular natural number K > can be generated by some TM; indeed, it's even computable by a > finite-state machine (with K states). Furthermore, the infinite sequence of all natural numbers together > (e.g., in base 1, 010110111011110111110...) can be generated by a > (non-halting) TM -- the algorithm repeatedly copies the last string of > 1's (which can be done finitely by temporarily replacing the most recent > 1 copied with some other symbol, to keep track of how much has already So in what sense isn't every natural number computable? Consider all TM's that write an initial, finite and contiguous strings of 1's and then halt. Let S be the set of all output tapes from these TM's. Define another TM I call a comparator (CTM). CTM compares the input tape to its output tape. If the input tape has more 1's, the CTM rewinds to the beginning of its output tape, copies the input tape, We let CTM read all the tapes in S. What is on the output tape of CTS? The output of CTM must have a finite number of 1's. Every tape read by CTM was finite in length. The output of CTM has exactly one more 1 than some input tape in S. S can not contain every possible string that represents a natural number. Russell - 2 many 2 count ==== comp.theory and sci.math need to come off the list of newsgroups for this thread. Those groups surely have more important things to discuss. : : Consider all TM's that write an initial, finite and contiguous strings of : 1's : and then halt. OK. There is one of these TMs for every natnum. : Let S be the set of all output tapes from these TM's. That's basically the set of all natural numbers. : Define another TM I call a comparator (CTM). : CTM compares the input tape to its output tape. CTM *DOES* *NOT* *HAVE* an output tape. CTM is a TM. In general, TMs DO NOT HAVE output tapes. TMs have ONE tape. They use it for input. They also write on it, but that does not make it an output tape. It is at best an input/output tape. : If the input tape has more 1's, This is ridiculous. The input tape IS the output tape, if CTM is a TM. It can NEVER have a DIFFERENT number of 1's from itself. : the CTM rewinds to the beginning of its output tape, copies the input tape, That is not how TMs work. What actually happens is that the TM reads at the end, and then halts. : We let CTM read all the tapes in S. No, we don't, because S has an infinite number of tapes, so CTM (if it really is a TM, which, as you have defined it, IT ISN'T, because it has an output tape) will never finish reading all these tapes. : What is on the output tape of CTS? That's a typo; CTS does not exist; do you mean CTM? : The output of CTM must have a finite number of 1's. Right. : Every tape read by CTM was finite in length. Right. : The output of CTM has exactly one more 1 than : some input tape in S. Wrong. S has infinitely many input tapes and there is no upper bound on how long they are. : S can not contain every possible string that : represents a natural number. As you defined it, it does indeed contain exactly that; it represents the natnums in unary. -- --- It's difficult ... you need to be united to have any strength, but internal issues have to be addressed. --- E. Ray Lewis, on liberalism in America <-sOdnVAq4PkNeXGiRVn-sA@comcast.com> <8aGdnb0k8ZEW73CiRVn-jA@comcast.com> <19idnd3Ywuot6HKiRVn-sA@comcast.com> ==== > [ Obviously, every natural number N is computable by an N-state FSM. ] > So in what sense isn't every natural number computable? Consider all TM's that write an initial, finite and contiguous strings of > 1's and then halt. Let S be the set of all output tapes from these TM's. S is an infinite set which is also countable. > Define another TM I call a comparator (CTM). > CTM compares the input tape to its output tape. > If the input tape has more 1's, the CTM rewinds > to the beginning of its output tape, copies the input tape, We let CTM read all the tapes in S. What do you mean read all the tapes in S? S has infinitely many members. If CTM tries to read all the members of S, it will never finish. > What is on the output tape of CTS? Nothing -- CTM never halts, as it never finishes reading its input tapes. > The output of CTM must have a finite number of 1's. > Every tape read by CTM was finite in length. > The output of CTM has exactly one more 1 than > some input tape in S. Non sequitur. > S can not contain every possible string that > represents a natural number. Non sequitur. Surely you've had this explained to you in the past, about how some infinities are bigger than others, and which ones, and how? Check Google Groups if you haven't, or don't remember -- I'm sure there's plenty of tutorial information on there. Check sci.math's archives, too. -Arthur ==== What are the parameteric coordinates of closed loops generated by intersection point F between b and c and intersection point I between a and c extensions in a quadrilateral of sides a,b,c,d ( side a smallest,side d largest, angle th variable/parameter between a and d )? These occur in a four bar linkage mechanism . The book 4 Bar Atlas(by Hrones and Nelson) does not include parameteric equations. TIA. ==== I need to know how to calculate x^y where y is not an integer. Any help in this matter would be appreciated.