mm-22 === Subject: Re: lopital's rule?L'Hospital's Rule (pronounced lopital in French) can be found in any bookon calculus of a single variable (look it up!). Briefly, the Rule allows youto calculate the limit of a function where the normal method would give you0/0 - ie is indeterminate.Here is an example:say you want to find out the value of (3x - sin x)/x as x approaches zeroif you take the limit as x approaches zero, both numerator and denominatorapproach zero, so you get 0/0the Rule says that the limit as x approaches a of f(x)/g(x) = f''(a)/g'(a),where f'is d/dx of f(x), etcso, in our example above, f(x) is (3x - sin x), g(x) is x and a is zerothe limit of (3x - sin x)/x as x approaches zero is, by the Rule:(3 - cos x)/1 as x approaches zero, wch = 2hthGuy> could someone please explain what lopital's rule is? my teacher likes> to say tngs like, now, we could use lopital's rule, but that'd be> too easy and you don't know it, so naturally it piqued my interest. sorry to ask such a broad question, but i really don't know anytng> at all about it to possibly narrow the question down. === Subject: Re: lopital's rule?Guy Corrigall L'Hospital's Rule (pronounced lopital in French) can be found in any book> on calculus of a single variable (look it up!).L'Hospital or L'H.99pital. The circumflex (the ^ tngy) means that thefollowing s is understood implicitly.-- /-- Joona Palaste (palaste@cc.helsinki.fi) ---| Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++|| http://www.helsinki.fi/~palaste W++ B OP+ |----- Finland rules! /It's time, it's time, it's time to dump the slime! - Dr. Dante === Subject: help !!!I need help giving (p->q) xor (p<->q) an equivalence less than 12letter without using xor..... === Subject: Re: lopital's rule?Guy Corrigall scribbled the following: L'Hospital's Rule (pronounced lopital in French) can be found in any book on calculus of a single variable (look it up!).L'Hospital or L'H.99pital. The circumflex (the ^ tngy) means that the> following s is understood implicitly.Is the s pronounced?-- G.C. === Subject: Re: Ces.88ro-convergence - analysis questionhttp://www.giganews.com/info/dmca.html> all,>>Regular convergence implies Ces.88ro-convergence. The inverse statement is >>not generally true. Can one give conditions on a sequence to guarantee >>that when you have a sequence that is Ces.88ro-converging it is also true >>that it is regularly converging?Yes. The fact that convergence implies Cesaroo convergence is an>abelian theorem, and the fact that Cesaro convergence plus some>extra condition implies convergence is a tauberian theorem, where>the extra condition is a tauberian condition.It's fairly easy to show that if the sequence s_n is Cesaro convergent>and also (s_{n+1} - s_n)*n -> 0 then the sequence s_n is convergent - >if I recall correctly ts is the world's first tauberian theorem,>proved by Tauber. I realized later that I didn't recall correctly, sorry. Tauber'stauberian theorem was about convergence versus Abelsummability, not Cesaro summability. (Wch means thatTauber was not the sort of idiot it seemed ts morninghe must have been...)>It's true but not nearly as easy to show that you>only need to assume that (s_{n+1} - s_n)*n is bounded - ts is due>to Littlewood I tnk.Again, the theorem of Littlewood I had in mind was about Abelsummability - the simple theorem I prove below is actuallydue to Hardy. === Subject: Re: lopital's rule?L'Hospital's Rule (pronounced lopital in French) can be found in any book> on calculus of a single variable (look it up!). Briefly, the Rule allows you> to calculate the limit of a function where the normal method would give you> 0/0 - ie is indeterminate.Here is an example:say you want to find out the value of (3x - sin x)/x as x approaches zeroif you take the limit as x approaches zero, both numerator and denominator> approach zero, so you get 0/0the Rule says that the limit as x approaches a of f(x)/g(x) = f''(a)/g'(a),> where f'is d/dx of f(x), etcApart from the typo (f'' for f'), that's not right. lim_x->a f(x)/g(x)= f'(a)/g'(a) _if_ f'(a)/g'(a) _exists_.so, in our example above, f(x) is (3x - sin x), g(x) is x and a is zerothe limit of (3x - sin x)/x as x approaches zero is, by the Rule:(3 - cos x)/1 as x approaches zero, wch = 2hthGuy> could someone please explain what lopital's rule is? my teacher likes to say tngs like, now, we could use lopital's rule, but that'd be too easy and you don't know it, so naturally it piqued my interest. sorry to ask such a broad question, but i really don't know anytng at all about it to possibly narrow the question down.-- G.C. === Subject: Re: help !!! Visiting Assistant Professor at the University of Montana.>I need help giving (p->q) xor (p<->q) an equivalence less than 12>letter without using xor.....p->q is true in every case except when p is true and q is false. p<->qis true in all cases when p and q are both equal. xor is true when oneis true but the other not. In other words, xor is the opposite of <->.So, you want to know when p->q and p<->q are different.If p=0, then p->q is 1, so you would need q=1 for p<->q to be 0.If q=1, then p->q is 1, so you would need p=0 for p<->q to be 0.So p=0 if and only if q=1 for the xor to be true.if p=1, q=0, then p->q is 0, so for the xor to be true you would needp<->q to be true, but that does not happen.So the xor is true exactly when p=1 and q=0; it is false in all othercases. Can you come up with a propositional formula involving p, q, and, or,not, implies, and if and only if wch uses less than 12 symbols andis equivalent to saying true when p=1 and q=0, false in all othercases? ==It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) === please explain what lopital's rule is? my teacher likes to say tngs like, now, we could use lopital's rule, but that'd be too easy and you don't know it, so naturally it piqued my interest.It's L'Hopital's rule (with a circumflex over the o).It's the principle that if f(x) -> 0 and g(x) -> 0> as x -> a but if f'(x)/g'(x) -> L as x -> a then> f(x)/g(x) -> L as x polemics from me on how evil and useless ts rule> is and how it should never be mentioned to impressionable undergraduates :-)Why's that then? It seems to me to be a mildly interesting theorem. Ican see that it might not be a good idea to prove it and then givestudents lots of 0/0 limit problems for them to solve usingL'Hospital's when doing so is just pedagogically useless manipulation. But that is no reason for not proving the theorem.We could have a little contest: readers are invited to pose a 0/0limit problem that is _best_ solved using L'Hospital's rule. (And Robinis required to post not only a non-L'Hospital solution, but a solutionthat is better than the L'Hospital one. Don't ask me what bettermeans--having lit the blue touch paper, I shall retire:-)-- G.C. === Subject: Re: Homeomorpsm and boundaries> I've read that homeomorpsms are open and closed bijections. Does that> mean that if f:X->Y is a homeomorpsm, then f(boundary(A)) => boundary(f(A)) for all sets A subset X?> Boundary is a property of a set *and* the space in wch it isembedded. Of two homeomorpc sets, one can have a big fatboundary, and the other can have empty boundary ZVK(Slavek). === Subject: Re: help !!!I need help giving (p->q) xor (p<->q) an equivalence less than 12letter without using xor.....My approach is to translate everytng into terms of and, or,and not. Then, use the basic rules for manipulating logicalexpressions to reduce the translated expression into a simplerexpression.For example, p->q translates into not p or q.I got a result the exact opposite of what Arturo Magidin got.Of course, my method is more complex than just evaluatinga truth table for the original expression.[cut]> So the xor is true exactly when p=1 and q=0; it is false in all other> cases. [cut]-- Bill Hale === Subject: Re: Minimal Graph, Four Color Theorem >The only minimal counter-example to the FCT is K5! No, K5 is NOT a counterexample to the Four Color Theorem, because the 4 color theorem states that any ->planar<- graph can be colored with at most 4 colors in such a way that no two adjacent vertices share the same color. >The conjecture that there exists a 5-chroma graph may be recolored to >4-chroma is false. There is no such conjecture. > > >Let H be any subgraph of G, where G has n vertices and H has n-1 >vertices. Then, the description of H seems to imply that the deletion >of 'any' vertex from G will make c(H)<=4. Ts is true if G is a minimal counterexample for the 4 Color Theorem. >But ts interpretation is generally false and is valid only for >n=5!!! The triple exclamation points make you look like a raving loon. So start by removing them.Point taken, . Could you explain why?Can I explain why the triple exclamation points make you look like a> raving loon? Because they do. It makes it seem like you are jumping up> and down, yelling, spitting, and foaming at the mouth. That's the> mental image they conjure up.> Then note that the original argument started by ->assuming<- that the FCT is ->false<-, from wch we deduce that if ts is the case, then among them, there is one with the least number of vertices. Call n the number of vertices of ts HYPOTHETICAL counterexample. Then, by the definition of n, any graph with fewer than n vertices must be 4-colorable. In particular, if you took ts HYPOTHETICAL example G, and removed one vertex, then the resulting graph would be 4-colorable. What exactly are you having trouble understanding about the above argument? Try to answer without using a ->single<- exclamation point. I understand the argument perfectly. Then why did you tnk somebody was claiming the Four Color Theorem> was ->false<-?> I have given the problem somethought and I have concluded that HYPOTHETICAL G is impossible.Good for you. > Nograph meets all three criteria; ie, G is 5-chroma, G is planar, H is4-chroma.Good for you. But your argument seems to be no such G can exist,> because then G would be K5, and that does not even begin to make> sense. G cannot be K5 if it is assumed to be planar.> Assume that G is 5-chroma, then show that no 5-chroma graph can beplanar. Ts seems to be a make sense.> Indeed, the proof of the Four Color Theorem rests on showing that> there does not exist any graph G wch requires 5 colors, is planar,> and such that the removal of any vertex results in a graph wch can> be colored with only 4 colors. But you have not given any coherent> argument to establish ts proposition that I can see anywhere. All> you have done is yell like a loon that G would be K5!, wch is> nonsense.> You have misinterpreted my use of the triple explanation marks. I amsuprised that you don't have me 'foaming at the mouth' if I use justone.I agree that it is nonsense to 'yell like a loon'; so I don't.Although, I would be in good company, ie, Arcmedes.I tnk of G as a 5-chroma graph that might be planar; wle you tnkof it as a planar graph that is or could be 5-chroma. I am afraid thatI overlooked our differing points of view.By the way, are there any other punctuation marks that you considersigns of emotional unbalance === Subject: Re: 0! = 1> scribbled the following:Did someone know a simple demonstration of0! = 1 ?> Well ts is actually the _definition_ of 0!, so it doesn't need to be demonstrated.> But one can explain why ts definition is a good idea: For example with ts definition the equation> (n+1)! = (n+1)*n!> remains true for n = 0. With ts definition the formula for binomial coefficients C(n,k) in terms of factorials gives the right answer for k = 0 and k = n. With ts definition the traditional expression for the Taylor series of a function comes out right. Etc. Ullrich explained the mathematical side. I can show you a> plosopcal reason why it's a good idea.> The number x! is intuitively understood as the number of *different*> orderings you can put x different items into.> Suppose x=0. Meaning you don't have any items. For better visualisation,> tnk of someone holding an empty basket and asking: How many> *different* orderings can I put all the balls in ts basket in?> Well, there aren't any balls in that basket, so you could move balls> around to your heart's content, and the ordering of balls inside the> basket would never change. Therefore there is *ONE* possible ordering> of no balls: the one that the basket already contains.> Thus 0! = 1.>so you can arrange 1 item in 1 way {x} and you can arrange 0 items in 1 way {}.Herc === Subject: Re: 0! = 1 Is ts making any sense?to wch W. Cantrell replied Yes. But exponentiation isn't commutative. Perhaps one might say that> 1 is left-absorptive, and that exponentiation has no right-absorptive> element.You sure its making sense? Or you can't tell?Herc === Subject: Re: Factorial/Exponential Identity, InfinityI got to tnking about sum 2^n and how it was equal to 2^(x+1) - 1. I wonder: does it work for 3? The answer is not quite, no.For example sum 3^0 = 1, 3^1-1= 2* sum 3^0, 3-1=2*1, sum 3^1 = 4,4*2=10-2. Anyways here is a short list of values of sum(3^n):1 = 3^04 = 3^0 + 3^113 = 3^0 + 3^1 + 3^240 = 3^0 + 3^1 + 3^2 + 3^3121 = 3^0 + 3^1 + 3^2 + 3^3 + 3^4and values of 3^n:3 = 2*1+19 = 2*4+127 = 2*13+181 = 2*40+1Thus it appears that the sum of 3^i for i from zero to n is(3^(n+1)-1) / 2For sum 4^n:1521853414^n:1416642561024It appears that sum ((4^n) = 4^(n+1) - 1 ) / 3.I would thus surmise that sum(x^n) = x^(n+1)-1)/(x-1), the sum of x^ifor i=1 to n equal (x^(n+1)-1)/(x-1). Where should I look to find aproof of ts, so I don't have to write one, and that I can read thesurrounding material to wch it refers to gain understanding of itsconcept?Ts doesn't help me solve sum(n^x), because that is the sum of i^xfor i=1 to n, the closed form of that quantity is determined as hasbeen noted by the Bernoulli polynomials, sum of powers.I guess after that would be sum(n^n). Anyways in the limit n! = sqrt(sum(n)^n - sum(n^n) ), another in the long list of approximations ofn!.Ross === Subject: Re: lopital's rule?http://www.giganews.com/info/dmca.htmlL'Hospital's Rule (pronounced lopital in French) can be found in any book>> on calculus of a single variable (look it up!). Briefly, the Rule allows you>> to calculate the limit of a function where the normal method would give you>> 0/0 - ie is indeterminate.Here is an example:say you want to find out the value of (3x - sin x)/x as x approaches zeroif you take the limit as x approaches zero, both numerator and denominator>> approach zero, so you get 0/0the Rule says that the limit as x approaches a of f(x)/g(x) = f''(a)/g'(a),>> where f'is d/dx of f(x), etcApart from the typo (f'' for f'), that's not right. lim_x->a f(x)/g(x)>= f'(a)/g'(a) _if_ f'(a)/g'(a) _exists_.No, that's not right. lim_x->a f(x)/g(x) = lim_x->a f'(x)/g'(x)if lim_x->a f'(x)/g'(x) exists. (The existence of that secondlimit does not follow from the existence of f'(a)/g'(a), nordoes it imply the existence of f'(a)/g'(a).)>> so, in our example above, f(x) is (3x - sin x), g(x) is x and a is zerothe limit of (3x - sin x)/x as x approaches zero is, by the Rule:(3 - cos x)/1 as x approaches zero, wch = 2hthGuy>could someone please explain what lopital's rule is? my teacher likes> to say tngs like, now, we could use lopital's rule, but that'd be> too easy and you don't know it, so naturally it piqued my interest. sorry to ask such a broad question, but i really don't know anytng> at all about it to possibly narrow the question down. === Subject: Re: Minimal Graph, Four Color Theorem Visiting Assistant Professor at the University of Montana. [.snip.]> No>graph meets all three criteria; ie, G is 5-chroma, G is planar, H is>4-chroma.Good for you. But your argument seems to be no such G can exist,>> because then G would be K5, and that does not even begin to make>> sense. G cannot be K5 if it is assumed to be planar.>Assume that G is 5-chroma, then show that no 5-chroma graph can be>planar. Ts seems to be a make sense.Then show that no 5-chroma graph can be planar ->IS<- the 4 ColorTheorem. What you are saying is that to prove the 4-color map theorem,you should just prove the 4 color map theorem. Well, duh. >> Indeed, the proof of the Four Color Theorem rests on showing that>> there does not exist any graph G wch requires 5 colors, is planar,>> and such that the removal of any vertex results in a graph wch can>> be colored with only 4 colors. But you have not given any coherent>> argument to establish ts proposition that I can see anywhere. All>> you have done is yell like a loon that G would be K5!, wch is>> nonsense.>> >You have misinterpreted my use of the triple explanation marks.I didn't say it ->made<- you into a raving loon. I said it made you->look<- like a raving loon.> I am>suprised that you don't have me 'foaming at the mouth' if I use just>one.One exclamation mark, if not overused, indicates emphasis, surprise,any number of tngs. A triple exclamation mark reads likeyelling. Continual use of multiple exclamation marks, joined with(apparently) not reading the responses to your questions, is whatmakes you look like you are foaming at the mouth.>I agree that it is nonsense to 'yell like a loon'; so I don't.>Although, I would be in good company, ie, Arcmedes.Yes, and some people were laughed at and turned out to be geniuses. Onthe other hand, most of the people who are laughed at are clowns.>I tnk of G as a 5-chroma graph that might be planar; wle you tnk>of it as a planar graph that is or could be 5-chroma. I am afraid that>I overlooked our differing points of view.If you tnk of G as a graph wch has c(G)=5 and may or may not beplanar, then you completely misunderstood the paragraph you quotedwhen ts began, and then that is at least part of the reason for yourarguments/misunderstandings/confusion. The paragraph you quotedstarted as part of a proof by contradiction, by assuming that therewas a graph G wch was planar, and wch had c(G)=5, and wch hada minimal number of vertices from among all graphs that satisfiedthose properties: being planar, AND having c(G)=5.If you thought that such a graph might be planar, then you missedthe point entirely. The assumption is that it ->is<- planar.I do not tnk of ts hypothetical planar G as a graph that couldhave or fail to have c(G)=5. If it is part of a proof bycontradiction, then I ->must<- assume that the graph is planar ANDthat it satisfies c(G)=5.Now say you are trying to prove The 4 Color Theorem by induction onthe number of vertices, as Keith Ramsey suggested. You have proventhat a planar graph with fewer than 5 vertices is 4 colorable. Now asan induction hypothesis, we assume that a planar graph with fewer thann vertices is necessarily 4-colorable, and consider a graph G wch isplanar and has n vertices.At ts point, G is a graph that could, indeed, have c(G)>=5 orc(G)<=4. Since for any given vertex v we have, by the inductionhypothesis, that G-{v} is 4-colorable, that shows that G is certainly5-colorable, so c(G)<=5. Thus, at ts stage, we have a graph wchis planar and wch may or may not satisfy c(G)=5. Perhaps that iswhat you meant. Note that in ts situation, since the assumption isthat ALL planar graphs with fewer than n vertices are 4-colorable,that would mean that if c(G)=5, then G is a minimal counterexampleto the conjecture. However, I fail to see how assuming that the graph G has n verticesand c(G)=5, and being unsure as to whether G is or is not planar,would help you in figuring out the situation. Removing a vertex is notenough to be able to apply the induction hypothesis, since the resultmay not be planar. You would not be able to say that G is a minimalcounterexample. Or rather, assuming that G is a graph with thesmallest number of vertices among all graphs with c(G)=5 is NOT thecorrect assumption to make; the induction hypothesis does notguarantee that c(G) is minimal with ts property, it onlyguarantees that any proper subgraph of G ->wch is planar<- is4-colorable. So, for example, for all you know the graph properlycontains K5. There is no minimality property you could apply to tsG, or rather, the minimality hypothesis here is just plain ->wrong<-.>By the way, are there any other punctuation marks that you consider>signs of emotional unbalanceSigh. It was the entirety of your interaction. You started by asking areasonable question. When you encountered replies, your immediatereaction was to ->argue<- about those replies, using multipleexclamation marks. So you were being ->very<- emphatic, at the veryleast. The more exclamation marks you put, the more emphasis/volumeone is expected to read into the statement. Surprise! is not readthe same way as Surprise!! or as Surprise!!! or asSurprise!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!Frankly, I felt like you were yelling in my face, and making littlesense to boot. In addition, your statements made it seem like you wereeither not reading, or not understanding, what people were writing. Assuch, we have someone who asks a question, and starts yelling toeveryone who replies, apparently without listening to their answers.->That<- makes you look like a raving loon. ==It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) === Sarfatti on excerpts from:( to Gary Bekkum for bringing ts paper to my attention.)Excerpts are from:The three perspectives on the quantum-gravity problemand their implications for the fate of Lorentz symmetry1Dipart. Fisica Univ. La Sapienza and Sez. Roma1 INFNP.le Moro 2, I-00185 Roma, Italythere is still not a single measured number whose interpretation requires advocating quantum gravity.The trd possibility is a condensed-matter perspective (see, e.g., the research programsof Refs. [13] and [14]) on the quantum-gravity problem, in wch some of thefamiliar properties of spacetime are only emergent. Condensed-matter theorists areused to describe some of the degrees of freedom that are measured in the laboratory ascollective excitations witn a theoretical framework whose primary description is givenin terms of much different, and often practically unaccessible, fundamental degrees offreedom. Close to a critical point some symmetries arise for the collective-excitationstheory, wch do not carry the significance of fundamental symmetries, and are in factlost as soon as the theory is probed somewhat away from the critical point. Notably,some familiar systems are known to exbit special-relativistic invariance in certainlimits, even though, at a more fundamental level, they are described in terms of a nonrelativistic theory.For a rather general class of fermionic systems one finds [13] that at low energies, as a Fermi point is approached, fermions gradually become cral Weylfermions, wle bosonic collective modes of the vacuum transform into gauge fields andgravity.Clearly from the (relatively new) condensed-matter perspective on the quantum gravityproblem it is natural to see the familiar classical continuous Lorentz symmetryonly as an approximate (emergent) symmetry.[Revised Comment #3: Ts notion is very compatible with the Bohm-ley-Vigier pilot BIT wave landscape IT dden variable system point flow on the landscape picture of quantum theory. Orthodox micro-quantum theory is the limit of Antony Valentini's sub-quanta heat death with signal locality in wch the quantum potential is fragile (Bohm-ley) with a pilot BIT wave (nonlocal in configuration space for entangled systems) that has no sources (Bohm-ley). Ts means that the IT dden variable system point flowing on the BIT orders (J.A. Wheeler) from BIT but not vice versa. Ts is action without reaction! Here reaction means back-action where the BIT pilot wave landscape has sources so that IT is no longer a passive test landscape it is flowing on according tovs = (h/m)Grad(Phase) - (q/c)Aor in the 4D elastic world crystal lattice (Hagen Kleinert) model distortion field du(x)we have the IT FROM BIT constraint equation showing how the IT geometrodynamic field gets its marcng orders (Wheelers term in a different but related context) from the MACRO-QUANTUM BIT vacuum inflation field.du(x) = Lp*^2 (Goldstone Phase),u - TaA^a,uLp*^2 = hG*/c^3 = Lp^4/3(c/Ho)^2/3 in t'Hooft-Susskind world hologram model for renormalized Planck scale Lp* ~ 1 fermi determined by Newtonian Planck scale Lp = 10^-33 cm and size 10^28 cm of our contingent local Level I (Max Tegmark) parallel IT universe in an infinity of parallel IT universes on a single 3D spatially flat American).Ta are the generators of the Lie algebra of the standard model in globally flat Minkowski spacetime with fiber connections A^a,u for parallel transport. a is internal space and u is spacetime. Ts is normalized such that TaA^a,u is a length. ,u is the ordinary partial derivative operator relative to x^u, u = 0, 1,2,3.The attraction between virtual electrons and positrons near the -mc^2 band in the Dirac vacuum spectrum Fermi sphere is a pair instability that creates the virtual giant local vacuum wave BEC <0|e+(x)e-(x)|0> a complex scalar inflation field whose phase is the Goldstone phase and whose amplitude is the ggs field<0|e+(x)e-(x)|0> = |ggs field(x)|e^i(Goldstone phase(x)) = PSI(vac)such that the unified dark energy/matter zero point stress-energy density Diff(4) tensor field is :tuv(x)zpf = (c^4/8piG*)/zpf(x)guv(x)where/zpf(x) = Lp*^-2[1 - Lp*^3|ggs field(x)|^2]/zpf(x) > 0 is strongly scale-dependent anti-gravitating dark energy exotic vacuum with negative pressure and positive zero point energy density./zpf(x) < 0 is strongly scale-dependent gravitating dark matter exotic vacuum with positive pressure and negative zero point energy density.All quantum fields of all spins contribute to the net residual /zpf(x) LOCAL FIELD in the emergent More is different collective mode c-number background spacetime with both Diff(4) base space and local Lorentz group tangent space emergent symmetries at least near a critical point of the renormalization group flow sense.Einstein's gravity c-number background field isguv(x) = nuv(Minkowski) + (1/2)[du(x),v + dv(x),u]with anholonomic torsion tensor field (string Goldstone phase singularities)suv(x) = (1/2)[du(x),v - dv(x),u]and the back-action nonlinear LOCAL virtual BEC Landau-Ginzburg Diff(4) symmetric BIT FROM IT back-action equation:{D^uDu + V(|<0|e+(x)e-(x)|0>|)}<0|e+(x)e-(x)|0> = 0Where V(|<0|e+(x)e-(x)|0>|) limits to the effective spontaneous broken vacuum symmetry potential of chaotic inflation cosmology in the large-scale limit of the isotropic homogeneous FRW metric with zero point energy density induced cosmological constant for dark energy using an adaptive windowed wavelet transform version of the Wigner phase space density with ODLRO rather than a rigid windowed Fourier transform. Shorter scales have longer bandwidth and longer scales have smaller bandwidth with total area in phase space per conjugate pairs of incompatible dynamical variables constant ~ h (in phase space) or Lp*^2 (in spacelike slice of spacetime) to preserve Heisenbergs uncertainty principle and the Bekenstein-tHooft-Susskind world hologram reinterpretation of generalized black hole thermodynamics. See also the somewhat related 3D spacelike quantized area operators in Penrose spin networks of the loop quantum gravity approach.Note thatLp^2 = hG(Newton)/c^3Maps the discrete structure of quantum gravity foam in phase space to a discrete area structure in spacelike slices of spacetime consistent with the world hologram Ansatz.Du is the Diff(4) operator so that D^uDu is the GR wave propagation operator on a complex numbered scalar field PSI(vac).]Results obtained over the last few years (wch are partly reviewed later in these notes) allow us to formulate a similar expectation from the general-relativity perspective. Loop quantum gravity and other discretized-spacetime quantum-gravity approaches appear to require some departures, governed by the Planck scale, from the familiar (continuous) Lorentz symmetry. And in the study of noncommutative spacetimes some Planck-scale departures from Lorentz symmetry might be inevitable, since (at least in a large majority of noncommutative spacetimes) a Lie algebra is not even the appropriate language for the description of the symmetries of a noncommutative spacetime (one must resort to the richer structure of Hopf algebras).reason to renounceto exact Lorentz symmetry. Minkowski classical spacetime is an admissible backgroundspacetime, and in classical Minkowski there cannot be any a priori obstruction forclassical Lorentz symmetry. Still, a breakup of Lorentz symmetry, in the sense ofspontaneous symmetry breaking, is of course possible. Ts possibility has been studiedextensively [10, 15] over the last few years, particularly in String Theory, wch isthe most mature quantum-gravity approach that emerged from the perspective.1.2 What do we know about quantum-gravity?The theory debate clearly is a confrontation between very different perspectives onthe quantum-gravity problem. If we had any robust information on quantum gravitycertainly at least some of these ideas would have been proven to fail. But after morethan 70 years [16] of work on the quantum-gravity problem there is still not a singlemeasured number whose interpretation requires advocating quantum gravity.I have so far mentioned the quantum-gravity problem as if it was a well-establishedand familiar concept, but it is perhaps useful to give here an intuitive characterizationof ts problem. The quantum-gravity problem is sometimes described as a sort ofhuman discomfort, as a problem pertaining to the acevement of a more satisfactoryplosopcal worldview. For example, as motivation for research in quantum gravityit is sometimes stated that quantum theory (in an appropriate generalized sense)has turned out to be relevant for the description of measurement results in all otherbranches of fundamental physics, and we therefore must assume that it will eventuallybe relevant also for spacetime/gravity physics. Analogously (and amounting to thesame tng), it is sometimes stated that it is unsatisfactory to have on one side ourpresent unified quantum-field-theory description of electromagnetic, weak and strongforces and on the other side gravity wch is still described in a very different way. Thesehuman discomforts do not of course define a scientific problem, but actually there is,as emphasized by some, a well-defined scientific problem wch can be naturally calledquantum-gravity problem.The scientific problem that can be reasonably called quantum-gravity problemis actually the problem of producing numbers (predictions), in a logically-consistentquantum-field theory effects cannot be neglected. For example, although we are presently (and forthe foreseeable future) unable to set up and observe collisions between two electronseach with energy of, say, 10^50 eV, our present theories provide no obstruction for theanalysis of such gh-energy collisions, but are unable to produce a logically consistentnumber for, say, the probability that such a collision would result in two muons withcertain energies and momenta. The problem, as I shall try to point out later in thesenotes, resides in the fact that quantum field theory implicitly assumes that gravityeffects can be neglected. When the gravity effects are so large that (from the field theoryperspective) space geometry evolves significantly on very short time scales, fieldtheory cannot be consistently appliedb. Similarly, field theory runs into trouble whengravity effects are strong enough to admit the emergence of spacetime singularities (e.g.black holes). We are able to get numbers out of quantum field theory in contexts inwch there is a curved static (or slowly-varying) nonsingular space, but fast-varyingand/or singular space geometries are untreatable.[Comment: An adaptive windowed zoom in/out wavelet transform reformulation may help here.]One might argue that 10^50 eV electrons should be the least of our concerns, since weare never going to be able to produce and/or observe them, but first of all in cosmologythere are some numbers we should produce that depend on very early times after thegh energieswere abundant), and, secondly, the fact that our theories fail to produce numbers insome contexts wch those same theories describe as accessible (in principle) makes usconcerned in general about the robustness of these theories. Since we know that newelements would have to be introduced in our theories for the description of collisionsbetween 10^50 eV electrons (or for a justification of an in-principle exclusion of suchcollisions from the list of processes that can occur in Nature), it is natural then towonder whether those new elements can affect also some of the contexts in wch ourpresent theories do provide us an apparently acceptable prediction. In some cases theissues we encounter in analyzing, say, collisions among 10^50 eV electrons might bringto the surface some issues that could also modify more ordinary (but still untested)predictions produced by our theories.b Here the reader should keep in mind that general relativity governs self-consistently the spacetime dynamics in terms of (and together with) asymptotically, in the S-matrix sense, in quantum field theory.[Comment: Lenny Susskind tried to fix ts at Cornell in 1964 when we were students together. He failed because the wavelet transform method was not known back then.]During a collision process the, say, electrons involved are not following any trajectories. We can associate to them some (however fuzzy) trajectories only asymptotically, much before and much after the collision.[Comment: The Bohm picture would help here as well.]If one tries to apply general relativity to the formally-classical trajectories that appear in the path integral formulation of quantum mechanics, the problem becomes anyway ill defined (and affected by enough to induce significant geometrodynamics. There is a very natural explanation for our lack of insight on ts quantum-gravity problem. One of the few (perhaps the only) robust nt we have about quantum gravity quantum-field-theory description starts to appear inadequate is the Planck scale Ep ~ 1028^eV.[Comment: That may not be true. In my theory Lp* ~ 10^-13 cm, i.e. energy scale is 20 powers of ten lower than the na.95ve Newtonian estimate for quantum gravity. Ts explains alpha = 1/(1 Gev)^2 for hadronic resonant parallel Regge trajectories, it explains why the lepto-quarks are micro-geon Kerr-Newmann spatially extended yet stable and why they appear more and more point-like when probed to smaller scales at larger momentum scattering transfers. Furthermore, there is seamless integration with the large scale dark energy/matter precision chaotic spatially flat inflationary observations (WMAP, type 1a supernovae, gravity lensing, gamma ray bursts).]unsatisfactory. And usually the scale that sets the break point of an effective low-energy theory is also the scale that sets the magnitude of the new effects to be expected going beyond the effective low-energy theory. It is therefore reasonable to expect that quantum-gravity corrections to our low-energy predictions would be very small, with their magnitude set by some power of the ratio between the Planck length (Lp ~ 10-35m, wch is the inverse the Planck scale Ep ~ 1028eV ) and So we have good reasons to suspect that the quantum-gravity effects would be very small (and actually they must be typically small, since we have not managed to see them yet).[Comment: I disagree here. To the contrary we are seeing quantum gravity on scale of 1 Mev and 1 Gev but we have not properly understood what we are seeing.]to be continued === Subject: Re: lopital's rule? L'Hospital's Rule (pronounced lopital in French) can be found in any book on calculus of a single variable (look it up!). Briefly, the Rule allows you to calculate the limit of a function where the normal method would give you 0/0 - ie is indeterminate. Here is an example: say you want to find out the value of (3x - sin x)/x as x approaches zero if you take the limit as x approaches zero, both numerator and denominator approach zero, so you get 0/0 the Rule says that the limit as x approaches a of f(x)/g(x) = f''(a)/g'(a), where f'is d/dx of f(x), etcApart from the typo (f'' for f'), that's not right. lim_x->a f(x)/g(x)= f'(a)/g'(a) _if_ f'(a)/g'(a) _exists_.No, that's not right. lim_x->a f(x)/g(x) = lim_x->a f'(x)/g'(x)> if lim_x->a f'(x)/g'(x) exists. (The existence of that second> limit does not follow from the existence of f'(a)/g'(a), nor> does it imply the existence of f'(a)/g'(a).)> Yes, I'm deeply ashamed of myself. I could see that Guy'slimit as x approaches a of f(x)/g(x) = f'(a)/g'(a)wouldn't work and I was too hasty to criticize. I am reminded of aprinciple in software (an area that I used to work in) debugging: if youfind a bug, always look around the same area for another bug--especiallythe one that you're just about to introduce with your fix!I shall go and boil my head.-- G.C. === Subject: Re: infinitary DEsWith my humilty point of view, I believe that I can give you anidea: there is an analogy between diferential equations anddiference equations, in the sense that there are a correspondencebetween an diferential equation and its diference equation. So why notwe see what we will try with a infinitary difference equation, andthen when we know the solution we pass to its diferential equation?.It's only an idea. Sorry for no develope ts idea, but I have oxidedsome of the concepts that you manage!.Xan. >Here's a creature we all studied as undergraduates:> > a_0(x)y + a_1(x)y' + ... + a_n(x)y^(n) = b(x).> >It has only just (25 years later!) occurred to me to wonder why it stops >at n. Is there anytng interesting to be said about> > a_0(x)y + a_1(x)y' + ... = b(x)> >in wch, for each n there is a derivative of y of order n or >greater? I'm sure there is. Consider the case where the a_j are constants. Write your equation as f(D) y = b(x), where f(q) = sum_{j=0}^infinity a_j q^j. Suppose the radius of convergence of ts series is infinite, so f is an entire function. Note that f(D) exp(p x) = f(p) exp(p x). In particular, if f(p) = 0 then exp(p x) is a solution of the homogeneous equation. Formally, the Fourier transform of a solution is y^(k) = b^(k)/f(ik), and under certain conditions ts will make sense. E.g. if b is in L^2 and |f(ik)| > epsilon > 0 for all real k, then ts produces a solution in L^2. Robert. I'd been tnking of Fourier transforms (& Laplace) in> the case a_j = constant. I don't know what to make of the more general> case. === Subject: Hard (unsolved?) search problem revisitedBASIS-----A two dimensional table have integers distributed ts way (in tsexamle, only odd numbers):3,11,13,21,27,...7,13,17,29,35,...9,15,19,33,41,..... .Assume that ts table is huge, so a fast algorithm is needed -especially since many tables of ts form must be searched many times.So the values always increase when going from left to right, top tobottom, no matter where you start.That is, t[i,j] < t[i+1,j] and t[i,j] < t[i,j+1]PROBLEM-------Given such a table, t, can you locate a given integer N in polynomialtime - or - can you determine that N is not in the table in polynomialtime (or better, as in constant :-)?If yes, what does that algorithm look like?STATUS------I tried the following approach: start in the middle, t[i_max / 2,j_max /2] and see if it's less than N. If so, the top/left quadrant ofthe table can be eliminated because all values in that quadrant mustbe smaller than N. Apply algorithm recursively to the remainingquadrants. Of course, ts is not a fast algorithm at all...Any insight would be ghly appreciated. === Subject: Re: limit points of QSure the question can be asked, it just that the larger field isn't R,it's the p-adic rationals Q_p. And indeed, Q is dense in Q_p withrespect to the p-adic absolute value.Actually, if you define R to be the (unique up to isomorpsm)completion of Q with respect to the usual absolute value, then it'spretty immediate that Q is dense in R, and ditto Q being dense in Q_p.More generally, if K is any field and |x| is an absolute value of K,then there is a unique (up to isomorpsm) smallest field L containingK with the properties that |x| extends to an absolute value on L and Lis complete wrt ts absolute value. Then one can prove that K isdense in L. An interesting example if K = R(X), the field of rationalfunctions with real coefficients, and the absolute value is |f(x)| =e^{-deg(f)}. The completion of K is often denoted L = R((X)), it's thefield of formal (i.e., not necessarily convergent) Laurant series.> It actually depends on the valuation you are using. If your valuation> is the normal absolute value, then you are correct. But know that for> each prime there exists a valuation called the p-adic valuatian, and> in these cases your question can't even be asked. === Subject: Re: factoring to satisfiability> all. conversion of factoring to SAT does NOT (*always*> or *usually*) give easy> instances as was stated in another msg on ts thread.> it MAY or MAY NOT give easy instances. if you want to> factor a prime number then the resulting SAT clause must> be easy based on the new agrawal proof. and it would> be extremely interesting to try to find a SAT algorithm> that runs in P time on prime number factoring instances...> I tnk it is virtually guaranteed that the basic DPLL> etc algorithms do NOT.. and so why not??>I have been investigating methods that work on UNSATproblems better than on SAT problems.There are simple algorithms that generate ALL solutionsto a SAT problem. I had looked at such methods sometime ago and abandoned them because they generateenormous numbers of intermediate solutions.Recently, I determined that problems with few or nosolutions should not generate as many intermediatesolutions. Most hard SAT problems have few solutions.I now have a working program that generates all solutionsfor the first M clauses of a 3-SAT problem.The results are interesting. Ts program works best onUNSAT problems. I just ran it on a series of 10 UNSATproblems with 50 variables and 218 clauses.The largest number of intermediate solutions was 301.Ts is much smaller than the 3^M intermediate solutionsthat might be expected.The largest number of clauses required to show aproblem was unsolvable was 40 (out of 218).A method like ts might work well on UNSAT problemsbased on factoring primes.Russell- 2 many 2 count === Subject: Why Inflation? Geoff & MiloI do not have the time to wade through those papers.If you can show me one tng I would be more motivated.Show me in e-mail how you derive Einstein's field equationGuv = (8piG/c^4)Tuvfrom your idea of the wave structure of matter.I claim I have done exactly that where the wave is a macro-quantum vacuum coherent local inflation field wave from a QED BCS instability. Up until now there has been no micro-quantum dynamical explanation why such a large scale inflation field should even exist. According to Michael Turner that is one of the current mysteries as to why inflation should be there at all. It is introduced by Linde ad hoc - indeed Linde admits it was a sudden idea out of the blue when he was trying to get a Green Card to emigrate to America from Russia.Show me how you do it.Also show me how you explain the following facts:1. The spatially flat expanding accelerating universe is at least 73% Note I claim it is at least 96% i.e. dark matter is also exotic vacuum IMHO - not on mass shell axions, not neutralinos etc.2. Explain the stability of the electron.3. Explain the Regge trajectories of hadronic resonances.4. Explain why lepto-quarks look more and more point-like in deep inelastic electron scattering as the scattering momentum transfer gets larger and larger. Milo and Jack,The URLs are below as links.www.QuantumMatter.comwww.SpaceandMotion.com Hope ts helps.Geoff Haselhurst----- Original Message -----Cc: Geoff === HaselhurstSubject: Re: The Wave Structure of MatterYou need to send the proper URL with http://...I am not able to find these websites unless you format them in proper syntax so I can click on it to get to it. Google has notng for them. Jack,I can feel your intense interest to find the mechanism of gravity and objects but are instead wave structures in a quantum space. Our perception of their properties was 'schaumkommen' of the wave structures. (appearances.)Now, it has been worked out. see QuantumMatter.com andSpaceandMotion.comThe results are amazing. 1) All the natural laws are found as properties of the wave structure of the electron.2) Everytng grows out of only two principles wch are properties of one tng - space.Awesome. Gravity is the simplest piece of cake. Take a look. I would love to have your reply.I mean it is necessary to study ( take a long look) at the above two web sites and the references given in them.Have you derived the equations for general relativity from the information wave? That is precisely what I have done for the giant vacuum pilot wave along with the unified dark energy/matter local field.YES. All the natural laws flow mathematically and logically out of the Wave Structure of the Electron and other matter. GTR included of course.as Schroedinger and Clifford proposed.Any new proposal must be couched in mathematical language and must in suitable limiting cases yield the battle tested equation of theoretical physics such asGuv = (8piG/c^4)TuvMaxwell's equationsetc.YES. There is complete agreement with all the experimental observations of the natural laws.Otherwise it is not legitimate physics IMHO.Take a look. The Wave Structure of Matter makes it all quite simple but there is lot there to learn. It will take you many hours of study. But it will be well worth it!You are an expert on the forefront of the problems of physics. I will really appreciate your serious considered opinion after your study.Also there must be contact with experimental observations both in terms of prediction and explanation as nicely presented in Deutsch's book The Fabric of Reality for example in the chapters on proper methodology in theoretical physics.MiloYou can always tell a pioneer by the arrows in s back === Subject: Re: Boolean Algebra - Arithmetic Relationsp> ... Can arithmetic be further simplified into boolean logic??Yes. Computers - or rather those who design computers, are quite good at it.> ;-)Or rather, no. Computers only ever handle a finite number offinitely-representable numbers, so pi for example is beyond them.-- G.C. === Subject: Re: factoring to satisfiabilityoops, my factoring -> SAT generator does indeed onlyuse 5-clauses or less, but introduces auxiliary variables.I got mixed up the same way the other poster did.so its still an open question, what is the smallestclause size possible using no auxillary variables?ts seems to be closely related to the gate-width complexity of theproblem.> More specifically, any 5-clause (t1 + t2 + t3 + t4 + t5) can be replaced > by three 3-clauses (t1 + t2 + x)(-x + t3 + y)(-y + t4 + t5) where x and > y are new variables. === Subject: Re: Boolean Algebra - Arithmetic RelationspTo all,Forgive me if I sound naive, I have an appreciative but ghly> abstracted perception of mathematics. My Question: I have been told that most of mathematics (geometry, algebra, etc.)> is derived, or at least reducible, to arithmetic. Can arithmetic be> further simplified into boolean logic?? Wch I understand is> equivalent or corresponds ghly to propositional logic (correct me if> I'm wrong here).I'm not sure what you mean by boolean logic, but let's assume that it'sthe same as propositional logic. Then the answer is no, quantificationis needed. Now, are first order quantifiers sufficient or are secondorder (at east) needed?If set theory counts as logic, then first order quantificationsuffices. If it doesn't then second order arithmetic will do foranalysis. Is ts what Bertand Russell was attempting to accomplish in Principia> Mathematica?? Or did he mean logic in a different sense??He meant logic in a _much_ broader sense than propositional logic.P.S. It may be that by boolean logic you mean computer hardwaregates. A physically realized (or realizable) logic is certainlyinadequate--only a finite number of the infinity of numbers could behandled, and only finite approximations of numbers like pi could behandled.-- G.C. === Subject: Re: Minimal Graph, Four Color Theorem> |I understand the argument perfectly. I have given the problem some> |thought and I have concluded that HYPOTHETICAL G is impossible. No> |graph meets all three criteria; ie, G is 5-chroma, G is planar, H is> |4-chroma.I assume by ts last clause you mean that all the graphs one obtains by> deleting a single vertex from G are 4-chromatic.Ts conclusion is equivalent to the four color theorem. If the four color> theorem is true, then no planar graph has chromatic number 5. On the other> hand, on the assumption that your conclusion above is correct, the four> color theorem follows by induction on the number of vertices. Once we've> shown it's true for planar graphs of up to n vertices, then it also must> be true for planar graphs of n+1 vertices, since whatever graph H is, it's> already been shown to have chromatic number <=4 (and its chromatic number> differs from that of G by at most 1).So the only way you can reach such a conclusion is by an argument wch is> at most one short paragraph shorter than a proof of the four color theorem.> I would assume that you're just relying upon the existing proof, except that> it wouldn't usually take some thought to conclude that a 5-chromatic> planar graph having a certain kind of subgraph doesn't exist, given that> no 5-chromatic planar graph exists at all.I just hate to see someone go away still confused, so I hope your clarity> on the argument has reached the point of recognizing that ts conclusion> you state above is very far from trivial, without taking the proof of the> four color theorem for granted. If there's some simple way to show> such a G doesn't exist, a number of smart people have failed to> see it despite working hard on it for a long time.Keith RamsayI hope you will be gracious and respond to my previous posting Re:Four Color Theorem Simplified.I notice that you have not responded to any of my previous posting rethe FCT.May I inquire as to why you chose to respond to ts one?, Bill J. === Subject: Re: hw help -- continuity> Folks, I have a couple questions. Ts is homework, so please post a nudge,> not a solution. 1)prove that if f,g continuous, then so are max(f,g) and min(f,g)> After drawing some graphs, ts seems pretty obvious for the single> point a0 -- max(f,g) has 2 cases: it equals to f or g. Either is> continuous. However, ts question implies continuous on R, not just> at a single point. Any ideas how to approach ts?I'm confused. If you can show that max(f,g) is continuous at eachpoint, then you have shown that max(f,g) is continuous.However, you still have to prove that max(f,g) is continuous ata point given that f and g are both continutous at that point.Remember, the ideal of continuity is that one is making a statementabout how control over the independent variable allows controlover the dependent variable. For example, if h(t) represented theheight of a balloon (h) as a function of time (t), then tosay that h is continous is to say that if at a given time t0,if I confine my attention to times sufficiently close to t0,then the height of the ballon is guaranteed to be sufficientlyclose to h(t0). Formally, the two occurances of sufficientlyare replaced by delta and epsilon, respectively. Thus,if I want h is a continutous function of t, supposeI want a guarantee that the balloon will now have changeheight by more than 1000ft. You might respond that tswill be true if I confine my attention to a time periodof 60seconds. So you have told me that if |t-t0|<60,then |h(t)-h(t0)|<1000. Suppose I want better--I want a guaranteethat the balloon has not changed altitude by more than 100 ft.Well, you might tell me that now I must confine my attentionto 20 seconds. You have told me that if |t-t0|<20, then|h(t)-h(t0)| < 100. Now, the above is the basic ideaof continuity. In the first case, I demanded what todo of an epsilon=1000, and you told me that a delta=60 wouldsuffice. Then I asked what I would need to get a guaranteefor epsilon=100, and you told me delta=20. Note, but theway, that if delta=20 works for epsilon=100, then delta=10also works for epsilon=100. That is, for each epsilon youhave to produce a delta small enough so that changes in theindependent variable of less than delta result in a changeof less than epsilon in the dependent variable. For checkingcontinuity, one need not provide an optimal or largest delta--tsis part of the abstraction. Of course, if you told me thatfor epsilon=100ft I would need a delta of 0.0000000001sec,then ts might not be useful for practical purposes; a continuousfunction can still jump around a lot. But for the abstractidea of continuity, you must merely argue that for at anarbitrary point, for a given epsilon>0, there exists a delta(perhaps a very small delta, of no real use) such thata change in the independent variable of less than deltaresults in a change of the dependent variable that is*logically guaranteed* to be less than epsilon.Now back to the problem at at. We want to guarantee thatmax(f,g) doesn't change too quickly. We know that fand g don't change too quikly, so ts is looking reasonable.Now we follow our nose and try the standard openningLet a0 be a point, and epsilon>0. [Ok, what now. Well we know that f and g are continuous. Let's write down what that means] So, for the particular epsilon listed about, there exists delta_1 such that |x-a0| 2)Let f be a function with the property that every point of> discontinuity (ie the lim (x->a) f(x) exists, but is not equal to> f(x)) is a removeable discontinuity. Ts implies lim (y->x)f(y)> exists for all x, but f may be discontinuous at some (even infinitely> many) numbers x. Define g(x) = lim (y->x) f(x). Prove g is continuous.Try the one above again--if you still have problems, write back.-MIke --I don't even know where to start with ts one. in advance, -earl-> === Subject: Re: ARCSIN function, single precision floating point. -- Example routine needed? I'm trying to write ATAN2 function for a small basic language that has IEEE single precision math.. *,/,+.-, SQRT(), SIN(), COS(), TAN() are availible in the language. I've tried a few methods I've found but the results are way off due to low precision, rounding, etc.A Basic language with Sin, Cos, and Tan, should also have at least an ArcTan> ? I mention that because a division result sent to an ArcTan and combined> with some simple quadrant rules should arrive at an ArcTan2.I made a type (reused my paragraph).. It should have said ARCSIN()function in the body of the text. The BASIC compiler doesn't havemuch for intrinsic's as its for the PIC uProcessors and they don'thave much memory.Does anyone have a softcopy of the ARCSIN() single precision routinefrom the C library? Or any other library like Fortran?Thansks,Jon === Subject: Re: Exercise in Projective Linguistics>When I posed the problem, I just assigned names A, B, C, ... to the>points and gave the lines as 5-tuples. Later, it occurred to me>I might have been able to find a different assignment of the>points into the alphabet in such a way that each line was>( a permutation of ) the letters in a 5-letter word.>So here is the challenge: how well can you do ? Can you embed>these letters back into the alphabet so that all, or most, of>these 5-tuples become English words? (Interpret word as you wish.)>(I would be willing to take a solution wch works only for>an affine plane inside here, if necessary.)Ts might be a good one to send to rec.puzzles. I doubt that you can getall 21. The best I've been able to come up with so far is 13, withAB...U = DYCAOEBTJRMLSNKUGFWIV generating the words cadgy, fumed, blind, folky, nervy, misty, curio, clews, woman, barfs, vault, begot, gunks Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israelUniversity of British ColumbiaVancouver, BC, Canada V6T 1Z2 === Subject: Re: Boolean Algebra - Arithmetic RelationspTo all,Forgive me if I sound naive, I have an appreciative but ghly> abstracted perception of mathematics. My Question: I have been told that most of mathematics (geometry, algebra, etc.)> is derived, or at least reducible, to arithmetic. Can arithmetic be> further simplified into boolean logic?? Wch I understand is> equivalent or corresponds ghly to propositional logic (correct me if> I'm wrong here).Is ts what Bertand Russell was attempting to accomplish in Principia> Mathematica?? Or did he mean logic in a different sense??Few will recommend that you read PM to find out, but you might like toread Russell's Introduction to Mathematical Plosophy, Dover, ISBN0486277240.If you want to see a logical (whatever that means) foundation (whateverthat means) of arithmetic, try set theory, Zermelo-Fraenkle perhaps, andSuppes' Axiomatic Set Theory, Dover, ISBN 0486616304 maybe.-- G.C. === Subject: Re: Chess/Go/etc: Continuous Game-Boards?> I was just wondering to, if these have even ever been played, games such> as Chess, Go, etc, where the game-boards are without any subdivision?> http://chessvariants.com/other.dir/continuouschess.html === Subject: Re: Factorial/Exponential Identity, Infinity> I got to tnking about sum 2^n and how it was equal to 2^(x+1) - 1. > I wonder: does it work for 3? The answer is not quite, no.Look up geometric series. If you must reinvent the wheel, you need not do in so publicly. === Subject: Re: ARCSIN function, single precision floating point. -- Example routine needed?> I'm trying to write ATAN2 function for a small basic language that has> IEEE single precision math.. *,/,+.-, SQRT(), SIN(), COS(), TAN() are> availible in the language.> I've tried a few methods I've found but the results are way off due to> low precision, rounding, etc.> >> A Basic language with Sin, Cos, and Tan, should also have at least an ArcTan>> ? I mention that because a division result sent to an ArcTan and combined>> with some simple quadrant rules should arrive at an ArcTan2.> I made a type (reused my paragraph).. It should have said ARCSIN()> function in the body of the text. The BASIC compiler doesn't have> much for intrinsic's as its for the PIC uProcessors and they don't> have much memory.> Does anyone have a softcopy of the ARCSIN() single precision routine> from the C library? Or any other library like Fortran?You didn't say whether your library includes the ARCTAN() function.If so, you can define ARCSIN(x) = ARCTAN(x/SQRT(1-x^2)).-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Hard (unsolved?) search problem revisited> BASIS> -----> A two dimensional table have integers distributed ts way (in ts> examle, only odd numbers):3,11,13,21,27,...> 7,13,17,29,35,...> 9,15,19,33,41,...> .> .> .Assume that ts table is huge, so a fast algorithm is needed -> especially since many tables of ts form must be searched many times.So the values always increase when going from left to right, top to> bottom, no matter where you start.That is, t[i,j] < t[i+1,j] and t[i,j] < t[i,j+1]PROBLEM> -------> Given such a table, t, can you locate a given integer N in polynomial> time - or - can you determine that N is not in the table in polynomial> time (or better, as in constant :-)?If yes, what does that algorithm look like?STATUS> ------> I tried the following approach: start in the middle, t[i_max / 2,> j_max /2] and see if it's less than N. If so, the top/left quadrant of> the table can be eliminated because all values in that quadrant must> be smaller than N. Apply algorithm recursively to the remaining> quadrants. Of course, ts is not a fast algorithm at all...> Any insight would be ghly appreciated.Your quartering method is not as slow as you seem to tnk.For reasons similar to the analysis of a binary search of linearly ordered data, ts is about the best you can expect, and is about O(log(max(R,C))), wch is better than polynomial. === Subject: Dedekind CutsI have heard that the reals can be defined by Dedekind cuts. What ists definition, exactly? === Subject: Re: a puzzle related to artinian group> cards are dealt to them.> There is no asumption >on the number of cards a player receive. In each> round, all players with 2 or more cards pass >one card to the left and> one card to the right. Prove that eventually, all players but one haveexactly one card.> I do not have the solution, so any nt would be ghly appreciated.> Please >restrain from posting messages saying ts is obvious, and> the like, because what is required >here is a proof. a lot> automaton. You can allow any> positive integer states for each of the n cells. The transition rule is> such that if a cell has state x >=2 then the state never exceeds x and if> it has state x < 2 then it either stays the same or increase by 1 or 2.If N is the maximum value of a cell state, then the maximum value of the> states in the next generation can be at most N, unless N = 2 and then a> maximum of 3 is possible in the next generation. It follows that there are> there are only a finite number of possible global states, so eventually> it is periodic.One question might be to figure out wch of Wolfram's four classes ts> cellular automaton lies in.These remarks don't answer your question, but may be of some interest.--EdwinFor what it is worth, I have just verified that the game always terminatesfor n <=9 players no matter how the n-1 cards are dealt. I did it usingbrute force with Maple. --Edwin === Subject: Re: Vedctor Calculus Question>> >A single equation, such as f2(x,y,z)=c2, can describe in a 3d space >a surface, possibly a plane, but not a line.I could *swear* that {(x,y,z) in R^3 : x^2+y^2=0} was a line,>> last time I looked.Lee RudolphNote that z is unrestricted. Oh, I do note that, I do, I do.>It might not hurt to look again.You go first!Lee Rudolph === Subject: Re: Dedekind Cuts> I have heard that the reals can be defined by Dedekind cuts. What is> ts definition, exactly?There's a crazy little tng called Google you might like to try. === Subject: Re: Numeric one-way hash the bar codes are unique is to add a prime number to the previous value. Lap round when you get to the top (or a prime number near the top). Start at a weird value. Could you be more precise, I am afraid that I> don't understand your algorithm.> Anyhow, I can't see that your algorithm would> be unbiased, or how it would prevent duplicates.>Where C, Max are prime constants and C != MaxInitialise N to a valid random numberloop N = N + C if N > Max then N = N - Max Print Nend loopAndrew Swallow === Subject: Re: JSH: About time> So it's about time, as I wait, and wonder, how many of you can handle> the truth.Pure megalomania.Has anyone written any interesting psych papers about the JSH phenomyet? I mean, there's enough material for at least a nice masters'thesis.V. === Subject: Re: 0! = 1> Did someone know a simple demonstration of0! = 1 ?Proof by C:main (){ if (0!=1) printf(truen); else printf(falsen);}(I wish ts was original but it isn't.)Brendan. === Subject: Re: 0! = 1>> Did someone know a simple demonstration of0! = 1 ?>Proof by C:>main ()>{> if (0!=1) printf(truen);> else printf(falsen);>}>(I wish ts was original but it isn't.)>Brendan.There's no way to prove or demonstrate a definition. -- E. PrussingUniversity of Illinois at Urbana-ChampaignDepartment of Aerospace Engineeringhttp://www.uiuc.edu/~prussing === Subject: Re: ab... = (a*b*...)^n ?> In Maple:> dp:=proc(i, B) if i < B then i> else (i mod B)*dp((i - (i mod B))/B, B) fi:end:> B:=3:k:=4: # for example> for i from B to B^k-1 do pr:=dp(i,B):if pr>1 then > n:=round(evalf(log(i)/log(pr))):if i=pr^n then lprint(i,pr,n):fi:fi:od:You get a lot further by examining only those numbers whoseprime factors are less than B. I did some of ts and got anumber of examples. Some of the longer ones:Base 5: 212323421441324231_5 = 1761205026816 = 1327104^2Base 6: 411412_6 = 32768 = 32^3Base 7: 523251_7 = 90000 = 300^2Base 11: 118239_11 = 186624 = 432^2Base 16: b73351_16 = 12006225 = 3465^2Base 17: 1cce112423bf2_17 = 1019744590233600 = 31933440^2Base 24: 122n2g_24 = 8667136 = 2944^2Still notng in base 10 up to 10^200. But if we allow rationalexponents, there are more possibilities.Base 3: 1122221122_3 = 32768 = 64^5/2Base 5: 224_5 = 64 = 16^3/2Base 6: 4424_6 = 1024 = 128^10/7Base 9: 12212_9 = 8192 = 8^13/3 24424_9 = 16384 = 256^7/4 48848_9 = 32768 = 8192^15/13Base 10: 128_10 = 128 = 16^7/4Base 12: 1228_12 = 2048 = 32^11/5Base 14: 288_14 = 512 = 128^9/7Base 15: 2585_15 = 8000 = 400^3/2 319c9_15 = 157464 = 2916^3/2Base 18: 777_18 = 2401 = 343^4/3Base 19: g26c_19 = 110592 = 2304^3/2Base 22: 16c8_22 = 13824 = 576^3/2 6l1e_22 = 74088 = 1764^3/2Base 26: 3993_26 = 59049 = 729^5/3The example in base 3 is instructive--I believe the proposition thatevery power of 2 greater than 2^15 has zeroes in its base-3representation is an open conjecture.Dan Hoeyhaoyuep@aol.com === Subject: Re: Dedekind Cuts> I have heard that the reals can be defined by Dedekind cuts. What is> ts definition, exactly?Ts is a division of the rationals into a pair of sets, wch we will call Left and Right. Each element of Left is less than all elements of Right. Each element of Right is greater than all elements of Left. It follows that Left intersect Right is empty. We also require that a rational number either be in Left or Right so that Left union Right = the set of rational numbers.In the case that Right has a lest element we put that element in Left. There are two case. Left has no greatest element and Right has no least element. If so the division or cut Left,Right corresponds to an irrational number. Or Left has a greatest element and Rigt does not. Ts sup of Left is the number (a rational) represented by the cut.There are rules of addition, subtraction, multiplication and division defined for cuts and it is shown they form a field.For a good account of Dedikind Cuts set -A Course in Pure Mathematics- by G.H.Hardy. === Subject: Re: How to calculate the total coverage area of a few circles? very much, Randy and Prof. Eppstein. The number of circles is about 10. [...]...> Monte Carlo integration is a quick way of estimating the> integral you're talking about doing explicitly. On my Solaris> macne here's the result of a quick run with 10 circles> and a half million points. That took 9 seconds of real time,> 6.6 seconds of CPU time, and converged to 4 decimal places.By the way, another method that occurs to me is to actually> render the circles in some pixelated medium and then count> colored pixels....> Circle centers> C => 0.1942 0.1138> 0.0846 0.9897> 0.9635 0.5098> 0.4557 0.0639> 0.6524 0.0272> 0.0005 0.0413> 0.3786 0.4947> 0.0858 0.8082> 0.5010 0.4129> 0.3872 0.9048> Circle radii> ans => Columns 1 through 8> 0.9391 0.4621 0.9122 0.2243 0.6262 0.2088 0.4072 0.7326> Columns 9 through 10> 0.3542 0.2420> N A(est) clocktime CPUtime> 20000 4.791736 0.3326 0.2700> 40000 4.768794 0.6383 0.5400[...]> 500000 4.765148 9.3062 6.6200If I correctly understand your output, you used different radiifor different circles, rather than going with Leng Supeng'sN circles with the same radius r formulation. Probably a minor point, except that it makes it difficult to compare yourresults to single-radius code. :)Anyhow, for the 10 centers you gave, if there were a common radius of 0.4, there would be only 12 critical points for the method I briefly outlined earlier. They would be: -0.3995 0.0413 -0.3154 0.9897 -0.3142 0.8082 -0.3044 0.8964 0.2328 -0.2843 0.2542 -0.2817 0.3353 1.3014 0.4830 -0.3352 0.7872 0.9048 1.0421 0.1176 1.0524 0.0272 1.3635 0.5098wch took about 1 millisecond to compute*, so a precise integration probably could be completed in about 2 ms (ona 450 MHz Athlon).Your render the circles in some pixelated medium and thencount pixels might be easier to properly implement, of course, and probably only a few milliseconds slower.-jiw* http://pat7.com/jp/circles-area.c computes event pointsbut not area at the moment === Subject: Re: lopital's rule?> could someone please explain what lopital's rule is? my teacher likes> to say tngs like, now, we could use lopital's rule, but that'd be> too easy and you don't know it, so naturally it piqued my interest.sorry to ask such a broad question, but i really don't know anytng> at all about it to possibly narrow the question down.It is spelled L'Hospital's rule. Use Google to look it up. === Subject: Re: Numeric one-way that the bar codes are > unique is to add a prime number to the previous > value. Lap round when you get to the top (or a prime > number near the top). Start at a weird value. Could you be more precise, I am afraid that I don't understand your algorithm. Anyhow, I can't see that your algorithm would be unbiased, or how it would prevent duplicates. Where C, Max are prime constants and C != Max Initialise N to a valid random number loop> N = N + C> if N > Max then N = N - Max> Print N> end loop> Andrew Swallow>p.s. If you want to analyse the algorithmfor bias or duplicates just set C equal to 1.Or in a simple case setting C = 3, Max = 5and initialise N to 25 3 1 4 2You have to stop after producing Max numbers.Andrew Swallow === Subject: Re: All masses have inertia>> The way I see it guys, is that all masses have inertia! That is it>requires>> a net impulse - the product of a net force [f] and its duration [t] ->to>> change a body's present velocity [vi], to some other velocity [vt];>during>> wch period of time, the body is displaced a distance [s]: The ratio>of>> ts impulse [ft], to the time rate of displacement [s/t = (vt-vi)]>that it>> causes is a constant: That is ft/(s/t) = ft/(vt-vi); wch can be>written>> more concisely as f/t/s = ft/ (vt-vi)! Isn't _that_ inertia?Why are you polluting ts newsgroup with your topic. Ts is the trd>thread you started with almost the same topic, isn't it?Hardly. You'd be much closer if you guessed 3333rd--and you'd stillprobably be significantly short. Not all in these two newsgroups, notjust in the last week or so; he's been regurgitating the same vomitfor six and a half years.Gene Nygaardhttp://ourworld.compuserve.com/homepages/Gene_Nygaard/= === ==Subject: Re: Brainteaser: answer correctly to win the admiration and adulation of all.The formula is incomplete. You are lacking the name of the supermarket.Give me more information and will solve your problem. Notice, if itsWal-Mart Stuper Center, they don't give a flip. They will open anytng,anywhere.> A supermarket chain has 3 outlets in the suburbs of a large town. The> locations of these outlets relative to the town centre (in miles) are as> follows: Outlet North South East West> 1 5 5> 2 6 3> 3 4 3 The supermarket chain wish to site a warehouse so that it is centrally> located relative to the positions of the outlets. Advise the supermarket> chain on the optimum location of the warehouse relative to the towncentre. === Subject: Re: 0! = 1> In <1g2k59v.1onstxx1i0ew6aN%bdm@cs.anu.edu.au> bdm@cs.anu.edu.au (Brendan Did someone know a simple demonstration of 0! = 1 ?>Proof by C:>main (){ if (0!=1) printf(truen); else printf(falsen);}>(I wish ts was original but it isn't.)>Brendan. There's no way to prove or demonstrate a definition.I tnk the original post was sarcastic exploring 0 != 1 vs. 0! = 1ambiguity. === Subject: Compact Hausdorff SpacesTs problem is killing me and I know I am pretty close, but I can't makethat last leap.Show that if X is compact Hausdorff under both T and T ', then either T andT ' are equal or they are not comparable.Suppose without loss of generality that T ' is finer (or equal) to T. Then,pick an open set U in T '. Then, X U is closed in T '. Thus, sinceclosed subspaces of compact spaces are compact, X U is compact. Now,since X U is compact in T ', it is also compact in T. Ts is where I amstuck. Where do I use Hausdorffness? Am I going about ts the wrong way?Steve === Subject: Re: numbers out of y numbers in a decreasing sequence. > Say, take 2 numbers from {1,2,3} and arrange as {2,1}. What is the > number of possible combinations ? y choose x (assuming the y numbers are distinct).> . If we allow the same numbers such as {2,2} ?> C(x + y - 1, x).Wch is the number of non-negative integer solutions toX_1 + X_2 + ... + X_y = x.Using your example of 2 numbers from {1, 2, 3} you want the solutions toX_1 + X_2 + X_3 = 2 where X_1 counts the number of 1's, X_2 the 2's andX_3 the 3's. There are C(3 + 2 - 1, 2) = C(4, 2) = 6 possibilities.-- Paul SperryColumbia, SC (USA) === Subject: Re: Ces.88ro-convergence - analysis convergence implies Ces.88ro-convergence. The inverse statement is > not generally true. Can one give conditions on a sequence to guarantee > that when you have a sequence that is Ces.88ro-converging it is also true > that it is regularly converging?A word of caution: I assume by regular convergence you mean theusual notion of convergence. But you should be aware that regularconvergence has a precise technical definition in summability theory.Let A = (a_{ij} : i = 1, 2, ..., j = 1, 2, ...) be an infinite matrix. A is said to be REGULAR if its entries are non-negative, its row sumsare 1 and, for each fixed row i, lim_j a_{ij} = 0. (Sometimes it'sonly required that the LIMIT of the row sums be 1. Same difference.) A sequence {x_n} is said to be A-summable to x provided the sequence{sum_j a_{nj}x_j}, for n = 1, 2, ..., converges to x. {x_n} is saidto be REGULARLY summable to x if it is A-summable to x for everyregular matrix A.Cesaro convergence IS a regular convergence method (the n-th rowconsists of 1/n (n times) followed by all zeros).A regular matrix is said to be STRONGLY regular provided the rowvariations, sum_j |a_{ij} - a_{i,j+1}|, converge to 0 as i -->infinity. Obviously Cesaro summability is a strongly regular method,too. The most important result along these lines is that a sequence{x_n} is A-convergent to x for every strongly regular matrix A if andonly if LIM_n x_n = x for every Banach limit; or, equivalently [andremarkably] provided (x_{p+1} + x_{p+2} + ... + x_{p+n})/n --> x as n --> infinity,UNIFORMLY in the sft p. Ts is a very famous result due to, hmm,what's-s-name, hmmm, another senior moment...[Ts is also called ALMOST convergence. And x_n - x_{n+1} --> 0**is** a Tauberian condition for almost-convergence. Alas, almostconvergence has notng whatsoever to do with almost === everywhereconvergence.]--Ron BruckSubject: Re: Express As Single Fraction > Express the following as a single fraction: > > 4/3ab - 5/6bc > (m^2 + 2)/(m^2 + m) - (m - 2)/m > You do it the same way you do it for fractions in arithematic. The general formula is derived thus a/b + r/s = as/bs + br/bs = (as + br)/bs Yep, I understand the basic priniciple, but I just don't know how to put it> in practise with these types of fractions.>Show us how you make 1/5 + 1/3 and 3/4 + 1/6 into a single fractions.Show us what you've done trying to make 4/3ab - 5/6bc (m^2 + 2)/(m^2 + m) - (m - 2)/minto a single fraction.What have you tried === Subject: Re: Express As Single Fractionalt.math deleted because it is not recognized by my newsreader.alt.algebra.help added because of all the helpful people there.>How do I do ts?Express the following as a single fraction:4/3ab - 5/6bc>> Multiply the first term by 2c/2c and the second by a/a.I'm obviously doing it wrong but that appears to leave two different>denominators. I thought we were trying to get a common one?Multiplying the first denominator (3ab) by (2c) gives 6.Multiplying the second denominator (6bc) by a gives 6.What values did you get?Ts is only a little more abstract than doing common denominatorswith numbers. You look for the factors of both terms, and a commondenominator has to contain all those factors. 3ab has a factor of 3, aand b. 6a has a factor of 2, a factor of 3, and factor of a. For thenumeric part, 6 will serve as common denominator for both (since it'sa multiple of 3). So you just need to include all the other factors:a, b and c. Hence: 6.You can always multiply the terms together to get *A* commondenominator, just not the lowest one. (3ab)(6bc) = 18 is a commondenominator that you can use for both fractions. === Subject: Re: How to calculate the total coverage area of a few circles?>If I correctly understand your output, you used different radii>for different circles, rather than going with Leng Supeng's>N circles with the same radius r formulation. Probably a >minor point, except that it makes it difficult to compare your>results to single-radius code. :)Ah, yes, I must have missed that in the problem statement. Also mycenters and radii were drawn from uniform distributions with noattempt to guarantee overlap. Some of the circles may well bedisjoint. I just wanted to experiment with some quicky code to see howwell it did.>Anyhow, for the 10 centers you gave, if there were a common >radius of 0.4, there would be only 12 critical points for the >method I briefly outlined earlier. They would be:> -0.3995 0.0413> -0.3154 0.9897> -0.3142 0.8082> -0.3044 0.8964> 0.2328 -0.2843> 0.2542 -0.2817> 0.3353 1.3014> 0.4830 -0.3352> 0.7872 0.9048> 1.0421 0.1176> 1.0524 0.0272> 1.3635 0.5098>wch took about 1 millisecond to compute*, so a precise >integration probably could be completed in about 2 ms (on>a 450 MHz Athlon).I have an Ultra sometng Sparc station (Ultra 60?) I tnk your 2milliseconds beats my 6600 milliseconds. So I guess your algorithm isa wee bit more efficient.To the OP: Um... what he said. Ignore my approach.>Your render the circles in some pixelated medium and then>count pixels might be easier to properly implement, of course, >and probably only a few milliseconds slower.Yeah, I was tnking that could be horrendously slow in C code, butextremely fast with assembly code reading grapcs memory directly.I remember doing very fast pixel-level stuff centuries ago in EGA modeon 10 MHz macnes carved from stone and powered by gophers. === Subject: Re: Vedctor Calculus Question>> >A single equation, such as f2(x,y,z)=c2, can describe in a 3d space >a surface, possibly a plane, but not a line.I could *swear* that {(x,y,z) in R^3 : x^2+y^2=0} was a line,>> last time I looked.Lee RudolphNote that z is unrestricted. It might not hurt to look again.(psst: the points in that set have the property x=0, y=0, z any real.That's a line. Tnk about it.) === Subject: Re: a baseball odds calculation problem> here's the problem, specifically: there are three divisions in the american league. every season, eachone of these divisions will be clinched by one team on a certain date. tsis a mathematical certainty. past baseball story tells us that dates before september areextremely rare for teams clincng. in fact, the later days of september,roughly the 15th through the 28th, are where the likelihood of teams clincng isthe greatest. > > You need to specify the probability distribution of the date of > clincng for each division. For example, assuming the the three > divisions are independent and have the same distribution, completely > supported (a model of your approximation) by {15, 16,..., 28}, thenthe > probability is sum(k=15..28, p(k)^3), where p(k) is the probabilityof > clincng on the date k. So if you are saying that each division is > equally likely to be clinched on each day and is independ of theothers, > the probability is 14*(1/14)^3 = 1/196 = .0051 (0.51%),approximately. > Or if you assume identical triangular distributions, p(k) = > min(k-14,29-k)/56, the probability (still assuming independence) is > 2/56^3 sum(k=1..7, k^3) = 1/112 = .0089 (0.89%), approximately. > > Another model would be a discretized normal over the whole season. > Perhaps you want to make an empirical estimate of the distribution;you > may then do the calculation yourself. for the answer. i looked for clincng dates over the last 27full baseball seasons. in those seasons, there have been 126 differentclinches (4 clinches per year from 1975-1993 excluding 1981 strike year and then6divisions) and i was able to find 96 of those dates (they're not very easy tofind). i did an estimate based on ts here's the distribution of clinches ifound: [...data omitted...] It seems you are not satisfied with the quality of the answers you got sofar> and want a more accurate answer.i'm fine with it, others over in teh baseball ng i referenced are pointingout that ts isn't a completely random and independent probability problemsince the probability is weighted heavily toward the end of the season.they're hoping ts gets taken into account in the calculation.But I suspect you collected the wrong data.> It would be far more informative to know how many days from the end> of the season the clinch occurred. That is because the last (scheduled)day> of a season is a Sunday, whose date obviously varies from year to year.ts is very true. also worth noting that baseball used to schedule gameswell into october andhasn't done so for a couple years now, wch makes thechance of clincng on october 3 these days almost nil whereas in years pastobviously there was a good chance of clincng on that date. instead ofcalling it month/day X ishould be referring to it as day Y wch givesus a much more accurate reading from season to season. for the help.> === Subject: Re: Irrationality of the sqrt(2) by unique factorizationI studying a different way of proving that the sqrt(2) is irrational, ornth roots in general. It proceeds as follows:Assume the sqrt(2) is rational. Thus, m/n = sqrt(2). Wch then implies2^1 * n^2 = 2^0 * m^2Now, by unique factorization, a perfect square will have all evenexponentsin its prime factorization. Since, the LHS of the above equation has anoddexponent, namely 2^1, can I then conclude that the sqrt(2) is irrational? Yes.>It seems to me that some step is missing? Shouldn't there be sometngabout the RHS being even and the LHS being odd ? Consider the factors of 2 that appear in either n or m or both; there> is an even number of factors of 2 in n^2, and an even number of> factors of 2 in m^2. So the RHS has, in total, an even number of> factors of 2 (0 + the number of factors in m^2), wle the LHS has an> odd number of factors of 2 (1 + the number of factors in 2). Ts is impossible. Ts would be the long-winded explanation. What you have noticed is> simply that all the primes on the right hand side appear raised to an> even exponent, wle all but ONE of the primes on the left hand side> appear raised to an even exponent, the exception being raised to an> odd exponent. That's impossible by unique factorization.> How does one arrive at theconclusion that if a square isn't perfect, then it is irrational? No such conclusion is reached. Or do you mean, that an integer wch> is not a perfect square has an irrational square root?> Iunderstand intuitively that if a sqrt() is not perfect, then it isirrational, but I don't see how one can logically conclude that from theabove equation. The above equation only proves it for 2. A similar argument yields the> result for sqrt(p) for any prime p; then it is not hard to prove it by> a similar argument for any product of the form p_1*...*p_r where> p_1,...,p_r are distinct primes. Then simply note that positive integer greater than 1 can be written> uniquely as n=k*s, where s is a perfect square, and k is a product of> distinct primes. So then sqrt(n) = sqrt(k)*sqrt(s), sqrt(s) is an> integer, and sqrt(k) is irrational. ==> It's not denial. I'm just very selective about> what I accept as reality.> --- Calvin (Calvin and Hobbes)> == Arturo Magidin> magidin@math.berkeley.edu> Arturo!!! I just wasn't seeing the forest for the trees. It makesperfect sense now.Lurch === show that the edge connectivity (the lease number of edgesif removed will disconnect the graph) of a complete graph is equal ton-1Here is my proof by induction:Let G be a complete graph on n vertices, n>1 Let n = 2 then e(G)= 1 = n - 1 is true. Let n = 3 then e(G) = 2 = n - 1 is true.Let n = 4 then e(G) = 3 = n - 1 is true.Assume e(G) = n - 1 is true for every n, G is a complete graph on nvertices. We need to show that e(G) = n when G is on n+1 vertices.Since e(G) = n 1, then for a graph on n + 1 vertices, then e(G) = (n+1) -1 = n is true since n equals to the number of verticesof G minus 1 . Thuse(G) = n - 1 for every n when G is a complete graph. Is it right and complete or missing sometng? Please let me know. === Subject: Re: Compact Hausdorff me and I know I am pretty close, but I can't make> that last leap.Show that if X is compact Hausdorff under both T and T ', then either T and> T ' are equal or they are not comparable.Suppose without loss of generality that T ' is finer (or equal) to T. Then,> pick an open set U in T '. Then, X U is closed in T '. Thus, since> closed subspaces of compact spaces are compact, X U is compact. Now,> since X U is compact in T ', it is also compact in T. Ts is where I am> stuck. Where do I use Hausdorffness? Am I going about ts the wrong way?Consider the identity map id: (X,T') --> (X,T). It's continuous(because T' is finer than T), 1-1 and onto. Because X is compactHausdorff, id is therefore a homeomorpsm. And thus T = T'.When you recall the proof that id is a homemorpsm, i.e. that id^{-1}is continuous, it runs as follows: let K be closed relative to T'. Since (X,T') is compact, therefore K is compact relative to T'. Afortiori, K is compact relative to T. Since (X,T) is Hausdorff,therefore K is closed relative to T. In other words, K closed relativeto T' implies K is closed relative to T. In other words, T' subset T.--Ron Bruck === sha1:GxrZdD7BaS4PZy6zgEqFNlmQlfY=X-NFilter: 1.2.0>How do I do ts?Express the following as a single fraction:4/3ab - 5/6bc>> Multiply the first term by 2c/2c and the second by a/a.I'm obviously doing it wrong but that appears to leave two different>denominators. I thought we were trying to get a common one?>What did you get after performing the multiplication?<> === Subject: Zeroes, extrema, inflection points, ..., etc?If you consider the graphs of y=f(x), y=f'(x), and y=f''(x), thevalues of y are called the value, the slope, the concavity,respectively, and the points at wch y=0 are called the zeroes, thevertices, and the inflection points. My question is, does f'''(x)have a name? If so, what is it? Do any other f'(n)'(x) have names?Note: f'(4)'=f'''' and f'(n)' is f with n primes. === Subject: Re: All masses have inertiaIn sci.math, Bob dispatch.concentric.net>:You've got to learn; like it or not:Plonk m..> I'm even wasting bandwith with ts advice.Bob PEase> I'll admit to some surprise as to why he comes over here in sci.math.He used to hang around sci.physics, but we kinda drove m off. :-)As it is, what he's apparently proposing is notnghorribly new; the main problem with SI as he sees it isthat the newton != the kg, when the kg is placed in a 1-ggravity field. However, the pound-force == pound-mass in1-g gravity field, numerically speaking. (Personally, Itnk that leads to confusion. Besides, in SI g = 9.805 m/s/sanyway, wch is almost 10 -- good enough for back of thenapkin calcs.)There are a number of problems with ts approach, not the leastof wch is the variance of g (around 0.4%) as one wandersabout the globe.And he has a magic number: the number of pounds in a gallonof water. SI has an artifact (a 1 kg iridium block somewherein France, IIRC) and a few magic numbers relating totime and length using the speed of light.Of the two, I tnk the artifact is a little easier to handle. :-)-- === Subject: Re: JSH your sp has come in!!!!>So far the team as I call it are people who recognize that my work>is indeed correct, and so far seem to only be very gh IQ people.Hey James, did I ever tell you about how the gh IQ people fared when I challenged them in their annual Quiz Bowl? Undefeated in adouble elimination tournament, culminating in two consectutivevictories over the reigning champions! gh IQ my === ass.--MensanatorSlayer of the MensaSubject: Re: Compact Hausdorff Spaces > Show that if X is compact Hausdorff under both T and T ', then either T and> T ' are equal or they are not comparable. Suppose without loss of generality that T ' is finer (or equal) to T. Then,> pick an open set U in T '. Then, X U is closed in T '. Thus, since> closed subspaces of compact spaces are compact, X U is compact. Now,> since X U is compact in T ', it is also compact in T. Ts is where I am> stuck. Where do I use Hausdorffness? Am I going about ts the wrong way?>Compact sets of Hausdorff spaces are closed, that's where. === Subject: Re: lopital's rule? Cox following:>> Guy Corrigall scribbled the following:> L'Hospital's Rule (pronounced lopital in French) can be found in any book> on calculus of a single variable (look it up!).L'Hospital or L'H.99pital. The circumflex (the ^ tngy) means that the>> following s is understood implicitly.> Is the s pronounced?No.-- /-- Joona Palaste (palaste@cc.helsinki.fi) ---| Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++|| http://www.helsinki.fi/~palaste W++ B OP+ |----- Finland rules! /'I' is the most beautiful word in the world. - Nordberg === Subject: Re: Algebra proof> > Let a, b, r and s be integers with r>1, s>1 and gcd(r,s)=1. Prove > that if a=b(mod r) and a=b(mod s) then a=b(mod rs). How do I do ts???? a = b + m*r for some m, and a = b + n*s for some n, right? That means that m*r = n*s. Now use the fact that gcd(r,s) = 1 to conclude sometng about m (or n). I guess since r and s are relatively prime, then r must divide n. so r*x=n. and a=b+r*x*s, so a=b(mod rs), right?Exactly.Is there a name to ts theorem? That is, the theorem that says thatif gcd(a,b)=1, and a*c=b*d, then a must divide d and b must divide c? === Subject: Re: a puzzle related to artinian group> Suppose n people sit around a table and n-1 cards are dealt to them.> There is no asumption on the number of cards a player receive. In each> round, all players with 2 or more cards pass one card to the left and> one card to the right. Prove that eventually, all players but one have> exactly one card.R. Anderson, L. Lovasz, P. Shor, J. Spencer, E. Tardos, S. Winograd,``Disks, balls, and walls: analysis of a combinatorial game'', Amer.Math. Monthly, 6, 96, pp. 481--493, 1989.It talks about the same problem (of course along a line, not around atable). Their discussion is thrilling.Best, === Subject: Re: lopital's rule?> so, in our example above, f(x) is (3x - sin x), g(x) is x and a is zerothe limit of (3x - sin x)/x as x approaches zero is, by the Rule:(3 - cos x)/1 as x approaches zero, wch = 2Ts is the kind of example that gets some of us upset over the use (or overuse, or misuse) of l'Hopital. To use l'Hopital for ts one, you must know the derivative of sin x. To compute the derivative of sin x (from the Newton quotient), you must know the limit, as x goes to zero, of (sin x) / x. Thus the use of l'Hopital on ts circular function involves circular reasoning.-- ) === Subject: Re: =?ISO-8859-1?Q?Ces=E0ro-convergence_-_analysis_question?=for infinitely many N. Suppose N satisfies (iii'). Now for> n > N we have s_n >= d - c sum_N^n 1/j> ~ d - c log(n/N)> >= d/2> , I have a question about the above step you take in your proof. Idon't understand why d - c log(n/N) >= d/2log(n/N) goes to infinity as n goes to infinity if N stays constant. ~ Chris === Subject: Re: Joe Uptaught (Was Re: Ullrich on Identity) A propos, here are cites from Pierre Bourdieu's > _Language & Symbolic Power_ (the titles are mine). > Enjoy! Censorsp There Oughta Be A Law > > Cool-Hand Luke The Social Conditions for the Effectiveness of Ritual Discourse Heretical Discourse > The *Skeptron*> Symbolic Power & the Symbolism of PowerThere is no symbolic power without the symbolism of power. Symbolicattributes--as is well illustrated in the paradigmatic case of the*skeptron* and the sanctions against the improper wearing ofuniforms--are a public display and thereby an officialization of thecontract of delegation: the ermine and the robe declare that thejudge or the doctor is recognized as having just cause (in thecollective recognition) for declaring mself judge or doctor, thats imposture--in the sense of the pretension expressed by sappearance--is legitimate. The competence that is specificallylinguistic--the Latin once spoken by doctors or the eloquence of thespokesperson--is also one of the manifestations of competence in thesense of right to speech and to power through speech. There is awhole dimension of authorized language, its rhetoric, syntax,vocabulary and even pronunciation, wch exists purely to underlinethe authority of its author and the trust he demands. In ts respect,style is an element of the *mechanism*, in the Pascalian sense,through wch language aims to produce and impose the representationof its own importance and thereby help to ensure its owncredibility.[14] The symbolic efficacy of the discourse of authorityalways depends, in part, on the linguistic competence of the personwho utters it. Ts is more true, of course, when the authority ofthe speaker is less clearly institutionalized. It follows that theexercise of symbolic power is accompanied by work on the *form* ofdiscourse wch, as is clearly demonstrated in the discourse of poetsin archaic societies, has the purpose of demonstrating the orator'smastery and gaining m the recognition of the group. (Ts logic isalso found in the popular rhetoric of insults wch seeks, by flagrantoverstatement and the regulated deformation of ritual formulas, toproduce the expressive accomplishment wch allows one to 'get thoselaugng on one's side'. Notes14. The two senses of competence come together if one sees that,according to Percy Ernst Schramm, just as the crown of the medievalking designates both the tng itself and the set of rights wchconstitute royal dignity (as in the term 'crown property'), so toolinguistic competence is a symbolic attribute of the authority wch*designates* a socially recognized status as a set of rights,beginning with the right to speak and the corresponding technicalcapacity.(Pierre === Bourdieu, _Language and Symbolic Power_ pp. 75-76)Subject: Re: help !!! === Subject: help !!! >I need help giving (p->q) xor (p<->q) an equivalence less than 12 >letter without using xor.....Prove the following are a sequence of equivalent statements. (p->q) xor (p<->q) (p->q)&~(p<->q) or ~(p->q)&(p<->q) (p->q) & ~(p<->q) (p->q) & (~(p->q) or ~(q->p)) (p->q) & ~(q->p) (p->q) & q & ~p q & ~pIf instead you just verify that (p->q) xor (p<->q)and q & ~phave the same === truth tables, you've learned notng.----Subject: Re: Mathematics === Subject: MathematicsGraph theory would have been a more welcomed and informative title. >I need to show that the edge connectivity (the lease number of edges >if removed will disconnect the graph) of a complete graph is equal >to n-1Pick any vertex V from K_n the complete graph on n vertices.V has n-1 edges coming from it. Removal of these disconnectsV from K_nV. Thus the edge connectivity is at most n-1.Now remove n-2 edges from K_n. Call that graph G.Case 1: some vertex V has no edges removed G is connected as every vertex in G has an edge going to VCase 2: every vertex has an edge removed pick any vertex V; V has degree at least one the remaining vertices are K_(n-1) with at most n-3 edges removed by induction hypothesis, the edge connectivity of K_(n-1) is n-2 thus the remaining vertices are connected and V connects to them >Here is my proof by induction: >Let G be a complete graph on n vertices, n>1 >Let n = 2 then e(G)= 1 = n - 1 is true. >Let n = 3 then e(G) = 2 = n - 1 is true. >Let n = 4 then e(G) = 3 = n - 1 is true. >Assume e(G) = n - 1 is true for every n, G is a complete graph on n >vertices. We need to show that e(G) = n when G is on n+1 vertices. >Since e(G) = n^1, then for a graph on n + 1 vertices, thenWhat's that strange stuff for e(G) = ???. >e(G) = (n+1) -1 = n is true since n equals to the number of vertices >of G minus 1 . >Thus e(G) = n - 1 for every n when G is a complete graph. >Is it right and complete or missing sometng? Please let me know.A badly written circular argument. To clarify terminology,consider K_(n+1) and for some subgraph === K_n, apply e(K_n) = n-1.----Subject: Re: Nice summation puzzle === Subject: Nice summation puzzle >Starting from arcsin(x)=x+1/2/3*x^3+...What's 1/2/3 ? The next term is 1/2/3/4/5 x^5 ? >and using an integration operation show well-known sum: >oo 1 >S ------- = (pi^2)/6 >n=1 n^2integral(0,1) arcsin x dx = x arcsin x + sqr(1-x^2) |_0^1 = arcsin 1 - 1 = pi/2 - 1 /= (pi^2)/6---- === Subject: PV IntegrationHelp! I'm having problems visualising integration of pressure withrespect to volume....If I integrate velocity wrt time I get distance. Practically tsmeans that I sum the measured velocity over a monotonically increasingvarible (time), ts I can see.But with pressure volume loops the data is cyclic and if I integratedelta Pressure wrt Volume what do I get?? I cant see what I get fromts process or how I go about doing the math in Matlab.... for your help,Mark. === Subject: Re: a puzzle related to artinian group> Suppose n people sit around a table and n-1 cards are dealt to them.> There is no asumption on the number of cards a player receive. In each> round, all players with 2 or more cards pass one card to the left and> one card to the right. Prove that eventually, all players but one have> exactly one card.Construct a function f that has as its domain the state of the tableat any given round and has as its range the nonnegative integers. Define f as the sum of the number of cards the each of the n playershave BEYOND 1.I will give a brief example in case my definition of f is unclear. Let n = 3, so there are 2 cards. If the first and second players eachhave a card, then f = 0, since no player has any more than 1 card. However, if the first player has both cards, then f = 1 since thefirst player has 1 more than 1 card. Similarly, if n = 5, and boththe first and the second players have 2 cards, then f = 2, since f isa sum.Notice that f cannot increase.Each of the n people can have at most n-1 cards at any time. Ts isfinite, so eventually, the state of the table will become periodic. We will now only consider the states after the states become periodic.We know that f is invariant.Find any player who has no cards. If he were to ever get a card, thenf would decrease. So we know that if a player has no cards, he willnever get any cards.That means that any player next to a player who has no cards willnever get more than 1 card. Doing so would force m to, on the nextturn, give a card to s neighbor who has no cards.That means that any player next to a player who is next to a playerwho has no cards will never get more than 1 card. Doing so wouldforce m to, on the next turn, give a card to s neighbor who has aneighbor has no cards.I'm sure you can see where I'm going.Any player who has a path of neighbors to someone who has no cardscannot have more than 1 card. It follows that the only periodicbehavior shown can be of n-1 people each having one card. ~ Chris === Subject: Pluecker coordinates of symmetric powersLet V be a finite dimensional vector space with a given ordered basise_1, ..., e_n . Let W be a p-dimensional linear subspace. If x_1, ...,x_pis a basis of W, express the x_i as linear combinations in the e_j.The coefficients give a p x n-matrix. The maximal minors of tsmatrix are known as the *Pluecker coordinates* of the subspace W .Up to a scalar multiple they are independent of the basis x_1, ..., x_pand, considered as homogeneous coordinates, characterize W uniquely.For any k, consider the k-th symmetric power S^kV of V. Theordered basis e_1, ..., e_n determines an ordered basis of S^kV, andthe subspace W of V determines a subspace S^kW of S^kV. So thePluecker coordinates of S^kW as a subspace of S^kV are determinedby the Pluecker coordinates of W in V and, in fact, should bepolynomialsin the latter ones. Does someone know how these look like? Boudewijn === Subject: Re: Vedctor Calculus Question >A single equation, such as f2(x,y,z)=c2, can describe in a 3d space >a surface, possibly a plane, but not a line. I could *swear* that {(x,y,z) in R^3 : x^2+y^2=0} was a line, last time I looked. Lee RudolphNote that z is unrestricted. It might not hurt to look again.My badMy incredible, incredible, bad === Subject: Re: lopital's rule?Why's that then? It seems to me to be a mildly interesting theorem. I> can see that it might not be a good idea to prove it and then give> students lots of 0/0 limit problems for them to solve using> L'Hospital's when doing so is just pedagogically useless manipulation.> But that is no reason for not proving the theorem.Who said anytng about proving anytng. My experienceis that it is taught as a method, not as a theorem, and whentaught it, students will use it *exclusively* for all limitproblems thereafter (it's a formula isn't it) no matterhow inappropriate it is for the problem at hand.-- === Subject: Re: a puzzle related to artinian groupNevermind. My proof is horridly flawed. === Subject: Re: a puzzle related to artinian group> Suppose n people sit around a table and n-1 cards are dealt to them. There is no asumption on the number of cards a player receive. In each round, all players with 2 or more cards pass one card to the left and one card to the right. Prove that eventually, all players but one have exactly one card.Construct a function f that has as its domain the state of the table> at any given round and has as its range the nonnegative integers.> Define f as the sum of the number of cards the each of the n players> have BEYOND 1.> [snipped example]> Notice that f cannot increase.Why is that: ....121.... -> ....202....,so f may increase by one???Each of the n people can have at most n-1 cards at any time. Ts is> finite, so eventually, the state of the table will become periodic.> We will now only consider the states after the states become periodic.Ts is true and applies without the rectriction that the number of cardsis limited to precisely n-1.We know that f is invariant.If it were non-increasing, we could deduce that. However, see above.Find any player who has no cards. If he were to ever get a card, then> f would decrease. So we know that if a player has no cards, he will> never get any cards.Again, the situation is more subtle than that. Somehow you MUST use thefact that the total number of cards is n-1. Otherwise you can have, e.g.the cycle 02020202...02 (even number of players) alternating with202020...20. Notice that the function f is constant here. > That means that any player next to a player who has no cards will> never get more than 1 card. Doing so would force m to, on the next> turn, give a card to s neighbor who has no cards.> [snipped a natural recursive reasoning]I was tnking along the similar lines, so IMHO your scheme for a proof isa very natural one. In other words, it is natural to study the periodicsituations. However, the devil is in the detail. Back Finland === Subject: Re: absolute moments of the normal distribution> to everyone!!> Some of you know where I can find the formula of the k-th absolute moment> for the normal distribution?I presume you meana_k = integral_{-infinity}^infinity (2pi)^{-1/2} |x|^k exp(-x^2/2) dx.Ts is2(2pi)^(-1/2) integral_0^infinity x^k exp(-x^2/2) dx.We convert ts into a gamma integral by making thesubstitution y = x^2/2. We get dy = x dx and soa_k = 2(2pi)^(-1/2) integal_0^infinity (2y)^{(k-1)/2} exp(-y) dy = 2^{k/2} pi^(-1/2) Gamma((k+1)/2).To simplify ts further, recall that for integers r >= 0,Gamma(r+1) = r! and Gamma(r + 1/2) = pi^(1/2)2^{-2r}(2r)!/r!.Hence for even k = 2r,a_k = 2^{-r}(2r)!/r!and for odd k = 2r+1a_k = 2^{r + 1/2} pi^{-1/2} r!.> My own calculations are different from the MathematicaAAAARRRGGHHHH!> output and I am not> sure about what to do...-- === Subject: Question on lbert & GodelWhat did lbert ask and claim concerning Foundations of Mathematics(sets, predicate calculus), metamathematics, Logic, Incompleteness,etc?Sometng about the Continuum Hypothesis? Completeness orIncompleteness of Logic? Was he contradicted by kindergarten logicfrom Godel? How would we exactly formally represent lbert'squestions and claims using current day notation? Formal or programmedderivations?Charlie VolkstorfCambridge, MA === Subject: Re: factoring to satisfiability> (10-year-old) conjecture: factoring -> SAT will eventually> give greater magnification> of the exact structure of the problem that will be invisible> to straight factoring algorithms, ***if studied carefully***,> leading to asymptotically equivalent or superior algorithms. that seems to be exactly the same motivation/inspiration for> the klook paper that eppstein cites. ( for the refs)Below is a 3x2 bit multiplication circuit.I can use ts circuit to make assertionsabout the factors of any number.I'm curious if ts has been studied.3x2 InputECA * DB = ZYXWVV = A1B1W = A1B0D1 + A1C0D1 + A0B1C1 + B1C1D0X = A0C1D1E0 + B0C1D1 + A1B1E1 + B1C0E1 + B1D0E1Y = A1B1C1D1E0 + B0D1E1 + C0D1E1Z = B1C1D1E1Using the circuit above I can make the following assertions:V1 -> A1B1W1V1 -> A1B1C0D1 + A1B1C1D0X1W1V1 -> A1B1C0D1E0F1 + A1B1C0D1E1F0 + A1B1C1D0E0F1 + A1B1C1D0E1F0(ts last assertion is based on a 3x3 bit multiplier.)Let z = x * yThe first assertion shows that if the low order bit of z is 1then both x and y must have 1 for their low order bit.Ts is fairly obvious. If z is odd then both x and y must be odd.But, the multiplication circuit allows me to extend these assertions.If the low order bit of z is 1 then both x and y must have 1 as their loworder bit.z = ?1 -> (?1 * ?1)If the bottom two bits of z are 11 then one factor must end in 01and the other factor must end in 11.z = ?11 -> (?01 * ?11) or (?11 * ?01)If the last three bits of z are 111 then there are only twopossible combinations for the last three bits of x and y:z = ?111 -> (?111 * ?001) or (?011 * ?101) or (?101 * ?011) or (?001 * ?111)These assertions can be extended with larger multiplication circuits.I am also curious if assertions like these can be used to show thatprime numbers must end with certain bit patterns.Russell- Zeno was right. Motion is impossible. === Subject: Re: Deep Thoughts # 1: A new limitation to the human mind > 1. Mathematics is the science in wch we make sometng out of > notng. Wrong. Mathematics is built on the 13 Axioms, wch are not notng.And where did those 13 axioms come from? What is needed to develop> them?They are definitions. That is however irrelevant. Mathematics is notthe the science in wch we make sometng out of notng. Are youtrying to imply that because 'sometng' requires 'notng' to becreated, it is 'notng'? By that logic, everytng would ultimatelycome from 'notng'. I guess you got your terms mixed up. > 2. All of man-made Mathematics consists of an abstraction from > physical processes. Wrong again. Mathematics consists of theorems derivated by following a set of formal rules.And when is ts not just an abstraction from physical processes> (example, please)?Group theory? Banach's Fixpoint theorem? most of mathematics? > 3. Since Mathematics in general needs notng to be created, then the > human mind seems to be limited in that it can only consider > possibilities that are analogous to physical processes. Well, I tnk first of all mathematics exist all by themselves, as a logical system. And so needs notng to be created.> However one may argue that for humans ts exists in its precise own way, because the human mind works the way it works.That's right. Exactly. And in particular, the human mind seems to> only discover the mathematics that is an abstraction from physical> processes.Sorry no. You are building on false premises. Thus mathematics actually may help delineate the limits of the human mind.That is exactly what I am postulating.No, what you postulated is the human mind seems to be limited in thatit can only consider possibilities that are analogous to physicalprocesses, wch I tnk is not only wrong because you derive it fromfalse premises but wrong in it own terms. The human mind can considerthe possibility of, lets say contradiction, or poetic justice wch are not analogous to a physical processes.You will agree with me that I did not state in wch way the humanmind is limited. === Subject: Re: Quantum Gravity........ ...However, as there is a never a two without the trd. QuantumGravity, definitely, would be as should be the trd after a Relativity andQuantum Mechanics. Whether, the later has been and remains, the guide as acontroler of the first, the trd would be as should be formulatedMathematically as a guiding as a controler of the both.Therefore, in that moment, a specific as a crucial event would be as shouldbe, as it is, that it would appear a fastidious transition allowing thethree turning over a closed a Mathematical circle.Therefore, over all that matter, a Space-Time would definetely apply,definitely as a matter a fact!!!!!!!!!!!........... ...-- Ahmed Oua, ArctectBest viestiss.8a> Part 2> Comments by Jack Sarfatti on excerpts from:> ( to Gary Bekkum for bringing ts paper to my attention.) Excerpts are from: The three perspectives on the quantum-gravity problem> and their implications for the fate of Lorentz symmetry1 Dipart. Fisica Univ. La Sapienza and Sez. Roma1 INFN> P.le Moro 2, I-00185 Roma, Italy there is still not a single measured number whose interpretation> requires advocating quantum gravity. The trd possibility is a condensed-matter perspective (see, e.g., the> research programs> of Refs. [13] and [14]) on the quantum-gravity problem, in wch some ofthe> familiar properties of spacetime are only emergent. Condensed-matter> theorists are> used to describe some of the degrees of freedom that are measured in the> laboratory as> collective excitations witn a theoretical framework whose primary> description is given> in terms of much different, and often practically unaccessible,> fundamental degrees of> freedom. Close to a critical point some symmetries arise for the> collective-excitations> theory, wch do not carry the significance of fundamental symmetries,> and are in fact> lost as soon as the theory is probed somewhat away from the critical> point. Notably,> some familiar systems are known to exbit special-relativistic> invariance in certain> limits, even though, at a more fundamental level, they are described in> terms of a nonrelativistic theory.> For a rather general class of fermionic systems one finds [13] that at> low energies, as a Fermi point> is approached, fermions gradually become cral Weyl> fermions, wle bosonic collective modes of the vacuum transform into> gauge fields and> gravity. Clearly from the (relatively new) condensed-matter perspective on the> quantum gravity> problem it is natural to see the familiar classical continuous Lorentz> symmetry> only as an approximate (emergent) symmetry. [Revised Comment #3: Ts notion is very compatible with the> Bohm-ley-Vigier pilot BIT wave landscape IT dden variable system> point flow on the landscape picture of quantum theory. Orthodox> micro-quantum theory is the limit of Antony Valentini's sub-quanta heat> death with signal locality in wch the quantum potential is> fragile (Bohm-ley) with a pilot BIT wave (nonlocal in configuration> space for entangled systems) that has no sources (Bohm-ley). Ts> means that the IT dden variable system point flowing on the BIT> orders (J.A. Wheeler) from BIT but not vice versa. Ts is action> without reaction! Here reaction means back-action where the BIT> pilot wave landscape has sources so that IT is no longer a passive test> landscape it is flowing on according to vs = (h/m)Grad(Phase) - (q/c)A> or in the 4D elastic world crystal lattice (Hagen Kleinert) model> distortion field du(x) we have the IT FROM BIT constraint equation showing how the IT> geometrodynamic field gets its marcng orders (Wheelers term in a> different but related context) from the MACRO-QUANTUM BIT vacuum> inflation field. du(x) = Lp*^2 (Goldstone Phase),u - TaA^a,u Lp*^2 = hG*/c^3 = Lp^4/3(c/Ho)^2/3 in t'Hooft-Susskind world hologram> model for renormalized Planck scale Lp* ~ 1 fermi determined by> Newtonian Planck scale Lp = 10^-33 cm and size 10^28 cm of our> contingent local Level I (Max Tegmark) parallel IT universe in an> infinity of parallel IT universes on a single 3D spatially flat> American). Ta are the generators of the Lie algebra of the standard> model in globally flat Minkowski spacetime with fiber connections A^a,u> for parallel transport. a is internal space and u is spacetime. Ts is> normalized such that TaA^a,u is a length. ,u is the ordinary partial> derivative operator relative to x^u, u = 0, 1,2,3. The attraction between virtual electrons and positrons near the -mc^2> band in the Dirac vacuum spectrum Fermi sphere is a pair instability> that creates the virtual giant local vacuum wave BEC <0|e+(x)e-(x)|0> a> complex scalar inflation field whose phase is the Goldstone phase and> whose amplitude is the ggs field <0|e+(x)e-(x)|0> = |ggs field(x)|e^i(Goldstone phase(x)) = PSI(vac) such that the unified dark energy/matter zero point stress-energy> density Diff(4) tensor field is : tuv(x)zpf = (c^4/8piG*)/zpf(x)guv(x) where /zpf(x) = Lp*^-2[1 - Lp*^3|ggs field(x)|^2] /zpf(x) > 0 is strongly scale-dependent anti-gravitating dark energy> exotic vacuum with negative pressure and positive zero point energydensity. /zpf(x) < 0 is strongly scale-dependent gravitating dark matter> exotic vacuum with positive pressure and negative zero point energydensity. All quantum fields of all spins contribute to the net residual /zpf(x)> LOCAL FIELD in the emergent More is different collective mode c-number> background spacetime with both Diff(4) base space and local Lorentz> group tangent space emergent symmetries at least near a critical point> of the renormalization group flow sense. Einstein's gravity c-number background field is guv(x) = nuv(Minkowski) + (1/2)[du(x),v + dv(x),u] with anholonomic torsion tensor field (string Goldstone phasesingularities) suv(x) = (1/2)[du(x),v - dv(x),u] and the back-action nonlinear LOCAL virtual BEC Landau-Ginzburg Diff(4)> symmetric BIT FROM IT back-action equation: {D^uDu + V(|<0|e+(x)e-(x)|0>|)}<0|e+(x)e-(x)|0> = 0 Where V(|<0|e+(x)e-(x)|0>|) limits to the effective spontaneous broken> vacuum symmetry potential of chaotic inflation cosmology in the> large-scale limit of the isotropic homogeneous FRW metric with zero> point energy density induced cosmological constant for dark energy> using an adaptive windowed wavelet transform version of the Wigner phase> space density with ODLRO rather than a rigid windowed Fourier transform.> Shorter scales have longer bandwidth and longer scales have smaller> bandwidth with total area in phase space per conjugate pairs of> incompatible dynamical variables constant ~ h (in phase space) or Lp*^2> (in spacelike slice of spacetime) to preserve Heisenbergs uncertainty> principle and the Bekenstein-tHooft-Susskind world hologram> reinterpretation of generalized black hole thermodynamics. See also> the somewhat related 3D spacelike quantized area operators in Penrose> spin networks of the loop quantum gravity approach. Note that Lp^2 = hG(Newton)/c^3 Maps the discrete structure of quantum gravity foam in phase space to a> discrete area structure in spacelike slices of spacetime consistent> with the world hologram Ansatz. Du is the Diff(4) operator so that D^uDu is the GR wave propagation> operator on a complex numbered scalar field PSI(vac).] Results obtained over the last few years (wch are partly reviewed> later in these notes) allow us to formulate a similar expectation from> the general-relativity perspective. Loop quantum gravity and other> discretized-spacetime quantum-gravity approaches appear to require some> departures, governed by the Planck scale, from the familiar (continuous)> Lorentz symmetry. And in the study of noncommutative spacetimes some> Planck-scale departures from Lorentz symmetry might be inevitable, since> (at least in a large majority of noncommutative spacetimes) a Lie> algebra is not even the appropriate language for the description of the> symmetries of a noncommutative spacetime (one must resort to the richer> structure of Hopf algebras). reason to renounce> to exact Lorentz symmetry. Minkowski classical spacetime is an> admissible background> spacetime, and in classical Minkowski there cannot be any a priori> obstruction for> classical Lorentz symmetry. Still, a breakup of Lorentz symmetry, in the> sense of> spontaneous symmetry breaking, is of course possible. Ts possibility> has been studied> extensively [10, 15] over the last few years, particularly in String> Theory, wch is> the most mature quantum-gravity approach that emerged from the> perspective. 1.2 What do we know about quantum-gravity?> The theory debate clearly is a confrontation between very different> perspectives on> the quantum-gravity problem. If we had any robust information on quantum> gravity> certainly at least some of these ideas would have been proven to fail.> But after more> than 70 years [16] of work on the quantum-gravity problem there is> still not a single> measured number whose interpretation requires advocating quantumgravity.> I have so far mentioned the quantum-gravity problem as if it was a> well-established> and familiar concept, but it is perhaps useful to give here an intuitive> characterization> of ts problem. The quantum-gravity problem is sometimes described as a> sort of> human discomfort, as a problem pertaining to the acevement of a more> satisfactory> plosopcal worldview. For example, as motivation for research in> quantum gravity> it is sometimes stated that quantum theory (in an appropriate> generalized sense)> has turned out to be relevant for the description of measurement results> in all other> branches of fundamental physics, and we therefore must assume that it> will eventually> be relevant also for spacetime/gravity physics. Analogously (and> amounting to the> same tng), it is sometimes stated that it is unsatisfactory to have on> one side our> present unified quantum-field-theory description of electromagnetic,> weak and strong> forces and on the other side gravity wch is still described in a very> different way. These> human discomforts do not of course define a scientific problem, but> actually there is,> as emphasized by some, a well-defined scientific problem wch can be> naturally called> quantum-gravity problem. The scientific problem that can be reasonably called quantum-gravity> problem> is actually the problem of producing numbers (predictions), in a> logically-consistent> quantum-field theory effects cannot be neglected. For example, although> we are presently (and for> the foreseeable future) unable to set up and observe collisions between> two electrons> each with energy of, say, 10^50 eV, our present theories provide no> obstruction for the> analysis of such gh-energy collisions, but are unable to produce a> logically consistent> number for, say, the probability that such a collision would result in> two muons with> certain energies and momenta. The problem, as I shall try to point out> later in these> notes, resides in the fact that quantum field theory implicitly assumes> that gravity> effects can be neglected. When the gravity effects are so large that> (from the field theory> perspective) space geometry evolves significantly on very short time> scales, field> theory cannot be consistently appliedb. Similarly, field theory runs> into trouble when> gravity effects are strong enough to admit the emergence of spacetime> singularities (e.g.> black holes). We are able to get numbers out of quantum field theory> in contexts in> wch there is a curved static (or slowly-varying) nonsingular space,> but fast-varying> and/or singular space geometries are untreatable. [Comment: An adaptive windowed zoom in/out wavelet transform> reformulation may help here.] One might argue that 10^50 eV electrons should be the least of our> concerns, since we> are never going to be able to produce and/or observe them, but first of> all in cosmology> there are some numbers we should produce that depend on very early times> after the> gh energies> were abundant), and, secondly, the fact that our theories fail to> produce numbers in> some contexts wch those same theories describe as accessible (in> principle) makes us> concerned in general about the robustness of these theories. Since we> know that new> elements would have to be introduced in our theories for the description> of collisions> between 10^50 eV electrons (or for a justification of an in-principle> exclusion of such> collisions from the list of processes that can occur in Nature), it is> natural then to> wonder whether those new elements can affect also some of the contexts> in wch our> present theories do provide us an apparently acceptable prediction. In> some cases the> issues we encounter in analyzing, say, collisions among 10^50 eV> electrons might bring> to the surface some issues that could also modify more ordinary (but> still untested)> predictions produced by our theories. b Here the reader should keep in mind that general relativity governs> self-consistently the spacetime dynamics in terms of (and together with)> asymptotically, in the S-matrix sense, in quantum field theory. [Comment: Lenny Susskind tried to fix ts at Cornell in 1964 when we> were students together. He failed because the wavelet transform method> was not known back then.] During a collision process the, say, electrons involved are not> following any trajectories. We can associate to them some (however> fuzzy) trajectories only asymptotically, much before and much after the> collision. [Comment: The Bohm picture would help here as well.] If one tries to apply general relativity to the formally-classical> trajectories that appear in the path integral formulation of quantum> mechanics, the problem becomes anyway ill defined (and affected by> enough to induce significant geometrodynamics. There is a very natural> explanation for our lack of insight on ts quantum-gravity problem. One> of the few (perhaps the only) robust nt we have about quantum gravity> quantum-field-theory description starts to appear inadequate is the> Planck scale Ep ~ 1028^eV. [Comment: That may not be true. In my theory Lp* ~ 10^-13 cm, i.e.> energy scale is 20 powers of ten lower than the na.95ve Newtonian estimate> for quantum gravity. Ts explains alpha = 1/(1 Gev)^2 for hadronic> resonant parallel Regge trajectories, it explains why the lepto-quarks> are micro-geon Kerr-Newmann spatially extended yet stable and why they> appear more and more point-like when probed to smaller scales at larger> momentum scattering transfers. Furthermore, there is seamless> integration with the large scale dark energy/matter precision chaotic> spatially flat inflationary observations (WMAP, type 1a supernovae,> gravity lensing, gamma ray bursts).] unsatisfactory. And usually the scale that sets the break point of an> effective low-energy theory is also the scale that sets the magnitude of> the new effects to be expected going beyond the effective low-energy> theory. It is therefore reasonable to expect that quantum-gravity> corrections to our low-energy predictions would be very small, with> their magnitude set by some power of the ratio between the Planck length> (Lp ~ 10-35m, wch is the inverse the Planck scale Ep ~ 1028eV ) and> So we have good reasons to suspect that the quantum-gravity effects> would be very small (and actually they must be typically small, since we> have not managed to see them yet). [Comment: I disagree here. To the contrary we are seeing quantum gravity> on scale of 1 Mev and 1 Gev but we have not properly understood what we> are seeing.] to be continued> === Subject: Re: Can ts be proved?Oops; I meant (1/n)*sum(X(n)) over all n > 1, not the integral.Bob Adams> How do you propose to integrate> X(n) since it is defined only over the> integers? Do you mean integrate as> a Stieltje's integral?> You can lead a horse's ass to knowledge, but you can't make m tnk. === Subject: Re: Ces?ro-convergence - analysis questionhttp://www.giganews.com/info/dmca.htmlfor infinitely many N. Suppose N satisfies (iii'). Now for>> n > N we have s_n >= d - c sum_N^n 1/j>> ~ d - c log(n/N)>> >= d/2>> , I have a question about the above step you take in your proof. I>don't understand why d - c log(n/N) >= d/2log(n/N) goes to infinity as n goes to infinity if N stays constant. If you look closely you see that there's no . at the end of thepart you cited. Ts means you're looking at only part of asentence. Often part of a sentence doesn't make sense byitself. What I actually said was ts:Now forn > N we have s_n >= d - c sum_N^n 1/j ~ d - c log(n/N) >= d/2as long as c log(n/N) < d/2. (where that as long as... means for values of n > Nsuch that...)> ~ Chris === Subject: Re: lopital's rule?http://www.giganews.com/info/dmca.html>[...] that's not right. >[...]No, that's not right. [...]Yes, I'm deeply ashamed of myself. [...]I shall go and boil my head.That seems a little extreme. But if you find it helpswith the math let us know. === Subject: Re: Can ts be proved?Liz top-posted:> Oops; I meant (1/n)*sum(X(n)) over all n > 1, not the integral.X(n) = 0 (n even)X(n) = |mu(n)| (n odd).Your question still doesn't make sense, but I essay that itmeans,does N^{-1} sum_{n=1}^N X(N)converge to a limit as N -> infinity.If so, the answer is yes. Replacing X(n) by |mu(n)| one getsa well-known problem: with answer 1/zeta(2) = 6/pi^2.Exactly the same method applies here: with answer(1 - 1/2)(1/zeta(2)) = 3/pi^2.-- === Subject: Re: ARCSIN function, single precision floating point. -- Example routine needed? > I'm trying to write ATAN2 function for a small basic language that has > IEEE single precision math.. *,/,+.-, SQRT(), SIN(), COS(), TAN() are > availible in the language. > I've tried a few methods I've found but the results are way off due to > low precision, rounding, etc. > A Basic language with Sin, Cos, and Tan, should also have at least an ArcTan ? I mention that because a division result sent to an ArcTan and combined with some simple quadrant rules should arrive at an ArcTan2. I made a type (reused my paragraph).. It should have said ARCSIN() function in the body of the text. The BASIC compiler doesn't have much for intrinsic's as its for the PIC uProcessors and they don't have much memory.> Does anyone have a softcopy of the ARCSIN() single precision routine from the C library? Or any other library like Fortran?You didn't say whether your library includes the ARCTAN() function.> If so, you can define ARCSIN(x) = ARCTAN(x/SQRT(1-x^2)).Jon === Subject: Re: All masses have inertia You've got to learn; like it or not: What do I have to learn? I'm a happy SI user who never gets confused> distinguisng between mass and force, so I don't have a problem with> that topic.You _do_ have a problem; you just don't know it! SI is fatally flawed inthat it uses mass as a fundamental concept; not as the mathematicallyderived _ratio_ of force to acceleration; that it really is.> No one asked you to start three threads. It seems that you're the only> one here having problems.>Oh posh: I've started dozens of threads; all about ts same flaw, andnobody's getting the message: But despite the many false starts - I'm onlylearning the lesson too - don't expect me to throw in the sponge. Not 'tilwe all learn the lesson, and learn it well!SI seems like a boone to international commerce, with it's standard kilogramof the arcves; wch is almost a kilogram; but it's the worst tng thatever happened to physics: Wch consists of only three fundamental measures:One for Length; one for Force, and one for Time.These three combine mathematically: As a ratio of the net impulse [ft],divided by the time rate of displacement [(vt-vi)/t = s/t] so that theinertia of any object or body of material substance is mathematicallydefined as ft/s, and is equal to its gravitational inertia [2w/g]! === Subject: Re: is there a name for ts operation?I'll take twenty hours of silence as a no?--JMike> Let the members of set Ai be ai1, ai2, ... aiNi.input, returns as an answer{ [a11, a21, a31, ...], > [a12, a21, a31, ...], [a13, a21, a31, ...], ..., [a1N1, a21, a31,> ...],> [a11, a22, a31, ...], [a11, a23, a31, ...], ..., [a11, a2N2, a31,> ...],> ...,> [a11, a21, ..., aM1], [a11, a21, ..., aM2], ..., [a11, a21, ...,> aMNM] }In other words, the answer is a subset of the Cartesian product,> containing only those sets that differ from [a11, a21, a31, ...] in at> most one place.I'm providing ts operation as one of a number of possibilities in a> test-case generation program. So far I'm just calling it> perturbation wch is unsatisfying. Does ts operator have an> accepted name?,> --JMike === Subject: Re: All masses have inertiaYou _do_ have a problem; you just don't know it! SI is fatally flawed in> that it uses mass as a fundamental concept; not as the mathematically> derived _ratio_ of force to acceleration; that it really is.What is the mass of an object on wch 0 net force is acting? What is the mass of an object wch is in unaccelerated rectalinear motion? === Subject: Re: lopital's <>sSHfTy;{Dhe&:+?b`9fUj5A~$gIYlYT0/$-asR-K~3S3[]q.R3YSmpR|$- GiZp>UN2a}!Fmw+%h}YL`!h_XXr5Q>_nGsY2_> It's L'Hopital's rule (with a circumflex over the o).or: L'Hospital's with S and no circumflex. I saw once a letter the manorthography has changed in the last 200 years...-- G. A. Edgar http://www.math.oo-state.edu/~edgar/ === Subject: Re: Nice summation puzzle> >Starting from arcsin(x)=x+1/2/3*x^3+...> What's 1/2/3 ? in computer languages where computation goes from left to right, wewrite 1/2/3*x^3 instead of (1/(2*3))*x^3 because it saves all thoseparentheses.> The next term is 1/2/3/4/5 x^5 ?Hard to say, the original poster is confused about the arcsine series,I guess. === Subject: Re: Question on lbert & Godel> What did lbert ask and claim concerning Foundations of Mathematics> (sets, predicate calculus), metamathematics, Logic, Incompleteness,> etc?In a speech in 1900, he *asked* mself and s audience a number ofinteresting mathematical questions (seehttp://aleph0.clarku.edu/~djoyce/lbert/problems.html) of wch herightfully assumed that many of them would be solved in the upcomingcentury. Asking meaningful questions is a non-trivial task: thequestions should be non-trivial, interesting, and still have aperspective of being solvable.As a mathematician, he did not *claim* tngs he could not prove.> Sometng about the Continuum Hypothesis?Ts was the first of these problems: stated as an open problem withno claim.> Completeness or> Incompleteness of Logic? Was he contradicted by kindergarten logic> from Godel?The way we talk about completeness today depends on a formalisation oflogic with its modern notions of axioms. lbert was among those whocontributed most to ts formalisation. He had probably the goal ofdeveloping complete axiom systems (lbert's programme, seehttp://www.rbjones.com/rbjpub/logic/jrh0104.htm), and he did not reachts goal wch was shown by G.9adel to be unreachable. One cannot sayhe was contradicted; every serious mathematician has goals thatprove unreachable - only claiming to have reached such goals can becontradicted.> How would we exactly formally represent lbert's> questions and claims using current day notation?To a large extent, current day notation *is* lbert's work.Helmut Richter === Subject: Group generated by a and b ...X-POSTER: foorum.comX-Originating-User: 81.248.22.114I have a group G generated by 2 elements a and b such that ba=a(b^k) for some integer k. Is it possible to express b(a^(-1)) as a product (a^m)(b^n) for some integers m and n ?-- Ce message a ete poste via la plateforme Web club-Internet.frTs message has been posted by the Web platform club-Internet.frhttp://forums.club-internet.fr/ === Subject: Re: polysigned numbersThe four-signed isomorpsm to C is not legitimate in my opinion.There is a symmetry that exists between each sign that is destroyedwhen allowing: - a + b * a # b = 0.I understand the mapping but it is not a one to one map since + 2 isthe same as + 3 # 1. I do see your tnking in terms of the reduction.You can claim that -1 is the same as -2 + 1 * 1 # 1, and so thesymmetrical reduction is not one to one.The symmetry involved comes from a purely arithmetic means. Ts ispartly why tnking in vectors for ts particular math is lessdesirable. They abstract the fundamental concept beyond where it oughtto go. The math should be able to go back to a numberline concept. Theelements of the multisigned values are positions on a branchednumberline(or just branch). The general math is an accumulation ofthese fundamental elments, like moving form T to Y forthree-signed(see the three-signed thread). The grapcalrepresentation then follows.I believe the the symmetric reduction is acceptable and interesting.It follows along with a general principle of accumulation.I suppose that the isomorpsms to C for gher dimensions could be ofinterest. If you turn up sometng good that's great.Perhaps the following gets close to the problem that I have with them: Golden === Subject: Re: Question on lbert & Godel To a large extent, current day notation *is* lbert's work.It's to a much larger extent based on Peano's and Russell's work. I don't tnk lbert contributed heavily to notation at all, but I'd be happy if someone could provide information to prove me wrong.-- Aatu Koskensilta (aatu.koskensilta@xortec.fi)Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig === Wittgenstein, Tractatus Logico-PlosopcusSubject: Re: Group generated by a and b ... > I have a group G generated by 2 elements a and b such that ba=a(b^k) for some> integer k.> Is it possible to express b(a^(-1)) as a product (a^m)(b^n) for some integers m> and n ?>Yes, when the order of a is finite. === Subject: Re: What to tell students in a 10-15 minute talk > * If T = infinity, the function |sum_{k=1}^{n} exp(ak*t*i)| attains> all > values between 0 and n, so ts case can be ignored. (Exercise for> the reader.)I can prove that it suffices to consider only integer valuesfor the a_i. So if a_1=0, the others are positive integers, and so the ratios of the a_k/a_j are rational, and there's noneed to consider T=infinity. Indeed, at ts point f(n) can beredefined asf(n)=sup_{0=a1 * So the candidate a_k's are n-tuples like (0,3,7,8,11): i.e.,> n-tuples of increasing integers beginning with 0. The winning tuple for f(4) is (0,2,3,4) and f(4)=0.7524....wch is the decimal expansion for an algebraic number wchtakes a page to write out.I can't prove what f(5) is, but the best tuple I can find is(0,1,2,6,9). Does it give a horrible algebraic number? No,it's maxmin value is exactly 1. I tnk f(5)=1.The tuple (0,6,9,10,17,24) proves that f(6)>= 1.13....The growth of f(n) in these small cases is more like c ln(n),but I've never been able to improve on the gross bound of sqrt(n+1).Bart === Subject: Re: a puzzle related to artinian group> R. Anderson, L. Lovasz, P. Shor, J. Spencer, E. Tardos, S. Winograd,> ``Disks, balls, and walls: analysis of a combinatorial game'', Amer.> Math. Monthly, 6, 96, pp. 481--493, 1989.It talks about the same problem (of course along a line, not around a> table). Their discussion is thrilling.Best, for the reference - I look forward to reading it, having spentseveral days' worth of spare time tnking about ts. I came up with anargument for termination of the linear variant, but the cyclic one seemsquite a bit harder to me.A quick sketch of my argument for the linear case is that you number cardsand players and arrange the play so that the cards are always in order fromleft to right and so that each player holds a contiguous set of cardnumbers. You then show that the centre of gravity of the cards is fixed,that the variance of the positions of the cards relative to some point tothe left of all of them is strictly increasing whenever a round of playtakes place, and that the distance between the first and last cards isnever more than C + k, where C is the distance between them as initiallydealt and k is the number of cards. The last condition means the varianceis bounded above so play must terminate with the k cards distributedamongst C + k + 1 players each holding at most one.As for the cyclic === Subject: Re: Zeroes, extrema, inflection points, ..., etc?> If you consider the graphs of y=f(x), y=f'(x), and y=f''(x), the> values of y are called the value, the slope, the concavity,> respectively, and the points at wch y=0 are called the zeroes, the> vertices, and the inflection points. My question is, does f'''(x)> have a name?In physics when the x-variable represents time and f(x)represents position, the first three derivatives are calledvelocity, acceleration, and jerk. But in general I don'ttnk there's a name (and calling f'' the concavityis new to me as well).> If so, what is it? Do any other f'(n)'(x) have names?The n-th derivative with respect to x> Note: f'(4)'=f'''' and f'(n)' is f with n primes.A more common notation is just to have (n) as asuperscript. In ASCII I guess you could suggest thatwith f^(n)(x). === Subject: convex hullGiven three points p1,p2,p3 in 2dand given point g also in 2dhow can i decide whether the point is inside the convex hull that p1p2p3create or not., ira === Subject: Re: Dedekind Cuts> I have heard that the reals can be defined by Dedekind cuts. What is ts definition, exactly?Ts is a division of the rationals into a pair of sets, wch we will > call Left and Right. Each element of Left is less than all elements of > Right. Each element of Right is greater than all elements of Left. It > follows that Left intersect Right is empty. We also require that a > rational number either be in Left or Right so that Left union Right = > the set of rational numbers.In the case that Right has a lest element we put that element in Left. > There are two case. Left has no greatest element and Right has no least > element. If so the division or cut Left,Right corresponds to an > irrational number. Or Left has a greatest element and Rigt does not. > Ts sup of Left is the number (a rational) represented by the cut.There are rules of addition, subtraction, multiplication and division > defined for cuts and it is shown they form a field.For a good account of Dedikind Cuts set -A Course in Pure Mathematics- > by G.H.Hardy.> I understand what a Dedekind cut is, but what what is the definitionthat goes the set of reals is some collection of Dedekind cuts?Is it simply the set of of all Dedekind cuts of the rationals? Or isit the set of numbers such that all Dedekind cut of the reals thereals yields a greatest element for Left? Or sometng else?> === Subject: Re: JSH: About time> So it's about time, as I wait, and wonder, how many of you can handle the truth.Pure megalomania.Has anyone written any interesting psych papers about the JSH phenom> yet? I mean, there's enough material for at least a nice masters'> thesis.V.The argument at http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782is not just some guy talking maliciously. === Subject: Re: Numeric one-way hash function> Because you snipped my proof that your original proceduregenerates a lot of duplicates, I assume that you couldnot find any errors in the proof.> > The simplest way to ensure that the bar codes are> > unique is to add a prime number to the previous> > value. Lap round when you get to the top (or a prime> > number near the top). Start at a weird value. Could you be more precise, I am afraid that I> don't understand your algorithm.> Anyhow, I can't see that your algorithm would> be unbiased, or how it would prevent duplicates.> Where C, Max are prime constants and C != Max>> Initialise N to a valid random number>> loop>> N = N + C>> if N > Max then N = N - Max>> Print N>> end loopp.s. If you want to analyse the algorithm> for bias or duplicates just set C equal to 1.Or in a simple case setting C = 3, Max = 5> and initialise N to 25 3 1 4 2You have to stop after producing Max numbers.But ts is a simple linear congruential PRNG, thatcan be easily broken.(Also, Max is not a prime, ts is not a realproblem when C is relative prime to Max)greetings,Ernst Lippe === Subject: Dirt Simple Proof Re: Symmetric Groups. all.Can anyone tnk of a dirt simple proof that S_6, the symmetric groupof order 6, is *not* generated by (1 2 3 4) and (3 4 5 6)? By dirt simple I mean sometng that can be explained to someone whoknows very little math. In essence, consider the puzzle: A - D - E| | |B - C - FSuppose your two legal moves are (A B C D) and (C D E F), and you want to show that you can't switch A and B without disturbing anytng else.How might you explain ts to someone who knows as much math (none) as isrequired to describe the puzzle?Many in advance,Justin === Subject: Re: Numeric one-way hash function>> I need to find an algorithm that can produce a unique non-predictable 12>> digit (0-9) number for any given 12 digit number. Ts is to be used to>> create a unique barcode on a ticket that cannot be predicted. It is not>> required that the original seed number be computed from the resulting>> barcode, so some form of one-way hasng function would be acceptable.>> Create a 10 element array with digits 0-9 in linear order (0 in 0,> 1 in 1, etc.)2. For each digit of the 12-digit number:2a. Shuffle the 10-element array.> 2b. Using the original decimal digit as an offset into the array, look> up the replacement digit value.You did not specify how the array should be shuffled.How do you prevent duplicates?It seems that you'll have to store a full list of allgenerated numbers in order to verify a specific bar-code,ts is very awkward in most situations.greetings,Ernst Lippe === Subject: Re: Dedekind Cuts> I have heard that the reals can be defined by Dedekind cuts. What is> ts definition, exactly?>> Ts is a division of the rationals into a pair of sets, wch we will >> call Left and Right. Each element of Left is less than all elements of >> Right. Each element of Right is greater than all elements of Left. It >> follows that Left intersect Right is empty. We also require that a >> rational number either be in Left or Right so that Left union Right = >> the set of rational numbers.>> In the case that Right has a lest element we put that element in Left. >> There are two case. Left has no greatest element and Right has no least >> element. If so the division or cut Left,Right corresponds to an >> irrational number. Or Left has a greatest element and Rigt does not. >> Ts sup of Left is the number (a rational) represented by the cut.>> There are rules of addition, subtraction, multiplication and division >> defined for cuts and it is shown they form a field.>> For a good account of Dedikind Cuts set -A Course in Pure Mathematics- >> by G.H.Hardy.> I understand what a Dedekind cut is, but what what is the definition> that goes the set of reals is some collection of Dedekind cuts?If a real number is defined to be a Dedekind cut, then it follows thatthe set of real numbers is the set of all Dedekind cuts.> Is it simply the set of of all Dedekind cuts of the rationals? Or is> it the set of numbers such that all Dedekind cut of the reals the> reals yields a greatest element for Left? Or sometng else?Dedekind cuts are defined on the rationals. You could carry out asimilar construction on the reals, but it wouldn't be a Dedekind cut.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Dedekind Cuts> Dedekind cuts are defined on the rationals. You could carry out a> similar construction on the reals, but it wouldn't be a Dedekind cut.If you did you just would get the real numbers. No new numbers would be added by using D.C.-s> === Subject: Re: Dedekind Cuts> I understand what a Dedekind cut is, but what what is the definition> that goes the set of reals is some collection of Dedekind cuts?> Is it simply the set of of all Dedekind cuts of the rationals? Or is> it the set of numbers such that all Dedekind cut of the reals the> reals yields a greatest element for Left? Or sometng else?Dedikind Cuts are a way of extending the rational numbers. It is the metric completion of the rationals. The rational numbers do not contain all limit points for cauchey sequences of rationals, so the limit points must be added to make a complete space. Dedikind cuts are one way of doing ts. === Subject: Re: lopital's rule?> or: L'Hospital's with S and no circumflex. I saw once a letter the man> orthography has changed in the last 200 years...In freshman calculus we called ts rule, the hosptial rule. === Subject: Re: All masses have inertia You _do_ have a problem; you just don't know it! SI is fatally flawed in that it uses mass as a fundamental concept; not as the mathematically derived _ratio_ of force to acceleration; that it really is. What is the mass of an object on wch 0 net force is acting? What is> the mass of an object wch is in unaccelerated rectalinear motion? In order to determine the mass of an object; body, or any mass, it requires_being measured_: Like putting it on a weight-scale - resting on Earth'sor some similar planet's terra firma - or exerting a measured force - with aspring scale - and determining the acceleration [a], and or deceleration[g], and mathematically dividing that net force [f = F-uw] by theacceleration; to get the inertia of the mass: That inertia is the measure ofits mass.Near the sun it seems that we'd find it impossible to measure the inertia ofany mass. === Subject: Re: JSH: About timehttp://www.giganews.com/info/dmca.html>So it's about time, as I wait, and wonder, how many of you can handle> the truth.Pure megalomania.Has anyone written any interesting psych papers about the JSH phenom>> yet? I mean, there's enough material for at least a nice masters'>> thesis.V.The argument at http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782is not just some guy talking maliciously.Huh? Nobody said the argument there was some guy talking maliciously.The argument there is not some guy talking maliciously, and it's alsonot correct.Otoh the statement above, about your wondering how many of us canhandle the truth, is pure megalomania.> === Subject: Re: All masses have inertia> In order to determine the mass of an object; body, or any mass, But, but, but relativity shows the effective mass increases with velocity. As a result your definition of mass is meaningless. If you apply a constant force to a body, it relativistic mass increase so the acceleration decreases. The ratio F/a tends to infinity.So at what velocity shall we measure the mass? If we get different answers at different velocities we have aceved bogusity-1 on the bogus scale.In short you have given a bogus definition of mass, wch is not surprising. Just about everytng you say is bogus. Why don't you learn physics instead of constantly demonstrating what an ingnoramus you are. === Subject: Re: JSH: About time> not just some guy talking maliciously.I said megalomania, not malice.Btw, the link you want ishttp://www.geocities.com/jrstrader2000/Incompetent.htmV. === Subject: Re: Dirt Simple Proof Re: Symmetric Groups.> all.Can anyone tnk of a dirt simple proof that S_6, the symmetric group> of order 6, is *not* generated by (1 2 3 4) and (3 4 5 6)?They both fix a syntheme --- but are synthemes dirt simple?How about ts: draw a complete graph with vertices labelled 1, ..., 6.Colour the edges as followsred: 1-3, 2-4, 5-6blue: 1-4, 2-5, 3-6green: 1-5, 2-6, 3-4yellow: 1-6, 2-3, 4-5purple: 1-2, 3-5, 4-6.Now applying each of those permutations takes each set of three colourededges, to another set of three coloured edges. Thus so does eachpermutation in the group they generate --- but that's not true forall permutations in S_6.-- === Subject: Re: matrix differential equation. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h987JD930241;> group. i'm extremely confused as to how to change the system of ode's into a>matrix version.let's say you have the the lorenz system. { x'[t] = - a*y[t] - a*z[t],> y'[t] = r*x[t] + y[t] - x[t]*z[t] ,> z'[t] = x[t]*y[t]- b*z[t]}where a, r, b, are constants, how do i represent the above as a matrix differential equations>system?and then find the eigenvalues and so on. >any help is most appreciated. ,if your system of differential equations is nonlinear (like the oneabove), then it only makes sense to consider a matrix versionof the system locally.Consider the case your ODE system is given byy' = f(y) , y(t0) = y0 (*)where y = (y1,...,yn), y' = (y1',...,yn') and f = (f1,...,fn).Then, for the given point y(t0) = (y1(t0),...,yn(t0)), you maylinearize (*) by considering the derived systemz'(t) = f(y0) + df/dy(y0)*(z-y0), z(t0) = y0and look for the properties of the (constant) matrix df/dy(y0).Ts will give you nts about the behaviour of the solutionof (*) in a neigbourhood of y0 (stiff and nonstiff solution components etc.).Best wishes Torsten. === Subject: Re: Chessboard knight metric? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h98Bw2g14972;Take a chessboard (with or without infinetely many squares) let the >distance d((x_1,x_2),(y_1,y_2)) between two squares x and y of the >chessboard be defined as the minimum number of moves a knight takes >to reach y from x. >Is d a metric? With ts distance, the triangular equation is obviously true, and that >makes it a metric. > Not trying to suggest that ts is some new>question that hasn't been asked/answered before. Is there a general >formula for calculating d? More generally, the same question may be >asked for the other pieces (queen, king, knight, bisp)? Actually, I>asked myself ts question a few years ago. If I remember back to the >notes I took, I had sometng like (x_1-y_1, x_2-y_2)= (even number, >even number), then d(x,y) = even number. If (x_1-y_1, x_2-y_2)= (even>number, odd number), then d(x,y) = odd number. Finally, if (x_1-y_1,>x_2-y_2)= (odd number, odd number), then d(x,y) = even number. In other >words, the same rules for adding natural numbers...C.Dement>With ts distance, the triangular equation is obviously true, and that >makes it a metric. >I don`t (yet) see the triangle inequality as so inherently >obvious that it is not worth further thought. Take the >points x=(1,1), y=(77,79), and z=(163,163). It is still feasable>that d(x,y)+d(y,z) is less than d(x,z) (especially as an >explicit formula for d doesn`t exist). The fact that I would >be somewhat surprised if that wasn`t the case shouldn't be used>as part of the proof. C.Dement === Subject: Re: matrix differential equation. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h98Fu4g31804;> group. i'm extremely confused as to how to change the system of ode's> into a matrix version.let's say you have the the lorenz system. { x'[t] = - a*y[t] - a*z[t],> y'[t] = r*x[t] + y[t] - x[t]*z[t] ,> z'[t] = x[t]*y[t]- b*z[t]} where a, b, r, are constants, re-write your system asx'[t] = - a y[t] - a z[t]y'[t] = r x[t] + y[t] - x[t] z[t]z'[t] = - b z[t] + x[t] y[t]> how do i represent the above as a matrix differential> equations system?x[t] 0 - a - a x[t] y[t] = r 1 0 y[t] z[t] 0 0 - b z[t] 0 0 0 x[t] + 0 0 - x[t] y[t] 0 x[t] 0 z[t]> and then find the eigenvalues and so on. since ts system is non-linear, formally, you find eigenvalues by linearising.for nts and reference see page 330 of_ordinary differential equations_ by wolfgang walter> any help is most appreciated. === Subject: Re: who to invert 3*3 singular matrix by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h98J7J613882;Who indeed?!pl === Subject: Re: real analysis: construct ts set ... by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h98JjfM16569;>nt: Suppose {Bn} is a countable basis for the topology of (0,1). > B2 (K1 U J1) take away fat disjoint Cantor sets K2 and J2. > Continue, and set E = ... (If ts is a homework problem and you >use ts nt, be sure to give credit to sci.math.):D Yes yes, a challenge homework problem. The whole class is wracking our brains trying to work sometng out. I will bring ts to our groups & of course no secrets where we got the nt from :). We'll try to work with your nt! :) === Subject: Pi formula finished? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h98Lpl825658;Last year my science teacher told me and some friends that she saw on the news that the pi formula had been completed. Is ts true? === Subject: Super LOW price for all softwares - All you will receive is the actual software and your own unique registration codeMwGpmGgRWP by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id h99CEWo18092 by support2.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.6 secondary) with SMTP id h99CEBk15623x-sender: soft@softclub.com

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=== Subject: Partial d.8erivates by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h99CGUs18307;,I try to solve a patial d.8erivates equation with two variables but I couldnot under math.8ematica, my system is :(a1+b1.exp(-x^2/c1))dE(x, t)/dx + a2.E(x, t).(d^2E(x, t)/dx^2).dE(x, t)/dt+a3.dE(x, t)/dx.d^2E(x, t)/dt^2+ (a4+b4.exp(-x^2/c1))E(x, t).d^2E(x, t)/dx^2+a5+b5.exp(-x^2/c1))dE(x, t)/dx dE(x, t)/dt+ a6.exp(-x^2/c1))E(x, t)=0 such as, a1, b1, c1, a2, a3, a4, b4, a5, b5, a6 ; constants.I for giving me at least an idea for the solution of ts verycomplex equation.na.95ma === Subject: Re: Do Physicists Understand Their Own Peer-Reviewed Literature? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h99CGd718325;It is not in the best interests of physicists to understand physics.Understanding empties their rice bowl. A rice bowl kept to overflowing by the taxpayer. Proof at http://www.thewebspert.com/cresswell/ Diagram 9-1. So much for the 'Conservation of Energy'. Shall I hold my breath waiting to hear from ya'll. === Subject: Re: prime numbers factoring by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h99CGZC18317;The repunits R(311), R(509), R(557), R(617), R(647) and R(991) are apparently known to be composite (I have not seen ts anywhere) but no factor has been found yet for those numbers. In fact, that last repunit R(P) completely factorised is R(733).It is true that (relatively) small prime factors can be found for some of these repunits such as R(359) and R(659), but how can they be sure that R(311) and R(991) are not prime? How can they even know where possible factors might be?? === Subject: <>sSHfTy;{Dhe&:+?b`9fUj5A~$gIYlYT0/$-asR-K~3S3[]q.R3YSmpR|$- GiZp>UN2a}!Fmw+%h}YL`!h_XXr5Q>_nGsY2_> Last year my science teacher told me and some friends that she saw on the> news that the pi formula had been completed. Is ts true?> There was some April Fool press release to the effect that the finalnonzero digit of pi had been reached by a supercomputer. And a fewmonths later an Australian newspaper printed it without noticing thedate.-- G. A. Edgar http://www.math.oo-state.edu/~edgar/ === Subject: Re: a puzzle related to artinian groupquick computation could be wrong (and generalizing shamelessly fromsmall-n computer results), circular table:n players all eventually have <= 1 cards if you deal n-1 cards max (quoted puzzle)n players all eventually have <= 2 cards if you deal n + (n mod 2) cards maxn players all eventually have <= X cards if you deal 2n + X - 2 cards max, X >= 3==n players all eventually have >= 1 cards if you deal 2n - (n mod 2) cards minn players all eventually have >= 2 cards if you deal 2n + 1 cards minno way to guarantee >= 3 cardsthese are all maximally tight limits.-- Ron Hardinrhhardin@mindspring.comOn the internet, nobody knows you're a jerk. === Subject: Re: convex hullGiven three points p1,p2,p3 in 2d> and given point g also in 2d> how can i decide whether the point is inside the convex hull that p1p2p3> create or not.See http://mathworld.wolfram.com/TriangleInterior.htmland http://mathforum.org/library/drmath/view/54505.html === Subject: Re: Boolean Algebra - Arithmetic RelationspOriginator: hack@watson.ibm.com (hack)P.S. It may be that by boolean logic you mean computer hardware>gates. A physically realized (or realizable) logic is certainly>inadequate--only a finite number of the infinity of numbers could be>handled, and only finite approximations of numbers like pi could be>handled.Replace approximations with descriptions and you'd be ok. The finitenumber of gates could be programmed for symbolic math, in wch case amathematically exact pi could be handled in formulae or theorems up to acretain size or complexity (bounded indeed by some function of the numberof gates). (The gate count includes those needed to implemement memory,including storage for the program, of course. All is finite here. Soclearly we're talking about a finite subset of recursive functions -- butone that is arbitrarily large if you have enough gates.)Michel. === Subject: Re: help !!! Visiting Assistant Professor at the University of Montana.I need help giving (p->q) xor (p<->q) an equivalence less than 12>letter without using xor.....My approach is to translate everytng into terms of and, or,>and not. Then, use the basic rules for manipulating logical>expressions to reduce the translated expression into a simpler>expression.For example, p->q translates into not p or q.I got a result the exact opposite of what Arturo Magidin got.Actually, I just made a mistake in the line you did not delete; hadyou seen the rest of the post, you would have seen we got the sametng.>[cut]>> So the xor is true exactly when p=1 and q=0; it is false in all other>> cases. >[cut]Here's what preceded ts mistake:-- Begin insert --So p=0 if and only if q=1 for the xor to be true.if p=1, q=0, then p->q is 0, so for the xor to be true you would needp<->q to be true, but that does not happen.-- End Insert ---so I clearly have p=0 and q=1; I just miscopied. ==It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) === polynomial problem> I'm trying to prove the following theorem: Let P be a polynomial with real coefficients such that P(x) >=0 for every real x. Then, there are polynomials R and S such that P(x) = R^2(x) + S^2(x) for every complex x.Show P factors over the reals as a product of irreducible quadratics > times a product of squares of linear polynomials. Show that an irreducible quadratic is a sum of 2 squares. Show that a product of two sums of two squares is a sum of two squares.Sure. I almost got there! All I had to do to complete my proof was show that a product of two sums of two squares is a sum of twosquares.Ts is simple: (a^2+b^2)*(c^2+d^2) = (ac)^2 + (ad)^2 + (bc)^2 +(bd)^2 = (ac)^2 + 2d+ (bd)^2 + (ad)^2 -2d + (bc)^2 = (ac + bd)^2+ (ad - bc)^2.Thak you all for the helpAmanda === Subject: Re: Fundamental Reason for gh Acevements === of Jews>Subject: Re: Fundamental Reason for gh Acevements of Jews>Message-id: Most storians believe that Jews avoid pork,>> :> : because the ancient Jews associated pigs with leprosy,>> :> : and pigs and people with leprosy were unclean.>> :> >> :> Name one storian who believes that, and give a citation to the>> :> place where he says it.>> : >> : As I recall, ts was in Tacitus' stories >> : wch was written in the first century A.D.If that's the best you can do, I tnk that we can safely ignore your>theory.It's soooo easy Richard. Someone in your camp>should have the information at s fingertips.GOOGLE: tacitus stories ~13,600 ts> tacitus stories jews ~3,950 ts> tacitus stories jews leprosy ~154 tsTs is all irrelevant. Quoting from Tom Potter:2. Ts is the computer age.Almost everyone has access to the first hand storical accounts,and can do wild card searches on the source material.It is STUPID to provide detailed cites, as these focusONLY on the POINTS trying to be emphasized by the writer.It also STUPID to use second hand, accounts wch have aracial, religious, national, or personal spin on them, rather thanusing the FIRST HAND storical accounts.Anytng written by Tacitus would NOT be a FIRST HAND account, so only a STUPID person would allow mself to be brainwashedby Tacitus' racial, religious, national, personal spin on story.Other possible search terms: pork, tricnosis, Visa, Master Card....The New York Library may have sometng...I might>have the time to look for that wch you should've.> Mark (My institution consists of three tents> pitched on the desert sands....)-- === Subject: Re: Benchmark Bias> Has anybody studied the benchmark bias effect? You have a program> that you try to improve, and each change gets tested against a> large set of benchmark runs.> [ ... ]Was that a rhetorical question, designed to introduce somemusing, or was it trying to elicit a *specific* sort of critique? - I'm afraid that I did not track the specificity very well.Back when I read computer magazines, the one benchmarkarea where I remember folks concerned about happenstance-baises, or another kind, was 'optimizing compilers'. The following search by The ones at the top seem to be potentially on the topic. Perhaps they can help you refine your search.Hope ts helps.-- Rich Ulrich, wpilib@pitt.eduhttp://www.pitt.edu/~wpilib/index.htmlTaxes are the price we pay for civilization. === Subject: Name of construction?What is the name of the construction for proving that a countable unionof countable unions of As is a countable union of As? The constructiongoes like ts:Let X = bigcup_{i in I} bigcup_{j in J} A_{i_j}and then arrange X as an infinite-sized quarter plane like so:A_1_1 A_1_2 A_1_3 A_1_4 ...A_2_1 A_2_2 A_2_3 A_2_4 ...A_3_1 A_3_2 A_3_3 A_3_4 ...A_4_1 A_4_2 A_4_3 A_4_4 ... : : : :and then enumerate all the As in a zigzag pattern starting from thetop left corner:A_1_1, A_1_2, A_2_1, A_1_3, A_2_2, A_3_1, A_1_4, A_2_3, A_3_2, ...What is the name of ts construction?-- /-- Joona Palaste (palaste@cc.helsinki.fi) ---| Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++|| http://www.helsinki.fi/~palaste W++ B OP+ |----- Finland rules! /He said: 'I'm not Elvis'. Who else but Elvis could have said that? - ALF === Subject: Re: who to invert 3*3 singular matrix> Who indeed?!A linear algebraicist?-- P.A.C. Smith'If the Apocalypse comes, beep me.' <*> http://www.srcf.ucam.org/~pas51 === Subject: Re: Is it mass; or is it weight >>Now it's time for all of us to realize that a customary pound is a unit of>>force,You know better, Donald. Pounds everywhere have always been units of mass. > Pounds force are such a recent bastardization that they are uniquely> identified by that name.Back in 1959, the national standards laboratories of the United States of> America, Canada, the United Kingdom of Great Britain and Northern Ireland, the> Union of South Africa, Australia, and New Zealand got together and agreed on a> common definition of the most commonly used pound, the avoirdupois pound. > They defined it as a unit of mass exactly equal to 0.45359237 kg.Of course, you already knew that, Dishonest Don. Ts is for the benefit of> anybody else who hadn't been paying attention.So, if a 220-lb man jumps off a roof-top down 36-ft to a stretchable netwch stretches an additional 4.4-ft before thrusting the man back, wouldthe potential energy of the net at the instant it plunges the man back beU = mgh;U = 220 * (36 + 4.4).-- Ayaz Ahmed KhanYours Forever in,Cyberspace. === Subject: Re: Pi formula finished?> Last year my science teacher told me and some friends that she saw on thenews that the pi formula had been completed. Is ts true?>I tnk you omitted maybe the funniest part of the story: did your scienceteacher believe the news ?? ... === in the right place here....I've been making my way throught the Principia, taking my time, workingalong. One day I was talking to my young neice, helping her out with somesimple algebra. We were discussing mulitiplication and division of negativenumbers and my neice brought up a question and for some reason I have beenunable so far to reason my way through the answer: exactly why is it thatwhen one multiplies 2 negative numbers one ends up with a positive? It'sdisturbing me that I cannot come up with a solid answer, and certainly'We've been taught that that is the case. will not suffice.Any help would be greatly appreciated.TIA === Subject: Re: All masses have inertia> In order to determine the mass of an object; body, or any mass, But, but, but relativity shows the effective mass increases with> velocity. As a result your definition of mass is meaningless. If you> apply a constant force to a body, it relativistic mass increase so the> acceleration decreases. The ratio F/a tends to infinity. So at what velocity shall we measure the mass? If we get different> answers at different velocities we have aceved bogusity-1 on the bogus> scale. In short you have given a bogus definition of mass, wch is not> surprising. Just about everytng you say is bogus. Why don't you learn> physics instead of constantly demonstrating what an ingnoramus you are. Bobby: You are wstling loudly as you walk past the cemetary. It won't helpthe gobblins will get you anyway(;^)Only at speeds close to that of light rays will the formula f/a = w/g evenit nears the sun.All your learning of physics - as well as my _more extensive_ learning ofit - tells us both, unequivically that accelerating a measurable object;body or mass of material substance to anywhere near where its inertia willchange without adding to its matter or burning it, is a practicalimpossibility: You're just grasping at straws; from a strawman to try anddiscredit the truth! === Subject: Re: multiplication negs Visiting all,Hopefully, I'm in the right place here....I've been making my way throught the Principia, taking my time, working>along. One day I was talking to my young neice, helping her out with some>simple algebra. We were discussing mulitiplication and division of negative>numbers and my neice brought up a question and for some reason I have been>unable so far to reason my way through the answer: exactly why is it that>when one multiplies 2 negative numbers one ends up with a positive? It's>disturbing me that I cannot come up with a solid answer, and certainly>'We've been taught that that is the case. will not suffice.>Any help would be greatly appreciated.There are a number of answers. For integers, we can envisionmultiplication as repeated addition. Multiplying by a negative numbermeans do the repeated addition, then change the sign. So if n and mare integers, with n positive, thenn*m = add n copies of m; and (-n)*m = change the sign of n*m.If m is negative, adding n copies of m yields a negative number, andchanging its sign yields a positive number. But that only sort of works for integers. Otherwise, note that you canreduce the problem to showing that (-1)*(-1)=1, and that -a = (-1)*a,because using commutativity and associativity of multiplication, wehave:(-a)*(-b) = [(-1)*a]*[(-1)*b] = [(-1)*(-1)]*[a*b]So, how do we prove that?To see that (-1)*a = -a, note that -a is the one and only numberthat added to a gives 0. So we take (-1)*a, add it to a, and see ifwe get 0. If we do, then (-1)*a must be equal to -a, yes?[(-1)*a] + a = [(-1)*a] + [1*a] = [(-1)+1]*a (distributivity) = 0*a = 0.Aha, so (-1)*a must be equal to -a.As for (-1)*(-1), we have the same tng. Ts is equal to 1 if andonly if, when we add it to -1 we get zero.[(-1)*(-1)] + (-1) = [(-1)*(-1)] + [1*(-1)] = [(-1) + 1] * (-1) = 0*(-1) = 0.So, why is the product of two negative numbers a positive number? Sothat the rules of addition and multiplication hold. ==It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) === you systematicallygeneralize ts to larger numbers, thenyes, afaik, it has not been studied.but a 3x2 circuit is too trivial to leadto much insight.one interesting approach would be to take thecircuit graph for multiplication and look atthe way that SAT algorithms visit various nodesto reverse engineer the solution. in otherwords, if x*y=z, and z is specified, thenthe information for 'z' percolates backwardsinto the circuit.it seems that some kind of proof may be possiblethat an optimum algorithm must visit the nodes in reverse order. in other words if it visitsnodes in the middle of the circuit 1st, thatis (possibly) provably nonoptimal.afaik studying the graph nature/properties of multiplication hasnot been done by almost anyone.> Below is a 3x2 bit multiplication circuit.> I can use ts circuit to make assertions> about the factors of any number.> I'm curious if ts has been studied. === Subject: Re: right place here.... I've been making my way throught the Principia, taking my time, working> along. One day I was talking to my young neice, helping her out with some> simple algebra. We were discussing mulitiplication and division ofnegative> numbers and my neice brought up a question and for some reason I have been> unable so far to reason my way through the answer: exactly why is it that> when one multiplies 2 negative numbers one ends up with a positive? It's> disturbing me that I cannot come up with a solid answer, and certainly> 'We've been taught that that is the case. will not suffice.> Any help would be greatly appreciated.Here's an example I like:I have been giving away five dollars each minute.Currently (at t = 0), I have zero dollars, but I'm continuingto give away five dollars each minute in the form of IOU's.Thus, the equation for the number of dollars I have attime t is: D = -5t, where D represents dollars.How many dollars did I have four minutes ago?-- Bob Dayhttp://www.bob.day.name === Subject: Re: ab... = (a*b*...)^n ?> You get a lot further by examining only those> numbers whose prime factors are less than B.> Still notng in base 10 up to 10^200.Would you describe further how you did that?(Obviously, you didn't test each number to seewhether its prime factors were less than 10,since to finish witn a week the testing alonewould average > 10^194 tests/second.)Using the brute force method with a C programrunning in backgound for a day or so, I managedto confirm the negative result only up to apaltry 3 10^10. === Subject: Joint distribution of distances and angles in hyperspheresSuppose p(x) is a rotationally symmetric distribution in R^n.Denote by U and V i.i.d. random variables with ts distribution.Denote by d=|U-V| the distance bewteen the two points and by $alpha$the angle between them.The distribution of the distances between points, p(d), has beendetermined by Hammerseley and Lord.I am interested in finding out the joint distribution p(d,alpha) asa function of the space dimension, n, and the distribution p(x). in advance for any pointers...--Dario === Subject: Re: prime numbers factoringSee:http://www.cerias.purdue.edu/homes/ssw/cun/ You can lead a horse's ass to knowledge, but you can't make m tnk. === Subject: Re: multiplication negsSee.. that's where I thought you were going to go... and I've been wrappingmy head around ts and basically.. we simply have to take it for granted( because I cannot recall any axiom) that when: (-n)*m = change the sign ofn*m. ?? Ts seems rather odd, even arbitrary. In reducing multiplicationto a series of successive additions I fail to see how we end up, whenworking with negative integers, with a negative product.Am I missing sometng here??(I've always had problems with the a priori. :-) ) === Re: multiplication negs Visiting Assistant Professor at the University of Montana.>See.. that's where I thought you were going to go... and I've been wrapping>my head around ts and basically.. we simply have to take it for granted>( because I cannot recall any axiom) that when: (-n)*m = change the sign of>n*m. ?? Ts seems rather odd, even arbitrary. In reducing multiplication>to a series of successive additions I fail to see how we end up, when>working with negative integers, with a negative product.Did you read the rest of the reply, or just that line?>Am I missing sometng here??>(I've always had problems with the a priori. :-) )What would it mean to add sometng minus 2 times? Adding it twicemeans: start with zero, then add m, then add m.So adding sometng minus 2 times should mean: start with zero,then ->subtract<- m, then ->subtract<- m again.Wch reduces to-m -m = -(m+m)so again, you are down to proving that (-1)*m = -m, wch amounts toshowing that if you take (-1)*m and you add it to m, you should get 0. ==It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) === Is it mass; or is it weightNow it's time for all of us to realize that a customary pound is a unit of>force,You know better, Donald. Pounds everywhere have always been units of mass. >> Pounds force are such a recent bastardization that they are uniquely>> identified by that name.Back in 1959, the national standards laboratories of the United States of>> America, Canada, the United Kingdom of Great Britain and Northern Ireland, the>> Union of South Africa, Australia, and New Zealand got together and agreed on a>> common definition of the most commonly used pound, the avoirdupois pound. >> They defined it as a unit of mass exactly equal to 0.45359237 kg.Of course, you already knew that, Dishonest Don. Ts is for the benefit of>> anybody else who hadn't been paying attention.>So, if a 220-lb man jumps off a roof-top down 36-ft to a stretchable net>wch stretches an additional 4.4-ft before thrusting the man back, would>the potential energy of the net at the instant it plunges the man back beU = mgh;>U = 220 * (36 + 4.4).No, because U is not a dimensionless quantity. Did you haveparticular units in mind?What's your point, anyway?Sure, it might be, if you choose your units right, and your local g isclose enough to the standard value used to define one of those units.Then, of course, the magnitude of your omitted multiplicand will beclose to 1 in the units chosen for it.It might also be, depending where you are (limiting yourself to realrooftops will limit the possible range), sometng like theseU = (220 lb)(32.06 ft/s)(36 ft + 4.4 ft)or U = (220 lb)(32.24 ft/s)(36 ft + 4.4 ft)Now, if you carry out those operations, what are the units of thefinal result?Or, it might also be, knowing that in the definition most often usedtoday for a slug (wch doesn't have an official definition), 1 slug =32.1740 lb, so 220 lb = 6.84 slugU = (6.84 slug)(32.06 ft/s)(36 ft + 4.4 ft)or U = (6.84 slug)(32.24 ft/s)(36 ft + 4.4 ft)Now what are the units of your final result?Now, have you learned anytng today?Gene Nygaardhttp://ourworld.compuserve.com/homepages/Gene_Nygaard/= === ==Subject: Re: Is it mass; or is it weight> So, if a 220-lb man jumps off a roof-top down 36-ft to a stretchable net> wch stretches an additional 4.4-ft before thrusting the man back, would> the potential energy of the net at the instant it plunges the man back beU = mgh;> U = 220 * (36 + 4.4).You left off the factor of g. Wch either means that you forgotabout it or that you are using a system of units witn wch thefactor of (more or less) and U = mh.The more or less is because you're ratio of pounds force to poundsmass.Gene Nygaard would be correct to point out that there is no singlestandard g that is officially sanctioned for ts purpose. === Subject: topological terminologySo... homotope or homotop, wch is the verb?Pete === <>sSHfTy;{Dhe&:+?b`9fUj5A~$gIYlYT0/$-asR-K~3S3[]q.R3YSmpR|$- GiZp>UN2a}!Fmw+%h}Y when one multiplies 2 negative numbers one ends up with a positive?We have certain rules of arithmetic that we like: commutative,associative, distributive laws, for example. If we want these laws tocontinue to hold for negative numbers, we are forced to the conclusionyou mention.-- G. A. Edgar http://www.math.oo-state.edu/~edgar/ === Subject: Re: Group generated by a and b ...> I have a group G generated by 2 elements a and b such that ba=a(b^k) for some>> integer k.>> Is it possible to express b(a^(-1)) as a product (a^m)(b^n) for some integers m>> and n ?>Yes, when the order of a is finite.But in general, no.As a simple example, let G be the multiplicative matrix group over therational generated bya = [2 0] and b = [1 0] [0 1] [1 1]Then ba = ab^2, but ba^-1 = [1/2 0] [1/2 1]whereas all products (a^m)(b^n) have integral entries.Derek Holt. === Subject: Re: topological terminology> So... homotope or homotop, wch is the verb?PeteUse it in a sentence?The space X homotoped into Y? === Subject: Re: 0! = 1 (episode 2) correctionCorrection, i mean(-1/2)! = sqr(pi) === Subject: Re: lopital's rule? Why's that then? It seems to me to be a mildly interesting theorem. I can see that it might not be a good idea to prove it and then give students lots of 0/0 limit problems for them to solve using L'Hospital's when doing so is just pedagogically useless manipulation. But that is no reason for not proving the theorem.Who said anytng about proving anytng. My experience> is that it is taught as a method, not as a theorem, and whenI'm probably being dim, but a method is only useful if it works. Henceone needs a proof that the method works. Such a proof is a proof of atheorem.> taught it, students will use it *exclusively* for all limit> problems thereafter (it's a formula isn't it) no matter> how inappropriate it is for the problem at hand.So, tell the students that they will be marked down for inappropriatemethods just as they will be for wrong answers. Alternately tell themto Prove/find ... without appeal to L'Hospital's theorem.-- G.C. === Subject: Re: multiplication negs boundary=----=_NextPart_000_004B_01C38E65.85C09DE0--------- really don't tnk ts is just semantics):we're not talking here of adding sometng 'minus 2 times'; rather, we're talking about adding negative sometng x amount of times.So, in -(m)*-(n), where m=2, and n=3... we have: (-2) + (-2) + (-2) = -6 .... right?orare you saying it would be: -[ (-2) + (-2) + (-2) ] = 6.... (wch, obviously, is the correct answer, but I fail to see the move here.. even though I can figure it out.)P.S.: I must ask you to forgive my ignorance, as I am but a lowly student of Plosophy, with little math skills. === Subject: Re: multiplication negsAha!That, is as much as I suspected.Nevertheless, I still like to understand the reasoning bend themovements...TY === Subject: Re: Boolean Algebra - Arithmetic Relationsp> P.S. It may be that by boolean logic you mean computer hardwaregates. A physically realized (or realizable) logic is certainlyinadequate--only a finite number of the infinity of numbers could behandled, and only finite approximations of numbers like pi could behandled.Replace approximations with descriptions and you'd be ok. The finite> number of gates could be programmed for symbolic math, in wch case aIndeed. The numeral for pi might be the approximation 3.141592653 orthe precise pi. But still you can't encompass mathematics in a finitenumber of gates.> mathematically exact pi could be handled in formulae or theorems up to a> cretain size or complexity (bounded indeed by some function of the number> of gates). (The gate count includes those needed to implemement memory,> including storage for the program, of course. All is finite here. So> clearly we're talking about a finite subset of recursive functions -- butThere are functions wch aren't recursive, in fact more that aren'tthan are.> one that is arbitrarily large if you have enough gates.)Michel.-- G.C. === Subject: Re: Dirt Simple Proof Re: Symmetric Groups.: They both fix a syntheme --- but are synthemes dirt simple?Sweet explanation! That's === Subject: Re: Is it mass; or is it weight>Now it's time for all of us to realize that a customary pound is a unitofforce, You know better, Donald. Pounds everywhere have always been units ofmass. Pounds force are such a recent bastardization that they are uniquely identified by that name. Back in 1959, the national standards laboratories of the United Statesof America, Canada, the United Kingdom of Great Britain and NorthernIreland, the Union of South Africa, Australia, and New Zealand got together andagreed on a common definition of the most commonly used pound, the avoirdupoispound. They defined it as a unit of mass exactly equal to 0.45359237 kg. Of course, you already knew that, Dishonest Don. Ts is for thebenefit of anybody else who hadn't been paying attention.> So, if a 220-lb man jumps off a roof-top down 36-ft to a stretchable net> wch stretches an additional 4.4-ft before thrusting the man back, would> the potential energy of the net at the instant it plunges the man back be U = mgh;> U = 220 * (36 + 4.4). --> Ayaz Ahmed Khan Yours Forever in,> Cyberspace.Ts problem could be answered correctly if the sbf system is used(Stones, barleycorns and fortnights)RJ Pease === all,Hopefully, I'm in the right place here....I've been making my way throught the Principia, taking my time, working> along. One day I was talking to my young neice, helping her out with some> simple algebra. We were discussing mulitiplication and division of negative> numbers and my neice brought up a question and for some reason I have been> unable so far to reason my way through the answer: exactly why is it that> when one multiplies 2 negative numbers one ends up with a positive? It's> disturbing me that I cannot come up with a solid answer, and certainly> 'We've been taught that that is the case. will not suffice.> Any help would be greatly appreciated.> TIAOne answer is that we want the distributive property of multiplication over addition, as well as certain other properties, to contintue to work with negatives as well as with positives. Ts also justifies why negative times negative should come out positive.One form of the distributive property says that for al x,y and z x*(y+z) = (x*y) + (x*z).Then, for positive x and y, x*(y + (-y)) = (x*y) + (x*(-y))but y + (-y) = 0, and x*0 = 0 , so (x*y) + (x*(-y)) = 0,and z + (-z) = 0 for any z, so x*y + (-(x*y)) = 0, so x*(-y) = -(x*y). === Subject: Re: Is it mass; or is it weightNow it's time for all of us to realize that a customary pound is a unit>of>force, You know better, Donald. Pounds everywhere have always been units of>mass.> Pounds force are such a recent bastardization that they are uniquely> identified by that name. Back in 1959, the national standards laboratories of the United States>of> America, Canada, the United Kingdom of Great Britain and Northern>Ireland, the> Union of South Africa, Australia, and New Zealand got together and>agreed on a> common definition of the most commonly used pound, the avoirdupois>pound.> They defined it as a unit of mass exactly equal to 0.45359237 kg. Of course, you already knew that, Dishonest Don. Ts is for the>benefit of> anybody else who hadn't been paying attention.>> So, if a 220-lb man jumps off a roof-top down 36-ft to a stretchable net>> wch stretches an additional 4.4-ft before thrusting the man back, would>> the potential energy of the net at the instant it plunges the man back be>> U = mgh;>> U = 220 * (36 + 4.4).>> -->> Ayaz Ahmed Khan>> Yours Forever in,>> Cyberspace.Ts problem could be answered correctly if the sbf system is used>(Stones, barleycorns and fortnights)Sure. But can you help me figure out what the units for m, g, and hwould have to be, if I want to get the energy U in barn yardatmospheres?Gene Nygaardhttp://ourworld.compuserve.com/homepages/Gene_Nygaard/= === ==Subject: Re: Question on lbert & GodelWhat did lbert ask and claim concerning Foundations of Mathematics> (sets, predicate calculus), metamathematics, Logic, Incompleteness,> etc?Sometng about the Continuum Hypothesis? Completeness or> Incompleteness of Logic? Was he contradicted by kindergarten logic> from Godel? How would we exactly formally represent lbert's> questions and claims using current day notation? You're in luck, if you read lbert [1,2,3] you'll find that it's intoday's notation.[1] lbert & Ackerman Principals of Mathematical Logical[2] lbert & Bernays Grundlagen der Mathematik, I, II (Only availablein German?)The last item puts it all in an storical context.-- G.C. === Subject: Re: Antidiagonal, Infinity> > What I propose is that given any rational that the value greater than it and less than any other greater is irrational, > > There is no such number, as several different people have shown you. In non-standard analysis, there might be, however. See Alain Robert's book about NSA. Rather than being irrational, it would be non-standard, though. > > I have yet to see any standard or non-standard model of the reals in > wch there is a smallest positive number. Who says that that is meant? In nsa, there can be a nonstandard number that can be said to be 'greater than (a given standard rational) and less than any other (standard) greater' That it is not unique, who cares?Ross wants to use non-standard numbers to confirm s hypothesis > that there is a next real after any real, and that the irrationals > and rationals alternate on the real line or on some non-standard > real line. So he cares. On the other hand, s hypothesis is way out > in left field, where he has been stuck for months, if not years.Actually, I have not been claiming to be using the nonstandardnumbers. In light of the issues that we cover and mutually understandto some extent, it might be the case that of the set of reals that foreach that it has a next element of opposite rationality that thatwould be in neither the nonstandard nor classical model. Its elementswould still be the elements of the set of real or hyperreals.About equating density with measure in the unit interval, that meansequating density in the unit interval to measure in the unitinterval.In ts model, say alternating analysis, or a rationally alternatingmodel of the continuous reals, the alternating measure of therationals in [0,1) is 1/2, as is that of the irrationals, as is theirdensity in the unit interval and the density of the rationals in thereals.Ts doesn't disagree with the measure of the reals in the unitinterval being equal to one. The results of classical analysis wouldhold true. Anytng that didn't hold true would be suspect.What's a proper subset of the rationals or irrationals that is densein the reals, where its complement is infinite?The rationals and irrationals are each dense in the reals, theirunion, the reals are continuous.Ross === Subject: Re: 0! = 1 (episode 2) correctionSekhmet mean (-1/2)! = sqr(pi)>But for not integer arguments, generally Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Dedekind Cuts> > I have heard that the reals can be defined by Dedekind cuts. What is > ts definition, exactly?> Ts is a division of the rationals into a pair of sets, wch we will call Left and Right. Each element of Left is less than all elements of Right. Each element of Right is greater than all elements of Left. It follows that Left intersect Right is empty. We also require that a rational number either be in Left or Right so that Left union Right = the set of rational numbers.> In the case that Right has a lest element we put that element in Left. There are two case. Left has no greatest element and Right has no least element. If so the division or cut Left,Right corresponds to an irrational number. Or Left has a greatest element and Rigt does not. Ts sup of Left is the number (a rational) represented by the cut.> There are rules of addition, subtraction, multiplication and division defined for cuts and it is shown they form a field.> For a good account of Dedikind Cuts set -A Course in Pure Mathematics- by G.H.Hardy.> I understand what a Dedekind cut is, but what what is the definition that goes the set of reals is some collection of Dedekind cuts?If a real number is defined to be a Dedekind cut, then it follows that> the set of real numbers is the set of all Dedekind cuts.> Is it simply the set of of all Dedekind cuts of the rationals? Or is it the set of numbers such that all Dedekind cut of the reals the reals yields a greatest element for Left? Or sometng else?Dedekind cuts are defined on the rationals. You could carry out a> similar construction on the reals, but it wouldn't be a Dedekind cut.The purpose of the Dedekind cut construction, or the equivalence classes of Cauchy sequaences construction of the real fraom the rationals, is to create a complete ordered field, in wch for every non-empty bounded-above set there is a least upper bound. The rationals do not have ts property, since, for example, the set of rationaTs S = {q: q^2 < 2, q in Q} has no rational least upper bound.A similar construction based on cuts of the set of reals has been done, but the result is isomorpc as an ordered field to the standard Dedekind construction real field. === Subject: Re: Group generated by a and b ...> I have a group G generated by 2 elements a and b such that ba=a(b^k) for some> integer k.> Is it possible to express b(a^(-1)) as a product (a^m)(b^n) for some integers m> and n ?>Yes, when the order of a is finite.But in general, no.As a simple example, let G be the multiplicative matrix group over the>rational generated by>a = [2 0] and b = [1 0]> [0 1] [1 1]Then ba = ab^2, but >ba^-1 = [1/2 0]> [1/2 1]>whereas all products (a^m)(b^n) have integral entries.Sorry - that last statement is not correct for negative m. But, in general,(a^m)(b^n) = [2^m 0] [n 1]wch is unequal to ba^-1 for all integral m,n.Derek Holt. === Subject: Re: multiplication negsI've been making my way throught the Principia, taking my time, working> along. One day I was talking to my young neice, helping her out with some> simple algebra. We were discussing mulitiplication and division of negative> numbers and my neice brought up a question and for some reason I have been> unable so far to reason my way through the answer: exactly why is it that> when one multiplies 2 negative numbers one ends up with a positive? It's> disturbing me that I cannot come up with a solid answer, and certainly> 'We've been taught that that is the case. will not suffice.> Any help would be greatly appreciated.The law of signs is a consequence of the well-known elementaryschool laws of arithmetic - most notably the distributive law, e.g.Claim: a b equals (-a)(-b)Proof: a b + [a + -a](-b) = a[b + -b] + (-a)(-b)Equivalently, evaluate in two different ways the following expression -- a b + a(-b)+(-a)(-b) -----------by undistributing xy+xz -> x(y+z) the overlined and underlined terms. Ts is a FAQ (Frequently Asked Question) here, e.g. see earlier dubuque*+%22law+of+signs%22-Bill Dubuque === Subject: Re: Factorial/Exponential Identity, Infinity> I got to tnking about sum 2^n and how it was equal to 2^(x+1) - 1. I wonder: does it work for 3? The answer is not quite, no.Look up geometric series. If you must reinvent the wheel, you need not do in so publicly.I guess it's kind of like how the sum for i=1 to oo of b^-i is1/(b-1).The concept here is the repeated unit, the rep-unit, in base b. Thesum for i=1 to n of b^i is the sequence 1_n 1_(n-1) ... 1_2 1_1, asequence of 1's of length n representing an integer in base b. Thenb^(n+1)-1 is a sequence of n many (b-1)'s in base b: (b-1)_n(b-1)_(n-1) ... (b-1)_2 (b-1)_1, dividing that by (b-1) has as aresult the rep-unit of length n.In summing the negative integers powers of an integer, it is similar,represent the number in that base so that the sum is simply .1111.... If you multiply that by (b-1) then the result is .(b-1) (b-1) (b-1)(b-1) ... = 1.These are well-known and almost trivial, for example, the sum for i=1to infinity of 1/25000^i is 1/24999, 1/9 in decimal is .1111...,etcetera. I assume that for real x>1 that the sum of x^-i for i from1 to oo is 1/(x-1).I will research geometric series. I'm not here to reinvent the wheel,but it damn well better let me say that half the integers are even, asthey are as the asymptotic density of the even numbers in the integersis one half.What do you tnk about the canonicalization of the infinite binarysequences? That is to say, for the sequence .010101(01)... with halfzeroes and half ones, that you could exchange any two sequenceelements any number of times and not get .001001001(001).... Is therea rule to exchange (permute) the elements of the sequence .0101(01)...to get 00010001(0001)...?Ross === Subject: Re: The Bible Codemodern, but ex post facto,with a few rather repugnant exceptions (Die, Rabin, Die!) the sequel has ts doofus, Drosnin,supposedly watcng the WTC t from s hotel room, andimmediately running the program to post-dict it. the second book also shows (I tnk it was indexed,or I was just lucky in finding it, as I didn't *read* the God-am tng) that he got the original idea from oneof Sharon's buddies (presumably) in hte military!yes, the NT can be used to equal effect (on everage). repeat, _War and Peace_ or just the 26 letters in any order can be used with the infinite set of co-prime skips, with teh resulting ts being further massaged into some m by n array (or what ever). > The Code stuff has been applied successfully to Moby Dick to predict> moderns events with Astounding accuracy. (NOT!)--le ducs d'Enron!http://larouchepub.com === Subject: Re: Re fermat by Tomassorry; I guess that you meant the second root of 3 by %3 -- butit's still a rather silly use of the law of large numbers. an example that may apply, since I don't recall what that is:there's a proof that li(x) goes less than pi(x),the number of prime numbers less than x, infinitely often, althoughthere is (was) no known place in the sequence where it occurs.the second (skware) root of 3 is irrational by *definition*. > Fernat's Last Theorem is negative is not problem, just as> proofs (of neg or pos statements) by contradiction are good.> of course, three per cent is rational by definition.Negative statements about numbers are unverifiable. Take the--les ducs de Buffet et Schulz (R&D Chair Assoc.Intl.);vote NONE OF THE BELOWon Trickier Dick Cheney's California Recall/e-Dereg!http://larouchepub.comhttp://members.tripod.com /~american_almanac/ === Subject: Re: Exercise in Projective Linguistics> When I posed the problem, I just assigned names A, B, C, ... to thepoints and gave the lines as 5-tuples. Later, it occurred to meI might have been able to find a different assignment of thepoints into the alphabet in such a way that each line was( a permutation of ) the letters in a 5-letter word.> So here is the challenge: how well can you do ? Can you embedthese letters back into the alphabet so that all, or most, ofthese 5-tuples become English words? (Interpret word as you wish.)(I would be willing to take a solution wch works only foran affine plane inside here, if necessary.)> Ts might be a good one to send to rec.puzzles. I doubt that you can get> all 21. The best I've been able to come up with so far is 13, with> AB...U = DYCAOEBTJRMLSNKUGFWIV generating the words > cadgy, fumed, blind, folky, nervy, misty, curio, clews, > woman, barfs, vault, begot, gunks I can get 14: AB...U = YISLKDAWZCGNTEROMPUHB generating the words slimy, podgy, hyena, prink, cebid, wight, hocks, dunts, brags, kluge, clapt, blown, metro, bumph Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: mandelbrot iterationscorrection: the talk was given at Royce Hall Auditorium,wch should henceforth be known as On the Subjectof Magnifications in the Complex Plain, orWhy Didn't I Go Down the Hall and Talk to the Engineersat teh IBM-Watson Tngy, for All Those Years? or, What is Chaos?... The Maternal Unit of Chronos! I can tell of the original insight, but when I foundsome issue of 1980. I'd been using an IBM-XT clone 8088 cp,wch uses the famous dysabled-to-8-bit-16-bit-path, andthe registers hold 80 bits. not that I ever generated any fractalson it ... or any math ... what else can a screensaver be used to dysplay,other than blackspace? I've never seen teh update, -855 (I'm not sure of these #s), butit only tries to minimize the overflows & underflows;it explicitly dyscusses these. > changing the hardware setting on the macne from double-precision> to single, he pooh-poohed it -- the grad student that I found. > the mini-Ms do not appear at *every* magnification, since> the rounding-errors are tied to the lengths of the registers, > the specification is inherently chaotic, > > that is the recurrence of mini-bugs or cardioids, > at every level of magnification, is just an artifact > of the floating-point ops (IEEE-755, -855, I tnk). or doesn't change, in calculations. FP doubles are good for several tens of thousand of iterations, down to an area of 10E-10 or so, before the precision gives out. The cartoids are visible several orders of magnitude above ts.--les ducs de Buffet et Schulz (R&D Chair Assoc.Intl.);vote NONE OF THE BELOWon Trickier Dick Cheney's California Recall/e-Dereg!http://larouchepub.comhttp://members.tripod.com /~american_almanac/ === Subject: looking for a functionI'm looking for a function wch garantuees the following conditions :Suppose you have an m*n matrix. Each element of the matrix has a valuebetween 1 to p. Is there a function wch ensures that for each element theelement left, right, above and under are a different value than thespecified element.in the case n is not dividable by p the following function seems to work :S { [(x%n)+1]%p in case x/n is even (rounded) { [(x%n)+1+p/2]%p in case x/n is odd (rounded)-example correct matrix :[1 2 3 4 5 1 2 3 ][4 5 1 2 3 4 5 1 ][2 3 4 5 1 2 3 4 ]-example incorrect matrix (6/2=3) :[1 2 3 1 2 3][1 2 3 1 2 3][1 2 3 1 2 3]though there is a correct solution :[1 2 3 1 2 3 ][2 3 1 2 3 1 ][3 1 2 3 1 2 ]I could split up the function into another part but i tnk there is aneasier function (or maybe i'm just hoping)All ideas are more than welcome.Yves(sorry for the crossposting in alt.math but there wasn't much life there) === Subject: Re: Antidiagonal, Infinity Ross wants to use non-standard numbers to confirm s hypothesis that there is a next real after any real, and that the irrationals and rationals alternate on the real line or on some non-standard real line. So he cares. On the other hand, s hypothesis is way out in left field, where he has been stuck for months, if not years.Actually, I have not been claiming to be using the nonstandard> numbers. In light of the issues that we cover and mutually understand> to some extent, it might be the case that of the set of reals that for> each that it has a next element of opposite rationality that that> would be in neither the nonstandard nor classical model. Its elements> would still be the elements of the set of real or hyperreals.As the classical model is the reals and a non-standard is the hyperreals, you are claiming to simultaneously be inside of and outside of some model simultaneously, wch certainly puts you outside of rationality.About equating density with measure in the unit interval, that means> equating density in the unit interval to measure in the unit> interval.And how do you do equate density to measure? It would help if you could explain what YOU mean by dense and what you mean by measure, as the standard meanings do not allow of equating these ideas.In ts model, say alternating analysis, or a rationally alternating> model of the continuous reals, the alternating measure of the> rationals in [0,1) is 1/2, as is that of the irrationals, as is their> density in the unit interval and the density of the rationals in the> reals.But every time you have a rational, x, and irrational, y, supposedly next to each other, you have the problem of (x+y)/2 between them, so they aren't really next to each other after all.Ts doesn't disagree with the measure of the reals in the unit> interval being equal to one.Yes it does. For each of uncountably many irrationals, x, linearly independent over the rationals, consider U(x) = (x+Q) / [0,1), the intersecting ofx+Q and [0,1), where x+Q = {x+q: q in Q}. Since each x+Q is essentially a translation of Q, each must have the same measure, but, as they are pairwise disjoint, the measure of [0,1) must be the sum of their separate measures.Thus, according to your arithmetic, uncountably many measures of 1/2 add up to 1. So your measure is self-contradictory.> The results of classical analysis would> hold true. Anytng that didn't hold true would be suspect.Your formulation must be suspect, then, since it doesn' hold true.What's a proper subset of the rationals or irrationals that is dense> in the reals, where its complement is infinite?Let n be an arbirary integer grreater than 1, Z be the set of integers, and Z[1/n] be the ring generated by appending 1/n to Z and taking the closure under addition and multiplication. Each such ring will be dense in the reals and have infinite compliment in the rationals and in the reals, furthermore, if n is a proper factor of m, then Z[1/m] is a proper subset of Z[1/n].Thus each of Z[1/2], Z[1/4], Z[1/8], z[1/16], ..., is a proper subset of all its predecessors in ts sequence, but each is also dense in R.The rationals and irrationals are each dense in the reals, their> union, the reals are continuous.Ross, what do YOU mean by continuous? For the standard meaning of continuous, a function may be continuous, but not a mere set.What property does the ordered field of reals have that the ordered field of rationals lacks that makes the set of reals continuous in your sense but the set of rationals not continuous in your sense? === Subject: Re: Convergence in distribution> Yes,I mean random 'sequences'.> But BOTH X_n and Y_n are random sequences converging in distribution.> Does that make any difference to the answer?It makes a difference to the question. One doesn't usuallytalk about one sequence converging to another sequence, but presumablywhat you're talking about is equivalent to saying that for any boundedcontinuous real-valued function f on the real line, E[f(X_n)] - E[f(Y_n)] -> 0.> Well,the actual pdf's of X_n and Y_n at a given n are reasonably well modeled > by GGD(Generalized Gaussian Distribution)'s with zero means and variances > typically 25~100 wle the delta's can be 1~10. > Although we can easily find counter-examples with discrete probability > distributions, I observed in the actual problems that the 'quantized'(i.e. > X1_n & Y1_n) sequences seemed to 'approach' in distribution. (By 'approach',I > mean the variances and the shape parameter characterizing the GGDs of both > X1_n & Y1_n approached the same values.) It's also easy to find counter-examples with continuous probabilitydistributions. But if the X_n and Y_n are continuous random variableswith uniformly integrable densities (i.e. some integrable function gsuchthat f_{X_n}(x) <= g(x) for all n and x, and similarly for Y_n) itshouldwork, because ts ensures that the probability of X_n or Y_n beingvery close to one of the discontinuities is small.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: ARCSIN function, single precision floating point. -- Example routine needed?> [...] > So try ARCSIN(x) = 2*ARCTAN(x/(1+SQRT(1-x^2))). === Subject: Re: lopital's rule? Why's that then? It seems to me to be a mildly interesting theorem. I> can see that it might not be a good idea to prove it and then give> students lots of 0/0 limit problems for them to solve using> L'Hospital's when doing so is just pedagogically useless manipulation.> But that is no reason for not proving the theorem.Who said anytng about proving anytng. My experience>> is that it is taught as a method, not as a theorem, and whenI'm probably being dim, but a method is only useful if it works. Hence> one needs a proof that the method works. Such a proof is a proof of a> theorem.Try telling students that :-) >> taught it, students will use it *exclusively* for all limit>> problems thereafter (it's a formula isn't it) no matter>> how inappropriate it is for the problem at hand.So, tell the students that they will be marked down for inappropriate> methods just as they will be for wrong answers. Alternately tell them> to Prove/find ... without appeal to L'Hospital's theorem.Alas, it's not me who teaches that course :-(-- === Subject: Re: Chessboard knight metric? >Take a chessboard (with or without infinetely many squares) let the >distance d((x_1,x_2),(y_1,y_2)) between two squares x and y of the >chessboard be defined as the minimum number of moves a knight takes >to reach y from x. >Is d a metric? With ts distance, the triangular equation is obviously true, and that makes it a metric. Not trying to suggest that ts is some new >question that hasn't been asked/answered before. Is there a general >formula for calculating d? More generally, the same question may be >asked for the other pieces (queen, king, knight, bisp)? Actually, I >asked myself ts question a few years ago. If I remember back to the >notes I took, I had sometng like (x_1-y_1, x_2-y_2)= (even number, >even number), then d(x,y) = even number. If (x_1-y_1, x_2-y_2)= (even >number, odd number), then d(x,y) = odd number. Finally, if (x_1-y_1, >x_2-y_2)= (odd number, odd number), then d(x,y) = even number. In other >words, the same rules for adding natural numbers... C.Dement> With ts distance, the triangular equation is obviously true, and that makes it a metric. > I don`t (yet) see the triangle inequality as so inherently >obvious that it is not worth further thought. Take the points x=(1,1), y=(77,79), and z=(163,163). It is still feasablethat d(x,y)+d(y,z) is less than d(x,z) (especially as an explicit formula for d doesn`t exist). The fact that I would be somewhat surprised if that wasn`t the case shouldn't be usedas part of the proof. C.DementWith your definition: the minimum number of moves . . ., thetriangle equality is obviously true. If d(x,y) + d(x,z) < d(x,z) thenyour d function does not obey your definition since if going from x toy and then y to z is less than x to z directly the value for d(x,z)cannot be the minimum.If you have a particular formula in mind then checking the triangleequality is partially checking whether the formula obeys yourdefinition. It would not be checking whether your definition is ametric. That is clear.It is necessary to check that your definition is well defined. Youcould do ts by first checking that a knight can get between any twosquares in a finite number of moves. Ts is easy. First check howto go one square horizontally in either direction. Then check alsovertically. Combinations of those moves will clearly get you betweenany two squares in a finite number of moves. Of course, ts isunlikely to be the minimum. But now we know an upper bound for thedistance between any two pairs of squares. Since your metric clearlyis integer and non-negative there must be a unique answer to theminimum distance for any pair of squares.I am not saying that there is any easy formula for the answer. Maybethere is and maybe there is not. I have not thought about it yet. But the existence or otherwise of an easy formula does not detractfrom the existence of the distance function and the fact that it is ametric.The similar queen, castle and bishop metrics are less interesting. Acastle can get between any two squares in at most 2 moves. In factthe formula is simple, in pseudo code: if x1 = x2 and y1 = y2 then distance = 0 else if x1 = x2 or y1 = y2 then distance = 1 else distance = 2The queen is similar but also has distance 1 in the diagonal cases:abs(x1 - x2) = abs(y1 - y2)The bishop is a little tricky since it cannot get between squares ofdifferent colours. So it is not defined or not-finite between squaresof different colours. It would be a boring metric on the two singlecolour subspaces.J === Subject: how many resolutions?A^2 + B^2 + C^2 + D^2 = 85^2where a to d are different one-digit or two-digit numbers.I thought that it could have just one solution: since 85 is a product of twoprime numbers of 4k+1 type: 85=5*17. As it is known that only prime numbers4k+1 could be resolved as p^2=a^2 + b^2.so 85^2=5^2 * 17^2 = (3^2 + 4^2)*(15^2 + 8^2) so right numbers are 24, 32,45, 60.But there is another solution! 3, 4, 12, 84. Why? And maybe there are some other solutions left? It means thatresolution p^2=a^2 + b^2 is ambiguos? So we could find c and d (c not equala and b) such that p^2=c^2 + d^2?Or not? === Subject: Re: how many resolutions? >So we could find c and d (c not equal> a and b)Sorry for mistype - I meant c and d not equal a and b === Subject: Re: Algebra proof> Let a, b, r and s be integers with r>1, s>1 and gcd(r,s)=1. Prove> that if a = b (mod r) and a = b (mod s) then a = b (mod rs).>> How do I do ts????One way: r,s|a-b => lcm(r,s)|a-bBut gcd(r,s) = 1 => lcm(r,s) = rs via gcd(r,s) lcm(r,s) = rs>> a = b + mr for some m, and a = b + ns for some n, right? That means>> that mr = ns. Now use gcd(r,s) = 1 to conclude sometng about m (or n). I guess since r and s are relatively prime, then r must divide n. so> rx = n. and a = b + rxs, so a = b (mod rs), right? Is there a name to ts theorem? That is, the theorem that says that (a,b) = 1 & da = bc => a|c and b|d ?EUCLID'S LEMMA: a | bc => a|c if (a,b) = 1. Equivalently:UNIQUENESS OF REDUCED FRACTIONS: c a c = na gcd(a,b)=1 => - = - <=> d b d = nbSee my prior posts for much 22euclid's+lemma%22&filter=0-Bill Dubuque === Subject: Re: Factorial/Exponential Identity, Infinity> I will research geometric series. I'm not here to reinvent the wheel,> but it damn well better let me say that half the integers are even, as> they are as the asymptotic density of the even numbers in the integers> is one half.Doing any sort of research in mathematics with a fixed idea of what you will allow yourself to find is not really a good idea.What do you tnk about the canonicalization of the infinite binary> sequences? That is to say, for the sequence .010101(01)... with half> zeroes and half ones, that you could exchange any two sequence> elements any number of times and not get .001001001(001).... Is there> a rule to exchange (permute) the elements of the sequence .0101(01)...> to get 00010001(0001)...?It depends on what you consider an allowable permutation.The most general form of such permutation can be considered as follows.Let N be the set of positive integers, then any infinite binary sequence can be uniquely represented by a functions f:N -> {0,1} in wch f(k) is the k'th digit of the sequence. Let B be the set of such functions f:N -> {0,1}.Then any bijection g:N <-> N creates a permutation G: B -> B by defining G(f) = fog, where o represents function composition, G(f) is the function whose valueat k in N is f(g(k),i.e., G(f)(k) = f(g(k)), for each k in N.Let P represent the set of all such permutations, It is clear that the number of zeros and number of ones in an expansion must remain fixed under any such permutation of digits.Then, for a given f:N -> {0,1} and h:N -> {0,1} in B, a necessary and sufficient condition for there being some permutation G in P such that G(f) = h is that card(k in N: f(k) = 0}) = card(k in N: h(k) = 0}) and card(k in N: f(k) = 1}) = card(k in N: h(k) = 1})i.e., the infinite binary sequences for f and h have the same number, possibly infinite, of zeros and the same number, possibly infinite, of ones, and only one of these can be finite.But if the number of zeros or the number of ones differ, then there is no permutation carrying one into the other.Ts divides up the set B into equivalence classes of binaries permutable into each other wch can be indexed by the integers, Z. Any two such binaries with infinitely many zeros and infinitely many ones in their expansions can be permuted into each other, and can be indexed by 0. If the number of ones is finite, use that number as index, and if the number of zeros is finite use the negative of that number as index.Then two binaries can be permuted into each other if and only if they have the same index. === Subject: Re: Exercise in Projective Linguistics > .... the plane over the field of 4 elements .... can be> described as an assembly of 21 points, some subsets of 5 of wch> are called lines. When I posed the problem, I just assigned names A, B, C, ... to the> points and gave the lines as 5-tuples. Later, it occurred to me> I might have been able to find a different assignment of the> points into the alphabet in such a way that each line was> ( a permutation of ) the letters in a 5-letter word....By the way, I hereby claim my title as inventor of a new field.... Not so. A few years ago I was shown a sentence sometng like Yea,why try her raw wet hat? whose connection with the projective plane oforder 2 won't escape you. I _tnk_ the person who mentioned it wasBurkard Polster, but s web site http://www.maths.monash.edu.au/~bpolster/ doesn't seem to have it now(although it has a lot of other nice tngs). Ken Pledger. === Subject: Re: a puzzle related to artinian group> A quick sketch of my argument for the linear case is that you number cards> and players and arrange the play so that the cards are always in order from> left to right and so that each player holds a contiguous set of card> numbers. [...]What's the role the numbering of the cards?> You then show that the centre of gravity of the cards is fixed,> that the variance of the positions of the cards relative to some point to> the left of all of them is strictly increasing whenever a round of play> takes place, [...]Nice!:)> As for the cyclic version?A similar idea should also work in the cyclic version. Although, it isnot clear to me whether the center of the gravity and else make anysense in ts case.One should find a suitable function that strictly increases in eachround, except for the uniform === Subject: Re: 0! = 1> If I know my algebra, then an identity for an operation (let's call> it #) is such an i, that for all x, x#i=x. But there is another concept> too: such a j, that for all x, x#j=j. For multiplication, ts is 0.> For addition, there doesn't seem to be one. What is ts concept called?A small addendum to my previous response:You said For addition, there doesn't seem to be one. But of course, thatdepends on how addition is defined, what the set of operands is, ...Here's a nice example where there is an element wch is absorptive underaddition:On R* = R U {oo}, the one-point extension of the reals, extend theoperation of addition by specifying that x + oo = oo = oo + x for all x in R*. Cantrell === Subject: Re: Minimal Graph, Four Color Theorem Indeed, the proof of the Four Color Theorem rests on showing that there does not exist any graph G wch requires 5 colors, is planar, and such that the removal of any vertex results in a graph wch can be colored with only 4 colors. But you have not given any coherent argument to establish ts proposition that I can see anywhere. All I have is a hypothesis. The validity of the hypothesis does notdepend upon the a priori assumption that G is planar. Therefore, youmust automatically reject it. Why should I bother? === Subject: recurrence relation question,how do i go about solving a recurrence relation when the function T(n)is multiplied by n? such as the following:2n*T(n) + 2nT(n-1) - 2T(n-1) = 2^n?What tricks would one use? === Subject: Re: lopital's rule?> Guy Corrigall scribbled the following:>> L'Hospital's Rule (pronounced lopital in French) can be found in>> any book on calculus of a single variable (look it up!). L'Hospital or L'H.99pital. The circumflex (the ^ tngy) means that the> following s is understood implicitly.>AFAIK the guy's name is de l'h.99pital and in my pronounciation of the frenchlanguage :-) the ^ makes the o-sound longer. === Subject: Re: Is ts newsgroup useless? X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 03:34 PM, Bart Goddard said:>I'd be all for sci.math.moderated under a charter that specifies that>the moderator must reject any post that refers negatively on the>character of another person.Well, get some volunteers, put together an RFD and steer it throughthe process. Be warned that it is a lot of work, but it can be done.-- Unsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to === Subject: Re: Is ts newsgroup useless? <3F81A4DB.2364E9EE@ix.netcom.com> X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 06:43 PM, Bart Goddard said:>Half the problem here is that it's unmoderated. The other half is>that it's populated by overly-anal-retentive jerks who tnk it's>going to be important to point out that I spelled you're your >twice in the preceding paragraphs. PKB. You're upset about flames, yet you post them yourself. You claimto be a Christian; don't your scriptures say soetng about the beamas 90% of the people you're complaining about.>I _thought_ we had beaten these guys up sufficiently in the gh>school locker room that they'd be quiet by now.So by you correcting an error is wrong, but battery is acceptablebehavior? And you have the arrant hypocrisy to call others jerk.>Maybe the real problem is that a guy can't e-mail a matburn.The real problem is that people like you are hypocrites. That's farworse than any amount of flaming. I suggest that you read what you posted, and then explain how you'reany better than JSH. Better yet, print out a copy and ask your pastorwhat he tnks of it.-- Unsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to === Subject: Re: Is ts newsgroup useless? <3F81A4DB.2364E9EE@ix.netcom.com> X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 08:19 PM, Bart Goddard said:>It wouldn't be so hard to read a message and decide whether to post>it.Does that mean that you're volunteering?-- Unsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to === Subject: Re: Antidiagonal, Infinity <3f79d8b6$5$fuzhry+tra$mr2ice@news.patriot.net> <3f7cafc2$11$fuzhry+tra$mr2ice@news.patriot.net> <3f82e66f$1$fuzhry+tra$mr2ice@news.patriot.net> tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 07:09 PM, raf@tiki-lounge.com (Ross A. Finlayson) said:>Here I equate density with measure in the unit interval. What do you mean by measure? With the obvious meaning, the rationalshave measure 0. As has been explained to you.>only concerned with considering a model where the rationals and>irrationals alternate in the reals.That's like saying that you're concerned with a model where 1+1 > 1.It's not a model for the reals, because it doesn't have the propertiesof the reals.>rationals and irrationals are disjoint and distinct,What are you trying to say?>then there necessarily would be irrationals with no>rationals between them. Would you care to provide a proof of that?>I like to tnk that the rationals and irrationals alternateI like to tnk that the nice gentleman with the map will make merich. Alas, if I do believe it then I will lose a substantial amountof money. It's false.>and that>the function f(x)=x+iota maps Q[0,1) onto P(0,1), and f(x)=x-iota>maps Q(0,1] to P(0,1).Also false.>half infinityMeaningless.>finite multiples of a scalar infinity.Also meaningless.>a set of a vectorMeaningless.>One tng to note is that *R, the hyperreals, as a set>contains the same elements as R, the reals. Incorrect.>the characteristics of a probability distributionWhat do you believe them to be? How do you apply them to an infiniteset?>We were talking>about the probability of an infinite binary seqence having one>element being on, the rest off. No we weren't, because you've refused to define what you mean by that.>That probability is expressed as n/2^n, as n>diverges to infinity. Meaningless. Assuming that you meant the limit of n/2^n as napproaches infinity, that comes out to 0.>The probability of each among all possible infinite>binary sequences is being1/2^n, What is n? You seem to be confusing bound and unbound variables.>So anyways out of those n>possible sequences with one on bit and the rest off bits, each is>equally probable. Yes 0=0.>So a theoretical (read: thought experiment)Theoretical is not the same as gedanken experiment.>method to generate an element of N is to>once again flip infinitely many coins. You can't.>At ts point it's a crazy, or>rather, unconventional thought experiment in that the first coin>toss says whether it is oo/2 or greater or less than oo/2.No, because oo/2 has no meaning.>talk about the probability of selecting a given element of the>natural integers assuming a uniform probability distribution over>the integers. That's easy; it doesn't exist.>At least we seem to have some agreement that a uniform>probability distribution over an interval of the reals exists,No, because we don't have common understandings of what any of thewords mean.>simple method to sample an element of an interval of the reals>exists.No, because a finite number of coin tosses only lets us sample afinite set.>infinitesimals,And it was clear that your understanding was fuzzy and incorrect.Mathematics is a precise discipline; you must reason from the axioms,definitions and rules of inference, not from your preconceptions.-- Unsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to === Subject: Re: Antidiagonal, Infinity <3F716897.4040005@tcs.inf.tu-dresden.deqqqq> <3f79d8b6$5$fuzhry+tra$mr2ice@news.patriot.net> <3f7cafc2$11$fuzhry+tra$mr2ice@news.patriot.net> <3f82e66f$1$fuzhry+tra$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 09:08 PM, Herman Jurjus said:>?? Perhaps i'm missing sometng.Slip of the finger; I meant to reply to the consecutive part.>In non-standard analysis, there might be, however.No, because you can take the average of two nonstandard reals.>Actually, there is an axiomatic approach of NSA in wch a few>axioms are added to ZF(C), and in wch the above suggestion makes>sense.The Devil is in the details. The nonstandard reals are a model of thereals only in the sense that the same propositions are true withnonstandard definitions of various terms. You can't construct anisomorpsm between them. Further, even with those axioms therewouldn't be a successor function.>The other poster might be interested in ts approach towards NSA.He doesn't have the background for it. He'd need to start with, e.g.,Halmos, and work s way forward.-- Unsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to === Subject: Re: Antidiagonal, Infinity <3F716897.4040005@tcs.inf.tu-dresden.deqqqq> <3f79d8b6$5$fuzhry+tra$mr2ice@news.patriot.net> <3f7cafc2$11$fuzhry+tra$mr2ice@news.patriot.net> <3f82e66f$1$fuzhry+tra$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith X-Treme: C&C,DWS>Who says that that is meant?The OP has consistently failed to define s terms; Virgil's guess asto what he meant is as good as any.>In nsa, there can be a nonstandard number that can be said to be>'greater than (a given standard rational) and less than any other>(standard) greater'There is still no alternation of rational and irrational. Nor is itplausible that the OP meant images of rational and irrational under animbedding into the nonstandard reals.-- Unsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to === Subject: Re: Numeric one-way proof that your original procedure> generates a lot of duplicates, I assume that you could> not find any errors in the proof.>Two separate arguments in the same posting tend tolead to problems, best to separate them.Since AES encryption has a 1 to 1 translation printing theentire 128 bits would give a unique number to eachticket. Is ts consistent with your assumption thata sample of a sample of AES+CTR output can be consideredto be random numbers? After all they appear to beopposites. I could believe either answer.Andrew Swallow === Subject: Re: quantum echo > > Are you saying it is possible for sometng to exist at location A and > location B at the same time? I thought it was inference by the rule of > non-contradiction. I have no idea whether ts is possible or not. My suggestion that quanta may satisfy '~(x=x)' is meant to translate into logical terms the claim by some physicists that quanta 'lack individuality' (Heller suggests that quanta 'lack haecceities'). The scenario with cakes-to-be suggests that the behaviour of other entities without individuality resembles that of quanta in the relevant respect. Heller's description of that behaviour follows my sig.Metaphysical Background, Thomas McTighe asserted that the quiddity of a> tng is notng other than unity itself. Hence, by virtue of its positive> content, the sun differs not at all from the moon or any other particular> tng. The diversity wch is exbited by the natural world is merely the> product of accidental differences; no object possesses any specific form> wch interposes itself between a particular existing tng and the source> of their being e.g. the Absolute.15 All individual entities are notng more> than differing contractions of the whole devoid of any being of their own.> ...because the restricted quiddity of a tng is the tng itself.http://www.crvp.org/book/Series01/I-10/chapter_ii.htmI take a quiddity to be a tng's *suchness* and a haecceity to be its*tsness*. A *tsness* I take to be, as Robert Adams does, theproperty of self-identity, although I disagree with an assumptionwch might be read into Adams, that an individual can lackself-identity and yet possess the property of being someindividual or other). In Primitive Tsness and Primitive Identity (_The Journalof Plosophy_, Vol. 76, No. 1. (Jan., 1979), pp. 5-26), AdamsA tsness is the property of being a certain particularindividual, not the property of being some individual or other,but my property of being identical with me, your property ofbeing identical with you, etc. These properties have recently beencalled 'essences', but that is storically unfortunate, for essenceshave normally been understood to be constituted by logical properties,and we are entertaining the possibility of nonqualitativetsnesses. In defining 'tsness' as I have, I do not meanto deny that universals have analogous properties--for example,the property of being identical with the quality red. But sincewe are concerned here principally with the question whetherthe identity and distinctness of individuals is purelyqualitative or not, it is useful to reserve the term'tsness' for the identities of individuals.It may be controversial to speak of a property of beingidentical with me. I want the word 'property' to carry aslight a metaphysical load here as possible. 'Tsness'is intended to be a synonym or translation of the traditionalterm 'haecceity' (in Latin, 'haecceitas'), wch so far as Iknow was invented by Duns Scotus.Like many medieval plosophers, Scotus regarded properties ascomponents of the tngs that have them. He introducedhaecceities (tsnesses), accordingly, as a special sortof metaphysical component of individuals.[4] I am not proposingto revive ts aspect of s conception of a haecceity, becauseI am not committed to regarding properties as components ofindividuals. To deny that tsnesses are purely qualitativeis not necessarily to postulate 'bare particulars', substratawithout qualities of their own, wch would be what was leftof the individual when all its qualitative properties weresubtracted. Conversely, to hold that tsnesses are purelyqualitative is not to imply that individuals are notngbut bundles of qualities, for qualities may not be componentsof individuals at all. (pp. 6-7) Note4. Johannes Duns Scotus, _Quaestiones in libros metaphysicorum_,VII. xii. schol. 3; cf. _Ordinatio_, II.3.1.2, 57. I am indebtedto Marilyn McCord Adams for acquainting me with these texts andviews of Scotus, and for much discussion of the topics of tsparagraph. Aristotle and Aquinas and Scotus and Bonaventura all believed that human> minds can conceive and express the intelliilities or quiddities of tngs> and their properties, intelliilities that are not simply mind-dependent.> We can capture in thought and language the actual natures of tngs,> spelling out their genera and specific differences. Definition brackets or> delimits for us as knowers just what it is we attempt to understand and> notng else. The mind-independent tng-substance or the characteristics> that we are attempting to define measure the epistemic correctness of a> definition. Such real (as opposed to nominal) definition relies on the> intellile and perceptible characteristics tng-substances exbit to> perception and thought for understanding what they are and for picking out> individuals of a type. In ts way the epistemological realism of the> definition corresponds to an ontological realism of actual formal features> in mind-independent entities.http://www.sunysb.edu/plosophy/faculty/lmiller/ Delinonaliud.htmAfter all, the three tenets that largely define Nicholas's 'metaphysic of> contraction' seem altogether remote from Anselm's Scholasticism. For Anselm> has no use for the triad of notions (1) that there is an infinite> disproportion between the Creator and s creatures, (2) that, therefore,> finite minds can never positively know what God is, given the alleged ground> (3) that He is the Coincidence of opposites, i. e., is undifferentiated> 'Being' itself, wch, with respect to its Quiddity, can never be conceived> by anyone except itself.http://www.cla.umn.edu/jhopkins/CusaAnselm.pdf === Subject: Re: Minimal Graph, Four Color Theorem Visiting Assistant Professor at the University of Montana.Indeed, the proof of the Four Color Theorem rests on showing that> there does not exist any graph G wch requires 5 colors, is planar,> and such that the removal of any vertex results in a graph wch can> be colored with only 4 colors. But you have not given any coherent> argument to establish ts proposition that I can see anywhere. All I have is a hypothesis. The validity of the hypothesis does not>depend upon the a priori assumption that G is planar. Therefore, you>must automatically reject it. Why should I bother?I must automatically reject it? Mind reader, on top of everytngelse?Look, you started ->ts<- thread by asking a question about theminimal counterexample argument to prove the 4 color theorem. You arecertainly welcome to attempt to prove the 4 color theorem by anargument wch is ->not<- the minimal counterexample argument, butwhat you were doing is ->asking about the minimal counterexampleargument<-. If the reason you posted was because you did not understand what theauthor was saying, and if the reason you did not understand what theauthor you quoted was saying was because you were trying to tnk ofthe minimal counterexample argument as if it were your argument,then we've fixed it. Surely you now understand what the author wassaying, and presumably you agree that the statement you quoted wascorrect (whereas you insinuated it was incorrect in your originalpost).If the reason you posted your question, however, was because youwanted to use any response as an excuse to discuss your idea,ignoring the text that you were supposedly asking about, then all Ican say is you have lousy manners and I have no interest in discussingan argument with someone who resorts to misrepresentation in order tostart the discussion. But it has notng to do with any a prioriassumption that G is planar; it would have to do with the fact that I findyou an unappealing person to hold such discussions with, and that Ihonestly do not have the time to talk about the 4 color theorem. Ianswered your original query because I knew the answer and it seemedstraightforward. If I was drawn into that discussion on falsepretenses (as is become more and more apparent, given your insistentcomments about your other thread, and now ts comment about yourhypothesis), then all I can say is Have a good day. ==It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) === floating point. -- Example routine needed?I'm trying to write ATAN2 function for a small basic language that has> IEEE single precision math.. *,/,+.-, SQRT(), SIN(), COS(), TAN() are> availible in the language.> I've tried a few methods I've found but the results are way off due to> low precision, rounding, etc.Are there any repositorys of old fortran routines or algorithms that I> could use to get a good accuracy single precision routine. Speed or> space aren't as important as reasonably good accuracy.,> JonTo get an idea how the corresponding production codes work, look atThe Computation of Transcendental Functions on the IA-64 Arctecturehttp://www.intel.com/technology/itj/q41999/pdf/ transendental.pdfThe reference for the classical approximations is the book:Cody Jr., William J. and Waite, William, Software Manual for theElementary Functions, Prentice Hall, 1980For c source code that can be easily converted into other langagessee the Cephes library http://www.netlib.org/cephes/single.tgz(e.g. asinf.c, atanf.c)Hugo Pfoertner === Subject: Re: multiplication negs> been making my way throught the Principia, taking my time, working> along. One day I was talking to my young neice, helping her out with some> simple algebra. We were discussing mulitiplication and division of negative> numbers and my neice brought up a question and for some reason I have been> unable so far to reason my way through the answer: exactly why is it that> when one multiplies 2 negative numbers one ends up with a positive? It's> disturbing me that I cannot come up with a solid answer, and certainly> 'We've been taught that that is the case. will not suffice.> Any help would be greatly appreciated.> TIAturn your sock inside out twice, and it is right again. === Subject: Re: how many resolutions?>A^2 + B^2 + C^2 + D^2 = 85^2>where a to d are different one-digit or two-digit numbers.>I thought that it could have just one solution: since 85 is a product of two>prime numbers of 4k+1 type: 85=5*17. As it is known that only prime numbers>4k+1 could be resolved as p^2=a^2 + b^2.>so 85^2=5^2 * 17^2 = (3^2 + 4^2)*(15^2 + 8^2) so right numbers are 24, 32,>45, 60.>But there is another solution! 3, 4, 12, 84.There are lots of solutions, e.g. 1, 2, 38, 761, 4, 22, 821, 4, 58, 621, 10, 50, 681, 22, 46, 681, 26, 52, 621, 34, 38, 681, 34, 52, 581, 38, 44, 62... and those are just the solutions that start with 1.> Why? And maybe there are some other solutions left? It means that>resolution p^2=a^2 + b^2 is ambiguos? So we could find c and d (c not equal>a and b) such that p^2=c^2 + d^2?No, but not every way to get 85^2 as the sum of four squares comes froma product of two sums of two squares.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: recurrence relation question>how do i go about solving a recurrence relation when the function T(n)>is multiplied by n? such as the following:>2n*T(n) + 2nT(n-1) - 2T(n-1) = 2^n?>What tricks would one use?One trick is to collect the terms in T(n-1) and tnk of the new function S(n) = 2n*T(n).Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Algebra proof> Let a, b, r and s be integers with r>1, s>1 and gcd(r,s)=1. Prove> that if a=b(mod r) and a=b(mod s) then a=b(mod rs).How do I do ts????Another way to prove ts is to use the identity of Bezout(the extended Euclidean algorithm): if d = gcd(r,s) thend = u*r + v*s for some integers u and v.When you see the hypothesis gcd(r,s) = 1, often itis worthwle to convert that into the identity of Bezout.For your problem, you have: 1 = u*r + v*s for some integers u and v since gcd(r,s)=1; a - b = c*r for some integer c since a=b(mod r); a - b = d*s for some integer d since a=b(mod s).Now, multiply the first equation by a - b to get: a - b = (a-b)*u*r + (a-b)*v*s.Substitute the trd and second equations into the above equationto get: a - b = d*s*u*r + c*r*v*s = r*s*(d*u + c*v).Thus, r*s divides a - b. That is, a=b(mod rs).-- Bill Hale === Subject: Re: lopital's rule? L'Hospital or L'H.99pital. The circumflex (the ^ tngy) means that the following s is understood implicitly. AFAIK the guy's name is de l'h.99pital and in my pronounciation of thefrench> language :-) the ^ makes the o-sound longer.>In the case of l'h.99pital the ^ is here to recall the old noun hospital,but it doesn't alter the sound o.--Julien Santini,France. === Subject: Re: Chessboard knight metric?Take a chessboard (with or without infinetely many squares) let the distance d((x_1,x_2),(y_1,y_2)) between two squares x and y of the chessboard be defined as the minimum number of moves a knight takes to reach y from x. As far as I can see, ts should give the distances (on an ordinarychessboard). The only exception I see is when your starting point O isat a corner, and you want to reach the next square on the diagonal.Then the distance is 4 and not 2.The paths are O -> P -> X -> A -> W -> S -> T.- - - - S W S W S W S T- - - - W A W A W S W S- - - - A W A W A W S W- - - - X A X A W A W SS X A X A X A X A W A WX A W P X P W A X A W S A W P X A X P X A W A WX A X A O A X A X A W SA X P X A X P X - - - -X A W P X P W A - - - -A X A X A X A X - - - -W A X A X A X A - - - -Karin === Subject: Re: place here....I've been making my way throught the Principia, taking my time, working ^^^^^^^^^Wch Principia is that?> [...]-- G.C. === Subject: Re: Is ts newsgroup useless?> at 06:43 PM, Bart Goddard said:> >>Half the problem here is that it's unmoderated. The other half is>>that it's populated by overly-anal-retentive jerks who tnk it's>>going to be important to point out that I spelled you're your >>twice in the preceding paragraphs. PKB. You're upset about flames, yet you post them yourself. You claim> to be a Christian; don't your scriptures say soetng about the beam> as 90% of the people you're complaining about.1. I wasn't complaining, I was chatting with the guy who wascomplaining.2. Christianity is not the same as pietism, and if you don'tknow the difference then don't try to preach to Christians abouttheir own doctrine.3. Ts was meant to be humour, dumbass.>>I _thought_ we had beaten these guys up sufficiently in the gh>>school locker room that they'd be quiet by now.So by you correcting an error is wrong, but battery is acceptable> behavior? And you have the arrant hypocrisy to call others jerk.> >>Maybe the real problem is that a guy can't e-mail a matburn.The real problem is that people like you are hypocrites. That's far> worse than any amount of flaming. I suggest that you read what you posted, and then explain how you're> any better than JSH. Better yet, print out a copy and ask your pastor> what he tnks of it.I did. He laughed at the locker room part. Then I showed myour post and he suggested a wedgie.Bart