when last we left Sarfatti X and the other great-great-great- grandkids of Jack's great-great-great-grandkids, they had smashed the shackles set on them by the Hawkings -- as well as any extant DVDs of Star Trek, The Movie -- and were rapidly tunnelling toward gran-pere, with a fierce vengeance in their hearts. JS: kids, you are violating the Bohmian ineffectuality this moment!... come back, when you have enough backgound cosmic anisotropy! > JS: You do not have enough background. > The giant DeBroglie MACRO-QUANTUM q-bit ground state condensate in the > living brain IS the MIND! > JS: Read p. 30 and 14.6 of the above. > Yes, no ACTION without REACTION. > This idea needs further development, i.e. putting Wheeler-Feynman > together with Bohm's ontology. > See Dick Bierman's experiments on presponse. > If I am correct, the presponse curve of the subject should weaken as the > subject gets increasing sedation with > anaesthetics. > The two act together in a self-organizing loop of conscious intent. > See first if you can understand that book, which is pre-requisite. > You must crawl before you can dance. > In the FRW metric define a GLOBAL cosmic time with global simultaneity, > i.e. a preferred 3D space-like foliation of 4D space-time in which all > points on the preferred space-like hypersurface see MAXIMALLY isotropic > cosmic blackbody radiation with SAME absolute temperature. This is a > practical way to navigate when you use star gate time travel tunnels and > weightless warp drive by metrically engineering the zero point quantum > pressures of the physical vacuum via a kind of Josephson effect IMHO, > i.e. interfering the virtual vacuum condensate with a real ground state > condensate using variations on the Bohm-Aharonov dynamical phase and > Berry topological phase effects. --Give the Gift of Trickier Dick Cheeny -- out of office, at last! http://www.benfranklinbooks.com/ http://www.wlym.com/pages/music.html http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac http://www.wlym.com/PDF-68-76/CAM7606.pdf ==== Yo, Brian, How are you dooen? What's your buddy, ugly LaRouge, your constant Usenet companion doing lately? Isn't he even Trickier [then] Dick Cheeny? Once every 4 years, I make it a point to listen to his 1 hour long, truly grand and outstanding rant, just before election. For that special election show I always have a bottle of Champaign and plenty of soft floor coverings in place for convenient ROTFLs. I love fanatics, any sort, any place, any time. ahahahaha.....ahahahanson > the great-great-grandkids of Jack's great-great-grandkids are coming > -- having freed themselves from Hawking's shackles -- > to put an end to this misery. woe, to future Trekkie conferees! >pilot wave. >exotic vacuum zero point energy density field. UFOs show us that someone >Out There knows how to do this trick. The control parameter for the >zero point energy density is vacuum coherence -- an idea missing from > JS: You mean Gaia. Yes, that's Hawking's Mind of God in the >Intelligent Conscious Universe of Star Maker, Solaris, Black Cloud, >VALIS, Cosmic Coincidence Control -- all that stuff from The Poets prior >to The Physicists. > http://www.classicalmidiconnection.com/cgibin/x.cgi?midi/n3/zsunrise.mid > Z.D. What I mean by this is a Wheeler-Feynmen absorber theory of the >wavefunction its self. Isn't the superposition of the advanced and >retarded potentials the equivalent to the Q.P.? > There is no such thing as unconscious intent. That is simply training >and any AI machine can do it. Much of our daily behavior is machine-like. > JS: Goldstein is attempting to use the presponse advanced micro-causal >influence to free Bohm's theory from its seemingly Galilean shackles. --Give the Gift of Trickier Dick Cheeny -- out of office, at last! > http://www.benfranklinbooks.com/ > http://www.wlym.com/pages/music.html > http://www.rand.org/publications/randreview/issues/rr.12.00/ > http://members.tripod.com/~american_almanac > http://www.wlym.com/PDF-68-76/CAM7606.pdf ==== message that keeps showing??? ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- ==== > On 7 Jan 2004 21:01:21 -0800, webmaster@elken.com (Chumpmeister) >How can I create a unique number from the two that will not be >produced by another pair? The classic method for this is through use of the chinese remainder > theorem. [example deleted] Who uses the classic method, and why? Any computer programmer given the task would most likely just interlace the digits of the two numbers (probably in binary) e.g. f(77777, 333) = 7070737373 I don't really know how to go about starting this proof. Does anyone have any ideas how this should go? Question: For any integer n>2, show that there are at least 2 elements in U(n) that satisfy x^2=1. I know it should be by induction but I don't know where to go after 3 and I'm not really sure how to set it up. U(n) is the set of all positive integers less than n and relatively prime to n. Is this proof correct? Question: Let a and b be elements of an abelian group and let n be any integer. Show that (ab)^n=a^nb^n. Proof: When n=1, (ab)^1=ab=a^1b^1 Assume (ab)^n=a^nb^n. Now (ab)^(n+1)=(ab)^n(ab)=a^nb^n(ab)=(a^n)a(b^n)b(since it is an abelian group). So (a^n*a)(b^n*b)=a^(n+1)b^(n+1). Thus, (ab)^n=a^nb^n. ==== > I don't really know how to go about starting this proof. Does anyone have > any > ideas how this should go? Question: For any integer n>2, show that there are at least 2 elements in > U(n) > that satisfy x^2=1. I know it should be by induction but I don't know where to go after 3 and I'm > not really sure how to set it up. U(n) is the set of all positive integers > less than n and relatively prime to n. It is easy to see that the residue classes, mod n, of 1 and n-1 are both in U(n), and both square to give the residue class of 1, mod n. I see no need of induction. ==== >It is easy to see that the residue classes, mod n, of 1 and n-1 are both >in U(n), and both square to give the residue class of 1, mod n. I see no need of induction. ==== > I don't really know how to go about starting this proof. Does anyone have any > ideas how this should go? Question: For any integer n>2, show that there are at least 2 elements > in U(n) that satisfy x^2=1. I know it should be by induction but I don't know where to go after 3 and I'm > not really sure how to set it up. U(n) is the set of all positive integers > less than n and relatively prime to n. > There are no integers x in U(n) with x^2 = 1, since 1 coprime to n. > Is this proof correct? Question: Let a and b be elements of an abelian group and let n be any > integer. Show that (ab)^n=a^nb^n. Proof: When n=1, (ab)^1=ab=a^1b^1 > Assume (ab)^n=a^nb^n. Now (ab)^(n+1)=(ab)^n(ab)=a^nb^n(ab)=(a^n)a(b^n)b(since it is an abelian > group). So (a^n*a)(b^n*b)=a^(n+1)b^(n+1). Thus, (ab)^n=a^nb^n. > I suppose so if one presumes you're using induction. If all the spaces weren't crammed out the equations they'd be easier to read and check for accuracy. So I'll just suppose they're correct. ==== > I am trying to write a piece of software quite like Granular Synthesis > Demo 1 in http://www.dcs.gla.ac.uk/~jhw/audioclouds/ . Hmm, that looks like a very interesting project! It looks difficult but well worth solving -- that's what I call a good problem. > I was wondering if anyone could explain the difference between a > probabilty distribution and a probabilty density? Well, a probability distribution is essentially a function that tells you how much stuff is within a given region. It turns out that a very convenient way to construct such functions is to work instead with the probability density -- that's a function that says here the stuff is piled up so deep, and there it's so deep, and stuff in a region -- voila, the probability distribution. > And if you have a probabilty distribution can a density be obtained > from it and if so how? Yes, you can get the density by differentiating the distribution. Whether it's easier to go from density to distribution or vv depends on the details of the problem. > Also the main bit i am stuck on is how a density can be conditioned > (in the example given it is conditioned by the x and y positions of a > mouse) how would you go about doing such a thing. By conditional on the mouse position they just mean that the density is a function of the mouse position -- conceptually there's a different density for every mouse position. Of course, for position that are close together, the densities will be very similar; the trick is to figure out just how the density should vary. Robert Dodier ==== Paul I just read Ohanian & Ruffini's 1.9 and it completely supports my position and not yours IMHO. PZ: My position on what? On our ability to measure tidal effects within an arbitrarily small region, down to a spacetime point? JS: The following statements are true: 1. To a good approximation the non-tensor connection field g-force on a where m(passive) = m(active) The approximation is two-fold A. Not near a space-time singularity, i.e. not falling behind the event horizon of a black hole. B. The scale is large enough so that quantum gravity metric fluctuations are ignorable. 2. To the same approximation the connection field g-force (which reduces to Newton's gravity force) is eliminated on a timelike geodesic. The connection field is non-zero only on time-like non-geodesics. Rotation 3. The curvature tensor is also a local observable. IF you define a true gravity field as one in which the tidal curvature tensor is NOT zero, then one can always locally, in principle, distinguish a true gravity field from a fake gravity field (where the tidal tensor vanishes in all frames geodesic LIF & non-geodesic LNIF) by definition. On the other hand, if you define a gravity field as the connection field, then you cannot make such a distinction! Therefore, the problem here is only a semantic problem from wavering between the two different definitions of gravity field. principle locally see tidal effects as a torque around the center of mass or a precession of a small spinning gyroscope of course. Also you can see deformations of shape of a geodesic droplet if the surface tension is small enough. 5. Since Einstein's GR comes from locally gauging ONLY the 4-parameter translation subgroup of the Poincare group, of course Einstein's early formulation of the equivalence principle was only an approximate rotational degrees of freedom. Einstein's guv field of curved space-time with the symmetric connection force field is simply the compensating gauge force field needed to restore the now local gauge symmetry which is equivalent to local general coordinate transformations. If, in addition, you locally gauge the 6-parameter Lorentz subgroup, then you get an additional torsion force field, i.e. an anti-symmetric piece of the connection field. This will obviously modify the predictions of GR for the tidal torques on extended test 6. Hagen Kleinet shows that tidal curvature, the basis for gravity waves, comes from stringy (vortex core if one puts in a vacuum coherence O(1) ODLRO parameter) disclination defects in a Planck lattice in 4D. There is no torsion field in 1915 GR, but if there was one in Nature, as Akimov & Shipov claim in Moscow, then it would correspond to dislocation lattice defects. The different 4D discrete world crystal symmetry groups of the tiled unit cell are different physical vacuum structures quite obviously. This is 4D (and maybe 11D) crystallography with additional supersymmetry matrix dimensions). 7. There are still two more subgroups left in the 15 parameter Conformal Group. You will get new compensating gauge force field physics if you locally gauge the 1-parameter dilation subgroup and also locally gauge the 4-parameter special conformal transformation of constant acceleration boosts for hyperbolic motion in Special Relativity e.g. MTW has a chapter on this. elegant fashion show of weaves of the fabric of reality from spin networks to pre-geometric spin foams to quantized Area operators and world holograms et-al I find Penrose saying: The algebra I have used to treat linear displacements and rotations together, or linear and angular momentum together, I call the algebra of twistors. I have used the term 'twistor' to denote a 'spinor' for the six-dimensional (++----) pseudo-orthogonal group O(2,4). 9. It is obvious to any School Boy that O(2,4) is simply Lorentz SR O(1,3) with string world sheet O(1,1) fiber. Therefore, string theory is inherent here. Penrose continues: The twistor group is the (++--) pseudo-unitary group SU(2,2) which is locally isomorphic with O(2,4). In turn O(2,4) is locally isomorphic with the fifteen-parameter (local) conformal group of space-time. Under a conformal transformation of space-time the twistors will transform linearly according to a representation of the group SU(2,2). p. 175 Combinatorial Space-Time Quantum Theory and Beyond (Cambridge, 1971). Suddenly the connections among 1. Quantum loop gravity of spin networks --> foams with quantized area etc. operators 2. Local gauge invariance 3. String theory 4. Twistors & Conformal Group are becoming clearer. ==== >what I think you are missing is that .999... (a zero, a dot and an >infinite amount of 9s) is equal to 1, so what you are asking is >whether 1^2, 1^3, 1^4, etc. is 'tending to' 1. i understand the premise. > Look at it this way: We can write a number like 0.9 as 1 - 0.1, and a >number like 0.99 as 1 - 0.01. In general we can write it down as: >f(n) = 1 - (1/(10^n)) for n>0 where n is the number of nines behind >the dot. >So the number 0.999... can be found by looking at: >lim{n->infinity}{f(n)} = >lim{n->infinity}{1 - (1/(10^n))} = >lim{n->infinity}{1} - lim{n->infinity}{1/(10^n)} but infinity means unlimited... limit{n->unlimited}{f(n)} = > limit{n->unlimited}{1 - (1/(10^n))} = > limit{n->unlimited}{1} - limit{n->unlimited}{1/(10^n)} ...which contradicts the premise. What premise does that contradict? If it is your premeise that 0.999... is not equal to 1, then good. > >Since the limit of the fraction-part goes to zero, the whole thing >becomes equal to 1. In other words 0.999... is equal to 1. only if the premise the unlimited has a limit is believed. If it does not, then unlimited decimals have no meaning in teh first place, and your 0.999... is NaN. But in every mathematician's world, every Cauchy sequence, of which (1 - 1/10^n) is an example, has a real number limit, however unlimited you seem to think it should be. in teh case of (1 - 1/10^n), that limit is 1, and 1 is the only possible numerical value that 0.999... can possibly mean, to anyone who pretends to any mathematical skills. If 0.999... represents a number the that number is 1. If 0.999... does not represent a number, then this whole discussion is moot. If you can't stand that heat, get out of the mathematical kitchen. garry denke, geologist > denoco inc. of texas ==== What is a double limit? lim(m,n->+oo) (1 - 1/10^m)^n Does the above have an assigned value? lim(m->+oo) (1-1/10^m)^n = 1 Why do you prefer the above as opposed to the below? lim(n->+oo) (1-1/10^m)^n = 0 Garry Denke, Geologist Denoco Inc. of Texas ==== > What is a double limit? lim(m,n->+oo) (1 - 1/10^m)^n Does the above have an assigned value? lim(m->+oo) (1-1/10^m)^n = 1 Why do you prefer the above as opposed to the below? lim(n->+oo) (1-1/10^m)^n = 0 > Garry Denke, Geologist > Denoco Inc. of Texas The double limit, lim_{(m,n)->(+oo,+oo)} (1 - 1/10^m)^n, does not exist. For it to exist it would be necessary at least for both the single limits above to have the same value, which they do not. The limit as m goes to +oo is relevant the meaning of 0.999..., whereas the limit as n goes to +oo is not. Just telling. Your welcome. ==== > What is a double limit? > lim(m,n->+oo) (1 - 1/10^m)^n > Does the above have an assigned value? No. If it had a value, say L, then L would have to satisfy the condition that fore every epsilon > 0, there exists M, N > 0 such that |(1-1/10^m)^n - L| < epsilon for every m > M and for every n > N. The fact that no such L exists follows from the fact that the iterated limits below have different values. > lim(m->+oo) (1-1/10^m)^n = 1 This holds for each n > 0, and therefore lim(n->+oo) lim(m->+oo) (1-1/10^m)^n = 1 (1) > Why do you prefer the above as opposed to the below? > lim(n->+oo) (1-1/10^m)^n = 0 This holds for each m > 0, and therefore lim(m->+oo) lim(n->+oo) (1-1/10^m)^n = 0 (2) Comparing (1) and (2), we see that the limits do not commute. It's not a matter of prefering one limit to the other. It's simply a matter of understanding what the notation 0.999... means. Since 0.999..., by definition, means lim (m->+oo) (1-1/10^m), it follows that lim (n->+oo) 0.999...^n = lim (n->+oo) lim (m->+oo) (1-1/10^m)^n, which is the limit that appears in (1). -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== How to integrate sinc(x) (without Fourier transformation) ?? > That integral is not an elementary function. It is known as the Sine integral function. ==== >How to integrate sinc(x) (without Fourier transformation) ?? That integral is not an elementary function. It is known as the Sine > integral function. > Nice link. Beautiful images. :-) 1. first of all, thank you for having confirmed me that it wasn't trivial 2. my purpose was to find the Fourier transformation of sinc, and not just to take it from a book. Is it possible by an easy process, or else... no way ?.. ==== >How to integrate sinc(x) (without Fourier transformation) ?? > > That integral is not an elementary function. It is known as the Sine >integral function. > Nice link. Beautiful images. :-) 1. first of all, thank you for having confirmed me that it wasn't trivial 2. my purpose was to find the Fourier transformation of sinc, and not just > to take it from a book. Is it possible by an easy process, or else... no way > ?.. > The infinite Fourier transform is fairly trivial. If you use the integral representation sinc(x) = {{sin(pi x)} over {pi x}} = (2pi )^{-1} int _{-pi} ^{pi) {dt e^{ixt} } it is easy to see that the Fourier transform (multiply by e^{-ikx} and integrate from -infty to infty ) is theta left({ pi ^2 - k^2}right) where the theta function is 1 for positive arguments, 0 for negative, and defined to be 1/2 for argument 0. -- Julian V. Noble Professor Emeritus of Physics jvn@lessspamformother.virginia.edu ^^^^^^^^^^^^^^^^^^ http://galileo.phys.virginia.edu/~jvn/ God is not willing to do everything and thereby take away our free will and that share of glory that rightfully belongs to us. -- N. Machiavelli, The Prince. 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 ==== In , on 01/08/2004 at 03:14 PM, Joona I Palaste said: >At least in the University of Helsinki, if there is a function >f:X->Y, and U is a subset of X, then fU means {f(x) | x in U}. Are you sure that you mean fU and not f_U (f subscript U) or f|U? >Could there be any functions f:X->Y and subsets U of X so that fU >and f(U) would both be defined but would not be equal? If you're using standard notation then the two are identical by definition. If you're using fU to mean f|U then the two are different by definition. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== Shmuel (Seymour J.) Metz scribbled the following: > In , on 01/08/2004 > at 03:14 PM, Joona I Palaste said: >>At least in the University of Helsinki, if there is a function >>f:X->Y, and U is a subset of X, then fU means {f(x) | x in U}. > Are you sure that you mean fU and not f_U (f subscript U) or > f|U? Yes, I am sure. f_U would mean a function f associated with the set U, not iterated over it. f|U would mean the function f constrained into the set U. >>Could there be any functions f:X->Y and subsets U of X so that fU >>and f(U) would both be defined but would not be equal? > If you're using standard notation then the two are identical by > definition. If you're using fU to mean f|U then the two are > different by definition. I can't see how using the notation I defined above, fU = {f(x) | x in U}, fU and f(U) could be identical by definition. -- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ A friend of mine is into Voodoo Acupuncture. You don't have to go into her office. You'll just be walking down the street and... ohh, that's much better! - Stephen Wright ==== > Shmuel (Seymour J.) Metz scribbled the following: >> In , on 01/08/2004 >> at 03:14 PM, Joona I Palaste said: >At least in the University of Helsinki, if there is a function >f:X->Y, and U is a subset of X, then fU means {f(x) | x in U}. >> Are you sure that you mean fU and not f_U (f subscript U) or >> f|U? > Yes, I am sure. f_U would mean a function f associated with the set U, > not iterated over it. f|U would mean the function f constrained into > the set U. >Could there be any functions f:X->Y and subsets U of X so that fU >and f(U) would both be defined but would not be equal? >> If you're using standard notation then the two are identical by >> definition. If you're using fU to mean f|U then the two are >> different by definition. > I can't see how using the notation I defined above, fU = {f(x) | x in > U}, fU and f(U) could be identical by definition. Because f(U) = { f(x) : x in U } is perfectly standard notation when U is a subset of the domain of f. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== Dave Seaman scribbled the following: >> Shmuel (Seymour J.) Metz scribbled the following: > In , on 01/08/2004 > at 03:14 PM, Joona I Palaste said: >>At least in the University of Helsinki, if there is a function >>f:X->Y, and U is a subset of X, then fU means {f(x) | x in U}. > Are you sure that you mean fU and not f_U (f subscript U) or > f|U? >> Yes, I am sure. f_U would mean a function f associated with the set U, >> not iterated over it. f|U would mean the function f constrained into >> the set U. >>Could there be any functions f:X->Y and subsets U of X so that fU >>and f(U) would both be defined but would not be equal? > If you're using standard notation then the two are identical by > definition. If you're using fU to mean f|U then the two are > different by definition. >> I can't see how using the notation I defined above, fU = {f(x) | x in >> U}, fU and f(U) could be identical by definition. > Because f(U) = { f(x) : x in U } is perfectly standard notation when U is a > subset of the domain of f. I see. However, in the question I asked, I use fU for that concept and f(U) very specifically for f applied to U itself, not to every element of U. I think David Ulrich got the right idea. -- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ Shh! The maestro is decomposing! - Gary Larson ==== Joona I Palaste scribbled the following: > I see. However, in the question I asked, I use fU for that concept and > f(U) very specifically for f applied to U itself, not to every element > of U. I think David Ulrich got the right idea. Typo, should be David Ullrich with two l's. -- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ He said: 'I'm not Elvis'. Who else but Elvis could have said that? - ALF 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 ==== In , on 01/08/2004 at 10:25 AM, Allan Adler said: >In Hebrew, the feminine plural of the foreign loan word, nilpotent, >when written without vowels or other punctation, is written >nylpw_tn_tywt >where _t is teth. That last t would have to stand for a Tav, not a Tet. I assume that the w stands for a Vav. Also, are you sure that what you are looking at is intended to be the plural of nilpotents (nilpotetioth) and not nilpotency (nilpotentiuth)? >What I would like to know is how it would be written >if one did write the vowels and other punctation. Well, transliterations of loan words, and transformations of transliterations, are always dicey. >(0) which of the consonants is dotted? I would Expect a Dagesh where you have p (otherwise it would be nilfotentioth >(1) is the y following the first n retained and the n endowed with a >chireq? ITYM hiriq. Yes to both. >(2) is there a schwa under the l? I would expect a schwa nach. >(3) is the w following the p replaced by a cholem or is it replaced >by a qamats qatan? I would expect a cholem in a transliteration of a worde with an o. >(5) is there a schwa under the second n? I would expect a schwa na'. >(6) is the y following the second teth retained and the teth endowed >with a chireq? >I would also like to know how I might figure out examples like this, >involving foreign loan words, on my own, instead of having to ask >people in every instance. Try to find niqud that matches the pronunciation of the original. Be glad that you recognized it as a transliteration; I once spent 5 minutes trying to figure out the shoresh of vatergate before realizing that there was none - it was a transliteration of Watergate. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== > In , on 01/08/2004 > at 10:25 AM, Allan Adler said: >In Hebrew, the feminine plural of the foreign loan word, nilpotent, >when written without vowels or other punctation, is written >nylpw_tn_tywt >where _t is teth. That last t would have to stand for a Tav, not a Tet. I had thought that words transliterated into Hebrew strictly used tet. YS --- ==== >In , on 01/08/2004 > at 10:25 AM, Allan Adler said: > >In Hebrew, the feminine plural of the foreign loan word, nilpotent, >>when written without vowels or other punctation, is written >>nylpw_tn_tywt >>where _t is teth. > > That last t would have to stand for a Tav, not a Tet. > I had thought that words transliterated into Hebrew strictly used tet. > Certainly not. For words of English origin, th becomes Tav and t becomes Teth. Case in point: Mathematics -> mathematika, mem-tav-mem-teth-yod-qoph-heh. Alan ==== Alan Greenwood schreef in bericht [ ... ] >I had thought that words transliterated into Hebrew strictly used tet. Certainly not. For words of English origin, th becomes Tav and t > becomes Teth. Case in point: Mathematics -> mathematika, > mem-tav-mem-teth-yod-qoph-heh. Alan But the old English of Euclid and Ptolemy is hard to read. ŗģįłäéļł .a6ōļģåķįé[IDouble Dot]ł ķįõčķįōé[EDoub leDot]č÷ ōåōņįāéāģ[IDo ubleDot]ł ółīōįźåö[Divide ] I would have expected õ becoming teth and ō becoming tav instead? ==== in >> In , on 01/08/2004 >> at 10:25 AM, Allan Adler said: >>In Hebrew, the feminine plural of the foreign loan word, nilpotent, >>when written without vowels or other punctation, is written >>nylpw_tn_tywt >>where _t is teth. >That last t would have to stand for a Tav, not a Tet. > > I had thought that words transliterated into Hebrew strictly used tet. Certainly not. For words of English origin, th becomes Tav and t > becomes Teth. Case in point: Mathematics -> mathematika, > mem-tav-mem-teth-yod-qoph-heh. Alan Are there any other examples of this occurance? YS --- windows-nt) Cancel-Lock: sha1:SHDIYfVy+L0N0jSittzBcKXLiKo= ==== >> That last t would have to stand for a Tav, not a Tet. I had thought that words transliterated into Hebrew strictly used > tet. They do--for the t sounds found in the borrowed word. The final Taf is part of the Hebrew feminine plural, so it is not changed to Tet. Len. ==== > That last t would have to stand for a Tav, not a Tet. > > I had thought that words transliterated into Hebrew strictly used >tet. They do--for the t sounds found in the borrowed word. The final Taf > is part of the Hebrew feminine plural, so it is not changed to Tet. Len. > YS <3ffd484c$2$fuzhry+tra$mr2ice@news.patriot.net> <20040108104625.582$w3@newsreader.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$oft ==== In <20040108104625.582$w3@newsreader.com>, on 01/08/2004 at 03:46 PM, David W. Cantrell said: > Or maybe dx post facto. e^x, dy, dx e^x, dx (Aristophanes) or perhaps d(hi/ho) -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org <3ffa32dd$8$fuzhry+tra$mr2ice@news.patriot.net> <8765fp8z1v.fsf@becket.becket.net> <3ffc9d5e$9$fuzhry+tra$mr2ice@news.patriot.net> <87lloj9qj9.fsf@becket.becket.net> <87hdz79ppk.fsf@becket.becket.net> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft ==== In <87hdz79ppk.fsf@becket.becket.net>, on 01/07/2004 at 05:47 PM, tb+usenet@becket.net (Thomas Bushnell, BSG) said: >Mathematics in antiquity and the middle ages (and even surprisingly >late) was thought in fairly concrete terms; consider Euclid's rule >that you can extend a line indefinitely. A line, for Euclid, is not >an infinite thing; it's a *finite* thing, but you can extend it as >much as you like--with it always remaining finite. And yet Euclid published a proof that there is no largest prime number. For that matter, the proof that 2^.5 is irrational would seem to be an acknowledgement of the infinite in another sense. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== > In <87hdz79ppk.fsf@becket.becket.net>, on 01/07/2004 > at 05:47 PM, tb+usenet@becket.net (Thomas Bushnell, BSG) said: >Mathematics in antiquity and the middle ages (and even surprisingly >late) was thought in fairly concrete terms; consider Euclid's rule >that you can extend a line indefinitely. A line, for Euclid, is not >an infinite thing; it's a *finite* thing, but you can extend it as >much as you like--with it always remaining finite. And yet Euclid published a proof that there is no largest prime > number. Oh, indeed! For us, that means there are an infinite number of primes. But for Euclid, it means that you can produce as many primes as you like. It's easy to miss the difference, because we have come to see such differences as (mathematically) pointless. > For that matter, the proof that 2^.5 is irrational would seem > to be an acknowledgement of the infinite in another sense. Um, no. The Greeks didn't have anything like decimal expansions. The proof was taken to establish that two particular lines are not commensurate, that is, there is no line segment which is an integral divisor of each. The usual Greek commentary on this discovery amounted to there being no common measure for all lines or something like that, not anything about infinity. Thomas <3ffa32dd$8$fuzhry+tra$mr2ice@news.patriot.net> <8765fp8z1v.fsf@becket.becket.net> <3ffc9d5e$9$fuzhry+tra$mr2ice@news.patriot.net> <87lloj9qj9.fsf@becket.becket.net> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft ==== In <87lloj9qj9.fsf@becket.becket.net>, on 01/07/2004 at 05:29 PM, tb+usenet@becket.net (Thomas Bushnell, BSG) said: >Of course they did; it means a thing without limit; in some >contexts, more exactly, without any limit in some regard. That's hand waving, not a definition. What did they mean by without limits? >Finite means whatever can be limited. Anything can be limited. >Well, it's my stock and trade, yes. I have a definition: the >infinite is whatever is not limited Again, that begs the question. >A very easy thing to say; a very hard thing to prove Count your wife's teeth. >But yes, he did. Then please explain what he meant by completed infinite and as at once, without handwaving or circular definitions. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== > In <87lloj9qj9.fsf@becket.becket.net>, on 01/07/2004 > at 05:29 PM, tb+usenet@becket.net (Thomas Bushnell, BSG) said: >Of course they did; it means a thing without limit; in some >contexts, more exactly, without any limit in some regard. That's hand waving, not a definition. What did they mean by without > limits? >Finite means whatever can be limited. Anything can be limited. I'm not sure how you find it to be vague to say without limits at the same time as you assert that anything can be limited. >A very easy thing to say; a very hard thing to prove Count your wife's teeth. So, Aristotle said something incorrect in one area, he must be crazy about math? (And everyone else too?) By this measure, I don't think Newton fares too well, not to mention Euler, Gauss... >But yes, he did. Then please explain what he meant by completed infinite and as at > once, without handwaving or circular definitions. I gave several examples: lines infinitely extended, for example, and as at once in terms of co-existence of physical things at the same moment in time. Thomas 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 ==== In <55135de4.0401072036.bb0fbff@posting.google.com>, on 01/07/2004 at 08:36 PM, diegoandresalvarez@lycos.co.uk (Diego Andres Alvarez Marin) said: >I think that you should teach her what are the mathematics useful >for few years I may have left, is an an understanding and love for the beauty of Mathematics. But I don't know how to impart that. If he wanted to impart an understanding and love for the beauty of art, would you tell him to teach how to use paintings to cover cracks in the wall? >Don't use the mathematician point of view, maybe the engineer point >of view... application. That's fine if he wants her to be an engineer who is blind to beauty. >Also, have you considered teach her to program?. That might be a good thing in it's own right, but has nothing to do with teaching an appreciation of Mathematics. Children are born with boundless curiosity and sense of play. The public school system carefully and systematically tries to destroy those. It is the job of parents and grandparents to protect their children as much as they can from the stultifying atmosphere of the schools. There will be time in later life for her to learn that Mathematics can be useful; now is the time for her to learn that Mathematics is fascinating and fun. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== >In <55135de4.0401072036.bb0fbff@posting.google.com>, on 01/07/2004 > at 08:36 PM, diegoandresalvarez@lycos.co.uk (Diego Andres Alvarez >Marin) said: >>I think that you should teach her what are the mathematics useful >>for >few years I may have left, is an an understanding and love for the >beauty of Mathematics. But I don't know how to impart that. If he >wanted to impart an understanding and love for the beauty of art, >would you tell him to teach how to use paintings to cover cracks in >the wall? Considering that mathematics was (and is) used to gain insight into practical problems (e.g. engineering), I don't see why it is wrong to teach applications. IMHO, applications reinforce the theory. >>Don't use the mathematician point of view, maybe the engineer point >>of view... application. >That's fine if he wants her to be an engineer who is blind to beauty. I strongly disagree. Being an engineer does not necessarily blind one to the beauty of mathematics. It may be the case, as I said above, that the engineer prides herself on the practical application of mathematical principles. Not everyone needs to be a theorist. Not everyone wants to be a theorist. >>Also, have you considered teach her to program?. >That might be a good thing in it's own right, but has nothing to do >with teaching an appreciation of Mathematics. Again, this is an opportunity for practical application of theory. What bothered me a bit about the OP's post was that it implied that there was something wrong with not studying higher math and studying molecular biology instead. There's nothing wrong with that, imho, and certainly any degree from Caltech is a considerable accomplishment. >Children are born with boundless curiosity and sense of play. The >public school system carefully and systematically tries to destroy >those. It is the job of parents and grandparents to protect their >children as much as they can from the stultifying atmosphere of the >schools. There will be time in later life for her to learn that >Mathematics can be useful; now is the time for her to learn that >Mathematics is fascinating and fun. Why not useful, fascinating, and fun? --gregbo gds at best dot 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 ==== In , on 01/09/2004 at 07:09 AM, Tim Smith said: >Is it just me, or does this happen to everyone? I find that when I read things that I believe I have completely forgotten, it all starts coming back to me. Also, I find that sometimes the complex things are easy but the simple things are difficult. YMMV. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org 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$oft X-Terminate: SPA(GIS) ==== In <200401071748.i07HmR628278@proapp.mathforum.org>, on 01/07/2004 at 05:50 PM, nico80@jazzfree.com (Niclas de la Foz) said: >Point 3: >As the one-to-one correspondence N <-> R is not possible (Point 1), >the criterion settled by Cantor (Point 2) is not valid to resolve >whether R is countable or not. No. If you stipulate that R is uncountable, then it immediately follows that R is uncountable. If you don't stipulate it, a reductio ad absurdum is perfectly reasonable, although not necessary. A reductio ad absurdum does not assume a false statement: it proves that the statement must be false by proving that it leads to a contradiction. >Point 4: >If a proof (any) comes to the conclusion that R is not countable >because it is not possible to accomplish the criterion established >in Point 2, then the proof is useless Incorrect. >because that criterion is not valid for that purpose (Point 3). No, it is your Point 3 that is not valid. >DEFINITION >Useless proof: A correct proof that proves nothing. What do you mean by proves nothing? >QUESTION >Does somebody know any other criteria (different from the one >established in Point 2) to resolve whether R is countable or not? Do you know a definition of frequency other than the number of cycles per unit of time? The definition is the definition, and any proof ultimately comes down to showing that it satisfies the definition. >NOTE: >Someone could interpret that if N <-> R is not possible, then R is >uncountable. This conclusion is true, but in this case, it will >imply that R is uncountable because we haven®t got a suitable tool >to count its elements, No. It will imply that no such tool exists. >since the properties of N and R are incompatible. You still haven't defined what you mean by that statement in Mathematical terms. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org <20040104102842.911$Mv@newsreader.com> <468305d6.0401051618.7985909e@posting.google.com> <3ffcc389$17$fuzhry+tra$mr2ice@news.patriot.net> <20040108120001.885$2g@newsreader.com> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft ==== In <20040108120001.885$2g@newsreader.com>, on 01/08/2004 at 05:00 PM, David W. Cantrell said: >(Or maybe you meant to say properly, rather than poperly. ;-) That depends on whether Gauss was speaking ex cathedra. Yes, I have a tendencey towards dropped, doubled and transposed letters; I meant properly, and I was also thinking of the issue of whether he allowed the one-point compactification. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft ==== In , on 01/08/2004 at 08:15 PM, Wolfram Research said: What was the upshot of your dispute with CRC a few years ago? -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== > >|> Infinitesimals are totally unintuitive and correspond to no physical >|> objects. Recall the history of infinitesimals in calculus/analysis. >|> They were widely dismissed and ridiculed (e.g. Bishop Berkeley). >| >|There will always be foot-draggers. I think Bishop Berkeley was using his criticism of infinitesimal > calculus as an indirect defense of theology. He who can digest > a second or third fluxion, a second or third difference, need not, > we think, be squeamish about any point of divinity. -- The Analyst It sounds like he was saying I know you are but what am I? -- http://hertzlinger.blogspot.com ==== >|> Infinitesimals are totally unintuitive and correspond to no physical >|> objects. Recall the history of infinitesimals in calculus/analysis. >|> They were widely dismissed and ridiculed (e.g. Bishop Berkeley). >| >|There will always be foot-draggers. >I think Bishop Berkeley was using his criticism of infinitesimal >calculus as an indirect defense of theology. He who can digest >a second or third fluxion, a second or third difference, need not, >we think, be squeamish about any point of divinity. -- The Analyst It sounds like he was saying I know you are but what am I? No, he was saying that you can't appeal to but I can't see it! in theology if you don't admit the same argument in physics. Thomas ==== > : o the answer appears to be > > : 1/[(999/1024) + 1] = .5061789... > > This doesn't have the right limiting behavior. It has. > If I toss the coin 1e6 > times and it still comes up heads every time, then I can be (almost) > certain that I have the two-headed coin. Right? Yes. Make n the number of tosses, the answer is: 2^n / (999 + 2^n) which goes to 1 as n goes to infinitity. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== > >Your start is already wrong. P(two headed) = 1/1000 by the definition >of the problem. Coming from the recesses of my memory I find Bayes' >theorem: > P(A|B) = P(A and B)/P(B) >Let's have A = two headed B = 10 heads in a row. >You want P(A|B) i.e. the probability of A (two headed) assuming that >B (10 heads in a row) did occur. Now P(A and B) = 1 > > How can P(A and B) be greater than P(A) = P(two headed) = 1/1000? mind. P(A and B) = P(A) - P(A and not B) = P(A) - 0 = 1/1000. Revise the remainder accordingly and I get: P(B) = P(A and B) + P(not A and B) = 1/1000 + 999/1024000. So P(A|B) = (1/1000)/(1/1000 + 999/1024000) = 1024/2023 ~ 50.61%. - The recesses of my mind are indeed pretty deep. The last time I used Bayes was about 30 years ago... -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== : One apparent way of avoiding the paradoxes of naive set theory is to : turn set-defining characteristic functions into partial functions from : sets to the three-valued logic {T,F,bot}. This three-valued logic : extends classical logic in the obvious way with : and(T,bot)=or(F,bot)=bot, and(F,x)=F, or(T,x)=T, not(bot)=bot, so that : every truth function has a fixed point. Very much Lukasiewicz' trivalent logic, then. That is good, and it allows easy extension to polyvalent logics. : Obviously the law of excluded middle does not hold: or(a,not(a)) only : implies that a is T or bot. This gives the logic a constructive : character. Constructivism has been one of the driving reasons behind the introduction of alternate logics. Removing the constraint of excluded middle generalises one into the realm of Heyting algebras, where constructive theories roam, while allowing the avoidance of the antinomies found in the Boolean. : In my formulation, I identify each set with its characteristic : function from sets to {T,F,bot}. Thus given s:set, the membership of : an element x can be tested with s(x). In the general case, this is a : partial function, returning bot for some values. I call such sets : partial sets, and sets whose characteristic function is in {T,F} : total sets. Every set of ZF and NF is a total set, with this theory : admitting a strictly larger class of sets than either. In some ways, this is the structure of a topoi, or at least a functor from a topoi category to something similar. Is the approach meant to be categorial (which, by the way, is a good thing in my eyes -- I'm just curious)? Then, your partial classification appears to mainly distinguish the topoi of sets from some of the many other topoi. : Russell's set R={s:set|not(s(s))} is then a partial set. All ZF sets : are elements of R, while some elements of NF are not elements of R, : while some new sets such as R itself are of undecidable membership. : : I've translated the ZF axioms to this set theory, rephrasing them in : terms of characteristic functions and new for-all and there-exists : logic operators performing logical conjunction and disjunctions across : all elements of a characteristic functions. Everything appears to be : sound and avoids known paradoxes. The categorial study of paradox is becoming a large field these days, and it appears you may be repeating some of the work already done (which can be soooo frustrating sometimes!). I don't mean to assume any level of study, but perhaps I might suggest that, if you haven't, you should check out some online I can suggest. : With the new axioms, it is easy to construct a bijection from the : universal set to its power set. Cantor's proof that |P(x)|>|x| for : all non-empty sets x proceeds by constructing C={a:x|not(P(x)(a)} and : using the law of excluded middle to derive a contradiction on its : membership in P(x). This goes away for lack of excluded middle, : leaving C a partial set which appears not to be constructively : contradictory. : : The one worrying aspect of this approach is that it identifies sets : with characteristic functions from sets to logic values: : Set=Set->{T,F,bot}. I have only been able to develop an intuition of : such sets in a purely constructive way, by writing down a finite list : of possibly self-referential equations defining sets, and convincing : myself that a unique solution exists. This is much in the style of : NF's axiom that every (possibly cyclic) graph corresponds to a set, : but I allow unlimited comprehension. : : Are there any known problems with this approach to set theory? Any : pointers to research on the topic? No known problems that I am aware of. In fact, it seems to me to be one of the more successful modern approaches for classifying paradox. However, if I could make one suggestion, it would be to not restrict yourself to your trivalent logic. Any Heyting algebra is possible, and expands your research into the much more fruitful world that all topoi present. In fact, because of natural distinctions that present themselves between propositions that take on the middle value, trivalent theories are often looked at only as summarisations of a more natural infinitely valent theory. Good resources for this can be found in intuitionist discussions, but it is more general. Also, may I ask why you posted to comp.lang.functional? This intrigues me because some of my own research has been around the evaluation of the lambda calculus and proof / evaluation theory in the context of non standard logics, but I do not see this approach explicitly stated in your message. Good luck with your researche! -- -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar ==== > I just read a section in a real analysis book that states that up to a set of > measure zero any L.measurable function on the line is equal to a function of > Baire class two. Really??? Where can one find a proof of this? Is it true for b.c. 1? I'm sure the answer is no, but I can't think of a counterexample. ==== Yeah, is that really true? Can anybody suggest any books, or (if its not too much to ask) an outline of a proof? That's really cool. Can't think of a counterexample for class one either... >I just read a section in a real analysis book that states that up to a set of >measure zero any L.measurable function on the line is equal to a function of >Baire class two. ==== > hi > thank you very much for your posting, but seem not to be able to find > in it the answer to my question - how they are the same 'entities' if > they are coded by different numbers? thank you > maxim They are not really the same entities. Signs of type >1 are variables of the same type number but as a sequence of one number. Just as 0 is not the same as {0}, the number X is not the same as the sequence containing exactly one number X. Signs of type >1 are then joined together to form elementary formulas. Godel defines the operation of joining sequences (such as signs) to form a new sequence. So you have to have sequences, not individual numbers, to join them to form a new sequence, an elementary formula, in definition (20) in his paper. Charlie Volkstorf Cambridge, MA >> hi, >> i have very technical and boring question, but may be somebody out >> there read the english translation of the Godel's original paper and >> could help me. >Boring? Quite the contrary. It is refreshing to see someone discuss >something of significance. (Too much time is spent quibbling over >exercises in elementary arithmetic!) >> My question is, in what sense 'sign of type n' and 'variable of type >> n' are the same ('dasselbe')? >Let represent the sequence of numbers a, b, c, . . . >Godel's system P (PM plus Peano's Axioms) contains: >variables of type 1: X1, Y1, Z1, . . . >variables of type 2: X2, Y2, Z2, . . . >variables of type 3: X3, Y3, Z3, . . . >etc. >signs of type 1: 0, S0, SS0, . . . , X1, SX1, SSX1, . . . , Y1, SY1, >SSY1, . . . , Z1, SZ1, SSZ1, . . . >signs of type 2: , , , . . >signs of type 3: , < , . . . >etc. >As you point out, the signs of type >1 are just the variables of type >>1 seen as sequences of length 1. The reason Godel defines signs of >type >1 is so that he can juxtapose these sequences to form elementary >formulas in (20) as: sign of type n+1(sign of type n). That is: >elementary formulas: X2(0), X2(S0), X2(SS0), . . . , X2(X1), X2(SX1), >X2(SSX1), . . . , Y2(0), Y2(S0), . . ., X3(X2), X3(Y2), . . . >Elementary formulas are relation variables with arguments. Subsequent >definitions include logical operators negation, disjunction and >quantification in order to build Godel numbers for all of (higher >order) Predicate Calculus theorem proving and show that the >corresponding relationships are recursive. >BTW: In (19) Godel includes v=R(x)=2^x so that v=also occurs in 11.) Why does he include this redundant clause? IMHO: >This shows that these functions and relations are indeed recursive >(see theorem IV.) >I use the same technique in my Program Synthesis system > (http://www.mathpreprints.com/math/Preprint/CharlieVolkstorf/20021008.1/1 >section IV Proof 1 Step 2 clause ~LT(J,A) in my synthesis of programs >that determine if one number is a factor of another.) Godel is just >using Logic as a programming language, just as Peano did before him to >write a program to demonstrate that the universal set of the natural >numbers is recursively enumerable, which is axiom 1 in Section VI of >http://www.arxiv.org/html/cs.lo/0003071 . >In fact, many of Godel's axioms and definitions of recursive functions >and relations are formal rules and theorems in this paper. This is >because a computer program is simply a constructive proof that a >particular set is recursive or recursively enumerable. Mathematicians >such as Peano, Godel and Turing for generations have been showing that >various sets and functions are, or are not, recursive or recursively >enumerable by constructing programs in various bases of computing. >Today we call that process Computer Programming! >Charlie Volkstorf >Cambridge, MA >> thank you >>maxim ==== so, when he says that they are the same (dasselbe), he is wrong? >hi >thank you very much for your posting, but seem not to be able to find >in it the answer to my question - how they are the same 'entities' if >they are coded by different numbers? >thank you >maxim They are not really the same entities. Signs of type >1 are variables > of the same type number but as a sequence of one number. Just as 0 is > not the same as {0}, the number X is not the same as the sequence > containing exactly one number X. Signs of type >1 are then joined > together to form elementary formulas. Godel defines the operation of > joining sequences (such as signs) to form a new sequence. So you have > to have sequences, not individual numbers, to join them to form a new > sequence, an elementary formula, in definition (20) in his paper. Charlie Volkstorf > Cambridge, MA >> hi, >>i have very technical and boring question, but may be somebody out >>there read the english translation of the Godel's original paper and >>could help me. >>Boring? Quite the contrary. It is refreshing to see someone discuss >> something of significance. (Too much time is spent quibbling over >> exercises in elementary arithmetic!) > My question is, in what sense 'sign of type n' and 'variable of type >>n' are the same ('dasselbe')? >>Let represent the sequence of numbers a, b, c, . . . >> Godel's system P (PM plus Peano's Axioms) contains: >>variables of type 1: X1, Y1, Z1, . . . >> variables of type 2: X2, Y2, Z2, . . . >> variables of type 3: X3, Y3, Z3, . . . >> etc. >> signs of type 1: 0, S0, SS0, . . . , X1, SX1, SSX1, . . . , Y1, SY1, >> SSY1, . . . , Z1, SZ1, SSZ1, . . . >> signs of type 2: , , , . . >> signs of type 3: , < , . . . >> etc. >>As you point out, the signs of type >1 are just the variables of type >>1 seen as sequences of length 1. The reason Godel defines signs of >> type >1 is so that he can juxtapose these sequences to form elementary >> formulas in (20) as: sign of type n+1(sign of type n). That is: >>elementary formulas: X2(0), X2(S0), X2(SS0), . . . , X2(X1), X2(SX1), >> X2(SSX1), . . . , Y2(0), Y2(S0), . . ., X3(X2), X3(Y2), . . . >>Elementary formulas are relation variables with arguments. Subsequent >> definitions include logical operators negation, disjunction and >> quantification in order to build Godel numbers for all of (higher >> order) Predicate Calculus theorem proving and show that the >> corresponding relationships are recursive. >>BTW: In (19) Godel includes v=> R(x)=2^x so that v=> also occurs in 11.) Why does he include this redundant clause? IMHO: >> This shows that these functions and relations are indeed recursive >> (see theorem IV.) >>I use the same technique in my Program Synthesis system >> (http://www.mathpreprints.com/math/Preprint/CharlieVolkstorf/20021008.1/1 >> section IV Proof 1 Step 2 clause ~LT(J,A) in my synthesis of programs >> that determine if one number is a factor of another.) Godel is just >> using Logic as a programming language, just as Peano did before him to >> write a program to demonstrate that the universal set of the natural >> numbers is recursively enumerable, which is axiom 1 in Section VI of >> http://www.arxiv.org/html/cs.lo/0003071 . >>In fact, many of Godel's axioms and definitions of recursive functions >> and relations are formal rules and theorems in this paper. This is >> because a computer program is simply a constructive proof that a >> particular set is recursive or recursively enumerable. Mathematicians >> such as Peano, Godel and Turing for generations have been showing that >> various sets and functions are, or are not, recursive or recursively >> enumerable by constructing programs in various bases of computing. >> Today we call that process Computer Programming! >>Charlie Volkstorf >> Cambridge, MA > thank you >maxim ==== ... >> If c and 7 are coprime in A then there are elements p and q in A >> such that cp + 7q = 1. > >Be careful here. In James' all-inclusive ring of two complex conjugates, >one can be a unit while the other is not a unit. A bit strange, and I >do not know whether his ring really does exists, but the proof >fails for that. Because if c is a unit, q = 0 and p = inv(c). But >p' = inv(c)' does not necessarily belong to the ring (conjugation is >not a ring operation). > >> So 1 = 1' = (cp + 7q)' = (cp)' + (7q)' = c'p' + 7'q' = c'p' + 7q', >> and since p' and q' are in A, c' and 7 are coprime in A. > >Indeed, a valid proof is conjugation is a valid operation in the ring. > > Actually, I thought I was careful to restrict my argument to the > algebraic integers, the ring in which James alleged coprimality. I do not know whether James alleges coprimality in that ring. He is pretty unclear about that. > I agree that such an argument might not go through in the Harris > ring, but might it not fail for a more basic reason? It is not > obvious that his ring would be a B'ezout domain. That is something I am also wondering about. But I think James' ring does not even need to be really be Bezout, but what he utters about his ring is not even sufficient to clear this. If I remember right, it has been shown that you can get rings with only one of two complex conjugates adjoined to the integers. > James has on several occasions lowered his shield of vagueness > sufficiently to allow identification of a specific element of his > ring that is not an algebraic integer. I have done a few times too, and each time I could not get through the conjugation. In James' case you need many complex conjugates of which only one is included in the ring. Does that lead to a contradiction? I have no idea. It may be that it is possible to make a choice in each case such that you get a complete ring. On the other hand, that might be impossible. > Each time, people have > jumped up and down pointing out the unintended consequences if that > number's (algebraic) conjugates were also included. Yup. > The usually > implicit appeal to the symmetry rendering conjugates algebraically > indistinguishable probably went way over James' head. Certainly, > he has always run away from the questions raised, but his silence > might be explained as withdrawal of his examples after realising > their inapplicability, perhaps due to some ineffable subtlety > visible only to himself. I think this is *not* the case. Note his current insistance where (I presume in his ring) one of two complex conjugate numbers is divisible by 7 and the other is coprime to 7, where you can not tell which is which due to the duality of the sqrt operator. > I am therefore reluctant to deduce any > properties of his ring from those examples. The problem with James' proposition is that he needs a ring where only two of three factors are divisible by 7 and the third is co-prime to it. It is my opinion that *if* such a ring does exist (i.e. where complex conjugates are not necessarily both element of the ring), that might be a proof of FLT for n=3 (I should read through James' proof to see whether that is true, but I am reluctant to do so until the existance of such a ring is established). The sad thing is that the inverse is not true (I think). So proving the existence of such a ring is harder than proving FLT for n = 3. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== >... >>If c and 7 are coprime in A then there are elements p and q in A >>such that cp + 7q = 1. >>Be careful here. In James' all-inclusive ring of two complex conjugates, >>one can be a unit while the other is not a unit. A bit strange, and I >>do not know whether his ring really does exists, but the proof >>fails for that. Because if c is a unit, q = 0 and p = inv(c). But >>p' = inv(c)' does not necessarily belong to the ring (conjugation is >>not a ring operation). >> So 1 = 1' = (cp + 7q)' = (cp)' + (7q)' = c'p' + 7'q' = c'p' + 7q', >>and since p' and q' are in A, c' and 7 are coprime in A. >>Indeed, a valid proof is conjugation is a valid operation in the ring. >Actually, I thought I was careful to restrict my argument to the >algebraic integers, the ring in which James alleged coprimality. I do not know whether James alleges coprimality in that ring. He is >pretty unclear about that. > For once, he was reasonably explicit, in the thread title and in the antepenultimate paragraph of the post starting this thread: >>Of course the problem with the ring of algebraic integers is the >>implication from previous interpretations (here I rely on what I've >>heard primarily from sci.math posters like Arturo Magidin, Nora >>Baron, and Dik winter) is that *both* roots have non-unit factors in >>common with 7 in the ring of algebraic integers. >I agree that such an argument might not go through in the Harris >ring, but might it not fail for a more basic reason? It is not >obvious that his ring would be a B'ezout domain. That is something I am also wondering about. But I think James' ring >does not even need to be really be Bezout, but what he utters about >his ring is not even sufficient to clear this. If I remember right, >it has been shown that you can get rings with only one of two complex >conjugates adjoined to the integers. > James has on several occasions lowered his shield of vagueness >sufficiently to allow identification of a specific element of his >ring that is not an algebraic integer. I have done a few times too, and each time I could not get through >the conjugation. In James' case you need many complex conjugates >of which only one is included in the ring. Does that lead to a >contradiction? I have no idea. It may be that it is possible to >make a choice in each case such that you get a complete ring. On >the other hand, that might be impossible. > Each time, people have >jumped up and down pointing out the unintended consequences if that >number's (algebraic) conjugates were also included. Yup. > The usually >implicit appeal to the symmetry rendering conjugates algebraically >indistinguishable probably went way over James' head. Certainly, >he has always run away from the questions raised, but his silence >might be explained as withdrawal of his examples after realising >their inapplicability, perhaps due to some ineffable subtlety >visible only to himself. I think this is *not* the case. Note his current insistance where >(I presume in his ring) one of two complex conjugate numbers is >divisible by 7 and the other is coprime to 7, where you can not tell >which is which due to the duality of the sqrt operator. I think I see what you mean. Perhaps James accepts that neither of the zeros, a and a', of x^2 - x + 42 is coprime to 7 in the ring A of algebraic integers, but is asserting that in his ring, H, a is divisible by 7 and a' is coprime to 7. In A, choose v = gcd {6, a} and w = gcd {7, a} so that a = vw. Then a' = v'w' and ww'=7u for some unit u in A. If A is a subring of H then a' = v'w' in H, and if a' is coprime to 7 in H then w' must be a unit in H and w/7 = u/w' is in H. It is then clear that 7 is not a unit in H, since 1 = a + a' = vw + v'w' = 7v(u/w') + v'w' and v'w' is coprime to 7 in H. At first sight, James would have no problems in working in one of the rings formed by adjoining a or a', but not both, to A. However, if he also adjoins quotients arising from equations other than the only one he is really interested in, all he achieves is to increase the potential for undesirable interactions. But it is typical of James that, instead of adjoining just the elements he needs, he throws in everything but the kitchen sink and no one can say what are the properties of the ensuing monstrosity. > I am therefore reluctant to deduce any >properties of his ring from those examples. The problem with James' proposition is that he needs a ring where >only two of three factors are divisible by 7 and the third is >co-prime to it. It is my opinion that *if* such a ring does >exist (i.e. where complex conjugates are not necessarily both >element of the ring), that might be a proof of FLT for n=3 (I should >read through James' proof to see whether that is true, but I am >reluctant to do so until the existance of such a ring is established). >The sad thing is that the inverse is not true (I think). So proving >the existence of such a ring is harder than proving FLT for n = 3. I have not had time to look at any cubic examples. John Roberts-Jones ==== >Message-id: The lopsided Collatz tree? > >Maybe this is known so just disregard! >I couldn't find anything about it on web searches about the >Collatz. > >Probably of little value for the overall conjecture but still a >curiosity! > >The Collatz tree level count is odd up to level 5 where level >1,2,3,4,5 [1,2,4,8,16] has a count of (1) for each level. >At level 6 the count becomes even and fall one for one on each >side of the tree up too and including level 16. >This changes @ level 17 where the left side of the tree adds one >more path. >This happens also for certain levels higher than 17 where the >left side adds 1 more path over the right side. > >I have denoted the 2 sides of the tree by their highest-ranking >sequence. > >Where [1,2,4,8,16] is the trunk and the first main branch >assigned to the RIGHT side of the tree is -- >32,64,128,256,512,1024,2048,4096...and with all applicable nodes >branching from it creating all branches on the RIGHT side of >the Collatz tree. > >Where [1,2,4,8,16] is the trunk and the first main branch >assigned to the LEFT side of the tree is -- >5,10,20,40,80,160,320,640,1280,2560,...and with all applicable >nodes branching from it creating all branches on the LEFT side of >the Collatz tree. > > 8) 20 3 21 128 > | | / > 7) 10 64 > / > 6) 5 32 > / > 5) 16 > 4) 8 > 3) 4 > 2) 2 > Levels 1) 1 > >Why does the left side have this property of adding (1) more >branch at certain levels =>17 then the right side? > >Will this mean, at a very high level there will be many more >branches on the left side of the tree than on the right side >of the tree making the tree lopsided? > >Putting the above question in another perspective, at some higher >level will the right side branching number eventually catch up and >equal the left side branching number? > >It could be that the left side of the tree is facing the sun giving >that side of the tree a more vigorous growth. ;-) > >Dan Personally, I don't like the way you draw the Collatz tree. My preference > is for vertical placement to always represent x*2 and for the diagonal > placement to always represent (x-1)/3: 3 20 128 21 > | | / > 10 64 > | | > 5 32 > | > 16 > | > 8 > | > 4 > | > 2 > | > 1 By your method, 32 is part of the right branch whereas I consider the right > branch > single left branch and right branch. Rather, there is the central trunk from > which > sprout an infinite number of branches on either side, so I don't see any > problem. Note that in my view each branch starts with an odd number and extends > vertically > to infinity, all of which are even. Does this mean there are infinitely more > even > numbers than odd? Infinity always catches up in the end. Mansenator, I know the way I did the tree was different but I used each sequence as a one to one correspondence (left,right) with each other. Where the (2) sequences, [3,6,12,24...]is the last sequence left of center and [21,42,84,168...] is the last sequence right of center. This contains the inner limits and actually separates the tree into two halfs because these two sequences are 0 (mod 3)'s and will continue to double and continue on vertically ----->oo. The other two sequences, [5,10,20,40,80,160...]is the last sequence or outer branch limit of the left side of tree and [32,64,128,256,...] is the last sequence or outer branch limit of the right side of tree where both sequences double ----->oo. Presenting the tree in this why, it just seemed neat, up to level 16 that you have this even looking tree and with boundry limits even after level 16. Using this representation of the tree it is obvious the level seed counts are the same as the standard tree layout. Right now it appears after level 16 there are increasingly more sequences with a (5) in its path other than a (32) in its path. True or false? This can be checked with the standard tree also proving or disproving my claim. Dan ==== >Message-id: >Message-id: >The lopsided Collatz tree? >Maybe this is known so just disregard! >>I couldn't find anything about it on web searches about the >>Collatz. >Probably of little value for the overall conjecture but still a >>curiosity! >> >>The Collatz tree level count is odd up to level 5 where level >>1,2,3,4,5 [1,2,4,8,16] has a count of (1) for each level. >>At level 6 the count becomes even and fall one for one on each >>side of the tree up too and including level 16. >>This changes @ level 17 where the left side of the tree adds one >>more path. >>This happens also for certain levels higher than 17 where the >>left side adds 1 more path over the right side. >I have denoted the 2 sides of the tree by their highest-ranking >>sequence. >Where [1,2,4,8,16] is the trunk and the first main branch >>assigned to the RIGHT side of the tree is -- >>32,64,128,256,512,1024,2048,4096...and with all applicable nodes >>branching from it creating all branches on the RIGHT side of >>the Collatz tree. >Where [1,2,4,8,16] is the trunk and the first main branch >>assigned to the LEFT side of the tree is -- >>5,10,20,40,80,160,320,640,1280,2560,...and with all applicable >>nodes branching from it creating all branches on the LEFT side of >>the Collatz tree. >> >> 8) 20 3 21 128 >> | | / >> 7) 10 64 >> / >> 6) 5 32 >> / >> 5) 16 >> 4) 8 >> 3) 4 >> 2) 2 >> Levels 1) 1 >Why does the left side have this property of adding (1) more >>branch at certain levels =>17 then the right side? >Will this mean, at a very high level there will be many more >>branches on the left side of the tree than on the right side >>of the tree making the tree lopsided? >Putting the above question in another perspective, at some higher >>level will the right side branching number eventually catch up and >>equal the left side branching number? >It could be that the left side of the tree is facing the sun giving >>that side of the tree a more vigorous growth. ;-) >Dan >> >> Personally, I don't like the way you draw the Collatz tree. My preference >> is for vertical placement to always represent x*2 and for the diagonal >> placement to always represent (x-1)/3: >> >> 3 20 128 21 >> | | / >> 10 64 >> | | >> 5 32 >> | >> 16 >> | >> 8 >> | >> 4 >> | >> 2 >> | >> 1 >> >> By your method, 32 is part of the right branch whereas I consider the right >> branch >no >> single left branch and right branch. Rather, there is the central trunk >from >> which >> sprout an infinite number of branches on either side, so I don't see any >> problem. >> >> Note that in my view each branch starts with an odd number and extends >> vertically >> to infinity, all of which are even. Does this mean there are infinitely >more >> even >> numbers than odd? >> >> Infinity always catches up in the end. Mansenator, I know the way I did the tree was different but I used each sequence >as a one to one correspondence (left,right) with each other. >Where the (2) sequences, [3,6,12,24...]is the last sequence left of >center and [21,42,84,168...] is the last sequence right of center. >This contains the inner limits and actually separates the tree into >two halfs because these two sequences are 0 (mod 3)'s and will >continue to double and continue on vertically ----->oo. >The other two sequences, [5,10,20,40,80,160...]is the last sequence >or outer branch limit of the left side of tree and [32,64,128,256,...] >is the last sequence or outer branch limit of the right side of tree >where both sequences double ----->oo. >Presenting the tree in this why, it just seemed neat, up to level >16 that you have this even looking tree and with boundry limits >even after level 16. >Using this representation of the tree it is obvious the level seed >counts are the same as the standard tree layout. Ok, my diagram is biased towards the way I use the tree, so I shouldn't complain if you're trying to prove a point. Right now it appears after level 16 there are increasingly more >sequences with a (5) in its path other than a (32) in its path. True or false? Well, I would say it's probably true. On my diagram, 5 is lower on the trunk than the next live branch (85, 21 being 0 mod 3 has no sub-branches). For any given level, we would then expect there to be more sub-branches leading to 5 since it got a head start in sprouting sub-branches. > >This can be checked with the standard tree also proving or disproving >my claim. > I don't know if you saw that message I posted about sci.math appreciation. If not, I want to thank you because that thread about twin primes in Collatz sequences led me to discover the solution to a problem that's had me stumped for 4 years. Sometimes the thinking out loud that takes place in these newsgroups is beneficial to those who are replying to other's queries. Dan -- Mensanator Ace of Clubs ==== I am obssesively addicted to working on hard problems, especially like the 3x+1(see http://www.cecm.sfu.ca/organics/papers/lagarias/.) Every time, I look at any mathematical text, or do anything else I get drawn back to hard mathematical problems. I could easily spend hours, days, weeks, or even years working on these problems. I am currently a Sophomore in College, but I did very poorly last sumester, and I don't know what to do. Peole have always though I should straight A's, but my grades have always muich worse. Many times this obsession is so strong I have trouble even reading or studying mathematics as soon a some new idea possibly applicable to some hard problem comes up. If I make a bit of an effort to stay away from math problems for a little while, I easily get addicted to internet poker, and get a bit of a gambling problem. Another issue I face is that I have trouble writing, especially if it relates novels or other fiction. I can do outlines, research, etc. but when it comes down to actually writing I get stuck, and can sit for hours and never be able to get started. I can write internet posts, but I can't write full papers. I talked to a psychiatrist, and I think I may have A.D.D. I guess I could get some drugs. Does anyone have any ideas of things I could do? I don't really this obssesion to go away, but I also don't really want to flunk out of college. Has anyone else experienced similar problems? Anyone have any advice, or know something I should do? ==== >I am obssesively addicted to working on hard problems, especially like >the 3x+1(see http://www.cecm.sfu.ca/organics/papers/lagarias/.) Every >time, I look at any mathematical text, or do anything else I get drawn >back to hard mathematical problems. I could easily spend hours, days, >weeks, or even years working on these problems. I am currently a Sophomore in College, but I did very poorly last >sumester, and I don't know what to do. Peole have always though I >should straight A's, but my grades have always muich worse. Many times >this obsession is so strong I have trouble even reading or studying >mathematics as soon a some new idea possibly applicable to some hard >problem comes up. If I make a bit of an effort to stay away from math problems for a >little while, I easily get addicted to internet poker, and get a bit >of a gambling problem. Another issue I face is that I have trouble writing, especially if it >relates novels or other fiction. I can do outlines, research, etc. but >when it comes down to actually writing I get stuck, and can sit for >hours and never be able to get started. I can write internet posts, >but I can't write full papers. I am a 1967 Putnam fellow who did poorly in most English classes. I aced the grammar rules in 8th grade, but detested fiction. A book review which criticized the characters for double negatives and other grammatical errors was not what the teachers wanted. Go to the college library and find journals such as American Mathematical Monthly, College Mathematics Journal, Mathematics Magazine, Fibonacci Quarterly with problem sections. Do as many problems as you can, but be sure to write up your solutions. Install (free) Tex (probably latex2e) on your system so you can format mathematical text. Review your solution a few days later to see whether it is clear and accurate, and to look for simplifications. When you are happy, submit it. In another year, look for some of your solutions to be published, Don't forget to try the Putnam in December, even if those solutions are written up by hand. Peter Montgomery -- Intelligence agents foresee foreign flights crashing into landmarks. Those flights are from earth to mars. pmontgom@cwi.nl Home: San Rafael, California Microsoft Research and CWI ==== > I am obssesively addicted to working on hard problems, especially like > the 3x+1(see . > I am currently a Sophomore in College, but I did very poorly last > this obsession is so strong I have trouble even reading or studying > mathematics as soon a some new idea possibly applicable to some hard > problem comes up. > If I make a bit of an effort to stay away from math problems for a > little while, I easily get addicted to internet poker, and get a bit > of a gambling problem. > Another issue I face is that I have trouble writing, especially if it > relates novels or other fiction. I can do outlines, research, etc. but > when it comes down to actually writing I get stuck, and can sit for > hours and never be able to get started. I can write internet posts, > but I can't write full papers. > or know something I should do? Just take courses that suit you and don't worry about a degree ? Or get into an engineering school where things like psychology-based analysis of literature is not so important ? Or just take one course per quarter ? Gambling ? Try trading options or futures. This high risk investing is actually less risk than gambling and thus can be reliable endeavor. ==== >I am obssesively addicted to working on hard problems, especially like >the 3x+1(see . >I am currently a Sophomore in College, but I did very poorly last >this obsession is so strong I have trouble even reading or studying >mathematics as soon a some new idea possibly applicable to some hard >problem comes up. >If I make a bit of an effort to stay away from math problems for a >little while, I easily get addicted to internet poker, and get a bit >of a gambling problem. >Another issue I face is that I have trouble writing, especially if it >relates novels or other fiction. I can do outlines, research, etc. but >when it comes down to actually writing I get stuck, and can sit for >hours and never be able to get started. I can write internet posts, >but I can't write full papers. >or know something I should do? Just take courses that suit you and don't worry about a degree ? Or get into an engineering school where things like psychology-based > analysis of literature is not so important ? Or just take one course per quarter ? Gambling ? Try trading options or futures. This high risk investing is > actually less risk than gambling and thus can be reliable endeavor. Math, I can't tell if your ADD, though your grades history indicates the posability. When you speak of uncontrolled urges that is usually called Obsessive Compulsive. I beleive I got my ADD from my mother. She had both gambling and alchohol problems as well. It is not unusual to have more than one condition. In any case I think seeking a professional opinion is an excellent idea for you. One man's opinion. Next! Ray ==== > I can write internet posts, but I can't write full papers. >.. know something I should do? Better to resolve emotional conflicts by oneself, taking out time for it. Else BM or Behaviour Modification methods can be tried on by trained psychologists. ==== Dear all, I have a question about integer optimization software/solvers: it turns out to me in NEOS there are two solvers for integer optimzation: I'd like to try but just don't know there capacity and where are they running... I previously had some experiences on running some large scale optimization problem on Sun, several weeks passed until I know it was just too slow... So I want to ask what are normally the capacity of the NEOS integer optimization solvers? Are they running fast, for hundreds variables integer optimization problems? I have also seen a number of solvers/softwares on Internet... it was just a question that which ones are better before I try one by one... I have only one computer to use, so if I spend several days on one solver, then turn to another it is bad, it will need me several months to at least get a little idea... Also there are a number of modeling languages, for example AMPL, MPS, etc... which one is the right one that one must learn in order to do my hundreds variable integer optimization problem? We have CPLEX 8.1 in our school, what is the relationship between this CPLEX 8.1 and other softwares on the Internet? Oooops, after several days diving on Internet, these stuff really confused me... Please help me to choose the right one to solve my problem! thanks a lot, -Walala ==== What is the smallest number of dimensions that can support the largest number (e.g >500,000)of objects meeting the following two conditions? Condition One: circle radius = x Condition Two: circles boundaries touch (and do not overlap) ==== What is the smallest number of dimensions that can support the largest > number (e.g >500,000)of objects meeting the following two conditions? Condition One: circle radius = x > Condition Two: circles boundaries touch (and do not overlap) If I understand your question correctly you are looking for Y. Edel, E. M. Rains and N. J. A. Sloane, On Kissing Numbers in Dimensions 32 to 128, Elect. J. Combinatorics 5(1), paper R22 (1998). http://www.combinatorics.org/Volume_5/PDF/v5i1r22.pdf For n=36 a kissing number of 438872 is obtained, and for n=40 991792. Hugo Pfoertner ==== Suppose A is a real square matrix whose entries are all in (0,1) and such that the sum of each of its columns is 1. It can be seen as a matrix whose entries are probabilities. Then, it's easy to see that A^2 satisfies these same properties and, therefore, so does A^n for every natural n. Using a spreadsheet I concluded that the sequence (A^n) converges to a matrix whose colums are identical, have sum 1 and correspond to an eigenvector of A associated with the eigenvalue 1. However, I couldn't give a formal proof for this fact. I tried using fixed points theorems (maybe Brower's could do), but got confused. I also tried using the minimal polynomial of A. Could anyone give an idea? In the convergence processe, I defined the norm of A as ||A|| = Sqrt(Sum(i=1 to n Sum (j=1 to n) (a_i_j)^2) (just and extension of the norm of a Euclidean vector - is this the usual definition of the norm of a matrix?) and, based on Cauchy criterion, tried to make ||A^m - A^n|| < eps for m and n sufficiently large. I noticed that if the entries a_i_j are allowed to be in [0,1] then the above conclusion may fail, and I get a permutation of A if one of the a_i_j's is 1 (which implies the others in the same column are 0) Artur ==== >Suppose A is a real square matrix whose entries are all in (0,1) and >such that the sum of each of its columns is 1. It can be seen as a >matrix whose entries are probabilities. Then, it's easy to see that >A^2 satisfies these same properties and, therefore, so does A^n for >every natural n. Using a spreadsheet I concluded that the sequence >(A^n) converges to a matrix whose colums are identical, have sum 1 and >correspond to an eigenvector of A associated with the eigenvalue 1. >However, I couldn't give a formal proof for this fact. I tried using >fixed points theorems (maybe Brower's could do), but got confused. I >also tried using the minimal polynomial of A. Could anyone give an >idea? The transpose of your matrix is an irreducible aperiodic stochastic matrix. Look up Markov chain or Perron-Frobenius Theorem. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 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 ==== In <20040109034857.12300.00002581@mb-m20.aol.com>, on 01/09/2004 at 08:48 AM, kramsay@aol.com (KRamsay) said: >I thought you said you thought >the set theory you had in mind was from the 1950s. What I saw was a text book, not one of the original papers. Now I'm going to have to see whether I can track down a copy. I should have bought it when I had the chance :-( I'll get back to you if I can find a copy and check whether I got any details wrong. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org <3ffc9a68$7$fuzhry+tra$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft ==== In , on 01/08/2004 at 12:01 AM, Dan Christensen said: >Only by doing things like arbitrarily banning self-membership of sets >or postulating some sort of infinite hierarchy of sets. What's wrong with an infinite hierarchy of sets? And how do you do Analysis without infinite sets? >To me, it all seems so unnecessary and so inelegant. De gustibus non diputandem est. >One way -- and I don't >know if it is really anything new -- is to have a simplified set >theory that does not postulate the existence of any particular set >or sets. That somewhat restricts its utility, doesn't it? >It seems to work and neatly avoid problems like RP that have plagued naive set theory. Problems that were dealt with nearly a century ago. Certainly GBN and ZF don't have those problems. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== > In , on 01/08/2004 > at 12:01 AM, Dan Christensen said: >Only by doing things like arbitrarily banning self-membership of sets >or postulating some sort of infinite hierarchy of sets. What's wrong with an infinite hierarchy of sets? Nothing. It just seems to be unnecessary. That's all. And how do you do > Analysis without infinite sets? > Who said anything about doing without infinite sets? If you want to develop number theory in my system, for example, you simply start with a premise corresponding to Peano's Axioms which includes an axiom of infinity. What do you do in ZF if you want to develop group theory? It's axioms are not built into ZF. Wouldn't you simply start by assuming the group axioms? >To me, it all seems so unnecessary and so inelegant. De gustibus non diputandem est. >One way -- and I don't >know if it is really anything new -- is to have a simplified set >theory that does not postulate the existence of any particular set >or sets. That somewhat restricts its utility, doesn't it? > Not really. >It seems to work and neatly avoid problems like RP that have plagued naive set theory. Problems that were dealt with nearly a century ago. Certainly GBN and > ZF don't have those problems. > Indeed. The only advantage my system offers is its simplicity. My purpose, however, is not to supplant ZF or GBN, but to provide a learning tool for the non-specialist taking introductory courses in courses in logic or pure math. While it may suit your purposes, for them, ZF is needlessly complicated and overly abstract. Dan Visit DC Proof Online at http://www.dcproof.com -- FREE download ==== > De gustibus non diputandem est. You should get your Latin right, if you are going to post it. Thomas <1001nipi5rcmc28@corp.supernews.com> ==== How about base -10. You get the 1's place, the -10's place, the 100's place, the -1000 place, and > so on. With this scheme you can write any number positive or negative > without using a minus sign. And in base -2, N bits can represent all the numbers in the range [-2/3*((2^N)-1), 1/3*((2^N)-1)], inclusive. Interesting. If only there were a way to do calculations with these numbers... :) Anyone have any references for this system? Looks interesting. [Crossposted to sci.math.] -Arthur (I guess that also implies that 3 | (2^N)-1, which looked interesting at first until I saw that 2^2k = 4^k = 1 (mod 3), so duh.) ==== , Arthur > > How about base -10. > > You get the 1's place, the -10's place, the 100's place, the -1000 place, >and >so on. With this scheme you can write any number positive or negative >without using a minus sign. You do need 10 digits, though. Continue with decimals, places worth -1/10, 1/100, -1/1000, etc. And in base -2, N bits can represent all the numbers in the range > [-2/3*((2^N)-1), 1/3*((2^N)-1)], inclusive. Interesting. Another interesting one for COMPLEX numbers uses base -1+i and digits {0,1}. Represent all complex numbers that way. > If only > there were a way to do calculations with these numbers... :) > Anyone have any references for this system? Knuth, The Art of Computer Programming. Vol 3 I think. > Looks interesting. -Arthur (I guess that also implies that 3 | (2^N)-1, which looked interesting > at first until I saw that 2^2k = 4^k = 1 (mod 3), so duh.) -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ ==== I was wondering, what are some of the criteria that mathmaticians use in order to figure ouf if there is (an) anlytical solution(s) to a given ODE or ODE system? Is there any textbook that will list such criteria if there is such thing? any thoughts are welcome. john ==== I was wondering, what are some of the criteria that mathmaticians use > in order to figure ouf if there is (an) anlytical solution(s) to a > given ODE or ODE system? analytical? Do you mean analytic in the sense the solution is given by a power series? Or do you mean closed form in some sense? The theory of integration in closed form has some things to say about solution of DEs in closed form. See M. Bronstein, _Symbolic Integration I: Transcendental Functions_ (Springer-Verlag 1997) -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ ==== >> I was wondering, what are some of the criteria that mathmaticians use >> in order to figure ouf if there is (an) anlytical solution(s) to a >> given ODE or ODE system? >analytical? Do you mean analytic in the sense the solution is >given by a power series? Or do you mean closed form in some sense? >The theory of integration in closed form has some things to say about >solution of DEs in closed form. See > M. Bronstein, _Symbolic Integration I: Transcendental Functions_ > (Springer-Verlag 1997) Closed-form solutions can often be found using symmetries. These methods are implemented in Maple's dsolve command. See e.g. Symmetry and Integration Methods for Differential Equations by George W. Bluman, Springer-Verlag 2002, ISBN 0387986545. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== >Closed-form solutions can often be found using symmetries. These methods >are implemented in Maple's dsolve command. See e.g. Symmetry and Integration >Methods for Differential Equations by George W. Bluman, Springer-Verlag >2002, ISBN 0387986545. Sorry, that should be George W. Bluman and Stephen C. Anco. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== > I was wondering, what are some of the criteria that mathmaticians use > in order to figure ouf if there is (an) anlytical solution(s) to a > given ODE or ODE system? If the coefficients are analytical, then there is an analytical solution. This is Cauchy-Kovalevskaya's theorem; see http://www-gap.dcs.st-and.ac.uk/~history/Projects/Ellison/Chapters/Ch5.html > Is there any textbook that will list such criteria if there is such > thing? I suggest the first chapter of A primer of real analytic functions, by Steven G. Krantz and Harold R. Parks. Jose Carlos Santos ==== So this is where George Cox hangs out. It actually seems as though he is reasonable and helpful with people in here. You must have a split personality George - you make an arse of yourself elsewhere with your trolling!! LOL LOL Russell -- You had better stop fighting by the time I get back,or you're ALL grounded. -God ==== >> How can i proove that equation x^3+2y^3+4z^3-34xyz=0 is true >> then x=y=z=0, if x,y,z are whole numbers? that x, y, and z are even, so x/2, y/2, z/2 is also a solution. Sorry, i made a mistake writing question, i need to prove that ONLY ONE solution exists (0,0,0) in whole numbers --------------------------------------------------------------------------- >John R Ramsden (jr@adslate.com) >--------------------------------------------------------------------------- >Eternity is a long time, especially towards the end. > Woody Allen ==== En el mensaje:btr2760295@drn.newsguy.com, Student escribi.97: > Ramsden says... >> On 10 Jan 2004 05:20:09 -0800, Student > How can i proove that equation x^3+2y^3+4z^3-34xyz=0 is true > then x=y=z=0, if x,y,z are whole numbers? >> that x, y, and z are even, so x/2, y/2, z/2 is also a solution. > Sorry, i made a mistake writing question, i need to prove that ONLY > ONE solution exists (0,0,0) in whole numbers It is the same yo has said before, no? Suppose that (x, y, z) is a solution. Then, x^3+2y^3+4z^3 = 34xyz ===> x must be even. Let x = 2x' 8x'^3 + 2y^3 + 4z^3 = 68x'yz ===> 4x'^3 + y^3 + 2z^3 = 34x'yz ===> y = 2y' ===> 4x'^3 + 8y'^3 + 2z^3 = 68x'y'z ===> 2x'^3 + 4y'^3 + z^3 = 34x'y'z ===> z = 2z' ===> 2x'^3 + 4y'^3 + 8z'^3 = 68x'y'z' ===> x'^3 + 2y'^3 + 4z'^3 = 34x'y'z' ===> (x', y', z') also is a solution! Then, if you have a solution (x, y, z), then (x/2, y/2, z/2) is a integer solution, then ((x/2)/2, (y/2)/2, (z/2)/2)) = (x/4, y/4, z/4) also is a integer solution, ...., as (x/2^k, y/2^k, z/2^k). You have infinitely many integer solutions, each one of them with less absolute value that the previous. But it is impossible, except that x/2 = x, y/2 = y and z/2 = z ===> x = y = z = 0. This method of prove, a particular case of reasoning by contradiction, is called 'infinite descend', ant it is due to Fermat. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com ==== thanx to both of you, now it's clear for me:) This method of prove, a particular case of reasoning by contradiction, is >called 'infinite descend', ant it is due to Fermat. >-- Ignacio Larrosa Ca.96estro >A Coru.96a (Espa.96a) >ilarrosaQUITARMAYUSCULAS@mundo-r.com >-- Ignacio Larrosa Ca.96estro >A Coru.96a (Espa.96a) >ilarrosaQUITARMAYUSCULAS@mundo-r.com ==== Aahh ! I got it !! > if lim [p(x)] = lim [q(x)] when x-> c, then lim[p'(x)] = lim > [q'(x)] when x-> c. when x-> c.[/quote:0fc2fa9f81] When c = oo ( oo = infinite ), if lim p(x) =oo, lim q(x) =oo when x->oo, then lim [p(x) /q(x)] = lim[p'(x) / q'(x)] ( that is L'Hopital rule ) let h(x) = ax + b, lim [f(x)/h(x)] = lim [f'(x)/h'(x)] = 1 when x-> oo ( because h(x) is asymptote of f(x). So lim f'(x) = lim h'(x) ( x-> oo ), or f'(oo) = h'(oo) = a Since g''(x) >0, we have g'(x) is an increasing function for x>0, so g'(x) < g'(oo) = f'(x) - a < f'(oo) - a = 0 ( because f''(x) > 0 for x>0 ) Thus g'(x) < 0 for all x>0 And so, g(x) is decreasing functioc for x>0, so g(x) > g(oo) = lim [f(x) - h(x)] = 0 when x->oo. Therefore [b:0fc2fa9f81]g(x) > 0 for all x>0[/b:0fc2fa9f81] Is my solution exact ? ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- ==== > The matter now is that I don't know this is right or wrong : > if lim [p(x)] = lim [q(x)] when x-> c, then lim[p'(x)] = lim > [q'(x)] when x-> c. > May anyome confirm that for me, plzzz. > If it's right, I can solve the problem above. 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 c = 0, p(x) = x and q(x) = x^2, to show your conjecture false. ==== Given a nonzero cardinal k, what is known about the maximal cardinality of all sets of pairwise non-homeomorphic topologies on k? For infinite k, is it equal to (2^k)^k? Is there a known combinatorial formula for finite k? How about the maximal cardinality of sets of non-homeomorphic T_x topologies (x in {0, 1, 2, 3, 3.5, 4, 5})? Of metric spaces? -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu ==== > Given a nonzero cardinal k, what is known about the maximal > cardinality of all sets of pairwise non-homeomorphic topologies on > k? For infinite k, is it equal to (2^k)^k? You probably mean 2^(2^k), seeing as (2^k)^k = 2^k. Anyway, that is the answer to your question: there are 2^(2^k) non-homeomorphic topologies on a set of cardinality k. There are 2^(2^k) topologies on a set of (infinite) cardinality k, and each homeomorphism class contains at most 2^k of those topologies. > Is there a known combinatorial formula for finite k? Beats me. See A001930 at topologies (x in {0, 1, 2, 3, 3.5, 4, 5})? Of metric spaces? ==== >Given a nonzero cardinal k, what is known about the maximal >>cardinality of all sets of pairwise non-homeomorphic topologies on >>k? For infinite k, is it equal to (2^k)^k? >> You probably mean 2^(2^k), seeing as (2^k)^k = 2^k. Anyway, that is >the answer to your question: there are 2^(2^k) non-homeomorphic >topologies on a set of cardinality k. There are 2^(2^k) topologies on >a set of (infinite) cardinality k, and each homeomorphism class >contains at most 2^k of those topologies. > Yes, sorry - I meant 2^(2^k), the cardinality of P(P(k)). Is the proof relatively easy? If so, hints? If not, reference? >Is there a known combinatorial formula for finite k? >> Beats me. See A001930 at > scribbled the following: >> Given a nonzero cardinal k, what is known about the maximal >> cardinality of all sets of pairwise non-homeomorphic topologies on >> k? For infinite k, is it equal to (2^k)^k? > You probably mean 2^(2^k), seeing as (2^k)^k = 2^k. Don't you mean (2^k)^k = 2^2k, assuming that ^ means exponentation? -- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ I said 'play as you've never played before', not 'play as IF you've never played before'! - Andy Capp ==== > Fred Galvin scribbled the following: >> Given a nonzero cardinal k, what is known about the maximal >> cardinality of all sets of pairwise non-homeomorphic topologies on >> k? For infinite k, is it equal to (2^k)^k? > You probably mean 2^(2^k), seeing as (2^k)^k = 2^k. Don't you mean (2^k)^k = 2^2k, assuming that ^ means exponentation? I meant (2^k)^k = 2^(k^2), but it's all the same since k is infinite (and I'm assuming the axiom of choice), k^2 = k = 2k. ==== A variable X follows an Ito Process if: dx=a(x,t)dt+b(x,t)dz Where dz is a wiener process. Ito's lemma shows that if a variable G is a function of x and t: G=g(x,t), it will follow the process described in Ito's differential. If G is a function of x, t, and G itself: G=g(x,t,G), can I use Ito lemma? if not, is there any alternative way to solve the problem? ==== Consider a simpler situation first. Suppose X_t = f(X_t). If this is to hold almost surely, then P(X_t in S) = 1, where S = {x: f(x) = x}. But then again, if X_t is in S almost surely, then we don't care about the values of f(x) outside of S, so we may as well assume f(x) = x. > A variable X follows an Ito Process if: > dx=a(x,t)dt+b(x,t)dz > Where dz is a wiener process. > Ito's lemma shows that if a variable G is a function of x and t: > G=g(x,t), it will follow the process described in Ito's differential. > If G is a function of x, t, and G itself: G=g(x,t,G), can I use Ito > lemma? if not, is there any alternative way to solve the problem? ==== I didn't see this thread till after I posted. I find that many of my questions have ongoing or recent threads. NSA was what I was thinking of but had forgotten what they called it. (Don't know how my message got posted twice.) Van ==== > Say we are working in the ring of differential operators... How would you go about simplifying the following... > [Using * for the ring multiplication] [cos(t) * d/dt * sin(t) * d/dt] Basically I am not sure if you can 'forget' the rightmost d/dt since > that is just to differentiate the function - i.e it does not > simplify... so would it then become [cos(t) * [d/dt * sin(t)] * d/dt] > [cos(t) * [cos(t) + Sin(t) * d/dt] * d/dt] > [cos^2(t) * d/dt + cos(t)sin(t) *d/dt ] etc... Or am I forgetting something ? > - Sandos Please this has me quite stuck... I am unsure how to proceed Cancel-Lock: sha1:Bw1Yvjh0UCGMreu/JoZsH9pgqyQ= ==== >> >> [cos(t) * d/dt * sin(t) * d/dt] >> >> Basically I am not sure if you can 'forget' the rightmost d/dt >> since that is just to differentiate the function - i.e it does not >> simplify... Please this has me quite stuck... I am unsure how to proceed You can't throw a differential operator away. Somebody else should pipe up if I'm wrong, but this looks like a chain rule. Put D for d/dt, and you get: cos(t) * D * sin(t) * D = cos(t) * D * (sin(t) * D) (by associativity) = cos(t) * (D*sin(t) + DD) (by the chain rule) = cos(t) * (cos(t) + D^2) = cos^2(t) + cos(t)*D^2 Len. ==== > I have to build a context free grammar that generates the following >language : >{ a^i b^j c^k | i,j,k >= 0 and k <= i+j } > [his attempt deleted] It looks like you're on the wrong track. First solve the problem of generating b^j c^k with k<=j, That's what was supposed to do S5. > and then try to generate a^i B c^k with k<=i, where > B is something you have already generated. DGA ==== >I have to build a context free grammar that generates the following language : >{ a^i b^j c^k | i,j,k >= 0 and k <= i+j } I take it you are unaware of the pumping lemma for CFL's? Yes, I am. I have been learning grammars for a week . > Perhaps you should do some of the exercises that appear in the > first hit at http://www.google.com/search?q=pumping+lemma+CFL DGA ==== Can You agree to the following? The important part of a function is the way from the independent variable in the domain to the dependent in the range (the arrow in f:R ->R )When i monitor the weather in my place, i 'll get a function, i attribute temperature to time. Temperature is not a term calculated from time. In math You have to be precise, so if You have a function, which is not term-defined, You have to attribute to every element of the domain an element of the range, the domain must be precise. Attribute to every nunber the amount of digits of this number in decimal notation, f.e. 3 ->1, 2.114 ->4,...This function is not term-defined, at least not for a simple mind like me. Choose the domain R, then You are not able to draw a graph,still You can choose a domain at Your choice. f(x)=1/x is a term-defined function - You can say it's a function R -> R, as everybody knows the rules of calculation of division (x must not be zero) and by these rules can determine the exact domain. The same with the function f:R -> R: x -> ln (x^3+x+3), as ln is defined for values greater zero. The algebraic structure in which You are calculating is important. f: R2 -> R2 : z=(x,y) -> (1/x, 0) is not term-defined. f(z) = ( 1/ ( z dot(1,0)) ) real scalar multiplication (1,0) is term defined and obviously in (R2, +, r.s.m., dot ), the euclidian vectorspace. And the term requires, that the first component x must not be zero. This function can not be defined by a term in (R2, +, * ), the commutative field, which is known under the name of complex numbers. Have fun Hero ==== Can You agree to the following? The important part of a function is the way from the independent variable in the domain to the dependent in the range (the arrow in f:R ->R )When i monitor the weather in my place, i 'll get a function, i attribute temperature to time. Temperature is not a term calculated from time. In math You have to be precise, so if You have a function, which is not term-defined, You have to attribute to every element of the domain an element of the range, the domain must be precise. Attribute to every nunber the amount of digits of this number in decimal notation, f.e. 3 ->1, 2.114 ->4,...This function is not term-defined, at least not for a simple mind like me. Choose the domain R, then You are not able to draw a graph,still You can choose a domain at Your choice. f(x)=1/x is a term-defined function - You can say it's a function R -> R, as everybody knows the rules of calculation of division (x must not be zero) and by these rules can determine the exact domain. The same with the function f:R -> R: x -> ln (x^3+x+3), as ln is defined for values greater zero. The algebraic structure in which You are calculating is important. f: R2 -> R2 : z=(x,y) -> (1/x, 0) is not term-defined. f(z) = ( 1/ ( z dot(1,0)) ) real scalar multiplication (1,0) is term defined and obviously in (R2, +, r.s.m., dot ), the euclidian vectorspace. And the term requires, that the first component x must not be zero. This function can not be defined by a term in (R2, +, * ), the commutative field, which is known under the name of complex numbers. Have fun Hero ==== On 10 Jan 2004 11:10:07 -0800, mogensn@mogensn.dk (Mogens Nielsen) >litterature on the real case. Actually come to think of it I'm not nearly as sure of what I said as I was yesterday... let's review: >>If you want me to be more precise, here is the problem: >>Let K be a compact subset of R^k. Let T:C(K) -> C(K) be an operator, >>where C(K) is the set of continuous functions from K to R. In the >>of T, without defining it. >> >> He's probably referring to the complex spectral radius, ie the sup >> of |lambda| for lambda in the (complex) spectrum. Because that's >> the _standard_ definition of the spectral radius, even for real >> Banach algebras. Come to think of it, if B is a real Banach algebra then it really makes no sense to talk about the complex spectrum: L is in the spectrum of x if x - Le is not invertible, but if L is complex there's no such thing as x - Le in the first place. What seems most likely to me is that C(K) is actually the space of continuous functions from K to C; that's the standard meaning, it seems to me. Are you sure he's talking about functions from K to R? Another possibility is that the standard definition of the spectrum of an element of a real Banach algebra is the spectrum in the complexification of the algebra. And another possibility is that he's really talking about the real spectrum (this last seems least likely to me). Just realized I'm not certain what's the standard definition of the spectrum in a real Banach algebra, how embarassing (never thought about real Banach algebras). But I bet it's supposed to be the spectrum in the complexification of the algebra - that would be consistent with the standard definition of the spectrum of a real matrix, for example. >Mogens ************************ David C. Ullrich ==== the smallest non-definable ordinal is definable. >why is this argument wrong? When you post the same message to two different groups you should _cross-post_ instead of posting the message separately! You would do that by putting sci.math,sci.logic in the newsgroups field (how you do that depends on what software you're using.) The reason is this: If you crosspost then readers in sci.logic will see replies from sci.math (I would not have bothered with giving an explanation in sci.logic if I'd known there were already correct explanations in sci.math. More important, if you're posting a question you assume that the readers will be interested in the answer - so you should assume that readers of sci.logic will benefit from replies made in sci.math.) ************************ David C. Ullrich ==== > the smallest non-definable ordinal is definable. > why is this argument wrong? > I don't think that it is definable, because it seems to me that definability is not a first order property. You presumably say a set x is definable if there is some first order statement p such that x = { y : p(y) } ?. If you try to turn this into a first order property then you end up trying to get your quantifiers to range over predicates, which dosn't work. So, as you can't talk about the property 'is definable' in a first order way, there certainly doesn't have to be a least such element with that property in a first order theory. Or did you mean something else by definability? The thing to be careful about in ZF and other axiomatic set theories is that, while most of the time intuitive and verbal reasoning about the axioms will do the job, they really are first order theories and you need to stay within the first order reasoning if you expect them to hold together. David ==== > the smallest non-definable ordinal is definable. > why is this argument wrong? > It's nothing to do with ordinals. The same this happens with: The least natural number not definable in fewer than twenty English words ==== > the smallest non-definable ordinal is definable. > why is this argument wrong? Because you cannot express the smallest non-definable ordinal using the language of first order logic within ZF. I think that what it boils down to is that it is not possible, within ZF, to show that the class of all sets is a model for set theory. Let me give more details. ZF is consistent if and only if NBG is consistent. NBG is a theory in which the basic objects are classes rather than sets. In NBG a class is defined to be a set if it is an element of another class. Amongst NBG's axioms is: if a is a set, and if F(x) is a well formed formula in which all quantifiers are over sets, then {x in a: F(x)} is also a set. What you would like to do is to show that the class of all sets forms a model for ZF. But if you could do that, this would contradict Goedel's Theorem. It turns out that in order to go through the usual construction to show that the class of all sets is a model of ZF that one would need to work in a richer theory, that is, with this axiom: if a is a set, and if F(x) is a well formed formula, then {x in a: F(x)} is also a set. This theory has to be richer, because it is possible to prove that ZF is consistent within this theory. In order to construct the smallest non-definable ordinal within your set theory, you would need this richer set theory. This is not a paradox - all you have done is to define the smallest ordinal not definable in ZF, which will certainly not be the smallest ordinal definable in this richer theory. I hope this helps. Stephen ==== >the smallest non-definable ordinal is definable. >why is this argument wrong? Because you cannot express the smallest non-definable ordinal using > the language of first order logic within ZF. I think that what it > boils down to is that it is not possible, within ZF, to show that the > class of all sets is a model for set theory. I think you have pulled the wool over giovanni's eyes (or tried to). Nothing in his question mentioned ZFC. Presumably he means something like definable in English. Of course, there is still an answer, and it proceeds similar to yours, but differently, in some interesting ways. Thomas ==== Thomas Bushnell, BSG ha scritto nel messaggio > Presumably he means something > like definable in English. no, I mean definable in the sense of the descriptive set theory!!!! ==== > Thomas Bushnell, BSG ha scritto nel messaggio >Presumably he means something >like definable in English. no, I mean definable in the sense of the descriptive set theory!!!! Oh, then the original answer seems entirely correct to me. ==== Now I have just solved the problem, I have understood. ==== > I think that what it boils down to > is that it is not possible, within ZF, > to show that the class of all sets > is a model for set theory. scuse me, can you explain to me what has that to do with it? > What you would like to do is to > show that the class of all sets > forms a model for ZF. why is it necessary for my argument? > It turns out that in order to go > through the usual construction to > show that the class of all sets is > a model of ZF that one would > need to work in a richer theory Bernays-Morse, for example. but in my opinion I don't want (and don't have) to show that the class of all sets is a model of ZF. > In order to construct the smallest > non-definable ordinal within your > set theory, you would need this > richer set theory. why? I think I can construct the smallest non-definable ordinal also in the ordinal c of the continuum. I hope you can explain to me why I'm wrong... ==== Now I have just solved the problem, I have understood. ==== > .9^2 = .81 > .99^2 = .9801 > .999^2 = .998001 > .9999^2 = .99980001 > .99999^2 = .9999800001 > .999999^2 = .999998000001 > .9999999^2 = .99999980000001 > .99999999^2 = .9999999800000001 > .999999999^2 = .999999998000000001 > .9999999999^2 = .99999999980000000001 > .99999999999^2 = .9999999999800000000001 > .999999999999^2 = .999999999998000000000001 > .9999999999999^2 = .99999999999980000000000001 > .99999999999999^2 = .9999999999999800000000000001 > .999999999999999^2 = .999999999999998000000000000001 > .9999999999999999^2 = .99999999999999980000000000000001 > .99999999999999999^2 = .9999999999999999800000000000000001 > .999999999999999999^2 = .999999999999999998000000000000000001 > .9999999999999999999^2 = .99999999999999999980000000000000000001 > .99999999999999999999^2 = .9999999999999999999800000000000000000001 Each position after the decimal point eventually becomes a 9 and stays > that way. Daniel W. Johnson has proven that 0.999... will never equal 1. I knew someone here could do it. Garry Denke, Geologist Denoco Inc. of Texas ==== > Daniel W. Johnson has proven that 0.999... will never equal 1. No, I've proven that the 8's, 0's, and 1's in your blitherings are irrelevant. Because no matter what number you choose less than 1, the squares of the original Cauchy sequence will eventually exceed it (and remain above it). In other words, the squares of 0.999... tend toward 1. -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W ==== In sci.math, Garry Denke on 11 Jan 2004 06:31:26 -0800 <54056cf9.0401110631.39cca721@posting.google.com>: >> .9^2 = .81 >> .99^2 = .9801 >> .999^2 = .998001 >> .9999^2 = .99980001 >> .99999^2 = .9999800001 >> .999999^2 = .999998000001 >> .9999999^2 = .99999980000001 >> .99999999^2 = .9999999800000001 >> .999999999^2 = .999999998000000001 >> .9999999999^2 = .99999999980000000001 >> .99999999999^2 = .9999999999800000000001 >> .999999999999^2 = .999999999998000000000001 >> .9999999999999^2 = .99999999999980000000000001 >> .99999999999999^2 = .9999999999999800000000000001 >> .999999999999999^2 = .999999999999998000000000000001 >> .9999999999999999^2 = .99999999999999980000000000000001 >> .99999999999999999^2 = .9999999999999999800000000000000001 >> .999999999999999999^2 = .999999999999999998000000000000000001 >> .9999999999999999999^2 = .99999999999999999980000000000000000001 >> .99999999999999999999^2 = .9999999999999999999800000000000000000001 >> >> Each position after the decimal point eventually becomes a 9 and stays >> that way. Daniel W. Johnson has proven that 0.999... will never equal 1. > I knew someone here could do it. All he's proven is that (.999...9)^2 leads to a Cauchy sequence. This is a good start, as Cauchy sequences inherently define real numbers. http://mathworld.wolfram.com/CauchySequence.html So what is the limit of that sequence? We know .999...9 = 1 - 10^(-n) for any finite expansion. Pick an epsilon > 0, and I can find an M = ceil(log(1/epsilon)) which is such that, for any N > M, 0 < (1 - 10^(-N) ) - 1 < epsilon. So I can get as close to 1 as desired. Squaring, cubing, fourthing, fifthing -- it doesn't matter too much as the only adjustment is that I might have to use a slightly bigger M: M = ceil(log(5/epsilon)) would probably work for fifth powers, for example. Of course things get a little weird if one uses variable powers. (1 - 1/n)^n = 1/e, for example, as n tends towards positive infinity. (In your case, that might equally easily be expressed as (1 - 10^(-n))^(10^n), leading to .9^10, .99^100, .999^1000, etc. Try it -- that series, at least, does *not* tend to 1.) Garry Denke, Geologist > Denoco Inc. of Texas -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== > In sci.logic, Dave Seaman >> Omega+1 has the same cardinality as omega, but they are different >> ordinals. If the (w+1)-strings of digits are ordered lexicographically, >> then there is no order-preserving map of such strings to the reals. > Interesting. So expressions such as omega+1 and 2*omega actually do > make sense? If so, mea culpa. >> Yes. In fact, w is identical to N, the set of natural numbers (including 0), >> and w+1 = w U {w} = { 0, 1, 2, 3, ..., w }. >> Addition and multiplication are noncommutative: >> 1+w = w < w+1, (w+1 has a last element but 1+w does not) >> 2*w = w < w*2 = w+w. (an w of pairs vs. a pair of w's) > Ugh. My brain's beginning to hurt now... :-) > So which one of these applies to such Garry-Denke-isque decimal > expansions as > (.999....)^2 = .999...99800...001 ? None of them, as far as I can see. > It feels like 8 is in the w+2 or w+3 position but I want to make sure. > 1 is presumably in the position w*2 + 2. That seems to be implied by the notation, but I'll resist any urge to try to make sense of it. > Presumably 1+w refers to things like > 9,(9,9,9,9,...) > where one tacks things onto the left side of an infinite sequence. > If one computes .999... / 10 + .8 = .8999..., one gets a similar > sequence. > However, if we do allow transfinites and infinitesimals into our > number system a la Garry Denke, I'm not sure how the rest of mathematics > is going to deal with it (the concept of limit in particular may have > to be redefined). > Hopefully it's a simple retrofit but I wonder. >> There's nothing wrong with considering transfinite ordinal sequences of >> digits in general, but as I said, there is no order-preserving map of >> such objects to the reals for ordinals > w. > That might be good enough. :-) Of course that's probably related > to the impossibility of reliably defining things such as > .999... - .999...99800...001 = .000...?!?!?!... >:-) I wouldn't go so far as to say it's impossible. After all, nonstandard analysis makes sense of infinitely large integers. But then, the hyperintegers of NSA are not the same as the transfinite ordinals, and I am not aware that anyone has tried to use digit strings indexed by the hyperintegers to define values in the hyperreals. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== In sci.logic, Dave Seaman on Sun, 11 Jan 2004 02:58:09 +0000 (UTC) : >> In sci.logic, Dave Seaman > > Omega+1 has the same cardinality as omega, but they are different > ordinals. If the (w+1)-strings of digits are ordered lexicographically, > then there is no order-preserving map of such strings to the reals. >> Interesting. So expressions such as omega+1 and 2*omega actually do >> make sense? If so, mea culpa. > > Yes. In fact, w is identical to N, the set of natural numbers (including 0), > and w+1 = w U {w} = { 0, 1, 2, 3, ..., w }. > > Addition and multiplication are noncommutative: > > 1+w = w < w+1, (w+1 has a last element but 1+w does not) > 2*w = w < w*2 = w+w. (an w of pairs vs. a pair of w's) > >> Ugh. My brain's beginning to hurt now... :-) > >> So which one of these applies to such Garry-Denke-isque decimal >> expansions as > >> (.999....)^2 = .999...99800...001 ? None of them, as far as I can see. > >> It feels like 8 is in the w+2 or w+3 position but I want to make sure. >> 1 is presumably in the position w*2 + 2. That seems to be implied by the notation, but I'll resist any urge to try > to make sense of it. Yeah, well, it doesn't make much sense anyway in light of the subtraction below. :-) As Barb has pointed out, 10^(-omega) = 0 by any reasonable metric, so if one states that 8 is in the w+2 position then the 8's value equals 10^(-(omega+2)) = 10^(-omega) * 10^2 = 0 * .01 = 0. Personally, I think .999... is a great example of a Cauchy sequence, with limit 1. :-) > >> Presumably 1+w refers to things like > >> 9,(9,9,9,9,...) > >> where one tacks things onto the left side of an infinite sequence. >> If one computes .999... / 10 + .8 = .8999..., one gets a similar >> sequence. >> However, if we do allow transfinites and infinitesimals into our >> number system a la Garry Denke, I'm not sure how the rest of mathematics >> is going to deal with it (the concept of limit in particular may have >> to be redefined). > >> Hopefully it's a simple retrofit but I wonder. > > There's nothing wrong with considering transfinite ordinal sequences of > digits in general, but as I said, there is no order-preserving map of > such objects to the reals for ordinals > w. >> That might be good enough. :-) Of course that's probably related >> to the impossibility of reliably defining things such as > >> .999... - .999...99800...001 = .000...?!?!?!... > >>:-) I wouldn't go so far as to say it's impossible. After all, nonstandard > analysis makes sense of infinitely large integers. But then, the > hyperintegers of NSA are not the same as the transfinite ordinals, and I > am not aware that anyone has tried to use digit strings indexed by the > hyperintegers to define values in the hyperreals. > NSA presumably routinely works with very large (1,024-bit) integers, but they *are* integers; the challenge is mostly in defining operations therewith in terms of arrays of smaller integers, sort of like defining multidigit arithmetic in terms of the times table, except that instead of a 10 [Times] 10 matrix one can memorize by rote, one has a 65536 [Times] 65536 affair that is implemented by the microprocessor's multiply instruction. It's not that difficult in principle but it's tricky. I have some ideas on how to compute M^e mod N, where all of M, N, and e are very large 1,024-bit integers, but have no idea how well they'd work. (The simplest method I can think of is to keep squaring M in an accumulator, multiplying the accumulator with M whenever there's a '1'. It would still take 1,024 to 2,048 multiplications of two multiprecision numbers. There are probably more elegant methods.) -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== > ... The sum of _many_ beta distributions converges to a gaussian distribution, > but neither my beloved book (and some hard core integral solvining) or extensive search on the internet has given me > anything on what the distribution of the sum of two beta functions is. ... The distribution of a sum of independent Beta random variables is close to normal if no variance is large compared to the sum, and the parameters do not get too extreme; The pdf of the sum of 2 Beta rv's (the question posed above) will generally not be close to Normal. One can derive the characteristic function of the sum, and then invert it numerically to derive the pdf, for given parameter values. For example, here is a quick 2-line derivation and pdf plot using mathStatica: http://www.mathstatica.com/Sumof2Betas/ Colin ______________________________ Dr Colin Rose mathStatica Pty Ltd Web: www.mathStatica.com ______________________________ ==== >... The sum of _many_ beta distributions converges to a gaussian distribution, >but neither my beloved book (and some hard core integral solvining) or extensive search on the internet has given me >anything on what the distribution of the sum of two beta functions is. ... > The distribution of a sum of independent Beta random variables > is close to normal if no variance is large compared to the sum, > and the parameters do not get too extreme; The pdf of the sum of 2 Beta rv's (the question posed above) will > generally not be close to Normal. One can derive the characteristic > function of the sum, and then invert it numerically to derive the pdf, > for given parameter values. For example, here is a quick 2-line derivation and pdf plot using > mathStatica: http://www.mathstatica.com/Sumof2Betas/ > Colin BTW: is there a 'standardized' way to measure how far a pdf is from being normal? I mean, hm, a practical one and may be only for some classes of pdf. ==== Concerning the results of the Gelfound-Schneider Theorem (http://mathworld.wolfram.com/GelfondsTheorem.html), which states that when A is an algebraic number, and B is an irrational and algebraic number, A^B is a transcendental number. So, given the contrapositive, which would state that if C is not a transcendental number, then the log(Base A) of C, where A is an algebraic number, would not be irrational and algebraic. And thus this leads me to believe, and please correct me if I've made a mistake in my logic, that a logarithm with any integer base, of another integer, is rational or transcendental. Does this mean that most logarithms with integer bases of integers are transcendental? Or could they all be rational? ==== >Concerning the results of the Gelfound-Schneider Theorem >(http://mathworld.wolfram.com/GelfondsTheorem.html), which states that >when A is an algebraic number, and B is an irrational and algebraic >number, A^B is a transcendental number. >So, given the contrapositive, which would state that if C is not a >transcendental number, then the log(Base A) of C, where A is an >algebraic number, would not be irrational and algebraic. >And thus this leads me to believe, and please correct me if I've made >a mistake in my logic, that a logarithm with any integer base, of >another integer, is rational or transcendental. Does this mean that >most logarithms with integer bases of integers are transcendental? Or >could they all be rational? Most such logarithms are transcendental. The lograithm of one integer with respect to another integer is only rational if they are both powers of the same integer. David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills. ==== >Concerning the results of the Gelfound-Schneider Theorem >(http://mathworld.wolfram.com/GelfondsTheorem.html), which states that >when A is an algebraic number, and B is an irrational and algebraic >number, A^B is a transcendental number. So, given the contrapositive, which would state that if C is not a >transcendental number, then the log(Base A) of C, where A is an >algebraic number, would not be irrational and algebraic. Yes. If C is algebraic then either A is transcendental or B is either rational or transcendental. >And thus this leads me to believe, and please correct me if I've made >a mistake in my logic, that a logarithm with any integer base, of >another integer, is rational or transcendental. Yes. >Does this mean that >most logarithms with integer bases of integers are transcendental? Or >could they all be rational? Let log_a(c) = n/m where n and m are relatively prime. Then c is integer only if a^(n/m) is integer which is true only if a = d^(km) where d and k are integers. If a is prime then the only solution is k = 1, d = a and m = 1. So any logarithm with a prime integer base will only have integer and transcendental values. ==== Download beta 1 - www.master-graph.com/mgraph20b1.exe (only 477KB). All who will help me to improve this program will get a free registration key. You can do the following: -find bugs -feedback about you impression by the program -advice me to change or add some new features -translate it to the other language (see www.master-graph.com/instructions/interface.html) Feel free to contact with me - roman@master-graph.com Sincerely, Roman. ==== >2. Why can symmetry be used in the solving procedure ? If an equation has a certain symmetry, one can show that the *set* of > its solutions has to have the same symmetry.... Do you know the name of the person who proved that ? >One see people >-expand the the assumed solution in a set of basis functions that have >the same symmetry as the system considered, Well, usually the basis functions have *not* the same symmetry as the > system... (for example, not all wave functions for the H atom are > spherically symmetric!) >and solve for the >coefficients as system of linear equations.(Gauss-elimination) > The basis functions of the one-dimesional representaion of the group >that the system belongs to, are the basis functions that are used. Why? Sorry, I'm not very familiar with this... >Because the symmtry operators and the hamiltonian operator commute >(?????????)Help ! When one can find operators for certain symmetries which commute with > the Hamiltonian, one usually chooses the wave functions to be > eigenfunctions to these operators, too. Again, this is done because it > usually simplifies finding the solutions (example: for solving the H > atom, it is *very* helpful to try to find a wave function which is an > eigenstate for angular momentum, too - because then one already knows > that the wave function has to be a multiple of a spherical harmonic!) Any references about the theory of the commutation ? >3. How does variational theory (and what is it ?) come into the >solving theory of Schr.9adinger eqn. ? AFAIK, AFAIK ?? You've lost me. the Schroedinger equation usually can't be solved by the > variational theory - one can only find *approximate* solutions. What it > is? Well, the idea (in the case of the Schroedinger equation) is to use > some test functions and calculate the expectation value for the energy > for them. It is easy to prove that these expectation values have to be > greater or equal to the ground state energy, hence by looking for the > lowest possible energy among all of the test functions, one gets an > upper bound for the ground state energy. (this is only the tip of the > iceberg, but I think you get the general idea...) > Easy introductory text? > 4. Perturbation theory in quantum mechanics. Is it one of these >none-proven things ? > >Well, I think in many mechanics classes, this is considered to be an > advanced topic. >When you suggest books, consider that I have the education of an >electronics engineer. No fancy mathematics courses like functional >analysis, etc. But apparently you have some experience with differential equations and > matrices, don't you? What about the general definition of vector > space? What about the notion of differential operators? > Sure Bjoern, I've read some intriductory quantum mechanics books, and they often have a section of functions whicj are orthogonal to each that I can write a diff. eqn. like (d^2/dt^2)y + y=0, like (D^2 +1)*y=0. I've calculated these type of equations in the calculus courses. But I have no clue of the behindlying theory. It is the theory I want to dig into.... > Lasse ==== Then the > branches depend more immediately on these cuts which define a domain > for them than on the branch points. Discussion of the consequences of > this requires a few more paragraphs and will be for another time. The following supplies what was left out of my Jan6 post. In order for branches to be fully defined a domain has to be provided for them. The use of cuts to produce a simply-connected region where the monodromy theorem applies is familiar and so single-valuedness of each branch on the domain is obtained. A branch is just a function from the cut plane to a region of the w-plane and is distinguished from its fellows by its initial value. If the given relation is linear in the independent variable z then the branches have disjoint ranges. For a proof let an ordinary point w have the single counter-image z and be in the ranges of the branches which have the initial values wh and wk. An arc in the cut plane say A from z0 to z is imaged by (among others) an arc from wh to w and an arc in the cut plane say B from z0 to z is imaged by an arc from wk to w. Then the circuit AinverseB is imaged by the arc wh to w to wk. By the cut construction and the monodromy theorem the circuit must be imaged by a circuit so wh=wk and so w cannot belong to two ranges. When f(w,z) is non-linear in z overlapping ranges can be expected. Permutations on branches are intrinsic to closed circuits but one wants to associate them with crossings of cuts. NB: A permutation cannot generally be associated consistently with just a branch-point. Put an arrow (orientation counts) across each cut. Now take a circuit from and to the initial point which crosses one cut only in the direction of its arrow. The permutation produced by the circuit is given to the arrow. Crossing the opposite way gives the inverse permutation. A general circuit crosses more than one cut. The permutation it produces is the same as the product of the assigned permutations taken in the order in which the cuts are crossed. This is shown by distorting the circuit back to the initial point in between crossings, Now translatability, to get permutations for a system of cuts from those assumed already known for another system. A circuit crossing one cut only in the system for determination may cross more than one cut in the known system. But its permutation is determinable by the previous paragraph and this permutation just has to be given to the crossed cut. It only needs another step to define a Riemann surface. But perhaps implicit functions can be studied just as well staying with branches and using the above theory. One advantage is that the z-plane and the w-plane can be represented on different parts of the blackboard and the details filled in without any strain on the visual imagination. Copyright Jan04. Quotation with acknowlegement permitted. ==== ... > Permutations on branches are intrinsic to closed circuits but one >wants to associate them with crossings of cuts. NB: A permutation >cannot generally be associated consistently with just a branch-point. >Put an arrow (orientation counts) across each cut. Now take a circuit >from and to the initial point which crosses one cut only in the >direction of its arrow. The permutation produced by the circuit is >given to the arrow. Crossing the opposite way gives the inverse >permutation. > A general circuit crosses more than one cut. The permutation it >produces is the same as the product of the assigned permutations taken >in the order in which the cuts are crossed. This is shown by >distorting the circuit back to the initial point in between crossings, ... I don't know whether you (or anyone here) took the (not inconsiderable) trouble to look up the references I gave, to several papers (two or three by Orevkov, one by Zoladek), in an earlier post. Within the a full-scale, at least a more detailed exposition of the point of view which is (somewhat sketchily) described in their papers (and, about as sketchily, in some earlier publications of mine). For the time being, I will simply point out here that (in the case where your Riemann surface is sitting inside C^2, as the set where some holomorphic function f(z,w) of two complex variables vanishes) you can do better than putting permutations on branch cuts: you can (if you chose the cuts canonically, to be where two distinct branches have equal real part) put *braids* on branch cuts. (And in this case the arrows are really important; in the generic situation where the permutation for each cut is just a transposition, and therefore self-inverse, the arrows are insignificant. One way to think of the braid group on n strings is, in fact, to start with the presentation of the permutation group on n letters which has has generators the transpositions of adjacent letters from 1 to n, and as relations the rules (1) (i i+1)(i+1 i+2)(i i+1) = (i+1 i+2)(i i+1)(i+1 i+2) for i from 1 to n-2, (2) (i i+1) commutes with (j j+1) for j>i+1, (3) the square of each generator is the identity, and throw away the relations (3).) When you do so you (potentially, and actually) learn more about f(z,w) than you do without braids. (And you learn some interesting things about braids, too.) Lee Rudolph ==== > Here is a simplified reformulation of my previous problem: > How to solve this integer matrix equation? > [ a1^2*u1, a1*a2*u2, a1*a2*u3, a2^2*u4] > [ a1*a3*u1, a1*a4*u2, a3*a2*u3, a2*a4*u4] > [ a1*a3*u1, a3*a2*u2, a1*a4*u3, a2*a4*u4] > [ a3^2*u1, a3*a4*u2, a3*a4*u3, a4^2*u4] > = > [ 16 16 -16 -4] > [ 4 -16 -4 4] > [ 4 4 16 4] > [ 1 -4 4 -4] a1, a2, a3, a4, u1, u2, u3, u4 should be 2's positive integer power, such > as -2, +1, etc. ... > Since there are more equations than unknowns, sometimes there are > conflicting assignment to the variables, so only approximated assignment can > be made to minimize the mean square error between the left hand side and > right hand side in the above equation... ... > The only problem is that one the matrix gets larger, with hundreds of > unknowns, [exhaustive search] cannot be used any more... Can anybody give me more ideas and pointers? I haven't followed your other posts so don't know how typical the above example is of the structures of the real problem; if it is at all typical, you would be better off treating it as a set of independent equations. Making obvious substitutions and cancellations appears to rapidly reduce the number of equations: -jiw ==== Hey, We all know the 3X+1 conjecture is unprooved to this moment, But I was wondering if we know by proof that f.i. the 5X+1 ( or 7X+1, ...) analogue problem has divergent sequences. My guess is that such a proof could be 'simpler' and maybe is 'already out there' but i don't seem to find much about these generalizations when I google .. . Does someone know of any proof of these generalizations or can help me with some links for more informatioon about this generalizations? Thx in advance ==== Hey, We all know the 3X+1 conjecture is unprooved to this moment, But I was wondering if we know by proof that f.i. the 5X+1 ( or 7X+1, ...) analogue problem has divergent sequences. My guess is that such a proof could be 'simpler' and maybe is 'already out there' but i don't seem to find much about these generalizations when I google .. . Does someone know of any proof of these generalizations or can help me with some links for more informatioon about this generalizations? Thx in advance Kristof Clevers ==== >Last night, at about 00.55 AM, I went up on terrace and with full moon >right over my head, I tossed one rupee coin. > >It was Head. > >Today morning, I tossed again one rupee coin in blue sky. > >Again, it was Head. > >So God is telling me to file one more patent application. I have >already filed 7 patent applications. > No, God is telling you that Rosencrantz and Guildenstern are dead. ----------------------------- In the end, R&G resign themselves to their fate, although Guildenstern says, There must have been a moment, at the beginning, when we could have said -- no. But somehow we missed it. Perhaps. ----------------------------- No? To what? Is it beginning? Or is it end? In the beginning, we are never aware of what is going to happen in future. By this time, I am sure that absolutely everything is setup. What is happening now, it could have happened when I was in Mumbai for about 50 days and it could have happened when I had about Rs.50,000/- I ran to Mumbai for eight times, filed 7 patent application. And now when, I am completely devastated, running Rs.40,000/- negative, I have got complete understanding of this mechanism of gravity. Destiny. It is already written, well scripted. This entire Universe is a stage and we are just Actors. Somekind of drama is taking place and it has been shown to me that, yes, everything is setup. But He will never tell me what is scripted in future. It is up to God now, I said it and I have dropped envelope containing my patent application no.8 in post box just short time ago. It is going by ordinary post. Patent office may never receive it or even if they receive, those people may never issue any receipt. I will never know. For my beloved Gravity... ------------------------------------ tadap tadapke is dil se Aah nikalti rahi mujhko saza di pyaar Ki aisa kya gunaah kiya to lut gaye haan lut gaye to lut gaye Hum teri mohabbat mein ----------------------------------- Tomorrow morning, I will start execution sequence. As soon as I see this Action Device hanging in air, I will disclose it and mechanism to local media, people. I am going to follow my deadline 14 January 04.14 PM(IST). I just can't hang anymore... Apocalypse NOW!! -Abhi. ==== >I ran to Mumbai for eight times, filed 7 patent application. And now >when, I am completely devastated, running Rs.40,000/- negative, I have >got complete understanding of this mechanism of gravity. Rs.? Are you talking about Indian Rupees? I hope not, because that works out to about US$880 that you're in debt. Not exactly the Enron collapse, Dude. -- V.G. People are more violently opposed to fur than leather, because it is easier to harrass rich women than it is motorcycle gangs. - Bumper Sticker (This sig file contains not less than 80% recycled SPAM) Sarcasm is my sword, Apathy is my shield. ==== I met the following problem i one book,which can be solved either by measure theory or by Lebesque Dominated Convergence theorem,but the author says it is too difficult to handle without these means .I think I need help on it. Let fn:[0,1]--->[0,1] are continuous functions and fn--->0 for each x in [0,1].Then Int(fn(x)dx,0,1)--->0! Mladen ==== I met the following problem i one book,which can be solved either by > measure theory or by Lebesque Dominated Convergence theorem,but the > author says it is too difficult to handle without these means .I think > I need help on it. Let fn:[0,1]--->[0,1] are continuous functions and fn--->0 for each x > in [0,1].Then Int(fn(x)dx,0,1)--->0! > I remember seeing a problem like this in one of the early chapters in Real and Complex Analysis by Rudin. My memory is that it also contained a reference to a paper which had the solution to this problem. But it was about 20 years ago when I saw this, so my memory might be faulty. ==== > I met the following problem i one book,which can be solved either by >measure theory or by Lebesque Dominated Convergence theorem,but the >author says it is too difficult to handle without these means .I think >I need help on it. Let fn:[0,1]--->[0,1] are continuous functions and fn--->0 for each x >in [0,1].Then Int(fn(x)dx,0,1)--->0! Define f_n for n >= 2 by 2 f (x) = n x for x in [0,1/n] n = n(2 - nx) for x in [1/n,2/n] = 0 for x in [2/n,1] Each f_n is continuous on [0,1] and f_n(x) -> 0 for each x in [0,1], yet, for all n >= 2, |1 | f (x) dx = 0 | 0 n All the f_n are dominated by 1/x, but 1/x is not in L^1[0,1]. There is no function in L^1[0,1] that dominates all the f_n. Rob Johnson take out the trash before replying ==== >> I met the following problem i one book,which can be solved either by >>measure theory or by Lebesque Dominated Convergence theorem,but the >>author says it is too difficult to handle without these means .I think >>I need help on it. >>Let fn:[0,1]--->[0,1] are continuous functions and fn--->0 for each x >>in [0,1].Then Int(fn(x)dx,0,1)--->0! >Define f_n for n >= 2 by > 2 > f (x) = n x for x in [0,1/n] > n > = n(2 - nx) for x in [1/n,2/n] = 0 for x in [2/n,1] He did say fn:[0,1]--->[0,1]. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== I met the following problem i one book,which can be solved either by > measure theory or by Lebesque Dominated Convergence theorem,but the > author says it is too difficult to handle without these means .I think > I need help on it. Let fn:[0,1]--->[0,1] are continuous functions and fn--->0 for each x > in [0,1].Then Int(fn(x)dx,0,1)--->0! It is false as stated. See if you can find a counterexample. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ ==== >> I met the following problem i one book,which can be solved either by >> measure theory or by Lebesque Dominated Convergence theorem,but the >> author says it is too difficult to handle without these means .I think >> I need help on it. Help on what? The measure theory, the Lebesgue Dominated Convergence Theorem, or the too difficult way? >> Let fn:[0,1]--->[0,1] are continuous functions and fn--->0 for each x >> in [0,1].Then Int(fn(x)dx,0,1)--->0! >It is false as stated. See if you can find a counterexample. I see nothing to complain about here except the grammar, unless you're reading ! as factorial, or taking fn as a net instead of a sequence. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== The page with links to mathematical departments in the world http://www.math.hr/~duje/mathdept.html now contains links to 1003 departments from 93 countries. Any comments and suggestions are very welcomed. Andrej Dujella ==== Is there an efficient way to solve the following problem: Given a,b,eps>0 find all positive n of being integer. For example, let a=sqrt(2), b=sqrt(3) and eps=10^-k. I've compiled a table of > n that work for various small values of k. Is it complete? [The numbers are shortened: 297..072(26) is the 26 digit number starting with > 296 and ending with 072] k=13 > 297..072(26) > 617..409(26) > 915..158(26) k=14 > 394..552(28) > 483..617(28) > 685..472(28) > 774..537(28) k=15 > 135..884(30) k=16 > 218..750(32) k=17 > 173..971(34) k=18 > 506..352(36) > 546..517(36) > 972..568(36) k=19 > 131..132(38) k=20 > 239..576(40) > 296..519(40) > 904..493(40) > 961..436(40) rich Just an idea: Use continued fractions, see Best Rational Approximations to Real Numbers by R. Knott http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/cfINTRO.html#converge nts Hugo Pfoertner ==== Jack Zamat > In the development of the calculus, notation was an important factor. The > typical notation of d/dt and the integral sign are from Liebntiz, and > because Newton had such jacked up notation, England fell behind the rest of > continental Europe in analysis for over 100 yrs. My question is this: I know > that Newton used the dot notation for differentiation, but what type of > notation did he use for the integral/integration? If anybody knows the anser > or can point me to a source, I would be most appreciative. A translation of Newton's Principia would tell the tale, assuming that the translator has not modernized the notation. If I remember correctly, he had no explicit notation for integrals. If he wanted to say (for example) that x^3 / 3 is the integral (inverse fluxion) of x^2, he would phrase it like x^2 will be the fluxion of x^3 / 3. I do know that he used a whole long series of ingenious but rather arbitrary synthetic constructions, in the style of Euclid and Archimedes, in his problems on integration. LH <3ffadc52$20$fuzhry+tra$mr2ice@news.patriot.net> <1g75muz.1d66n4i14gjvdlN%panoptes@iquest.net> <1g75yr4.1si6z861lg1554N%panoptes@iquest.net> <87brpe1wtx.fsf@phiwumbda.org> <878ykhzlss.fsf@phiwumbda.org> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft ==== In , on 01/09/2004 at 11:31 AM, Russell Easterly said: >The obvious flaw in this argument What is obvious to you and what is true are two very different things. >is that every real number has a definition. You're making a pun, not presenting a logical argument. The statement that there are only a finite number of definable real numbers refers to a specific definition of definable, and it is not the definition that you are using. >It is hard to imagine That is a flaw in your imagination. >Doesn't this mean that every real number is computable? No. Again, you're making a pun instead of an argument. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org <3ffd3edf$21$fuzhry+tra$mr2ice@news.patriot.net> <61706c80.0401100837.7618ab15@posting.google.com> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft X-Terminate: SPA(GIS) ==== In <61706c80.0401100837.7618ab15@posting.google.com>, on 01/10/2004 at 08:37 AM, hack@watson.ibm.com (Michel Hack) said: >I enjoy Shmuel's postings -- usually he sets people straight. In >this case however I think he misremembers what Goedel did. G.9adel did several things. One of them was to translate staements about apparently stronger systems into statements about naturals. That lead, among other things, to his incompleteness result, but I'm pretty sure that he also proved con(PA) => con(ZF). -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org <3ffd3edf$21$fuzhry+tra$mr2ice@news.patriot.net> <61706c80.0401100837.7618ab15@posting.google.com> <40019929$48$fuzhry+tra$mr2ice@news.patriot.net> ==== > In <61706c80.0401100837.7618ab15@posting.google.com>, on 01/10/2004 > at 08:37 AM, hack@watson.ibm.com (Michel Hack) said: >I enjoy Shmuel's postings -- usually he sets people straight. In >>this case however I think he misremembers what Goedel did. G.9adel did several things. One of them was to translate staements about > apparently stronger systems into statements about naturals. That lead, > among other things, to his incompleteness result, but I'm pretty sure > that he also proved con(PA) => con(ZF). Pretty sure? As a result of pretty sure, you advised me thus? ,---- | Well, it takes more machinery. Read a Scientific American papaer | called Goedel's Proof for background, then read On the Consistency | of the Axiom of Choice and the Generalized Continuum Hypotheses. `---- This after saying (in the same post), [...] but it's old news. Google for G.9adel. I suppose it's good that you're stating your opinions with a bit of humility now, but it would've been better if you had expressed that you're pretty sure all along. I made the mistake of taking your word for this claim in a couple of posts (although, I always made clear it was hearsay for me). I did this despite your annoying arrogance and condescension just because you seemed confident of the result of which you're now pretty sure. I suppose I learned my lesson. references unless you are certain that they contain the result of interest. For extra points, it would be swell if you were certain that *some* paper contain the result of interest. If I hadn't been too damn lazy to look up Goedel's original, I'd be a touch pissed. Aside: You write, One of them was to translate staements about apparently stronger systems into statements about naturals. That lead, among other things, to his incompleteness result. What apparently stronger system was relevant for his proof of the incompleteness theorems? Unless you take his proof as involving a translation of metamathematical statements into arithmetic, I don't know what you mean here. Personally, I wouldn't use the term translation there, since the metamathematics were not treated as a formal theory as far as I recall. Maybe you have something else in mind. -- Jesse F. Hughes [Mathematical] society has evolved far enough away from mainstream society that it has become rogue, and now is willing to push its needs against that of the majority. -- James S. Harris ==== >>Rationals are Uncountable > Then how do you explain the numerous _injective_ mappings one may > construct from Q to N? For example: An arbitary rational may be represented uniquely as p/q, where, WLOG, we > may require that: > p is an integer, and > q is a positive integer, and > p and q have no common factors except 1 and > if p = 0 then q = 1 Define f : Q -> N by > f(p,q) = 2^p*5^q if p is positive or zero, and > f(p/q) = 3^(-p)*5^q if p is negative. It is trivial to show that f is injective, thus the cardinality of the > rationals cannot be larger than that of the naturals. And, of course, the trivial injective mapping g:N -> Q : n -> n/1 > completes the issue. We already know the rationals are countable. The idea is to find the defect(s) in his proof. ==== >Also, thank you to R3769, Michel Hack and Jesse F. Hughes for correcting >>my sloppy thinking concerning the notion of inconsistency with regard to >>mathematics. I now understand that an inconsistency in ZFC does not >>mean an inconsistency in mathematics. >Well, now we must be a bit careful. Shmuel Metz informs me that >Goedel proved the relative consistency of ZFC to PA --- so that *if* >PA is consistent, *then* ZFC is consistent. Hence, an inconsistency >in ZFC would in fact yield an inconsistency in PA --- not due to the >simple fact that PA has a model in ZFC, but to the more surprising >(reported) fact that ZFC has a model in PA. >>Umm. Are you sure you're remembering that correctly? > I'm fairly sure that I'm remember what Shmuel said and am stating it > correctly. I don't know anything about the claim than I've said > above. My apologies to Shmuel if I've misrepresented things. > Having discussed this with a friend who knows more about the subject than me, we've concluded that the statement is wrong if one assumes the consistency of ZF(C) Here's a proof: Suppose you can construct a model of ZFC in PA. There is a set in ZFC whic his a model of PA, so in that set we can then construct a model of ZFC. This means we have some set in ZFC which is itself a model of ZFC, so ZFC must be consistent. This contradicts one of Goedel's theorems, as a system strong enough to prove it's own consistency must be inconsistent (if it has a model of PA in it). David ==== >> In <87smiudtrj.fsf@phiwumbda.org>, on 01/05/2004 >> at 09:29 PM, jesse@phiwumbda.org (Jesse F. Hughes) said: > I don't see that an inconsistency in ZFC would necessarily yield an > inconsistency in, say, Peano arithmetic. >> Because, having modelled ZFC in PA, an inconsistency in ZFC can be >> translated back into an inconsistency in PA. ZF with the axiom of infinity replaced by its negation is equivalent > to PA. You are helpful too!! Russell -- Luck is when the paths of opportunity and preparation cross. ==== : Its the same machine. > : Assume I have a TM that always finds a natural number larger than > : any natural number on any given finite tape. > : This same TM will try to find the largest natural on an infinite tape. > : The TM has no way of knowing the tape is infinitely long. > : This TM will always say there is a natural number larger than > : any natural it has found on the tape. What is the output of the following program if it never reaches > the end of input? I am assuming the computer can perform an infinite number of operations in finite time. > while (!end_of_input) > { > input x; > if (x>max) > max=x; > } > print max; It will print a natural number. We are assuming that every input is a natural number. max must also be a natural number. > What do you think the output of this program would be? while (true) > { > n=n+1; > } > print done; > It will print an infinitely long string of numbers followed by the word done. Russell - 2 many 2 count ==== I am assuming the computer can perform an infinite number of >operations in finite time. Yes. I suppose there is nothing *wrong* with making this assumption provided you are willing to live with it's consequences. Clearly, you will need to be more careful about what infinite number of operations means. It's unlikely that you will find many followers of your beliefs without first demonstrating some very interesting (and useful) consequences of your assumption. It's also possible you will deduce from your assumption something that is untenable, and thus have to abandon it. In any event you will have learned something, and that is always a very good thing. When I last asked about whether there were any models of computation that allowed an infinite number of operations in finite time, the answer was: no, but there are things called oracles you may want to look at. I suppose google for oracle is the best (however lame) advice I can offer. Good Luck! rich ==== > > : Its the same machine. >: Assume I have a TM that always finds a natural number larger than >: any natural number on any given finite tape. >: This same TM will try to find the largest natural on an infinite tape. >: The TM has no way of knowing the tape is infinitely long. >: This TM will always say there is a natural number larger than >: any natural it has found on the tape. > > What is the output of the following program if it never reaches >the end of input? I am assuming the computer can perform an infinite number of > operations in finite time. > while (!end_of_input) >{ >input x; >if (x>max) >max=x; >} >print max; It will print a natural number. > We are assuming that every input is a natural number. > max must also be a natural number. But, from a tape containing all natural numbers, it will never reach an end of input condition, so it will never print anything. > What do you think the output of this program would be? > > while (true) >{ >n=n+1; >} >print done; > > It will print an infinitely long string of numbers followed by > the word done. (1) It will not print any numbers, since there is no instruction to do so, and (2) it will never print done since it will be locked in an endless loop not containing the print command. > Russell > - 2 many 2 count you mean - 2 dumb 2 count <3ffadc52$20$fuzhry+tra$mr2ice@news.patriot.net> ==== > What is the output of the following program if it never reaches >the end of input? I am assuming the computer can perform an infinite number of > operations in finite time. You assume wrongly. PLEASE PLEASE PLEASE think before you post! -Arthur <3ffadc52$20$fuzhry+tra$mr2ice@news.patriot.net> <1g75muz.1d66n4i14gjvdlN%panoptes@iquest.net> <1g75yr4.1si6z861lg1554N%panoptes@iquest.net> <97adneZXNdeRbWCi4p2dnA@comcast.com> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft ==== In <97adneZXNdeRbWCi4p2dnA@comcast.com>, on 01/08/2004 at 04:04 PM, Russell Easterly said: >There is no theoretical limit on how fast something can be computed. There is, however, a theoretical limit one whether something that doesn't satisfy the definition of a TM is a TM. In <462dnZeiCeE1bGCi4p2dnA@comcast.com>, on 01/08/2004 at 04:11 PM, Russell Easterly said: >I am assuming that a TM can perform an infinite >number of operations in finite time. Such a device might have a role in the theory of computation, but it would not be a turing machine and theorems about Turing Machines would not apply to it. >It appears that induction and infinite computation come to different >conclusions on some problems. No. The different conclusions come from applying things to contexts where they are inapplicable. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org 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 ==== In <20040109140413.01689.00002352@mb-m18.aol.com>, on 01/09/2004 at 07:04 PM, r3769@aol.com (R3769) said: >So every theorem of ZFC is also a theorem of ZF. But not the same theorem. >At this point the only real problem is that we all >know the C in ZFC stands for Choice, and not, say, Cohen. The flaw is that you are assuming that the corresponding theorems are in fact identical, which they are not. Theorems in ZFC translate into very different, and more complex, theorems in ZF. >However, as near as I can figure, yes it is true that you can encode >statements about ZFC as statements about the naturals but NOT in the >way Shmuel Metz suggests. That there is an interpretation, f, from >the theory cn(PA) to the theory ZFC is a fact. You've got it backwards. What is at issue is translating statements about ZFC into statements about PA. You need to be very careful in looking at this, because the translations are not what you would naively expect. >Because interpretations are, by definition, 1-1. No. All that is required is that the interpretation satisfies the same axioms. For a more recent, and perhaps more understandable, example of that, google for Nonstandard Analysis. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== > at 07:04 PM, r3769@aol.com (R3769) said: >So every theorem of ZFC is also a theorem of ZF. But not the same theorem. >At this point the only real problem is that we all >>know the C in ZFC stands for Choice, and not, say, Cohen. The flaw is that you are assuming that the corresponding theorems are >in fact identical, which they are not. Theorems in ZFC translate into >very different, and more complex, theorems in ZF. > Yes, I agree the argument is flawed. What I can't figure out is how theorems in ZFC translate into theorems of ZF, yet axioms can't be so translated. >>However, as near as I can figure, yes it is true that you can encode >>statements about ZFC as statements about the naturals but NOT in the >>way Shmuel Metz suggests. That there is an interpretation, f, from >>the theory cn(PA) to the theory ZFC is a fact. You've got it backwards. What is at issue is translating statements >about ZFC into statements about PA. You need to be very careful in >looking at this, because the translations are not what you would >naively expect. > True enough. I would add that translating theorems in ZFC to theorems in PA was also an issue. >>Because interpretations are, by definition, 1-1. No. All that is required is that the interpretation satisfies the same >axioms. Ok. I clearly need to think about this some more. . rich 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 X-Terminate: SPA(GIS) ==== In <20040108220055.26331.00002340@mb-m15.aol.com>, on 01/09/2004 at 03:00 AM, r3769@aol.com (R3769) said: >Another question: Do you mean to say that ZFC is inconsistent implies >ZF is inconsistent or that ZF can't be modelled in PA? Kurt G.9adel proved that. Likewise if ZF plus the Generalized Continuum Hypothesis is inconsistent then ZF is inconsistent. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org 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$oft ==== In <20040108180252.29555.00001953@mb-m02.aol.com>, on 01/08/2004 at 11:02 PM, r3769@aol.com (R3769) said: >However, the *relative* inconsistency of >PA isn't really what we are concerned about. For mathematics to be >doomed, you still need to show the inconsistency of ALL other set >theories, ST, where ST can be modelled in PA. If PA is inconsistent, then anything more powerful than PA is inconsistent. So proving the inconsistency of anything modeled in PA would indeed be a fatal blow. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org <87smiudtrj.fsf@phiwumbda.org> <3FF9DD0F.CC42A336@mdli.com> <87isjmyzmw.fsf@phiwumbda.org> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft ==== In <87isjmyzmw.fsf@phiwumbda.org>, on 01/08/2004 at 09:05 PM, jesse@phiwumbda.org (Jesse F. Hughes) said: >I won't get into that can of worms. I resolve that by claiming to be a Formalist on odd numbered days and a Platonist on even numbered days; I'm not really happy with either position. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org <87smiudtrj.fsf@phiwumbda.org> <3ffadc1f$19$fuzhry+tra$mr2ice@news.patriot.net> <87smisrg0b.fsf@phiwumbda.org> <3ffd41a6$22$fuzhry+tra$mr2ice@news.patriot.net> <871xqa1n9s.fsf@phiwumbda.org> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft ==== In <871xqa1n9s.fsf@phiwumbda.org>, on 01/08/2004 at 04:19 PM, jesse@phiwumbda.org (Jesse F. Hughes) said: >I know who Goedel is, thanks. This is not sci.math.biography; I was suggesting that you read about his work, not his identity. >What does the consistency of the axiom of choice have to do with >this? Nothing. Why do you assume that a paper will contain only the result summarized in its title? >Is this where he models ZFC in PA? papers, Paul Cohen's book may still be in print. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org <87smiudtrj.fsf@phiwumbda.org> <3ffadc1f$19$fuzhry+tra$mr2ice@news.patriot.net> <87smisrg0b.fsf@phiwumbda.org> <3ffd41a6$22$fuzhry+tra$mr2ice@news.patriot.net> <871xqa1n9s.fsf@phiwumbda.org> <40009ae6$11$fuzhry+tra$mr2ice@news.patriot.net> ==== > In <871xqa1n9s.fsf@phiwumbda.org>, on 01/08/2004 > at 04:19 PM, jesse@phiwumbda.org (Jesse F. Hughes) said: >What does the consistency of the axiom of choice have to do with >>this? Nothing. Why do you assume that a paper will contain only the result > summarized in its title? I assume that every result in the paper has some relation to the main topics, even if it is only illustrative of the methods. Or does this paper also have a section on the mating habits of aardvarks? >>Is this where he models ZFC in PA? papers, Paul Cohen's book may still be in print. How is it that, of all the participants in this thread, you are the only one who claims Goedel proved Con(PA) -> Con(ZFC)? This is an honest question. This result, if true, is surely *at least* as important and interesting as the consistency of the axiom of choice. Maybe you should be less condescending and ask yourself if you're absolutely sure that this result is found in Goedel's work. If it is, it would be very enlightening if you could sketch the development, as I surely cannot spend time on researching it myself at present. Even barring that, a reference to any other work (preferably academic) which claims Con(PA) -> Con(ZFC) is contained, even implicitly, in the publications of Goedel would be useful. As it is, I think I can't take your assertions at face value, although I did at first. I did so because I'm admittedly not familiar with the Goedel paper to which you refer, and you seemed pretty certain of the result. However, given that others have questioned this claim, I'll simply wait until more evidence comes in. A more ambitious lad than I am, or one with more time or for whom these claims are close to his current research interests, would look up the original sources, as you suggest, but I won't at present. -- All intelligent men are cowards. The Chinese are the world's worst fighters because they are an intelligent race[...] An average Chinese child knows what the European gray-haired statesmen do not know, that by fighting one gets killed or maimed. -- Lin Yutang ==== On Sun, 11 Jan 2004 10:35:20 +0100, jesse@phiwumbda.org (Jesse F. > In <871xqa1n9s.fsf@phiwumbda.org>, on 01/08/2004 >> at 04:19 PM, jesse@phiwumbda.org (Jesse F. Hughes) said: >What does the consistency of the axiom of choice have to do with >this? >> Nothing. Why do you assume that a paper will contain only the result >> summarized in its title? I assume that every result in the paper has some relation to the main >topics, even if it is only illustrative of the methods. Or does this >paper also have a section on the mating habits of aardvarks? >>Is this where he models ZFC in PA? >> papers, Paul Cohen's book may still be in print. How is it that, of all the participants in this thread, you are the >only one who claims Goedel proved Con(PA) -> Con(ZFC)? This is an >honest question. This result, if true, is surely *at least* as >important and interesting as the consistency of the axiom of choice. Maybe you should be less condescending and ask yourself if you're >absolutely sure that this result is found in Goedel's work. If it is, >it would be very enlightening if you could sketch the development, as >I surely cannot spend time on researching it myself at present. Even >barring that, a reference to any other work (preferably academic) >which claims Con(PA) -> Con(ZFC) is contained, even implicitly, in the >publications of Goedel would be useful. I've been waiting for someone who really knows this stuff to speak up, but nobody has (or if anyone has, he's also included disclaimers about how he could be wrong...) So, although I'm not an expert, I'll say what I think happened here. Note that anything I say could be wrong, but I think I can explain how it could be that Shmuel came to believe that Goedel proved Con(PA) implies Con(ZF), even though Godel proved no such thing. Note that the last clause is supposed to be within the scope of the it could be that - I'm not _asserting_ that Goedel did no such thing, although I'm pretty sure he didn't. It seems clear he didn't, since Con(PA) _is_ a theorem of ZF (I think, in a suitable sense), while Godel _did_ prove that Con(ZF) is not a theorem of ZF... but there are subtleties in all this sort of thing so I could be wrong. One can give formulas of ZF Godel numbers, formalize the notion of proof, etc. So the statement that a given sequence of formulas is a valid proof in ZF is equivalent to a certain statement about the natural numbers. Ie, although of course it doesn't work out this way: Could be that n is the Godel number of an axiom of ZF if and only if n is congruent to 6 mod 10, while A follows from B and C by modus ponens if and only if the corresponding Goedel numbers satisfy b = a*c. So saying that a sequence of formulas is a valid proof in ZF is the same as saying that a certain sequence of integers has the property that [*] each one is either congruent to 6 mod 10 or is the quotient of two preceding integers on the list. Of course that's not how it works, but there _is_ a true statement that's conceptually the same as the preceding paragraph, just _much_ more complicated. So in some sense we've reduced statements about sets to statements about natural numbers. If one is not careful about various things one could easily jump from here to the statement that if PA is consistent then so is ZF, since in some sense we're reduced ZF to arithmetic. But Con(PA) implies Con(ZF) certainly does not follow from the above arithmetization of ZF. Because exactly the same arithmetization as above can be performed with _any_ theory in place of ZF, consistent or not! For those of us who still don't get it: Suppose for the sake of argument that the Godel number of not P is 7 times the Goedel number of P. If ZF is inconsistent then there is a _valid_ proof in ZF of P and also a valid proof of not P, for some P. After arithmetization as above, that means there is a valid sequence of Goedel numbers that ends with 42, and another valid sequence of Godel numbers that ends with 7*42. The existence of those two sequences of numbers, both satisfying [*], shows that ZF is inconsistent, but it does not show that PA is inconsistent - they _are_ sequences of integers which satisfy [*], so what? The fact that we have two sequences satisfying [*], one of which ends in 42 and one of which ends in 7*42, is not a proof of 0 = 1 from the axioms of PA. >As it is, I think I can't take your assertions at face value, although >I did at first. I did so because I'm admittedly not familiar with the >Goedel paper to which you refer, and you seemed pretty certain of the >result. However, given that others have questioned this claim, I'll >simply wait until more evidence comes in. A more ambitious lad than I >am, or one with more time or for whom these claims are close to his >current research interests, would look up the original sources, as you >suggest, but I won't at present. ************************ David C. Ullrich ==== On Sun, 11 Jan 2004 06:40:08 -0600, David C. Ullrich >On Sun, 11 Jan 2004 10:35:20 +0100, jesse@phiwumbda.org (Jesse F. >> In <871xqa1n9s.fsf@phiwumbda.org>, on 01/08/2004 > at 04:19 PM, jesse@phiwumbda.org (Jesse F. Hughes) said: >What does the consistency of the axiom of choice have to do with >>this? >> Nothing. Why do you assume that a paper will contain only the result > summarized in its title? >>I assume that every result in the paper has some relation to the main >>topics, even if it is only illustrative of the methods. Or does this >>paper also have a section on the mating habits of aardvarks? >>Is this where he models ZFC in PA? >> papers, Paul Cohen's book may still be in print. >>How is it that, of all the participants in this thread, you are the >>only one who claims Goedel proved Con(PA) -> Con(ZFC)? This is an >>honest question. This result, if true, is surely *at least* as >>important and interesting as the consistency of the axiom of choice. >>Maybe you should be less condescending and ask yourself if you're >>absolutely sure that this result is found in Goedel's work. If it is, >>it would be very enlightening if you could sketch the development, as >>I surely cannot spend time on researching it myself at present. Even >>barring that, a reference to any other work (preferably academic) >>which claims Con(PA) -> Con(ZFC) is contained, even implicitly, in the >>publications of Goedel would be useful. I've been waiting for someone who really knows this stuff to speak >up, but nobody has (or if anyone has, he's also included disclaimers >about how he could be wrong...) So, although I'm not an expert, >I'll say what I think happened here. Note that anything I say could >be wrong, but I think I can explain how it could be that >Shmuel came to believe that Goedel proved Con(PA) implies >Con(ZF), even though Godel proved no such thing. the it could be that - I'm not _asserting_ that Goedel did no >such thing, although I'm pretty sure he didn't. It seems clear >he didn't, since Con(PA) _is_ a theorem of ZF (I think, >in a suitable sense), while Godel _did_ prove that Con(ZF) >is not a theorem of ZF... but there are subtleties in all >this sort of thing so I could be wrong. >One can give formulas of ZF Godel numbers, formalize the >notion of proof, etc. So the statement that a given sequence >of formulas is a valid proof in ZF is equivalent to a certain >statement about the natural numbers. Ie, although of course it doesn't work out this way: >Could be that n is the Godel number of an axiom >of ZF if and only if n is congruent to 6 mod 10, while >A follows from B and C by modus ponens if and >only if the corresponding Goedel numbers satisfy >b = a*c. So saying that a sequence of formulas is >a valid proof in ZF is the same as saying that a >certain sequence of integers has the property that [*] each one is either congruent to 6 mod 10 or is >the quotient of two preceding integers on the list. Of course that's not how it works, but there _is_ a >true statement that's conceptually the same as the >preceding paragraph, just _much_ more complicated. >So in some sense we've reduced statements about >sets to statements about natural numbers. If one >is not careful about various things one could easily >jump from here to the statement that if PA is >consistent then so is ZF, since in some sense >we're reduced ZF to arithmetic. But Con(PA) implies Con(ZF) certainly does not >follow from the above arithmetization of ZF. >Because exactly the same arithmetization as above >can be performed with _any_ theory in place of >ZF, consistent or not! For those of us who still don't get it: Suppose for >the sake of argument that the Godel number of >not P is 7 times the Goedel number of P. If >ZF is inconsistent then there is a _valid_ proof >in ZF of P and also a valid proof of not P, for >some P. After arithmetization as above, that >means there is a valid sequence of Goedel >numbers that ends with 42, and another valid >sequence of Godel numbers that ends with >7*42. The existence of those two sequences >of numbers, both satisfying [*], shows that >ZF is inconsistent, but it does not show that >PA is inconsistent - they _are_ sequences >of integers which satisfy [*], so what? The >fact that we have two sequences satisfying >[*], one of which ends in 42 and one of which >ends in 7*42, is not a proof of 0 = 1 from the >axioms of PA. And then come to think of it, carrying this a step farther leads to something which someone might even more plausibly misunderstand or misremember as saying that Con(PA) implies Con(ZF): Encoding sequences of integers as integers, the statement ZF is not consistent is equivalent to [**] There exists a positive integer n such that [some complicated condition], _where_ the condition only involves existential quantifiers - that's not precisely stated, but in fact that condition satisfies some technical condition, so that in fact if there is an integer satisfying that condition then this fact can be proved in PA just by exhibiting n and verifying that it satisfies the condition. So it's correct to say There is no proof in PA of not Con(ZF) implies Con(ZF). >>As it is, I think I can't take your assertions at face value, although >>I did at first. I did so because I'm admittedly not familiar with the >>Goedel paper to which you refer, and you seemed pretty certain of the >>result. However, given that others have questioned this claim, I'll >>simply wait until more evidence comes in. A more ambitious lad than I >>am, or one with more time or for whom these claims are close to his >>current research interests, would look up the original sources, as you >>suggest, but I won't at present. >************************ David C. Ullrich ************************ David C. Ullrich <87smiudtrj.fsf@phiwumbda.org> <3ffadc1f$19$fuzhry+tra$mr2ice@news.patriot.net> <87smisrg0b.fsf@phiwumbda.org> <3ffd41a6$22$fuzhry+tra$mr2ice@news.patriot.net> <871xqa1n9s.fsf@phiwumbda.org> <40009ae6$11$fuzhry+tra$mr2ice@news.patriot.net> <87smim7rqf.fsf@phiwumbda.org> ==== > I've been waiting for someone who really knows this stuff to speak > up, but nobody has (or if anyone has, he's also included disclaimers > about how he could be wrong...) So, although I'm not an expert, > I'll say what I think happened here. Note that anything I say could > be wrong, but I think I can explain how it could be that > Shmuel came to believe that Goedel proved Con(PA) implies > Con(ZF), even though Godel proved no such thing. [snip rest] I think that you're exactly right about how Shmuel may have misunderstood matters, assuming he did so. I was fairly explicit about what I think is necessary[1] for the translation of proofs in ZF to following: ,----[ <87u135nv7i.fsf@phiwumbda.org> ] | PA, presumably in a natural way, so that any proof of P & ~P in ZFC | would yield a proof of Q & ~Q for some formula Q of PA. `---- Unless a proof of P & ~P in ZFC translates to a proof of Q & ~Q (or something similar) in PA, the mere existence of a translation of proofs in ZFC to proofs in PA does not appear to yield Con(PA) -> Con(ZFC). Shmuel referred satisfied this condition, but I was explicit about that assumption in using presumably. I don't assume any such thing any longer. Footnotes: [1] Well, not *really* necessary in the technical sense, but (a) sufficient and (b) the most likely way that such a proof would proceed. Another way it could proceed, of course, is by proving first that Con(PA) is false, but I think I would've heard of that proof. -- Jesse Hughes Time and again, history has shown that people who think their beliefs trump reality lose, and lose badly. Luckily, I don't have to listen to you. -- James Harris on reality avoidance <87smiudtrj.fsf@phiwumbda.org> <3ffadc1f$19$fuzhry+tra$mr2ice@news.patriot.net> <87smisrg0b.fsf@phiwumbda.org> <3ffd41a6$22$fuzhry+tra$mr2ice@news.patriot.net> <871xqa1n9s.fsf@phiwumbda.org> <40009ae6$11$fuzhry+tra$mr2ice@news.patriot.net> ==== > In <871xqa1n9s.fsf@phiwumbda.org>, on 01/08/2004 > at 04:19 PM, jesse@phiwumbda.org (Jesse F. Hughes) said: >>I know who Goedel is, thanks. > This is not sci.math.biography; I was suggesting that you read about > his work, not his identity. Son, you shouldn't be insulting. I am familiar with some of his work, too, but nothing that supports your claim. -- [I]t's good for the economy to charge for intellectual property, so open source software cannot be good, while Microsoft is the most far-thinking company around and is doing it all for the good of the public. -- Linus Torvalds paraphrases Microsoft VP Craig Mundie <3ffadc52$20$fuzhry+tra$mr2ice@news.patriot.net> <1g75muz.1d66n4i14gjvdlN%panoptes@iquest.net> <3ffd4579$1$fuzhry+tra$mr2ice@news.patriot.net> <1g78quy.1a1bzig1e5uy44N%panoptes@iquest.net> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft ==== In <1g78quy.1a1bzig1e5uy44N%panoptes@iquest.net>, on 01/08/2004 at 11:55 AM, panoptes@iquest.net (Daniel W. Johnson) said: >He described a bijection between the rationals in [0,1) and the >naturals. He claimed to. Hew was wrong. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== > He claimed to. Hew was wrong. While there are several inconsistencies among his claims, I don't think that was one of them. Any non-negative integer can be written as a finite sum a1*1! + a2*2! + a3*3! + ... + an*n! where each ak is an integer such that 0 <= ak <= k. If trailing zeroes are removed from the sum, this representation is unique. (Note that the case of 0 ends up with zero terms.) One way to prove the necessary things about this representation is by induction on the maximum number of terms; the numbers from 0 to (n+1)!-1 can be represented uniquely (up to trailing zeroes) with at most n terms. Any rational in [0,1) can be written as a finite sum b1/2! + b2/3! + b3/4! + ... + bn/(n+1)! where each bk is an integer such that 0 <= bk <= k. If trailing zeroes are removed from the sum, this representation is unique. (Note that the case of 0 ends up with zero terms.) This proof is trickier. It would probably be necessary to use induction on n to show that all multiples of 1/(n+1)! in [0,1) can be represented uniquely (up to trailing zeroes) with at most n terms, and note that any rational in [0,1) has some factorial that can be used as a denominator. (Uniqueness would require either induction on the trailing terms or a combinatorial argument.) And then the bijection consists of transferring the coefficients from one sum to the other. -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W ==== In particular, you simply > misstate Turing's definition here. Turing defines a computable number as any number that can be > represented by a computable sequence. A computable sequence is the > output tape of a TM that has an infinite number of 0's and/or 1's. Turing did *not* allow an infinite number of 1's on the output tape, > so this is simply not his definition. I recently read Turing's ON COMPUTABLE NUMBERS, WITH AN APPLICATION TO THE ENTSCHEIDUNGSPROBLEM http://www.cs.umass.edu/~immerman/cs601/turingReference.html#section-2 I may have mis-read it, but it appears this is how he defined computable numbers in this paper. Computable numbers may have been defined differently in later papers. 2. Definitions. Automatic machines. If at each stage the motion of a machine (in the sense of ¤1) is completely determined by the configuration, we shall call the machine an automatic machine (or a-machine). For some purposes we might use machines (choice machines or c-machines) whose motion is only partially determined by the configuration (hence the use of the word possible in ¤1). When such a machine reaches one of these ambiguous configurations, it cannot go on until some arbitrary choice has been made by an external operator. This would be the case if we were using machines to deal with axiomatic systems. In this paper I deal only with automatic machines, and will therefore often omit the prefix a-. Computing machines. If an a-machine prints two kinds of symbols, of which the first kind (called figures) consists entirely of 0 and 1 (the others being called symbols of the second kind), then the machine will be called a computing machine. If the machine is supplied with a blank tape and set in motion, starting from the correct initial m-configuration, the subsequence of the symbols printed by it which are of the first kind will be called the sequence computed by the machine. The real number whose expression as a binary decimal is obtained by prefacing this sequence by a decimal point is called the number computed by the machine. At any stage of the motion of the machine, the number of the scanned square, the complete sequence of all symbols on the tape, and the m-configuration will be said to describe the complete configuration at that stage. The changes of the machine and tape between successive complete configurations will be called the moves of the machine. {233} Circular and circle-free machines. symbols of the first kind it will be called circular. Otherwise it is said to be circle-free. A machine will be circular if it reaches a configuration from which there is no possible move, or if it goes on moving, and possibly printing symbols of the second kind, but cannot print any more symbols of the first kind. The significance of the term circular will be explained in ¤8. Computable sequences and numbers. A sequence is said to be computable if it can be computed by a circle-free machine. A number is computable if it differs by an integer from the number computed by a circle-free machine. We shall avoid confusion by speaking more often of computable sequences than of computable numbers. I understand this to mean that a circle free machine must write an infinite number of symbols of the first type. > By the way, induction over all natural numbers has nothing to do with > feasibility and everything to do with the definition of N and with the > meaning of universal quantification. Induction does not require that > one actual go through all of the natural numbers and ensure that the > next one satisfies the proposition. The system I am describing is similar to hypercomputation. http://www.alanturing.net/turing_archive/pages/Reference%20Articles/hypercom putation/hypercomputation.html Do we need induction if we assume we can actually go through all of the natural numbers and ensure that they all satisfy the proposition? Would induction and hypercomputation always give the same results? Russell - 2 many 2 count 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 ==== In <40005FD8.9080008@comcast.net>, on 01/10/2004 at 03:26 PM, Tom Adams said: >He then shows how the metric can define a topological space. Presumably he defines the topology by taking the d-open sets as a basis. >That sounds right, but I don't see why it must be so. The >definitions of T-limit points and d-limit points are so different. If p is a d-limit point of S, p in U and U in T, then there is an epsilon>0 such that B(p,epsilon) C U. By the definition of d-limit, that ball intersects S. But then U intersects S. Conversely, is p is a T-limit, then for every epsilon>0, B(p,epsilon) in T, and therefor intersects S. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== > I'm reading an introductory book on topology and the author defines a > d-limit point p of a set S in a pseudometric space as a point having the > property that for every eps>0 there is in S a point s different than p > such that the metric d(p,s) property that d(x,y)=0 only if x=y.) In the topological space (X, T) he defines a T-limit point p of a set S > as a point having the property that every element of T containing p > meets S in a point other than p. He then shows how the metric can define a topological space. He then says when we begin with a pseudometric d on a set X, we obtain > a collection of d-open sets that is a topology T, and the T-limit points > of a set coincide with the d-limit points of a set. That sounds right, but I don't see why it must be so. The definitions > of T-limit points and d-limit points are so different. We might argue > that every open set S is the union of some collection of open sets > (which should be the same for the metric and topological spaces), but I > don't see how that shows that the T-limit points and d-limit points of S > are the same. > If x is a T-limit pt of S, then for all open U nhood x, Ux / S is nonnul. As any d-ball containing x is open nhood of x in T, it quickly follows x is a d-limit pt of S. Conversely if x is a d-limit point of S and U an open nhood x, then there's some d-ball B with x in B subset U. As x is d-limit pt of S, Bx / S is nonnul, so also Ux / S is nonnul. ==== >>I'm reading an introductory book on topology and the author defines a >>d-limit point p of a set S in a pseudometric space as a point having the >>property that for every eps>0 there is in S a point s different than p >>such that the metric d(p,s)>property that d(x,y)=0 only if x=y.) >>In the topological space (X, T) he defines a T-limit point p of a set S >>as a point having the property that every element of T containing p >>meets S in a point other than p. >>He then shows how the metric can define a topological space. >>He then says when we begin with a pseudometric d on a set X, we obtain >>a collection of d-open sets that is a topology T, and the T-limit points >>of a set coincide with the d-limit points of a set. >>That sounds right, but I don't see why it must be so. The definitions >>of T-limit points and d-limit points are so different. We might argue >>that every open set S is the union of some collection of open sets >>(which should be the same for the metric and topological spaces), but I >>don't see how that shows that the T-limit points and d-limit points of S >>are the same. If x is a T-limit pt of S, then for all open U nhood x, Ux / S is > nonnul. As any d-ball containing x is open nhood of x in T, it quickly > follows x is a d-limit pt of S. > It seems the key point is switching between the T and d point of view by noticing that any d-ball is T-open (because the metric d determines openness in the metric space and in the derived topology T). > Conversely if x is a d-limit point of S and U an open nhood x, then > there's some d-ball B with x in B subset U. As x is d-limit pt of S, > Bx / S is nonnul, so also Ux / S is nonnul. Here the key point seems to be that a d-open set is a T-open set (is an element of T) so the d-type operations on d-open sets give us information about T-limit pts. explain this. I looked at another text and this interesting topic wasn't even discussed. Maybe I need a better book. Tom Adams <40014E86.4060308@comcast.net> ==== > >>I'm reading an introductory book on topology and the author defines a >>d-limit point p of a set S in a pseudo-metric space as a point having the >>property that for every eps>0 there is in S a point s different than p >>such that the metric d(p,s)>property that d(x,y)=0 only if x=y.) >In the topological space (X, T) he defines a T-limit point p of a set S >>as a point having the property that every element of T containing p >>meets S in a point other than p. >He then shows how the metric can define a topological space. >>He then says when we begin with a pseudometric d on a set X, we obtain >>a collection of d-open sets that is a topology T, and the T-limit points >>of a set coincide with the d-limit points of a set. >That sounds right, but I don't see why it must be so. The definitions >>of T-limit points and d-limit points are so different. We might argue >>that every open set S is the union of some collection of open sets >>(which should be the same for the metric and topological spaces), but I >>don't see how that shows that the T-limit points and d-limit points of S >>are the same. > > If x is a T-limit pt of S, then for all open U nhood x, Ux / S is >nonnul. As any d-ball containing x is open nhood of x in T, it quickly >follows x is a d-limit pt of S. It seems the key point is switching between the T and d point of > view by noticing that any d-ball is T-open (because the metric d > determines openness in the metric space and in the derived topology T). > Conversely if x is a d-limit point of S and U an open nhood x, then >there's some d-ball B with x in B subset U. As x is d-limit pt of S, >Bx / S is nonnul, so also Ux / S is nonnul. Here the key point seems to be that a d-open set is a T-open set (is an > element of T) so the d-type operations on d-open sets give us > information about T-limit pts. explain this. I looked at another text and this interesting topic > wasn't even discussed. Maybe I need a better book. > You're welcome. Don't despair quite yet, did he leave such proofs as exercises for the student in the problem section? Did the author explain why the d-open balls for metric or pseudo-metric is a topology? The hard part is to show the intersection of two balls is open. ==== > >>I'm reading an introductory book on topology and the author defines a >>d-limit point p of a set S in a pseudo-metric space as a point having the >>property that for every eps>0 there is in S a point s different than p >>such that the metric d(p,s)>property that d(x,y)=0 only if x=y.) >>In the topological space (X, T) he defines a T-limit point p of a set S >>as a point having the property that every element of T containing p >>meets S in a point other than p. >>He then shows how the metric can define a topological space. >>He then says when we begin with a pseudometric d on a set X, we obtain >>a collection of d-open sets that is a topology T, and the T-limit points >>of a set coincide with the d-limit points of a set. >>That sounds right, but I don't see why it must be so. The definitions >>of T-limit points and d-limit points are so different. We might argue >>that every open set S is the union of some collection of open sets >>(which should be the same for the metric and topological spaces), but I >>don't see how that shows that the T-limit points and d-limit points of S >>are the same. >>If x is a T-limit pt of S, then for all open U nhood x, Ux / S is >nonnul. As any d-ball containing x is open nhood of x in T, it quickly >follows x is a d-limit pt of S. >>It seems the key point is switching between the T and d point of >>view by noticing that any d-ball is T-open (because the metric d >>determines openness in the metric space and in the derived topology T). >Conversely if x is a d-limit point of S and U an open nhood x, then >there's some d-ball B with x in B subset U. As x is d-limit pt of S, >Bx / S is nonnul, so also Ux / S is nonnul. >>Here the key point seems to be that a d-open set is a T-open set (is an >>element of T) so the d-type operations on d-open sets give us >>information about T-limit pts. >>explain this. I looked at another text and this interesting topic >>wasn't even discussed. Maybe I need a better book. You're welcome. Don't despair quite yet, did he leave such proofs as > exercises for the student in the problem section? > Don't see it in the exercises for this and the next section. He may get to it later. I'll be OK with the generous help I get from this group. > Did the author explain why the d-open balls for metric or pseudo-metric is > a topology? Yup, very interesting, too. > The hard part is to show the intersection of two balls is > open. He does that indirectly by showing every set is T-open iff it is an element of the topology T. So d-open balls are T-open, and are members of the topology T; and the intersection of two of them is still a member of the topology T and therefore open. Tom Adams ==== > >>the form Adot = f(x,p)..where x and p are the states of this dynamical >>system and f is a nonlinear function in x, p. If everything is real you might want to write A as tilde B B , namely as > the product of a matrix and its transpose. Whatever the resulting equations, > A will be guranteed symmetric positive-definite. Perhaps you could make > B upper-triangular for simplicity. > And most likely, you can rewite your differential equation for A into one for B; then you can work with B only, never need any checks, and form A at the very end... Arnold Neumaier ==== > Let me try to pose the problem, I am trying to solve. I have a matrix, A (for simplicity let us assume it is a 2 X 2 matrix), that > is supposed to be symmetric and positive definite. There is a differential > equation the numerical solution of which gives me A at a current instant, in > the form Adot = f(x,p)..where x and p are the states of this dynamical > system and f is a nonlinear function in x, p. Since A has only three unique > parameters for this case, I actually obtain, theta_dot = g(x,p), where theta > A(t) thereby enforcing symmetry..... To enforce positive definiteness, I > need to do the following, theta(1) > 0, theta(2) > 0 and -sqrt(theta(1)*theta(2)) < theta(3) < > sqrt(theta(1)*theta(2))... Then by using some sort of a projection algorithm, I could enforce positive > definiteness in this way, however, this doesnt scale well for higher > dimensioned matrices. If the indefiniteness is simply caused by roundoff, multiplying the diagonal after each integration step by 1.00000001 should cure the problem (assuming double precision). If not, something is likely to be wrong with your modeling... Arnold Neumaier ==== > En el mensaje:f0e1779.0401100411.20dab4d@posting.google.com, > shoe escribi.97: >I'm doing som math and came to this >question about graphs. > > The Graph y= f(x) has a symmetry line x=3 > > a) decide f(7), if f(-1) = 0 >b) decide f(5), if f(1) = 3 >c) Great the Graph > > I have calculated that a) should be 0 >and b) should be 3. > > what I wonder is how do I create a graph? (question (c)) >How do I create the equation like (y= -2x^2 +2x +1)? >I do need that one I think to calculate all the Points in the graph. > > Please advice a newbe from Sweden! I think that in some way it is implicit that you can seek a quadratic > function. Then, as x = 3 is a simetry line, the equation must be y = a(x - 3)^2 + b Replace x by 7 or -1 (it is the same) and y by 0, and you get a eqution in a > and b. Do the same with x = 5 or 1 and y = 3, and you get a system with two > equations with two unknowns, a and b. I'm still a bit blur about how to precede. You say that I get two unknowns, a and b. what I wanted was the equation to put into my calculator to get the Graph written. I dont understand this high match. Can you explain a bit more? However I did do what you suggested and calculatet a bit with your equation. y = a(x - 3)^2 + b dunno how to get it right however. ==== En el mensaje:f0e1779.0401111209.1d82cf5d@posting.google.com, shoe escribi.97: > Ignacio Larrosa Ca.96estro En el mensaje:f0e1779.0401100411.20dab4d@posting.google.com, >> shoe escribi.97: > I'm doing som math and came to this > question about graphs. >> The Graph y= f(x) has a symmetry line x=3 >> a) decide f(7), if f(-1) = 0 > b) decide f(5), if f(1) = 3 > c) Great the Graph > I have calculated that a) should be 0 > and b) should be 3. >> what I wonder is how do I create a graph? (question (c)) > How do I create the equation like (y= -2x^2 +2x +1)? > I do need that one I think to calculate all the Points in the > graph. >> Please advice a newbe from Sweden! >> I think that in some way it is implicit that you can seek a quadratic >> function. Then, as x = 3 is a simetry line, the equation must be >> y = a(x - 3)^2 + b >> Replace x by 7 or -1 (it is the same) and y by 0, and you get a >> eqution in a and b. Do the same with x = 5 or 1 and y = 3, and you >> get a system with two equations with two unknowns, a and b. > I'm still a bit blur about how to precede. You say that I get two unknowns, a and b. > what I wanted was the equation to put into > my calculator to get the Graph written. I dont understand this high match. Can you > explain a bit more? However I did do what you suggested > and calculatet a bit with your equation. > y = a(x - 3)^2 + b > dunno how to get it right however. If you must to look for a quadratic function f(x) = p*x^2 + q*x + r, and it must be simmetric with respect to the line x = 3, it must be f(x) = a(x - 3)^2 + b To see that, equate f(3 - t) = f(3 + t), as x = 3 is an axis of simmetry. f(3 - t) = p(3 - t)^2 + q(3 - t) + r = p*t^2 - (6p + q)t + 9p + 3q + r f(3 + t) = p(3 + t)^2 + q(3 + t) + r = p*t^2 + (6p + q)t + 9p + 3q + r f( 3 - t) = f(3 + t) for all t ===> -(6p + q) = 6p + q ==> q = - 6p ==> f(x) = p*x^2 - 6p*x + r = p*x^2 - 6p*x + 9p + r - 9p = p(x - 3)^2 + (r - 9p) = a*(x - 3)^2 + b Then, As f(-1) = 0 ===> 16a + b = 0 ====> b = -16a As f(1) = 3 ===> 4a + b = 3 ===> 4a - 16a = -12a = 3 ===> a = -1/4 , b = 4 f(x) = (-1/4)(x - 3)^2 + 4 -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com ==== Imagine yourself in space and look at this third rock from Sun, Earth. Known most intelligent living being named human being rules this Planet, 6 billion in number. For millions of years, we lived in jungle, caves. We had to go through endless sufferings. When we gained consciousness, we began to ask ourselves, where from we came, what is going on around us. Our quest to understand nature began. We have come a long way in this quest of understanding our universe, God. When I am typing this, Spirit is exploring Mars. But if radius of our Universe is 1 trillion Kilometer, we can not go a single millimeter in our entire lifetime using Newton's Third Law. If Einstein is right, we can not travel FTL. Then why creator of this universe has created those billions of Galaxies? Where things have gone wrong? You are trying to understand quantum mechanics, Newtonian mechanics, Relativity rather than trying to understand our Nature and God. This is the biggest mistake you are doing. You are caught in jungle of knowledge and you have abandoned that supreme force, God. You are physics theory, you must not talk about God! Why not? Because there is no proof of such supernatural thing. No proof? Things are hidden, life is created and we are forced to find those hidden things. This is happened thousands of times, in every discovery, every invention. We can find something only when someone hides it or lost by ourselves. Certainly every discovery or invention was not lost by us in past. So definitely, someone has hidden it long before creation of Life and Universe. Why don't you people understand such simple logic. entire knowledge built over thousands of years comes down crashing and you will be forced to spend your days and nights on streets without shelter, like those people surviving Earthquake. Like those homeless people, you will be theoryless. You are going to see awesome power of Gravity. Through this Action Device, you are going to see none other than that Supreme Force, Gravity, God in Action. Watch Out... -Abhi. (BTW, why don't you see a thread, Action Device Working Model, Part II in sci.math NG. I learned about this thread in the night of January 06. I really never see any message in sci.math NG. Some text from this thread is given below.) ------------------------------------------------------ This is what has happened until now... > called action device, I decided to give it a try in all fairness. > After all, everyone is saying it should be so easy to build. So I got > myself some wood, steel angles, springs and bolts and built it. And > guess what, it really works! > I climbed on the roof (I didn't want the thing sticking to the > ceiling and then having to do a lot of explaining), I stretched my > arms, and let go. Wow, it imediately started moving in an orthogonal > angle to the plain, as had been described. The stupid thing is I must > have been holding it upside down or something, because instead of > going off into space, it shot straight downwards and landed on the > driveway. At that moment my mom's truck drove over it and reduced it > to a miserable heap of splinters... So I'm going to have to wait to > get new materials and give it a new try. I'll give you the news as > soon as I get to it. PART II accident, I got fresh materials and fresh insights. I built a new action device with a few changes however. First the angle of the device is a lot sharper, about 45 degrees, this should provide for greater push, then I painted marks on the thing to make sure which part is up and which one is down (figuring that out wasn't easy though). I must tell you of something I had already observed the first time around which I had kept to myself because it was an unconfirmed observation: When I tried to install the springs, for some strange reason, they resisted being stretched! I felt there where undiscovered forces in action even before completion of the device! After climbing to the roof I gave the thing a nice shove, just to make sure it landed on the lawn if something went wrong this time. Again, it worked ! the thing started moving up and away from me. Then fate struck again, I haven't figured out what exactly happened, my guess is that point d detacthed itself and continued the trip without the rest of the device, or something. I'll tell you when I find out. However there is something else I wan't to share with you, I carefully observed the trajectory of the device from the time I released it to the time it landed and found it was a parable. I'm going to try and figure the implicatins this has, but I already consider it a very interesting and revolutionary fact in itself. Greetings Guenther. ------------------------------------------------------------------- ==== Dear Abhi: ... > millimeter in our entire lifetime using Newton's Third Law. If > Einstein is right, we can not travel FTL. Then why creator of this > universe has created those billions of Galaxies? Where things have gone wrong? The mobile over a crib is out of reach too... Maybe it is a way of teaching us to extend our reach. David A. Smith ==== It actually wasn't a homework problem, in case anybody was wondering. I'm past the geometry stage of math (currently in second year calculus), but my geometry has gotten a little rusty. Actually, I'm not sure we ever learned that in Geometry class. Typical American Public Schools, I suppose. (On the other hand, I wasn't the best Geometry student. My teacher gave liberal amounts of extra credit for doing math team, and so I got A's all year without doing the homework...) The question arose when I calculated (I was bored, away from home, with nothing to do) that the koch snowflake occupies exactly 4/5 of the bounding hexagon, if that makes any sense. So then I was wondering if that was the only way to occupy exactly 4/5 of a regular hexagon, and then I was wondering if it was actually possible to draw a koch snowflake with only compass adn straightedge, and the only step I was unsure of was trisecting a line. Anyway, today I realised that you can occupy 4/5 of a regular hexagon with simpler, finite triangles. Split the hexagon into 6 triangles, mark off 20 equal segments on each side of each of the triangles, draw small equilateral triangle with that as the side length, which will fill the entire hexagon up with 2400 small triangles. Then just slect 4/5 of them. So then I felt a bit stupid. But anyway... Jonathan Christensen Is there an easy way to trisect a line segment, i.e. split it into > three equal portions, with just straightedge and compass? If not an easy way, is there any way at all? Jonathan Christensen ==== > Is there an easy way to trisect a line segment, i.e. split it into > three equal portions, with just straightedge and compass? If not an easy way, is there any way at all? Jonathan Christensen In 1995 two high school freshmen discovered a construction different from the usual one dating back to Euclid. They named it the GLaD Construction -- Goldenheim, Litchfield and Dietrich. Look at their web page: http://www.gfacademy.org/GLaD/glad.html At the bottom of this page is a link to a MS Powerpoint presentation Fibonacci Sketchpad Euclid http://www.gfacademy.org/GLaD/presentation/Presentation.ppt explaining their method. Hugo Pfoertner ==== > Is there an easy way to trisect a line segment, i.e. split it into > three equal portions, with just straightedge and compass? Define easy. Here's one approach that avoids the tedium of constructing a couple parallel lines: Call the endpoints of the segment P1 and P2. Draw a line through P2 (preferably close to perpendicular). Use the compass with the center at P2 to mark two points P3 and P4 on the new line which are equidistant from P2 (preferably about as far away as P1). Draw the line P1P3. Bisect the segment P1P3 and call its midpoint P5. Draw the segment P4P5. Call the intersection of P4P5 and P1P2 P8. Draw a circle with its center at P8 and passing through P2; call its other intersection with P1P2 P9. P8 and P9 are the desired trisection points. -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W ==== Is there an easy way to trisect a line segment, i.e. split it into > three equal portions, with just straightedge and compass? > Yes. Draw a ray such that it shares an endpoint (A) with the segment. Make three points on the ray B, C, and D such that AB = BC = CD. Draw a line from D to the other endpoint of the segment (point E). Now draw a line through B parallel to DE. Repeat at point C. The details can be filled in by any high school geometry book. -Tralfaz ==== > > Is there an easy way to trisect a line segment, i.e. split it into >three equal portions, with just straightedge and compass? > > Yes. Draw a ray such that it shares an endpoint (A) with the segment. Make > three points on the ray B, C, and D such that AB = BC = CD. Draw a line > from D to the other endpoint of the segment (point E). Now draw a line > through B parallel to DE. Repeat at point C. The details can be filled in by any high school geometry book.... or from the fountainhead: Euclid VI.9. Ken Pledger. ==== > Is there an easy way to trisect a line segment, i.e. split it into > three equal portions, with just straightedge and compass? Yes. In fact, there is an easy way of dividing a segment into N congruent sub-segments. (Assuming this is homework) Suppose you had a different segment, already divided into three equal parts. Can you see how to use that other segment together with some facts about parallels and about similar triangles to trisect the original segment? Can you easily construct such a helper segment, in a position that's convenient (say, one side of a triangle ...)? ==== Yes, each of the sequences except for 0.9999....^2 tends toward 0. It just takes longer to do so as you add more 9's. It is what happens at infinity that makes this a special case....Take a look. Lets look at 0.99999.... Let's just prove that is 1. Now, in an earlier post you defined infinity, so you completely agree that it goes on forever. That is important to understand for this small proof. Let x = 0.999999999...... Then 10x = 9.9999999.... Now lets take 10x - x. 10x = 9.999999.... - x = 0.999999.... Remember you said the 9's go on forever so they cancel each other out all the way to infinity. So, 9x = 9. So, x=1. But we said x was equal to 0.999999.... That means 0.999999.... = 1. Since 0.9999.... = 1 then... 0.999999...^2 = 1^2 = 1. Hope that explains it. Markis >> Squares of 0.999... tend away from 1 > > There is only one square of 0.999...; perhaps you can tell me why you >used the plural? Square, cube, etc. > But your numerical evidence below supports that the squares of >0.9, 0.99, 0.999, ... tend TOWARD 1, not AWAY. To see this, all you >need to do is arithmetic. Sequence 0.9^2, 0.9^3, 0.9^4, ... tends toward 0 > Sequence 0.99^2, 0.99^3, 0.99^4, ... tends toward 0 > Sequence 0.999^2, 0.999^3, 0.999^4, ... tends toward 0 > etc., etc., etc., Sequence 0.999...^2, 0.999...^3, 0.999...^4, ... tends toward 0 > Garry Denke, Geologist > Denoco Inc. of Texas ==== >> Squares of 0.999... tend away from 1 >There is only one square of 0.999...; perhaps you can tell me why you >used the plural? Square, cube, etc. > Ok. You should have said powers instead of squares. But now I understand your statement. >But your numerical evidence below supports that the squares of >0.9, 0.99, 0.999, ... tend TOWARD 1, not AWAY. To see this, all you >need to do is arithmetic. Sequence 0.9^2, 0.9^3, 0.9^4, ... tends toward 0 agreed. > Sequence 0.99^2, 0.99^3, 0.99^4, ... tends toward 0 agreed. > Sequence 0.999^2, 0.999^3, 0.999^4, ... tends toward 0 agreed. > etc., etc., etc., agreed. Sequence 0.999...^2, 0.999...^3, 0.999...^4, ... tends toward 0 > I do not agree. Why do you think this is true? Yes, I can. Proof: 0.999... = 9/10 + 9/100 + 9/1000 + ... = (9/10)/(1-1/10) = 1. So, 0.999...^n = 1^n = 1 for all positive integers n. So, the sequence 0.999...^2, 0.999...^3, 0.999...^4,... is the sequence 1, 1, 1, ... which tends toward 1. So this is > > Garry Denke, Geologist > Denoco Inc. of Texas ==== >Does sequence .999...^2, .999...^3, .999...^4, ... tend to 0 or to 1? Since .999... = 1 in your question, your sequence is 1,1,1,..., and > doesn't tend anywhere, it stays put at 1. Virgil assumes that the unlimited value .999... equals the limited value 1, but fails to prove that the powers of the unlimited value .999... tend toward the limited value 1. For all values less than the limited value 1 in the sequence a^2, a^3, a^4, ... the tendancy is toward 0, yet Virgil hangs on to the misconception that the unlimited is limited ... Main Entry: unįlimįitįed Pronunciation: -'li-m&-t&d Function: adjective 1 : lacking any controls : UNRESTRICTED 2 : BOUNDLESS, INFINITE 3 : not bounded by exceptions : UNDEFINED - unįlimįitįedįly adverb Main Entry: limįitįed Function: adjective 1 a : confined within limits : RESTRICTED b of a train : offering faster service especially by making a limited number of stops 2 : characterized by enforceable limitations prescribed (as by a constitution) upon the scope or exercise of powers 3 : lacking breadth and originality - limįitįedįly adverb - limįitįedįness noun ... Virgil's logic is limited. Garry Denke, Geologist Denoco Inc. of Texas ==== >>Does sequence .999...^2, .999...^3, .999...^4, ... tend to 0 or to 1? >Since .999... = 1 in your question, your sequence is 1,1,1,..., and >doesn't tend anywhere, it stays put at 1. Virgil assumes that the unlimited value .999... equals the limited > value 1, but fails to prove that the powers of the unlimited value > .999... tend toward the limited value 1. On the contrary, provided the unlimited value of .999... means the limit lim_{m -> oo} (1-1/10^m), the continuity of f(x) = x^n and the fact of lim_{m -> oo} (1-1/10^m) = 1, both of which I presented in this thread before but which Denke then ignored, proves it. > For all values less than the > limited value 1 in the sequence a^2, a^3, a^4, ... the tendancy is > toward 0, yet Virgil hangs on to the misconception that the > unlimited is limited ... The only misconception is Denke's of what others post. Main Entry: unįlimįitįed > Pronunciation: -'li-m&-t&d > Function: adjective > 1 : lacking any controls : UNRESTRICTED > 2 : BOUNDLESS, INFINITE > 3 : not bounded by exceptions : UNDEFINED > - unįlimįitįedįly adverb Main Entry: limįitįed > Function: adjective > 1 a : confined within limits : RESTRICTED b of a > train : offering faster service especially by making a limited number > of stops > 2 : characterized by enforceable limitations prescribed (as by a > constitution) upon the scope or exercise of powers monarchy 3 : lacking breadth and originality head -- Virginia Woolf - limįitįedįly adverb > - limįitįedįness noun I do not recall ever hearing of Virginia Woolf's expertize in mathematics so I cannot accept such definitions as relevant to my meanings. The only dictionaries relevant in a discussion of mathematical meanings are mathematical dictionaries. ... Virgil's logic is limited. So is the logic of every sane person limited by the laws of logic, but Denke's willful misunderstandings appear unlimited. A sequence may be unlimited in its number of terms and simultaneously limited by being Cauchy sequence, as is (1-1/10^m), in which case the sequence converges to a limit, as does (1-1/10^m). Garry Denke, Geologist > Denoco Inc. of Texas ==== > Let a_n = 1 - 10^(-n). Then a_0 = 0, a_1 = .9, a_2 = .99, etc. If we > take the sequence {(a_0)^2, (a_1)^2, (a_2)^2, (a_3)^2...} we can > examine what happens to the squares of .9999...999 as n grows to > infinity. We have that lim (n -> infinity) (a_n)^2 = lim (n -> infinity) 1 - 2 * > 10^(-n) + 10^(-2n) = 1 - lim [2 * 10^(-n) + 10^(-2n)] = 1 - 0 = 1. Notably, you don't seem to be interested in only the squares of the > numbers, since you include also the fourth power. Maybe you should > state more clearly precisely what you want. I am interested in the sequence a^2, a^3, a^4, ... for a = .9, a = .99, a = .999, through a = .999... Garry Denke, Geologist Denoco Inc. of Texas ==== In sci.math, Garry Denke on 11 Jan 2004 07:40:17 -0800 <54056cf9.0401110740.73489454@posting.google.com>: >> Let a_n = 1 - 10^(-n). Then a_0 = 0, a_1 = .9, a_2 = .99, etc. If we >> take the sequence {(a_0)^2, (a_1)^2, (a_2)^2, (a_3)^2...} we can >> examine what happens to the squares of .9999...999 as n grows to >> infinity. >> >> We have that lim (n -> infinity) (a_n)^2 = lim (n -> infinity) 1 - 2 * >> 10^(-n) + 10^(-2n) = 1 - lim [2 * 10^(-n) + 10^(-2n)] = 1 - 0 = 1. >> >> Notably, you don't seem to be interested in only the squares of the >> numbers, since you include also the fourth power. Maybe you should >> state more clearly precisely what you want. I am interested in the sequence a^2, a^3, a^4, ... > for a = .9, a = .99, a = .999, through a = .999... lim(n->+oo) (1 - 10^(-n))^n = lim(n->+oo) (1 - exp(-n * log(10))^n = lim(n->+oo) (1 - n*exp(-n * log(10)) + (n)(n-1)/2!*exp(-2*n*log(10)) - (n)(n-1)(n-2)/3!*exp(-3*n*log(10)) ± ...) lim(n->+oo) n*exp(-n * log(10)) = n - n^2*log(10)/1! + n^3*log(10)^2/2! - n^4*log(10)^3/3! ± ... + (-1)^k*n^(k+1)/k! ± ... It's a bit difficult to prove but this series is in fact absolutely convergent (since k! > n^(k+1) for sufficiently large k) and in fact the limit is 0. Therefore, lim(n->+oo) (1 - 10^(-n))^n = 1. Or one can look at lim(n->+oo) log(n * exp(-n * log(10)) = lim(n->+oo) log(n) - n*log(10) If one sets f(x) = x and g(x) = log(n), then f'(x) = 1 and g'(x) < 1 for all x > 1. (g'(x) = 1/x, in fact.) Therefore h(x) = log(x) - x*log(10) is such that h'(x) = (1/x) - log(10). For x > log(10), h'(x) is negative; therefore for sufficiently large x, h(x) < K for some K. In fact, for sufficiently large x, h(x) < 0 -- certainly for x > 10, h(x) < h(10) = -9*log(10) < 0, as is easy to see. It turns out that lim(n->+oo) log(n * exp(-n * log(10)) = -oo though I am not entirely sure of the easiest method of proof. > Garry Denke, Geologist > Denoco Inc. of Texas -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== >Let a_n = 1 - 10^(-n). Then a_0 = 0, a_1 = .9, a_2 = .99, etc. If we >take the sequence {(a_0)^2, (a_1)^2, (a_2)^2, (a_3)^2...} we can >examine what happens to the squares of .9999...999 as n grows to >infinity. >We have that lim (n -> infinity) (a_n)^2 = lim (n -> infinity) 1 - 2 * >10^(-n) + 10^(-2n) = 1 - lim [2 * 10^(-n) + 10^(-2n)] = 1 - 0 = 1. >Notably, you don't seem to be interested in only the squares of the >numbers, since you include also the fourth power. Maybe you should >state more clearly precisely what you want. I am interested in the sequence a^2, a^3, a^4, ... > for a = .9, a = .99, a = .999, through a = .999... > Garry Denke, Geologist > Denoco Inc. of Texas Been there, done that, and have the t-shirt. To repeat for the terminally illiterate, let f(n) = (1-1/10^m)^n, for fixed m, giving the series you cite an interest in above, and let g(m) = (1-1/10^m)^n, for fixed n, giving the series that everyone else is interested in . Then lim_{n -> oo} f(n) = 0 but lim_{m -> oo} g(m) = 1 and for suitable relations between m and n, lim_{(m,n) ->(oo,oo)} (1-1/10^m)^n can be any value in [0,1] 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 ==== In <210a83f4.0401081034.24b607ce@posting.google.com>, on 01/08/2004 Rather misleading subject, what? What you're writing about is not squares but arbitray powers. > .9^4 = .6561 Is not a square of .99 . . . -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== Let's use w as a stand-in for the Greek character omega here. As many here know, we have the following recursive definition for w_a, also called aleph_a, for all ordinals a. 1) w_0 = w 2) w_S(a) = the least cardinal greater than w_a, where S(a) = a+1 is the successor of a. 3) w_a = sup {w_b: b < a} when a is a limit ordinal. It is easy to show that a <= w_a for all a. Can one show in ZF that the inequality is always strict? How about in ZFC? In ZF+GCH? It is clear that if a = w_a, then a is a limit cardinal. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu ==== >Let's use w as a stand-in for the Greek character omega here. As many >here know, we have the following recursive definition for w_a, also >called aleph_a, for all ordinals a. >1) w_0 = w >2) w_S(a) = the least cardinal greater than w_a, where S(a) = a+1 is >the successor of a. >3) w_a = sup {w_b: b < a} when a is a limit ordinal. >It is easy to show that a <= w_a for all a. Can one show in ZF >that the inequality is always strict? How about in ZFC? In ZF+GCH? >It is clear that if a = w_a, then a is a limit cardinal. Define the sequence a_0 = w and a_S(n) = w_{a_n} for all natural numbers n, and let b = sup {a_n : n < w}, then b = w_b. David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills. ==== > Let's use w as a stand-in for the Greek character omega here. As many > here know, we have the following recursive definition for w_a, also > called aleph_a, for all ordinals a. > 1) w_0 = w > 2) w_S(a) = the least cardinal greater than w_a, where S(a) = a+1 is > the successor of a. > 3) w_a = sup {w_b: b < a} when a is a limit ordinal. It is easy to show that a <= w_a for all a. Can one show in ZF > that the inequality is always strict? How about in ZFC? In ZF+GCH? It is clear that if a = w_a, then a is a limit cardinal. > I believe that the inequality is not strict. Consider the ordinal defined recursively: a_0 = 0, a_n = w_{a_{n+1}} Then I believe that a = a_w satisfies a = w_a. I recall seeing somewhere some kind of fixed point theorem for ordinals, something to the effect that if f is an increasing function on the class of ordinals, perhaps satisfying some other axioms as well, then there exists an ordinal a = f(a). ==== I recall seeing somewhere some kind of fixed point theorem for ordinals, > something to the effect that if f is an increasing function on the class > of ordinals, perhaps satisfying some other axioms as well, then there > exists an ordinal a = f(a). > Now I think about it, I think that the additional axiom would be something like f(a) = sup{f(b):bI came across 2 questions I had that maybe someone can answer : >1) If A is an n x n matrix, and ||A|| is defined as the smallest number c >such that ||Ax|| <= c||x|| for all x in C^n, the n-fold complex numbers, >then why is it true that a sequence of matrices A_m converges to a matrix A >if and only if ||A_m - A|| --> 0? by definition. > verify that ||.|| above is a norm. > 2) Consider an infinite series whose terms are matrices : A_0 + A_1 + A_ 2 >+ ... If the sum (from m = 0 to infinity) of ||A_m|| < infinity, then >the infinite series of matrices is said to converge absolutely. this is an extension of a calculus theorem. it's proof is similar. > Why is it true that if a series converges absolutely, then the partial >sums of the series form a cauchy sequence? let space V possess a norm ||.||, and let epsilon > 0. we know that if S = Sum fj converges, then > ||Sn - S|| < epsilon/2 for some n large enough, > where Sn = Sum fj, j = 1 to infty. oops, j = 1 to n (NOT infty) > ||Sn - Sm|| loe ||Sn - S|| + ||Sm - S|| < epsilon > for m and n large enough. > Moshe ==== >Well, ||A_m - A|| -> 0 is often the definition of the convergence of a >sequence of matrices. What's your definition? > > Let A_n be a sequence of complex matrices. A_n converges to a matrix A if > each entry of A_n converges to the corresponding entry of A. So it's enough to show || (jth column of A_n) - (jth column of A) || -> 0, in the sense of vectors in C^m. But this can be written as || (A_n - A)(e_j) ||, where e_j is the jth standard basis vector. ==== My latest brainchild, as minor as most of the others, is this mini-theorem. Let s_0 and s_1 be nonnegative integers, not both zero. Define inductively s_{n+2} = | s_{n+1} - s_n | for n >= 0. Show that the sequence (s_n) eventually looks like 0, x, x, 0, x, x, 0, x, x, ... and x = gcd(s_0,s_1). For example, if s_0 = 69 and s_1 = 39, the sequence runs 69, 39, 30, 9, 21, 12, 9, 3, 6, 3, 3, 0, 3, 3, 0, ... LH ==== Dear all, Is there any function that relates log(x+y) with log(x) and log(y)... In an optimization problem, I basically need to seek the minization of the function of the form log(x+y)... but in order for further mathematical manipulation, I need to break log(x+y) up into log(x) and log(y), right? -Walala ==== >Dear all, Is there any function that relates log(x+y) with log(x) and log(y)... In an optimization problem, I basically need to seek the minization of the >function of the form log(x+y)... but in order for further mathematical >manipulation, I need to break log(x+y) up into log(x) and log(y), >right? >-Walala A simple-minded idea: minimizing x+y will minimize log(x+y). - Ken ==== > Dear all, Is there any function that relates log(x+y) with log(x) and log(y)... In an optimization problem, I basically need to seek the minization of the > function of the form log(x+y)... but in order for further mathematical > manipulation, I need to break log(x+y) up into log(x) and log(y), > log(x+y)=log(x)+log(1+y/x) might help. But for minimization it is better to set the derivatives to zero rather than do such transformations... Arnold Neumaier 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 ==== In <1073775886.602228@athnrd02.forthnet.gr>, on 01/11/2004 at 01:04 AM, Ioannis said: >Then we went over what appeared to be a slight variant of Zeno's >paradox, with him insisting that the passage of time cannot be >trully analog, since if time was continuous, the vector of time There is no such thing. >would have to have infinite speed There is no such thing. >speed as it passed over ONE particular real second How does time pass over anything? >The above was quite interesting for me. It's all handwaving. Nothing that you quoted has any meaning, Mathematical or physical. >If the passing seconds are represented as real numbers, then indeed, >a _specific_ second in the future (which is a specific real number) >approaching the present and becoming past, How does a second approach a time? >passes from the state future to the state past in 0 time. Why? Why don't say that it takes future_time - past_time to pass? >Thus the speed vector of time as it passes over >a _specific_ (real) second, Again, that whole phrase has no meaning. >In other words, the rate of conversion How do you convert seconds, and how do you measure the rate? >In mathematical terms, No. The one thing that you have *NOT* done is to use anything vaguely resembling Mathematical terms. >The above is related to Zeno's paradox, No. >The above smells fishy to me. It's total balderdash. >Indeed the quantifier for every, forces the brain to consider >mathematical continuity here, but in practical terms isn't always a >_specific_ epsilon that's being tested? What good would that do? If you want to prove continuity then you must deal with the universal quantifier. >So, although the quantifier intuitively (and >mathematically) satisfies this crave for continuity, in actual >practice, the available epsilons are discreetized in our minds No. >We certainly cannot >practically test EVERY available epsilon > 0. Irrelevant. Even if we could it would be a brute force proof and you'd be asked to find a more elegant proof. >It appears though that the notion of philosophical continuity is >still somehwat elusive. (to me at least). It's a notion that has nothing to do with Mathematics and may even have nothing to do with Physics. >To top the cake, if one considers any possible action inside this >universe, One would not be considering Mathematics, but Physics. >every such action is quantized. That's an assumption. It is certainly *NOT* required by Quantum Mechanics. >Following this, does it make sense to talk >about the philosophical notion of having time being continuous? Sure, in a Philosophy news group. Here, time is off topic. In sci.physics, the issue of quantized time is very much an open question. >Time itself might well exist outside our existence continuously, ObMinkowsky If we take current physical theory seriously, time does not exist, any more than space does. Only spacetime exists. >But if consciousness is quantized, >then the term continuous time is nonsense. To a solipsist. Not to one who believes that there is a universe beyond the perceptions. >It appears as though the mathematician's main stance 1) above is >invalid. Irrelevant; it is a stance that relates to Biology, Psychology and Physics, not to Mathematics. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== > It appears as though the mathematician's main stance 1) above is invalid. > Read Immanuel Kant's Critique of Pure Reason. Ignore Dedekind/Peano succession. The concept of number in mathematics is associated with a closure property in the fundamental theorem of algebra. Geometric intuitions rule. Kant treated time and space in ideality rather than absolutely. Kant's notion of infinity precludes Cantor's use of the successor function to obtain transfinite numbers. It can only be interpreted in terms of single-point compactification, whence you get the Riemann sphere, stereoscopic projections, and the spherical encodings and quantizations discussed by Conway and Sloane in Sphere Packings, Lattices, and Groups. Quantization is an artifact of serializing sense data into grammatical forms used to talk about sense data. You will get far better answers on a philosophy group. :-) mitch ==== > Two days ago I had an interesting discussion with my coworker at the Math *snip* That was a long post & I am sure it was very interesting. To answer the question posed in the subject line of your post; read last month's issue of Scientific American. If loop quantum gravity is correct, then both space & time are discrete. ==== > The above was quite interesting for me. It appears as though it is true. If >the passing seconds are represented as real numbers, then indeed, a >_specific_ second in the future (which is a specific real number) >approaching the present and becoming past, passes from the state future to >the state past in 0 time. Thus the speed vector of time as it passes over >a _specific_ (real) second, must be infinite. I do not know what a speed vector of time is supposed to be. Best measurements to date place it at about 1 billion nanoseconds per second in the future direction. There are other directions, such as the past direction and dream directions. There are times when the magnitude of the time vector slows down by a factor of perhaps 100. Exact measurement of the factor is difficult to determine as such times occur only in moments of unusual alert or duress. Sometimes of timelessness have also be noted. When awake, time is continuous. At other times, time is quantized into waking periods and dreams. > Do you view the present moment as the origin and specific moments in the > future as moving backward toward the origin? If so, I see no infinite. > I see the constant vector -1. > In no time at all, time had begun, for it's time had come. ==== ėÄ William Elliot ė[CapitalEth]ėŅģŖ[EDouble Dot]±ģĆė[Micro] ģ°ģ.b3ėĄ ė.b9ėØėøģ·[EDo ubleDot].b9ė± [snip] >I do not know what a speed vector of time is supposed to be. Best measurements to date place it at about > 1 billion nanoseconds per second > in the future direction. There are other directions, such as the past > direction and dream directions. There are times when the magnitude of the time vector slows down by a > factor of perhaps 100. Exact measurement of the factor is difficult to > determine as such times occur only in moments of unusual alert or duress. > Sometimes of timelessness have also be noted. When awake, time is continuous. At other times, time is quantized > into waking periods and dreams. > Do you view the present moment as the origin and specific moments in the >future as moving backward toward the origin? If so, I see no infinite. >I see the constant vector -1. > In no time at all, time had begun, for it's time had come. -- Ioannis Galidakis http://users.forthnet.gr/ath/jgal/ ------------------------------------------ Eventually, _everything_ is understandable ==== > Then we went over what appeared to be a slight variant of Zeno's paradox, > with him insisting that the passage of time cannot be trully analog, since > if time was continuous, the vector of time would have to have infinite > speed as it passed over ONE particular real second (represented as a real > number on the Real line of seconds). > Nonsense. ==== >Then we went over what appeared to be a slight variant of Zeno's paradox, >with him insisting that the passage of time cannot be trully analog, since >if time was continuous, the vector of time would have to have infinite >speed as it passed over ONE particular real second (represented as a real >number on the Real line of seconds). Nonsense. I know it probably is. You could have done us all the courtesy of showing us exactly where the nonsense lies though, genius. But, what should I expect with a name like fishfry? -- Ioannis Galidakis http://users.forthnet.gr/ath/jgal/ ------------------------------------------ Eventually, _everything_ is understandable <1073788832.691260@athnrd02.forthnet.gr> ==== > >Then we went over what appeared to be a slight variant of Zeno's >> paradox, with him insisting that the passage of time cannot be >> trully analog, since if time was continuous, the vector of time >> would have to have infinite speed as it passed over ONE particular >> real second (represented as a real number on the Real line of >> seconds). > > Nonsense. I know it probably is. You could have done us all the courtesy of showing us > exactly where the nonsense lies though, genius. > Not before long, the number of Zeno steps would be infinitely increasing to swallow the same eternally unvarying constant quantum of time. > But, what should I expect with a name like fishfry? Sole food? ==== 1073788832.691260@athnrd02.forthnet.gr... > >Then we went over what appeared to be a slight variant of Zeno's > paradox, >> with him insisting that the passage of time cannot be trully analog, > since >> if time was continuous, the vector of time would have to have infinite >> speed as it passed over ONE particular real second (represented as a > real >> number on the Real line of seconds). > > Nonsense. I know it probably is. So why publish it? You could have done us all the courtesy of showing us > exactly where the nonsense lies though, genius. Well, the same argument say that speed over any paricular point in any continuous motion is infinite. This is not the *definition* of speed But, what should I expect with a name like fishfry? > -- > Ioannis Galidakis > http://users.forthnet.gr/ath/jgal/ > ------------------------------------------ > Eventually, _everything_ is understandable > ==== Ć Denis Feldmann Ūēņįćå óōļ ķÜīłķį [snip] >> Nonsense. > > I know it probably is. So why publish it? I guess because I don't quite understand the notion of rate of continuous time flow. After thinking about Wade's answer for a while, (constant vector of -1), I think maybe what he meant was: One could define rate of time flow as 1 sec/sec, which indeed is the opposite of what Wade suggested (because he assumed that I was looking at time backwards). I don't understand in what sense this can be a vector in the conventional sense though. This vectorappears to be dimensionless. Correct me if I am wrong. > You could have done us all the courtesy of showing us >exactly where the nonsense lies though, genius. Well, the same argument say that speed over any paricular point in any > continuous motion is infinite. This is not the *definition* of speed Quite so. -- Ioannis Galidakis http://users.forthnet.gr/ath/jgal/ ------------------------------------------ Eventually, _everything_ is understandable ==== Ć The World Wide Wade Ūēņįćå óōļ >The above was quite interesting for me. It appears as though it is true. If >the passing seconds are represented as real numbers, then indeed, a >_specific_ second in the future (which is a specific real number) >approaching the present and becoming past, passes from the state future to >the state past in 0 time. Thus the speed vector of time as it passes over >a _specific_ (real) second, must be infinite. I do not know what a speed vector of time is supposed to be. Do you view > the present moment as the origin and specific moments in the future as > moving backward toward the origin? If so, I see no infinite. I see the > constant vector -1. I can't define it mathematically, cause I would need to define time in terms of time again. Think of it as follows: Imagine (your) consciousness (being a point) moving along the time line, forward, with time being continuously mapped onto the Real line. In a sense, you are continuously moving along the reals in the forward direction. A future point in time has an exact real representation somewhere in front of you. It hasn't become present yet. Call this time-point t_0. Now it approaches and it transits in front of you, becoming past the moment your clock ticks t_0. What is the transition speed of t_0, as it passes from being a future time point to being a past time point RELATIVE to your point-consciousness? If it was any finite number k, after, say, 2 whole seconds, there would be 2*k points having been transitioned from futurepoints to past points. But, say, 2 whole seconds worth of time, have a finite measure, contradicting k finite, since 2*k time-points have measure 0. So k must be infinite (in fact uncountable), otherwise your consciousness couldn't possibly have moved any finite amount of time forward. Can you see it? -- Ioannis Galidakis http://users.forthnet.gr/ath/jgal/ ------------------------------------------ Eventually, _everything_ is understandable Let p and q be distinct primes. Suppose that H is a proper subset of the integers and H is a group under addition that contains exactly 3 elements of the set {p, p+q, pq, p^q, q^p}. Determine which fothe following are the 3 elements in H. a) pq, p^q, q^p b) p+q, pq, p^q c) p, p+q, pq d) p, p^q, q^p e)p, pq, p^q The answer turns out to be e) but it took me forever to figure this out. Since it was on a GRE math practice exam, it must be fair game for a regular exam. My question is this: Is there any fast way to know that the answer is e) as opposed to the others? I would have had to skip the question on a real exam because it took me far too long. ==== >Let p and q be distinct primes. Suppose that H is a proper subset of the >integers and H is a group under addition that contains exactly 3 elements of >the set {p, p+q, pq, p^q, q^p}. Determine which fothe following are the 3 >elements in H. >a) pq, p^q, q^p >b) p+q, pq, p^q >c) p, p+q, pq >d) p, p^q, q^p >e)p, pq, p^q >The answer turns out to be e) but it took me forever to figure this out. Since >it was on a GRE math practice exam, it must be fair game for a regular exam. >My question is this: Is there any fast way to know that the answer is e) as >opposed to the others? I would have had to skip the question on a real exam >because it took me far too long. Since H is a group under addition, then H is the set of the integral multiples of some integer. Since H is a proper subset of the integers, then H is the set of integral multiples of some nonnegative integer distinct from 1. This means that the gcd of a set of elements of H is a nonnegative integer distinct from 1. In all cases a)-d), the gcd of the three in the list is 1, thus eliminating them as possibilities. This leaves only e), where the gcd is p, and H is the set of integral multiples of p. David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills. ==== I want some of what he is smoking! Damn > Dear Jack Sarfatti What if as hypothesized by the likes of Prof Frank Tipler, David Deutsh, > Prof Paul Davies and Nick Bostrum that we are all living within a > 'computer programme of an ancestral simulation' created by an advanced > super-civilisation within the far-flung future! JS: Like The Matrix films. BTW I will be in London at least March 8 - 13. MD: What if the 'simulation' so unaware at first becomes aware of its > own predicament by becoming frustrated at not being able to 'break > through the wall of light' and therefore assumes true sentience by > understanding that it is being caged within a loop of an enfolding > programmed cave of information! Would this then make the 'Godmind' of the programmer very irate who then > decides to wipe the programme clean to then start again from scratch! > Perhaps this has happened many a time already and that each time of a > new Alpha of a begining when reaching an Omega point foreseen, the > programme contains elements of primary memories previously attained that > avoid being wiped clean, which survive unseen within our collective myths! Myths of previous worlds remembering giant birds that flew across the > oceans spitting fire at cities to therefore destroy and of arrows that > could reach the moon from a heroe's bow. Perhaps, even to become aware > of ones part to live all over again at certain points in ones own life > reenacting and to foresee future events about to happen one already has > lived! Perhaps such memories are left there to drive us forwards towards a > future pre-written looped back! The Mayans have 2012 and the Chinese also have the same mythic symbolic > time indicator that has now become as a collective focus pertaining to > the end of 'time.' This shared time between America and China may > indicate that the scientists of these two nations prove something that > shocks humanity into awakening only then to initiate a conflaguration of > an apocalypse as the Godmind wipes the simulation clean! Maybe this time indicator indicates our collective awareness of the > predicament and thereby the attainment of true sentience, which > invariably leads to the simulation being wiped again by the Godmind of > the programmer for it might be somewhat jealous and does not want > another to rival it. But, what would happen if the simulation finds a way of not being wiped > clean and that it breaks free through the wall of light to strike the > Godmind in the eye? Perhaps the simulation might then be welcomed as a prodigal child, but > would such a child stand the pace, and if not, only then to reel back > into an autohypnotic state of self induced trauma remembering fractured > memories becoming as mythic symbols to rise again when the child starts > to heal slowly from the future shock! Of course there is another perspective and that this reality is a > 'self-generated simulation' amongst a myriad number of others! Then we have the somewhat mundane dilemma of dealing with certain past > individuals who have molded human society by their influence and were in > fact influenced by a signal looping back from the future! Should such an influence be proved that their electron dreams were > merely spun by a charged signal from the future it would have far > reaching implications that might set the very world on fire. Were such individuals such as Christ, Muhammad and Abraham for example > aware of the nature of the charged signal's true origin that spun their > electron dreams? Perhaps not, since the signal was filtered through their culturally and > socially imprinted neural-nets and therefore saw only what they were > expecting to see pertaining to whatever internal dialogue they ascribed to! Of course the signal is on going as to imprint certain individuals at > certain times whom see whatever they are imprinted to see, such as UFO's > and Grays where as before long ago they perceived Angels and Demons for > whatever is seen is always one step ahead as an unknowable! Why an unknowable, which is masked by imprinted collective dialogue of a > symbol? It is because a prodigal child who reaches into 'breaking through the > wall of light' will invariably reel back from the shock of it all into > an autohypnotic state of trauma that will mask the truer experince that > rises as a black tidal wave into consuming a fragile mind! What be this truer experience? That we are all sustaining a self-created reality of a particular brane > and that we are all unconsciously shifting between alternate branes all > of the time without even realising it and the proof of it lies within > the nature of synchronicity subtle, for each synchronicity is a stargate > into an alternate brane next door! This particular brane is just one of a myriad number of others tuned > into and that we can consciously engineer the tuning into others should > we desire! But if this be true that a subtle synchronicity indicates a shift into > an alternate brane next door one would then invariably find that the > true nature of 'Magick' and its power is very much alive and that this > power resides within each and every one of us all! The super technology of a computer and of walking a myriad worlds is > here already...for it is us, self-created! Yours Sincerely Mark Dunn JS: Well you have a germ of an idea for a Sci Fi film. Dear Schwann many minds cohered by emotively charged symbols that spin the electron > dreams of those zapped by lightening winding back as in a loop! How many a Moses, Christ's, Abraham's, Muhammad's, Pacal Votan's, Buddha's > etc does it take to change a lightbulb as they all stand one legged on a > tightrope of a spun super-stringed lightening zap of a telephone line? What can we make of those who wander wind swept fields in the dead of night > dressed in black creating crop-circles and when caught out they say that > the > aliens in their heads got them to do it? Of course such individuals may just go and phone up certain people making > out that they are an alien or two as they speak like a computer and again > driven to do so by the voices strobing their brains! And when they are not about to get spun into the Wyrd of the web > trembled by > a possible consciously engineered directed signal of an emotive charge > certain predisposed individuals see dancing lights that hum a vortex into > opening up on a side of the hill as the eye of the moon hides its self away > behind a looming triangular cloud! As the lights do dance to hypnotise into spinning those into seeing and > without realising they are spun into an alternate brane next door! Perhaps a certain individual whose eyes becomes glazed over as if in trance > not themselves for a moment becomes as a window to look through by a time > traveller of a tourist who zaps their neural-net by spinning their electron > dream as a stargate to open to thereby perceive when not in trance a crop > circed mandala that suddenly appears within another brane verily spun into! The reasoning behind all of this perhaps being that to engineer fractal > realities of alternate branes next door to light up like so many > lightbulbs! > Yours Sincerely Mark Dunn > X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i08GkkK10487; ==== Sorry, I need to make yet more corrections. The property (1) B_n(r) subset B_{n+1}(r) listed below should read: If B_n(r) = (x_1, x_2, ...,x_n) and B_{n+1}(r) = (y_1, y_2, ...,y_{n+1}), then {x_1, x_2, ...,x_n } subset {y_1, y_2, ...,y_{n+1}} As for the sequence (which was supposed to be an 8-gon) (x, (x+y)/2, y, (y-x)/2, -x, (-x - y)/2, -y, (-y +x)/2) 2 should be replaced with sqrt(2) to make all vectors of the same norm. Note: Presumably, showing s_n -> 0 shouldn't be too hard since every bounded sequence in a finite dimensional space has a convergent subsequence. >Three corrections to my first message: >1. dim X >= 2 should be prerequisited. >2. lim n -> infty (n*s_n/2r) >= 2 does not follow > from the triangle inequality as shown. It follows > from norm homogenity: > > For x != 0, take the 2-gon (x, -x) > Then the distance from x to -x,and from -x back > to x, divided by twice the radius r = ||x|| is > (||x - (-x)|| + ||-x - x||)/(2||x||) = 4||x||/(2||x||) = 2 3. Not so much a correction, as (hopefully helpful) statement > I forgot to include: should be expected. If I remember correctly from my (lost) > notes, one can go from the square 4-gon (x, y, -x, -y) > where ||x+y|| = ||x-y|| to an 8-gon (x, (x+y)/2, y, (y-x)/2, -x, (-x - y)/2, -y, (-y +x)/2) Increasing the lower bound estimate of lim n -> infty (n*s_n/2r) > from ~2.6 to something closer to 3. Of course, one > first has to assume that the limit exists, i.e. (n*s_n/2r) > is a Cauchy-sequence. >>The kite equation thread inspired me to think back >>to another property I«ve had in the back of my mind for a >>while. I think I can be fairly sure someone has thought of >>it before. >>Let X be a complex normed space. Let B_n(r) = (x_1, x_2, ...,x_n) >>(a finite sequence of elements of X) be an n-gon of radius >>r; this means >>(a) x_i != x_j for all 1 <= i,j <= n >>(b) dim (lin {x_1, x_2, ...,x_n}) = 2 >>(c) ||x_1|| = ||x_2|| = ... ||x_n|| = r > 0 >> >>(d) ||x_1 - x_2|| = ||x_2 - x_3|| = ... = ||x_n - x_1|| = s_n > 0 >>Call (B_n(r)) a sequence of n-gons if: >>for all n in naturals: >>(1) B_n(r) subset B_{n+1}(r) >>(2) B_n(r) != B_{n+1}(r) >>(3) B_{n+1}(r) is an n-gon of radius r >>For all n in naturals, define s_{n+1} to be as in (d), above, >>for B_{n+1}(r). >> >>What if all we know about X is that it is a normed space and: >>lim n -> infty (n*s_n/2r) -> pi = 3,14...) for every n-gon sequence >>(or use for every two dimensional subspace, there exists an n-gon >>sequence... - which I believe should be equivilant to the above) >> >>Notice that if X is a normed space and ||x|| = ||y|| > 0, >>then ||x-y||/||x|| <= (||x|| + ||y||)/||x|| = 2||x||/||x|| = 2 >>Thus, by requiring X be normed space, we are actually >>stating that lim n -> infty (n*s_n/2r) >= 2 for X. It is also >>the property that there are nonzero vectors orthogonal to each >>other (-> ||x+y|| = ||x-y||) (this may be the case in every >>lim n -> infty (n*s_n/2r >= 2,6 just by using the parallelogram >>equation (hint: make a square). >>Note: actually, I forgot exactly what the number was (2,6 or >>something close) since it«s been a while and I don«t know where my >>notes are on that anymore. However, it shouldn«t take long to >>recheck. >>For me, it would be interesting to know any of the two >>C. Dement >> X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i08I3IK16839; ==== In Hebrew, the feminine plural of the foreign loan word, nilpotent, when >written without vowels or other punctation, is written nylpw_tn_tywt where _t is teth. What I would like to know is how it would be written >if one did write the vowels and other punctation. For example: >(0) which of the consonants is dotted? >(1) is the y following the first n retained and the n endowed with a chireq? >(2) is there a schwa under the l? >(3) is the w following the p replaced by a cholem or is it replaced by > a qamats qatan? >(4) what is under the first teth: a schwa, a segol or a chataf segol? >(5) is there a schwa under the second n? >(6) is the y following the second teth retained and the teth endowed with > a chireq? I would also like to know how I might figure out examples like this, involving >foreign loan words, on my own, instead of having to ask people in every >instance. Ignorantly, >Allan Adler >ara@zurich.ai.mit.edu *************************************************************************** * >* * >* Intelligence Lab. My actions and comments do not reflect * >* in any way on MIT. Moreover, I am nowhere near the Boston * >* metropolitan area. * >* * >*************************************************************************** * That doesn't look like the feminine plural, but rather an abstraction, like nilpotency or nilpotentiation. phil X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i08K3AS26439; ==== What (if at all) the existence of higher ordinal infinities (such as strongly inaccessible cardinals and higher) imply in terms of CH or GCH ? ==== > What (if at all) the existence of higher ordinal infinities (such as > strongly inaccessible cardinals and higher) imply in terms of CH or > GCH ? The general rule so far is that adding large cardinal axioms does not decide CH one way or the other. Of course, some as-yet-unconsidered large cardinal axiom might do so. Thomas X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i08K3AG26434; ==== >> Mathematicians are literate guys, so who ever first used homogeneous meant >> of the same form i.e. ax +by+cz+b isn't homogoneous since b isn't of the >> same form as ax,by,cz? According to Jeff Miller's nice webpage at http://members.aol.com/ jeff570/mathword.htmlthe first use is in: >HOMOGENEOUS EQUATIONS is found in 1815 in the second edition of Hutton's >mathematics dictionary: Homogeneous Equations ... in which the sum of the >dimensions of x and y... rise to the same degree in all the terms (OED2). > That definition of homogeneous is still used in many d.e. books where it causes enormous confusion with homogeneous linear d. e. which is completely different. That last use of the word homogenous is more in keeping with what, in modern linear algebra would just be a linear transformation. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i08K3Bw26471; ==== > Let p(z) be polynomial of degree n such that >> >> |p(z)|<= M for all z in the unit disc i.e. |z|<1 >> >> Prove that |p(z)|<= M|z|^n for all |z|>1 > this function f brings sex in the city. How can I prove now |p(z)|<=M|z|^n? Could you explain little more? X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0933el25569; ==== > > In general is there an agreed concensus as to naming rights in > mathematics, and/or is there a body that governs naming rights? >> >> No. Most things are named after totally irrelevant people >> (e.g. Pell's equation) :-) And: during a representation theory conference some speaker talks of using >Brauer's trick. A hand goes up. 'What's Brauer's trick?' Asks Richard >Brauer. third hand anecdote, so open to corrections that actually it was Witten, >or someone else entirely. Herr Weyl you must explain me one thing-what is What about Chapman inequality?Is it named after Robin Chapman? X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0933dF25511; ==== >> 1)The function f(x) has f ''(x) > 0 for all x 0 and the graph is asymptotic to y = ax + b as xā. >> Show that f(x) - ax - b has negative derivative for all x > 0, >and that f(x) > ax + b for all x 0.[/quote:b29fe8cde7] (1) Suppose f'(p) - a > 0 for some x. Then, since f'(x) is increasing (f''(x) > 0), f'(x) > f'(p) for all x >= p and hence: f(x) > f(p) + (x - p) f'(p) = = f(p) - p f'(p) + f'(p) x for all t >= x. Thus: lim f(x)/x >= f'(p) > a, which contradicts our assumption that f is assimptotic to y = ax + b. Hence f'(x) <= a. But f' is strictly increasing, so f'(x) < a. (2) Function f(x) - ax - b is strictly decreasing (since it has negative derivative) and has a limit 0 when x tends to infinity. Hence it is (strictly) positive. Hope this helps. Mateusz X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0933dI25515; ==== >Really ! Who? Who is inventor of multiplication table? >What exactly is the multiplication table? -- >glenn > X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0933dF25523; ==== Could a proof of the Goldbach Conjecture be worth of a Fields prize? P.S. Since the proofs for higer dimensional cases of Poincare Conjecture could be worth the prize, I think the proof of Poincare Conjecture could also be. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0933de25519; ==== >
I had written a program to find Pythagorean Triples years ago. I have
>>now restructured it to find quadruples with power 3 as well - there are
>>a lot, even a few starting with 1^3.
>>But there are no quintruples (is that word correct?) with power 4, no
>>hexuples (is that word correct?) with power 5, etc. (Not even triples or
>>quadruples with higher powers.)
>>What I haven't tried yet is quadruples or quintruples with powers 2 and
>>3...
-- 

>http://www.tarpley.net
Could you send me that program ASAP? X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0933e525581; ==== Anyone know of any recent work on the odd perfect number question? Ciao, Mark X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0933du25543; ==== Wouldn't a dictionary without vowels be a dctnry (or sometimes, dctnr)? You said you are using a dictionary, right? What does it say? >If it too has no vowels, you probably ought to invest in one >that has vowels. > >> In Hebrew, the feminine plural of the foreign loan word, nilpotent, when >> written without vowels or other punctation, is written >> >> nylpw_tn_tywt >> >> where _t is teth. What I would like to know is how it would be written >> if one did write the vowels and other punctation. For example: >> (0) which of the consonants is dotted? >> (1) is the y following the first n retained and the n endowed with a chireq? >> (2) is there a schwa under the l? >> (3) is the w following the p replaced by a cholem or is it replaced by >> a qamats qatan? >> (4) what is under the first teth: a schwa, a segol or a chataf segol? >> (5) is there a schwa under the second n? >> (6) is the y following the second teth retained and the teth endowed with >> a chireq? >> >> I would also like to know how I might figure out examples like this, involving >> foreign loan words, on my own, instead of having to ask people in every >> instance. >> >> Ignorantly, >> Allan Adler >> ara@zurich.ai.mit.edu >> >> **************************************************************************** >> * * >> * Intelligence Lab. My actions and comments do not reflect * >> * in any way on MIT. Moreover, I am nowhere near the Boston * >> * metropolitan area. * >> * * >> **************************************************************************** X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i09DLVu01322; ==== >Squares of 0.999... tend toward 0, not toward 1 .9^2 = .81 > .9^3 = .729 > .9^4 = .6561 > > .99^2 = .9801 > .99^3 = .970299 > .99^4 = .96059601 .999^2 = .998001 > .999^3 = .997002999 > .999^4 = .996005996001 > > .9999^2 = .99980001 > .9999^3 = .999700029999 > .9999^4 = .9996000599960001 .99999^2 = .9999800001 > .99999^3 = .999970000299999 > .99999^4 = .99996000059999600001 .999999^2 = .999998000001 > .999999^3 = .999997000002999999 > .999999^4 = .999996000005999996000001 .9999999^2 = .99999980000001 > .9999999^3 = .999999700000029999999 > .9999999^4 = .9999996000000599999960000001 > > .99999999^2 = .9999999800000001 > .99999999^3 = .999999970000000299999999 > .99999999^4 = .99999998000000059999999600000001 .999999999^2 = .999999998000000001 > .999999999^3 = .999999997000000002999999999 > .999999999^4 = .999999996000000005999999996000000001 .9999999999^2 = .99999999980000000001 > .9999999999^3 = .999999999700000000029999999999 > .9999999999^4 = .9999999996000000000599999999960000000001 .99999999999^2 = .9999999999800000000001 > .99999999999^3 = .999999999970000000000299999999999 > .99999999999^4 = .99999999996000000000059999999999600000000001 .999999999999^2 = .999999999998000000000001 > .999999999999^3 = .999999999997000000000002999999999999 >.999999999999^4 = .999999999996000000000005999999999996000000000001 [snip rest for brevity] Prove squares of 0.999... tend toward 1 >or stop claiming that 0.999... = 1 > Garry Denke, Geologist >Denoco Inc. of Texas You're a geologist? Don't you need to know arithmetic for that? In the first place since 0.999... is a single number it makes no sense to talk about squares plural. What you (perhaps) mean is that squares of the partial sums, 0.9, 0.99, 0.999, etc. I say (perhaps) because you appear to be really looking at cubes, fourth powers, etc. The squares YOU give are: .9^2= 0.81, .99^2= 0.9801, .999^2= 0.998001 ... to .999999999999^2 = .999999999998000000000001. And you claim that these tend to 0??? What IS true is that if a< 1 then a^2, a^3, a^4 ... tends to 0- but that's NOT the same as saying squares of 0.999.... tend to 0. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i09DLVV01334; ==== X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i09EwYE08366; ==== >Rationals are Uncountable Let S be the set of all rational numbers [0,1). >s is a member of S if 0.000... <= s < 1.000... >and s is rational. Assume s is represented in base factorial (!). >Base ! is used because every rational number >has a finite representation in base !. In base ! the allowable digits for >position k are (0,1,...,k). >(k starts at 1) Every position, k, represents 1 / (k+1)!. k >1 1/2! = 1/2 >2 1/3! = 1/6 >3 1/4! = 1/24 .123 (base !) = 1/2 + 2/6 + 3/24 = 0.958333... (base 10). Every rational number has an unique finite base ! >representation. Any finite base ! number is rational. We can create the set S defined above by taking the >set produced by counting in base !. .0 >.1 >.01 >.11 >.02 >.12 >.001 >.101 >... There exists a rational number, x, not in S. If S(i) is of the form .111...1 and >its length is equal to or greater than x >then set x to a string of 1's one longer than S(i). > If I understand correctly, you seek an x of the form .111...1 one digit longer than the maximum x of this form. Since you list all of them (all numbers of this form are rationals in the [0,1) range) there's no such maximum, and therefore no such x. >x differs from every member of S. >x is a rational number because it has a finite number >of digits. The length of x is exactly one greater >than some member of S. 0.0 <= x < 1 x = 1/2! + 1/3! + ... + 1/k! >and equals the largest rational number >less than the fractional part of e. Actually, it is the smallest rational approximation of e that is not in set >S >where a rational approximation of e is any number of the form .111...111. >Russell >- Zeno is right. Motion is impossible. > X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i09EwZO08407; ==== I have a problem to solve where I stuck with this subproblem: Say m(x) is a certain function, for which transformation T(m(x)) = k(x)can be said that: Sum(i=1 to (p-1)){(-1)^(i+1)*(diff^i(k(x))/(diff(x))^i))} is always larger than (-1)^(p+1)*( diff^p(k(x))/(diff(x))^p) ) could anyone help me whith this. Kees X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i09FdvP11549; ==== A MATHEMATICAL PLAY IN THREE ACTS PROLEGOMENON [An informal definition of naturals] · an infinite set of whole numbers...., etc. A direct outcome of this definition is that * Point 1: There are not naturals with infinite nonzero digits. * Point 2. As a result of Point 1, it is not possible to assume the bijection N <-> R. - Proof (see below). # FIRST ACT [Cantor and Human Curiosity] HC: Will it be possible to count the elements of R? C: Perhaps. HC: But how? [Cantor thinks for a moment, and states a criterion] C: Well, I think it is easy. Without getting into extreme formalities, we would be able to say that an infinite set would be countable if and only if we can establish a bijection between K and N, i.e. N <-> K. HC: Thatās great! It sounds elegant, logical and intuitive. However, I think we will have a little problem using it with R. C: Why? HC: Because Point 2 tells us that we cannot propose that bijection with R. So, your criterion doesnāt work in this case. We will have to look for another criterion. Itās a pity, because it was nice. C: Yes, thatās true. # SECOND ACT [Mrs. Perfect Proof1, Mrs. Perfect Proof2, Mrs. Perfect Proof_n and Human Curiosity] answer to your curiosity is that the reals are uncountable, because it is no possible to achieve the bijecci.97n N <-> R. PP2: Yes I have reached the same conclusion. HC: Why? PP2: For the same reason that PP1. HC: And you? PP_n: Exactly the same that my colleagues. HC: Well ladies, thanks a lot for your interest, but itās not possible to conclude that R is not countable from that criterion, because it cannot be used with R. PP1: Oh, we are very sorry. We did our best. # FINAL ACT [Mr. Na.95ve and Human Curiosity] MR. NAIVE: Do you know, Mr. HC? If we admit Cantorās criterion to count R, as we positively know that N <-> R is not possible (it doesnāt matter why), it is clear that R will be necessarily uncountable. What do you think about it? HC: Well Mr. Na.95ve, we can do so, but it will not convince anybody. The problem is that we know that Point 2 invalidates Cantorās criterion. Therefore, R would be certainly uncountable, but no because R be intrinsically uncountable, but because we donāt have a valid criterion to count it. MR. NAIVE: Well then, perhaps there are other different criteria to count the reals. HC: I donāt know it, but if we put a simple analogy perhaps we could clarify our thoughts. Letās try it. We may take the following equivalence table N <==> M = A set of linear units of measure R <==> A = The air that there is in a room Being our goal to find out whether A is measurable or not, we have the following analogy: POINT 1: A cannot be measured with linear unities, so we cannot assume M -> A. CRITERION: A set D will be lineally measurable if and only if it is possible M -> D. POINT 2: We canāt use CRITERION to find out whether A is measurable or not, because POINT 1 does not allow it. PROOFS: We have found out that A is not measurable, because CRITERION case. NOSENSE: We can admit CRITERION to measure A, and consequently A is not measurable. Well, clearly it seems a foolish thing to say. In the first place because POINT 1 tells us that we cannot apply CRITERION to measure A, and in the second place because A would be able to be measured by other means. Good, Mr. Na.95ve, this is the end of our analogy. MR. NAIVE: Which is the result? Are there other criteria to find out whether R is countable or not? HC: Well, in our analogy A cannot be measured with linear units, but if we arrange orthogonally three of them, then we can. By analogy, using two or more bijections N <-> R at once perhaps we could count R. Think about this, MR. Na.95ve, think about this. MR. NAIVE: Iāll do it. *************************** Proof of Point 2 INITIAL ASSERTIONS 1) We can represent any real number using a given positional system of numeration. For ease we will use the decimal notation. 2) Cantor uses a transfinite method (the diagonal argument) to prove that the premise N <-> R is false, and we will use another transfinite method to prove that the transfinite construction N <-> R is impossible, and therefore we cannot assume the bijection N <-> R. Proposition: The transfinite construction N <-> R is not possible Proof: Every real number within the interval (0, 1) has as first decimal digit one of the ten digits of the decimal system of numeration, i. e. it must be of the form 0.0, 0.1, 0.2, ·, 0.9. As you can see for this first digit there are 10 ^ 1 possibilities. For the second digit we have 0.00, 0.01, 0.02, ·, 0.99. Consequently, there are 10 ^2 possible combinations for the two first digits, 10 ^3 for the first three digits, and so on. When the number of digits is infinite we get R. Now, when we have only a digit of the decimal expansion of the reals, we make the one-to-one correspondence 0 <-> 0.0, 1 <-> 0.1, 2 <-> 0.2, ·, 9 <-> 0.9. Next, with two digits we begin the 1-1 correspondence 0 <-> 0.00, 1 <-> 0.01, ·, 10 <-> 0.10, 11 <-> 0.11, ·, 99 <-> 0.99. With three digits, we continue doing the same, and so on. And now is when the impossible transfinite construction appears. While we keep doing the 1-1 correspondence within the finite decimal expansion, everything will go fine. But, what natural would correspond to the decimal expansion of the number e? By induction it is obvious that it should be 71828182·, i. e., a number with infinite nonzero digits, which doesnāt belong to N (Point 1). Conclusion: The transfinite construction N <->R is not possible if we do not admit naturals with an infinite number of nonzero figures. If we admit that, then the reals would be countable. Nicolas de la Foz X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i09EwZA08393; ==== okay the equations can be modified to 1.(sig)^2 = E[a0^2] + E[a1^2] + E[a2^2] + ..... +E[aN^2] 2.(sig)^2 = E[a0^2] + 2^2.E[a1^2] + 2^4.E[a2^2] + ..... +2^2N.E[aN^2] ......................... which reduces the problem to finding a closed-form expression for the matrix > -- -- >| 1 1 1 . . . 1 | >| 1 2^2 2^4 . . . 2^2n | >| . . . . . . | >| 1 n+1^2 n+1^4 . . . n+1^2n | > -- -- which is definitely a vandermonde matrix...i have tried the routines given in the numerical recipes in C book and the ways to invert i found on the net..i want to solve it for n as large as 1024..a routine written on the computer gets out of hand very fast as the intermediate steps in the matrix inversion entail finding factorials of numbers as large as n..is there a routine/method/algorithm that circumvents this problem of intermediate steps requiring huge numbers... please help!!!!! Ankit Seedher Texas Instruments, India >>Could you describe what problem you are trying to solve? The problem is the following: >An electronic circuit i am trying to analyze has an output of the form: >y(x) = a1.x + a2.x^2 + a3.x^3 + .... + aN. x^N The inputs {x} are discrete points from the set {1,2,3,4,...N} The ideal output is y = x => a1 = 1, aj = 0 for j ~=1 >However, the actual outputs are such that yj (output for x = j) is a gaussian random variable with mean j and variance j(sig)^2. All yj s are independent RVs. i.e. cov(yi,yj) = 0 for all i,j. >It can also be assumed that aj (for j ~=1) is a zero mean r.v. I basically want the variances of aj given that yj is known to be an r.v. with gaussian pdf N(j, j*(sig)^2). Also cov(ai, aj) = 0 for all i,j. So we have, >y1 = a1 + a2 + ....... + aN >y2 = 2.a1 + 2^2.a2 + ...... + 2^N.aN > ...................... >yN = N.a1 + N^2.a2 + .......+ N^N.aN Now if we take the mean square expectation on this equation we get 1.(sig)^2 = E[a1^2] + E[a2^2] + ..... +E[aN^2] >2.(sig)^2 = 2^2.E[a1^2] + 2^4.E[a2^2] + ..... +2^2N.E[aN^2] >......................... So we get a set of N linear equations. We wish to solve for E[a1^2], >E[a2^2],....,E[aN^2]. Basically, we wish to obtain a closed form expression for E[aj^2] which boils down to finding the inverse of the matrix: > -- -- >| 1 1 . . . 1 | >| 2^2 2^4 . . . 2^2n | >| . . . . . . | >| n^2 n^4 . . . n^2n | > -- -- I had expressed this matrix in a general from in my post yesterday. So we need to invert this matrix and get a closed form general expression for E[aj^2] in terms of (sig)^2. I shall try to dig up some info about Lagrange interpolation as u suggested to see if it can be applied here. Please let me know if you have some other suggestions now that I have described the problem. I would greatly appreciate any help. >Ankit Seedher >Texas Instruments, India. >> I am working on an engineering problem that has got stuck at a point where >>I need to find a closed form expression for the INVERSE of the following >>n-by-n matrix: > -- -- > | 1 1 . . . 1 | > | x1 x1^2 . . . x1^n | > | x2 x2^2 . . . x2^n | > | . . . . . . | > | xn-1 xn-1^2 . . . xn-1^n| > -- -- >>Could you describe what problem you are trying to solve? >>It looks like you are trying to interpolate a polynomial, perhaps finding >>coefficients of a polynomial through given points. Well, this can easily be >>accomplished using Lagrange interpolation. >>However, often you don't need the explicit values of the coefficients, but >>merely an efficient means of evaluating the polynonial at arbitrary points. >>In that case you might consider Chebyshev polynomials. >>I'll happily expand on these matters if you are interested. >>If you really want an analytic expression for the inverse of the matrix, I'd >>guess a simple formula could be given, but I don't have it here. I suggest >>considering a 3x3 matrix and see if you can determine a pattern by visual >>inspection. >>-Michael. >> X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i09HoAO22250; ==== > 1)The function f(x) has f ''(x) > 0 for all x > 0 > and the graph is asymptotic to y = ax + b as x -> oo. > Show that f(x) - ax - b has negative derivative for all x > 0, > and that f(x) > ax + b for all x > 0. I have an eyeball approach to the problem . . . not much rigor. Restricting our discussion to x > 0, the curve is always concave UP. Since it approaches the line y = ax + b asymptotically, the distance f(x) - (ax + b) is constantly decreasing. Therefore, its derivative will be constantly negative. To be concave UP and approaching a line (slanted or horizontal), the curve must be ABOVE the line. (Make a sketch.) Therefore, f(x) > ax + b. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i09HoAq22246; ==== >Let sigma(n) be the sum of the positive divisors of n. There is a term(Hardy ?; Ramanujan?) for sigma(n): sigma(n)=Pi^2n/6( 1+(-1)^n/4 + 2cos((2/3)nPi)/9 + > 2cos((1/2)nPi)/16 + 2(cos((2/5)nPi)+cos((4/5)nPi))/25 + > 2cos((1/3)nPi)/36 + ...) for which I have lost the reference. Is there a general expression for this term or how can I calculated the next elements? GG > You are still interesting? X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i09KfO603630; ==== >
> 
>Sorry, I meant to ask: what day of the year is the longest?
>This question is usually given a wrong or incomplete answer.
> 
> All the days are 24 hours except when a leap second is added.
> If you mean to ask which day has the longest daylight at a given
> location, that would be the day of the summer solstice.  
>>That is true if you define the day to be from sunrise to sunset.
>>But if you define the day to be from sunrise to sunrise, then
>>the day is longest when the Earth is closest to the Sun, which
>>happens around December 23 (a little after winter solstice).
I define the day to be from midnight to midnight.  Perhaps you meant to
>ask about the solar day, which is the time between transits of the
>sun.  However, your answer is incorrect.  Perihelion occurs in early
>January (around January 8, IIRC), and therefore the answer is *not* the
>perihelion date.  In general, the solar day is longer than average near
>the solstices (both of them) and shorter than average near the
>equinoxes.
The length of the solar day is determined by the sun's daily movement
>in right ascension, which determines the difference between the
>sidereal day and the solar day.  What's confusing you is that the sun
>moves fastest *in longitude* at perihelion, but fastest movement in
>longitude does not equal fastest movement in right ascension.  The
>obliquity of the ecliptic causes the fastest movement in right
>ascension to occur near the two solstices, despite the fact that one of
>them happens to be near aphelion.
>
Why is Dec. 19 the longest solar day instead of perihelion Jan. 4th?
>The longest solar day is presently achieved near the December solstice,
>because that one is closer to perihelion than the June solstice is.
>Because of precession of the equinoxes, there will come a time when
>perihelion is not particularly close to either solstice, and therefore
>nowhere near the longest solar day.
-- 
>Dave Seaman                    dseaman@purdue.edu
>      ++++ stop the execution of Mumia Abu-Jamal ++++
>    ++++ if you agree copy these lines to your sig ++++
>++++ see http://www.xs4all.nl/
~tank/spg-l/sigaction.htm ++++

>
X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i09JbDx30828; ==== >Okay I have never been very good with math and hopfully someone out >there can help. Here is the situation: I Have a total of five different items: I am shipping out bowes with a >combination of these different items and need to prepare all the >different combinations before hand. So lets name the items: >1. tshirt >2. glove >3. hat >4. ring >5. necklace now a package can have either all five items or a choice of lets say >number 1,2,4 or 1,2 or 1,4,5 or just 2 and so on... I think you probly >get the jist of what im asking. What are the total number of combinations for the items obviously >without just rearanging thier order. and what is the formula to do so? Happy New Year Assume that each item can be either in or out of the box that you are working on. Assume that the choices are independent. Assume that the items are unique (not diplicated). Then the number of possible boxes is 2 x 2 x 2 x 2 x 2 = 32 , but one is an empty box. So your answer is 31. phil X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i09LKeV06757; ==== >It took about 22 seconds to print out 1,000,000 primes. How much faster is >it able to be? also the program produces 10,000,000 primes in 22.27 seconds The program stores all the primes in a bit array and can be accessed to see if it is prime or not. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0A3Bq119159; ==== I am working on the program where is necessary to calculate huge >amount of numbers in always the same sequence, before the program >actually opens. In other words, these number have to be calculated and saved into >memory before the program starts. I was trying to use number pi, but to calculate 100 million decimal >places of number pi would take forever and no one would want to wait >that long. On the other hand I can't afford to include the whole 100 >million dec. places along with the software, that would take too much >of the space and too much time to download. Is there any other number like PI, which I could calculate faster? >Serge >---------------------------------------------------------- > ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** >---------------------------------------------------------- > http://www.usenet.com What do you mean when you say like pi? Transcendental? Irrational? Real between 3 and 4? ... phil X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0A3Btg19224; ==== >Greetings math lovers, I was given the following probability problem as part of a >job-interview pre-screen. After giving a totally wrong answer ('cause >I wasn't thinking), I gave the following answer. However, the >interviewer said that my answer was still incorrect. Fair enough; I find probability unintuitive. But now my curiosity is >peaked: What is the right answer? Question: You have a jar with 1000 coins in it, 999 are normal, but >one is two-headed (i.e both sides are heads). You choose a coin at >random and toss it 10 times. With each toss the coin comes up heads, >so you get 10 consecutive heads. What is probability that you have >choosen the two-headed coin? My answer: P(two headed) = 1 - P(normal coin *and* 10 heads in a row) > = 1 - .999 * (1/2)^10 > = 1 - .999 * (1/1024) > = 1 - .000976 > = 0.999024 > = 99.9024% If this is wrong, what is the right answer? How do you calculate it? Stuart Nice math, Stuart, but wrong. Instead of saying there is one two-headed coin, say there is one red coin. If you pick one coin at random out of 1000 (without looking), the probability that you picked the red coin is 1/1000. You can toss the coin all day long and this doesn't change. Get it? phil the probability it is the X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0A3BqJ19163; ==== I have just attended a course on scheduling for project managers and there >the beta distribution comes up in relation to estimation of task durations. I have looked up in my old probability theory book regarding this and found >the proof for the fact that the sum of _many_ beta distributions converges >to a gaussian distribution, but neither my beloved book (and some hard core >integral solvining) or extensive search on the internet has given me >what I have found (this includes characteristic function of the beta >distribution and it's density function) I am beginning to believe that there >is no distribution defined/named for the sum of two beta distributions, but >I might be wrong. Can anyone answer this one? /Torben >-- P.S. The views expressed above are my own. > Are you sure you want the sum of two beta distributions? Perhaps you want the pdf for the sum of two random variables, each independently distributed according to beta distribution, eh? phil X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0A3Bqd19155; ==== i have a rough draft. the subjects/approach are these: 1. using ternary logic to 2. axiomatize a universal set into existence while 3. avoiding russell's paradox by 4. modifying the subsets and foundation axioms in a way that theoretically shouldn't change anything when those axioms are applied to crisp sets. in other words, this is a strict extension of zfc. the main idea is that in russell's paradox, if S={x in U: x is not an element of x}, then the truth value of the statement S is in S is the third truth value. the details are in my paper. as i alluded to, when applied to crisp sets, the new subsets axioms and new foundation axioms work the same way. i have only found two fuzzy sets so far and they are both related to S. part of the idea is to construct some new membership symbols; there are four: 1. a membership symbol to mean membership is ambiguous, unknown, or general 2. a membership symbol to mean membership is true 3. a membership symbol to mean membership has the third truth value 4. a memberhsip symbol to mean membership is false. since symbols mentioned in 2-4 are crisp objects (eg, for 2, either it is true that membership is true or it's false that membership is true), standard deduction, contradiction, etc, applies. in fact, if all other axioms are taken to mean 2 when the usual symbol is written and 4 when the negation of the usual symbol is written, then it should work out. in the subsets axiom, i used a new biconditional that i defined by a truth table. this biconditional (and the new conditional it's based on) is an extension of the usual (bi)conditional, i.e., when the truth values are limited to T and F, it is the same. here is a link to it: http://www.physicsforums.com/attachment.php?s=&postid=124723 this is a zipped pdf located at physicsforums. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0AD3T816500; ==== >
>Anyway, I see no reason to stop there. Since
>>the system already uses deci- and centi-,
>>why not use all metric fractional prefixes? This
>>would give: 
>decillion =3D 10^33
>>centillion =3D 10^303
>>millillion =3D 10^3003 =3D 10^(1000^1 + 3) 
>>micrillion =3D 10^3000003 =3D 10^(1000^2 + 3)
>>nanillion =3D 10^(1000^3 + 3) 
>>picillion =3D 10^(1000^4 + 3)
>>femtillion =3D 10^(1000^5 + 3)
>>attillion =3D 10^(1000^6 + 3)
>>zeptillion =3D 10^(1000^7 + 3)
>>yoctillion =3D 10^(1000^8 + 3) 
>Uh, oh... ran out of prefixes! :-( =A0 BTW, the
>>American version of that last one has
>>1,000,000,000,000,000,000,000,003 zeroes,
>>and the British version (10^(1,000,000^8)) has
>>1,000,000,000,000,000,000,000,000,000,000
>>000,000,000,000,000,000 zeroes (somewhat
>>difficult to write out in full). 
Ooops... that should read:
decillion =3D 10^33
>centillion =3D 10^303
>millillion =3D 10^3003 =3D 10^(3*1000^1 + 3) 
>micrillion =3D 10^3000003 =3D 10^(3*1000^2 + 3)
>nanillion =3D 10^(3*1000^3 + 3) 
>picillion =3D 10^(3*1000^4 + 3)
>femtillion =3D 10^(3*1000^5 + 3)
>attillion =3D 10^(3*1000^6 + 3)
>zeptillion =3D 10^(3*1000^7 + 3)
>yoctillion =3D 10^(3*1000^8 + 3) 
where yoctillion has 3,000,000,000,000,000,000,000,003 zeroes
>(American), or 6,000,000,000,000,000,000,000,000 zeroes (British).
> 
....

DWIII

>
X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0AD3V716544; ==== I'm writing because a search showed that my infinite-precision calculator (http://medialab.dyndns.org/bignum) was cited here by uncle al. Unfortunately, dyndns.org have blocked our site without any warning, making it point to 10.0.1.128 instead (whoch doesn't exist, claiming that one of the 400 users on the site has published something that is covered by copyright. Yeah, right. Anyway, fortunately we have another name, medialab.freaknet.org, for the same computer, so my calculator can now only be reached as http://medialab.freaknet.org/bignum Please update your bookmarks as they say. All the best Martin X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0AKqwU16572; ==== >>3.14159265358979323... >>3,1,4,5,?,?,.... >How many digits of pi have you memorized? I did 100. The world record > > I knew about 1500 digits at one time. A. Say > I HAVE A NUMBER > to remember PI as 3.1416 B. Desire > MAY I HAVE A LARGE CONTAINER OF COFFEE > to remember PI as 3.1415926 C. Chant > GOPEEBHAGYA MADHURVAATHA SHRADGI SHODHADHI SANDHIGA I > KHALAJEEVITHA KHAATHAAVA GALAHAALAARA SANDHARA II > a sanskrit verse from Atharvana veda,which is in praise of > Lord Vishnu and Lord Shankara.If this is deciphered as per > language of Vedic Mathematics,it denotes the value of PI/10 to > 32 decimal places as below: > .31415926535897932384626433832792 > You need to know/learn which of the alphabets represent each of > the numerals 1,2,3,4 .... > Krsna Tirthaji Maharaja(1884-1960)Sankaracharya of Govardhana > Matha,Puri,Orissa State,INDIA) D. Professor Yasumasa Kanada and nine other researchers at The > Information Technology Centre at Tokyo University,calculated the > value of PI to 1.2411 trillion places,in September 2002. > which computed this value working continuously for over > 400 hours. > - bsr A should be read as 'YES,I HAVE A NUMBER' X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0AKr2B16618; ==== >> what I think you are missing is that .999... (a zero, a dot and an >> infinite amount of 9s) is equal to 1, so what you are asking is >> whether 1^2, 1^3, 1^4, etc. is 'tending to' 1. i understand the premise. > Look at it this way: We can write a number like 0.9 as 1 - 0.1, and a >> number like 0.99 as 1 - 0.01. In general we can write it down as: >> >> f(n) = 1 - (1/(10^n)) for n>0 where n is the number of nines behind >> the dot. >> >> So the number 0.999... can be found by looking at: >> >> lim{n->infinity}{f(n)} = >> lim{n->infinity}{1 - (1/(10^n))} = >> lim{n->infinity}{1} - lim{n->infinity}{1/(10^n)} but infinity means unlimited... limit{n->unlimited}{f(n)} = >limit{n->unlimited}{1 - (1/(10^n))} = >limit{n->unlimited}{1} - limit{n->unlimited}{1/(10^n)} ...which contradicts the premise. > >> Since the limit of the fraction-part goes to zero, the whole thing >> becomes equal to 1. In other words 0.999... is equal to 1. only if the premise the unlimited has a limit is believed. garry denke, geologist >denoco inc. of texas Now you seem to be playing games with words. The fact that n is unlimited does not mean the sequence is. It is easy to show that the limit limit{n->unlimited}{1} - limit{n->unlimited}{1/(10^n)} is 1. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0AKr5116644; ==== If I post it here is anyone here willing to look over an outline argument that there are no odd perfect numbers and point out the flaw/s? Any flaw shouldn't be too hard to spot: it's a mainly geometrical argument. Many thanks in advance, Mark G. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0AKrBi16725; ==== I have this question here that I attempted and failed. some pointer toward >this problem is greatly appreciated. The golden ratio is the ratio b:a so that b:a = (a+b):b. The ratio of the >golden ratio was considered by the ancient Greeks to be the most perfect >ratio for the lengths of the sides of rectangles, such as portraits. Show >that if b:a is the golden ratio, then b/a = (1+ sqrt(5))/2. I guess the question is really saying to derive b = (1+sqrt(5)), and a = 2 >from b/a = (a+b)/b (*). So I tried to manipulate with the algebra of (*), and failed. >Is there anything i am doing wrong? > Gavin , [ b+a]/b=b/a , or ab+a^2=b^2 , or b^2-ab-a^2=0 giving b=[a+(or-)sqrt{a^2+4a^2}]/2 =a[1+-sqrt(5)]/2 or b/a=[1+sqrt(5)] or b/a=[1-sqrt(5)]/2 But do it simply (the Classical Greek-UNGRADUATED STRAIGHT EDGE AND COMPASS mrthod): Draw sqrt(5),as the hypotenuse[AC] of the orthogonal triangle[ABC] with horizontal[AB] side 2, and vertical[BC] side 1. Etend AC by 1 unit to D to btain line ACD=sqrt(5)+1 . Use compass and divide line AD, at point E so that AE=(AC)/2 or AE = [sqrt(5)+1]/2. Panagiotis Stefanides http://www.stefanides.gr http://www.stefanides.gr/piquad.htm X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0B0PWM27730; ==== > >> There is another slightly subtle flaw with your argument. You know >> that a_1(x,y) and a_2(x,y) are roots of >> >> a^2 - (x - y) + 7(x^2 + xy). Typo, this should be a^2 - (x - y)a + 7(x^2 + xy). > You're right of course, thanks. >> >> There are two roots. Perhaps you choose as the 'negative' root, >> >> a_1(x,y) = ((x - y) - sqrt((x - y)^2 - 28*(x^2 + xy)))/2 >> >> and the other as the 'positive' root, >> >> a_2(x,y) = ((x - y) + sqrt((x - y)^2 - 28*(x^2 + xy)))/2. >> >> The choice is arbitrary. Your subsequent argument makes no >> use of whether the negative or positive root is chosen. No. At x=0, one of these roots ( the 'positive' root) is 0, >the other (the 'negative' root) is -y. >James chooses a_1(0,y) to equal 0 and a_2(0,y) = -y. Values for >other x can be specified by chosing a branch cut and insisting on >continuity. There is no ambiguity as long as y is not equal to 0. > Yes, you're right again. I knew this and had forgotten it. And similarly for the cubic. Careless thinking on my part. Other arguments however obviously still apply. I am wondering what error Harris himself thinks he has found in his choice of adding the variable y to the polynomial. Nora B. >> >> The same problem actually occurs with your argument about >> your cubic polynomial, except there you are arbitrarily choosing >> one of three roots a_1, a_2, and a_3 instead of one of two. You >> conclude that two of the three must be divisible by 7, and the >> other coprime to 7. Again James choses a_1 and a_2 to be 0 at 0. Again we can resolve the >ambiguity at other values of x by using continuity and branch cuts. >The double root at x=0 (and possible other double or triple roots for >other values of x) leads to a minor ambiguity, but not an important one. - William Hughes X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0B386L00778; ==== >>How to integrate sinc(x) (without Fourier transformation) ?? >> That integral is not an elementary function. It is known as the Sine >> integral function. >> <http://mathworld.wolf ram.com/SineIntegral.htmlto take it from a book. Is it possible by an easy process, or else... no way >?.. The Fourier transform of sinc(x) is a rectangle function in the frequency domain. You can derive this directly. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0B389E00789; ==== >If I post it here is anyone here willing to look over an outline argument that there are no odd perfect numbers and point out the flaw/s? Any flaw shouldn't be too hard to spot: it's a mainly geometrical argument. Many thanks in advance, Mark G. So post already! X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0B5tJS13327; ==== >>Message-id: >The lopsided Collatz tree? >Maybe this is known so just disregard! >>I couldn't find anything about it on web searches about the >>Collatz. >Probably of little value for the overall conjecture but still a >>curiosity! >> >>The Collatz tree level count is odd up to level 5 where level >>1,2,3,4,5 [1,2,4,8,16] has a count of (1) for each level. >>At level 6 the count becomes even and fall one for one on each >>side of the tree up too and including level 16. >>This changes @ level 17 where the left side of the tree adds one >>more path. >>This happens also for certain levels higher than 17 where the >>left side adds 1 more path over the right side. >I have denoted the 2 sides of the tree by their highest-ranking >>sequence. >Where [1,2,4,8,16] is the trunk and the first main branch >>assigned to the RIGHT side of the tree is -- >>32,64,128,256,512,1024,2048,4096...and with all applicable nodes >>branching from it creating all branches on the RIGHT side of >>the Collatz tree. >Where [1,2,4,8,16] is the trunk and the first main branch >>assigned to the LEFT side of the tree is -- >>5,10,20,40,80,160,320,640,1280,2560,...and with all applicable >>nodes branching from it creating all branches on the LEFT side of >>the Collatz tree. >> >> 8) 20 3 21 128 >> | | / >> 7) 10 64 >> / >> 6) 5 32 >> / >> 5) 16 >> 4) 8 >> 3) 4 >> 2) 2 >> Levels 1) 1 >Why does the left side have this property of adding (1) more >>branch at certain levels =>17 then the right side? >Will this mean, at a very high level there will be many more >>branches on the left side of the tree than on the right side >>of the tree making the tree lopsided? >Putting the above question in another perspective, at some higher >>level will the right side branching number eventually catch up and >>equal the left side branching number? >It could be that the left side of the tree is facing the sun giving >>that side of the tree a more vigorous growth. ;-) >Dan >> >> Personally, I don't like the way you draw the Collatz tree. My preference >> is for vertical placement to always represent x*2 and for the diagonal >> placement to always represent (x-1)/3: >> >> 3 20 128 21 >> | | / >> 10 64 >> | | >> 5 32 >> | >> 16 >> | >> 8 >> | >> 4 >> | >> 2 >> | >> 1 >> >> By your method, 32 is part of the right branch whereas I consider the right >> branch >> single left branch and right branch. Rather, there is the central trunk from >> which >> sprout an infinite number of branches on either side, so I don't see any >> problem. >> >> Note that in my view each branch starts with an odd number and extends >> vertically >> to infinity, all of which are even. Does this mean there are infinitely more >> even >> numbers than odd? >> >> Infinity always catches up in the end. Mansenator, I know the way I did the tree was different but I used each sequence >as a one to one correspondence (left,right) with each other. >Where the (2) sequences, [3,6,12,24...]is the last sequence left of >center and [21,42,84,168...] is the last sequence right of center. >This contains the inner limits and actually separates the tree into >two halfs because these two sequences are 0 (mod 3)'s and will >continue to double and continue on vertically ----->oo. >The other two sequences, [5,10,20,40,80,160...]is the last sequence >or outer branch limit of the left side of tree and [32,64,128,256,...] >is the last sequence or outer branch limit of the right side of tree >where both sequences double ----->oo. >Presenting the tree in this why, it just seemed neat, up to level >16 that you have this even looking tree and with boundry limits >even after level 16. >Using this representation of the tree it is obvious the level seed >counts are the same as the standard tree layout. Right now it appears after level 16 there are increasingly more >sequences with a (5) in its path other than a (32) in its path. True or false? > >This can be checked with the standard tree also proving or disproving >my claim. >Dan With further checking I believe the statement I made above about more sequenses with a (5) in them instead of a (32) is false. It does appear to go back and forth where sometimes the count for (32) sequences is higher than the (5) sequences and visa versa or somtimes the count is the same at any given level of the Collatz tree. Checking higher levels, what I thought was happening was not do to a programming error. Please forgive me for the blunder. Dan X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0B5tK213336; ==== what is inertia? is it measurable?how? X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0BDiDk23427; ==== I would like to know if this is a mathematical proof of the Cantorās goof, or conversely it is just another mathematical goof about the Cantorās proof. A MATHEMATICAL PROOF IN THREE ACTS PROLEGOMENON [The naturals] A direct outcome of the definition of the natural numbers is that * Point 1: There are not naturals with infinite nonzero digits. * Point 2. As a result of Point 1, it is not possible to assume the bijection N <-> R. - Proof: (see below). # FIRST ACT [Cantor and Human Curiosity] HC: Will it be possible to count the elements of R? C: Perhaps. HC: But how? [Cantor thinks for a while, and states a criterion] C: Well, I think it is easy. Without getting into extreme formalities, we would be able to say that an infinite set K would be countable, if and only if we would be able to establish a bijection between K and N, i.e. N <-> K. HC: Thatās great! It sounds elegant, logical and intuitive. However, I think we will have a little problem using it with R. C: Why? HC: Because Point 2 tells us that we cannot propose that bijection with R. So, your criterion doesnāt work in this case. We will have to look for another criterion. Itās a pity, because it was nice. C: Yes, thatās true. # SECOND ACT [Mrs. Perfect Proof1, Mrs. Perfect Proof2, Mrs. Perfect Proof_n and Human Curiosity] answer to your curiosity is that the reals are uncountable, because it is no possible to achieve the bijecci.97n N <-> R. PP2: Yes I have reached the same conclusion. HC: Why? PP2: For the same reason that PP1. HC: And you? PP_n: Exactly the same that my colleagues. HC: Well ladies, thanks a lot for your interest, but itās not possible to conclude that R is not countable from that criterion, because it cannot be used with R. PP1: Oh, we are very sorry. We did our best. # FINAL ACT [Mr. Mathman and Human Curiosity] MR. MATHMAN: Do you know, Mr. HC? If we admit Cantorās criterion to count R, as we positively know that N <-> R is not possible (it doesnāt matter why), it is clear that R will be necessarily uncountable. What do you think about it? HC: Well Mr. Mathman, we can do so, but it will not convince anybody. The problem is that we know that Point 2 invalidates Cantorās criterion. Therefore, R would be certainly uncountable, but no because R be intrinsically uncountable, but because we donāt have a valid criterion to count it. MR. MATHMAN: Well then, perhaps there are other different criteria to count the reals. HC: I donāt know it, but if we put a simple analogy perhaps we could clarify our thoughts. Letās try it. We may take the following equivalence table N <==> M = A set of linear units of measure R <==> A = The air in a room Being our goal to find out whether A is measurable or not, we have the following analogy: POINT 1: A cannot be measured with linear unities, so we cannot assume M -> A. CRITERION: A set D will be lineally measurable if and only if it is possible M -> D. POINT 2: We canāt use CRITERION to find out whether A is measurable or not, because POINT 1 tells us that A canāt be measured in this way. PROOFS: We have found out that A is not measurable, because CRITERION case. PROPOSAL: We can admit CRITERION to measure A, and consequently A is not measurable. Well, no one will accept this, because POINT 1 tells us that we cannot apply CRITERION to measure A, and also because A would be able to be measured by other means. Good, Mr. Mathman, this is the end of our analogy. MR. MATHMAN: And? HC: What? MR. MATHMAN: Are there other criteria to find out whether R is countable or not? HC: Well, in our analogy A cannot be measured with linear units, but if we arrange orthogonally three of them, then we can. Perhaps the reals have an undiscovered property which may be used as criterion to count them. Think about it Mr. Mathman, think about it. MR. MATHMAN: Iāll do it. *************************** Proof of Point 2 INITIAL ASSERTIONS 1) We can represent any real number using a given positional system of numeration. For ease we will use the decimal notation. 2) Cantor uses a transfinite method (the diagonal argument) to prove that the premise N <-> R is false, and we will use another transfinite constructive method to prove that the transfinite construction N <-> R is impossible, and therefore we cannot assume the bijection N <-> R. Proposition: The transfinite construction N <-> R is not possible Proof: Every real number within the interval (0, 1) has as first decimal digit one of the ten digits of the decimal system of numeration, i. e. it must be of the form 0.0, 0.1, 0.2, ·, 0.9. As you can see for this first digit there are 10 ^ 1 possibilities. For the second digit we have 0.00, 0.01, 0.02,·, 0.99. Consequently, there are 10 ^2 possible combinations for the two first digits, 10 ^3 for the first three digits, and so on. When the number of digits is infinite we get R. Now, when we have only a digit of the decimal expansion of the reals, we make the one-to-one correspondence 0 <-> 0.0, 1 <-> 0.1, 2 <-> 0.2, ·, 9 <-> 0.9. Next, with two digits we begin the 1-1 correspondence 0 <-> 0.00, 1 <-> 0.01,·, 10 <-> 0.10, 11 <-> 0.11, ·, 99 <-> 0.99. With three digits, we continue doing the same, and so on. And now is when the impossible transfinite construction appears. While we keep doing the 1-1 correspondence within the finite decimal expansion, everything will go fine. But, what natural number would correspond to the decimal expansion of the number e? By induction it is obvious that it should be 71828182·, i. e., a number with infinite nonzero digits, which doesnāt belong to N (Point 1). Conclusion: The transfinite construction N <->R is not possible if we do not admit naturals with an infinite number of nonzero digits. If some day we admit that, then the reals would be countable. Nicolas de la Foz X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0BDiGY23463; ==== X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0BDiIE23563; ==== >I think we have to center the tread, because at this moment it looks >>like the chats in the auditorium before the concert. >POINTS >>Point 1: >>The surjection f: N -> R is not possible because the definition of >>naturals and reals do not allow it. So what is it about the definitions of the naturals and the reals that >allows you to confidently state that such a surjection is not possible? This question has been answered in my latest version. >Obviously we cannot use the definitions of the naturals and the rationals >to prove that there is no surjection f : N -> Q, since we know that there >is such a surjection. This means that you cannot use the argument thatr >N is a proper subset of R (N is a proper subset of Q and there exists a >surjection f : N -> Q). > There is a thread dedicated to this stuff and, as you can see, it is not so clear that Q is countable. I also have some arguments against this proof, but if you don't mind I will leave untreated the theme at moment. >>Point 2: >>Cantor clearly stated that we can say that an infinite set K is >>countable if and only if it is possible to establish a one-to-one >>correspondence between K and N, i.e. f: N <-> K. >Point 3: >>As the one-to-one correspondence N <-> R is not possible (Point 1), Your claim is to the effect that it is trivial to demonstrate that >such a one-to-one correspondence does not exist. Let's see your >trivial proof. > As probably you have already seen, it is a really trivial proof. You can find a less annoying version of the whole proof in the thread *A mathematical proof of the Cantorās goof?* There you donāt need to play the role of Mr. Na.95ve if you want to refute it. >>the criterion settled by Cantor (Point 2) is not valid to resolve >>whether R is countable or not. Point 1 means that R is not countable, so it does provide an answer, >except that you have not deemed it necessary to provide the trivial >proof that makes the non-existence of a surjection trivially obvious >to you, but not to others. Point 2 is valid. It is the DEFINITION >of countability of a set. If you can show that R does not satisfy >Cantor's criterion for countability as espoused in Point 2, then R >is not countable. What is so difficult about this point? > Perhaps the analogy in the latest version it will help you to grasp the idea I want to transmit. >> In other words, we cannot use this >>criterion with R. (See note below) Point 1 implies that R is uncountable. What more do you want? > Once again, the analogy may help you to see the difference. >>Point 4: >>If a proof (any) comes to the conclusion that R is not countable >>because it is not possible to accomplish the criterion established in >>Point 2, In other words, we conclude that R is uncountable because it fails >the criterion for countability. > YES >> then the proof is useless because that criterion is not >>valid for that purpose (Point 3). Yes, it is. It is the DEFINITION of countability of a set. It lays >down the one and only criterion for countability. R fails that >criterion, so R does not satisfy the definition for countability, >so R is not countable. NO. If for you *the money only can be counted if and only is it is counted by hand* is a definition, for me is a CRITERION that obviously fails in some cases. > Note that the proof could be >>correct and the conclusion too. >DEFINITION >>Useless proof: A correct proof that proves nothing. You have not demonstrated why we should accept that there is a >trivial proof of the nonexistence of a surjection f : N -> R. > Now I did >>QUESTION >>Does somebody know any other criteria (different from the one >>established in Point 2) to resolve whether R is countable or not? It is Cantor's definition for countability. This means that any >infinite set which satisfies the criterion is countable, and any >infinite set which does not satisfy the criterion is uncountable. > NO. It is Cantorās criterion. If there are not other criteria, then this is another problem. >> If >>you know one of them, I would like to know it please (but first be >>sure that it doesn't involve implicitly the criterion of Point 2). >NOTE: >>Someone could interpret that if N <-> R is not possible, then R is >>uncountable. This conclusion is true, but in this case, it will imply >>that R is uncountable because we haven't got a suitable tool to count >>its elements, since the properties of N and R are incompatible. It >>will never imply that R has more elements than N. We know that R has more elements than N since there exists an injection >f : N -> R, but there is no surjection f : N -> R. That is ALL that >you need. > Probably you are already thinking that this not as obvious as it looks like. Nicolas de la Foz >David McAnally Despite anything you may have heard to the contrary, > the rain in Spain stays almost invariably in the hills. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0BDiJG23577; ==== >In <200401071748.i07HmR628278@proapp.mathforum.org>, on 01/07/2004 > at 05:50 PM, nico80@jazzfree.com (Niclas de la Foz) said: >Point 3: >>As the one-to-one correspondence N <-> R is not possible (Point 1), >>the criterion settled by Cantor (Point 2) is not valid to resolve >>whether R is countable or not. No. If you stipulate that R is uncountable, then it immediately >follows that R is uncountable. If you don't stipulate it, a reductio >ad absurdum is perfectly reasonable, although not necessary. A >reductio ad absurdum does not assume a false statement: it proves that >the statement must be false by proving that it leads to a >contradiction. > It is not the same to be INTRINSICALY uncountable, that to be uncountable because we havenāt the suitable means to do it. Please, read carefully my latest version of the proof, and if necessary have a look to the analogy just to grasp the idea. (you can find this poof in the tread *A mathematical proof of the Cantorās goof?*) >>Point 4: >>If a proof (any) comes to the conclusion that R is not countable >>because it is not possible to accomplish the criterion established >>in Point 2, then the proof is useless Incorrect. >because that criterion is not valid for that purpose (Point 3). No, it is your Point 3 that is not valid. >DEFINITION >>Useless proof: A correct proof that proves nothing. What do you mean by proves nothing? > An useless job. >>QUESTION >>Does somebody know any other criteria (different from the one >>established in Point 2) to resolve whether R is countable or not? Do you know a definition of frequency other than the number of cycles >per unit of time? The definition is the definition, and any proof >ultimately comes down to showing that it satisfies the definition. > Why do you always take out the things out of context? As G.9adel and Cohen established, there are undecidable statements. Concluding whether the reals are countable or not it probably is one of them. By the way, someone asked what the heck G.9adel has to do with this. Here you can find the answer. I hope you don't have to regret it >>NOTE: >>Someone could interpret that if N <-> R is not possible, then R is >>uncountable. This conclusion is true, but in this case, it will >>imply that R is uncountable because we haven®t got a suitable tool >>to count its elements, No. It will imply that no such tool exists. > Sorry but, is it not the same? If such a tool doesnāt exist, then we havenāt got a suitable one to count R. >>since the properties of N and R are incompatible. You still haven't defined what you mean by that statement in >Mathematical terms. > Statement: sentence or affirmation that can be true, false or undecidable (within the current mathematics). Nicolas de la Foz >-- > Shmuel (Seymour J.) Metz, SysProg and JOAT >not reply to spamtrap@library.lspace.org X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0BDiKC23600; ==== > I think we have to center the tread, because at this moment it looks >> like the chats in the auditorium before the concert. >> >> >> POINTS >> Point 1: >> The surjection f: N -> R is not possible because the definition of >> naturals and reals do not allow it. This statement reqires proof, as it is not self evident. See latest version. >> >> Point 2: >> Cantor clearly stated that we can say that an infinite set K is >> countable if and only if it is possible to establish a one-to-one >> correspondence between K and N, i.e. f: N <-> K. More precisely, a set K is countable if there is a surjective map from N >to K and countably infinite (or some say denumerable) if there is a >bijective map. Good mathematician! >> >> Point 3: >> As the one-to-one correspondence N <-> R is not possible (Point 1), >> the criterion settled by Cantor (Point 2) is not valid to resolve >> whether R is countable or not. In other words, we cannot use this >> criterion with R. (See note below) You assert it to be impossible, but you provide no proof. In >mathematics, no assertion need be accepted without proof. Since you >provide no proof, we may assume that you do not know whether the reals >are countable. See latest version. >> >> Point 4: >> If a proof (any) comes to the conclusion that R is not countable >> because it is not possible to accomplish the criterion established in >> Point 2, then the proof is useless because that criterion is not >> valid for that purpose (Point 3). Note that the proof could be >> correct and the conclusion too. There were, at last count, well over 100 proofs of the Pythagorean >theorem. Which ones are invalid? If all of them are correct, all of them are valid. Why? >> DEFINITION >> Useless proof: A correct proof that proves nothing. > Obviously you never have done an useless work. Otherwise you would know what a useless proof is. >This is an empty definition, since it can never be instantiated. A >correct proof, by its own definition, proves something. You are defining >the memebers of the empty set. This Futz is nearly as wacky as JSH. This statement requires proof, as it is not self evident. Nicolas de la Foz X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0BDiKh23618; ==== >On Wed, 7 Jan 2004 17:50:17 +0000 (UTC), nico80@jazzfree.com (Niclas >I think we have to center the tread, because at this moment it looks >>like the chats in the auditorium before the concert. >>POINTS >>Point 1: >>The surjection f: N -> R is not possible because the definition of >>naturals and reals do not allow it. This needs proof. Although it can easily be proved, so in some >sense it's true. In exactly the same way it is true to say that >sqrt(2) is irrational because the defintion of the rationals >do not allow it to be rational. So if Point 1 makes the standard >proof of the uncountability of R invalid it also makes the standard >proof of the irrationality of sqrt(2) invalid. (And it makes _any_ >proof by contradiction invalid.) > If sqrt(2) cannot be rational by definition, what else it can be? Irrational. Good. And now If N <-> R cannot be by definition, what else it can be? NOTHING. Go with an ćoptionlessä premise to a proof by contradiction and youāll get NOTHING (or rubbish). >>Point 2: >>Cantor clearly stated that we can say that an infinite set K is >>countable if and only if it is possible to establish a one-to-one >>correspondence between K and N, i.e. f: N <-> K. >>Point 3: >>As the one-to-one correspondence N <-> R is not possible (Point 1), >>the criterion settled by Cantor (Point 2) is not valid to resolve >>whether R is countable or not. In other words, we cannot use this >>criterion with R. (See note below) Huh? This makes just as much sense as saying because sqrt(2) >rational is not possible, it is not valid to resolve whether sqrt(2) >is rational by using the definition of the word 'rational'. See the analogy in the latest version to grasp the concept. >You really are making no sense - these rules you state about >what's valid are simply not correct, they're just things you made up. > Please, prove it in the latest version. Nicolas de la Foz X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0BGfL131857; ==== >So what you're saying is the sequence .9^2, .9^3, .9^4 ... .9^2 = .81 >.9^3 = .729 >.9^4 = .6561 >etc > >the sequence .99^2, .99^3, .99^4 ... .99^2 = .9801 >.99^3 = .970299 >.99^4 = .96059601 >etc the sequence .999^2, .999^3, .999^4 ... .999^2 = .998001 >.999^3 = .997002999 >.999^4 = .996005996001 >etc > >tend to 1 making the sequence .999...^2, .999...^3, .999...^4 tend to 1. Is that what you are saying? Garry Denke, Geologist >Denoco Inc. of Texas No, she is not saying anything like that. She is saying, as is clear, that you do not know basic arithmetic! In the first place, >.999^4 = .996005996001 is not a SQUARE as you keep asserting. Your own example show that the SQUARES do tend to 1: .9^2= .81, 0.99^2= 0.9801, 0.999^2= 0.998001, 0.9999^2= 0.99980001, etc. Any school boy with a calculator should know that, as long as a< 1, different powers of a, a^2, a^3, a^4, ... tend to 0. But that tells you NOTHING about what happens as a itself goes to 1. If a-> 1 (for example, .9, .99, .999, .9999, etc.) the a^n for any FIXED n goes to 1. This has been pointed out to you repeatedly but you don't seem to understand simple arithmetic. Are you really a geologist? Don't geologist have to know SOME arithmetic? X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0BIiYE31471; ==== >I would like to know if this is a mathematical proof of the Cantorās >goof, or conversely it is just another mathematical goof about the >Cantorās proof. A MATHEMATICAL PROOF IN THREE ACTS PROLEGOMENON [The naturals] A direct outcome of the definition of the natural numbers is that * Point 1: There are not naturals with infinite nonzero digits. >* Point 2. As a result of Point 1, it is not possible to assume the >bijection N <-> R. >- Proof: (see below). # FIRST ACT [Cantor and Human Curiosity] >HC: Will it be possible to count the elements of R? >C: Perhaps. >HC: But how? >[Cantor thinks for a while, and states a criterion] >C: Well, I think it is easy. Without getting into extreme >formalities, we would be able to say that an infinite set K would be >countable, if and only if we would be able to establish a bijection >between K and N, i.e. N <-> K. >HC: Thatās great! It sounds elegant, logical and intuitive. However, >I think we will have a little problem using it with R. >C: Why? >HC: Because Point 2 tells us that we cannot propose that bijection >with R. So, your criterion doesnāt work in this case. We will have to >look for another criterion. Itās a pity, because it was nice. >C: Yes, thatās true. # SECOND ACT >[Mrs. Perfect Proof1, Mrs. Perfect Proof2, Mrs. Perfect Proof_n and >Human Curiosity] >answer to your curiosity is that the reals are uncountable, because >it is no possible to achieve the bijecci.97n N <-> R. >PP2: Yes I have reached the same conclusion. >HC: Why? >PP2: For the same reason that PP1. >HC: And you? >PP_n: Exactly the same that my colleagues. >HC: Well ladies, thanks a lot for your interest, but itās not >possible to conclude that R is not countable from that criterion, >because it cannot be used with R. >PP1: Oh, we are very sorry. We did our best. # FINAL ACT >[Mr. Mathman and Human Curiosity] >MR. MATHMAN: Do you know, Mr. HC? If we admit Cantorās criterion to >count R, as we positively know that N <-> R is not possible (it >doesnāt matter why), it is clear that R will be necessarily >uncountable. What do you think about it? >HC: Well Mr. Mathman, we can do so, but it will not convince anybody. >The problem is that we know that Point 2 invalidates Cantorās >criterion. Therefore, R would be certainly uncountable, but no >because R be intrinsically uncountable, but because we donāt have a >valid criterion to count it. >MR. MATHMAN: Well then, perhaps there are other different criteria to >count the reals. >HC: I donāt know it, but if we put a simple analogy perhaps we could >clarify our thoughts. Letās try it. >We may take the following equivalence table N <==> M = A set of linear units of measure > R <==> A = The air in a room Being our goal to find out whether A is measurable or not, we have >the following analogy: POINT 1: A cannot be measured with linear unities, so we cannot >assume M -> A. >CRITERION: A set D will be lineally measurable if and only if it is >possible M -> D. >POINT 2: We canāt use CRITERION to find out whether A is measurable >or not, because POINT 1 tells us that A canāt be measured in this way. >PROOFS: We have found out that A is not measurable, because CRITERION >case. >PROPOSAL: We can admit CRITERION to measure A, and consequently A is >not measurable. Well, no one will accept this, because POINT 1 tells >us that we cannot apply CRITERION to measure A, and also because A >would be able to be measured by other means. >Good, Mr. Mathman, this is the end of our analogy. >MR. MATHMAN: And? >HC: What? >MR. MATHMAN: Are there other criteria to find out whether R is >countable or not? >HC: Well, in our analogy A cannot be measured with linear units, but >if we arrange orthogonally three of them, then we can. Perhaps the >reals have an undiscovered property which may be used as criterion to >count them. Think about it Mr. Mathman, think about it. >MR. MATHMAN: Iāll do it. *************************** Proof of Point 2 >INITIAL ASSERTIONS >1) We can represent any real number using a given positional system >of numeration. For ease we will use the decimal notation. >2) Cantor uses a transfinite method (the diagonal argument) to prove >that the premise N <-> R is false, and we will use another >transfinite constructive method to prove that the transfinite >construction N <-> R is impossible, and therefore we cannot assume >the bijection N <-> R. Proposition: The transfinite construction N <-> R is not possible Proof: >Every real number within the interval (0, 1) has as first decimal >digit one of the ten digits of the decimal system of numeration, i. >e. it must be of the form 0.0, 0.1, 0.2, ·, 0.9. As you can see for >this first digit there are 10 ^ 1 possibilities. For the second digit >we have 0.00, 0.01, 0.02,·, 0.99. Consequently, there are 10 ^2 >possible combinations for the two first digits, 10 ^3 for the first >three digits, and so on. When the number of digits is infinite we get >R. >Now, when we have only a digit of the decimal expansion of the reals, >we make the one-to-one correspondence 0 <-> 0.0, 1 <-> 0.1, 2 <-> >0.2, ·, 9 <-> 0.9. Next, with two digits we begin the 1-1 >correspondence 0 <-> 0.00, 1 <-> 0.01,·, 10 <-> 0.10, 11 <-> 0.11, ·, >99 <-> 0.99. With three digits, we continue doing the same, and so on. >And now is when the impossible transfinite construction appears. >While we keep doing the 1-1 correspondence within the finite decimal >expansion, everything will go fine. But, what natural number would >correspond to the decimal expansion of the number e? By induction it >is obvious that it should be 71828182·, i. e., a number with infinite >nonzero digits, which doesnāt belong to N (Point 1). Conclusion: >The transfinite construction N <->R is not possible if we do not >admit naturals with an infinite number of nonzero digits. If some day >we admit that, then the reals would be countable. >Nicolas de la Foz If you want us to comment on a proof, then you will have to show us a proof. I saw a lot of words, a few terms that might have been intended as mathematics but were not defined (for example measurable by linear units). There are quite a lot of non-sequiturs (for example, you define countable set as meaning that there exist a one to one correspondence between the set and N (which is correct) and then assert that that criterion doesnāt work for the real numbers since it is said in Point 2 that no such correspondence. That criterion works quite nicely, thank you: it says that the set of real numbers is NOT countable. What did you think a criterion was for?) While Point 2, that there is no one to one correspondence between the set of real numbers and the set of positive integers is true, your purported proof, that (to summarize) the set of real numbers contains number with an infinite number of nozero digits while no integer does, therefore, there can be no such correspondence, is completely invalid. The set of all rational numbers also includes numbers with an infinite number of non-zero digits (1/3 for example) but there IS such a correspondence- the set of rational numbers IS countable. Finally, I can't see what this has to do with Cantor's proof. You don't actually say anything about Cantor's proof. By the way, you end with Conclusion: >The transfinite construction N <->R is not possible if we do not >admit naturals with an infinite number of nonzero digits. If some day >we admit that, then the reals would be countable. Yes, and if we define countable to mean simply is infinite then the reals would be countable- and if we defined blue to mean is infinite then the reals would be blue- but we wouldn't have learned anything important in either case. It's not a matter of admitting naturals with an infinite number of nonzero digits. The naturals, as defined in our basic arithmetic, do not have such. If you want to try to change elementary school arithmetic just to make the reals countable (and so make mathematics trivial and unable to do most of the things we rely on it for) I don't think you will be very sucessful.